A Fixed Point Theorem in the Lebesgue Spaces of Variable Integrability L^p(·)

Abstract

We establish a fixed point property for the Lebesgue spaces with variable exponents $L^{p(·)}$, focusing on the scenario where the exponent closely approaches 1. The proof does not impose any additional conditions. In particular, our investigation centers on $\rho$-non-expansive mappings defined on convex subsets of $L^{p(·)}$, satisfying the “condition of uniform decrease” that we define subsequently.

1. Introduction

Sequence spaces $l^{p_n}$ of variable exponent $\left(p_n\right)$ were introduced in 1931 by W. Orlicz [1] when addressing a question related to Fourier series. In the same work, Orlicz generalized the idea of Lebesgue spaces of variable integrability by defining the class of measurable functions f such that ${\int }_0^1{\left|f\left(x\right)\right|}^{p\left(x\right)}dx<\infty .$

Lebesgue spaces of variable exponent arise naturally in the study of hydrodynamic equations that describe the behavior of non-Newtonian fluids [2,3]. Electrorheological fluids, characterized by dramatic and sudden changes in viscosity when exposed to an electric or magnetic field, are typical examples. A vigorous mathematical research effort is being devoted to electrorheological fluids and their applications to civil engineering, military science and medicine, among others [4,5,6,7].

Though variable exponent Lebesgue spaces appeared for the first time in [1], they were first studied as Banach spaces in [8]. The core observation of our approach is the immediate effect the variability of the exponent p on the topology of $L^{p(·)}$, namely the modular structure engenders a topology that differs fundamentally from that induced by the Luxemburg norm. This phenomenon is only visible when the exponent p is not constant (in the classical case of constant p, the modular is topologically equivalent to the norm) and is particularly striking in the endpoint cases, namely when the exponent is finite everywhere but is either unbounded or takes up values arbitrarily close to 1.

Modular uniform convexity and its applications to fixed point theory and proximinality are now well-understood in the case when the exponent p is unbounded, i.e., if $\mathrm{ess}\underset{x\in \Omega }{sup}p\left(x\right)=p_{+}=\infty$, [9], but remains unexplored in the remaining endpoint situation, namely the case when $\mathrm{ess}\underset{x\in \Omega }{inf}p\left(x\right)=p_{-}=1$. This work aims at remedying this situation. Proposition 2 and Theorems 3 and 4 are our main results and fill the existing gap in the literature for $p_{-}=1$.

This work of exploiting the modular structure of $L^{p(·)}$, new convexity properties of the modular, are discussed in the endpoint case $p_{-}=1$ and concrete applications of these new properties to modular fixed point theory are presented. The reader interested in the investigation of partial differential equations in the variable exponent spaces may consult the books [10,11].

2. Modular Vector Spaces

Orlicz’s ideas have inspired the research of many mathematicians. In particular, his work published in 1931 [1] inspired Nakano to introduce the concept of modular vector spaces.

Definition 1.

([12]). Let X be a linear vector space over the field $\mathbb{R}$. A modular on X is a function $\rho :X\to [0,\infty ]$ satisfying the following conditions:

(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)

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

If (3) is replaced by

$$\rho (\alpha x+(1-\alpha \left)y\right)\le \alpha \rho \left(x\right)+(1-\alpha )\rho \left(y\right)$$

for any $\alpha \in [0,1]$ and $x,y\in X$, then ρ is called a convex modular. In addition, ρ is said to be left-continuous if $\underset{r\to 1-}{lim}\rho \left(rx\right)=\rho \left(x\right)$ for any $x\in X$.

It is possible to associate the idea of convergence to any modular $\rho$ on a vector space, in the following fashion:

Definition 2.

A sequence $\left\{x_n\right\}\subset X$ is said to ρ-converge to $x\in X$ if

$$\underset{n\to \infty }{lim}\rho (x-x_n)=0.$$

Let $x\in X$, $y\in X$ and let $\left\{y_n\right\}$ be $\rho$-convergent to y. If the inequality

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

holds, $\rho$ is said to satisfy the Fatou property.

A modular function on a vector space X engenders a norm in a natural fashion.

Definition 3.

([13]). Given a convex modular ρ defined on the vector space X, the modular space generated by ρ is the set

$$X_{\rho }=\{x\in X;\underset{\alpha \to 0}{lim}\rho \left(\alpha x\right)=0\}.$$

The functional on X, ${\parallel .\parallel }_{\rho }:X_{\rho }\to [0,\infty )$, defined by

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

is a norm, called the Luxemburg norm.

The following major point to be underlined at this juncture is as follows: though convergence in the sense of the Luxemburg norm implies modular convergence (introduced in Definition 2), the converse fails. Deep problems arise in the study of modular spaces in which both notions of convergence are different. One concrete such instance are the Lebesgue spaces of variable, unbounded exponents, which will be considered in the present work.

