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

Authors: Rida Outass; Karim Chaira; El Miloudi Marhrani; Nour-eddine El Harmouchi
Journal: Journal of Function Spaces
 (2020)

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## 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.


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## Body

## 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 C or Suzuki generalize nonexpansive mapping, as follows:Definition 1.
LetK be a subset of a Banach space X. A mapping T:K⟶K is said to satisfy condition C if for all x,y∈K,(1)12x−Tx≤x−yimpliesTx−Ty≤x−y.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.
LetK be a nonempty subset of a Banach space X. Let γ∈0,1 and μ∈0,1/2 such that 2μ≤γ. A mapping T:K⟶K is said to satisfy the condition Bγ,μ on K if for all x,y∈K,(2)γx−Tx≤x−y+μy−TyimpliesTx−Ty≤1−γx−y+μx−Ty+Tx−y.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 C 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

LetX,d be a metric space. A path joining x∈X to y∈X is a map c from a closed interval 0,l⊂ℝ to X such that c0=x, cl=y, and dct,ct′=t−t′ for all t,t′∈0,l. In particular, c is an isometry and dx,y=l. The image of c is called a metric segment joining x and y. When it is unique, the metric segment is denoted by x,y. We shall denote by 1−βx⊕βy the unique point z of x,y which satisfies(3)dx,z=λdx,y,dz,y=1−λdx,y.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 tripletX,d,H is said to be a hyperbolic metric space if X,d is a metric space and H:X×X×0,1⟶Xis a function such that for all x,y,w,z∈X and β,γ∈0,1, the following hold:(i)
dz,Hx,y,β≤1−βdz,x+βdz,y(ii)
dHx,y,β,Hx,y,γ=∣β−γ∣dx,y(iii)
Hx,y,β=Hy,x,1−β(iv)
dHx,y,β,Hw,z,β≤1−βdx,w+βdy,zThe setx,y=Hx,y,β,β∈0,1 is called the metric segment with endpoints x and y. Now onwards, we write Hx,y,β=1−βx⊕βy. A subset Y of X is said to be convex if 1−βx⊕βy∈Y for all x,y∈Y and β∈0,1.When there is no ambiguity, we writeX,d for X,d,H.LetX be a partially ordered set with a partial order “≤”, and let X,d,≤ be a partially ordered hyperbolic metric space. We say that x,y∈X are comparable whenever x≤y or y≤x. Throughout, we will assume that order intervals are closed convex subsets of hyperbolic metric space X. We denote these intervals as follows:(4)a,⟶=x∈X,a≤x,⟵,b=x∈X,x≤b,for any a,b∈X.Definition 4 (see [16, 17]).
LetX,d be a hyperbolic metric space. We say that X is uniformly convex if for any a∈X, r>0 and ε>0:(5)δr,ε=inf1−1rd12x⊕12y,a;dx,a≤r,dy,a≤r,dx,y≥rε>0.Proposition 5 (see [18]).
LetX,d be a uniformly convex hyperbolic metric space and Ann be a decreasing sequence of nonempty closed, bounded, and convex subsets of X. Then, ⋂n≥1An≠∅.Lemma 6 (see [19]).
LetE,d be a uniformly convex hyperbolic metric space with a monotone modulus of uniform convexity δ. Let z∈E and αnn be a sequence such that 0<a≤αn≤b<1. If unn and vnn are two sequences in Esuch that limsupn→∞dun,z≤r, limsupn→∞dvn,z≤r, and limn→∞dαnvn⊕1−αnun,z=r, for some r≥0, then(6)limn→∞dun,vn=0.Definition 7.
LetX,d,≤ be a metric space endowed with a partial order. A mapping T:X⟶X is said to be monotone nondecreasing, if T satisfies(7)Tx≤Tywheneverx≤y,for all x,y∈X.Definition 8 (see [14]).
LetX,d,≤ be a partially ordered metric space. A mapping T:X⟶X is called monotone nonexpansive, if T is monotone and(8)dTx,Ty≤dx,y,for all x,y∈X such that x and y are comparable.Definition 9 (see [14]).
LetK be a nonempty subset of a hyperbolic metric space X,d. A function τ:K⟶0,∞ is said to be a type function, if there exists a bounded sequence xnn in X such that(9)τx=limsupn→∞dxn,x,for any x∈K.Lemma 10 (see [14]).
