# A New Best Approximation Result in (S) Convex Metric Spaces

Authors: M. Sabiri; J. Mouline; A. Bassou; T. Sabar
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

Consider a self-mappingT defined on the union of p subsets of a metric space, and T is said to be p cyclic if TAi⊆Ai+1 for i=1,…,p with Ap+1=A1. In this article, we introduce the notion of S convex structure, and we acquire a best proximity point for p cyclic contraction in S convex metric spaces.


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## Body

## 1. Introduction

LetAii=1p be nonempty subsets of a metric space X,d. A mapping T:∪i=1pAi⟶∪i=1pAi is said to be cyclic if TAi⊆Ai+1 for i=1,…,p with Ap+1=A1.In 2003, Kirk et al. [1] proved that if T:∪i=1pAi⟶∪i=1pAi is cyclic and for some k∈0,1, dTx,Ty≤kdx,y for all then ∩i=1pAi≠∅ and T has a unique fixed point in ∩i=1pAi. In the case where ∩i=1pAi=∅, Eldred and Veeramani [2] introduced the existence of the best proximity point for the map T in setting of uniformly convex Banach spaces.Theorem 1.
(see [2]). Let A and B be nonempty closed and convex subsets of a uniformly convex Banach space. Suppose T:A∪B⟶A∪B is a cyclic contraction map, then there exists a unique best proximity point x in A (that is, with x−Tx=distA,B). Further, if x0∈A and xn+1=Txn, then x2n converges to the best proximity point.
This result received considerable interest by many authors recently and more results have been obtained, see for example [3, 4].
In 2009, Al-Thagafi and Shahzad [5] studied convergence and existence results of best proximity points for φ−contraction mappings, and in 2011, Sadiq Basha [6] stated some best proximity point results. We can find other results on best proximity points in [7] by Felhi and Aydi. In [8], results of the best proximity point for cyclic Meir–Keeler contraction mappings were found.
In 2017, T. Sabar et al. [9] studied convergence and existence results of best proximity points for tricyclic contraction.Theorem 2.
(see [10]). Let A, B, and C be nonempty closed, bounded, and convex subsets of a S convex metric space X,d,W which has the C property; suppose A, B, and C are disjoint subsets of a,b where a,b∈X, let T:A∪B∪C⟶A∪B∪C be a tricyclic contraction map. Then, T has a best proximity point.
In this work, we introduce new results of the best proximity points for a self-mapping defined on the union ofp nonempty subsets of a S convex metric space X,d,W.
LetAii=1p be nonempty subsets of a metric space X,d, then we shall adopt the following notations throughout this paper:(1)Dp:Xp⟶ℝ+,x1,x2,…,xp⟼Dpx1,x2,…,xp=∑i<jdxi,xjfor1≤i,j≤p,δΠi=1pAi=infDpx1,x2,…,xp:xi∈Ai,i=1,…,p,ΔΠi=1pAi=supDpx1,x2,…,xp:xi∈Ai,i=1,…,p,Δx1,x2,…,xp−1Ap=supDpx1,x2,…,xp:xp∈Ap,for all xi∈Ai,i=1,…,p−1.Definition 1.
LetAii=1p be nonempty subsets of a metric space X,d, and a mapping T:∪i=1PAi⟶∪i=1PAi is said to be p cyclic contraction if(1)
TAi⊆Ai+1 for 1≤i≤p, with Ap+1=A1(2)
DpTx1,Tx2,…,Txp≤kDpx1,x2,…,xp+1−kδΠi=1pAi for some k∈0,1 and all xi∈Ai,i=1,…,p


## 2. Preliminaries

Definition 2.
LetAii=1p be nonempty subsets of a metric space X,d.
T
:
∪
i
=
1
P
A
i
⟶
∪
i
=
1
P
A
i is a p cyclic contraction; let x∈∪i=1PAi, where x is said to be a best proximity point for T if(2)Dpx,Tx,sT2x,…,Tp−1x=δΠi=1pAi.Definition 3.
