Fixed Point Results of a General Class of Monotone Nonexpansive Mappings in Hyperbolic Metric Spaces

Outass, Rida; Chaira, Karim; Marhrani, El Miloudi; El Harmouchi, Nour-eddine

doi:https://doi.org/10.1155/2020/5854674

Journal of Function Spaces

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Research Article | Open Access

Volume 2020 | Article ID 5854674 | https://doi.org/10.1155/2020/5854674

Fixed Point Results of a General Class of Monotone Nonexpansive Mappings in Hyperbolic Metric Spaces

Rida Outass,¹Karim Chaira,¹El Miloudi Marhrani,¹and Nour-eddine El Harmouchi¹

Academic Editor: Pasquale Vetro

Received09 Jan 2020

Revised24 Mar 2020

Accepted06 Apr 2020

Published21 Jul 2020

Abstract

In this paper, we introduce and study some properties of a new class of generalized monotone nonexpansive mappings in hyperbolic metric spaces. Further, we give some results on the convergence of the Mann iterative process for this class of mappings in hyperbolic ordered metric spaces with some interesting examples.

1. Introduction

Nonexpansive mappings appear in many nonlinear problems modeled by integral equations arising from different scientific fields as mechanics, electricity, and population dynamics. Recall that a nonexpansive mapping is 1-Lipschitzien mapping.

The study of the existence of fixed points for nonexpansive mappings was initiated by Browder [1, 2], Gôhde [3], and Kirk [4]. In the last few years, a number of extensions and generalizations of nonexpansive mappings in different spaces were obtained by many mathematicians. In 2008, Suzuki [5] introduced a generalized definition of nonexpansive mappings, named condition or Suzuki generalize nonexpansive mapping, as follows:

Definition 1. Let be a subset of a Banach space . A mapping is said to satisfy condition if for all ,

He proved also some existence and convergence results for such type of mappings. Patir et al. [6] defined the following general class of nonexpansive-type mappings, and they presented some fixed point results for this class of operators.

Definition 2. Let be a nonempty subset of a Banach space . Let and such that . A mapping is said to satisfy the condition on if for all

Otherwise, the structure of spaces plays an important role in the study of the existence of fixed points. It is well known in the literature that Banach spaces have been studied extensively, because Banach spaces always have convex structures. However, metric spaces do not naturally benefit from this structure. Therefore, there is a need to introduce and define convex structures.

The study of fixed point theory for nonexpansive mappings in the framework of hyperbolic metric spaces was initiated by Takahashi [7]. He formulated some fixed point results in convex metric spaces. Goebel and Kirk [8] used hyperbolic-type spaces, which contain hyperbolic metric spaces (see also [9]).

Reich and Shafrir [10] introduced hyperbolic metric spaces as an appropriate background to study the existence and the approximation of fixed points for nonexpansive mappings. Accommodating previous definitions of hyperbolic metric spaces, Kohlenbach [11] introduced a more general definition of hyperbolic metric spaces and he proved the existence of fixed points for nonexpansive mappings. This class of spaces includes all normed linear spaces and Busemann spaces [12]. Leustean [13] showed that CAT(0) spaces are uniformly convex hyperbolic metric spaces. In 2016, Bin Dehaish and Khamsi [14] gave an analogue theorem to Browder and Göhde’s fixed point theorem for monotone nonexpansive mappings in uniformly convex hyperbolic metric spaces. Recently, Shukla et al. [15] present some existence and convergence results for a monotone mapping satisfying condition in partially ordered hyperbolic metric spaces.

Motivated by all these facts, we introduce in this paper a general class of nonexpansive-type mappings. We prove some fixed point existence results in uniformly convex hyperbolic metric spaces. Further, we study the approximation of fixed points for such type of mapping using the Mann iterative process defined in hyperbolic metric spaces with some illustrative examples. The results obtained in this work extend and generalize corresponding results on uniformly convex Banach spaces and CAT(0) spaces and many other results in this direction.

2. Preliminaries

Let be a metric space. A path joining to is a map from a closed interval to such that , , and for all . In particular, is an isometry and . The image of is called a metric segment joining and . When it is unique, the metric segment is denoted by . We shall denote by the unique point of which satisfies

Such metric spaces are usually called convex metric spaces (see Takahashi [7])

Let us recall the following definition which is due to Kohlenbach [11].

Definition 3 (see [11]). A triplet is said to be a hyperbolic metric space if is a metric space and is a function such that for all and , the following hold:(i)(ii)(iii)(iv)

The set is called the metric segment with endpoints and . Now onwards, we write . A subset of is said to be convex if for all and .