3. Variable Exponent Lebesgue Spaces ${\mathbf{L}}^{\mathbf{p}(·)}$

Due to their applications to the hydrodynamics of smart fluids, partial differential equations with non-standard growth have been intensively studied in the past two decades (see [2,3,8,14] and the references therein). Problems of interest include the existence and uniqueness of solutions and the corresponding regularity of the solutions, when they exist. The simplest differential operator with non-standard growth is the variable exponent p-Laplacian. Specifically, let $\Omega \subset {\mathbb{R}}^n$ be open and connected with boundary $\partial \Omega$ and let $\mathcal{P}(\Omega )$ stand for the class of measurable functions $p:\Omega \to [1,\infty )$. The $p\left(x\right)$-Laplacian operator is defined as

$${\Delta }_{p\left(x\right)}u=\mathrm{div}\left({|\nabla u|}^{p\left(x\right)-2}\nabla u\right).$$

Not surprisingly, the behavior of ${\Delta }_{p\left(x\right)}$ is highly sensitive to the quality of the function p. Existence and uniqueness results for the Dirichlet problem

$$\left\{\begin{array}{c}{\Delta }_{p\left(x\right)}u=f\hfill \\ {u|}_{\partial \Omega }=\phi \hfill \end{array}\right.$$

(1)

are known [15] under certain conditions for f and $\phi$ under the assumption that p is bounded away from 1 and ∞, that is, assuming

$$1<p_{-}={\mathrm{essinf}}_{\Omega }p\le {\mathrm{esssup}}_{\Omega }p=p_{+}<\infty .$$

(2)

The Restriction (2) is standard in the literature due to the inherent limitations of standard methods in the cases $p_{-}=1$ or $p_{+}=\infty$. In an ongoing project, two of the authors of this article showed that Theorem 2 (see below) and its implications are the right tool to handle the case of the exponent $p\left(x\right)$ being unbounded (but still with $p_{-}>1$) on the domain $\Omega$. In fact, the authors show there that modular uniform convexity lies at the heart of the solvability of the Dirichlet problem (1) for $p_{+}=\infty$ as long as $p_{-}>1$.

The ultimate aim of this work is to present a theory that includes the limit case $p_{-}=1$, similar to the one developed in [9] for $p_{+}=\infty$, having in mind the goal of solving the boundary value problem (1) in the still open case $p_{-}=1$. Substantial progress is being made in this direction for the results in the present article to be viewed as the fundamental mathematical foundation of such endeavor.

Denote by $\mathcal{M}(\Omega )$ the vector space of all real-valued, Borel-measurable functions defined on an open and connected domain $\Omega \subset {\mathbb{R}}^n$ with boundary $\partial \Omega$. Recall

$$\mathcal{P}(\Omega )=\{p\in \mathcal{M}(\Omega ):1\le p<\infty \}.$$

The Lebesgue measure of a subset $A\subset {\mathbb{R}}^n$ will be denoted by $\left|A\right|$. For each such p define the following sets:

$${\Omega }_0=\left\{x\in \Omega :1<p\left(x\right)<\infty \right\}\mathrm{and}{\Omega }_1=\left\{x\in \Omega :p\left(x\right)=1\right\}.$$

Theorem 3.

([8,10]). Fix $p(·)$ in $\mathcal{P}(\Omega )$. The function $\rho :\mathcal{M}(\Omega )⟶[0,\infty ]$ defined by

$$\rho \left(u\right)=\underset{\Omega }{\int }{\left|u\left(x\right)\right|}^{p\left(x\right)}d\mu ,$$

is a convex, continuous modular on $\mathcal{M}(\Omega )$.

For $p\in \mathcal{P}(\Omega )$, we set

$$p_{-}:=\underset{t\in {\Omega }_0}{ess\; inf}p\left(t\right)\mathrm{and}{p}_{+}:=\underset{t\in {\Omega }_0}{ess\; sup}p\left(t\right).$$

The space $L^{p(·)}(\Omega )$ is defined as

$$L^{p(·)}(\Omega )=\{u\in \mathcal{M}(\Omega );\exists \lambda >0:\rho (u/\lambda )<\infty \},$$

endowed with the Luxemburg norm, i.e., for $u\in L^{p(·)}(\Omega )$,

$${\parallel u\parallel }_{p(·)}=inf\left\{\lambda >0:\rho \left(\frac{u}{\lambda }\right)\le 1\right\}.$$

(3)

It has been proved in [8,14] that $L^{p(·)}(\Omega )$ is a Banach space and that, if p is constant on $\Omega ,$ then $L^{p(·)}(\Omega )$ coincides with the original Lebesgue space $L^p(\Omega )$.

The interested reader can refer to [8,10,14] for a detailed treatment of these generalized Lebesgue spaces. A further note is in order at this point: $L^{p(·)}(\Omega )$ is the Musielak–Orlicz space corresponding to the Musielak–Orlicz function $\phi :\Omega \times [0,\infty )⟶[0,\infty )$ given by

$$\phi (x,t)=t^{p\left(x\right)}.$$

The Musielak–Orlicz spaces were introduced by Nakano in 1950 [12]; the works [13,14] contain further information on this area of mathematics.

The modular $\rho$ associated to a constant exponent p is none other than the p-th power of the Luxemburg norm; it is thus clear that the obstacles found in the normed-space structure of $L^p(\Omega )$ are essentially the same as those found in its modular structure. However, if p is non-constant, the handling of the geometry of the norm presents serious technical difficulties, especially in the end-point cases for the exponent p. For this reason, the modular structure presents a viable alternative in this case.

The following lemma, of a technical nature, plays a key role in our development.

Lemma 1.

The following inequalities hold:

(i)

[16] 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)

[17] If $1\le p\le 2$, 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 $\left|a\right|+\left|b\right|\ne 0$.

We refer the reader to [9] for a detailed proof of the following result:

Theorem 2.

