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

Authors: Rameshwar Pandey; Rajendra Pant; Ahmed Al-Rawashdeh
Journal: Journal of Function Spaces
 (2018)

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## 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.


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## Body

## 1. Introduction

A mappingp from the set of reals R to a metric space (X,ρ) is said to bemetric embedding if ρ(p(m),p(n))=|m-n| for all m,n∈R. The image of R under a metric embedding is called a metric line. The image of a real interval [a,b]={t∈R:a≤t≤b} under metric embedding is called ametric segment. Assume that (X,ρ) has a family F of metric lines such that, for each pair u,v∈X(u≠v), there is a unique metric line in F which passes through u and v. This metric line determines a unique metric segment joining u and v. This segment is denoted by [u,v] and this is an isometric image of the real interval [0,ρ(u,v)]. We denote by γu⊕(1-γ)v the unique point w of [u,v] which satisfies(1)ρu,w=1-γρu,vandρw,v=γρu,v,where γ∈[0,1]. Such a metric space with a family of metric segments is called aconvex metric space [1]. Moreover, if we have(2)ργu⊕1-γv,γw⊕1-γz≤γρu,w+1-γρv,zfor all u,v,w,z∈X, then X is said to be ahyperbolic 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 mappingT:X→X is said to be nonexpansive if ρ(T(u),T(v))≤ρ(u,v), for all u,v∈X. 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 T:X→X be a mapping such that, for all u,v∈X,(3)ρTu,Tv≤aρu,v+bρTu,u+cρTv,v,where a, b, and c are nonnegative constants such that a+b+c=1. A mapping satisfying (3) is also known as Reich type nonexpansive mapping. A mapping satisfying (3) also satisfies the following condition:(4)ρTu,Tv≤kρTu,u+kρTv,v+1-2kρu,v,wherek∈[0,1).Suzuki [26] introduced the following class of nonexpansive type mappings.Definition 1 (see [26]).
LetY be a nonempty subset of a Banach space X. A mapping T:Y→Y is said to satisfy condition (C) if, for all u,v∈Y,(5)12u-Tu≤u-vimpliesTu-Tv≤u-v.
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(X,ρ,H) is said to be a hyperbolic metric space if (X,ρ) is a metric space and H:X×X×[0,1]→X is a function such that, for all u,v,w,z∈X and β,γ∈[0,1], the following hold: (K1)
ρ(z,H(u,v,β))≤(1-β)ρ(z,u)+βρ(z,v);(K2)
ρ(H(u,v,β),H(u,v,γ))=|β-γ|ρ(u,v);(K3)
H(u,v,β)=H(v,u,1-β);(K4)
ρ(H(u,z,β),H(v,w,β))≤(1-β)ρ(u,v)+βρ(z,w).The setseg[u,v]≔{H(u,v,β);β∈[0,1]} is called the metric segment with endpoints u and v. Now onwards, we write H(u,v,β)=(1-β)u⊕βv. A subset Y of X is said to be convex if (1-β)u⊕βv∈Y for all u,v∈Y and β∈[0,1]. When there is no ambiguity, we write (X,ρ) for (X,ρ,H).LetE be a partially ordered set with a partial order ‘⪯’ and let (E,ρ,⪯) be a partially ordered hyperbolic metric space. We say that u,v∈E are comparable whenever u⪯v or v⪯u. Throughout, we will assume that order intervals are closed convex subsets of hyperbolic metric space E. We denote these intervals as follows:(6)a,⟶≔u∈E;a⪯uand⟵,b≔u∈E;u⪯b,for any a,b∈E (cf. [8]).Definition 3 (see [28, 29]).
Let(E,ρ) be a hyperbolic metric space. For any a∈E, r>0 and ϵ>0. Set(7)δr,ϵ=inf1-12ρ12u⊕12v,a;ρu,a≤r,ρv,a≤r,ρu,v≥rϵ.
