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

In this work, we prove a fixed-point theorem in the variable exponent spaces ℓp(.), 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 ρ-nonexpansive mappings defined on convex subsets of ℓp(.) that satisfy a specific condition which we call “condition of uniform decrease”.


Introduction
Variable exponent spaces first appeared in a work of Orlicz in 1931 [1] (see also [2]), where he defined the following space: |λx n | p(n) < ∞, f or some λ > 0 .
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 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 p (.) .
In this work, we will introduce a class of subsets of p(.) that have some interesting geometrical properties. This will allow us to prove a new fixed-point theorem concerning
For any subset I of N, we consider the functional If I = ∅, we set ρ I (x) = 0. We define on modular spaces a modular topology which is similar to the topology induced by a metric.
(a) We say that a sequence {x n } ⊂ p(.) is ρ-convergent to x ∈ p(.) if and only if ρ( Note that ρ satisfies the Fatou property, i.e., holds whenever {y n } ρ-converges to y, for any x, y, y n ∈ p(.) . Throughout, we will use the notation B ρ (x, r) to denote the ρ-ball with radius r ≥ 0 centered at x ∈ p(.) and defined as Note that Fatou property holds if and only if the ρ-balls are ρ-closed. That is, all ρ-balls are ρ-closed in p(.) .

Definition 2.
Let C ⊂ p(.) be a nonempty subset. A mapping T : C −→ C is called ρ-Lipschitzian if there exists a constant K ≥ 0 such that The concept of modular uniform convexity was first introduced by Nakano [11], but a weaker definition of modular uniform convexity called (UUC2) was introduced in [9] and seems to be more suitable to hold in 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 p(.) . Let C be a nonempty subset of p(.) .
In [13], the authors proved a similar fixed-point theorem in the case where {n ∈ 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 p(.) spaces.
In this work, using a different approach, we obtain some fixed-point results when p − = 1 without the known conditions on the function p(·).

Uniform Decrease Condition
First, we introduce an interesting class of subsets of p(.) , which will play an important part in our work. In particular, they enjoy similar modular geometric properties as p(.) when p − > 1. Before, let us introduce the following notations: where a ∈ [1, +∞).

Definition 4.
Consider the vector space p(.) . A nonempty subset C of p(.) is said to satisfy the uniform decrease condition (in short (UD)) if for any α > 0, there exists a > 1 such that Obviously the condition (UD) passes from a set to its subsets. Moreover, if p(·) is identically equal to 1, then the only (UD) subset is C = {0}. 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 p(.) satisfies the condition (UD). Indeed, let C be a nonempty subset of p(.) and α > 0. Let a ∈ (1, p − ). Then J a = ∅ which implies Therefore, the condition (UD) 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
Consider the subset C is nonempty, convex and ρ-closed. Let us show that it satisfies the condition (UD). Indeed, fix for all x ∈ C, which proves our claim that C is (UD).
Before we give a characterization of subsets which satisfy the condition (UD), we need to introduce a new class of subsets of p(.) .

Definition 5.
Consider the vector space p(.) such that p − = 1 and p(·) not identically equal to 1. Let f : (0, +∞) → (1, 2] be a nondecreasing function. Define the set C f to be Note that C f is never empty since 0 ∈ C f . Some of the basic properties of C f are given in the following lemma.

Lemma 2.
Consider the vector space p(.) such that p − = 1 and p(·) not identically equal to 1. Let f : (0, +∞) → (1, 2] be a non-decreasing function. Then the following properties hold: The Fatou property implies easily that C f is ρ-closed as a subset of p(.) which in turn implies that C f is ρ-complete. non-decreasing such that C ⊂ C f .
Proof. First, we prove that C f satisfies the condition (UD). Fix α > 0. If we take a = f (α), we obtain sup which proves our claim. Clearly, any subset C of C f will also satisfy the condition (UD).
Proposition 2 allows us to focus on the subsets C f instead of subsets which satisfy the condition (UD). 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.
Since ρ is convex, we have which implies Next, set Since I f (α) = K ∪ L, we obtain ρ I f (α) (z) = ρ K (z) + ρ L (z), for all z ∈ C f . From our assumptions, we have Assume first that Using Lemma 1, we obtain Using the convexity of the modular, we have For the second case, assume Set c = ε 8 , L 1 = n, |x n − y n | ≤ c |x n | + |y n | and L 2 = L\L 1 .
Since c < 1, we obtain Hence Our assumption on ρ L x − y 2 implies For any n ∈ L 2 , we have Using Lemma 1, we obtain x n + y n 2 p(n) for any n ∈ L 2 . Hence Both cases imply that ρ is (UC2) on C f with since f (a) > 1, for any a > 0. Since f (·) is nondecreasing, we may set to see that in fact ρ is (UUC2) on C f which completes the proof of Theorem 2.
The following lemma will be useful: Consider the vector space p(.) such that p − = 1 and p(·) not identically equal to 1.
In the next section, we will prove a fixed-point theorem for modular nonexpansive mappings.