For a bounded domain $\Omega \subset {\mathbb{R}}^n$, let $p\in \mathcal{P}(\Omega )$ satisfy $p_{-}>1$. Fix $r>0$, $\epsilon >0,$$f\in L^{p(·)}(\Omega )$ and $g\in L^{p(·)}(\Omega )$ such that $\rho \left(f\right)\le r$, $\rho \left(g\right)\le r$ and ${\displaystyle \rho \left(\frac{f-g}{2}\right)\ge \epsilon r}$. Then, it holds the bound

$$\rho \left(\frac{f+g}{2}\right)\le r\left(1-min\left\{\frac{\epsilon }{2},(p_{-}-1)\frac{{\epsilon }^2}{2}\right\}\right).$$

4. Uniform Decrease Condition

In this section a class of subsets of $L^{p(·)}(\Omega )$ is introduced. Subsets in this class satisfy modular geometric properties similar to those of $L^{p(·)}$ when $p_{-}>1$. The following notations will be used:

$$J_a=\left\{x\in \Omega ;p\left(x\right)<a\right\}and{I}_a=\Omega \setminus J_a=\left\{x\in \Omega ;p\left(x\right)\ge a\right\},$$

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

For any measurable subset I of $\Omega$ and $f\in L^{p(·)}(\Omega )$, define the functional

$${\rho }_I\left(f\right)={\int }_{x\in I}{\left|f\left(x\right)\right|}^{p\left(x\right)}dx.$$

Here, it is assumed that ${\rho }_I\left(f\right)=0$ if $\left|I\right|=0$.

Definition 4.

For $\varnothing \ne C\subset L^{p(·)}(\Omega )$, C satisfies the uniform decrease condition $\left(UD\right)$ (or C is a $\left(UD\right)$ set) if for each $\alpha >0$, there exists $a>1$ such that

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

It is clear that the $\left(UD\right)$ condition is inherited by the subsets of a given $\left(UD\right)$ set and it is easy to verify that if $|{\Omega }_0|=0$ (that is $p(·)\equiv 1$), then $L^{p(·)}(\Omega )$ has only one $\left(UD\right)$ subset, namely $C=\left\{0\right\}$. This case, however, is not interesting and it will be assumed henceforth that $|{\Omega }_0|>0$.

On another note, it is not hard to prove that if $p_{-}>1$, then any $\varnothing \ne C\subset L^{p(·)}(\Omega )$ satisfies the condition $\left(UD\right)$. To see this, let C be such set, fix $\alpha >0$ and set $a\in (1,p_{-})$. Then, $|J_a|=0$, from which it follows that

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

It transpires from the above that the condition $\left(UD\right)$ is only interesting when $p_{-}=1$ and $p(·)$ is not identically equal to 1. These two conditions will be assumed from now on.

Example 1.

Set $\Omega =[1,+\infty )$. Consider the function $p(·)$ defined by

$$p\left(x\right)=1+\frac{1}{x+1}.$$

Note that $p_{-}=1$ and $p_{+}<\infty$. Consider the set

$$C=\left\{f\in L^{p(·)}(\Omega );\left|f\left(x\right)\right|\le \frac{1}{{(x+1)}^2}a.e.\right\}.$$

It is easy to verify that C is nonempty, convex, and ρ-closed. In addition, C satisfies $\left(UD\right)$. This can be shown by fixing $\alpha >0$. Then, there exists $A\ge 1$ for which

$${\int }_{[A,+\infty )}\frac{1}{{(1+x)}^2}dx<\alpha .$$

Set $a=1+1/(1+A)$. Note that $p\left(x\right)<a$ if and only if $A<x$. Let $f\in C$. We have

$$\begin{array}{ccc}{\rho }_{J_a}\left(f\right)\hfill & =\hfill & {\displaystyle {\int }_{J_a}{\left|f\left(x\right)\right|}^{p\left(x\right)}dx}\hfill \\ & =\hfill & {\displaystyle {\int }_{[A,+\infty )}{\left|f\left(x\right)\right|}^{p\left(x\right)}dx}\hfill \\ & \le \hfill & {\displaystyle {\int }_{[A,+\infty )}\frac{1}{{(1+x)}^{2p\left(x\right)}}dx}\hfill \\ & \le \hfill & {\displaystyle {\int }_{[A,+\infty )}\frac{1}{{(1+x)}^2}dx}\hfill \\ & <\hfill & \alpha ,\hfill \end{array}$$

which proves our claim that C is $\left(UD\right)$.

Yet another class of subsets of $L^{p(·)}(\Omega )$ has to be introduced in order to proceed to the characterization of $\left(UD\right)$ sets.

Definition 5.

Let $L^{p(·)}(\Omega )$ with $p_{-}=1$ and $|{\Omega }_0|>0$. Consider a non-decreasing function $\Phi :(0,+\infty )\to (1,+\infty )$ and set $C_{\Phi }$ to be

$$C_{\Phi }=\left\{f\in L^{p(·)}(\Omega );{\rho }_{J_{\Phi \left(\alpha \right)}}\left(f\right)\le \alpha ,forall\alpha >0\right\}.$$

We remark that $C_{\Phi }\ne \varnothing$, since $0\in C_{\Phi }$. In the following lemma, some elementary properties of $C_{\Phi }$ are presented.

Lemma 2.

In the notation of Definition 5, one has the following:

(1)

$C_{\Phi }$ is convex.

(2)

$C_{\Phi }$ is symmetric, i.e., $-f\in C_{\Phi }$ whenever $f\in C_{\Phi }$.

(3)

The set $C_{\Phi }$ is ρ-closed. In particular, this implies that $C_{\Phi }$ is ρ-complete.

Proof. 

