Mann iteration for monotone nonexpansive mappings in ordered CAT(0) space with an application to integral equations

Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 https://doi.org/10.1186/s13660-018-1925-2 RESEARCH Open Access Mann iteration for monotone nonexpansive mappings in ordered CAT(0) space with an application to integral equations Izhar Uddin1, Chanchal Garodia1* and Juan Jose Nieto2 *Correspondence: c.gar[email protected] 1Department of Mathematics, Faculty of Natural Sciences, Jamia Millia Islamia, New Delhi, India Full list of author information is available at the end of the article Abstract In this paper, we establish some convergence results for a monotone nonexpansive mapping in a CAT(0) space. We prove the - and strong convergence of the Mann iteration scheme. Further, we provide a numerical example to illustrate the convergence of our iteration scheme, and also, as an application, we discuss the solution of integral equation. Our results extend some of the relevant results. MSC: 47H09; 47H10 Keywords: CAT(0) space; Fixed point; -convergence; Monotone nonexpansive mapping 1 Introduction TheBanachcontractionprinciple[1]isoneofthemostfundamentalresultsinfixedpoint theoryandhasbeenutilizedwidelyforprovingtheexistenceofsolutionsofdifferentnonlinear functional equations. In the last few years, many efforts have been made to obtain fixedpointsinpartiallyorderedsets.In2004,RanandReurings[2]generalizedtheBanach contraction principle to ordered metric spaces. Later on, in 2005, Nieto and Rodriguez [3] used the same approach to further extend some more results of fixed point theory in partially ordered metric spaces and utilized them to study the existence of solutions of differential equations. Note that the Banach contraction principle is no longer true for nonexpansive mappings, thatis, a nonexpansivemappingneednotadmitafixedpointonacompletemetric space. Also, Picard iteration need not converge for a nonexpansive map in a complete metric space. This led to the beginning of a new era of fixed point theory for nonexpansive mappings by using geometric properties. In 1965, Browder [4], Göhde [5], and Kirk [6] gave three basic existence results for nonexpansive mappings. With a view to locating fixed points of nonexpansive mappings, Mann [7]andIshikawa[8]introducedtwobasic iteration schemes. Now,fixedpointtheoryofmonotonenonexpansivemappingsisgainingmuchattention amongtheresearchers.Recently,BacharandKhamsi[9],Abdullatifetal.[10],andSonget ©The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 2 of 13 al.[11]proved some existenceand convergenceresults for monotonenonexpansivemappings. Dehaish and Khamsi [12] proved the weak convergence of the Mann iteration for a monotone nonexpansive mapping. In 2016, Song et al. [11] considered the weak convergence of the Mann iteration scheme for a monotone nonexpansive mapping Tunder some mild different conditions in a Banach space. Theaimofthispaperistostudytheconvergencebehaviorofthewell-knownManniter- ation [7]inaCAT(0) space for a monotone nonexpansive mapping. Further, we provide a numericalexampleandapplicationrelatedtosolutionofanintegralequation.Ourresults generalize and improve several existing results in the literature. 2Preliminaries To make our paper self-contained, we recall some basic definitions and relevant results. AmetricspaceXis a CAT(0) space if it is geodesically connected and if every geodesic triangleinXisatleastasthinasitscomparisontriangleintheEuclideanplane.Forfurther informationabout thesespaces andthe fundamentalrole theyplayinvarious branchesof mathematics, we refer to Bridson and Haefliger [13] and Burago et al. [14]. Every convex subsetofEuclideanspaceRnendowedwiththeinducedmetricisaCAT(0)space.Further, the class of Hilbert spaces are examples of CAT(0) spaces. The fixed point theory in CAT(0) spaces is gaining attention of researchers, and many results have been obtained for single- and multivalued mappings in a CAT(0) space. For differentaspectsoffixedpointtheoryinCAT(0)spaces,wereferto[15–24].Thefollowing few results are necessary for our subsequent discussion. Lemma 2.1 ([21]) Let (X,d)be a CAT(0) space.For e,f∈Xandz∈[0,1], there exists a unique h∈[e,f]such that d(e,h)=zd(e,f)and d(f,h)=(1–z)d(e,f). We use the notation (1–z)e⊕zf for the unique point hof the lemma. Lemma2.2([21]) Let (X,d)be a CAT(0) space.For e,f,h∈Xandz∈[0,1], we have d(1–z)e⊕zf,h≤(1–z)d(e,h)+zd(f,h). Lemma2.3([21]) Let X be a CAT(0) space.Then d(1–z)e⊕zf,h2≤(1–z)d(e,h)2+zd(f,h)2–z(1–z)d(e,f)2 for all e,f,h∈Xandz∈[0,1]. Let {un}beaboundedsequenceinacompleteCAT(0) space X.Foru∈X,wedenote ru,{un}=limsup n→∞ d(u,un). The asymptotic radius r({un})isgivenby r{un}=infr(u,un):u∈X, Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 3 of 13 and the asymptotic center A({un})of{un}is defined as A{un}=u∈X:r(u,un)=r{un}. It is known that in a CAT(0) space, A({un}) consists of exactly one point [25,Proposition 5]. In 1976, Lim [26] introduced the concept of -convergence in a metric space. Later on, Kirk and Panyanak [22]provedthatCAT(0) spaces presented a natural framework for Lim’s concept and provided precise analogs of several results in Banach spaces involving weak convergence in CAT(0) space setting. Definition 2.4 Asequence{un}in Xis said to be -convergent to u∈Xif uis the unique asymptotic center of {vn}for every subsequence {vn}of {un}.Inthiscase,wewrite -limnun=uand say that uis the -limit of {un}. Definition 2.5 A Banach space Xis said to satisfy Opial’s condition if for any sequence {un}in Xwith unu(denotes weak convergence), we have limsupn→∞ un–u< limsupn→∞un–vfor all v∈Xwith v=u. Examples of Banach spaces satisfying this condition are Hilbert spaces and all lpspaces (1<p<∞). On the other hand, Lp[0,2π]with1<p=2 fail to satisfy Opial’s condition. Notice that if given a sequence {un}in Xsuch that {un}-converge to u,thenforv∈X with v=u,wehave limsup n→∞ un–u<limsup n→∞ un–v. So,every CAT(0) space satisfies Opial’s property. Lemma 2.6 ([22]) Every bounded sequence in a complete CAT(0) space admits a - convergent subsequence. Lemma 2.7 ([21]) If G is a closed convex subset of a complete CAT(0) space X and if {un} is a bounded sequence in G,then the asymptotic center of {un}is in G. Next, we introduce the concept of partial order in the setting of CAT(0) spaces. Let Xbe a complete CAT(0) space endowed with partial order “”. An order inter v al is any of the subsets [a,→)={u∈X;au}or (←,a]={u∈X:ua} for any a∈X. So, an order interval [u,v] for all u,v∈Xis given by [u,v]={w∈X:uwv}. Throughout we will assume that the order intervals are closed and convex subsets of an ordered CAT(0) space (X,). Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 4 of 13 Definition 2.8 Let Gbe a nonempty subset of an ordered metric space X. A mapping P:G→Gis said to be: (i) monotone if PuPvfor all u,v∈Gwith uv, (ii) monotone nonexpansive if P is monotone and d(Pu,Pv)≤d(u,v) for all u,v∈Gwith uv. Now we present the Mann iteration scheme in the setting of ordered CAT(0) spaces (X,). Let Gbe anonemptyconvexsubset of a CAT(0) space X. Then the Mann iteration is as follows: u1∈G, un+1 =(1–κn)un⊕κnPun,n∈N,(2.1) where {κn}⊂[0,1]. In this paper, we prove some -convergence and strong convergence results in CAT(0) spaces. 