(see [11]). Let X,d be metric space, then a mapping W:X×X×I⟶X is to be a convex structure on X provided that(3)du,Wx,y,λ≤λdu,x+1−λdu,y,for allu,x,y∈Xandλ∈I≔0,1.
A metric spaceX,d along with a convex structure W is called a convex metric space and is denoted by X,d,W, and we denote by x,y the set Wx,y,λ:λ∈I. A subset C of a convex metric space is said to be a convex if x,y⊆C for all x,y∈C. The closed and convex hull of set A will be denoted by
c
o
¯
A
≔
∩
C
:
C is a closed and convex subset of X that contains A.Definition 4.
(see [12]). A normed linear space X is said to have the property C if every bounded decreasing net of nonempty closed and convex subsets of X has a nonempty intersection.
For example, a reflexive Banach space has the propertyC, so does every weakly compact convex subset of Banach.Definition 5.
Letx1,x2,…,xp−1 be points of a convex metric space X,d,W and r>0 the closed ball (resp. opened) of focuses x1,x2,…,xp−1,then ray r is defined by(4)Bx1,x2,…,xp−1,r=x∈X:Dpx1,x2,…,xp−1,x0≤resp<r.Remark 1.
B
x
1
,
x
2
,
…
,
x
p
−
1
,
r is bounded, closed, and convex.Definition 6.
(see [4]). Let X,d,W be a convex metric space, in which W is said to be a strict convex structure if it has the property that whenever w∈X there is x,y,λ∈X×X×I for which(5)du,w≤λdu,x+1−λdu,y,for all u∈X then w=Wx,y,λ. If W is a strict convex structure on X,d, the X,d,W is called a strictly convex metric space.Lemma 1.
(see [4]). Let X,d,W be a strictly convex metric space; then, for every x,y,λ1,λ2∈X2×I2 we have WWx,y,λ1,y,λ2=Wx,y,λ1λ2.Definition 7.
(see [4]). Let X,d,W be a convex metric space. W is to be called a S convex structure on X provided that whenever x,y∈X such that x=Wa,b,α and y=Wa,b,β where α>β and a,b∈X, we have x=Wa,y,α−β/1−β and y=Wx,b,β/α.
IfW is a S convex structure on X, then X,d,W is called a S convex metric space.Proposition 1.
LetX,d,W be a S convex metric space and let x,y∈X such that x=Wa,b,α,y=Wa,b,β, and z=Wa,b,γ where a,b∈X and α>β>γ, then dx,y=α−βda,b and y=Wx,z,β−γ/α−γ.Proof.
By theS property, we have x=Wa,y,α−β/1−β,(6)dx,y=dWa,y,α−β1−β,y=α−β1−βda,y=α−β1−βda,Wa,b,β=α−βda,b.
Sinceα>γ and β>γ, then x=Wa,z,α−γ/1−γ and y=Wa,z,β−γ/1−γ and put α′=α−γ/1−γ and β′=β−γ/1−γ, then we have y=Wx,z,β′/α′=Wx,z,β−γ/α−γ.
Now, letXi,dXi,WXi, i=1,…,p−1 be convex metric spaces, then the mapping(7)d1:Πi=1p−1Xi×Πi=1p−1Xi⟶ℝ+,defined by(8)d1x1,x2,….,xp−1,y1,y2,…,yp−1=∑i=1p−1dXixi,yi,is a distance on Πi=1p−1Xi.


## 3. Main Results

Before presenting our results, we give the following lemma.Lemma 2.
The mapping(9)WΠi=1p−1Xi:Πi=1p−1Xi×Πi=1p−1Xi×I⟶Πi=1p−1Xi,defined by(10)WΠi=1p−1Xix1,x2,…,xp−1,y1,y2,…,yp−1,λ=WX1x1,y1,λ,WX2x2,y2,λ,…,WXp−1xp−1,yp−1,λ,is a convex structure on the metric space Πi=1p−1Xi,d1.Proof.
