# Best Proximity Point for Generalized andS-Geraphty Contractions

Authors: Abdelhafid Bassou; Taoufik Sabar; Mohamed Aamri
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

This paper introduces a new class of mappings calledS-Geraphty-contractions and provides sufficient conditions for the existence and uniqueness of a best proximity point for such mappings. It also presents the best proximity point result for generalized contractions as well. Our results extend and generalize some theorems in the literature.


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## Body

## 1. Introduction and Preliminaries

The center of interest of fixed point theory is the solving of the equationTx=x where T is a mapping defined on a subset of a metric space, a normed linear space, or a topological vector space. Ever since its appearance, the well-known Banach contraction principle has been extensively studied, and the literature contains numerous interesting extensions and generalizations of the aforementioned result, in particular, Geraphty’s generalization of the Banach contraction principle.In 1973, Geraphty ([1]) introduced the class S of functions β:0,∞⟶0,1 satisfying the following condition:
(1)βtn⟶1impliestn⟶0.And proved the following result.Theorem 1.
LetX,dbe a complete metric space andT:X⟶Xbe an operator satisfying the the following inequality for someβ∈S,(2)dTx,Ty≤βdx,y.dx,y,forallx,y∈X.Then,T has a unique fixed point.SinceS contains the class of constant functions βt=k∈0,1, the previous theorem extends that of Banach.Another interesting extension of the Banach contraction principle is due to Kirk et al. ([2]). The authors introduced the class of cyclic mappings, i.e., T:A∪B⟶A∪B such that TA⊆B and TB⊆A. And, under a suitable condition on T, proved a fixed point theorem which extends that of Banach. Interestingly, a more important problem than the extension of the Banach principle arose.Whereas a cyclic mapping does not necessarily have a fixed point, it is desirable to determine an elementx which is somehow closest to Tx. More precisely, an element x for which the error dx,Tx assumes the least possible value distA,B where distA,B=infdx,y:x∈A,y∈B, such a point is called a best proximity point of the cyclic mapping T. Since 2003, research on best proximity points of cyclic mapping became an important topic in nonlinear analysis and has been studied by many authors [2–8].In 2012, Caballero and al, introduced the following contraction.Definition 2 (see [9]).
LetA,Bbe nonempty subsets of a metric spaceX,d.A nonself mappingT:A⟶Bis said to be a Geraphty-contraction if there exists aβ∈Ssuch that(3)dTx,Ty≤βdx,ydx,yforallx,y∈A.Just like cyclic mappings, nonself mappings may not have fixed points and a best proximity point for a nonself mappingT:A⟶B is a point x∈A such that dx,Tx=distA,B.Later on, the current authors ([10]) introduced the notion of tricyclic mappings and the best proximity point thereof. Let A,B, and C be nonempty subsets of a metric space X,d.A mapping T:A∪B∪C⟶A∪B∪C is said to be tricyclic provided that TA⊆B,TB⊆C, and TC⊆A. A best proximity point of T is a point x∈A∪B∪C such that Dx,Tx,T2x=δA,B,C, where the mapping D:X×X×X⟶0,+∞ is defined by Dx,y,z=dx,y+dy,z+dz,x, and
(4)δA,B,C=infDx,y,z:x∈A,y∈Bandz∈C.Some results about the best proximity points of tricyclic mappings can be found in [10–15].In the next section of this paper, taking inspiration from our recent works, we introduce the new class ofS−cyclic mappings, which stands somewhere between the two classes of cyclic and tricyclic mappings. We define S−Geraphty-contractions and establish a best proximity point theorem for such mappings. As a special case, we obtain the best proximity point and fixed point theorem for cyclic mapping.To describe our results, we need some definitions and notations. Given a triadA,B,C of nonempty subsets of a metric space X,d, then, the proximal pair A0,B0 of A,B is given by:
(5)A0≔x∈A:dx,y′=distA,Bforsomey′∈B,B0≔y∈B:dx′,y=distA,Bforsomex′∈A.A pairx,y∈A×B is said to be proximal in A,B if dx,y=distA,B. We subsequently use the following notations:
(6)A00=x∈A:Dx,y′,z′′=δA,B,Cforsomey′∈Bandz′′∈C,B00=y∈B:Dx′′,y,z′=δA,B,Cforsomex′′∈Aandz′∈C,C00=z∈C:Dx′,y′′,z=δA,B,Cforsomex′∈Aandy′′∈B,A×B0=x,y∈A×B,Dx,y,z=δA,B,Cforsomez∈C,B×C0=y′,z′∈B×C,Dx′,y′,z′=δA,B,Cforsomex′∈C,C×A0=z′′,x′′∈C×A,Dx′′,y′′,z′′=δA,B,Cforsomey′′∈C.Note thatB×C0 is included in B00×C00 but the inverse does not always hold, and it is obvious that δA00,B00,C00=δA,B,C. Let us illustrate the cases B×C0⊂B00×C00 and B×C0=B00×C00 with simple examples.