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 ρ-closed convex subsets which satisfies the condition (UD). Proposition 3. Consider the vector space p(.) such that p − = 1 and p(·) not identically equal to 1. Let f : (0, +∞) → (1, 2] non-decreasing. Any nonempty ρ-closed convex subset C of C f is proximinal, i.e., for any x ∈ C f such that there exists a unique c ∈ C such that d ρ (x, C) = ρ(x − c).
Proof. Without loss of generality, we assume that x / ∈ C. Since C is ρ-closed we have, R = d ρ (x, C) > 0. For any n ≥ 1, there exists y n ∈ C such that ρ(x − y n ) < R(1 + 1/n). We claim that {y n /2} is ρ-Cauchy. Assume not. Then there exists a subsequence {y φ(n) } of {y n } and ε 0 > 0 such that for any n > m ≥ 1. According to Lemma 3, Since , and using Theorem 2, we obtain Since y φ(n) and y φ(m) are in C and C is convex, we obtain If we let m → +∞, we obtain This contradiction implies that {y n /2} is ρ-Cauchy. Since p(·) is ρ-complete, there exists y ∈ p(·) such that {y n /2} ρ-converges to y. Since C is convex and ρ-closed, we conclude that 2y ∈ C. Using the Fatou property, we have If we set c = 2y, we obtain d(x, C) = ρ(x − c). The uniqueness of the point c comes from the fact that ρ is strictly convex on C g since it is (UUC2).
The next result discusses an intersection property known as the property (R) [9]. Recall that a nonempty ρ-closed convex subset C of p(·) is said to satisfy the property (R) if for any decreasing sequence of nonempty ρ-closed ρ-bounded convex subsets of C have a nonempty intersection.
Proof. Let {C n } be a decreasing sequence of nonempty ρ-closed ρ-bounded convex subsets of C f . Let x ∈ C 1 . We have Since {C n } is decreasing, the sequence {d ρ (x, C n )} is increasing bounded above by δ ρ (C 1 ).
If R = 0, then x ∈ C n for any n ≥ 1, which will imply n≥1 C n = ∅. Otherwise, assume R > 0. Using Proposition 3, there exists c n ∈ C n such that d ρ (x, C n ) = ρ(x − c n ), for any n ≥ 1. Similar argument as the one used in the proof of Proposition 3 will show that {c n /2} is ρ-Cauchy and converges to c ∈ p(·) . Since {C n } is a decreasing sequence of ρ-closed subsets, we conclude that 2c ∈ n≥1 C n . Again this will show that n≥1 C n = ∅ which completes the proof of Proposition 4. Moreover, using Fatou property, we note that which will imply

