# Some Results on the Best Proximity Pair

Authors: Mohammad Reza Haddadi; S. Mohammad Moshtaghioun
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We give some new conditions for existence and uniqueness of best proximity point. We also introduce the concept of strongly proximity pair and give some interesting results.


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## Body

## 1. Introduction

LetX be a metric space and A and B nonempty subsets of X. If there is a pair (x0,y0)∈A×B for which d(x0,y0)=d(A,B), that d(A,B) is distance of A and B, then the pair (x0,y0) is called a best proximity pair for A and B. Best proximity pair evolves as a generalization of the concept of best approximation, and reader can find some important results of it in [1–4].Now, as in [5] (see also [6–14]), we can find the best proximity points of the sets A and B by considering a map T:A∪B→A∪B such that T(A)⊂B and T(B)⊂A. We say that the point x∈A∪B is a best proximity point of the pair (A,B), if d(x,Tx)=d(A,B), and we denote the set of all best proximity points of (A,B) by PT(A,B), that is, (1.1)PT(A,B)∶={x∈A∪B:d(x,Tx)=d(A,B)}.
Best proximity pair also evolves as a generalization of the concept of fixed point of mappings, because if A∩B≠∅, every best proximity point is a fixed point of T.The concept ofapproximate best proximity pair on metric space was introduced in [10, Definition 1.1], but it is clear that PTa(A,B)=A×B. Now, in section two of this paper, we give some conditions that guarantee the existence, uniqueness, or compactness of the set PT(A,B). Then, in section three, by introducing the concepts of T-approximatively compact pair and T-strongly compact pair, we give some characterizations of a subclass of the best proximity points, namely, the strongly proximity pairs of sets.


