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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# Cyclic pairs and common best proximity points in uniformly convex Banach spaces

Moosa Gabeleh
/ P. Julia Mary
/ A.Anthony Eldred Eldred
/ Olivier Olela Otafudu
Published Online: 2017-06-09 | DOI: https://doi.org/10.1515/math-2017-0059

## Abstract

In this article, we survey the existence, uniqueness and convergence of a common best proximity point for a cyclic pair of mappings, which is equivalent to study of a solution for a nonlinear programming problem in the setting of uniformly convex Banach spaces. Finally, we provide an extension of Edelstein’s fixed point theorem in strictly convex Banach spaces. Examples are given to illustrate our main conclusions.

MSC 2010: 90C48; 47H09; 46B20

In 2003, an interesting extension of Banach contraction principle was given as below.

#### Theorem 0.1

([1]). Let A and B be nonempty and closed subsets of a complete metric space (X, d) . Suppose that T : ABAB is a cyclic mapping, that is, T(A) ⊆ B and T(B) ⊆ A, such that d(Tx, Ty) ≤ α d(x, y) for some α ∈ [0, 1 [and for all xA, yB. Then AB is nonempty and T has a unique fixed point in AB.

If in Theorem 0.1 AB = ∅, then the fixed point equation Tx = x does not have any solution. In this situation we have the following notion. We should mention that the existence of best approximant for non-self mappings was first studied by Ky Fan as below.

#### Theorem 0.2

([2]). Let A be a nonempty, compact and convex subset of a normed linear space X and T : AX be a continuous mapping. Then there exists a point x*A such that $∥x∗−Tx∗∥=dist({Tx∗},A).$

We now state the notion of best proximity points for cyclic mappings.

#### Definition 0.3

Let A, B be nonempty subsets of a metric space (X, d) and T : ABAB be a cyclic mapping. A point pAB is called a best proximity point of T if $d(p,Tp)=dist(A,B),$ where dist (A, B) : = inf{d(x, y) : xA, yB}.

Notice that best proximity point results have been studied to find necessary conditions such that the minimization problem $minx∈A∪Bd(x,Tx),$(1) has at least one solution, where T is a cyclic mapping defined on AB.

In 2006, a class of cyclic mappings was introduced in [3] as follows.

#### Definition 0.4

([3]). Let A and B be nonempty subsets of a metric space (X, d). A mapping T : ABAB is said to be a cyclic contraction provided that T is cyclic on AB and $d(Tx,Ty)≤αd(x,y)+(1−α)dist(A,B)$ for some α ∈ [0, 1 [and for all (x, y) ∈ A × B.

After that in 2009, a generalized class of cyclic contractions was introduced as below.

#### Definition 0.5

([4]). Let A and B be nonempty subsets of a metric space (X, d). A mapping T : ABAB is said to be a cyclic φ-contraction if T is cyclic on AB and φ : [0, ∞) → [0, ∞) is a strictly increasing function and $d(Tx,Ty)≤d(x,y)−φ(d(x,y))+φ(dist(A,B)),$ for all (x, y) ∈ A × B.

It is remarkable to note that the class of cyclic φ-contraction mappings contains the class of cyclic contractions as a subclass by considering φ(t) = (1 − α)t for t ≥0 and for some α ∈ [0, 1 [.

Next theorem guarantees the existence, uniqueness and convergence of a best proximity point for cyclic φ-contractions in uniformly convex Banach spaces.

#### Theorem 0.6

(Theorem 8 of [4]). Let A and B be nonempty subsets of a uniformly convex Banach space X such that A is closed and convex, and let T : ABAB be a cyclic φ-contraction mapping. For x0A, define xn + 1 : = Txn for each n ≥ 0. Then there exists a unique pA such that x2np andpTp∥ = dist(A, B).

Recently, many authors have studied the existence of best proximity points for various classes of cyclic mappings which one can refer to [518] for more information.

In the current paper, we discuss sufficient conditions which ensure the existence and uniqueness of a solution for a nonlinear programming problem. Then we obtain a similar result of Theorem 0.6 for another class of cyclic mappings in uniformly convex Banach spaces. We also study the existence of best proximity pairs for noncyclic contractive mappings in strictly convex Banach spaces and so we present a generalization of Edelstein’s fixed point theorem.

## 1 Preliminaries

In this section, we recall some notions which will be used in our main discussions.