Let(11)u1,u2,…,up−1∈Πi=1p−1Xi,x1,x2,…,xp−1,y1,y2,…,yp−1∈Πi=1p−1Xi,λ∈I,(12)d1u1,u2,…,up−1,Wπp−1Xii=1x1,x2,…,xp−1,y1,y2,….,yp−1,λ=d1u1,u2,…,up−1,WX1x1,y1,λ,WX2x2,y2,λ,…,WXp−1xp−1,yp−1,λ=∑i=1p−1dXiui,WXixi,yi,λ≤∑i=1p−1λdXiui,xi+∑i=1p−11−λdXiui,yi≤λ∑i=1p−1dXiui,xi+1−λ∑i=1p−1dXiui,yi=λd1u1,u2,…,up−1,x1,x2,…,xp−1+1−λd1u1,u2,…,up−1,y1,y2,…,yp−1.Definition 8.
A subsetE of the convex metric space Πi=1p−1Xi,d1,WΠi=1p−1Xi is a convex if(13)WΠi=1p−1Xix1,x2,….,xp−1,y1,y2,….,yp−1,λ∈E,for all x1,x2,….,xp−1,y1,y2,….,yp−1∈E and λ∈I.
We now state our first main result.Theorem 3.
LetAii=1p be nonempty, closed, bounded, and convex subsets of a S convex metric space X,d,W which has the C property; suppose Aii=1p are disjoint subsets of a,b where a,b∈X, let a map T:∪i=1pAi⟶∪i=1pAi be p cyclic contraction map. Then, T has a best proximity point.Proof.
We denote by∑ the set of all nonempty, bounded, closed, and convex subset B1,B2,…,Bp⊆A1,A2,…,Ap such that T is p cyclic on ∪i=1PBi. Then, ∑ is nonempty and partially ordered by reverse inclusion, that is,(14)B1i≤B2i⟺B2i⊆B1i.Fori=1,2,…,p.
LetBijj∈Ii=1,…,p be an increasing chain of ∑, and since X has the property C, ∩j∈IBij for i=1,…,p are bounded, closed, and convex. We have(15)Bij≤∩j∈IBijfori=1,…,pfor allj∈I.
By Zorn’s lemma, we have a maximal element sayC1,C2,……Cp∈∑. We have(16)co¯TCi⊆Ci+1,fori=1,2,…,pwithCp+1=C1,hence, Tco¯TCi⊆TCi+1⊆co¯TCi+1; we have T which is p cyclic on ∪i=1pco¯TCi, and by the maximality of C1,C2,…,Cp we have(17)Ci+1=co¯TCi,fori=1,2,…,p.
Now letxi∈Ci for i=1,2,…,p, k∈0,1:(18)DpTx1,….,Txp≤kDpx1,…,xp+1−kδΠi=1pAi≤kΔC1,…,Cp+1−kδΠi=1pAi=Λ.
Then,Txp∈BTx1,…,Txp−1,Λ ∀xp∈Cp⇒TCp⊂BTx1,…,Txp−1,Λ. C1=co¯TCp⊂BTx1,…,Txp−1,Λ so ΔTx1,…,Txp−1C1≤Λ.
Put(19)B1=x2,…,xp∈Πi=2pCi:Δx2,…,xpC1≤Λ,Bi=xi+1,…..xp+i−1∈Πj=i+1p+i−1Cj:Δxi+1,…..xp+i−1Ci≤Λ,for i=2,…,p with Cp+j=Cj for j=1,…,p. Bi is nonempty, bounded, and closed for i=1,2,…,p.
PutΨx1:Πi=2pCi⟶ℝ+ and let x1∈C1 such that Ψx1x2,….,xp=Dpx1,….,xp, then we have(20)B1=∩x1∈C1Ψx1−10,Λ.