We say thatE is uniformly convex if δ(r,ϵ)>0, for any r>0 and ϵ>0.Definition 4 (see [30]).
A hyperbolic metric space(E,ρ) is said to satisfy property (R) if, for each decreasing sequence {Fn} of nonempty bounded closed convex subsets of E, ⋂n=1∞Fn≠∅.Uniformly convex hyperbolic spaces satisfy the propertyR; see [8].Definition 5 (see [8]).
Let(E,ρ,⪯) be a metric space endowed with a partial order. A mapping T:X→X is said to be monotone(8)Tu⪯Tvwheneveru⪯v,for all u,v∈E.Definition 6 (see [8]).
Let(E,ρ,⪯) be a metric space endowed with a partial order. A mapping T:X→X is said to be monotone nonexpansive if T is monotone and(9)ρTu,Tv≤ρu,v,for all u,v∈E such that u,v are comparable.Definition 7 (see [31]).
LetK be a subset of a metric space (E,ρ). A mapping T:K→K is said to satisfy condition (I) if there exists a nondecreasing function g:[0,∞)→[0,∞) satisfying g(0)=0 and g(r)>0 for all r∈(0,∞) such that ρ(u,T(u))≥g(dist(u,F(T))) for all u∈K, and here dist(u,F(T)) denotes the distance of u from F(T).LetK be a nonempty subset of a hyperbolic metric space (E,ρ) and {un} a bounded sequence in E. For each u∈E, define(i)
asymptotic radius of{un} at u as r({un},u)≔limsupn→∞ρ(un,u);(ii)
asymptotic radius of{un} relative to K as(10)run,K≔inf⁡run,u;u∈K;(iii)
asymptotic centre of{un} relative to K by(11)Aun,K≔u∈K;run,u=run,K.Lim in [32] introduced the concept of Δ-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{un} in E is said to Δ-converge to a point u∈E, if u is the unique asymptotic centre of every subsequence {unk} of {un}.Definition 9 (see [8]).
LetK be a nonempty subset of a hyperbolic metric space (E,ρ). A function τ:K→[0,∞) is said to be a type function, if there exists a bounded sequence {un} in E such that(12)τu=limsupn→∞ρun,u,for any u∈K.We know that every bounded sequence generates a unique type function.Lemma 10 (see [8]).
Let(E,ρ,⪯) be a uniformly convex hyperbolic metric space and K a nonempty closed convex subset of E. Let τ:K→[0,∞) be a type function. Then τ is continuous. Moreover, there exists a unique minimum point z∈K such that τ(z)=inf⁡{τ(u);u∈K}.Now, we rephrase the concept ofΔ-convergence in hyperbolic metric spaces.Definition 11.
A bounded sequence{un} in E is said to Δ-converge to a point z∈E if z is unique and the type function generated by every subsequence {unk} of {un} attains its infimum at z.Abbas and Nazir [27] introduced an iteration process which can be defined in the framework of hyperbolic metric spaces as follows:(13)u1∈Kun+1=1-αnTvn⊕αnTwnvn=1-βnTun⊕βnTwnwn=1-γnun⊕γnTun;n∈N,where {αn}, {βn}, and {γn} are real sequences in (0,1).


## 3. Existence Results

In this section, we define a new class of nonexpansive type mappings and present some auxiliary and an existence result (see Theorem16 below). We also discuss a couple of illustrative examples.Definition 12.
Let(E,ρ,⪯) be a partially ordered metric space and let T:E→E be a monotone mapping. Then T is calledmonotone Reich-Suzuki type nonexpansive mapping, if there exists a k∈[0,1) such that(14)12ρu,Tu≤ρu,vimpliesρTu,Tv≤kρTu,u+kρTv,v+1-2kρu,vfor all u,v∈E such that u and v are comparable.Lemma 13.