Remark 1.
Let us note that under the assumptions of Proposition 4, the conclusion still holds when we consider any family {C α } α∈Γ of nonempty, convex, ρ-closed subsets of C, where (Γ, ≺) is upward directed, such that there exists x ∈ C which satisfies sup . Without loss of generality, we may assume d > 0. For any n ≥ 1, there exists Since (Γ, ≺) is upward directed, we may assume α n ≺ α n+1 which implies C α n+1 ⊂ C α n . Proposition 4 implies C 0 = n≥1 C α n = ∅. Clearly C 0 is ρ-closed and using the last noted point in the proof of Proposition 4, we obtain . We claim that c 0 ∈ C α , for any α ∈ Γ. Indeed, fix α ∈ Γ. If for some n ≥ 1 we have α ≺ α n , then obviously we have c 0 ∈ C α n ⊂ C α . Therefore let us assume that α ≺ α n , for any n ≥ 1. Since Γ is upward directed, there exists β n ∈ Γ such that α n ≺ β n and α ≺ β n , for any n ≥ 1. We can also assume that β n ≺ β n+1 for any n ≥ 1. Again we have C 1 = n≥1 C β n = ∅. Since C β n ⊂ C α n , for any n ≥ 1, we obtain C 1 ⊂ C 0 . Moreover we Hence, d ρ (x, C 1 ) = d which implies the existence of a unique point c 1 ∈ C 1 such that d ρ (x, C 1 ) = ρ(x − c 1 ) = d. Since ρ is (SC) on C f , we obtain c 0 = c 1 . In particular, we have c 0 ∈ C β n , for any n ≥ 1. Since α ≺ β n , we conclude that C β n ⊂ C α , for any n ≥ 1, which implies c 0 ∈ C α . Since α was taking arbitrary in Γ, we obtain c 0 ∈ α∈Γ C α , which implies α∈Γ C α = ∅ as claimed.
The next result is necessary to obtain the fixed-point theorem sought for ρ-nonexpansive mappings.
Proposition 5. Consider the vector space p(.) such that p − = 1 and p(·) are not identically equal to 1. Let f : (0, +∞) → (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 f not reduced to one point, there exists x ∈ C such that sup y∈C ρ(x − y) < δ ρ (C).
Proof. Let C be a ρ-closed convex ρ-bounded subset C of C f not reduced to one point. Since C is not reduced to one point, we have δ ρ (C) > 0. Let x, y ∈ C such that x = y. Set Fix c ∈ C. Using Lemma 3, we have x − c and y − c are in Since c was taken arbitrary in C, we conclude that 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 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 p(.) , which satisfies the condition (UD). Any ρ-nonexpansive mapping T : C → C has a fixed point.
Proof. Since C satisfies the condition (UD), Proposition 2 secures the existence of a nondecreasing function f : (0, +∞) → (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., δ ρ (C) > 0. Consider the family The family F is not empty since C ∈ F . Since C is bounded, we use Remark 1 to be able to use Zorn's lemma and conclude that 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(T(K 0 )) to be the intersection of all ρ-closed convex subset of C containing T(K 0 ). Hence co(T(K 0 )) ⊂ K 0 since K 0 ∈ F . Moreover, we have T(co(T(K 0 ))) ⊂ T(K 0 ) ⊂ co(T(K 0 )), which implies that co(T(K 0 )) ∈ F . K 0 being a minimal element of F we deduce that K 0 = co(T(K 0 )). Using Proposition 5, we deduce the existence of x 0 ∈ K 0 such that Define the subset K = x ∈ K 0 , sup y∈K 0 ρ(x − y) ≤ r 0 . K is not empty since x 0 ∈ K. Note that we have K = y∈K 0 B ρ (y, r 0 ) ∩ K 0 . Using the properties of modular balls, K is a ρ-closed and convex subset of K 0 . Next, we prove that T(K) ⊂ K. Indeed, let x ∈ K. Since T is ρ-nonexpansive, we have ρ(T(x) − T(y)) ≤ ρ(x − y) ≤ r 0 , for all y ∈ K 0 . So we have T(y) ∈ B ρ (T(x), r 0 ) ∩ K 0 , which implies T(K 0 ) ⊂ B ρ (T(x), r 0 ). Since K 0 = co(T(K 0 )), we conclude that K 0 ⊂ B ρ (T(x), r 0 ), which implies ρ(T(x) − y) ≤ r 0 , for all y ∈ K 0 . Hence T(x) ∈ K. Since x was taken as arbitrary in K, we obtain T(K) ⊂ K. The minimality of K 0 will force K = K 0 . Hence This is a contradiction. Therefore, K 0 is reduced to one point and it is a fixed point of T because T(K 0 ) ⊂ K 0 .

Remark 2.
In Theorem 3, the condition (UD) can be replaced by the following condition which is slightly more general: there exists x 0 ∈ p(.) such that x 0 + C satisfies the condition (UD).

Conflicts of Interest:
The authors declare no conflict of interest.

Abbreviations
The following abbreviations are used in this manuscript: