Best proximity point results in geodesic metric spaces

Abstract

In this paper, the existence of a best proximity point for relatively u-continuous mappings is proved in geodesic metric spaces. As an application, we discuss the existence of common best proximity points for a family of not necessarily commuting relatively u-continuous mappings.

1 Introduction

Let A be a nonempty subset of a metric space $(X,d)$ and $T:A\to X$. A solution to the equation $Tx=x$ is called a fixed point of T. It is obvious that the condition $T(A)\cap A\ne \mathrm{\varnothing }$ is necessary for the existence of a fixed point for T. But there occur situations in which $d(x,Tx)>0$ for all $x\in A$. In such a situation, it is natural to find a point $x\in A$ such that x is closest to Tx in some sense. The following well-known best approximation theorem, due to Ky Fan [1], explores the existence of an approximate solution to the equation $Tx=x$.

Theorem 1 [1]

Let A be a nonempty compact convex subset of a normed linear space X and $T:A\to X$ be a continuous function. Then there exists $x\in A$ such that $\parallel x-Tx\parallel =dist(Tx,A)=inf\{\parallel Tx-a\parallel :a\in A\}$.

The point $x\in A$ in Theorem 1 is called a best approximant of T in A. Let A, B be nonempty subsets of a metric space X and $T:A\to B$. A point $x_0\in A$ is called a best proximity point of T if $d(x_0,T{x}_0)=dist(A,B)$. Some interesting results in approximation theory can be found in [2–8].

Eldred et al. [2] defined relatively nonexpansive mappings and used the proximal normal structure to prove the existence of best proximity points for such mappings.

Definition 2 [2]

Let A, B be nonempty subsets of a metric space $(X,d)$. A mapping $T:A\cup B\to A\cup B$ is said to be a relatively nonexpansive mapping if

• (i) $T(A)\subseteq B$, $T(B)\subseteq A$;

• (ii) $d(Tx,Ty)\le d(x,y)$, for all $x\in A$, $y\in B$.

Theorem 3 [2]

Let $(A,B)$ be a nonempty, weakly compact convex pair in a Banach space X. Let $T:A\cup B\to A\cup B$ be a relatively nonexpansive mapping and suppose $(A,B)$ has a proximal normal structure. Then there exists $(x,y)\in A\times B$ such that

$$\parallel x-Tx\parallel =\parallel Ty-y\parallel =dist(A,B).$$

Remark 4 [2]

Note that every nonexpansive self-map is a relatively nonexpansive map. Also, a relatively nonexpansive mapping need not be continuous.

In [8], Sankar Raj and Veeramani used a convergence theorem to prove the existence of best proximity points for relatively nonexpansive mappings in strictly convex Banach spaces.

Recently, Elderd, Sankar Raj and Veeramani [9] introduced a class of relatively u-continuous mappings and investigated the existence of best proximity points for such mappings in strictly convex Banach spaces.

Definition 5 [9]

Let A, B be nonempty subsets of a metric space X. A mapping $T:A\cup B\to A\cup B$ is said to be a relatively u-continuous mapping if it satisfies:

• (i) $T(A)\subseteq B$, $T(B)\subseteq A$;

• (ii) for each $\epsilon >0$, there exists a $\delta >0$ such that $d(Tx,Ty)<\epsilon +dist(A,B)$, whenever $d(x,y)<\delta +dist(A,B)$, for all $x\in A$, $y\in B$.

Theorem 6 [9]

Let A, B be nonempty compact convex subsets of a strictly convex Banach space X and $T:A\cup B\to A\cup B$ be a relatively u-continuous mapping. Then there exists $(x,y)\in A\times B$ such that

$$\parallel x-Tx\parallel =\parallel y-Ty\parallel =dist(A,B).$$

Remark 7 [9]

Every relatively nonexpansive mapping is a relatively u-continuous mapping, but the converse is not true.

Example 8 [9]

Let $(X={\mathbb{R}}^2,{\parallel \cdot \parallel }_2)$ and consider $A=\{(0,t):0\le t\le 1\}$ and $B=\{(1,s):0\le s\le 1\}$. Define $T:A\cup B\to A\cup B$ by

$$T(x,y)=\{\begin{array}{cc}(1,\sqrt{y})\hfill & \text{if }x=0,\hfill \\ (0,\sqrt{y})\hfill & \text{if }x=1.\hfill \end{array}$$

Then T is relatively u-continuous, but not relatively nonexpansive.