To prove $\left(3\right)$, let $\left\{f_i\right\}$ be in $C_{\Phi }$ with $f_i\stackrel{\rho }{\to }f$. Fix $\epsilon >0$ and $\alpha >0$, select $\theta :e^{\frac{ln(1+\epsilon )}{1-\Phi \left(\alpha \right)}}<\theta <1.$ It follows that

$$\begin{array}{cc}\hfill {\int }_{J_{\Phi \left(\alpha \right)}}{\left|f\left(x\right)\right|}^{p\left(x\right)}dx& ={\int }_{J_{\Phi \left(\alpha \right)}}{|f\left(x\right)-f_i\left(x\right)+f_i\left(x\right)|}^{p\left(x\right)}dx\hfill \\ & ={\int }_{J_{\Phi \left(\alpha \right)}}{\left|(1-\theta )\frac{f\left(x\right)-f_i\left(x\right)}{1-\theta }+\theta \frac{f_i\left(x\right)}{\theta }\right|}^{p\left(x\right)}dx\hfill \\ & \le (1-\theta ){\int }_{J_{\Phi \left(\alpha \right)}}{\left|\frac{(f-f_i)\left(x\right)}{1-\theta }\right|}^{p\left(x\right)}dx+\theta {\int }_{J_{\Phi \left(\alpha \right)}}{\left|\frac{f_i\left(x\right)}{\theta }\right|}^{p\left(x\right)}dx\hfill \\ & \le \frac{1}{{(1-\theta )}^{\Phi \left(\alpha \right)}}{\int }_{J_{\Phi \left(\alpha \right)}}{\left|(f-f_i)\left(x\right)\right|}^{p\left(x\right)}dx+{\int }_{J_{\Phi \left(\alpha \right)}}\frac{|f_i{\left(x\right)|}^{p\left(x\right)}}{{\theta }^{p\left(x\right)-1}}dx\hfill \\ & \le \frac{1}{{(1-\theta )}^{\Phi \left(\alpha \right)}}{\int }_{J_{\Phi \left(\alpha \right)}}|(f-f_i)\left(x\right){|}^{p\left(x\right)}dx+(1+\epsilon ){\int }_{J_{\Phi \left(\alpha \right)}}{\left|f_i\left(x\right)\right|}^{p\left(x\right)}dx\hfill \\ & \le \frac{1}{{(1-\theta )}^{\Phi \left(\alpha \right)}}{\int }_{J_{\Phi \left(\alpha \right)}}{\left|(f-f_i)\left(x\right)\right|}^{p\left(x\right)}dx+(1+\epsilon )\alpha .\hfill \end{array}$$

The claim follows from the arbitrariness of $\epsilon$ and the fact that the left term on the right-hand side tends to zero as $i\to \infty$. □

Proposition 1.

If $p_{-}=1$, $|{\Omega }_0|>0$, and $C\subset L^{p(·)}(\Omega )$. Then, the following conditions are equivalent:

(i)

C satisfies the condition $\left(UD\right)$.

(ii)

There exists $\Phi :(0,+\infty )\to (1,2]$ non-decreasing, such that $C\subset C_{\Phi }$.

Proof. 

$\left(i\right)\Rightarrow \left(ii\right)$ Let $\varnothing \ne C\subset L^{p(·)}(\Omega )$ satisfy the condition $\left(UD\right)$. For any $\alpha >0$, there exists $a>1$ such that $\underset{f\in C}{sup}{\rho }_{J_a}\left(f\right)\le \alpha$. Set

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

Define

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

It follows easily that $\Phi$ is well defined and that, for each $\alpha >0,$$\Phi \left(\alpha \right)\in (1,2]$. Pick $0<\alpha \le \beta$. It will be proved that $\Phi \left(\alpha \right)\le \Phi \left(\beta \right)$. To this end, note that $\left[\alpha \right]\subset \left[\beta \right]$ and that if $\left[\alpha \right]\cap (1,2]\ne \varnothing$, then it necessarily follows that $\left[\beta \right]\cap (1,2]\ne \varnothing$. This yields $\Phi \left(\alpha \right)\le \Phi \left(\beta \right)$. On the other hand, if $\left[\alpha \right]\subset [2,+\infty )$,one can select $a\in \left[\alpha \right]$. Clearly, $a\ge 2$ and $a\in \left[\beta \right]$. It follows by definition of J, that it holds the inclusion $J_2\subset J_a$. Consequently, from ${\rho }_{J_2}\left(f\right)\le {\rho }_{J_a}\left(f\right)$, for all $f\in L^{p(·)}(\Omega )$, one infers that

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

that is, $2\in \left[\beta \right]$. This fact will force $\Phi \left(\beta \right)=2$. Thus, one concludes $\Phi \left(\alpha \right)\le \Phi \left(\beta \right)$ in both cases, that is, $\Phi :(0,+\infty )\to (1,2]$ is non-decreasing.

It will next be shown that, for $\Psi \left(\alpha \right)=(1+\Phi (\alpha \left)\right)/2$, $\alpha >0$, it holds that $C\subset C_{\Psi }$. Notice that for all $\alpha >0$, one has

$$1<\Psi \left(\alpha \right)<\Phi \left(\alpha \right),$$

(4)

which is a direct consequence of the inequality $1<\Phi \left(\alpha \right)$. The proof will follow from the consideration of two mutually exclusive cases, namely $\left[\alpha \right]\subset [2,+\infty )$ and $\left[\alpha \right]\cap (1,2]\ne \varnothing$.