## 2. Some Existence Theorems

In this section, we will consider the existence of the best proximity points, by considering some sequences which converge to that best proximity point. At first, we generalize some result of Eldred and Veeramani [6].Theorem 2.1.
LetA and B be nonempty closed subsets of a complete metric space X. Suppose that the mapping T:A∪B→A∪B satisfying T(A)⊂B, T(B)⊂A, and
(2.1)d(Tx,Ty)≤αd(x,y)+β[d(x,Tx)+d(y,Ty)]+γd(A,B),
for all x,y∈A∪B, where α,β,γ≥0 and α+2β+γ<1. If A (or B) is boundedly compact, then there exists x∈A∪B with d(x,Tx)=d(A,B).Proof.
Supposex0 is an arbitrary point of A∪B and define xn+1=Txn. Now,
(2.2)d(xn+1,xn+2)=d(Txn,Txn+1)≤αd(xn,xn+1)+β[d(xn,Txn)+d(xn+1,Txn+1)]+γd(A,B).
So
(2.3)d(xn+1,xn+2)≤α+β1-βd(xn,xn+1)+γ1-βd(A,B),
which implies that
(2.4)d(xn+1,xn+2)≤kd(xn,xn+1)+(1-k)d(A,B),
where k=((α+β)/(1-β))<1. Hence, inductively, we have
(2.5)d(xn+1,xn)≤knd(x1,x0)+(1-kn)d(A,B),
and so
(2.6)d(xn+1,xn)⟶d(A,B).
Therefore, by Proposition 3.3 of [6], both sequences {x2n} and {x2n+1} are bounded. Now, since A (or B) is boundedly compact then {x2n} has a convergent subsequence, and so, by Proposition 3.2 of [6], there exists x∈A such that d(x,Tx)=d(A,B).Now, we show that the mappingT, which satisfies (2.1), has a unique best proximity point in the uniformly convex Banach space X.Theorem 2.2.
LetA and B be two nonempty closed and convex subsets of a uniformly convex Banach space X. Suppose that the mapping T:A∪B→A∪B satisfyingT(A)⊂B, T(B)⊂A and the condition (2.1). Then, there exists a unique element x∈A such that ∥x-Tx∥=d(A,B). Further, if x0∈A and xn+1=Txn, then {x2n} converges to the above unique element.Proof.
One can prove this theorem by the method of the Proposition 3.10 of [6].Note that the uniform convexity of the Banach spaceX is necessary for uniqueness of PT(A,B); for instance, let X=ℝ2 with ∥·∥∞, A=x0+BX, and B=-x0+BX where x0=(2,0). If T:A∪B→A∪B by T(a,b)=(-a,b), then one can easily see that PT(A,B) is an infinite set.Corollary 2.3.
LetA and B be two nonempty closed and convex subsets of a uniformly convex Banach space X. Suppose that the mapping T:A∪B→A∪B satisfying T(A)⊂B, T(B)⊂A, and
(2.7)‖Tx-Ty‖≤a1‖x-y‖+a2‖x-Tx‖+a3‖y-Ty‖+a4d(A,B)
for all x,y∈A∪B, where ai≥0,i=1,2,3,4 and ∑i=14ai<1. Then, there exists a unique element x∈A∪B with ∥x-Tx∥=d(A,B).Proof.
Interchange the roles ofx and y in (2.7); then add the new inequality with (2.7).In the following, we present some new conditions on the mappingT, such as weak closedness, such that it has a best proximity point in the uniformly convex Banach space X. We remember that the mapping T:A∪B→A∪B is said to be weakly closed if xn⇀x weakly in A∪B and Txn⇀y weakly, then Tx=y.Theorem 2.4.
LetA and B be two nonempty closed and convex subsets of a uniformly convex Banach space X such that A is bounded. Suppose that the mapping T:A∪B→A∪B satisfying T(A)⊂B, T(B)⊂A, and
(2.8)‖Tx-Ty‖≤α‖x-y‖+β[‖x-Tx‖+‖y-Ty‖],
for all x,y∈A∪B, where α,β≥0 and α+2β≤1. If one of the following conditions:(i)
T is weakly closed and T(A) is bounded,(ii)
T is weakly sequentially continuous,satisfies, then there exists x∈A with ∥x-Tx∥=d(A,B).Proof.
Let(2.9)A0:={x∈A:‖x-y‖=d(A,B),forsomey∈B}.
By Lemma 3.2 of [5], A0 is nonempty; hence, there are x0∈A and y0∈B such that ∥x0-y0∥=d(A,B). For every positive integer n∈ℕ, define
(2.10)Tn(x)={1ny0+(1-1n)Tx,x∈A,1nx0+(1-1n)Tx,x∈B.
Then, for every x,y∈A∪B,
(2.11)‖Tnx-Tny‖≤(1-1n)‖Tx-Ty‖+1nd(A,B)≤(1-1n)α‖x-y‖+(1-1n)β(‖x-Tx‖+‖y-Ty‖)+1nd(A,B).
Therefore, by Theorem 2.2, for every n∈ℕ, there exists xn∈A such that
(2.12)‖xn-Tnxn‖=d(A,B).
Since A is bounded and closed, there exist x∈A such that xn⇀x (by passing to a subsequence, if necessary). If (i) holds, then the sequence {Txn} has a weakly convergent subsequence Txnk⇀Tx, thanks to the weak closedness of T. So xnk-Txnk⇀x-Tx. On the other hands, since ∥Tnxn-Txn∥=(1/n)∥y0-Txn∥→0, we have
(2.13)‖xnk-Txnk‖≤‖xnk-Tnkxnk‖+‖Tnkxnk-Txnk‖⟶d(A,B).
Therefore,
(2.14)‖x-Tx‖≤liminf⁡k→∞‖xnk-Txnk‖≤d(A,B).
The proof of the statement in the case (ii) is even simpler and is a part of the above proof.Theorem 2.5.
LetA and B be nonempty subsets of a metric space X. Suppose that the mapping T:A∪B→A∪B satisfying T(A)⊂B, T(B)⊂A, and
(2.15)d(Tx,T2x)≤kd(x,Tx)+(1-k)d(A,B),
for all x∈A∪B, where 0≤k<1. If there are u∈A∪B and n∈ℕ such that Tnu=u, then d(u,Tu)=d(A,B).Proof.
Suppose there areu∈A∪B and n∈ℕ such that Tnu=u. If d(A,B)<d(u,Tu), since T satisfies (2.15), we have
(2.16)d(u,Tu)=d(T(Tn-1u),T2(Tn-1u))≤kd(Tn-1u,T(Tn-1u))+(1-k)d(A,B)≤k2d(Tn-2u,Tn-1u)+(1-k2)d(A,B)⋮≤knd(u,Tu)+(1-kn)d(A,B)<d(u,Tu),
which is a contradiction, so d(u,Tu)=d(A,B).Corollary 2.6.
LetA and B be nonempty subsets of a metric space X. Suppose that the mapping T:A∪B→A∪B satisfying T(A)⊂B, T(B)⊂A, and (2.1). If there are u∈A∪B and n∈N such that Tnu=u, then d(u,Tu)=d(A,B).Proof.
Ify=Tx in (2.1), then
(2.17)d(Tx,T2x)≤αd(x,Tx)+β[d(x,Tx)+d(Tx,T2x)]+γd(A,B).
So
(2.18)d(Tx,T2x)≤α+β1-βd(x,Tx)+γ1-βd(A,B).
Hence, by Theorem 2.5, we have d(u,Tu)=d(A,B).


