Some applications of Caristi s fixed point theorem in metric spaces

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1 Khojaseh e al. Fixed Poin Theory and Applicaions 2016) 2016:16 DOI /s z R E S E A R C H Open Access Some applicaions of Carisi s fixed poin heorem in meric spaces Farshid Khojaseh 1*,ErdalKarapinar 2 and Hassan Khandani 3 * Correspondence: f-khojase@iau-arak.ac.ir 1 Young Researcher and Elie Club, Arak-Branch, Islamic Azad Universiy, Arak, Iran Full lis of auhor informaion is available a he end of he aricle Absrac In his work, parial answers o Reich, Mizoguchi and Takahashi s and Amini-Harandi s conjecures are presened via a ligh version of Carisi s fixed poin heorem. Moreover, we inroduce he idea ha many of known fixed poin heorems can easily be derived from he Carisi heorem. Finally, he exisence of bounded soluions of a funcional equaion is sudied. MSC: Primary 47H10; secondary 54E05 Keywords: Carisi s fixed poin heorem; Hausdorff meric; Mizoguchi-Takahashi; Reich s problem; Boyd and Wong s conracion 1 Inroducion and preliminaries In he lieraure, he Carisi fixed poin heorem is known as one of he very ineresing and useful generalizaions of he Banach fixed poin heorem for self-mappings on a complee meric space. In fac, he Carisi fixed poin heorem is a modificaion of he ε-variaional principle of Ekeland [1, 2]), which is a crucial ool in nonlinear analysis, in paricular, opimizaion, variaional inequaliies, differenial equaions, and conrol heory. Furhermore, in 1977 Wesern [3] proved ha he conclusion of Carisi s heorem is equivalen o meric compleeness. In he las decades, Carisi s fixed poin heorem has been generalized and exended in several direcions see e.g.,[4, 5] and he relaed references herein). The Carisi s fixed poin heorem assers he following. Theorem 1.1 [6]) Le X, d) be a complee meric space and le T : X X be a mapping such ha dx, Tx) ϕx) ϕtx) 1) for all x X, where ϕ : X [0, + ) is a lower semiconinuous mapping. Then T has a leas a fixed poin. Le us recall some basic noaions, definiions, and well-known resuls needed in his paper. Throughou his paper, we denoe by N and R he ses of posiive inegers and real numbers, respecively. Le X, d) be a meric space. Denoe by CBX) he familyof all nonempy, closed, and bounded subses of X. A funcion H : CBX) CBX) [0, ) 2016 Khojaseh e al. This aricle is disribued under he erms of he Creaive Commons Aribuion 4.0 Inernaional License hp://creaivecommons.org/licenses/by/4.0/), which permis unresriced use, disribuion, and reproducion in any medium, provided you give appropriae credi o he original auhors) and he source, provide a link o he Creaive Commons license, and indicae if changes were made.

2 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 2 of 10 defined by { HA, B)=max sup x B } dx, A), sup dx, B) x A is said o be he Hausdorff meric on CBX) inducedbyhemericd on X. Apoinv in X is a fixed poin of a map T if v = Tv when T : X X is a single-valued map) or v Tv when T : X CBX) is a muli-valued map). Le X, d) be a complee meric space and T : X X amap.supposehereexissa funcion φ :[0,+ ) [0, + )saisfyingφ0) = 0, φs)<sfor s >0,andsupposehaφ is righ upper semiconinuous such ha dtx, Ty) φ ), x, y X. Boyd-Wong [7] showed ha T has a unique fixed poin. In 1972, Reich [8] inroduced he following open problem. Problem 1.1 Le X, d) beacompleemericspaceandlet : X CBX) beamulivalued mapping such ha HTx, Ty) μ ) 2) for all x, y X, whereμ : R + R + is coninuous and increasing map such ha μ)<, for all >0.DoesT have a fixed poin? Some parial answers o Problem 1.1 were given by Daffer e al. 1996) [9]andJachymski 1998) [10]. In hese works, he auhors consider addiional condiions on he mapping μ o find a fixed poin. Daffer e al. assumed ha μ : R + R + is upper righ semiconinuous, μ)< for all >0,and μ) a b,wherea >0, 1<b <2on some inerval [0, s], s >0. Jachymski assumed ha μ : R + R + is superaddiive, i.e., μx + y)>μx)+μy), for all x, y R +,and μ) is nondecreasing. In 1983, Reich [11], inroduced anoher problem as follows. Problem 1.2 Le X, d) be a complee meric space and le T : X CBX) be a mapping such ha HTx, Ty) η ) 3) for all x, y X, whereη :0,+ ) [0, 1) is a mapping such ha lim sup r + ηr)<1,for all r 0, + ). Does T have a fixed poin? In 1989, Mizoguchi and Takahashi [12], gave a parial answer o Problem 1.2 as follows.