If $\left[\alpha \right]\subset [2,+\infty )$, let $a\in \left[\alpha \right]$; clearly $\Psi \left(\alpha \right)=3/2<a$, which yields $J_{\Psi \left(\alpha \right)}\subset J_a$. Consequently, one has

$${\rho }_{J_{\Psi \left(\alpha \right)}}\left(f\right)\le {\rho }_{J_a}\left(f\right),forallf\in C,$$

which in turn yields $\underset{f\in C}{sup}{\rho }_{J_{\Psi \left(\alpha \right)}}\left(f\right)\le \underset{f\in C}{sup}{\rho }_{J_a}\left(f\right)\le \alpha$. Assume $\left[\alpha \right]\cap (1,2]\ne \varnothing$; then, $\Phi \left(\alpha \right)=sup\left(\left[\alpha \right]\cap (1,2]\right)$. It follows from (4) that there exists $a\in \left[\alpha \right]$ such that $\Psi \left(\alpha \right)<a\le \Phi \left(\alpha \right)$. An analogous reasoning shows that

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

Hence, $\underset{f\in C}{sup}{\rho }_{J_{\Psi \left(\alpha \right)}}\left(f\right)\le \alpha$, for all $\alpha >0$, i.e., $C\subset C_{\Psi }$ as claimed.

To show the implication $\left(ii\right)\Rrightarrow \left(i\right)$, let $\alpha >0$. Taking $a=\Phi \left(\alpha \right)$ it is easy to see that

$$\underset{f\in C_{\Phi }}{sup}{\rho }_{J_a}\left(f\right)\le \alpha ,$$

which proves that $C_{\Phi }$ satisfies the condition $\left(UD\right)$. Clearly, any subset $C\subset C_{\Phi }$ also satisfies $\left(UD\right)$. □

By virtue of Proposition 1, the study of $\left(UD\right)$ subsets of $L^{p(·)}(\Omega )$ can be reduced to the consideration of the sets of the form $C_{\Phi }$. In this connection, the next result is of deep importance. Plainly, it states that subsets of the type $C_{\Phi }$ satisfy a well-known modular geometric property known as $\left(UUC2\right)$ (see [18]), even in the unfavorable case when $p_{-}=1$.

Theorem 3.

Let $L^{p(·)}(\Omega )$ with $p_{-}=1$ and $|{\Omega }_0|>0$. Let $\Phi :(0,+\infty )\to (1,2]$ be a non-decreasing function. For all $f,g\in C_{\Phi }$ such that $\rho \left(f\right)\le r$, $\rho \left(g\right)\le r$ and $\rho \left({\displaystyle \frac{f-g}{2}}\right)\ge r\epsilon$, we have the estimate

$$\rho \left(\frac{f+g}{2}\right)\le r\left(1-min\left(\frac{\epsilon }{4},\frac{{\displaystyle \left(\Phi \left(\frac{r\epsilon }{2}\right)-1\right){\epsilon }^2}}{128}\right)\right),$$

for all $r>0$ and $\epsilon >0$.

Proof. 

Fix $r>0$, $\epsilon >0$, $f,g\in C_{\Phi }$ such that $\rho \left(f\right)\le r$, $\rho \left(g\right)\le r$ and $\rho \left({\displaystyle \frac{f-g}{2}}\right)\ge r\epsilon$. It follows from the convexity of $\rho$ that $\epsilon \le 1$. Fix ${\displaystyle \alpha =\frac{r\epsilon }{2}}$. On account of the properties $C_{\Phi }$, one has ${\displaystyle \frac{f-g}{2}\in C_{\Phi }}$. Therefore,

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

which implies

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

Next, define

$$K=I_{\Phi \left(\alpha \right)}\cap \left\{x,p\left(x\right)\ge 2\right\}andL=I_{\Phi \left(\alpha \right)}\cap \left\{x,p\left(x\right)<2\right\}$$

and observe that $I_{\Phi \left(\alpha \right)}=K\cup L$. Consequently, ${\rho }_{I_{\Phi \left(\alpha \right)}}\left(h\right)={\rho }_K\left(h\right)+{\rho }_L\left(h\right)$, for all $h\in C_{\Phi }$.

From our assumptions, we have

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

Assume first that

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

Using Lemma 1, we obtain

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

which implies

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

Using the convexity of the modular, we have

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

which implies

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

For the second case, assume

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

Set ${\displaystyle c=\frac{\epsilon }{8}}$ and

$$L_1=\left\{x\in L,|f\left(x\right)-g\left(x\right)|\le c\left(\left|f\right(x\left)\right|+\left|g\right(x\left)\right|\right)\right\}and{L}_2=L\setminus L_1.$$

Since $c<1$, we obtain

$$\begin{array}{ccc}{\displaystyle {\rho }_{L_1}\left(\frac{f-g}{2}\right)}\hfill & =\hfill & {\displaystyle {\int }_{L_1}{\left|\frac{f\left(x\right)-g\left(x\right)}{2}\right|}^{p\left(x\right)}dx}\hfill \\ & & \\ & \le \hfill & {\displaystyle {\int }_{L_1}{c}^{p\left(x\right)}{\left(\frac{\left|f\right(x\left)\right|+\left|g\right(x\left)\right|}{2}\right)}^{p\left(x\right)}dx}\hfill \\ & & \\ & \le \hfill & {\displaystyle \frac{c}{2}{\int }_{L_1}\left({\left|f\left(x\right)\right|}^{p\left(x\right)}+{\left|g\left(x\right)\right|}^{p\left(x\right)}\right)dx}\hfill \\ & & \\ & =\hfill & {\displaystyle \frac{c}{2}\left({\rho }_{L_1}\left(f\right)+{\rho }_{L_1}\left(g\right)\right)}\hfill \\ & & \\ & \le \hfill & {\displaystyle \frac{c}{2}\left(\rho \left(f\right)+\rho \left(g\right)\right)}\hfill \\ & & \\ & \le \hfill & {\displaystyle \frac{cr}{2}.}\hfill \end{array}$$