Also, in [9], the authors proved the existence of common best proximity points for a family of commuting relatively u-continuous mappings.

The aim of this paper is to discuss the existence of a best proximity point for relatively u-continuous mappings in the frameworks of geodesic metric spaces. As an application, we investigate the existence of common best proximity points for a family of not necessarily commuting relatively u-continuous mappings.

2 Preliminaries

In this section, we give some preliminaries.

Definition 9 [10]

A metric space $(X,d)$ is said to be a geodesic space if every two points x and y of X are joined by a geodesic, i.e., a map $c:[0,l]\subseteq \mathbb{R}\to X$ such that $c(0)=x$, $c(l)=y$, and $d(c(t),c(t^{\mathrm{\prime }}))=|t-t^{\mathrm{\prime }}|$ for all t, $t^{\mathrm{\prime }}\in [0,l]$. Moreover, X is called uniquely geodesic if there is exactly one geodesic joining x and y for each $x,y\in X$.

The midpoint m between two points x and y in a uniquely geodesic metric space has the property $d(x,m)=d(y,m)=\frac{1}{2}d(x,y)$. A trivial example of a geodesic space is a Banach space with usual segments as geodesic segments.

A point $z\in X$ belongs to the geodesic segment $[x,y]$ if and only if there exists $t\in [0,1]$ such that $d(z,x)=td(x,y)$ and $d(z,y)=(1-t)d(x,y)$. Hence, we write $z=(1-t)x+ty$.

A subset A of a geodesic metric space X is said to be convex if it contains any geodesic segment that joins each pair of points of A.

The metric $d:X\times X\to \mathbb{R}$ in a geodesic space $(X,d)$ is convex if

$$d(z,(1-t)x+ty)\le (1-t)d(z,x)+td(z,y)$$

for any $x,y,z\in X$ and $t\in [0,1]$.

Definition 10 [10]

A geodesic metric space X is said to be strictly convex if for every $r>0$, a, x and $y\in X$ with $d(x,a)\le r$, $d(y,a)\le r$ and $x\ne y$, it is the case that $d(a,p)<r$, where p is any point between x and y such that $p\ne x$ and $p\ne y$, i.e., p is any point in the interior of a geodesic segment that joins x and y.

Remark 11 [10]

Every strictly convex metric space is uniquely geodesic.

In [10], Fernández-León proved the existence and uniqueness of best proximity points in strictly convex metric spaces. For more details about geodesic spaces, one may check [11–13].

In the particular framework of geodesic metric spaces, the concept of global nonpositive curvature (global NPC spaces), also known as the $CAT(0)$ spaces, is defined in [13] as follows.

Definition 12 A global NPC space is a complete metric space $(X,d)$ for which the following inequality holds true: for each pair of points $x_0$, $x_1\in X$ there exists a point $y\in X$ such that for all points $z\in X$,

$$d^2(z,y)\le \frac{1}{2}{d}^2(z,x_0)+\frac{1}{2}{d}^2(z,x_1)-\frac{1}{4}{d}^2(x_0,x_1).$$

Proposition 13 [13]

If $(X,d)$ is a global NPC space, then it is a geodesic space. Moreover, for any pair of points $x_0,x_1\in X$ there exists a unique geodesic $\gamma :[0,1]\to X$ connecting them. For $t\in [0,1]$ the intermediate points ${\gamma }_t$ depend continuously on the endpoints $x_0$, $x_1$. Finally, for any $z\in X$,

$$d^2(z,{\gamma }_t)\le (1-t)d^2(z,x_0)+t{d}^2(z,x_1)-t(1-t)d^2(x_0,x_1).$$

Corollary 14 [13]

Let $(X,d)$ be a global NPC space, $\gamma ,\eta :[0,1]\to X$ be geodesics and $t\in [0,1]$. Then

$$d^2({\gamma }_t,{\eta }_t)\le (1-t)d^2({\gamma }_0,{\eta }_0)+t{d}^2({\gamma }_1,{\eta }_1)-t(1-t){[d({\gamma }_0,{\gamma }_1)-d({\eta }_0,{\eta }_1)]}^2$$

and

$$d({\gamma }_t,{\eta }_t)\le (1-t)d({\gamma }_0,{\eta }_0)+td({\gamma }_1,{\eta }_1).$$

Corollary 14 shows that the distance function $(x,y)\mapsto d(x,y)$ in a global NPC space is convex with respect to both variables. Consequently, all balls in a global NPC space are convex.