(i)
ConsiderA=0,1×0,B=0,1×1, and C=0,1×2. Clearly δA,B,C=4,0,1∈B00 and 1,2∈C00. For all t,0∈A, we have(7)Dt,0,0,1,1,2=t2+1+2+1−t2+4>4.Example 3.
LetXbeℝ2endowed with its Euclidean distance.Hence,0,1,1,2∉B×C0 and so B×C0⊂B00×C00.(ii)
LetA=−3,−2×0,B=−1,1×−1,1 and C=2,3×0. Then δA,B,C=8, and(8)A00=−2,0,B00=−1,1×0,C00=2,0.Lety,0∈B00,(9)D−2,0,y,0,2,0=8.And then,B×C0=B00×C00.(iii)
In the special case whereC=A, we get A00=A0,B00=B0 and(10)A×B0=x,y∈A×B:x,yisproximalinA,B.LetA and B be defined as
(11)A=x,x:x∈−3,−1∪1,3andB=x,−x:x∈−3,−1∪1,3.Clearly,A0=−1,−1,1,1,B0=−1,1,1,−1, and
(12)A×B0=−1,−1,−1,1,−1,−1,1,−1,1,1,−1,1,1,1,1,−1=A0×B0.Definition 4.
LetA,B, andCbe nonempty subsets of a metric spaceX,d.A mappingT:A∪B⟶B∪Cis said to be aS−cyclicifTA⊆BandTB⊆C.A best proximity point forTis a pointx∈Aprovided thatDx,Tx,T2x=δA,B,C.Definition 5.
LetA,B,Cbe nonempty subsets of a metric spaceX,d.AS−cyclicmappingT:A∪B⟶B∪Cis said to be aS-Geraphty-contraction if there exists aβ∈Ssuch that:(13)DTx,Ty,Tz≤βDx,y,z.Dx,y,zforallx,y,z∈A.Notice that since theβ is strictly smaller than one, we have
(14)DTx,Ty,Tz≤βDx,y,z.Dx,y,z<Dx,y,zforallx,y,z∈A.In the special case wherex=z, we obtain
(15)DTx,Ty,Tx<Dx,y,x,which means
(16)dTx,Ty<dx,yforallx,y∈A.Thus, everyS-Geraphty-contraction is continuous on A.Example 6.
Consider a mappingT:A∪B⟶B∪Csuch thatTA⊆BandTB⊆C.SupposeT/A:A⟶Bis Geraphty-contraction for some nondecreasingβ∈S.We have(17)dTx,Ty≤βdx,ydx,y,for all x,y∈A. Then
(18)DTx,Ty,Tz≤maxβdx,y,βdy,z,βdz,x.Dx,y,z≤βDx,y,z.Dx,y,z.
Which means,T:A∪B⟶B∪C is a S-Geraphty-contraction.Definition 7 (see [16]).
LetAandBbe nonempty subsets of a metric spaceX,d.The pairA,Bis said to have theP−propertyif and only if(19)dx1,y1=dA,Bdx2,y2=dA,B⟹dx1,x2=dy1,y2,where x1,x2∈A and y1,y2∈B.The concept ofP−property of pairs can naturally be extended to triads.Definition 8.