#### Definition 1.1

A Banach space X is said to be

1. uniformly convex if there exists a strictly increasing function δ : [0, 2] →[0, 1] such that for every x, y, pX, R > 0 and r ∈ [0, 2R], the following implication holds: $∥x−p∥≤R,∥y−p∥≤R,⇒∥x+y2−p∥≤(1−δ(rR))R;∥x−y∥≥r$

2. strictly convex if for every x, y, pX and R > 0, the following implication holds: $∥x−p∥≤R,∥y−p∥≤R,⇒∥x+y2−p∥

It is well known that Hilbert spaces and lp spaces (1 < p < ∞) are uniformly convex Banach spaces. Also, the Banach space l1 with the norm $|x|=∥x∥1+∥x∥2,∀x∈l1,$ where, ∥.∥1 and ∥.∥2 are the norms on l1 and l2, respectively, is strictly convex which is not uniformly convex (see [19] for more details).

Let A and B be nonempty subsets of a normed linear space X. We shall say that a pair (A, B) satisfies a property if both A and B satisfy that property. For example, (A, B) is convex if and only if both A and B are convex. We define $Ψ:={ψ:[0,∞[→[0,∞[;ψ is upper semi-continuous from the right and 0≤ψ(t)0},∥x−y∥∗:=∥x−y∥−dist(A,B),∀(x,y)∈A×B,A0:={x∈A:∥x−y∥=dist(A,B), for some y∈B},B0:={y∈B:∥x−y∥=dist(A,B), for some x∈A}.$

Notice that if (A, B) is a nonempty, bounded, closed and convex pair in a reflexive Banach space X, then (A0, B0) is also nonempty, closed and convex pair in X. We say that the pair (A, B) is proximinal if A = A0 and B = B0. Also, the metric projection operator 𝓟A : X → 2A is defined as 𝓟A(x) : = {yA : ∥xy∥= dist({x}, A)} where 2A denotes the set of all subsets of A. It is well known that if A is a nonempty, bounded, closed and convex subset of a uniformly convex Banach space X, then the metric projection 𝓟A is single valued from X to A.

#### Definition 1.2

([20]). Let (A, B) be a pair of nonempty subsets of a metric space (X, d) with A0 ≠ ∅. The pair (A, B) is said to have P-property if and only if $d(x1,y1)=dist(A,B)d(x2,y2)=dist(A,B)⇒d(x1,x2)=d(y1,y2),$ where x1, x2A0 and y1, y2B0.

It was announced in [21] that every nonempty, bounded, closed and convex pair in a uniformly convex Banach space X has the P-property.

Next two lemmas will be used in the sequel.

#### Lemma 1.3

([3]). Let A be a nonempty, closed and convex subset and B be a nonempty and closed subset of a uniformly convex Banach space X. Let {xn} and {zn} be sequences in A and let {yn} be a sequence in B such that

1. znyn∥ → dist(A, B),

2. for every ε > 0, there exists N0 ∈ ℕ so that for all m > n > N0, ∥xmyn∥ ≤ dist(A, B) + ε.

Then for every ε > 0, there exists N1 ∈ ℕ such thatxmzn∥ ≤ ε for any m > n > N1.

#### Lemma 1.4

([3]). Let A be a nonempty, closed and convex subset and B be a nonempty and closed subset of a uniformly convex Banach space X. Let {xn} and {zn} be sequences in A and let {yn} be a sequence in B satisfying

1. xnyn∥ → dist(A, B),

2. znyn∥ → dist(A, B).

Thenxnzn∥ → 0.

## 2 A nonlinear programming problem: common best proximity point

Let (A, B) be a nonempty pair in a normed linear space X and T, S : ABAB be two cyclic mappings. A point pAB is called a common best proximity point for the cyclic pair (T;S) provided that $∥p−Tp∥=dist(A,B)=∥p−Sp∥.$

In view of the fact that $min{∥x−Tx∥,∥x−Sx∥}≥dist(A,B),∀x∈A∪B,$ the optimal solution to the problem of $minx∈A∪B{∥x−Tx∥,∥x−Sx∥}$(2) will be the one for which the value dist(A, B) is attained. Thereby, a point pAB is a common best proximity point for the cyclic pair (T;S) if and only if that is a solution of the minimization problem (2).

In this section, we provide some sufficient conditions in order to study the existence of a solution for (2). We begin with the following result.

#### Theorem 2.1

Let (A, B) be a nonempty, closed, and convex pair in a uniformly convex Banach space X and (T;S) be a cyclic pair defined on AB such that

1. S(A) ⊆ T(A) ⊆ B and S(B) ⊆ T(B) ⊆ A,

2. SxSy*ψ(∥TxTy*), for all (x, y) ∈ A × B where ψ ∈ Ψ,

3. S and T commute,

4. T|A is continuous.

Then (T;S) has a unique common best proximity point in B.