3Some-convergence and strong convergence theorems We begin with the following important lemma. Lemma3.1 LetGbeanonemptyclosedconvexsubsetofacompleteordered CAT(0)space (X,), and let P :G→G be a monotone nonexpansive mapping.Fix u1∈Gsuchthat u1Pu1.If {un}is defined by (2.1)with condition ∞ n=1 κn(1–κn)=∞,then we have: (i) unun+1 Punfor any n≥1, (ii) unu,provided that {un}-converges to a point u∈G. Proof (i) We will prove the result by induction on n.Notethatifq1,q2∈Gare such that q1q2,thenq1λq1+(1–λ)q2q2for any λ∈[0,1]. This is true because we have assumedthatorderintervalsareconvex.ThusweonlyneedtoshowthatunPunforany n≥1. We have already assumed that u1Pu1, and hence the inequality holds for n=1. Assume that unPunfor n≥2. Since κn∈[0,1] for all n,wehave un(1–κn)un⊕κnPunPun, that is, unun+1 Pun.SincePis monotone, we have PunPun+1. By using the transitivity of the order we get un+1 Pun+1. Thus by induction the inequality is true for any n≥1. (ii) Let ube the -limit of {un}.Frompart(i)wehaveunun+1 for all n≥1since{un} is increasing and the order interval [um,→) is closed and convex. Therefore u∈[um,→) for a fixed m∈N;otherwise,ifu/∈[um,→), then we could construct a subsequence {ur} of {un}by leaving the first m–1termsofthesequence{un}, and then the asymptotic center of {ur}would not be u, which contradicts the assumption that uis the -limit of the sequence {un}. This completes the proof of part (ii).  Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 5 of 13 Lemma3.2 Let G be a nonempty closed convex subset of a complete CAT(0) space (X,), and let P :G→G be amonotone nonexpansivemapping.Fix u1∈Gsuchthatu 1Pu1.If {un}is a sequence described as in (2.1)and F(P)=∅with r ∈F(P)such that r u1,then: (i) limn→∞d(un,r)exists,and (ii) limn→∞d(Pun,un)=0. Proof (i)Since ru1,usingpart(i)of Lemma3.1,wehaveunun+1 Pun.Inparticular, for n=1,wehaveu1u2Pu1. Using the transitivity of the order, we get ru2.By mathematical induction we have runfor all n≥1. Now we have d(un+1,r)=d(1–κn)un⊕κnPun,r ≤(1–κn)d(un,r)+κnd(Pun,r) =(1–κn)d(un,r)+κnd(Pun,Pr). Since Pis a monotone map and runfor all n≥1, we have d(un+1,r)≤(1–κn)d(un,r)+κnd(un,r) =d(un,r). Thus we have d(un+1,r)≤d(un,r) for all n≥1. So {d(un,r)}is a decreasing real sequence bounded below by zero. Hence limn→∞ d(un,r)exists. (ii) First, consider d(Pun+1,un+1)=dPun+1,(1–κn)un⊕κnPun ≤(1–κn)d(Pun+1,un)+κnd(Pun+1,Pun) ≤(1–κn)d(Pun+1,un)+κnd(un+1,un) ≤(1–κn)d(Pun+1,Pun)+d(Pun,un)+κnd(un+1,un) ≤(1–κn)d(un+1,un)+d(Pun,un)+κnd(un+1,un) =(1–κn)d(Pun,un)+d(un+1,un) =(1–κn)d(Pun,un)+d(1–κn)un⊕κnPun,un ≤(1–κn)d(Pun,un)+(1–κn)d(un,un)+κnd(Pun,un) =d(Pun,un). So limn→∞d(Pun,un)exists. Since ru1, using the Lemma 3.1,wehaveru1unfor all n≥1. Then, since Pis a nonexpansive map and ris a fixed point of P,wehave d(un+1,r)2=d(1–κn)un⊕κnPun,r2 ≤(1–κn)d(un,r)2+κnd(Pun,r)2–(1–κn)κnd(un,Pun)2 =(1–κn)d(un,r)2+κnd(Pun,Pr)2–(1–κn)κnd(un,Pun)2 Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 6 of 13 ≤(1–κn)d(un,r)2+κnd(un,r)2–(1–κn)κnd(un,Pun)2 =d(un,r)2–(1–κn)κnd(un,Pun)2. From this we get ∞  n=1(1–κn)κnd(un,Pun)2≤d(u1,r)2<∞. (3.1) Since ∞ n=1(1–κn)κn=∞,thereexistsasubsequence{unk}of {un}such that lim n→∞d(Punk,unk)=0. Since limn→∞d(Pun,un) exists, it follows that limn→∞d(Pun,un) = 0, and this proves the result.  