LetX,d,≤ be a uniformly convex hyperbolic metric space and K a nonempty closed convex subset of X. Let τ:K⟶0,∞ be a type function. Then, τ is continuous. Moreover, there exists a unique minimum point z∈K such that τz=infτx;x∈K.


## 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.
LetX,d,≤ be a partially ordered metric space, and let γ∈0,1 and μ∈0,1/2 such that 2μ≤γ. Let T:X⟶X be a monotone mapping. Then, T is said to satisfy the condition Bγ,μ on X if for all x,y∈X such that x≤y,(10)γdx,Tx≤dx,y+μdy,TyimpliesdTx,Ty≤1−γdx,y+μdx,Ty+dTx,y.Clearly, this class includes the class of monotone nonexpansive mappings (forγ=μ=0).Example 1.
LetX=ℝ be equipped with the usual ordering and the distance dx,y=∣x−y∣. Let K=0,1 and T:K⟶K be a mapping defined by(11)Tx=x+35,ifx∈0,15,x+45,ifx∈15,1,(1)
T is not a monotone nonexpansive mapping(2)
T is monotone and satisfies condition Bγ,μ
Settingx=18/100 and y=1/5, we obtain(12)dTx,Ty=6100>2100=dx,y.
However,T is not a nonexpansive mapping. Now, we show that T is monotone and satisfies condition Bγ,μ with γ=3/4 and μ=3/8. In fact, we have the following three cases:(i)
Letx,y∈0,1/5 such that x≤y, and we have(13)1−γdx,y+μdTx,y+dx,Ty=14x−y+38x+35−y+x−y−35≥14x−y+382x−2y=x−y=dTx,Ty.(ii)
Letx,y∈1/5,1 such that x≤y, and we have(14)1−γdx,y+μdTx,y+dx,Ty=14x−y+38x5+45−y+x−y5−45≥14x−y+386x5−6y5≥14+94x5−y5=104x5−y5≥x5−y5=dTx,Ty.(iii)
Letx∈0,1/5 and y∈1/5,1, and we have(15)1−γdx,y+μdx,y+dx,Ty=14x−y+38x+35−y+y5+45−x≥14x−y+3875−4y5=−14x−120y+2140.
Otherwise, we have(16)dTx,Ty=x+35−y5−45=15−x+y5,since x−y/5−1/5≤0. From (19) and (20), we have(17)dTx,Ty−1−γdx,y−μdx,y+dx,Ty=15−x+y5+14x+120y−2140=−1340−34x+14y.
Since−11/40≤1/4y−13/40≤−3/40, then(18)dTx,Ty≤1−γdx,y+μdx,y+dx,Ty.Note that there is no relationship between the conditionC and the class of mapping satisfying the condition Bγ,μ, 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 mappingT:0,3⟶0,3 by(19)Tx=0ifx≠3,1ifx=3.
The monotone mappingT satisfies the condition C (for details, see [5]) but does not satisfy the condition Bγ,μ for γ=μ=0. In fact, for x=5/2 and y=3, we have(20)dTx,Ty=1>dx,y=12.
Therefore, the conditionBγ,μ 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 Bγ,μ.Definition 12.
LetX,d,≤ be an ordered hyperbolic metric space. Let K be a nonempty convex subset of X and T:K⟶K a nondecreasing mapping on X,≤. Fix x0∈K. The Mann iteration process is the sequence xnn∈ℕ defined by(21)xn+1=1−tnxn⊕tnTxn,for all n≥0, where tnn≥0⊂0,1.Proposition 13 (see [8]).
LetX,d,≤ be an ordered hyperbolic metric space. Let xnn≥0 and ynn≥0 be two sequences in X and tn∈0,1 which satisfy for all n∈ℕ,(i)
xn+1∈xn,yn with dxn,xn+1=tndxn,yn(ii)
dyn+1,yn≤dxn+1,xn
Then, for alli≥0 and n≥1,(22)1+∑s=ii+n−1tsdyi,xi≤dyi+n,xi+∏s=ii+n−11−ts−1dyi,xi−dyi+n,xi+n.The next lemma is a technical tool which will be used to prove thatxnn is an almost fixed point sequence.Lemma 14.