Our assumption on ${\rho }_L\left((f-g)/2\right)$ implies

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

For any $x\in L_2\subset I_{\Phi \left(\alpha \right)}$, we have $\Phi \left(\alpha \right)\le p\left(x\right)$, which implies

$$\Phi \left(\frac{r\epsilon }{2}\right)-1=\Phi \left(\alpha \right)-1\le p\left(x\right)-1\le p\left(x\right)(p\left(x\right)-1).$$

Let $x\in L_2^{*}=\{x\in L_2;\left|f\left(x\right)\right|+\left|g\left(x\right)\right|\ne 0\}$, we have

$$c\le c^{2-p\left(x\right)}\le {\left(\frac{\left|f\right(x)-g(x\left)\right|}{\left|f\left(x\right)\right|+\left|g\left(x\right)\right|}\right)}^{2-p\left(x\right)}.$$

Using Lemma 1, we obtain

$${\left|\frac{f\left(x\right)+g\left(x\right)}{2}\right|}^{p\left(x\right)}+\frac{(\Phi (\alpha )-1)}{2}c{\left|\frac{f\left(x\right)-g\left(x\right)}{2}\right|}^{p\left(x\right)}\le \frac{1}{2}\left({\left|f\left(x\right)\right|}^{p\left(x\right)}+{\left|g\left(x\right)\right|}^{p\left(x\right)}\right),$$

for any $x\in L_2^{*}$. Since ${\rho }_{L_2}\left(f\right)={\rho }_{L_2^{*}}\left(f\right)$, ${\rho }_{L_2}\left(g\right)={\rho }_{L_2^{*}}\left(g\right)$ and

$${\rho }_{L_2}\left(\frac{f\left(x\right)+g\left(x\right)}{2}\right)={\rho }_{L_2^{*}}\left(\frac{f\left(x\right)+g\left(x\right)}{2}\right),$$

we obtain

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

which implies

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

Putting both cases together, we obtain

$$\rho \left(\frac{f+g}{2}\right)\le r\left(1-min\left(\frac{\epsilon }{4},\frac{{\displaystyle \left(\Phi \left(\frac{r\epsilon }{2}\right)-1\right){\epsilon }^2}}{128}\right)\right),$$

as claimed. □

The following lemma will lead to the main result of this work.

Lemma 3.

Let $L^{p(·)}(\Omega )$ with $p_{-}=1$ and $|{\Omega }_0|>0$. Consider a non-decreasing function $\Phi :(0,+\infty )\to (1,2]$ and define ${\displaystyle \Psi \left(\alpha \right)=\Phi \left(\frac{\alpha }{4}\right)}$, for $\alpha >0$. Then, it holds

$$C_{\Phi }+C_{\Phi }=\left\{f+g;f,g\in C_{\Phi }\right\}\subset C_{\Psi }.$$

Proof. 

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

$${\left|\frac{f\left(x\right)+g\left(x\right)}{2}\right|}^{p\left(x\right)}\le \frac{1}{2}\left({\left|f\left(x\right)\right|}^{p\left(x\right)}+{\left|g\left(x\right)\right|}^{p\left(x\right)}\right),$$

which implies ${|f\left(x\right)+g\left(x\right)|}^{p\left(x\right)}\le 2^{p\left(x\right)-1}\left({\left|f\left(x\right)\right|}^{p\left(x\right)}+{\left|g\left(x\right)\right|}^{p\left(x\right)}\right)$. Hence,

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

Therefore, ${\rho }_{J_{\Psi \left(\alpha \right)}}(f+g)\le \alpha$, that is $f+g\in C_{\Psi }$; this is the desired result. □

The next section is devoted to the discussion of a fixed point theorem for modular non-expansive mappings.

5. Application

As an application to Theorem 3, a modular version of a fixed point result for non-expansive mappings will be discussed. For an extensive discussion on the metric fixed point theory, the interested readers are referred to [18]. Recall that, throughout this work, we assume that the exponent function $p(·)$ is not identically equal to 1.

First, a proximinality property of $\rho$-closed convex $\left(UD\right)$ subsets will be presented.

Proposition 2.

Let $L^{p(·)}(\Omega )$ such that $p_{-}=1$. Let $\Phi :(0,+\infty )\to (1,2]$ be non-decreasing and let $C\subset C_{\Phi }$ be ρ-closed and convex. Then, C is proximinal. In other words, for any $x\in C_{\Phi }$ satisfying

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

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

Proof. 

It is clear that no generality is lost by assuming that $x\notin C$. It holds $R=d_{\rho }(x,C)>0$, since C is $\rho$-closed. Now, a sequence $\left\{y_n\right\}$ will be constructed in the following way: For any $n\ge 1$, let $y_n\in C$ be defined by the condition $\rho (x-y_n)<R(1+1/n)$. It will next be shown by contradiction that $\{y_n/2\}$ is $\rho$-Cauchy. Assume otherwise; then, for some $\epsilon >0$ one can construct a subsequence $\left\{y_{\varphi \left(n\right)}\right\}$ of $\left\{y_n\right\}$ such that

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

for any $n>m\ge 1$. On account of Lemma 3, $\{x-y_{\varphi \left(n\right)}\}$ is in $C_{\Psi }$, where $\Psi \left(\alpha \right)=\Phi (\alpha /4)$ for $\alpha >0$. For fixed $n>m\ge 1$, it holds

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

Next, observe that

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

with ${\displaystyle {\epsilon }_1=\frac{{\epsilon }_0}{2R}}$; by virtue of Theorem 3, it follows that

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

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 $y_{\varphi \left(n\right)}\in C$ and $y_{\varphi \left(m\right)}\in C$ the convexity of C yields

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

Letting $m\to +\infty$, one easily concludes that

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

Due to the above contradiction, it follows that $\{y_n/2\}$ is $\rho$-Cauchy; in conjunction with the $\rho$- completeness of $L^{p(·)}(\Omega )$, one concludes that the sequence $\{y_n/2\}$$\rho$-converges to $y\in L^{p(·)}(\Omega )$.