Example 15 Every Hilbert space is a global NPC space.

Example 16 Every metric tree is a global NPC space.

Example 17 A Riemannian manifold is a global NPC space if and only if it is complete, simply connected, and of nonpositive curvature.

More details about global NPC spaces can be found in [13–16].

We need the following notations in the sequel. Let $(X,d)$ be a metric space and A, B be nonempty subsets of X. Define

Given C a nonempty subset of X, the metric projection $P_C:X\to 2^C$ is the mapping

$$P_C(x)=\{y\in C:d(x,y)=dist(x,C)\},\text{for every }x\in X,$$

where $2^C$ denotes the set of all subsets of C.

Definition 18 [9]

Let A, B be nonempty convex subsets of a geodesic metric space. A mapping $T:A\cup B\to A\cup B$ is said to be affine if

$$T(\lambda x+(1-\lambda )y)=\lambda Tx+(1-\lambda )Ty,$$

for all $x,y\in A$ or $x,y\in B$ and $\lambda \in (0,1)$.

Definition 19 [8]

Let X be a metric space. A subset C of X is called approximatively compact if for any $y\in X$ and for any sequence $\{x_n\}$ in C such that $d(x_n,y)\to dist(y,C)$ as $n\to \mathrm{\infty }$, $\{x_n\}$ has a subsequence which converges to a point in C.

In [13], Sturm presented the following result which ensures the existence and uniqueness of the metric projection on a global NPC space.

Proposition 20

• (i) For each closed convex set C in a global NPC space $(X,d)$, there exists a unique map $P_C:X\to C$ (projection onto C) such that

  $$d(P_C(x),x)=\underset{y\in C}{inf}d(x,y)\mathit{\text{for every}}x\in X;$$

• (ii) $P_C$ is orthogonal in the sense that

  $$d^2(x,y)\ge d^2(x,P_C(x))+d^2(P_C(x),y)$$

for every $x\in X$, $y\in C$;

• (iii) $P_C$ is nonexpansive,

  $$d(P_C(x),P_C(z))\le d(x,z)\mathit{\text{for every}}x,z\in X.$$

Remark 21 Note that the existence of a unique metric projection does not need the compactness of C.

Remark 22 [13]

For any subset A of a global NPC space $(X,d)$, there exists a unique smallest convex set $\mathit{co}(A)={\bigcup }_{n=0}^{\mathrm{\infty }}{A}_n$, containing A and called convex hull of A. Where $A_0=A$, and for $n\in \mathbb{N}$, the set $A_n$ consists of all points in global NPC space X which lie on geodesics which start and end in $A_{n-1}$.

Based on Proposition 20, Niculescu and Roventa [17] proved the Schauder fixed point theorem in the setting of a global NPC space.

Theorem 23 Let C be a closed convex subset of a global NPC space X with the property that the closed convex hull of every finite subset of X is compact. Then every continuous map $T:C\to C$, whose image $T(C)$ is relatively compact, has a fixed point.

3 Main results

In this section, we will prove the existence of best proximity points for a relatively u-continuous mapping. Also, we obtain a result on the existence of common best proximity points for a family of not necessarily commuting relatively u-continuous mappings.

Proposition 24 Let A, B be nonempty subsets of a metric space X with $A_0\ne \mathrm{\varnothing }$ and $T:A\cup B\to A\cup B$ be a relatively u-continuous mapping. Then $T(A_0)\subseteq B_0$ and $T(B_0)\subseteq A_0$.