Letx2,…,xp,y2,…,yp∈B1,x1∈ C1 and λ∈I. We have(21)DpWx2,y2,λ,Wx3,y3,λ,….,Wxp,yp,λ,x1=∑2≤i<j≤pdWxi,yi,λ,Wxj,yj,λ+∑i=2pdWxi,yi,λ,x1≤∑2≤i<j≤pλdxi,Wxj,yj,λ+1−λ∑2≤i<j≤pdyi,Wxj,yj,λ+λ∑i=2pdxi,x1+1−λ∑i=2pdyi,x1≤λ∑i=2pdxi,x1+1−λ∑i=2pdyi,x1++λ∑2≤i<j≤pλdxi,xj+1−λdxi,yj+1−λ∑2≤i<j≤pλdyi,xj+1−λdyi,yj≤λ∑i=2pdxi,x1+1−λ∑i=2pdyi,x1+λ2∑2≤i<j≤pdxi,xj+λ1−λ∑2≤i<j≤pdxi,yj+dyi,xj+1−λ2∑2≤i<j≤pdyi,yj.
Sincexi,yi∈a,b for i=1,…,p implies there exist αi,βi∈I, such that dxi,yj=αi−βjda,b, dxi,xj=αi−αjda,b, and dyi,yj=βi−βjda,b.
Ifαi>βi, αj∉]αi,βi[ and βi∉]αj,βj[ for i≠j.
Suppose for exampleα1<α2<β1, then we have x2=Wy1,x1,α2−α1/β1−α1⟹x2∈y1,x1 since C1 is convex then y1,x1⊆C1⇒x2∈C1, which is a contradiction, C1∩C2=∅.
Supposeαi>αj and βi>βj.
We havedxi,yj+dyi,xj=dxi,xj+dyi,yj.
So(22)DpWx2,y2,λ,Wx3,y3,λ,…,Wxp,yp,λ,x1≤λ∑i=2pdxi,x1+1−λ∑i=2pdyi,x1+λ2∑2≤i<j≤pdxi,xj+λ1−λ∑2≤i<j≤pdxi,yj+dyi,xj+1−λ2∑2≤i<j≤pdyi,yj≤λ∑1≤i,j≤pdxi,xj+1−λ∑i=2pdyi,x1+∑2≤i<j≤pdyi,yj=λDpx1,x2,…,xp+1−λDpx1,y2,…,yp≤Λ.
That means(23)Wx2,y2,λ,Wx3,y3,λ,…Wxp,yp,λ=WΠi=2pCix2,x3,…,xp,y2,y3,…yp,λ∈B1,for all x2,x3,…,xp,y2,y3,…,yp∈B1 and λ∈I.
Define(24)T˜:∪i=1pΠj=ip+i−2Aj⟶∪i=1pΠj=ip+i−2Aj,by(25)T˜x1,x2,…,xp−1=Tx1,Tx2,…,Txp−1.
We haveT˜ which is pcyclic:(26)T˜Πi=1p−1Ai⊆Πi=2pAi,T˜Πi=2pAi⊆Πi=3pAi×A1,…,T˜AP×Πi=1p−2Ai⊆Πi=1p−1Ai.
Let(27)x1,x2,…,xp−1∈Πi=1p−1Ci,T˜x1,x2,…,xp−1=Tx1,Tx2…,Txp−1∈Πi=2pCi.
We have(28)ΔTx1,Tx2,…,Txp−1C1≤Λ⇒T˜x1,x2,…,xp−1∈B1⇒T˜Πi=1p−1Ci⊆B1.