On the other hand, C is convex and $\rho$-closed: it follows from these observations that $2y\in C$ and by virtue of Fatou property, it is concluded that

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

Writing $c=2y$, it follows $d(x,C)=\rho (x-c)$. The uniqueness of the point c can be readily obtained from the strict-convexity of $\rho$ on $C_{\Psi }$, which follows from $\left(UUC2\right)$. □

The next analysis concerns the intersection property known as the property $\left(R\right)$. Property $\left(R\right)$ was first introduced in the context of metric spaces. Specifically,

Definition 6.

([18]). A nonempty ρ-closed convex subset C of ${}^{p(·)}(\Omega )$ is said to satisfy the property $\left(R\right)$ if for any decreasing sequence $\left(C_i\right)$ of nonempty, ρ-closed, ρ-bounded, and convex subsets of C, we have ${\displaystyle \bigcap _{i=1}^{\infty }}{C}_i\ne \varnothing .$

Proposition 3.

Let $L^{p(·)}(\Omega )$ such that $p_{-}=1$. Consider a non-decreasing function $\Phi :(0,+\infty )\to (1,2]$. Then, $C_{\Phi }$ has the property $\left(R\right)$.

Proof. 

Let $\left\{C_n\right\}$ be a decreasing sequence of nonempty $\rho$-closed $\rho$-bounded convex subsets of $C_{\Phi }$. For any $x\in C_1$. For each $n\in \mathbb{N}$, the $\rho$-distance from x to $C_n$ is subject to the bound

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

Hence, the sequence $\left\{d_{\rho }(x,C_n)\right\}$ is increasing (since $C_n$ is decreasing) and it is bounded above by ${\delta }_{\rho }\left(C_1\right)$; let $R=\underset{n\to +\infty }{lim}{d}_{\rho }(x,C_n)=\underset{n}{sup}{d}_{\rho }(x,C_n)$. One either has $R=0$ or $R>0$. In the first case, it follows that $x\in C_n$ for any $n\ge 1$; this in turns yields ${\displaystyle \bigcap _{n\ge 1}}{C}_n\ne \varnothing$.

If $R>0$, on account of Proposition 2, one can construct a sequence $\left\{w_n\right\}$, $n\ge 1$, such that $d_{\rho }(x,C_n)=\rho (x-w_n)$ with $w_n\in C_n$. As in the proof of Proposition 2, it follows that $\{w_n/2\}$ is $\rho$-Cauchy; let $w\in L^{p(·)}$ be its $\rho$-limit of $\{w_n/2\}$. By definition of $\left\{C_n\right\}$, it follows that $2c\in {\displaystyle \bigcap _{n\ge 1}}{C}_n$, which shows that ${\displaystyle \bigcap _{n\ge 1}}{C}_n\ne \varnothing$, as claimed. On another note, Fatou property yields the following inequality:

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

In conclusion, one has

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

□

Remark 1.

Under the assumptions of Proposition 3, the conclusion still holds for any family ${\left\{C_{\alpha }\right\}}_{\alpha \in \Gamma }$ of nonempty, convex, $\rho$-closed subsets of C, where $(\Gamma ,\prec )$ is an upward directed index set, as long as there exists $x\in C$ with ${\displaystyle {\displaystyle \underset{\alpha \in \Gamma }{sup}}{d}_{\rho }(x,C_{\alpha })<\infty }$.

For the proof of this generalized version, let ${\displaystyle d={\displaystyle \underset{\alpha \in \Gamma }{sup}}{d}_{\rho }(x,C_{\alpha })}$ and observe that there is no loss of generality by assuming $d>0$. For $n\in \mathbb{N}$, let ${\alpha }_n\in \Gamma$ be defined by the condition

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

Since $(\Gamma ,\prec )$ is upward-directed, it can be assumed that ${\alpha }_n\prec {\alpha }_{n+1}$. This yields that $C_{{\alpha }_{n+1}}\subset C_{{\alpha }_n}$ and according to Proposition 3, it follows that $C_0={\displaystyle \bigcap _{n\ge 1}}{C}_{{\alpha }_n}\ne \varnothing$. It is obvious that $C_0$ is $\rho$-closed and by virtue of the last statement in the proof of Proposition 3, it is clear that

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

Select $c_0\in C_0$ such that $d_{\rho }(x,C_0)=\rho (x-c_0)$. It will be proved that, for any $\alpha \in \Gamma$, $c_0\in C_{\alpha }$. Indeed, for fixed $\alpha \in \Gamma$, if for some $n\ge 1$, $\alpha \prec {\alpha }_n$, then it obviously holds that $c_0\in C_{{\alpha }_n}\subset C_{\alpha }$. It is thus sufficient to assume that, for any $n\ge 1$, $\alpha \nprec {\alpha }_n$. $\Gamma$ is upward-directed; hence, there exists ${\beta }_n\in \Gamma$ such that for any $n\ge 1$${\alpha }_n\prec {\beta }_n$ and $\alpha \prec {\beta }_n$. It can further be assumed that ${\beta }_n\prec {\beta }_{n+1}$ for any $n\ge 1$. It holds that $C_1={\displaystyle \bigcap _{n\ge 1}}{C}_{{\beta }_n}\ne \varnothing$. Since for all n, $C_{{\beta }_n}\subset C_{{\alpha }_n}$, it follows that $C_1\subset C_0$. Furthermore, it holds that