3 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 3 of 10 Theorem 1.2 Le X, d) be a complee meric space and le T : X CBX) be a mapping such ha HTx, Ty) η ) 4) for all x, y X, where η :0,+ ) [0, 1) is a mapping such ha lim sup r + ηr)<1,for all r [0, + ). Then T has a fixed poin. Anoher analogous open problem was raised in 2010 by Amini-Harandi [13], which we sae afer he following noaions. In he following, γ :[0,+ ) [0, + ) is subaddiive, i.e. γ x+y) γ x)+γ y), for each x, y [0, + ), a nondecreasing coninuous map such ha γ 1 {0})={0},andleƔ consis of all such funcions. Also, le A be he class of all maps θ :[0,+ ) [0, + ) forwhich here exiss an ɛ 0 >0suchha θ) ɛ 0 θ) γ ), where γ Ɣ. Problem 1.3 Assume ha T : X CBX) is a weakly conracive se-valued map on a complee meric space X, d), i.e., HTx, Ty) θ ) for all x, y X,whereθ A.DoesT have a fixed poin? The answer is yes if Tx is compacforevery x Amini-Harandi [13], Theorem3.3). In his work, we show ha many of he known Banach conracion generalizaions can be deduced and generalized by Carisi s fixed poin heorem and is consequences. Also, parial answers o he menioned open problems are given via our main resuls. For more deails as regards a fixed poin generalizaion of muli-valued mappings we refer o [14]. 2 Main resul In his secion, we show ha many of he known fixed poin resuls can be deduced from he following ligh version of Carisi s heorem. Corollary 2.1 Le X, d) be a complee meric space, and le T : X X be a mapping such ha ϕx, y) ϕtx, Ty) 5) for all x, y X, where ϕ : X X [0, ) is lower semiconinuous wih respec o he firs variable. Then T has a unique fixed poin. Proof For each x X,ley = Tx and ψx)=ϕx, Tx). Then for each x X dx, Tx) ψx) ψtx)

4 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 4 of 10 and ψ is a lower semiconinuous mapping. Thus, applying Theorem 1.1 leadsusoconclude he desired resul. To see he uniqueness of he fixed poin suppose ha u, v are wo disinc fixed poins for T.Then du, v) ϕu, v) ϕtu, Tv)=ϕu, v) ϕu, v)=0. Thus, u = v. Corollary 2.2 [15], Banach conracion principle) Le X, d) be a complee meric space and le T : X X be a mapping such ha for some α [0, 1) dtx, Ty) α 6) for all x, y X. Then T has a unique fixed poin. Proof Define ϕx, y)= dx,y) 1 α.then6)showsha 1 α) dtx, Ty). 7) I means ha 1 α dtx, Ty) 1 α 8) and so ϕx, y) ϕtx, Ty), 9) and so by applying Corollary 2.1,onecanconcludehaT has a unique fixed poin. Corollary 2.3 Le X, d) be a complee meric space and le T : X X be a mapping such ha dtx, Ty) η ), 10) where η :[0,+ ) [0, ) is a lower semiconinuous mapping such ha η)<, for each >0,and η) is a nondecreasing map. Then T has a unique fixed poin. Proof Define ϕx, y)= 1 I means ha dx,y) 1 ηdx,y)) dx,y),ifx y and oherwise ϕx, x)=0.then10)showsha ) η) dtx, Ty). 11) 1 ηdx,y)) dx,y) dtx, Ty). 12) 1 ηdx,y)) dx,y)

5 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 5 of 10 Since η) is nondecreasing and dtx, Ty) <, 1 ηdx,y)) dx,y) dtx, Ty) 1 ηdtx,ty)) dtx,ty) = ϕx, y) ϕtx, Ty), 13) and so by applying Corollary 2.1,onecanconcludehaT has a unique fixed poin. The following resuls are he main resuls of his paper and play a crucial role o find he parial answers for Problem 1.1,Problem1.2, and Problem 1.3. Comparing he parial answers for Reich s problems, our answers include simple condiions. Also, he compacness condiion on Tx is no needed. Theorem 2.1 Le X, d) be a complee meric space, and le T : X CBX) be a nonexpansive mapping such ha, for each x X, and for all y Tx, here exiss z Ty such ha ϕx, y) ϕy, z), 14) where ϕ : X X [0, ) is lower semiconinuous wih respec o he firs variable. Then Thasafixedpoin. Proof Le x 0 X and le x 1 Tx 0.Ifx 0 = x 1 hen x 0 is a fixed poin and we are hrough. Oherwise, le x 1 x 0. By assumpion here exiss x 2 Tx 1 such ha dx 0, x 1 ) ϕx 0, x 1 ) ϕx 1, x 2 ). Alernaively, one can choose x n Tx n 1 such ha x n x n 1 and find x n+1 Tx n such ha 0<dx n 1, x n ) ϕx n 1, x n ) ϕx n, x n+1 ), 15) which means ha {ϕx n 1, x n )} n is a non-increasing sequence, bounded below, so i converges o some r 0. By aking he limi on boh sides of 15) wehave lim n dx n 1, x n ) = 0. Also, for all m, n N wih m > n, m dx n, x m ) dx i 1, x i ) i=n+1 m ϕx i 1, x i ) ϕx i, x i+1 ) i=n+1 ϕx n, x n+1 ) ϕx m, x m+1 ). 16) Therefore, by aking he limsup on boh sides of 16)wehave { lim sup dxn, x m ):m > n }) =0. n