LetA,B, andCbe nonempty subsets of a metric spaceX,d.The triadA,B,Cis said to have theP−propertyif and only if(20)Dx1,y1,z1=δA,B,CDx2,y2,z2=δA,B,CDx3,y3,z3=δA,B,C⟹Dx1,x2,x3=Dy1,y2,y3=dz1,z2,z3,where x1,x2,x3∈A, y1,y2,y3∈B, and z1,z2,z3∈C.Example 9.
Letx1,x2,x3∈A, y1,y2,y3∈B, andz1,z2,z3∈Csuch that(21)Dx1,y1,z1=Dx2,y2,z2=Dx3,y3,z3=δA,B,C.(1)
SupposeδA,B,C=distA,B+distB,C+distC,A. We necessarily have(22)dx1,y1=dx2,y2=dx3,y3=distA,B,dy1,z1=dy2,z2=dy3,z3=distB,C,dz1,x1=dz2,x2=dz3,x3=distC,A.
It is plain to see that if the pairsA,B,B,C, and C,A have the P−property, then, the triad A,B,C does as well.
(2)
IfA=C, then(23)δA,B,C=δA,B,A=2distA,B.
Which impliesx1=z1,x2=z2,x3=z3, and
(24)dx1,y1=dx2,y2=distA,B.Thus, the triadA,B,A has the P−property if the pair A,B does.


## 2.S-Geraphty-Contractions

Next, we state the main result of this section.Theorem 10.
LetA,B,Cbe a triad of nonempty closed subsets of a complete metric spaceX,dsuch thatA00is nonempty andB00×C00=B×C0.LetT:A∪B⟶B∪Cbe aS-Geraphty-contraction satisfyingTA00⊆B00andTB00⊆C00.Assume the triadA,B,Chas theP-property. Then, there exists a uniquex′∈Asuch thatDx′,Tx′,T2x′=δA,B,C.Proof.
Letx0∈A00, since TA00⊆B00 and TB00⊆C00,(25)Tx0,T2x0∈B00×C00=B×C0.
There then existsx1∈A00 such that Dx1,Tx0,T2x0=δA,B,C. By the same argument, we obtain some x2∈A00, for which Dx2,Tx1,T2x1=δA,B,C.Keeping on this process, we can get a sequence xn in A00 satisfying
(26)Dxn+1,Txn,T2xn=δA,B,C,foralln∈ℕ.
SinceA,B,C has the P-property, we have
(27)Dxn,xn+1,xn+2=DTxn−1,Txn,Txn+1.
Taking into consideration thatT is a S-Geraphty-contraction, we have
(28)Dxn,xn+1,xn+2=DTxn−1,Txn,Txn+1≤βDxn−1,xn,xn+1.Dxn−1,xn,xn+1<Dxn−1,xn,xn+1.
In the case whereDxn0,xn0+1,xn0+2=0, for some n0∈ℕ, we have
(29)0=Dxn0,xn0+1,xn0+2=DTxn0−1,Txn0,Txn0+1.
Hence,Txn0−1=Txn0=Txn0+1. Consequently
(30)δA,B,C=Dxn0,Txn0−1,T2xn0−1=Dxn0,Txn0,T2xn0.
And the desired result then follows. Therefore, we suppose the contrary case, that is,Dxn,xn+1,xn+2>0, for every n∈ℕ. We note that Dxn,xn+1,xn+2 is a decreasing sequence of positive real numbers, hence, there exists r≥0 such that
(31)limn→∞Dxn,xn+1,xn+2=r.
Supposer>0, we have
(32)0<Dxn,xn+1,xn+2Dxn−1,xn,xn+1≤βDxn−1,xn,xn+1<1foralln∈ℕ.
Which implies that(33)limn→∞βDxn−1,xn,xn+1=1.
And sinceβ∈F, we get r=0 and this contradicts our assumption. Thus, limn→∞Dxn,xn+1,xn+2=0.
In the sequel, we show thatxn is a Cauchy sequence. Notice that Dxp+1,Txp,T2xp=Dxq+1,Txq,T2xq=δA,B,C for any p,q∈ℕ, and since A,B,C satisfies the P-property, Dxp+1,xq+1,xp+1=Dxp,xq,xp. Suppose xn is not a Cauchy sequence, hence,
(34)limsupDxp,xq,xp>0.