#### Proof

Choose x0A. Since S(A) ⊆ T(A), there exists x1A such that Sx0 = Tx1. Again, by the fact that S(A) ⊆ T(A), there exists x2A such that Sx1 = Tx2. Continuing this process, we can find a sequence {xn} in A such that Sxn = Txn+1. It follows from the conditions (ii) and (iii) that $∥Sxn−SSxn∥∗≤ψ(∥Txn−TSxn∥∗)=ψ(∥Txn−STxn∥∗)=ψ(∥Sxn−1−SSxn−1∥∗)≤∥Sxn−1−SSxn−1∥∗,$ that is, {∥ SxnSSxn*} is a decreasing sequence of nonnegative real numbers and hence it converges. Let r be the limit of ∥ SxnSSxn*. We claim that r = 0. Suppose that r > 0. Then $lim supn→∞∥Sxn−SSxn∥∗≤lim supn→∞ψ(∥Sxn−1−SSxn−1∥∗),$ which implies that rψ(r) which is a contradiction. So, ∥ SxnSSxn∥→ dist(A, B). Moreover, $∥Sxn+1−SSxn∥∗≤ψ(∥Txn+1−TSxn∥∗)=ψ(∥Sxn−SSxn−1∥∗),∥Sxn−SSxn+1∥∗≤ψ(∥Txn−TSxn+1∥∗)=ψ(∥Sxn−1−SSxn∥∗).$ By a similar argument we conclude that $∥Sxn+1−SSxn∥→dist(A,B),∥Sxn−SSxn+1∥→dist(A,B).$ Let us prove that for any ε > 0 there exists N0 ∈ ℕ such that for all m > n > N0, $∥Sxm−SSxn∥∗<ε.$

Suppose the contrary. Then there exists ξ > 0 such that for all k ∈ ℕ there exist mk > nkk for which $∥Sxmk−SSxnk∥∗≥ε,∥Sxmk−1−SSxnk∥∗<ε.$ We now have $ε≤∥Sxmk−SSxnk∥∗≤∥Sxmk−SSxmk−1∥+∥Sxmk−1−SSxnk∥∗.$

Letting k → ∞ we obtain ∥ SxmkSSxnk*ε. Besides, $∥Sxmk−SSxnk∥∗≤∥Sxmk−Sxmk+1∥+∥Sxmk+1−SSxnk+1∥∗+∥SSxnk+1−SSxnk∥≤∥Sxmk−Sxmk+1∥+ψ(∥Txmk+1−TSxnk+1∥∗)+∥SSxnκ+1−SSxnk∥=∥Sxmk−Sxmk+1∥+ψ(∥Sxmk−SSxnk∥∗)+∥SSxnk+1−SSxnk∥.$ Therefore, $lim supk→∞∥Sxmk−SSxnk∥∗≤lim supk→∞ψ(∥Sxmk−SSxnk∥∗),$ and from the upper semi-continuity of ψ we have εψ(ε)which is a contradiction. Using Lemma 1.3, {Sxn} is a Cauchy sequence and converges to qB. So Txnq. By this reality that T|A is continuous, STxn = TSxnTq and so TTxnTq. We have $∥STxn−Sxn∥∗≤ψ(∥TTxn−Txn∥∗).$

Letting lim sup in above relation when n → ∞, then by the fact that ψ is upper semi-continuous from the right, we obtain ∥ Tqq*ψ(∥ Tqq*), which implies that ∥ qTq* = 0. Also, $∥Sxn−Sq∥∗≤ψ(∥Txn−Tq∥∗),$ which concludes that ∥ qSq*ψ(∥ qTq*) = ψ(0) = 0. Thus $∥q−Tq∥=dist(A,B)=∥q−Sq∥,$ and so qB is a common best proximity point for the cyclic pair (T; S). Now assume that q′ ∈ B is another common best proximity point for the cyclic pair (T; S). Then ∥ q′−Tq′∥ = dist(A, B) = ∥ q′−Sq′∥ By the fact that (A, B) has the P-property, Tq = Sq and Tq′ = Sq′. We have $∥STq−Sq′∥∗≤ψ(∥TTq−Tq′∥∗)=ψ(∥TSq−Tq′∥∗)=ψ(∥STq−Sq′∥∗),$ which implies that ∥ STqSq′∥* = 0. Equivalently, ∥ STqSq* = 0. Therefore, $∥q−Sq∥=∥STq−Sq∥=dist(A,B)=∥q′−Sq′∥=∥STq−Sq′∥.$

Again since (A, B) has the P-property, q = STq = q′ and the proof is complete. □

The following corollary is the main result of [22].