6 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 6 of 10 I means ha {x n } is a Cauchy sequence and so i converges o u X. Now we show ha u is a fixed poin of T.Wehave du, Tu) du, x n+1 )+dx n+1, Tu) = du, x n+1 )+HTx n, Tu) du, x n+1 )+dx n, u). 17) By aking he limi on boh sides of 17), we ge dx, Tx)=0andhismeanshax Tx. The following heorem is a parial answer o Problem 1.1. Theorem 2.2 Le X, d) be a complee meric space, and le T : X CBX) be a mulivalued funcion such ha HTx, Ty) η ) for all x, y X, where η :[0, ) [0, ) is a lower semiconinuous map such ha η)<, for all 0, + ), and η) is nondecreasing. Then T has a fixed poin. Proof Le x X and y Tx.Ify = x hen T has a fixed poin and he proof is complee, so we suppose ha y x.define θ)= η)+ 2 for all 0, + ). We have HTx, Ty) η) < θ) <. Thus here exiss ɛ 0 >0suchha θ) = HTx, Ty)+ɛ 0.Sohereexissz Ty such ha dy, z)<htx, Ty)+ɛ 0 = θ ) <. 18) We again suppose ha y z; herefore θ) dy, z)orequivalenly < 1 θdx,y)) dx,y) dy, z) 1 θdx,y)) dx,y), since θ) is also a nondecreasing funcion and dy, z) < wege < 1 θdx,y)) dx,y) dy, z) 1 θdy,z)) dy,z). Define x, y)= dx,y) 1 θdx,y)) dx,y) < x, y) y, z). if x y, oherwise 0 for all x, y X.Imeansha Therefore, T saisfies 14) oftheorem2.1 and so we conclude ha T has a unique fixed poin u and he proof is compleed. The following heorem is a parial answer o Problem 1.2.

7 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 7 of 10 Corollary 2.4 [12], Mizoguchi-Takahashi s ype) Le X, d) be a complee meric space and le T : X CBX) be a muli-valued mapping such ha HTx, Ty) η ) 19) for all x, y X, where η :[0,+ ) [0, 1) is a lower semiconinuous and nondecreasing mapping. Then T has a fixed poin. Proof Le θ)=η), θ)< for all R +,and θ) = η) is a nondecreasing mapping. By he assumpion dtx, Ty) η) =θ) for all x, y X, herefore by Theorem 2.2 T has a fixed poin. Noe ha if η :[0, ) [0, 1) is a nondecreasing map hen for all s [0, + ) lim sup s + η)=inf sup η) δ>0 s <s+δ = lim sup δ 0 s <s+δ η) ηs + δ)<1. I means ha Corollary 2.4 comes from he Mizoguchi-Takahashi s resuls direcly and here we deduce i from our resuls [12]. The following heorem is a parial answer o Problem 1.3. Corollary 2.5 Le X, d) be a complee meric space, and le T : X CBX) be a mulivalued funcion such ha HTx, Ty) θ ) for all x, y X, where θ :0, ) 0, ) is an upper semiconinuous map such ha, for all 0, + ), θ) is non-increasing. Then T has a fixed poin. Proof Le η)= θ), for each >0.Thenη)<, foreach >0,and η) nondecreasing. Thus, he desired resul is obained by Theorem 2.2. =1 θ) 3 Exisence of bounded soluions of funcional equaions Mahemaical opimizaion is one of he fields in which he mehods of fixed poin heory are widely used. I is well known ha dynamic programming provides useful ools for mahemaical opimizaion and compuer programming. In his seing, he problem of dynamic programming relaed o a mulisage process reduces o solving he funcional equaion is { ))} px)=sup f x, y)+i x, y, p ηx, y), x Z, 20) y T where η : Z T Z, f : Z T R, andi : Z T R R. We assume ha M and N are Banach spaces, Z M is a sae space, and T N is a decision space. The sudied process consiss of asaespace, which is he se of he iniial sae, acions, and a ransiion model of he process, and a decision space, which is he se of possible acions ha are allowed for he process.