When there is no ambiguity, we write for .

Let be a partially ordered set with a partial order “”, and let be a partially ordered hyperbolic metric space. We say that are comparable whenever or . Throughout, we will assume that order intervals are closed convex subsets of hyperbolic metric space . We denote these intervals as follows:for any .

Definition 4 (see [16, 17]). Let be a hyperbolic metric space. We say that is uniformly convex if for any , and :

Proposition 5 (see [18]). Let be a uniformly convex hyperbolic metric space and be a decreasing sequence of nonempty closed, bounded, and convex subsets of X. Then,

Lemma 6 (see [19]). Let be a uniformly convex hyperbolic metric space with a monotone modulus of uniform convexity . Let and be a sequence such that . If and are two sequences in such that , , and , for some , then

Definition 7. Let be a metric space endowed with a partial order. A mapping is said to be monotone nondecreasing, if satisfiesfor all .

Definition 8 (see [14]). Let be a partially ordered metric space. A mapping is called monotone nonexpansive, if is monotone andfor all such that and are comparable.

Definition 9 (see [14]). Let be a nonempty subset of a hyperbolic metric space . A function is said to be a type function, if there exists a bounded sequence in such thatfor any .

Lemma 10 (see [14]). Let be a uniformly convex hyperbolic metric space and a nonempty closed convex subset of . Let be a type function. Then, is continuous. Moreover, there exists a unique minimum point such that .

3. Main Results

Before we give the existence and convergence theorems, we present a class of monotone nonexpansive mappings and some auxiliary results.

Definition 11. Let be a partially ordered metric space, and let and such that . Let be a monotone mapping. Then, is said to satisfy the condition on if for all such that ,

Clearly, this class includes the class of monotone nonexpansive mappings (for ).

Example 1. Let be equipped with the usual ordering and the distance . Let and be a mapping defined by(1) is not a monotone nonexpansive mapping(2) is monotone and satisfies condition Setting and , we obtainHowever, is not a nonexpansive mapping. Now, we show that is monotone and satisfies condition with and . In fact, we have the following three cases:(i)Let such that , and we have(ii)Let such that , and we have(iii)Let and , and we haveOtherwise, we havesince . From (19) and (20), we haveSince , then

Note that there is no relationship between the condition and the class of mapping satisfying the condition , which contradicts with the statement of Patir et al. in the paper [6]. To justify this, we give the following example.

Example 2. Define the mapping byThe monotone mapping satisfies the condition (for details, see [5]) but does not satisfy the condition for . In fact, for and , we haveTherefore, the condition is not satisfied.

The Mann [20] iterative scheme has been used to study the approximation of fixed points of nonexpansive mappings by many authors [21–26]. Following the same line, we define the Mann process in the framework of hyperbolic metric spaces in order to prove its convergence to fixed points for monotone mapping satisfying condition .

Definition 12. Let be an ordered hyperbolic metric space. Let be a nonempty convex subset of and a nondecreasing mapping on . Fix . The Mann iteration process is the sequence defined byfor all , where .

Proposition 13 (see [8]). Let be an ordered hyperbolic metric space. Let and be two sequences in and which satisfy for all ,(i) with (ii)Then, for all and ,

The next lemma is a technical tool which will be used to prove that is an almost fixed point sequence.

Lemma 14. Let be a nonempty closed convex subset of a partially ordered hyperbolic metric space and be a monotone mapping satisfying the condition on , with . Assume that there exists such that . Let be the Mann iterative sequence defined by (25) such that with and . Then, for any and , we have

Proof. Let such that . Using the convexity of the order interval and the monotonicity of , we obtainThus, by induction, we getfor any . On the other hand, we haveSince for all , we haveBy the condition , for and , we haveSince , then . Hence,for all . According to (29), (30), and (33), it is concluded that all the assumptions of Proposition 13 are satisfied, where , which impliesfor any and .