We have(35)Dxp,xq,xp=2dxp,xq≤2dxp,xp+1+2dxp+1,xq+1+2dxq+1,xq≤Dxp,xp+1,xp+2+Dxp+1,xq+1,xp+1+Dxq,xq+1,xq+2≤Dxp,xp+1,xp+2+βDxp,xq,xp.Dxp,xq,xp+Dxq,xq+1,xq+2.
Which means(36)Dxp,xq,xp≤Dxp,xp+1,xp+2+Dxq,xq+1,xq+21−βDxp,xq,xp.
SincelimsupDxp,xq,xp>0 and limsupDxp,xp+1,xp+2=0. Then, by passage to the limit in the last inequality, we obtain
(37)limsup11−βDxp,xq,xp=∞.
Therefore,limsup1−βDxp,xq,xp=0.But, β∈S, thus, limsupDxp,xq,xp=0, which is contradictory with our assumption. Consequently xn is a Cauchy sequence.
Sincexn⊂A and A is a close subset of the complete metric space X,d,xn⟶x′∈A. Since T is continuous,Txn⟶Tx and T2xn⟶T2x. Which implies, Dxn,Txn,T2xn⟶Dx′,Tx′,T2x′. On the other hand, Dxn,Txn,T2xn is a constant sequence with the value δA,B,C. Thus
(38)Dx′,Tx′,T2x′=δA,B,C.
That meansx′ is a best proximity point of T. As for the uniqueness, suppose that x1 and x2 are two distinct best proximity points of T. That is
(39)Dx1,Tx1,T2x1=Dx2,Tx2,T2x2=δA,B,C.
Taking into account thatA,B,C has the P-property, we get
(40)Dx1,x2,x2=DTx1,Tx2,Tx2.
Which is contradictory with the fact thatT is a S-Geraphty-contraction. Indeed, we have
(41)Dx1,x2,x2=DTx1,Tx2,Tx2≤βDx1,x2,x2.Dx1,x2,x2<Dx1,x2,x2.
And the proof is completed.Example 11.
ConsiderX=ℝ2with its usual metric. LetA,B, andCbe defined by(42)A=0×0,∞,B=1×0,∞andC=2,∞×0.It is easy to see thatA00=0,0,B00=1,0 and C00=2,0. Hence, A00 is nonempty and B00×C00=B×C0.Letk∈0,1 and let Tk:A∪B⟶B∪C be the mapping defined as
(43)Tk0,x=1,kxforall0,x∈A,Tk1,y=2+ky,0forall1,y∈B.Let0,x1,0,x2,0,x3∈A,(44)DTk0,x1,Tk0,x2,Tk0,x3=D1,kx1,1,kx2,1,kx3≤kDx1,x2,x3.Since the constant functionsβt=k where k∈0,1, belong to S,T is a S-Geraphty-contraction. Also, A,B,C has the P-property. Indeed, if
(45)Dx1,y1,z1=δA,B,C=4,Dx2,y2,z2=δA,B,C=4,Dx3,y3,z3=δA,B,C=4.Thenx1=x2=x3=0,0, y1=y2=y3=1,0, and z1=z2=z3=2,0. Thus, Dx1,x2,x3=Dy1,y2,y3=Dz1,z2,z3=0. All things considered, T has a unique best proximity point, clearly 0,0.Corollary 12.
LetA,Bbe a pair of nonempty closed subsets of a complete metric spaceX,dsuch thatA0is nonempty.LetT:A∪B⟶B∪Abe aS-Geraphty-contraction satisfyingTA0⊆B0andTB0⊆A0.Assume the triadA,Bhas theP-property. Then there exists a uniquex′∈Athat is both a best proximity point for the cyclic mappingTand a fixed point for the self-mappingT2.Proof.
The triadA,B,A satisfies the condition of the previous theorem. Hence, there exists a unique x′∈A such that Dx′,Tx′,T2x′=2distA,B, which implies
(46)dx′,Tx′=distA,BandT2x′=x′,and that is the wanted result.


## 3.S-Min–Max Condition and Generalized Contractions

In this section, using theS-min–max condition, we obtain a best proximity point result for nonself generalized contractions. First, we recall and fix some notions and notations that will subsequently be used. Given mappings R:A⟶B,S:B⟶C, and T:C⟶A, where A,B, and C are nonempty subsets of a metric space X,d.Definition 13 (see [6]).