LetK be a nonempty closed convex subset of a partially ordered hyperbolic metric space X,d,≤ and T:K⟶K be a monotone mapping satisfying the condition Bγ,μ on K, with 2μ<γ<1. Assume that there exists x0∈K such that x0≤Tx0. Let xnn∈ℕbe the Mann iterative sequence defined by (25) such that tnn≥0⊂a,b with a=supγ,2μ/γ and b<1. Then, for any i≥0 and n≥1, we have(23)1+∑s=ii+n−1tsdxi,Txi≤dxi,Txi+n+∏s=ii+n−11−ts−1dxi,Txi−dxi+n,Txi+n.Proof.
Letx0∈K such that x0≤Tx0. Using the convexity of the order interval x0,Tx0 and the monotonicity of T, we obtain(24)x0≤x1≤Tx0≤Tx1.
Thus, by induction, we get(25)xn≤xn+1≤Txn≤Txn+1,for any n∈ℕ. On the other hand, we have(26)dxn,xn+1=dxn,1−tnxn⊕tnTxn=tndxn,Txn.
Sinceγ≤tn for all n∈ℕ, we have(27)γdxn,Txn≤tndxn,Txn=dxn,xn+1.
By the conditionBγ,μ, for x=xn and y=xn+1, we have(28)dTxn,Txn+1≤1−γdxn,xn+1+μdxn,Txn+1+dTxn,xn+1≤tn1−γdxn,Txn+μdxn,Txn+dTxn,Txn+1+1−tndTxn,xn≤tn1−γ+μ2−tn1−μdxn,Txn=tn1−γ+μ2−tntn1−μdxn,xn+1.
Sincetn≥2μ/γ, then tn1−γ+μ2−tn/tn1−μ≤1. Hence,(29)dTxn,Txn+1≤dxn,xn+1,for all n∈ℕ. According to (29), (30), and (33), it is concluded that all the assumptions of Proposition 13 are satisfied, where yn=Txn, which implies(30)1+∑s=ii+n−1tsdxi,Txi≤dxi,Txi+n+∏s=ii+n−11−ts−1dxi,Txi−dxi+n,Txi+n,for any i≥0 and n≥1.Theorem 15.
LetXbe as above. LetK be a nonempty, closed, convex, and bounded subset of X and T:K⟶K be a monotone mapping satisfying the condition Bγ,μ on K, with 2μ<γ<1. Assume that there exists x0∈K such that x0≤Tx0 and the sequence xnn defined by (25) such that tnn≥0⊂a,b, with a=supγ,2μ/γ and b<1. Then, we have limn→∞dxn,Txn=0.Proof.
First we show that the sequencedxn,Txnn≥0 is decreasing. Indeed, by the triangle inequality, we have(31)dxn+1,Txn+1≤dxn+1,Txn+dTxn,Txn+1=d1−tnxn⊕tnTxn,Txn+dTxn,Txn+1=1−tndxn,Txn+dTxn,Txn+1.
From the proof of Lemma14, we have(32)dTxn,Txn+1≤dxn,xn+1,for all n∈ℕ. Hence,(33)dxn+1,Txn+1≤1−tndxn,Txn+dTxn,Txn+1≤dxn,Txn,for all n∈ℕ. Consequently, dxn,Txnn is a decreasing and bounded sequence in ℝ+. Thus, there exists r≥0 such that(34)limn→∞dxn,Txn=r≥0.
On the other hand, we have(35)1+nγ≤1+∑s=ii+n−1ts,∏s=ii+n−11−ts−1≤1−b−n,dxi,Txi+n≤δK=supdx,y:x,y∈K,since supγ,2μ/γ≤tn and tn≤b<1 for all n≥0. Therefore, the inequality obtained in Lemma 14 implies(36)1+nγdxi,Txi≤δK+1−b−ndxi,Txi−dxi+n,Txi+n,for all i≥0 and n≥1. If we let i→+∞, we get 1+nγr≤δK, for any n≥1. Then,(37)r≤δK1+nγ,holds for all n≥1. Clearly, this is possible only for r=0. Thus,(38)limn→∞dxn,Txn=0.Lemma 16.
LetX,d,≤ be a partially ordered hyperbolic metric space and K a nonempty subset of E. Let T:K⟶K be a monotone mapping which satisfies the Bγ,μ condition. If z is a fixed point of T, then for all x∈K, x≤z,(39)dz,Tx≤dz,x.Proof.