#### Corollary 2.2

Let (A, B)be a nonempty, closed, and convex pair in a uniformly convex Banach space X and (T; S)be a cyclic pair defined on AB such that

1. S(A) ⊆ T(A) ⊆ B and S(B) ⊆ T(B) ⊆ A,

2. SxSy*kTxTy*, for some k∈ [0, 1) and for all (x, y) ∈ A × B,

3. S and T commute,

4. T|A is continuous.

Then (T; S)has a unique common best proximity point in B.

#### Proof

It is sufficient to consider ψ(t) = kt in Theorem 2.1. □

#### Remark 2.3

We mention that Theorem 2.1 can be proved in complete metric spaces by using a geometric notion of property UC on closed pairs, which is a property for closed and convex pairs in uniformly convex Banach spaces (see Theorem 3.9 of [22]). Since we will use the other geometric notions of uniformly convex Banach spaces, we prefer to prove Theorem 2.1 in uniformly convex Banach spaces.

The following best proximity point theorem is a different version of Theorem 0.6.

#### Theorem 2.4

Let (A, B)be a nonempty, bounded, closed and convex pair in a uniformly convex Banach space X and S : ABAB be a cyclic mapping such that $∥Sx−Sy∥∗≤ψ(∥x−y∥∗),$ for all (x, y)∈ A × B where ψΨ. Then S has a unique best proximity point in B.

#### Proof

As we mentioned, (A0, B0)is nonempty, closed and convex. Note that the mapping S is cyclic on A0B0. Indeed, if xA0 then there exists a unique yB0 such that ∥xy∥ = dist(A, B)or ∥ xy* = 0. Thus ∥ SxSy*ψ(∥ xy*) = ψ(0) = 0 and so, ∥ SxSy∥ = dist(A, B)which implies that SxB0, that is, S(A0) ⊆ B0. Similarly, S(B0) ⊆ A0. Now consider the mapping 𝓟 : A0B0A0B0 defined with $Px=PB0(x)if x∈A0,PA0(x)if x∈B0.$ It is clear that 𝓟 is cyclic on A0B0. We have two following observations.

• 𝓟 is surjective:

Let yB0. Then there exists a unique element xA0 such that ∥ xy∥ = dist(A, B). Therefore, $∥x−y∥=dist(A,B)≤∥x−Px∥=∥x−PB0x∥,$ which implies that y = 𝓟x by the uniformly convexity of X. Thus 𝓟(A0) = B0. Similarly, we can see that 𝓟(B0) = A0.

• 𝓟 is an isometry:

Assume that (x, y)∈ A0× B0. Then we have ∥ x−𝓟x∥ = dist(A, B) = ∥ y−𝓟y In view of the fact that (A, B) has the P-property, ∥ xy∥ = ∥ 𝓟x−𝓟y∥ and the result follows.

• S and 𝓟 commute on A0B0:

Suppose xA0. Then there exists a unique yB0 such that ∥ xy∥ = dist(A, B). Thus x = 𝓟y and y = 𝓟x. Hence, ∥ SxSy∥ = dist(A, B) which implies that Sy = 𝓟Sx and so, S 𝓟x = 𝓟Sx. Similar argument holds when xB0, that is, S and 𝓟 are commuting.

• 𝓟|A0 is continuous:

Let {xn} be a sequence in A0 such that xnxA0. We have $∥xn−Px∥≤∥xn−x∥+∥x−Px∥→dist(A,B),∥xn−Pxn∥=dist(A,B),∀n∈N.$ Now using Lemma 1.4 we conclude that ∥ 𝓟xn−𝓟x∥→ 0, or 𝓟xn→ 𝓟x.

Finally, we note that $∥Sx−Sy∥∗≤ψ(∥x−y∥∗)=ψ(∥Px−Py∥∗),$ for any (x, y)∈ A0× B0. Thereby, all of the assumptions of Theorem 2.1 are satisfied and then the cyclic pair (S; 𝓟) has a unique common best proximity point such as qB0 and this completes the proof. □

Let us illustrate Theorem 2.1 with the following example.