The following lemma is an analogue of Theorem 3.7 of [22]. Lemma3.3 Let G be a nonempty closed convex subset of a complete CAT(0) space (X,), and let P :G→G be a monotone nonexpansive mapping.Fix u1∈Gsuchthatu 1 Pu1.If {un}is a sequence described as in (2.1), then the conditions -limnun=uand limn→∞d(Pun,un)=0imply that u is a fixed point of P. Proof Since -limnun=u, by Lemma 3.1 we get unufor all n≥1. Then from the nonexpansiveness of Pand limn→∞ d(Pun,un)=0 it follows that d(Pu,un)≤d(Pu,Pun)+d(Pun,un), limsup n→∞ d(Pu,un)≤limsup n→∞ d(Pu,Pun)+d(Pun,un) =limsup n→∞ d(Pu,Pun) ≤limsup n→∞ d(u,un). Thus by the uniqueness of asymptotic center we get Pu =u, which proves the desired result.  Theorem3.4 LetGbeanonemptyclosedconvexsubsetofacompleteCAT(0)space(X,), and let P :G→G be a monotone nonexpansive mapping with F(P)=∅.Fix u1∈Gsuch that u1Pu1.If {un}is a sequence described as in (2.1), then {un}-converges to a fixed point of P. Proof From Lemma 3.2 we have that limn→∞ d(un,r) exists for each r∈F(P), so the sequence {un}is bounded, and limn→∞ d(un,Pun)=0. Let Wω({un})=:X({vn}), where the union is taken over all subsequences {vn}over {un}. To show the -convergence of {un}to a fixed point of P,wewillfirstprovethat Wω({un})⊂F(P) and thereafter argue that Wω({un}) is a singleton set. To show that Wω({un})⊂F(P), let y∈Wω({un}). Then there exists a subsequence {yn}of {un}such Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 7 of 13 that X({yn})=y. By Lemmas 2.6 and 2.7 there exists a subsequence {zn}of {yn}such that -limnzn=zandz∈G.Sincelimn→∞d(Pun,un)=0and{zn}isasubsequenceof{un},we havethatlimn→∞d(zn,Pzn)= 0.InviewofLemma3.3,wehavez=Pz,andhencez∈F(P). Nowwewishtoshowthatz=y. If, on the contrary, z=y, then we would have limsup n→∞ d(zn,z)<limsup n→∞ d(zn,y) ≤limsup n→∞ d(yn,y) <limsup n→∞ d(yn,z) =limsup n→∞ d(un,z) =limsup n→∞ d(zn,z), whichisacontradictionsince Xsatisfies theOpial condition andhence z=y∈F(P).Now it remains to show that Wω({un}) consists of a single element only. For this, let {yn}be a subsequence of {un}. Again, using Lemmas 2.6 and 2.7,wecanfindasubsequence{zn}of {yn}such that -limnzn=z.LetX({yn})=yand X({un})=u. Previously, we have already proved that y=z; therefore, it suffices to show that z=u.Ifz=u,thensincez∈F(P), {d(un,z)}is convergent by Lemma 3.2, By the uniqueness of asymptotic center we have limsup n→∞ d(zn,z)<limsup n→∞ d(zn,u) ≤limsup n→∞ d(un,u) <limsup n→∞ d(un,z) =limsup n→∞ d(zn,z), which gives a contradiction. Therefore we must have z=u,whichprovesthatWω({un}) is a singleton set and that a particular element is a fixed point of P.Hencetheconclusion follows.  Theorem 3.5 Let X be a complete CAT(0) space endowed with partial ordering ,and let G be a nonempty closed convex subsetof X.Let P :G→G beamonotone nonexpansive mapping such that F(P)=∅.Fix u1∈Gsuchthatandu 1Pu1.If {un}is a sequence described as in (2.1)such that ∞ n=1 κn(1–κn)=∞,then {un}converges to a fixed point of P if and only if liminfn→∞d(un,F(P))=0. Proof If the sequence {un}converges to a point u∈F(P), then it is obvious that liminfn→∞d(un,F(P))=0. For the converse part, assume that liminfn→∞ d(un,F(P)) = 0. From Lemma 3.2(i) we have d(un+1,r)≤d(un,r) for any r∈F(P), so that dun+1,F(P)≤dun,F(P). Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 8 of 13 Thus {d(un,F(P))}forms a decreasing sequence that is bounded below by zero, so limn→∞d(un,F(P)) exists. As liminfn→∞ d(un,F(P))=0, we have limn→∞d(un,F(P))=0. Now we prove that {un}is a Cauchy sequence in G.Let>0 be arbitrary. Since liminfn→∞d(un,F(P))= 0, there exists n0such that, for all n≥n0,wehave dun,F(P)< 4. In particular, infd(un0,r):r∈F(P)< 4, so there must exist r∈F(P)suchthat d(un0,r)< 2. Thus, for m,n≥n0,wehave d(un+m,un)≤d(un+m,r)+d(un,r)<2d(un0,r)<2 2=, which shows that {un}is a Cauchy sequence. Since Gis a closed subset of a complete metric space X,soGitself is a complete metric space, and therefore {un}must converge in G.Letliminfn→∞ un=q. Now Pis a monotone nonexpansive mapping, and from Lemma 3.3(i) we have limn→∞d(Pun,un)=0.Also,fromtheproofofLemma3.1in[12] we can easily deduce that unqfor any n≥1. Therefore we have d(q,Pq)≤d(q,un)+d(un,Pun)+d(Pun,Pq) ≤d(q,un)+dun,P(un)+d(un,q) →0asn→∞, and hence q=Pq.Thusq∈F(P).  4 Numerical example In this section, we present a numerical example to illustrate the convergence behavior of our iteration scheme (2.1). Let X=[0,+∞) be a complete metric space with the metric d(u,v)=|u–v|,u,v∈X. Now, consider the order relation uvas u,v∈[0,1] and u≤vor u,v∈(n,n+1] forsomen=1,2,... and u≤v. Uddin et al. Journal of Inequalities and Applications (2018) 2018:339 Page 9 of 13 Let Pbe defined by P(0)= 0, P(u)=n 2+u 2,u∈(n,n+1],n=0,1,2,.... Then, clearly, Pis not continuous at v=n+1forn=0,1,2,...,since Pn+1–=n+1 2=n+1=Pn+1+. Also, if uv,thenu,v∈[0,1] or u,v∈(n,n+1]forsomen=1,2,...,and dP(u),P(v)=dn 2+u 2,n 2+v 2=1 2d(u,v). So, Pis a monotone nonexpansive map but not a nonexpansive map, and 0 is the unique fixed point of P. Now, we show the convergence of (2.1) using two different sets of values. It is evident from the tables (Table 1and Table 2)andgraphs(Fig.1and Fig. 2)thatour sequence (2.1) converges to 0, which is a fixed point of P. Table 1 (κn=2n 5n+2 for all n∈N) Step When u1=0.25 u1=0.45 u1=0.65 1 0.25 0.45 0.65 2 0.1607142857142857 0.2892857142857142 0.4178571428571429 3 0.1071428571428571 0.1928571428571428 0.2785714285714286 4 0.07247899159663865 0.1304621848739496 0.1884453781512605 5 0.04941749427043545 0.0889514896867838 0.1284854851031322 6 0.03386013496307614 0.06094824293353705 0.088036350903998 7 0.02327884278711485 0.04190191701680672 0.0605249912464986 8 0.01604352678571429 0.02887834821428571 0.04171316964285715 9 0.01107767325680272 0.0199398118622449 0.02880195046768708 10 0.007660093209491246 0.01378816777708424 0.01991624234467724 11 0.005303141452724708 0.00954565461490447 0.01378816777708424 12 0.003674983989168876 0.006614971180503976 0.00955495837183908 13 0.002548779218294543 0.004587802592930178 0.006626825967565813 14 0.001768928860458153 0.003184071948824675 0.004599215037191199 15 0.001228422819762606 0.002211161075572691 0.003193899331382777 16 0.000853514556588304 0.001536326201858948 0.002219137847129592 17 0.0005932967039699188 0.001067934067145854 0.00154257143032179 18 0.0004125798918411505 0.0007426438053140709 0.001072707718786992 19 0.0002870120986721047 0.0005166217776097884 0.0007462314565474726 20 0.0001997249140244027 0.0003595048452439249 0.0005192847764634474 21 0.0001390242048601234 0.0002502435687482223 0.0003614629326363212 22 0.0000967972267484037 0.0001742350081471267 0.0002516727895458498 23 0.00006741235434263828 0.0001213422378167489 0.0001752721212908597 24 0.0000469581784523506 0.0000845247212142311 0.0001220912639761116 25 0.00003271676367581804 0.0000588901746164725 0.000085063585557127 26 0.00002279868964810942 0.00004103764136659699 0.00005927659308508453 27 0.000015889995815349 0.00002860199246762819 0.00004131398911990741 28 0.00001107660292237831 0.00001993788526028096 0.00002879916759818363 29 7.722420347291922 ×10–6 0.00001390035662512546 0.00002007829290295901 30 5.384680854404231 ×10–6 9.69242553792 ×10–6 0.00001400017022145 31 3.755106385308214 ×10–6 6.759191493554787 ×10–6 9.76327660180 ×10–6