## 3. Strongly Proximity Pairs

LetA and B be nonempty subsets of a metric space X, δ>0, and T:A∪B→A∪B such that T(A)⊂B and T(B)⊂A. Put (3.1)PTδ(A,B)∶={x∈A∪B:d(x,Tx)<d(A,B)+δ}.
We say that the pair (A,B) is a strongly proximity pair, if it is proximity pair, and, for any neighborhood V of 0 in X there exists δ>0 such that PTδ(A,B)⊆PT(A,B)+V.For example, if(3.2)A={(x,y):(x-2)2+y2≤1},B={(x,y):(x+2)2+y2≤1},
and T(x,y)=(-x,y), then for every ϵ>0, PTδ(A,B)⊆PT(A,B)+V, where δ=4+ϵ2-2 and V is the sphere with radius ϵ and center of zero. Hence the pair (A,B) is a strongly proximity pair.Also, if(3.3)A={(0,y):-1≤y≤1},B={(x,0):1≤x≤2or-2≤x<-1}
and T:A∪B→A∪B such that(3.4)T(x,y)={(y+1,x)y≥0,(x,y)∈A,(y-1,x)y<0,(x,y)∈A,(y,x-1)x≥1,(x,y)∈B,(y,x+1)x<-1,(x,y)∈B,
Therefore T(A)⊂B, and T(B)⊂A and PT(A,B)={(0,0),(1,0)}, but the pair (A,B) is not a strongly proximity pair, while it is a proximity pair.Here, by introducing the concepts ofT-approximatively compact pair and T-strongly compact pair, we give some characterizations of the strongly proximity pairs of sets.Definition 3.1.
LetA and B be nonempty subsets of a metric space X and T:A∪B→A∪B such that T(A)⊂B and T(B)⊂A. We say the following.(i)
The sequence{zn}⊆A∪B is T-minimizing if
(3.5)lim⁡n→∞d(zn,Tzn)=d(A,B).(ii)
The pair(A,B) is T-approximatively compact pair (T-a.c.p.) if every T-minimizing sequence {zn}⊆A∪B has convergent subsequence.(iii)
The pair(A,B) is T-strongly compact pair (T-s.c.p.) if every T-minimizing sequence {zn}⊆A∪B is convergent.In the last section, we find some conditions onT such that PT(A,B)≠∅, and so in this section, we can always suppose that PT(A,B)≠∅. At the first, we state an elementary lemma, which can be used in the proof of the main theorems that follow.Lemma 3.2.
LetA and B be nonempty subsets of a metric space X,T:A∪B→A∪B such that T(A)⊂B and T(B)⊂A, and the pair (A,B) is T-s.c.p. Then, PT(A,B) is singleton.Proof.
Letx,y∈PT(A,B), hence,
(3.6)d(y,Ty)=d(x,Tx)=d(A,B).
Now, define
(3.7)zn={x,nodd,y,neven.
Then, the sequence {zn} is T-minimizing but is not convergent provided that x≠y and so a contradiction.Now, we can prove the main theorems of this section.Theorem 3.3.
LetA and B be nonempty closed subsets of a normed space X, and T:A∪B→A∪B is a continuous function, such that T(A)⊂B and T(B)⊂A and PT(A,B)≠∅. Then, the pair (A,B) is T-a.c.p. if and only if the pair (A,B) is strongly proximity pair and PT(A,B) compact.Proof.
Let the pair(A,B) be T-a.c.p., and {xn}⊆PT(A,B) is an arbitrary sequence. Then for each n, d(xn,Txn)=d(A,B), and, by hypothesis, the sequence {xn} has a convergent subsequence to an element of PT(A,B). Thus, PT(A,B) is compact.
Also, if(A,B) is not strongly proximity pair, then there exist a neighborhood V of 0 and a T-minimizing sequence {zn}⊆A∪B with zn not belonging to PT(A,B)+V for all n≥1. Since (A,B) is T-a.c.p., there is a subsequence {znk} such that znk→z0. Then, z0∈PT(A,B), and so zn∈z0+V⊆PT(A,B)+V for sufficiently large n, that is a contradiction.