#### Example 2.5

Suppose X = l2 and let A = {te1+e2 : 0 ≤ t$\begin{array}{}\frac{1}{4}\end{array}$ } and B = {e2 + se3 : 0 ≤ s$\begin{array}{}\frac{1}{4}\end{array}$ }. Define the cyclic pair (T; S)as below $S(te1+e2)=e2+t2e3andS(e2+se3)=s2e1+e2,t,s∈[0,14],T(te1+e2)=e2+te3andT(e2+se3)=se1+e2,t,s∈[0,14].$ It is clear that S(A) ⊆ T(A) = B and S(B) ⊆ T(B) = A. Also, $TS(te1+e2)=T(e2+t2e3)=t2e1+e2=S(e2+te3)=ST(te1+e2),TS(e2+se3)=T(s2e1+e2)=e2+s2e3=T(s2e1+e2)=TS(e2+se3),$ that is, T and S are commuting. Now define the function ψΨ with $ψ(r)=r20≤r<1,rr+11≤r.$

For x: = te1 + e2A and y: = e2+se3B we have $∥Sx−Sy∥∗=s4+t4≤s2+t2=ψ(s2+t2)=ψ(∥Tx−Ty∥∗).$

Therefore, all of the assumptions of Theorem 2.1 hold and so, the cyclic pair (T; S)has a unique common best proximity point in B and this point is p = e2 which is a common fixed point of the mappings T and S in this case.

The following example shows that the uniformly convexity condition of the Banach space X in Theorem 2.4 is sufficient but not necessary.

#### Example 2.6

Let X be the real Banach space l2 renormed according to $∥x∥=max{∥x∥2,2∥x∥∞},$ where, ∥ x denotes the lnorm andx2 the l2 norm. Assume {en} is a canonical basis of l2. Note that for any xX we havex2≤∥ x ∥≤ $\begin{array}{}\sqrt{2}\end{array}$x2 which implies ∥.∥ that is equivalent to ∥.∥2 and so, (X, ∥.∥) is a reflexive Banach space. Moreover, in view of the fact that l is not strictly convex, X is not uniformly convex. Put $A={x=(xn);x2=1,∥x∥≤2}andB={y:=2e2}.$ Then (A, B)is a bounded, closed and convex pair in X and dist (A, B) = $\begin{array}{}\sqrt{2}\end{array}$ . Moreover, A0 = A and B0 = B. Define the cyclic mapping S : ABAB with $Sx=y(∀x∈A)andSy=12e1+e2.$

For all xA and r ∈(0, 1)we have $∥Sx−Sy∥=∥e2−12e1∥=max{14+1,2}=2≤r∥x−y∥+(1−r)dist(A,B),$ that is, S is cyclic contraction. We note that y is a unique best proximity point of S in B.

## 3 A generalization of Edelstein fixed point theorem

We begin the main results of this section by stating the well known Edelstein’s fixed point theorem.

#### Theorem 3.1

([23]). Let (X, d) be a compact metric space and T be a mapping on X such that $d(Tx,Ty) Then T has a unique fixed point and for any x0X the iterate sequence {Tn x0} converges to the fixed point of T.

Let (A, B) be a nonempty pair in a normed linear space X. A mapping T : ABAB is said to be noncyclic provided that T(A) ⊆ A and T(B) ⊆ B. A point (p, q) ∈ A × B is said to be a bestproximity pair for the noncyclic mapping T if $p=Tp,q=Tqand∥p−q∥=dist(A,B).$

It is interesting to note that the existence of best proximity pairs for noncyclic mappings is equivalent to the existence of a solution of the following minimization nonlinear problem: $minx∈A∥x−Tx∥,miny∈B∥y−Ty∥,andmin(x,y)∈A×B∥x−y∥.$(3)

The existence of best proximity pairs was first studied by Eldred et al. in [24] using a geometric notion of proximal normal structure on nonempty, weakly compact and convex pairs in strictly convex Banach spaces for noncyclic relatively nonexpansive mappings.

#### Definition 3.2

([24]). A convex pair (A, B) in a Banach space X is said to have proximal normal structure if for any bounded, closed, convex and proximinal pair (K1, K2) ⊆ (A, B) for which δ(K1, K2) > dist (K1, K2) and dist (K1, K2) = dist (A, B), there exits (x1, x2) ∈ K1 × K2 such that $max{δx1(K2),δx2(K1)}<δ(K1,K2).$

Since every nonempty, compact and convex pair in a Banach space X has proximal normal structure (Proposition 2.2 of [24]), the following result concludes.

#### Theorem 3.3

(Theorem 2.2 of [24]). Let (A, B) be a nonempty compact and convex pair in a strictly convex Banach space X and T be a noncyclic relatively nonexpansive mapping, that is, T is noncyclic andTxTy∥ ≤ ∥xyfor all (x, y) ∈ A × B. Then T has a best proximity pair.