SoT˜ is p cyclic on ∪i=1PBi. Πi=1p−1Ci,Πi=2pCi,…,CP×Πi=1p−2Ci is maximal in ∑∼ where(29)∑∼=Πi=1p−1Mi,Πi=2pMi,….,MP×Πi=1p−2Mi⊆Πi=1p−1Ai,Πi=2pAi,….,AP×Πi=1p−2Ai:Πi=1p−1Mi,Πi=2pMi,….,MP×Πi=1p−2Miare nonempty,closed,bounded, and convexwithT˜ispcyclic onπp−1Mii=1∪πpMii=2∪…∪MP×πp−2Mii=1,∑∼ is partially ordered by M1i≤˜M2i⇔M2i⊆M1i. For i=1,…,p. Let Mijj∈I i=1,…..,p be an increasing chain of ∑∼, and since X has a the C property, ∩j∈IMij for i=1,…,p are bounded, closed, and convex subsets. We have(30)Mij≤˜∩j∈IMijfori=1,…,pforallj∈I.
So(31)B1=C2,…,Cp,Bi=Ci+1,…,Cp+i−1fori=2,…,p.
Inductively,(32)Δx1,x2,…xp−1Cp≤Λ,∀x1,x2,…,xp−1∈Πi=1p−1Ci.
So, we have(33)Δx1,x2,…,xp−1Cp−kΔΠi=1pCi≤1−kσΠi=1pAi,∀x1,x2,…,xp−1∈Πi=1p−1Ci⇒1−kΔΠi=1pCi≤1−kσΠi=1pAi.
So(34)ΔΠi=1pCi≤σΠi=1pAi.
Now letr1,r2,…,rp∈Πi=1pCi, then we have(35)σΠi=1pAi≤Dpr1,Tr1,T2r1,…,Tp−1r1,DpTp−1r2,r2,Tr2,T2r2,…,Tp−2r2,…,DpTrp,T2rp,…,Tp−1rp,rp≤ΔΠi=1pCi≤σΠi=1pAi,thenΔΠi=1pCi=σΠi=1pAi,which finishes the proof of the theorem.
Now, we give some examples forp=4.Example 1.
LetX be ℝ2 normed by the norm x,y=x+y, and let A=1,2×0, B=0×−2,−1,C=−2,−1×0, and D=0×1,2, then σA,B,C,D=D41,0,0,−1,−1,0,0,1=12. We define W:X×X×I⟶X by Wx,y,λ=λx+1−λy for all x,y∈X,λ∈I, then X,d,W is a complete convex metric space. W is a S convex structure on X. Let a,b∈X, x=Wa,b,α, and y=Wa,b,β with α>β. We have(36)Wa,y,α−β1−β=α−β1−βa+1−α−β1−βy=α−β1−βa+1−α1−βy=α−β1−βa+1−α1−βWa,b,β=α−β1−βa+1−α1−ββa+1−βb=11−βα1−βa+1−αb=αa+1−αb=x,Wx,b,βα=βαx+1−βαb=βαWa,b,α+1−βαb=βααa+1−αb+1−βαb=βa+βα1−αb+1−βαb=βa+1−βb=y.
Put(37)T:A∪B∪C∪D⟶A∪B∪C∪D,such that(38)Tx,0=0,−x+12ifx∈1,2,T0,y=y−12,0ify∈−2,−1,Tz,0=0,1−z2ifz∈−2,−1,T0,t=t+12,0ift∈1,2.
We have(39)TA⊆B,TB⊆C,TC⊆D,TD⊆A,(40)D4Tx,0,T0,y,Tz,0,T0,t=x+12+1−y2+x+12+1−z2+x+12+t+12+1−z2+1−y2+t+12+1−y2+t+12+1−z2=12x−y+x−z+x+t+−y−z+t−y+t−z+6=12D4x,0,0,y,z,0,0,t+6=12D4x,0,0,y,z,0,0,t+1−12σA,B,C,D.
SoT is a quadricyclic contraction. Then, T has 1,0 a best proximity point, since(41)D41,0,T1,0,T21,0,T31,0=D41,0,0,−1,−1,0,0,1=12=σA,B,C,D.Example 2.