Let(E,ρ,⪯) be a uniformly convex partially ordered hyperbolic metric space and K a nonempty closed convex subset of E. Let T:K→K be a monotone mapping. Let u1∈K be such that u1⪯T(u1)(T(u1)⪯u1). Then, for sequence {un} defined by (13), we have (a)
un⪯T(un)⪯un+1 (or un+1⪯T(un)⪯un);(b)
un⪯p (or p⪯un), provided {un}Δ-converges to a point p∈K, for all n∈N.Proof.
We shall use induction to prove (a). By assumption, we haveu1⪯T(u1). By the convexity of order interval [u1,T(u1)] and (13), we have(15)u1⪯w1⪯Tu1.As T is monotone, T(u1)⪯T(w1). Again, by convexity of order interval [T(u1),T(w1)] and (13), we have(16)Tu1⪯v1⪯Tw1.Combining (15) and (16), we get(17)u1⪯w1⪯Tu1⪯v1.As T is monotone, T(w1)⪯T(v1). Again, by convexity of order interval [T(w1),T(v1)] and (13), we have(18)Tw1⪯u2⪯Tv1.By (15), (16), and (18), we get(19)u1⪯Tu1⪯Tw1⪯u2⪯Tv1.Thus (a) is true for n=1. Now, suppose it is true for n; that is,(20)un⪯Tun⪯un+1.By convexity of order interval [un,T(un)] and (13), we have(21)un⪯wn⪯Tun.As T is monotone, T(un)⪯T(wn). Again, by convexity of order interval [T(un),T(wn)] and (13), we have(22)Tun⪯vn⪯Twn.From (21) and (22), we have(23)wn⪯Tun⪯vn⪯Twn,so T(wn)⪯T(vn). By convexity of order interval [T(wn),T(vn)], (23), and (13), we have(24)Tun⪯Twn⪯un+1⪯Tvn.From (21), (23), and (24), we have(25)un⪯Tun⪯vn⪯Twn⪯un+1⪯Tvn,and then T(vn)⪯T(un+1). From (25), we have(26)un+1⪯Tun+1.By convexity of order interval [un+1,T(un+1)] and (13), we have(27)un+1⪯wn+1⪯Tun+1,so T(un+1)⪯T(wn+1). By convexity of order interval [T(un+1),T(wn+1)] and (13), we have(28)un+1⪯Tun+1⪯vn+1⪯Twn+1.By (27) and (28), wn+1⪯vn+1. By convexity of order interval we get [T(wn+1),T(vn+1)]. Hence, from (13), we have(29)Twn+1⪯un+2⪯Tvn+1.From (28) and (29), we have(30)un+1⪯Tun+1⪯Twn+1⪯un+2⪯Tvn+1;that is(31)un+1⪯Tun+1⪯un+2.Suppose p is a Δ-limit of {un}. Here the sequence {un} is monotone increasing and the order interval [um,→) is closed and convex. We claim that p∈[um,→) for a fixed m∈N. If p∉[um,→), then the type function generated by subsequence {ur} of {un} defined by leaving first m-1 terms of the sequence {un} will not attain an infimum at p, which is a contradiction to the assumption that p is a Δ-limit of the sequence {un}. This completes the proof.Lemma 14.
Let(E,ρ,⪯) be a partially ordered hyperbolic metric space and K a nonempty subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping. Then, for each u,v∈K, (i)
ρ(T(u),T2(u))≤ρ(u,T(u));(ii)
Either1/2ρ(u,T(u))≤ρ(u,v) or 1/2ρ(T(u),T2(u))≤ρ(T(u),v);(iii)
Eitherρ(T(u),T(v))≤kρ(T(u),u)+kρ(v,T(v))+(1-2k)ρ(u,v) or ρ(T2(u),T(v))≤kρ(T2(u),T(u))+kρ(T(v),v)+(1-2k)ρ(T(u),v),
whereu and v are comparable.Proof.