Motivated by Theorem 3.3, we study the convergence results of best proximity pairs for noncyclic contractive mappings in strictly convex Banach spaces.

#### Definition 3.4

Let (A, B) be a nonempty pair in a normed linear space X. A mapping T : ABAB is said to be a noncyclic contractive mapping if T is noncyclic on AB and $∥Tx−Ty∥<∥x−y∥,forall(x,y)∈A×B,with∥x−y∥>dist(A,B).$

Next lemma describes the relation between noncyclic relatively nonexpansive mappings and noncyclic contractive mappings in uniformly convex Banach spaces.

#### Lemma 3.5

Let (A, B) be a nonempty compact and convex pair in a strictly convex Banach space X and T : ABAB be a noncyclic contractive mapping. Then T is noncyclic relatively nonexpansive.

#### Proof

We only have to prove that ∥TxTy∥ = dist (A, B) whenever ∥xy∥ = dist (A, B). So let ∥xy∥ = dist (A, B). Choose a sequence ({xn}, {yn}) in A × B such that ∥xnyn∥ > dist (A, B) and xnx, yny for any nN. By the compactness condition of the pair (A, B), we may assume that limn→∞xn = xA and limn → ∞yn = yB. Then limn → ∞xnyn∥ = dist(A; B). Notice that if ∥xn0y∥ = dist (A, B) for some n0 ∈ ℕ, then by the strictly convexity of X we must have xn0 =x which is a contradiction. Thus $dist(A,B)≤∥PA(Ty)−Ty∥≤∥Txn−Ty∥<∥xn−y∥.$

Therefore, ∥TxnTy∥ → dist(A; B). Since ∥𝓟A(Ty)−Ty∥≤∥TxnTy∥ and TyB0, $Txn→PA(Ty).$

Similarly we can see that Tyn → 𝓟B(Tx). In view of the fact that ∥TxnTyn∥ → dist(A; B), we obtain ∥𝓟A(Ty) − 𝓟B(Tx)∥ = dist (A, B). Again, using the strict convexity of X, $Tx=PA(Ty),andTy=PB(Tx).$

Thereby, ∥TxTy∥ = dist (A, B) and the result follows.□

Next example shows that the strictly convexity of the Banach space X in Lemma 3.5 is a necessary condition.

#### Example 3.6

Let X = {ℝ2,∥.∥} and let A = {(0, s) : 0 ≤ s ≤ 2} and B = {(1, t) : 0 ≤ t ≤ 2}. It is clear that dist (A, B) = 1. Define the noncyclic mapping T : ABAB by $T(0,s)=(0,s1+s),T(1,t)=(1,t1+t).$

For (x, y) ∈ A × B ifxy > dist (A, B), then $∥Tx−Ty∥∞=∥(0,s1+s)−(1,t1+t)∥∞=max{1,|s−t|(1+s)(1+t)}<|s−t|=∥x−y∥∞,$

which implies that T is noncyclic contractive. Besides, ∥TxTy = dist (A, B), whenxy = dist (A, B). Hence, T is noncyclic relatively nonexpansive.

But if we modify T : AA as $T\left(0,\phantom{\rule{thinmathspace}{0ex}}s\right)=\left(0,\phantom{\rule{thinmathspace}{0ex}}\frac{s}{1+s}\right)\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}for\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0 and T(0, 0) = (0, 1.1), then T is still noncyclic contractive. To see T this, let x : = (0,0). Since ${y∈B:∥x−y∥∞>1}={(1,t):1

we have $∥Tx−Ty∥=max{1,1.1−t1+t}=1<∥x−y∥∞,$

whenever 1 < t ≤ 2. On the other hand, if v : = (1,0), then $∥Tx−Tv∥=max{1.1,1}>1=∥x−v∥∞,$

that is, T is not a noncyclic relatively nonexpansive mapping.

The following theorem is an extension of Edelstein’s fixed point theorem in strictly convex Banach spaces.

#### Theorem 3.7

Let (A, B) be a nonempty compact and convex pair in a strictly convex Banach space X and T : ABAB be a noncyclic contractive mapping. Then T has a unique best proximity pair Moreover for any (x0, y0) ∈ A0 × B0 if we define xn+1 := Txn and yn+1 := Tyn then the sequence {(xn, yn)} converges to the best proximity pair of T.