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

Thus, $d_{\rho }(x,C_1)=d$, which yields the existence of a unique point $c_1\in C_1$ such that $d_{\rho }(x,C_1)=\rho (x-c_1)=d$. Given that $\rho$ is $\left(SC\right)$ on $C_{\Phi }$, it is concluded that $c_0=c_1$. In particular, it follows that for any $n\ge 1$, $c_0\in C_{{\beta }_n}$. It follows from the fact that $\alpha \prec {\beta }_n$, that for any $n\ge 1$, $C_{{\beta }_n}\subset C_{\alpha }$; in turn this yields $c_0\in C_{\alpha }$. From the arbitrariness of $\alpha$ in $\Gamma$, one must conclude that $c_0\in {\displaystyle \bigcap _{\alpha \in \Gamma }}{C}_{\alpha }$, so that ${\displaystyle \bigcap _{\alpha \in \Gamma }}{C}_{\alpha }\ne \varnothing$, as claimed.

The following proposition is a further accessory to the fixed point theorem for $\rho$-non-expansive mappings. The following proposition is in order:

Definition 7.

A set $C\subset L^{p(·)}(\Omega )$ is said to have ρ-normal structure if for any $\rho -$closed, convex, $\rho -$bounded W, $\varnothing \ne W\subseteq C$, that contains more than one point, there exists $x\in W$ such that

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

Proposition 4.

Let $L^{p(·)}(\Omega )$ such that $p_{-}=1$. For a non-decreasing function $\Phi :(0,+\infty )\to (1,2]$, $C_{\Phi }$ has ρ-normal structure.

Proof. 

Let $C\subseteq C_{\Phi }$ be as in Definition 7. Since C consists of more than one point, ${\delta }_{\rho }\left(C\right)>0$. Let $x,y\in C$, $x\ne y$ and set

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

For $\alpha >0$, set $\Psi \left(\alpha \right)=\Phi (\alpha /4)$. For fixed $c\in C$, Lemma 3 yields that $x-c$ and $y-c$ are both in $C_{\Phi }-C_{\Phi }\subset C_{\Psi }$. 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 3 implies

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

Due to the arbitrariness of $c\in C$, it follows that

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

The latter completes the proof of Proposition 4. □

The central result of this work can now be proved.

Theorem 4.

Let $L^{p(·)}(\Omega )$ such that $p_{-}=1$. Let C be a nonempty ρ-closed, convex, and ρ-bounded subset of $L^{p(·)}(\Omega )$ and assume that C satisfies $\left(UD\right)$. Then, any ρ-non-expansive mapping $T:C\to C$ has a fixed point.

Proof. 

On account of Proposition 1, since C is $\left(UD\right)$, there exists a non-decreasing function $\Phi :(0,+\infty )\to (1,2]$ such that $C\subseteq C_{\Phi }$. The conclusion is trivial if C consists of just one point; it will thus be assumed that C contains at least two distinct points, i.e., suppose that ${\delta }_{\rho }\left(C\right)>0$. Define

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

$C\in \mathfrak{F}$, which yields $\mathfrak{F}\ne \varnothing .$ Remark 1 in concert with the boundedness of C and Zorn’s lemma rapidly yields a minimal element of $\mathfrak{F}$, say $K_0$. In what follows, it will be shown that $K_0$ consists of exactly one point. This follows by contradiction. If $K_0$ consisted of more than one point, one could set $co\left(T\left(K_0\right)\right)$ to be the intersection of all $\rho -$closed convex subset of C containing $T\left(K_0\right)$. Clearly, $co\left(T\left(K_0\right)\right)\subset K_0$ since $K_0\in \mathfrak{F}$. Furthermore,

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

which, in turn, yields $co\left(T\left(K_0\right)\right)\in \mathfrak{F}$. Since $K_0$ is a minimal element of $\mathfrak{F}$ it is easy to see that $K_0=co\left(T\left(K_0\right)\right)$. According to Proposition 4, there exists $x_0\in K_0$ such that

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

Define the subset $K=\left\{x\in K_0,{\displaystyle \underset{y\in K_0}{sup}}\rho (x-y)\le r_0\right\}$. It is obvious that $x_0\in K$, also, $K={\displaystyle \bigcap _{y\in K_0}}{B}_{\rho }(y,r_0)\cap K_0$. A straightforward reasoning using modular balls shows that $K\subseteq K_0$ is $\rho -$closed and convex. In fact, it follows that $T\left(K\right)\subset K$. Indeed, let $x\in K$. Since T is $\rho -$non-expansive, for all $y\in K_0$ it holds

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

Thus, $T\left(y\right)\in B_{\rho }(T\left(x\right),r_0)\cap K_0$, which yields $T\left(K_0\right)\subset B_{\rho }(T\left(x\right),r_0)$. The equality $K_0=co\left(T\left(K_0\right)\right)$, implies that $K_0\subset B_{\rho }(T\left(x\right),r_0)$, which in turn yields

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

for all $y\in K_0$. Thus, $T\left(x\right)\in K$, but x is an arbitrary element of K, so that in all, $T\left(K\right)\subset K$. The minimality of $K_0$ forces $K=K_0$. In all,

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

This is a contradiction. It follows that $K_0$ consists of only one point, which must be a fixed point of T because $T\left(K_0\right)\subset K_0$. □

Remark 2.

The condition $\left(UD\right)$ in Theorem 4 can be replaced with the slightly more general condition:

$$There exists x_0\in L^{p(·)}(\Omega ) such that x_0+C satisfies the condition \left(UD\right).$$