As1/2ρ(u,T(u))≤ρ(u,T(u)) by definition of monotone Reich-Suzuki type nonexpansive mapping, we have(32)ρTu,T2u≤kρTu,u+kρTu,T2u+1-2kρu,Tu,since 1-k>0, so(33)ρTu,T2u≤ρu,Tu.To prove (ii), arguing by contradiction, we suppose that(34)12ρu,Tu>ρu,vand12ρTu,T2u>ρTu,v.By (i) and triangle inequality, we have (35)ρu,Tu≤ρu,v+ρTu,v<12ρu,Tu+12ρTu,T2u≤ρu,Tu,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(E,ρ,⪯) be a partially ordered hyperbolic metric space and K a nonempty subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping. Then, for all u,v∈K such that u and v are comparable, we have(36)ρu,Tv≤3+k1-kρu,Tu+ρu,v.Theorem 16.
Let(E,ρ,⪯) be a uniformly convex partially ordered hyperbolic metric space and K a nonempty closed convex subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists u1∈K such that u1 and T(u1) are comparable. Let the sequence {un} defined by (13) be bounded, and there exists a point w∈K such that every point of the sequence {un} is comparable with w and liminfn→∞ρ(T(un),un)=0. Then T has a fixed point.Proof.
Suppose{un} is a bounded sequence and liminfn→∞ρ(T(un),un)=0. Then there exists a subsequence {unj} of {un} such that(37)limj→∞ρTunj,unj=0.By Lemma 13, we have u1⪯unj⪯unj+1. Define Kj={s∈K:unj⪯s} for all j∈N. Clearly, for each j∈N, Kj is closed convex and w∈Kj so Kj is nonempty. Set(38)K∞=⋂j=1∞Kj≠∅.Then, K∞ is closed convex subset of K. Let u∈K∞; then unj⪯u for all j∈N. Since T is monotone, for all j∈N,(39)unj⪯Tunj⪯Tu.This implies that T(K∞)⊆K∞. Let τ:K∞→[0,∞) be a type function generated by unj; that is,(40)τu=limsupj→∞ρunj,u.From Lemma 10, there exists a unique element z∈K∞ such that(41)τz=infσu;u∈K∞.By definition of type function,(42)τTz=limsupj→∞ρunj,Tz.Using Lemma 15, we get (43)τTz=limsupj→∞ρunj,Tz≤3+k1-klimsupj→∞ρunj,Tunj+limsupj→∞ρunj,z=τz.By the uniqueness of minimum point this implies that T(z)=z, and hence the proof is completed.Now, let us illustrate the following examples.Example 17.
LetE=R (the set of reals) be equipped with the usual ordering and standard norm u=|u|. Let K=[0,1]⊂R and T:K→K be a mapping defined by(44)Tu=1-u2,ifu∈0,14u4+12,ifu∈14,1.Then(1)
T is not a nonexpansive mapping,(2)
T is monotone Reich-Suzuki type nonexpansive mapping.
Foru=0 and v=1/4, we have T(u)-T(v)=7/16>1/4=u-v. Therefore T is not a nonexpansive mapping.
Now, we show thatT is monotone Reich-Suzuki type nonexpansive mapping with k=1/2. We consider the following three cases:(i)
Letu,v∈[0,1/4), and we have (45)12Tu-u+Tv-v=123u2-1+3v2-1≥123u2-1-3v2-1=34u-v≥12u-v=Tu-Tv.(ii)
Letu,v∈[1/4,1], and we have (46)12Tu-u+Tv-v=123u4-12+3v4-12≥123u4-12-3v4-12=38u-v≥14u-v=Tu-Tv.(iii)
Letu∈[0,1/4) and v∈[1/4,1], and we have (47)12Tu-u+Tv-v=123u2-1+3v4-12≥123u2-1+3v4-12=382-2u-v≥142-2u-v=Tu-Tv.
Therefore,T is monotone Reich-Suzuki type nonexpansive mapping with unique fixed point 2/3.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.