#### Proof

It follows from Lemma 3.5 that T is a noncyclic relatively nonexpansive mapping. Since the pair (A, B) is compact and convex, the existence of a best proximity pair for the mapping T is concluded from Theorem 3.3. Suppose (p, q) ∈ A × B is a best proximity pair of the mapping T. Then p = Tp, q = Tq and ∥pq∥ = dist (A, B). It is worth noticing that the fixed point sets of T in A0 and B0 are singleton. Indeed, if p′ ∈ A0 such that p′ = Tp′ and pp′ then from the strictly convexity of X we have ∥p′ − q∥ > dist (A, B). Therefore, $∥p′−q∥=∥Tp′−Tq∥<∥p′−q∥,$

which is impossible. Equivalently, we can see that the fixed point set in B0 is singleton. This implies that T has a unique best proximity pair in A × B. Let x0A0 and xn+1 = Txn. Assume that {xnk} is a subsequence of {xn} such that xnkzA0. Thus $d(xn,PBp)=d(Txn−1,PB(Tp))=d(Txn−1,T(PBp))

Hence, d(z,PBp) = limk → ∞d(xnk, 𝓟Bp). From Proposition 3.4 of [25] T is continuous on A0B0. Suppose d(z,𝓟Bp) > dist(A, B). We now have $d(z,PBp)=limk→∞d(xnk+1,PBp)=limk→∞d(Txnk,PBp)=d(Tz,PBp)=d(Tz,PB(Tp))=d(Tz,T(PBp)) which is a contradiction and so we must have d(z,𝓟Bp) = dist(A, B). Then z = p. Since any convergent subsequence of {xnk} converges to p, the sequence itself converges to p. Similarly we can prove the convergence of {yn} to the point q and this competes the proof.□

#### Remark 3.8

The notion of noncyclic contractive mappings was introduced in [25] as below (see Definition 3.2 and Theorem 4.6 of [25]): Let (A, B) be a nonempty pair in a metric space (X, d). A mapping T : ABAB is called noncyclic contractive if T is noncyclic on AB and

1. d(Tx,Ty) < d(x, y) whenever d(x, y) > dist(A, B) for xA and yB,

2. d(Tx,Ty) = d(x, y) whenever d(x, y) = dist(A, B) for xA and yB.

Then the existence result of a unique best proximity pair for such mappings was established using a notion of projectional property (Theorem 4.6 of [25]). It is remarkable to note that under the assumptions of Theorem 3.7 the condition (ii) on the noncyclic mapping T holds naturally.

At the end of this section, we study the existence of a unique common best proximity point for a cyclic pair of commuting mappings under a contractive condition.

We begin with the following lemma.

#### Lemma 3.9

Let (A, B) be a nonempty closed, and convex pair in a normed linear space X and (T;S) be a cyclic pair defined on AB such that

1. S(A) ⊆ T(A) ⊆ B and S(B) ⊆ T(B) ⊆ A,

2. T(A) and T(B) are compact subsets of B and A respectively.

3. SxSy∥ < ∥TxTy∥, for all (x, y)∈ A × B such thatSxSy∥ > dist(A, B),

Then $dist(S(A),S(B))=dist(T(A),T(B))=dist(A,B).$

#### Proof

Clearly $dist(S(A),S(B))≥dist(T(A),T(B))≥dist(A,B).$(4)

If dist(S(A), S(B)) = dist(A, B), then there is nothing to prove. Suppose dist(S(A), S(B)) > dist(A, B). By the assumption (iii), $dist(S(A),S(B))≤dist(T(A),T(B)).$

Therefore, dist(S(A), S(B)) = dist(T(A),T(B)) and so dist(T(A),T(B)) > dist(A, B). Let a′ ∈ T(B), b′ ∈ T(A) be such that dist(T(A),T(B)) = ∥a′ − b′∥ > dist(A, B). Assume a′ = T(b) and b′ = T(a) for some (a,b)∈ A × B. Since $∥S(a)−S(b)∥≥dist(S(A),S(B))>dist(A,B),$ we have $∥S(a)−S(b)∥<∥T(a)−T(b)∥=dist(T(A),T(B)),$ and this is a contradiction with (4) and the result follows.□

#### Theorem 3.10

Let (A, B) be a nonempty, closed and convex pair in a strictly convex Banach space X. Let (T;S) be a cyclic pair defined on AB such that

1. S(A) ⊆ T(A) ⊆ B and S(B) ⊆ T(B) ⊆ A

2. T(A) and T(B) are compact and convex subsets of B and A respectively.

3. SxSy∥ > ∥TxTyfor all (x, y) ∈ A × B such thatSxSy∥ > dist(A; B)

4. S and T commute.

Then (T;S) has a unique common best proximity point in B.