Letz be a fixed point of T and x∈K such that x≤z.
We have(40)0=dz,Tz≤dx,z≤dx,z+μdx,Tx.
By the conditionBγ,μ,(41)dTx,Tz=dz,Tx≤1−γdx,z+μdx,Tz+μdz,Tx.
Hence,(42)1−μdz,Tx≤1−γ+μdz,x,which implies(43)dz,Tx≤1−γ+μ1−μdz,x.
Since2μ≤γ, then(44)dz,Tx≤dz,x.In the sequel, we will use the fixed point set with the partial ordersFx0≥T given by(45)Fx0≥T=p∈K:Tp=p,x0≤p.Definition 17.
LetX,≤ be a partially ordered set and xnn be a sequence in X. The sequence xnn is said to be ≤-bounded if there exists a point w∈X such that xn≤w, for all n≥0.Lemma 18.
LetK be a nonempty closed and convex subset of a partially ordered hyperbolic metric space X,d≤. Let T:K⟶K be a monotone mapping satisfying the condition Bγ,μ on K, for x0∈K such that x0≤Tx0 and Fx0≥T is nonempty. Let xnn be a sequence defined by (25). Then, the following assertions hold:(i)
The sequencexnn is ≤-bounded(ii)
limn→∞dxn,z exists for z∈Fx0≥T, and limn→∞distxn,Fx0≥T exists, where distx,Fx0≥T denotes the distance from x to Fx0≥TProof.
(i)
LetFx0≥T≠∅, and fix z∈Fx0≥T. From the proof of Lemma 14, we have(46)xn≤xn+1≤Txn,foralln∈ℕ.
SinceT is monotone and Tx0≤Tz=z, we have(47)x0≤x1≤Tx0≤z.
So,x1≤z. By monotonicity of T,(48)x1≤x2≤Tx1≤z.
Applying the same argument, we get(49)xn≤Txn≤z,foralln≥0.
Thus,xnn is ≤-bounded.(ii)
By Lemma16, we have(50)dxn+1,z=d1−tnxn⊕tnTxn,z≤1−tndxn,z+tndTxn,z≤1−tndxn,z+tndxn,z=dxn,z.
Hence,(51)dxn+1,z≤dxn,z,for all n≥0. Consequently, dxn,zn is a decreasing and bounded sequence in ℝ+. Thus, limndxn,z exists. For each z∈Fx0≥T and n∈ℕ we have(52)dxn+1,z≤dxn,z.
Taking infimum over allz∈Fx0≥T, we get(53)distxn+1,Fx0≥T≤distxn,Fx0≥T,for all n∈ℕ. So the sequence distxn,Fx0≥Tn is bounded and decreasing. Therefore, limn→∞distxn,Fx0≥T exists.Theorem 19.
LetK be a nonempty closed convex and bounded subset of a hyperbolic ordered metric space X and T:K⟶K be a monotone mapping satisfying the condition Bγ,μ on K, with 2μ<γ<1. Assume that there exists x0∈K such that x0≤Tx0 and Fx0≥T is nonempty. Let xnn be a sequence defined by (25) where tnn≥0⊂a,b, with a=supγ,2μ/γ and b<1. Then, xnn converges to a fixed point of T.Proof.
Assume thatFx0≥T is nonempty, and let z∈Fx0≥T. Then, from the proof of Lemma 18, we have(54)xn≤Txn≤z,foralln∈ℕ.
By Lemma18, there exists r≥0 such that(55)limn→∞dxn,z=r≥0.
Assume thatr>0. By Theorem 15, one has(56)limn→∞dxn,Txn=0.
This leads todxn,Txn≤dxn,z, for large sufficient n≥0. Then,(57)γdxn,Txn≤dxn,z+μdz,Tz.
By the conditionBγ,μ, we get(58)dTxn,Tz≤1−γdxn,z+μdTxn,z+dxn,z≤1−γ+μdxn,z+μdTxn,z≤1−γ+μ1−μdxn,z.
Otherwise,(59)dxn,z≤dxn,Txn+dTxn,z≤dxn,Txn+1−γ+μ1−μdxn,z.
Taking limit asn goes to infinity, we get(60)r≤1−γ+μ1−μr.
Since1−γ+μ/1−μ<1, we get a contradiction. Therefore, xnn converges to z.Example 3.
