Journal of Function Spaces

Volume 2018, Article ID 5850181, 11 pages

https://doi.org/10.1155/2018/5850181

## Fixed Point Results for a Class of Monotone Nonexpansive Type Mappings in Hyperbolic Spaces

^{1}Department of Mathematics, Visvesvaraya National Institute of Technology, Nagpur 440010, India^{2}Department of Mathematical Sciences, UAEU, P.O. Box 15551, Al-Ain, UAE

Correspondence should be addressed to Ahmed Al-Rawashdeh; ea.ca.ueau@hedhsawarlaa

Received 1 March 2018; Revised 24 April 2018; Accepted 30 May 2018; Published 2 July 2018

Academic Editor: Giuseppe Marino

Copyright © 2018 Rameshwar Pandey 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.

#### Abstract

We consider a wider class of nonexpansive type mappings and present some fixed point results for this class of mapping in partially hyperbolic spaces. Indeed, first we obtain some existence results for this class of mappings. Next, we present some convergence results for an iteration algorithm for the same class of mappings. Some illustrative nontrivial examples have also been discussed. Finally, we provide an application of our results to nonlinear integral equations.

#### 1. Introduction

A mapping from the set of reals to a metric space is said to be* metric embedding* if for all The image of under a metric embedding is called a metric line. The image of a real interval under metric embedding is called a* metric segment*. Assume that has a family of metric lines such that, for each pair , there is a unique metric line in which passes through and This metric line determines a unique metric segment joining and This segment is denoted by and this is an isometric image of the real interval We denote by the unique point of which satisfieswhere Such a metric space with a family of metric segments is called a* convex metric space* [1]. Moreover, if we havefor all , then is said to be a* hyperbolic metric space* [2].

Hyperbolic spaces are more general than normed spaces and CAT(0) spaces. These spaces are nonlinear. Indeed, all normed linear and CAT(0) spaces are hyperbolic spaces (cf. [3–5]). As nonlinear examples, one can consider the Hadamard manifolds [6] and the Hilbert open unit ball equipped with the hyperbolic metric [7].

A mapping is said to be nonexpansive if , for all The study of existence of fixed point of nonexpansive mappings has been of great interest in nonlinear analysis. Fixed point theory of nonexpansive mappings in hyperbolic spaces has been extensively studied (cf. [2, 8–14]). Bin Dehaish and Khamsi [8] obtained a fixed point theorem for a monotone nonexpansive mapping in the setting partially ordered hyperbolic metric spaces.

On the other hand, a number of extensions and generalizations of nonexpansive mapping has been considered by many authors (cf. [15–26]). Greguš [21] considered the following class of nonexpansive type mappings. Let be a mapping such that, for all ,where , , and are nonnegative constants such that A mapping satisfying (3) is also known as Reich type nonexpansive mapping. A mapping satisfying (3) also satisfies the following condition:

where

Suzuki [26] introduced the following class of nonexpansive type mappings.

*Definition 1 (see [26]). *Let be a nonempty subset of a Banach space A mapping is said to satisfy condition if, for all ,Notice that the class of mappings satisfying (3)–(5) properly contains the class of nonexpansive mappings and need not be continuous.

Motivated by the above fact and works of Greguš [21], Suzuki [26], and others, in this paper, we make an attempt to define a wider class of nonexpansive type mapping which properly contains nonexpansive, Reich type nonexpansive, and Suzuki type nonexpansive mappings. In particular, in Section 3, we study some existence results in partially ordered hyperbolic space for this class of nonexpansive type mapping and some illustrative nontrivial examples have also been discussed. In Section 4, we present some convergence results for an iteration algorithm due to Abbas and Nazir [27]. In Section 5, we discuss an application of our results to nonlinear integral equations.

#### 2. Preliminaries

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

*Definition 2 (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: (K1);(K2);(K3);(K4)

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 (cf. [8]).

*Definition 3 (see [28, 29]). *Let be a hyperbolic metric space. For any , and SetWe say that is uniformly convex if , for any and

*Definition 4 (see [30]). *A hyperbolic metric space is said to satisfy property if, for each decreasing sequence of nonempty bounded closed convex subsets of ,

Uniformly convex hyperbolic spaces satisfy the property ; see [8].

*Definition 5 (see [8]). *Let be a metric space endowed with a partial order. A mapping is said to be monotonefor all

*Definition 6 (see [8]). *Let be a metric space endowed with a partial order. A mapping is said to be monotone nonexpansive if is monotone andfor all such that are comparable.