#### Proof

Let $[T(A)]0={y∈T(A):d(x,y)=dist(T(A),T(B)), for somex∈T(B)},$ and $[T(B)]0={x∈T(B):d(x,y)=dist(T(A),T(B)), for somey∈T(A)}.$

Notice that from Lemma 3.9, dist(T(A),T(B)) = dist(A, B). To show ST− 1 is singleton, let x ∈[T(B)]0. Then there exists y ∈[T(A)]0 such that ∥xy∥ = dist(A; B). For any zST− 1 xT(B) and wST− 1yT(A), $∥z−w∥=∥Sx′−Sy′∥$

where z = Sx′ and w = Sy′ for some x′ ∈ T−1x and y′ ∈ T−1y. If ∥Sx′ − Sy′ ∥> dist (A, B), then $∥Sx′−Sy′∥<∥Tx′−Ty′∥=∥TT−1x−TT−1y∥=∥x−y∥=dist(A,B),$ which is impossible. So $∥z−w∥=∥Sx′−Sy′∥=dist(A,B),$ for any zST−1x and wST−1y. It now follows from the strict convexity of X that ST−1x and ST−1y are singleton. Also, it is clear that ST−1([T(A)]0) ⊆[T(A)]0 and ST−1([T(B)]0) ⊆[T(B)]0, that is, ST−1 : [T(A)]0∪[T(B)]0→[T(A)]0∪[T(B)]0 is a noncyclic mapping. Let (x, y)∈[T(A)]0 ×[T(B)]0 be such that ∥xy∥ > dist (T(A),T(B))(= dist(A, B)). If ∥ ST−1xST−1y ∥ > dist (A, B), then $∥ST−1x−ST−1y∥<∥TT−1x−TT−1y∥=∥x−y∥,$ which implies that ST−1 is a noncyclic contractive mapping on a compact and convex pair ([T(A)]0, [T(B)]0). Now using Theorem 3.7 for any x ∈[T(A)]0∪[T(B)]0 the sequence {(ST−1)nx}n≥1 converges to the unique fixed point z of ST−1. Since S and T commutes, $Tz=T(ST−1z)=Sz=S(ST−1z)$ Suppose ∥SzT(Sz)∥ > dist (A, B). Thus $∥Sz−T(Sz)∥=∥Sz−S(Tz)∥<∥Tz−T(Tz)∥=∥Sz−T(Sz)∥,$ which is a contradiction. Hence ∥SzT(Sz)∥= dist (A, B). On the other hand, if ∥SzS(Sz)∥ > dist (A, B), then $∥Sz−S(Sz)∥<∥Tz−T(Sz)∥=∥Sz−T(Sz)∥=∥Sz−S(Tz)∥=∥Sz−S(Sz)∥,$ which is impossible. Thereby ∥SzS(Sz)∥= dist (A, B) and so the point Sz is a common best proximity point for the cyclic pair (T; S). Uniqueness of the common best proximity point follows as in the proof of Theorem 2.1. □

#### Example 3.11

Let X = {ℝ2, ∥.∥2} and let $\begin{array}{}A=\left\{\left(0,s\right):0\le s\le \frac{1}{2}\right\}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}B=\left\{\left(1,t\right):0\le t\le \frac{1}{2}\right\}.\end{array}$ Thus dist(A, B) = 1. Define the cyclic pair (T ; S) on AB as follows: $S(0,s)=(1,s3),S(1,t)=(0,t3)andT(0,s)=(1,s2),T(1,t)=(0,t2).$ Then $S(A)={(1,s):0≤s≤18},S(B)={(0,t):0≤t≤18},T(A)={(1,s):0≤s≤14},T(B)={(0,t):0≤t≤14}.$

Thereore, S(A) ⊆ T(A) ⊆ B and S(B) ⊆ T(B) ⊆ A. Also, T(A) and T(B) are compact and convex subsets of B and A, respectively. Moreover, $ST(0,s)=S(1,s2)=(0,s6)=T(1,s3)=TS(0,s),ST(1,t)=S(0,t2)=(1,t6)=T(0,t3)=TS(1,t),$ and so, S and T are commuting. Finally ifS(0, s)−S(1, t)∥ > dist(A, B), we conclude that st. Hence, $∥S(0,s)−S(1,t)∥=1+(|s3−t3|)2<1+(|s2−t2|)2=∥T(0,s)−T(1,t)∥,$ whenever s, t ≥0 and s + t ≤ 1. Thereby all of the assumptions of Theorem 3.10 hold and the cyclic pair (T; S) has a unique common best proximity point in B and this point is p = (0,0).

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Accepted: 2017-03-21

Published Online: 2017-06-09

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 711–723, ISSN (Online) 2391-5455,

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