8 Khojaseh eal. Fixed Poin Theory and Applicaions 2016) 2016:16 Page 8 of 10 Here, we sudy he exisence of he bounded soluion of he funcional equaion 20). Le BZ) denoe he se of all bounded real-valued funcions on W and, for an arbirary h BZ), define h = sup x Z hx). Clearly, BW), ) endowed wih he meric d defined by dh, k) =sup hx) kx) 21) x Z for all h, k BZ), is a Banach space. Indeed, he convergence in he space BZ) wih respec o is uniform. Thus, if we consider a Cauchy sequence {h n } in BZ), hen {h n } converges uniformly o a funcion, say h, ha is bounded and so h BZ). We also define S : BZ) BZ)by { ))} Sh)x)=sup f x, y)+i x, y, h ηx, y) y T 22) for all h BZ)andx Z. We will prove he following heorem. Theorem 3.1 Le S : BZ) BZ) be an upper semiconinuous operaor defined by 22) and assume ha he following condiions are saisfied: i) f : Z T R and I : Z T R R are coninuous and bounded; ii) for all h, k BZ), if ) ) 0<dh, k) <1 implies I x, y, hx) I x, y, kx) 1 2 d2 h, k), ) ) dh, k) 1 implies I x, y, hx) I x, y, kx) 2 dh, k), 3 23) where x Z and y T. Then he funcional equaion 20) has a bounded soluion. Proof Noe ha BZ),d) is a complee meric space, where d is he meric given by 21). Le μ be an arbirary posiive number, x Z,andh 1, h 2 BZ), hen here exis y 1, y 2 T such ha Sh 1 )x)<fx, y 1 )+I x, y 1, h 1 ηx, y1 ) )) + μ, 24) Sh 2 )x)<fx, y 2 )+I x, y 2, h 2 ηx, y2 ) )) + μ, 25) Sh 1 )x) f x, y 1 )+I x, y 1, h 1 ηx, y1 ) )), 26) Sh 2 )x) f x, y 2 )+I x, y 2, h 2 ηx, y2 ) )). 27) Le ϱ :[0, ) [0, )bedefinedby 1 2 ϱ)= 2, 0< <1, 1, 1. 2 Then we can say ha 23)isequivaleno I x, y, hx) ) I x, y, kx) ) ϱ dh, k) ) 28)

9 Khojaseh eal.fixed Poin Theory and Applicaions 2016) 2016:16 Page 9 of 10 for all h, k BZ). I is easy o see ha ϱ) <, for all >0,and ϱ) funcion. Therefore, by using 24), 27), and 28), i followsha is a nondecreasing Sh 1 )x) Sh 2 )x)<i x, y 1, h 1 ηx, y1 ) )) I x, y 2, h 2 ηx, y2 ) )) + μ I x, y1, h 1 ηx, y1 ) )) I x, y 2, h 2 ηx, y2 ) )) + μ ϱ dh 1, h 2 ) ) + μ. Then we ge Sh 1 )x) Sh 2 )x)<ϱ dh 1, h 2 ) ) + μ. 29) Analogously, by using 25)and26), we have Sh 2 )x) Sh 1 )x)<ϱ dh 1, h 2 ) ) + μ. 30) Hence, from 29)and30) we obain Sh2 )x) Sh 1 )x) < ϱ dh1, h 2 ) ) + μ, ha is, d Sh 1 ), Sh 2 ) ) < ϱ dh 1, h 2 ) ) + μ. Since he above inequaliy does no depend on x Z and μ > 0 is aken arbirary, we conclude immediaely ha d Sh 1 ), Sh 2 ) ) ϱ dh 1, h 2 ) ), so we deduce ha he operaor S is a ϱ-conracion. Thus, due o he coninuiy of S, Theorem 2.2 applies o he operaor S, which has a fixed poin h BZ), ha is, h is a bounded soluion of he funcional equaion 20). Compeing ineress The auhors declare ha hey have no compeing ineress. Auhors conribuions All of he auhors have made equal conribuions. Auhor deails 1 Young Researcher and Elie Club, Arak-Branch, Islamic Azad Universiy, Arak, Iran. 2 Deparmen of Mahemaics, Ailim Universiy, İncek, Düsernbrooker Weg 20, Ankara, 06836, Turkey. 3 Deparmen of Mahemaics, Mahabad-Branch, Islamic Azad Universiy, Mahabad, 06836, Iran. Acknowledgemens The firs auhor would like o hank Prof. S. Mansour Vaezpour for valuable suggesions. Received: 27 July 2015 Acceped: 13 January 2016

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