LetX=C00,1×C00,1. Define d:X×X⟶ℝ+ by(61)df1,f2,g1,g2=f2−g2∞+f1F2−g1G2∞,such that F2x=1+∫0xf2tdt and G2x=1+∫0xg2tdt, for all x∈0,1. It is obvious to see that X,d is a metric space.
First step. Let K=f1,f2∈X/1/2≤f1x≤1,0≤f2x≤1,forallx∈0,1⊂X. We can easily verify that K is a nonempty, bounded, and convex subset of X. We show that K is closed in X. Indeed, let fn=f1,n,f2,nn∈ℕ be a sequence in X and f=f1,f2∈X such that(62)f1,n,f2,n→dn→+∞f1,f2.
Then, for allε>0, there exists an Nε∈ℕ such that(63)dfn,f=f2,n−f2∞+f1,nF2,n−f1F2∞≤ε,∀n≥Nε.
This implies(64)f2,n−f2∞≤ε,f1,nF2,n−f1F2∞≤ε,∀n≥Nε.
Thus,(65)f2,n→unif.n→+∞f2∈C00,1,(66)f1,nF2,n→unif.n→+∞f1F2∈C00,1.
Using (69), it is easy to verify that(67)F2,n→unif.n→+∞F2∈C00,1.
On the other hand, fixx∈0,1, and we have(68)f1,nx−f1x≤f1,nx−f1xF2xF2,n−1x+f1x−f1xF2xF2,n−1x≤F2,n−1xf1,nxF2,nx−f1xF2x+f1xF2xF2−1x−F2,n−1x≤F2,n−1xf1,nxF2,nx−f1xF2x+f1xF2xF2,nxF2xF2,nx−F2x.
Since1≤F2,nx≤2 and 1/2≤f1,nx≤1, for all n∈ℕ, then(69)f1,n−f1∞≤f1,nF2,n−f1F2∞+2F2,n−F2∞,∀n≥Nε.
By (70) and (71), we conclude that(70)f1,n→unif.n→+∞f1∈C00,1.
Therefore,f1,f2∈K.
Second step. Define a function H:X×X×0,1⟶X by(71)Hf,g,β=1−βf1F2+βg1G21−βF2+βG2,1−βf2+βg2,for all f=f1,f2 and g=g1,g2 in X. We show that X,d,H is a hyperbolic metric space. For f=f1,f2, g=g1,g2, h=h1,h2, v=v1,v2∈X, and β, λ∈0,1, we have the following:(i)
dh,Hf,g,β=h2−1−βf2−βg2∞+h1H2−1−βf1F2−βg1G2∞≤1−βh2−f2∞+βh2−g2∞+1−βh1H2−f1F2∞+βh1H2−g1G2∞=1−βdh,f+βdh,g(ii)
dHf,g,β,Hf,g,λ=d1−βf1F2+βg1G21−βF2+βG2,1−βf2+βg2,1−λf1F2+λg1G21−λF2+λG2,1−λf2+λg2=λ−βf2+β−λg2∞+λ−βf1F2+β−λg1G2∞=λ−βdf,g(iii)
Hf,g,β=1−βf1F2+βg1G21−βF2+βG2,1−βf2+βg2=βg1G2+1−βf1F2βG2+1−βF2,βg2+1−βf2=Hg,f,1−β(iv)
dHf,g,β,Hh,v,β=1−βf2−h2+βg2−v2∞+1−βf1F2−h1H2+βg1G2−v1V2∞≤1−βdf,h+βdg,v
Consider the partial ordering defined onX as follows:(72)f≤g⇔f1t≥g1t,f2t≥g2t,∀t∈0,1,where f=f1,f2 and g=g1,g2 in X.
Thus,X,d,H is a partially ordered hyperbolic metric space, but not a normed space.
Third step. Let T:X⟶X be a mapping defined by(73)Tf1,f2=y1,y2iff1,f2≠12,0,z1,z2iff1,f2=12,0,where y1,y2 and z1,z2 designate constant functions, such that(74)y1t,y2t=12,12,z1t,z2t=12,511forallt∈0,1.