LetE={(u1,u2)∈R2;u1,u2>0}. Define ρ:E×E→[0,∞) by(48)ρu,v=u1-v1+u1u2-v1v2for all u=(u1,u2) and v=(v1,v2) in E. Then it can be easily seen that (E,ρ) is a metric space. Now, for β∈[0,1], define a function H:E×E×[0,1]→E by(49)Hu,v,β=1-βu1+βv1,1-βu1u2+βv1v21-βu1+βv1.We show that (E,ρ,H) is a hyperbolic metric space. For u=(u1,u2), v=(v1,v2), z=(z1,z2), and w=(w1,w2) in E, consider the following:(K1)
(50)dz,Hu,v,β=z1-1-βu1-βv1+z1z2-1-βu1u2-βv1v2≤1-βz1-u1+βz1-v1+1-βz1z2-u1u2+βz1z2-v1v2=1-βdz,u+βdz,v.(K2)
(51)dHu,v,β,Hu,v,γ=1-βu1+βv1-1-γu1-γv1+1-βu1u2+βv1v2-1-γu1u2-γv1v2=β-γu1-v1+u1u2-v1v2=β-γdu,v.(K3)
(52)Hu,v,β=1-βu1+βv1,1-βu1u2+βv1v21-βu1+βv1=βv1+1-βu1,βv1v2+1-βu1u2βv1+1-βu1=Hv,u,1-β.(K4)
(53)dHu,z,β,Hv,w,β=1-βu1+βz1-1-βv1-βw1+1-βu1u2+βz1z2-1-βv1v2-βw1w2≤1-βu1-v1+u1u2-v1v2+βz1-w1+z1z2-w1w2=1-βdu,v+βdz,w.
Therefore,(E,ρ,H) is a hyperbolic metric space but not a normed linear space. Now, let us define an order on E as follows: for u=(u1,u2) and v=(v1,v2), u⪯v if and only if u1<v1 or u1=v1 and u2≤v2. Thus (E,ρ,⪯) is an ordered hyperbolic metric space.
LetK≔[1,4]×[1,4]⊂E and T:K→K be a mapping defined by(54)Tu1,u2=1,1,ifu1,u2≠4,432,32,ifu1,u2=4,4.First we show that T is not a nonexpansive mapping on K. Let u=(1,1) and v=(4,4). Then(55)ρTu,Tv=74>18=ρu,v.Now, we show that T is monotone Reich-Suzuki type nonexpansive mapping for k=1/2. We consider the following cases.Case i.
Ifu=(u1,u2) and v=(v1,v2)≠(4,4), then(56)ρTu,Tv=0≤12ρu,Tu+ρv,Tv.Case ii.
Ifu=(u1,u2)≠(4,4) and v=(v1,v2)=(4,4), then (57)12ρu,Tu+ρv,Tv=12u1-1+u1u2-1+1252+554=12u1-1+u1u2-1+658>74=ρTu,Tv.Therefore T is monotone Reich-Suzuki type nonexpansive mapping. The only fixed point of T is (1,1).


## 4. Convergence Results

In this section, we discuss some strong convergence andΔ-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(E,ρ,⪯) be a partially ordered hyperbolic metric space and K a nonempty subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping with a fixed point z∈K. Then T is quasi-nonexpansive; that is, ρ(T(u),z)≤ρ(u,z) for all u∈K and z∈F(T) such that u and z are comparable.Lemma 20 (see [34]).
Let(E,ρ) be a uniformly convex hyperbolic metric space with monotone modulus of uniform convexity δ. Let z∈E and {αn} be a sequence such that 0<a≤αn≤b<1. If {un} and {vn} are sequences in E such that limsupn→∞ρ(un,z)≤r, limsupn→∞ρ(vn,z)≤r, and limn→∞ρ(αnvn⊕(1-αn)un,z)=r for some r≥0, then we have limn→∞ρ(un,vn)=0Theorem 21.