Proof Choose $x\in A_0$, then there exists $y\in B$ such that $d(x,y)=dist(A,B)$. But T is a relatively u-continuous mapping, then for each $\epsilon >0$, there exists a $\delta >0$ such that

$$d(p,q)<\delta +dist(A,B)\text{implies}d(Tp,Tq)<\epsilon +dist(A,B)$$

for each $p\in A$, $q\in B$. Since $d(x,y)<\delta +dist(A,B)$ for any $\delta >0$, hence

$$dist(A,B)\le d(Tx,Ty)<\epsilon +dist(A,B)$$

for each $\epsilon >0$. Therefore, $d(Tx,Ty)=dist(A,B)$ and then $T(x)\in B_0$. This shows that $T(A_0)\subseteq B_0$. Similarly, it can be seen that $T(B_0)\subseteq A_0$. □

Proposition 25 Let A, B be nonempty closed convex subsets of a global NPC space X with $A_0\ne \mathrm{\varnothing }$, $T:A\cup B\to A\cup B$ be a relatively u-continuous mapping, and $P:A\cup B\to A\cup B$ be a mapping defined by

$$P(x)=\{\begin{array}{cc}{P}_B(x),\hfill & \mathit{\text{if}}x\in A,\hfill \\ P_A(x),\hfill & \mathit{\text{if}}x\in B.\hfill \end{array}$$

Then $TP(x)=P(Tx)$ for all $x\in A_0\cup B_0$, i.e., $P_A(T(x))=T(P_B(x))$ for $x\in A_0$ and $T(P_A(y))=P_B(T(y))$ for $y\in B_0$.

Proof Choose $x\in A_0$, then there exists $y\in B$ such that $d(x,y)=dist(A,B)$. According to Proposition 20, since the metric projection is unique, we have $y=P_B(x)$ and $x=P_A(y)$. Recalling that T is relatively u-continuous, therefore, as in the proof of Proposition 24, $d(Tx,Ty)=dist(A,B)$. Thus, it follows that $T(x)\in B_0$ and $T(y)\in A_0$. Again, in view of the uniqueness of the projection operator, we have

$$P_A(T(x))=T(y)=T(P_B(x)).$$

So, $P_A(T(x))=T(P_B(x))$ for any $x\in A_0$. Similarly, it can be shown that $T(P_A(y))=P_B(T(y))$ for any $y\in B_0$. □

By an analogous argument to the proof of Theorem 3.1 [8], we can prove the following theorem.

Theorem 26 Let A, B be two nonempty subsets of a global NPC space X such that A is closed convex and B is closed. If $A_0$ is approximatively compact and $\{x_n\}$ is a sequence in $A_0$, and $y\in B$ such that $d(x_n,y)\to dist(A,B)$, then $x_n\to P_A(y)$.

Proof Assume the contrary, then there exists $\epsilon >0$ and a subsequence $\{x_{n_m}\}$ of $\{x_n\}$ such that

$$d(x_{n_m},y)\to dist(A,B)\text{but}d(x_{n_m},P_A(y))\ge \epsilon .$$

Since $A_0$ is approximatively compact, there exists a subsequence $\{x_{n_m^{\mathrm{\prime }}}\}$ of $\{x_{n_m}\}$ which converges to a point $x\in A$. Hence,

$$d(x_{n_m^{\mathrm{\prime }}},y)\to d(x,y).$$

Also,

$$d(x_{n_m^{\mathrm{\prime }}},y)\to dist(A,B).$$

Thus, $d(x,y)=dist(A,B)$. By Proposition 20, it follows that $x=P_A(y)$. Finally, we obtain

$$d(x_{n_m^{\mathrm{\prime }}},P_A(y))\to d(x,P_A(y))\ge \epsilon ,$$

which implies that $x\ne P_A(y)$. This leads to a contradiction and therefore $x_n\to P_A(y)$. □

The following theorem guarantees the existence of best proximity points for a relatively u-continuous mapping in a global NPC space.

Theorem 27 Let A, B be nonempty closed convex subsets of a global NPC space X with the property that the closed convex hull of every finite subset of X is compact. Let $A_0$, $B_0$ be nonempty compact convex and $T:A\cup B\to A\cup B$ be a relatively u-continuous mapping. Then there exist $x_0\in A$, $y_0\in B$ such that

$$d(x_0,T{x}_0)=d(y_0,T{y}_0)=dist(A,B).$$

Proof By Proposition 24, since T is a relatively u-continuous mapping, we have $T(A_0)\subseteq B_0$ and $T(B_0)\subseteq A_0$. The result follows from Theorem 23 once we show that $P_A\circ T:A_0\to A_0$ is a continuous mapping, where $P_A:X\to A$ is a metric projection operator.