Theorem 15. Let be as above. Let be a nonempty, closed, convex, and bounded subset of and be a monotone mapping satisfying the condition on , with . Assume that there exists such that and the sequence defined by (25) such that , with and . Then, we have

Proof. First we show that the sequence is decreasing. Indeed, by the triangle inequality, we haveFrom the proof of Lemma 14, we havefor all . Hence,for all . Consequently, is a decreasing and bounded sequence in . Thus, there exists such thatOn the other hand, we havesince and for all . Therefore, the inequality obtained in Lemma 14 impliesfor all and . If we let , we get , for any . Then,holds for all . Clearly, this is possible only for . Thus,

Lemma 16. Let be a partially ordered hyperbolic metric space and a nonempty subset of . Let be a monotone mapping which satisfies the condition. If is a fixed point of , then for all , ,

Proof. Let be a fixed point of and such that .
We haveBy the condition ,Hence,which impliesSince , then

In the sequel, we will use the fixed point set with the partial orders given by

Definition 17. Let be a partially ordered set and be a sequence in . The sequence is said to be -bounded if there exists a point such that , for all .

Lemma 18. Let be a nonempty closed and convex subset of a partially ordered hyperbolic metric space . Let be a monotone mapping satisfying the condition on , for such that and is nonempty. Let be a sequence defined by (25). Then, the following assertions hold:(i)The sequence is -bounded(ii) exists for , and exists, where denotes the distance from to

Proof. (i)Let , and fix . From the proof of Lemma 14, we haveSince is monotone and , we haveSo, . By monotonicity of ,Applying the same argument, we getThus, is -bounded.(ii)By Lemma 16, we haveHence,for all . Consequently, is a decreasing and bounded sequence in . Thus, exists. For each and we haveTaking infimum over all , we getfor all . So the sequence is bounded and decreasing. Therefore, exists.

Theorem 19. Let be a nonempty closed convex and bounded subset of a hyperbolic ordered metric space and be a monotone mapping satisfying the condition on , with . Assume that there exists such that and is nonempty. Let be a sequence defined by (25) where , with and . Then, converges to a fixed point of .

Proof. Assume that is nonempty, and let . Then, from the proof of Lemma 18, we haveBy Lemma 18, there exists such thatAssume that . By Theorem 15, one hasThis leads to , for large sufficient . Then,By the condition , we getOtherwise,Taking limit as goes to infinity, we getSince , we get a contradiction. Therefore, converges to .

Example 3. Let . Define bysuch that and , for all . It is obvious to see that is a metric space.
First step. Let . We can easily verify that is a nonempty, bounded, and convex subset of . We show that is closed in . Indeed, let be a sequence in and such thatThen, for all , there exists an such thatThis impliesThus,Using (69), it is easy to verify thatOn the other hand, fix , and we haveSince and , for all , thenBy (70) and (71), we conclude thatTherefore, .
Second step. Define a function byfor all and in . We show that is a hyperbolic metric space. For , , , , and , , we have the following:(i)(ii)(iii)(iv)Consider the partial ordering defined on as follows:where and in .
Thus, is a partially ordered hyperbolic metric space, but not a normed space.
Third step. Let be a mapping defined bywhere and designate constant functions, such thatIt is obvious that the mapping is monotone.(i) is not nonexpansive. In fact, for and , we have(ii) satisfies the condition for and . In fact, for and such that , we haveTherefore,Otherwise, if and (or equal to (1/2, 0)), it is quite easy to verify that the mapping satisfies the condition for and .
Therefore, is monotone and satisfies the condition and has a unique fixed point .
Fourth step. Suppose that and , then andwhere , for all . Indeed, assume that (82) is true for and let us prove it for . We haveSince , thenSince is monotone and satisfies condition and there exists such that , then all the conditions of Theorem 19 are satisfied. Consequently, it is obvious to see thatThus, converges to the fixed point .

Theorem 20. Let be a nonempty closed convex and bounded subset of a uniformly convex partially ordered hyperbolic metric space and be a monotone mapping satisfying the condition on , with . Assume that there exists such that and the sequence defined by (25) be -bounded such that with and . Then, has a fixed point , , and .

Proof. As is -bounded, then there exists a point such that , for all .
Let for all . It is clear that is nonempty, since , and is a closed and convex subset of , for all , and the sequence is decreasing, since for all . Hence, by Proposition 5,is a nonempty, closed, and convex subset of . Let ; then, by the definition of and the monotonicity of , we getHence, . Consider the type function generated by , that is, , for all From Lemma 10, there exists a unique element such thatMoreover,We assert that . Assume that this is not the case.
By Theorem 15, we have thenSince , then for all . By the condition , for and , we haveBy the triangle inequality,we obtainwhich impliesHence,By the uniqueness of the minimum point, we conclude a contradiction with our hypothesis . Then, we have .
is nonempty, because it contains . Following the same proof of Theorem 15, we obtain .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright © 2020 Rida Outass et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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