Let(E,ρ,⪯) be a uniformly convex partially ordered hyperbolic metric space and K a nonempty closed convex subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists u1∈K such that u1 and T(u1) are comparable. Suppose F(T) is nonempty, and u1 and z are comparable for every z∈F(T). Let {un} be a sequence defined by (13). Then the following assertions hold: (i)
The sequence{un} is bounded.(ii)
max{ρ(un+1,z),ρ(vn,z),ρ(wn,z)}≤ρ(un,z) for all n∈N.(iii)
limn→∞ρ(un,z) exists and limn→∞dist(un,F(T)) exists, where dist(u,F(T)) denotes the distance from u to FT.(iv)
limn→∞ρ(T(un),un)=0.Proof.
Without loss of generality, we may assume thatu1⪯z. Since T is monotone, T(u1)⪯T(z)=z. By (13) (as in (16) and (17)), we have(58)u1⪯w1⪯Tu1⪯v1⪯Tw1.So, w1⪯z. By monotonicity of T,(59)Tw1⪯Tz=z,and also, from (58), we have v1⪯z. By monotonicity of T,(60)Tv1⪯Tz=z.By (13) (as in (18)),(61)Tw1⪯u2⪯Tv1.From (60) and (61), we have u2⪯z, so, by monotonicity of T,(62)Tu2⪯Tz=z.Again from (31) and (62), for n=2, we have(63)u2⪯Tu2⪯z.Applying the same argument, we get(64)un⪯Tun⪯z.Now, by (13) and Proposition 19, we have(65)ρwn,z=ρ1-γnun⊕γnTun,z≤1-γnρun,z+γnρTun,z≤1-γnρun,z+γnρun,z=ρun,z.By (13), (65), and Proposition 19, we have(66)ρvn,z=ρ1-βnTun⊕βnTwn,z≤1-βnρTun,z+βnρTwn,z≤1-βnρun,z+βnρwn,z≤1-βnρun,z+βnρun,z=ρun,z.By (13), (65), (66), and Proposition 19, we have(67)ρun+1,z=ρ1-αnTvn⊕αnTwn,z≤1-αnρTvn,z+αnρTwn,z≤1-αnρvn,z+αnρwn,z≤1-αnρun,z+αnρun,z=ρun,z.Thus, the sequence {ρ(un,z)} is bounded and decreasing, so, limn→∞ρ(un,z) exists. For each z∈F(T) and n∈N we have ρ(un+1,z)≤ρ(un,z). Taking infimum over all z∈F(T), we get dist(un+1,F(T))≤dist(un,F(T)) for all n∈N. So the sequence dist(un,F(T)) is bounded and decreasing. Therefore limn→∞dist(un,F(T)) exists. Suppose(68)limn→∞ρun,z=r.From (68) and Proposition 19, we have(69)limsupn→∞ρTun,z≤r.By (65) and (68), we have(70)limsupn→∞ρwn,z≤r.From (70) and Proposition 19, we have(71)limsupn→∞ρTwn,z≤r.By (66) and (68), we have(72)limsupn→∞ρvn,z≤r.By (72) and Proposition 19, we have(73)limsupn→∞ρTvn,z≤r.By (68) and (13), we have(74)r=limn→∞ρun+1,z=limn→∞ρ1-αnTvn⊕αnTwn,z.In view of (71), (73), (74), and Lemma 20, we have(75)limn→∞ρTvn,Twn=0.By (13), (76)ρun+1,Twn=ρ1-αnTvn⊕αnTwn,Twn≤1-αnρTvn,Twn,letting n→∞ and using (75), we have(77)limn→∞ρun+1,Twn=0.By triangle inequality and Proposition 19, we have (78)ρun+1,z≤ρun+1,Twn+ρTwn,z≤ρun+1,Twn+ρwn,z,and as n→∞, we have(79)r≤liminfn→∞ρwn,z.By (70) and (79), we have (80)r=limn→∞ρwn,z,or(81)r=limn→∞ρwn,z=limn→∞ρ1-γnun⊕γnTun,z.Finally, from (68), (69), (81), and Lemma 20, we conclude that limn→∞ρ(T(un),un)=0.Now, we present the following main result aboutΔ-convergence.Theorem 22.