LetA,Bbe nonempty subsets of a metric spaceX,d.The mappingR:A⟶Bis called a generalized contraction if, given real numbersaandbwith0<a≤b, there exists a real numberαa,b∈0,1such that(47)a≤dx1,x2≤b⟹dRx1,Rx2≤αa,bdx1,x2,for all x1,x2 in A.Obviously, every generalized contraction is a contractive mapping.Definition 14 (see [17]).
LetA,Bbe nonempty subsets of a metric spaceX,d.It is stated that the pairR,Ssatisfies the min-max condition if, for allx∈Aandy∈B, we have(48)dA,B<dx,y⟹maxRx,Sy≠minRx,Sy,where minRx,Sy and maxRx,Sy are defined as
(49)minRx,Sy=mindx,y,dx,Rx,dy,Sy,dRx,Sy,dx,RSy,dy,SRx,dRx,SRx,dSy,RSy,dSRx,RSy,maxRx,Sy=maxdx,y,dx,Rx,dy,Sy,dx,Sy,dy,Rx,dRx,Sy,dx,SRx,dy,RSy,dx,RSy,dy,SRx,dRx,SRx,dSy,RSy,dSRx,RSy.Definition 15.
LetA,B,Cbe nonempty subsets of a metric spaceX,d.The triadR,S,Tis said to satisfyS-min–max condition if for allx∈A, y∈B, andz∈C,
(50)distA,B<dx,yandTz=x⟹SminRx,Sy,Tz≠SmaxRx,Sy,Tz,where
(51)SminRx,Sy,Tz=mindx,y,z,dx,Rx,z,dy,Sy,TSy,dRx,Sy,TSy,dx,RTSy,z,dy,TSRx,SRTz,dRx,SRx,TSRx,dSy,TSy,RTSy,dTSRx,RTSy,SRTz,SmaxRx,Sy,Tz=maxdx,y,z,dx,Rx,z,dy,Sy,TSy,dx,Sy,z,dy,Rx,RTz,dRx,Sy,TSy,dx,TSRx,z,dy,RTSy,y,dx,RTSy,z,dy,TSRx,SRTz,dRx,SRx,SRTz,dSy,TSy,RTSy,dTSRx,RTSy,SRTz.Now, we are at liberty to state the main result of this section.Theorem 16.
SupposeA,B, andCare nonempty closed subsets of a metric spaceX,dand the mappingsR,S, andTverify the following conditions:(52)Risageneralizedcontraction.SandTarenonexpansivemappings.ThetriadR,S,TsatisfiestheS−min−maxcondition.For a fixed elementx0 in A, let
(53)x3n+1=Rx3n,x3n+2=Sx3n+1andx3n=Tx3n−1.Then, the sequencex3n must converge to a best proximity point x∗of R and the sequence x3n+1 must converge to a point y∗ such that
(54)dx∗,y∗=distA,B.Proof.
Let a sequencesn of real numbers be defined as follows:
(55)sn≔dx3n,x3n+3forn≥0.
From the fact thatR is a generalized contraction and from the nonexpansiveness of S and T, it follows that sn is a bounded below, decreasing sequence of nonnegative real numbers and hence converges to some nonnegative real number, say s.We show that s must be nil. If not, then we choose a positive integer N such that
(56)s≤sn≤b+1foralln≥N.
Therefore,(57)sN+1=dx3N+3,x3N+6=dTx3N+2,Tx3N+5≤dx3N+2,x3N+5=dSx3N+1,Sx3N+4≤dx3N+1,x3N+4=dRx3N,Rx3N+3≤αs,s+1sN.
Similarly, we get(58)sN+2≤αs,s+1sN+1≤αs,s+12sN.
Keeping on that process, we obtain(59)sN+k≤αs,s+1ksN.
Whenk⟶∞, we deduce that s=0 and this is a contradiction. Hence, it can be deduced that sn⟶0 as n⟶∞. Next, we shall prove that x3n is a Cauchy sequence. Let ε>0 be given. Since sn⟶0, it is possible to choose a positive integer N such that
(60)sN=dx3N,x3N+3≤ε21−αε2,ε.