Letk∈0,1, and let A,B,C, and D the four subsets of lp, 1≤p≤∞, defined by(42)A=1+k4ne4n,n∈ℕ,B=1+k4m−3e4m−3,m∈ℕ,C=1+k4s−2e4s−2,s∈ℕ,D=1+k4t−1e4t−1,t∈ℕ.
Put(43)T:A∪B∪C∪D⟶A∪B∪C∪D,such that T1+kpep=1+kp+1ep+1 for all p∈ℕ.
We haveTA⊂B,TB⊂C,TC⊂D, and TD⊂A.
Sincek∈0,1, limn⟶∞1+k4n−ip+1+k4n−jp1/p=21/p for i,j=1,2,3.
SoσA,B,C,D=6.21/p.(44)D4T1+k4n,T1+k4m−3,T1+k4s−2,T1+k4t−1=D41+k4n+1,1+k4m−2,1+k4s−1,1+k4t=1+k4n+1p+1+k4m−2p1/p+1+k4n+1p+1+k4s−1p1/p+1+k4n+1p+1+k4tp1/p+1+k4m−2p+1+k4s−1p1/p+1+k4m−2p+1+k4tp1/p+1+k4s−1p+1+k4tp1/p=k1+k4n+1−kp+k1+k4m−3+1−kp1/p+k1+k4n+1−kp+k1+k4s−2+1−kp1/p+k1+k4n+1−kp+k1+k4t−1+1−kp1/p+k1+k4m−3+1−kp+k1+k4s−2+1−kp1/p+k1+k4m−3+1−kp+k1+k4t−1+1−kp1/p+k1+k4s−2+1−kp+k1+k4t−1+1−kp1/p≤k1+k4np+k1+k4m−3p1/p+21−kp1/p+k1+k4np+k1+k4s−2p1/p+21−kp1/pk1+k4np+k1+k4t−1p1/p+21−kp1/p+k1+k4m−3p+k1+k4s−2p1/p+21−kp1/p+k1+k4m−3p+k1+k4t−1p1/p+21−kp1/p+k1+k4s−2p+k1+k4t−1p1/p+21−kp1/p≤k1+k4np+k1+k4m−3p1/p+k1+k4np+k1+k4s−2p1/pk1+k4np+k1+k4t−1p1/p+k1+k4m−3p+k1+k4ns−2p1/p+k1+k4m−3p+k1+k4t−1p1/p+k1+k4s−2p+k1+k4t−1p1/p+6.21/p1−k=kD41+k4n,1+k4m−3,1+k4s−2,1+k4t−1+1−kσA,B,C,D.
T is a quadricyclic contraction.
PutW: lp×lp×I⟶lp such that Wx,y,λ=λx+1−λy/λx+1−λy.
Let(45)u∈lp,x,y∈lp,du,Wx,y,λ=du,λx+1−λy/λx+1−λy=1λx+1−λydu,λx+1−λy=1λx+1−λy∑ui+λxi+1−λyip1/p=1λx+1−λy∑λui+1−λui+λxi+1−λyip1/p=1λx+1−λy∑λui+xi+1−λui+yip1/p≤1λx+1−λyλp∑ui+xip+1−λp∑ui+yip1/p≤1λx+1−λyλ∑ui+xip1/p+1−λ∑ui+yip1/p≤λdu,x+1−λdu,y.
Then,lp,d,W is a complete convex metric space for all 1≤p≤∞.
Forp=2, let a=1,0,0,0,…∈lp, b=0,1,0,0,…∈lp,(46)x=Wa,b,12=12,12,0,0,0,…,y=Wa,b,14=110,310,0,0,0,….
We have(47)Wa,y,1/3=1/3+2/310,2/10,0,0,…≠x,Wx,b,1/2=1/22,1/22+1/2,0,0,0,…≠y.
W is not a (S) convex structure on X. Then, T has not a best proximity point.



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*Source: 4367482-2020-03-16.xml*