Let(E,ρ,⪯) be a uniformly convex partially ordered hyperbolic metric space and K a nonempty closed convex subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists u1∈K such that u1 and T(u1) are comparable and F(T) is nonempty and totally ordered. Then the sequence {un} defined by (13) Δ-converges to a fixed point of T.Proof.
By Theorem21, the sequence {un} is bounded. Therefore there exists a subsequence {unk} of {un} such that {unk}Δ-converges to some p∈K. By Lemma 13, we have(82)u1⪯unk⪯p∀k∈N.Now, we show that every Δ-convergent subsequence of {un} has a unique Δ-limit in F(T). Suppose {un} has two subsequences {unk} and {unr}Δ-converging to l and m, respectively. By Theorem 21, {unk} is bounded and(83)limk→∞ρTunk,unk=0.We claim that l∈F(T). Let τ:K→[0,∞) be a type function generated by unk; that is,(84)τu=limsupk→∞ρunk,l.By Lemma 15 and (83), we have (85)τTl=limsupk→∞ρunk,Tl≤3+k1-klimsupk→∞ρunk,Tunk+limsupk→∞ρunk,l≤τl.By the uniqueness of the element l and definition of Δ-convergence, T(l)=l. Similarly T(m)=m. By the definition of Δ-convergence and Lemma 10, we have (86)limsupn→∞ρun,l=limsupk→∞ρunk,l<limsupk→∞ρunk,m=limsupn→∞ρun,m=limsupr→∞ρunr,m<limsupr→∞ρunr,l=limsupn→∞ρun,l,which is a contradiction, unless l=m.Next, we present a strong convergence theorem.Theorem 23.
Let(E,ρ,⪯) be a uniformly convex partially ordered hyperbolic metric space and K, T, and {un} be the same as in Theorem 21 with F(T)≠∅. The sequence defined by (13) converges strongly to a fixed point of T if and only if liminfn→∞dist(un,F(T))=0, provided F(T) is a totally ordered set.Proof.
Suppose thatliminfn→∞dist(un,F(T))=0. From Theorem 21, limn→∞dist(un,F(T)) exists, so(87)limn→∞distun,FT=0.First we show that the set F(T) is closed. For this, let {zn} be a sequence in F(T) converging to a point v∈K. Since 1/2ρ(zn,T(zn))=0≤ρ(zn,v) for all n∈N, we have (88)ρTzn,Tv=ρzn,Tv≤kρTzn,zn+kρTv,v+1-2kρzn,v≤kρTv,zn+kρzn,v+1-2kρzn,v.As 1-k>0,(89)ρzn,Tv≤ρzn,v,and hence(90)limsupn→∞ρzn,Tv≤limsupn→∞ρzn,v=0.Then, {zn} converges strongly to T(v). This implies that T(v)=v. Therefore F(T) is a closed set. In view of (87), let {unk} be a subsequence of sequence {un} such that ρ(unk,zk)≤1/2k for all k≥1, where {zk} is a sequence in F(T). By Theorem 21, we have(91)ρunk+1,zk≤ρunk,zk≤12k.Now, by the triangle inequality and (91), we have (92)ρzk+1,zk≤ρzk+1,unk+1+ρunk+1,zk≤12k+1+12k<12k-1.A standard argument shows that {zk} is a Cauchy sequence. Since F(T) is closed, {zk} converges to some point v∈F(T). Now(93)ρunk,v≤ρunk,zk+ρzk,v.Letting k→∞ implies that {unk} converges strongly to v. By Theorem 21, limn→∞ρ(un,v) exists. Hence {un} converges strongly to v. The converse is obvious; hence the proof is completed.Theorem 24.
Let(E,ρ,⪯) be a uniformly convex partially ordered hyperbolic metric space and , T, and {un} be the same as in Theorem 21. Let T satisfy the condition (I) and F(T)≠∅. Then {un} converges strongly to a fixed point of T.Proof.