It is obvious that the mappingT is monotone.(i)
T is not nonexpansive. In fact, for f=1/2,1/33 and g=1/2,0, we have(75)dTf,Tg=344>122=df,g.(ii)
T satisfies the condition Bγ,μ for γ=10/21 and μ=5/44. In fact, for f=f1,f2≠1/2,0 and g=g1,g2=1/2,0 such that f≤g, we have(76)dTf,Tg=d12,12,12,511=344,dTf,g=12∞+12Id2∞=34.
Therefore,(77)1−γdf,g+μdTf,g+dTg,f=1121df,g+15176+544dTg,f≥15176>344=dTf,Tg.
Otherwise, iff=f1,f2 and g=g1,g2≠1/2,0 (or equal to (1/2, 0)), it is quite easy to verify that the mapping T satisfies the condition Bγ,μ for γ=10/21 and μ=5/44.
Therefore,T is monotone and satisfies the condition Bγ,μ and T has a unique fixed point h∗=1/2,1/2.
Fourth step. Suppose that x0=1,1/2 and tn=2μ/γ=21/44≥γ, then x0≤Tx0 and(78)xn=12+1−αn2,12,where α=21/44, for all n∈ℕ. Indeed, assume that (82) is true for n and let us prove it for n+1. We have(79)xn+1=1−αxn+αTxn.
Sincexn≠1/2,0, then(80)xn+1=1−αxn+α12,12=121−αn+1+121−α,121−α+12α,12α=121−αn+1+12,12.
SinceT is monotone and satisfies condition Bγ,μ and there exists x0∈K such that x0≤Tx0, then all the conditions of Theorem 19 are satisfied. Consequently, it is obvious to see that(81)limn→∞dxn,12,12=0.
Thus,xnn converges to the fixed point h∗=1/2,1/2.Theorem 20.
LetK be a nonempty closed convex and bounded subset of a uniformly convex partially ordered hyperbolic metric space X,d,≤ and T:K⟶K be a monotone mapping satisfying the condition Bγ,μ on K, with 2μ<γ<1. Assume that there exists x0∈K such that x0≤Tx0 and the sequence xnn defined by (25) be ≤-bounded such that tnn≥0⊆a,b with a=supγ,2μ/γ and b<1. Then, T has a fixed point z∈K, x0≤z, and limn→∞dxn,z=0.Proof.
Asxnn is ≤-bounded, then there exists a point w∈X such that xn≤w, for all n≥0.
LetKn=t∈K:xn≤t for all i∈ℕ. It is clear that Ki is nonempty, since w∈Kn, and is a closed and convex subset of K, for all n∈ℕ, and the sequence Knn is decreasing, since x0≤xn≤xn+1 for all n∈ℕ. Hence, by Proposition 5,(82)K∞=⋂n≥0Kn,is a nonempty, closed, and convex subset of K. Let p∈K∞; then, by the definition of K∞ and the monotonicity of T, we get(83)xn≤Txn≤Tp.
Hence,TK∞⊆K∞. Consider the type function τ:K∞⟶0,∞ generated by xnn, that is, τv=limsupn→∞dxn,v, for all v∈K∞. From Lemma 10, there exists a unique element z∈K∞ such that(84)τz=infτv;v∈K∞.
Moreover,(85)τTz=limsupn→∞dxn,Tz.
We assert thatTz=z. Assume that this is not the case.
By Theorem15, we have limn→∞dxn,Txn=0, then(86)dxn,Txn≤μdz,Tz,forlargeintegersn.
Sincez∈K∞, then xn≤z for all n≥1. By the condition Bγ,μ, for x=xn and y=z, we have(87)dTxn,Tz≤1−γdxn,z+μdxn,Tz+dz,Txn≤1−γ+μdxn,z+μdTxn,Tz+2dxn,Txn≤1−γ+μ1−μdxn,z+2μ1−μdxn,Txn.
By the triangle inequality,(88)dxn,Tz≤dxn,Txn+dTxn,Tz,we obtain(89)dxn,Tz≤1−γ+μ1−μdxn,z+1+μ1−μdxn,Txn,which implies(90)limsupn→∞dxn,Tz≤1−γ+μ1−μlimsupn→∞dz,xn.
Hence,(91)τTz≤τz.
By the uniqueness of the minimum point, we conclude a contradiction with our hypothesisTz≠z. Then, we have Tz=z.
Fx0≥T is nonempty, because it contains z. Following the same proof of Theorem 15, we obtain limn→∞dxn,z=0.



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*Source: 5854674-2020-07-21.xml*