To prove this, first notice that $P_A(B_0)\subseteq A_0$. Since X is a global NPC space, by Proposition 20, we obtain that $P_A:X\to A$ is a continuous mapping. In what follows, we see that the mapping T is continuous on $A_0$, In fact, let $\{x_n\}$ be a sequence in $A_0$ such that $x_n\to x_0$ for some $x_0\in A_0$. From Proposition 25, we have

$$P_B(P_A(T{x}_0))=P_B(T(P_B{x}_0))=T(P_A(P_B{x}_0))=T{x}_0.$$

Notice that

$$\begin{array}{rcl}d(x_n,P_B(x_0))& \le & d(x_n,x_0)+d(x_0,P_B(x_0))\\ =& d(x_n,x_0)+dist(A,B)\to dist(A,B)\end{array}$$

(3.1)

as $n\to \mathrm{\infty }$. Since T is relatively u-continuous, for each $\epsilon >0$, there exists a $\delta >0$ such that $d(x,y)<\delta +dist(A,B)$ implies $d(Tx,Ty)<\epsilon +dist(A,B)$ for all $x\in A$, $y\in B$. From (3.1), with this $\delta >0$, it follows that there is $N\in \mathbb{N}$ such that $d(x_n,P_B(x_0))<\delta +dist(A,B)$ for all $n\ge N$. This implies

$$d(T(x_n),T(P_B(x_0)))<\epsilon +dist(A,B)$$

for all $n\ge N$. Therefore,

$$d(T(x_n),P_A(T{x}_0))=d(T(x_n),T(P_B(x_0)))\to dist(A,B).$$

This together with Theorem 26 implies that $T{x}_n\to P_B(P_A(T{x}_0))=T{x}_0$. Thus, T is continuous on $A_0$.

Now, since $P_A\circ T$ is a continuous mapping of $A_0$, by the Schauder fixed point theorem for a global NPC space, Theorem 23, $P_A\circ T$ has a fixed point $x_0\in A_0$. From $P_A(T{x}_0)=x_0$, we find that $d(x_0,T{x}_0)=dist(T{x}_0,A)$. But since $T{x}_0\in B_0$, there is $x^{\mathrm{\prime }}\in A_0$ such that $d(x^{\mathrm{\prime }},T{x}_0)=dist(A,B)$. Consequently,

$$dist(A,B)\le dist(T{x}_0,A)\le d(T{x}_0,x^{\mathrm{\prime }})=dist(A,B),$$

which gives

$$dist(T{x}_0,A)=dist(A,B).$$

Thus, $d(x_0,T{x}_0)=dist(A,B)$. This completes the proof. □

Next, we will show that Theorem 27 is also true for an appropriate family of relatively u-continuous mappings. The following notations define the set of all best proximity points of a relatively u-continuous mapping:

Theorem 28 Let A, B be nonempty closed convex subsets of a global NPC space X with the property that the closed convex hull of every finite subset of X is compact. Let $A_0$, $B_0$ be nonempty compact convex and $T:A\cup B\to A\cup B$ be a relatively u-continuous mapping. Let T be affine. Then $F_A(T)$ is a nonempty compact convex subset of $A_0$ and $F_B(T)$ is a nonempty compact convex subset of $B_0$.

Proof It is obvious that $F_A(T)$ is a nonempty subset of $A_0$ by Theorem 27. Assume that $\{x_n\}$ is a sequence in $F_A(T)$ such that $x_n\to x_0$ for some $x_0\in A_0$. By the continuity of T on $A_0$, we have $x_0\in F_A(T)$. Therefore, $F_A(T)$ is closed and then compact. Now we claim that $F_A(T)$ is convex. In fact, let $\lambda \in [0,1]$, $x_1$, $x_2\in F_A(T)$, and $z=(1-\lambda )x_1+\lambda x_2$. Since the distance function d is convex with respect to both variables, by Corollary 14, we have

$$\begin{array}{rcl}dist(A,B)& \le & d(z,Tz)\\ =& d((1-\lambda )x_1+\lambda x_2,T((1-\lambda )x_1+\lambda x_2))\\ =& d((1-\lambda )x_1+\lambda x_2,(1-\lambda )T{x}_1+\lambda T{x}_2)\\ \le & (1-\lambda )d(x_1,T{x}_1)+\lambda d(x_2,T{x}_1)\\ =& dist(A,B).\end{array}$$