*Definition 7 (see [31]). *Let be a subset of a metric space A mapping is said to satisfy condition if there exists a nondecreasing function satisfying and for all such that for all , and here denotes the distance of from

Let be a nonempty subset of a hyperbolic metric space and a bounded sequence in For each , define(i)asymptotic radius of at as ;(ii)asymptotic radius of relative to as (iii)asymptotic centre of relative to by

Lim in [32] introduced the concept of -convergence in a metric space. Kirk and Panyanak in [33] used Lim’s concept in CAT(0) spaces and showed that many Banach space results involving weak convergence have precise analogs in this setting.

*Definition 8 (see [33]). *A bounded sequence in is said to -converge to a point , if is the unique asymptotic centre of every subsequence of

*Definition 9 (see [8]). *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

We know that every bounded sequence generates a unique type function.

Lemma 10 (see [8]). *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 *

Now, we rephrase the concept of -convergence in hyperbolic metric spaces.

*Definition 11. *A bounded sequence in is said to -converge to a point if is unique and the type function generated by every subsequence of attains its infimum at

Abbas and Nazir [27] introduced an iteration process which can be defined in the framework of hyperbolic metric spaces as follows:where , , and are real sequences in

#### 3. Existence Results

In this section, we define a new class of nonexpansive type mappings and present some auxiliary and an existence result (see Theorem 16 below). We also discuss a couple of illustrative examples.

*Definition 12. *Let be a partially ordered metric space and let be a monotone mapping. Then is called* monotone Reich-Suzuki type nonexpansive mapping,* if there exists a such thatfor all such that and are comparable.

Lemma 13. *Let be a uniformly convex partially ordered hyperbolic metric space and a nonempty closed convex subset of Let be a monotone mapping. Let be such that . Then, for sequence defined by (13), we have *(a)* (or );*(b)* (or ), provided -converges to a point , for all *

*Proof. *We shall use induction to prove (a). By assumption, we have By the convexity of order interval and (13), we haveAs is monotone, Again, by convexity of order interval and (13), we haveCombining (15) and (16), we getAs is monotone, Again, by convexity of order interval and (13), we haveBy (15), (16), and (18), we getThus (a) is true for Now, suppose it is true for ; that is,By convexity of order interval and (13), we haveAs is monotone, Again, by convexity of order interval and (13), we haveFrom (21) and (22), we haveso By convexity of order interval , (23), and (13), we haveFrom (21), (23), and (24), we haveand then From (25), we haveBy convexity of order interval and (13), we haveso By convexity of order interval and (13), we haveBy (27) and (28), . By convexity of order interval we get Hence, from (13), we haveFrom (28) and (29), we havethat isSuppose is a -limit of Here the sequence is monotone increasing and the order interval is closed and convex. We claim that for a fixed If , then the type function generated by subsequence of defined by leaving first terms of the sequence will not attain an infimum at , which is a contradiction to the assumption that is a -limit of the sequence This completes the proof.

*Lemma 14. Let be a partially ordered hyperbolic metric space and a nonempty subset of Let be a monotone Reich-Suzuki type nonexpansive mapping. Then, for each , (i);(ii)Either or ;(iii)Either or , where and are comparable.*

*Proof. *As by definition of monotone Reich-Suzuki type nonexpansive mapping, we havesince , soTo prove (ii), arguing by contradiction, we suppose thatBy (i) and triangle inequality, we have which is a contradiction. Hence (ii) holds. Condition (iii) follows directly from condition (ii).

*The following lemma is a consequence of the above lemma.*

*Lemma 15. Let be a partially ordered hyperbolic metric space and a nonempty subset of Let be a monotone Reich-Suzuki type nonexpansive mapping. Then, for all such that and are comparable, we have*

*Theorem 16. Let be a uniformly convex partially ordered hyperbolic metric space and a nonempty closed convex subset of Let be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists such that and are comparable. Let the sequence defined by (13) be bounded, and there exists a point such that every point of the sequence is comparable with and Then has a fixed point.*

*Proof. *Suppose is a bounded sequence and Then there exists a subsequence of such thatBy Lemma 13, we have Define for all Clearly, for each , is closed convex and so is nonempty. SetThen, is closed convex subset of Let ; then for all Since is monotone, for all ,This implies that Let be a type function generated by ; that is,From Lemma 10, there exists a unique element such thatBy definition of type function,Using Lemma 15, we get By the uniqueness of minimum point this implies that , and hence the proof is completed.

*Now, let us illustrate the following examples.*

*Example 17. *Let (the set of reals) be equipped with the usual ordering and standard norm Let and be a mapping defined byThen(1) is not a nonexpansive mapping,(2) is monotone Reich-Suzuki type nonexpansive mapping. For and , we have Therefore is not a nonexpansive mapping.