From Theorem21, it follows that(94)liminfn→∞ρTun,un=0.As T satisfies condition (I), we have ρ(T(un),un)≥g(dist(un,F(T))). From (94), we have(95)liminfn→∞gdistun,FT=0.Since g:[0,∞)→[0,∞) is a nondecreasing function with g(0)=0 and g(r)>0 for all r∈(0,∞), therefore we have(96)liminfn→∞distun,FT=0.Therefore all the assumptions of Theorem 23 are satisfied and {un} converges strongly to a fixed point of T.


## 5. Applications to Nonlinear Integral Equations

In this section, we present an application of our results to nonlinear integral equations.We give an existence theorem in an ordered Banach space. These spaces are a special case of partially ordered hyperbolic spaces.Theorem 25.
LetE be an ordered Banach space and K be a nonempty compact subset of E. Let T:K→K be a monotone Reich-Suzuki type nonexpansive mapping. Assume that there exists an a.f.p.s. {un}. Then T has a fixed point.Proof.
It may be completed following Theorem16 and [26, Theorem 4].Consider the integral equation(97)ut=ht+λ∫01kt,sfs,usds,t∈I=0,1,λ≥0.Now, we present the following result (see also [35]).Theorem 26.
LetX=C[0,1], the space of continuous functions on I=[0,1] with ordered relation ‘⪯’ in X defined as, for all u,v∈X, u⪯v if and only if u(t)≤v(t), ∀t∈[0,1]. Suppose X is equipped with supremum norm defined by u-v=supt∈I|u(t)-v(t)| and F is a compact subset of X. Assume that the following conditions hold: (a)
h:I→R is continuous.(b)
f:I×F→F is continuous, f(t,u)≥0, and there exists a constant L≥0 such that, for all u,v∈F,(98)ft,u-ft,v≤Lut-vt.(c)
k:I×F→R is continuous such that k(t,u)≥0, and ∫01‍k(t,s)ds≤K for all t,u∈I×F.(d)
λKL=1.(e)
T:F→F is a mapping defined by(99)Tut=ht+λ∫01kt,sfs,usds,t∈I=0,1,λ≥0.
Then, the nonlinear integral equation (97) has a solution in C[0,1], if T admits an a.f.p.s.Proof.
Foru,v∈F such that u⪯v, we have (100)Tut-Tvt=ht+λ∫01kt,sfs,usds-ht+λ∫01kt,sfs,vsds=λ∫01kt,sfs,us-fs,vsds≤λ∫01kt,sfs,us-fs,vsds≤λ∫01kt,sLvs-usds.Taking the supremum norm on both sides, we have (101)Tu-Tv≤λKLu-v=u-v.i.e., T is monotone nonexpansive. Hence, T is monotone Reich-Suzuki type nonexpansive mapping with k=0. Therefore all the assumptions of Theorem 25 have been satisfied, so (97) has a solution in F⊆C[0,1].Now, let us introduce the following computational example.Example 27.
Consider the following functional integral equation:(102)ut=cosπ2t-8π-13π2t+209∫013t1+s5us2ds,t∈0,1.It is observed that the above integral equation is a special case of (97) with(103)ht=cosπ2t-8π-13π2t;kt,s=3t1+s5andft,u=ut2.Now, for arbitrary u,v∈C(I,R) with u(t)≤v(t) for t∈I=[0,1], we have(104)ft,u-ft,v=12ut-vt≤12ut-vt.As h is continuous and for t∈[0,1], we have(105)∫01kt,sds=∫013t1+s5ds=910t≤910.Consequently, all the conditions of Theorem 26 are satisfied with L=1/2, K=9/10, and λ=20/9 (LKλ=1), so there exists a solution of integral equation (102). It can be easily verified that u(t)=cos⁡π/2t is a solution of nonlinear integral equation (102).



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