This implies that $d(z,Tz)=dist(A,B)$, i.e., $z\in F_A(T)$. Therefore, $F_A(T)$ is convex. Similarly, it can be shown that $F_B(T)$ is a nonempty compact convex subset of $B_0$. □

Lemma 29 Let A, B be nonempty closed convex subsets of a global NPC space X with the property that the closed convex hull of every finite subset of X is compact. Let $A_0$, $B_0$ be nonempty compact convex and $T,S:A\cup B\to A\cup B$ be relatively u-continuous mappings such that S and T are commuting on $F_A(T)\cup F_B(T)$. Then $S(F_A(T))\subseteq F_B(T)$ and $S(F_B(T))\subseteq F_A(T)$.

Proof For each $x\in F_A(T)$, we have $d(x,Tx)=dist(A,B)$. Since S is a relatively u-continuous mapping, then for $\delta >0$,

$$d(x,Tx)<\delta +dist(A,B)\text{implies}d(S(x),S(Tx))<\epsilon +dist(A,B)$$

for each $\epsilon >0$. Therefore, $d(S(x),S(Tx))=dist(A,B)$. The commutativity for S and T on $F_A(T)$ implies that $d(S(x),T(Sx))=dist(A,B)$. Thus, we deduce that $Sx\in F_B(T)$. This shows that $S(F_A(T))\subseteq F_B(T)$. Also, we can prove that $S(F_B(T))\subseteq F_A(T)$. □

Now, we define a new class of mappings called cyclic Banach pairs.

Definition 30 Let A, B be nonempty subsets of a metric space $(X,d)$ and let $T,S:A\cup B\to A\cup B$ be mappings. The pair $\{S,T\}$ is called a cyclic Banach pair if $S(F_A(T))\subseteq F_B(T)$ and $S(F_B(T))\subseteq F_A(T)$.

The following is an example of a pair of non-commuting mappings that are relatively u-continuous and that are a cyclic Banach pair.

Example 31 Let $X={\mathbb{R}}^2$ with the Euclidean metric and consider (as in [18])

Let $T,S:A\cup B\to A\cup B$ be defined as

Then T and S are relatively u-continuous mappings. Since

$$TS(0,x)\ne ST(0,x),$$

T and S are non-commuting mappings. Also, $dist(A,B)=1$. It is easy to verify that

$$F_A(T)=\{(0,0)\}\text{and}{F}_B(T)=\{(1,0)\}$$

and

$$S(F_A(T))\subseteq F_B(T),S(F_B(T))\subseteq F_A(T).$$

Therefore, $\{S,T\}$ is a cyclic Banach pair.

The following theorem proves that two relatively u-continuous mappings which are not necessarily commuting have common best proximity points.

Theorem 32 Let A, B be nonempty closed convex subsets of a global NPC space X with the property that the closed convex hull of every finite subset of X is compact. Let $A_0$, $B_0$ be nonempty compact convex and $T,S:A\cup B\to A\cup B$ be affine relatively u-continuous mappings. If $\{S,T\}$ is a cyclic Banach pair, then $F_A(T)\cap F_A(S)\ne \mathrm{\varnothing }$.

Proof By Theorem 28, $F_A(T)$ is a nonempty compact convex subset of $A_0$ and $F_B(T)$ is a nonempty compact convex subset of $B_0$. For each $x\in F_A(T)$, we have

$$dist(A,B)\le dist(F_A(T),F_B(T))\le d(x,Tx)=dist(A,B),$$

which implies that $dist(F_A(T),F_B(T))=dist(A,B)$. By the definition of cyclic Banach pairs $S:F_A(T)\cup F_B(T)\to F_A(T)\cup F_B(T)$. Since $\{S,T\}$ is a cyclic Banach pair and since for each $\epsilon >0$ there exists a $\delta >0$ such that

$$d(x,y)<\delta +dist(A,B)\text{implies}d(S(x),S(y))<\epsilon +dist(A,B)$$

for all $x\in F_A(T)$, $y\in F_B(T)$, hence S is a relatively u-continuous mapping on $F_A(T)\cup F_B(T)$. The conditions of Theorem 27 are satisfied, so there exists $x_0\in F_A(T)$ such that

$$d(x_0,S{x}_0)=dist(F_A(T),F_B(T))=dist(A,B).$$

Thus, $x_0\in F_A(S)$. This implies that $F_A(T)\cap F_A(S)\ne \mathrm{\varnothing }$. □