Conversely, suppose that(A,B) is a strongly proximity pair and PT(A,B) compact, but (A,B) is not T-a.c.p. Then, there is a T-minimizing sequence {zn}⊆A∪B without any convergent subsequence. It follows that, for any x∈PT(A,B), there is a neighborhood Ux of x such that, for sufficiently large n, zn does not belong to Ux. Since PT(A,B) is compact, one can cover PT(A,B) by finitely many Uxi,i=1,2,...,n. So there is a neighborhood V of 0 and n0∈N such that for all n≥n0, zn does not belong to PT(A,B)+V. Since PT(A,B) is strongly proximity pair, there exists δ>0 such that PTδ(A,B)⊆PT(A,B)+V. Since {zn} is a T-minimizing sequence, zn∈PTδ(A,B) for sufficiently large n∈N and this is a contradiction.Corollary 3.4.
LetA and B be nonempty subsets of normed space X such that A∪B is compact and T:A∪B→A∪B is continuous such that T(A)⊂B and T(B)⊂A. Then, the pair (A,B) is strongly proximity pair and PT(A,B) is compact.Proof.
SinceA∪B is compact, it is obvious that the pair (A,B) is T-a.c.p. Now, apply Theorem 3.3.Theorem 3.5.
LetA and B be nonempty closed subsets of a normed space X, and T:A∪B→A∪B is continuous such that T(A)⊂B and T(B)⊂A and PT(A,B)≠∅. Then, the pair (A,B) is a T-s.c.p. if and only if the pair (A,B) is strongly proximity pair and PT(A,B) singleton.Proof.
Suppose that(A,B) is T-s.c.p. By Theorem 3.3, (A,B) is strongly proximity pair, and, by Lemma 3.2, PT(A,B) is singleton.
Conversely, suppose(A,B) is strongly proximity pair and PT(A,B)={z0}. Let V be a neighborhood of 0. Since (A,B) is strongly proximity pair, there exists δ>0 such that PTδ(A,B)⊆z0+V. Thus, for any T-minimizing sequence {zn}⊆A∪B, zn∈PTδ(A,B)⊆z0+V for sufficiently large n. Hence, zn→z0.Theorem 3.6.
LetA and B be nonempty closed and convex subsets of a uniformly convex Banach space X. Suppose that the mapping T:A∪B→A∪B satisfying T(A)⊂B, T(B)⊂A, and, for every x,y∈A(3.8)‖Tx-Ty‖≤α‖x-y‖(0<α<1).
Then, the pair (A,B) is a T-s.c.p. if and only if PT(A,B) is singleton.Proof.
The necessary condition follows from Theorem3.5.
For the proof of sufficient condition, suppose thatPT(A,B)={z0} but (A,B) is not a T-s.c.p. Then, there is a T-minimizing sequence {zn}⊆A∪B that is not convergent. It follows that there exists a subsequence {znk} of {zn} and a scaler d>0 such that for all integer k,
(3.9)‖znk-z‖≥d.
By uniform convexity of X, there exists ϵ>0 such that
(3.10)(d(A,B)+ϵ)[1-δX(d(1-α)d(A,B)+ϵ)]<d(A,B).
Since limn→∞∥zn-Tzn∥=d(A,B), there exists k such that
(3.11)‖znk-Tznk‖<d(A,B)+ϵ.
Also,
(3.12)‖z0-Tz0‖=d(A,B)<d(A,B)+ϵ.
But
(3.13)‖z0-Tz0-(znk-Tznk)‖≥‖z0-znk‖-‖Tz0-Tznk‖≥‖z0-znk‖-‖z0-ynk‖α=‖z0-znk‖(1-α)>d(1-α),
Because A and B are convex, ((z+znk)/2)∈A, and Tz+Tznk/2∈B, the following inequality leads to a contradiction:
(3.14)‖z+znk2-Tz+Tznk2‖≤(d(A,B)+ϵ)[1-δX(d(1-α)d(A,B)+ϵ)]<d(A,B).



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