Now, we show that is monotone Reich-Suzuki type nonexpansive mapping with We consider the following three cases:(i)Let , and we have (ii)Let , and we have (iii)Let and , and we have Therefore, is monotone Reich-Suzuki type nonexpansive mapping with unique fixed point

*Notice that the space considered in the above example was a linear space. Now we present an example of a hyperbolic space which is not linear. Therefore it is a nontrivial example of a hyperbolic space.*

*Example 18. *Let Define byfor all and in Then it can be easily seen that is a metric space. Now, for , define a function byWe show that is a hyperbolic metric space. For , , , and in , consider the following:(K1)(K2)(K3)(K4)Therefore, is a hyperbolic metric space but not a normed linear space. Now, let us define an order on as follows: for and , if and only if or and Thus is an ordered hyperbolic metric space.

Let and be a mapping defined byFirst we show that is not a nonexpansive mapping on Let and ThenNow, we show that is monotone Reich-Suzuki type nonexpansive mapping for We consider the following cases.

*Case i. *If and , then

*Case ii. *If and , then Therefore is monotone Reich-Suzuki type nonexpansive mapping. The only fixed point of is

*4. Convergence Results*

*4. Convergence Results*

*In this section, we discuss some strong convergence and -convergence results in a partially ordered hyperbolic space for Abbas and Nazir iteration algorithm [27]. Our results are prefaced by the following proposition and lemma.*

*Proposition 19. Let be a partially ordered hyperbolic metric space and a nonempty subset of Let be a monotone Reich-Suzuki type nonexpansive mapping with a fixed point Then is quasi-nonexpansive; that is, for all and such that and are comparable.*

*Lemma 20 (see [34]). Let be a uniformly convex hyperbolic metric space with monotone modulus of uniform convexity Let and be a sequence such that If and are sequences in such that , , and for some , then we have *

*Theorem 21. Let be a uniformly convex partially ordered hyperbolic metric space and a nonempty closed convex subset of Let be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists such that and are comparable. Suppose is nonempty, and and are comparable for every Let be a sequence defined by (13). Then the following assertions hold: (i)The sequence is bounded.(ii) for all (iii) exists and exists, where denotes the distance from to (iv)*

*Proof. *Without loss of generality, we may assume that Since is monotone, By (13) (as in (16) and (17)), we haveSo, By monotonicity of ,and also, from (58), we have By monotonicity of ,By (13) (as in (18)),From (60) and (61), we have , so, by monotonicity of ,Again from (31) and (62), for , we haveApplying the same argument, we getNow, by (13) and Proposition 19, we haveBy (13), (65), and Proposition 19, we haveBy (13), (65), (66), and Proposition 19, we haveThus, the sequence is bounded and decreasing, so, exists. For each and we have Taking infimum over all , we get for all So the sequence is bounded and decreasing. Therefore exists. SupposeFrom (68) and Proposition 19, we haveBy (65) and (68), we haveFrom (70) and Proposition 19, we haveBy (66) and (68), we haveBy (72) and Proposition 19, we haveBy (68) and (13), we haveIn view of (71), (73), (74), and Lemma 20, we haveBy (13), letting and using (75), we haveBy triangle inequality and Proposition 19, we have and as , we haveBy (70) and (79), we have orFinally, from (68), (69), (81), and Lemma 20, we conclude that

*Now, we present the following main result about -convergence.*

*Theorem 22. Let be a uniformly convex partially ordered hyperbolic metric space and a nonempty closed convex subset of Let be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists such that and are comparable and is nonempty and totally ordered. Then the sequence defined by (13) -converges to a fixed point of .*

*Proof. *By Theorem 21, the sequence is bounded. Therefore there exists a subsequence of such that -converges to some By Lemma 13, we haveNow, we show that every -convergent subsequence of has a unique -limit in Suppose has two subsequences and -converging to and , respectively. By Theorem 21, is bounded andWe claim that Let be a type function generated by ; that is,By Lemma 15 and (83), we have By the uniqueness of the element and definition of -convergence, Similarly By the definition of -convergence and Lemma 10, we have which is a contradiction, unless

*Next, we present a strong convergence theorem.*

*Theorem 23. Let be a uniformly convex partially ordered hyperbolic metric space and , , and be the same as in Theorem 21 with The sequence defined by (13) converges strongly to a fixed point of if and only if , provided is a totally ordered set.*

*Proof. *Suppose that From Theorem 21, exists, soFirst we show that the set is closed. For this, let be a sequence in converging to a point Since for all , we have As ,and henceThen, converges strongly to This implies that Therefore is a closed set. In view of (87), let be a subsequence of sequence such that