Next, we will extend Theorem 32 to the case of a countable family of not necessarily commuting relatively u-continuous mappings. Let $\mathrm{\Omega }=\{T_i:i\in \mathbb{N}\}$ be a family of relatively u-continuous mappings. Define

for each $i=1,\dots ,n$.

Definition 33 Let A, B be nonempty subsets of a metric space $(X,d)$ and let $T,S:A\cup B\to A\cup B$. The pair $\{S,T\}$ is called a symmetric cyclic Banach pair if $\{S,T\}$ and $\{T,S\}$ are cyclic Banach pairs, that is, $S(F_A(T))\subseteq F_B(T)$, $S(F_B(T))\subseteq F_A(T)$, $T(F_A(S))\subseteq F_B(S)$ and $T(F_B(S))\subseteq F_A(S)$.

Theorem 34 Let A, B be nonempty closed convex subsets of a global NPC space X with the property that the closed convex hull of every finite subset of X is compact. Let $A_0$, $B_0$ be nonempty compact convex and Ω a countable family of affine relatively u-continuous mappings such that $\{T_i,T_j\}$ is a symmetric cyclic Banach pair for each $i,j\in \mathbb{N}$. Then Ω has a common best proximity in A.

Proof First, we prove that $F_A(T_1)\cap F_A(T_2)\cap F_A(T_3)\ne \mathrm{\varnothing }$. By an analogous argument to the proof of Theorem 32, $F_A(T_i)$ is a nonempty compact convex subset of $A_0$, $F_B(T_i)$ is a nonempty compact convex subset of $B_0$ and $dist(F_A(T_i),F_B(T_i))=dist(A,B)$, for $i=1,2,3$. So, we have $F_A(T_1)\cap F_A(T_2)$ and $F_B(T_1)\cap F_B(T_2)$ are nonempty compact convex with

$$dist(F_A(T_1)\cap F_A(T_2),F_B(T_1)\cap F_B(T_2))=dist(A,B).$$

Suppose that $T_3$ is a mapping on $(F_A(T_1)\cap F_A(T_2))\cup (F_B(T_1)\cap F_B(T_2))$. Since both of $\{T_3,T_1\}$ and $\{T_3,T_2\}$ are cyclic Banach pairs, $T_3$ is a relatively u-continuous mapping on $(F_A(T_1)\cap F_A(T_2))\cup (F_B(T_1)\cap F_B(T_2))$. From Theorem 27, $T_3$ has a best proximity point $z\in F_A(T_1)\cap F_A(T_2)$. This shows that $F_A(T_1)\cap F_A(T_2)\cap F_A(T_3)\ne \mathrm{\varnothing }$.

By induction, for a finite symmetric cyclic Banach family ${\mathrm{\Omega }}^{\mathrm{\prime }}=\{T_1,T_2,\dots ,T_n\}$ of affine relatively u-continuous mappings, there exists $x_0\in {\bigcap }_{i=1}^n{F}_A(T_i)$.

Now, let $\mathrm{\Omega }=\{T_i:i\in \mathbb{N}\}$. For each $T_i$, $F_A(T_i)$ is a nonempty compact convex of $A_0$, and for $i=1,\dots ,n$, we have

$$\bigcap _{i=1}^n{F}_A(T_i)\ne \mathrm{\varnothing }.$$

This shows that the set $\{F_A(T_i):i\in \mathbb{N}\}$ has a finite intersection property. Thus, we have

$$\bigcap _{i=1}^{\mathrm{\infty }}{F}_A(T_i)\ne \mathrm{\varnothing },$$

i.e., Γ has a common best proximity point in A. □