It is sufficient to prove that wheneverx is an element of the closed ball Bx3N,ε then so is TSRx. If x satisfies dx3N,x≤ε/2, then
(61)dTSRx,x3N≤dTSRx,x3N+3+dx3N+3,x3N≤dRx,Rx3N+sN≤dx,x3N+sN≤ε.
And, ifx satisfies ε/2<dx3N,x≤ε, then
(62)dTSRx,x3N≤dTSRx,x3N+3+dx3N+3,x3N≤dRx,Rx3N+sN≤αε2,εdx,x3N+sN≤ε.
We conclude thatx3n∈Bx3N,ε for all n≥N and that means x3n is a Cauchy sequence. Taking the completeness of the space under consideration, x3n⟶x∗∈A and from the continuity of R,S, and T, we get
(63)x3n⟶x∗⟹x3n+1=Rx3n⟶Rx∗=y∗,x3n+1⟶y∗⟹x3n+2=Sx3n+1⟶Sy∗=z∗,x3n+2⟶z∗⟹x3n+3=Tx3n+2⟶Tz∗=x∗,for some y∗∈B and z∗∈C. Then, we deduce that TSRx∗=x∗,RTSy∗=y∗, and SRTz∗=z∗. We also have
(64)SminRx∗,Sy∗,Tz∗=Dx∗,y∗,z∗=SmaxRx∗,Sy∗,Tz∗.
And since the triadR,S,T satisfies the S-min–max condition, we obtain
(65)dx∗,y∗=distA,BorTz∗≠x∗.
Then, we necessarily have(66)dx∗,y∗=distA,B.
And that finishes the proof.As a special case of our result, we get the following best proximity point theorem, which was proved in [18].Corollary 17 (see [18]).
LetA and B be nonempty, closed subsets of a complete metric space. Let R:A⟶B and S:B⟶A satisfy the following conditions:
(67)Risageneralizedcontraction.Sisanonexpansivemapping.ThepairR,Ssatisfiesthemin−maxcondition.
Further, for a fixed elementy0 in A, let
(68)y2n+1=Ry2nandy2n=Sy2n−1.
Then, the sequencey2n must converge to a best proximity point x∗of R and the sequencey2n+1 must converge to a best proximity point y∗of S such that
(69)dx∗,y∗=dA,B.Further, ifS has two distinct best proximity points, then dA,B does not vanish and hence the sets A and B should be disjoint.Proof.
The triadR,S,T, where T:A⟶A is the identity mapping, satisfies the S-min–max condition. Indeed, let x,y,z∈A×B×A such that dx,y>distA,B and Tz=z=x. We have
(70)SminRx,Sy,Tz=min2dx,y,2dx,Rx,2dy,Sy,2dRx,Sy,2dx,RSy,2dy,SRx,2dRx,SRx,2dSy,RSy,2dSRx,RSy.=2minRx,Sy,
and(71)SmaxRx,Sy,Tz=max2dx,y,2dx,Rx,2dy,Sy,2dx,Sy,2dy,Rx,2dRx,Sy,2dx,SRx,2dy,RSy,2dx,RSy,2dy,SRx,2dRx,SRx,2dSy,RSy,2dSRx,RSy.=2maxRx,Sy.
And sinceR,S satisfies the min-max condition,
(72)SminRx,Sy,Tz=2minRx,Sy≠2maxRx,Sy=SmaxRx,Sy,Tz.
Thus, the triadR,S,T fulfills all the conditions of the previous theorem. For x0=y0∈A, define
(73)x3n+1=Rx3n,x3n+2=Sx3n+1andx3n=Tx3n−1=x3n−1.
Then,x3n must converge to a best proximity point x∗of R, the sequence x3n+1 must converge to a point y∗ and sequence x3n−1 must converge to z∗such that
(74)Rx∗=y∗,Sy∗=z∗=x∗anddx∗,y∗=distA,B.
It suffices to notice that(75)y2n+1=x3n+1andy2n=x3n=x3n−1.
Now, ifR has two distinct best proximity points a,b∈A, then
(76)da,b≤da,Sa+dSa,Sb+db,Sb<da,b+distA,B.
Which meansA and B are disjoint.



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