On best proximity points of upper semicontinuous multivalued mappings

Abstract

In this paper we study the existence of best proximity points of a nonself upper semicontinuous multivalued mapping $T:A\rightarrow2^{B}$ in a strictly convex Banach space. This multivalued mapping commutes with affine, noncyclic, and relatively u-continuous single-valued mapping $f:A \cup B\rightarrow A\cup B$. Also, we study the case when T commutes with a family of commuting, affine, noncyclic, and relatively u-continuous single-valued mappings on $A\cup B$. Moreover, we present some examples to illustrate our results.

1 Introduction

Let A, B be nonempty subsets of a metric space $(X, d)$ and $T:A\rightarrow2^{B}$, where $2^{B}$ is the family of all nonempty subsets of B. If $A\cap B=\emptyset$, the operator inclusion $x\in T(x)$ has no solution. In this case, it is logical to look for a point $x\in A$ such that $\operatorname{dist}(x,T(x))$ is minimum. Because $\operatorname{dist}(x,T(x))$ is at least $\operatorname{dist}(A,B)$, the point x is the solution of the equation $\operatorname{dist}(x,T(x))=\operatorname{dist}(A,B)=\inf\{d(x, y):x\in A, y\in B\}$. This point is called the best proximity point of T. Indeed, best proximity point theorems examine the existence of such optimal approximate solutions of the operator inclusion $x\in T(x)$ when there is no exact solution. If $A\cap B \neq \emptyset$, the best proximity point is the fixed point of T.

For multivalued mappings, the existence of best proximity points was established by many authors, e.g., Abkar and Gabeleh in [1] and [2], Al-Thagafi and Shahzad in [3], Amini-Harandi in [4], De la Sen in [5], Kirk et al. in [6] and Włodarczyk et al. in [7]. Best proximity point theorems for relatively nonexpansive single-valued mapping were studied in [8] in 2005. Since then there has been a lot of activity in this area and a number of results appeared by various authors. Best proximity point theorems for relatively u-continuous mapping were proved in [9] and [10]. For other related results, we refer the reader to [11–16] and [17]. In this paper, we study the existence of best proximity points for an upper semicontinuous multivalued mapping with nonempty, compact, and convex values $T:A \rightarrow2^{B}$ which commutes with an affine and relatively u-continuous single-valued mapping $f:A \cup B:\rightarrow A\cup B$ such that $f(A)\subseteq A$ and $f(B)\subseteq B$ (noncyclic). In addition, we present some support examples for our results and we also give an example showing that the condition ‘$T(x) \cap B_{0} \neq\emptyset$ for each $x\in A_{0}$’ is necessary. Moreover, we add a similar theorem for a multivalued mapping which commutes with a family of commuting, affine, noncyclic, and relatively u-continuous single-valued mappings on $A\cup B$.

2 Preliminaries

Definition 2.1

[9]

Let A, B be nonempty subsets of a metric space X. A mapping $f:A \cup B\rightarrow A\cup B$ is said to be relatively u-continuous if for each $\epsilon>0$ there exists $\delta>0$ such that $d(f(x),f(y))<\epsilon+\operatorname{dist}(A,B)$ whenever $d(x,y)<\delta+\operatorname{dist}(A,B)$ for each $x\in A$, $y\in B$.

Definition 2.2

[8]

Let A, B be nonempty subsets of a metric space X. A mapping $f:A\cup B \rightarrow A\cup B$ is called relatively nonexpansive if $d(f(x),f(y)) \leq d(x,y)$ for each $x\in A$, $y\in B$.

Every relatively nonexpansive mapping is relatively u-continuous. However, the converse is not true (see [9]).

Definition 2.3

[3]

Let A, B be nonempty subsets of a metric space X and $T:A\rightarrow2^{B}$ a multivalued mapping. A point $x\in A$ is called a (i) fixed point of T if $x\in T(x)$ and (ii) best proximity point of T if $\operatorname{dist}(x,T(x))=\operatorname{dist}(A,B)$. Note that if $\operatorname{dist}(A,B)=0$, then we get a fixed point of T.

Definition 2.4

Let A, B be nonempty subsets of a metric space X. A multivalued mapping $T:A\rightarrow2^{B}$ is called upper semicontinuous if $T^{-1}(C)=\{x\in A: T(x)\cap C\ne\emptyset\}$ is closed in A whenever C is closed in B.

Proposition 2.5

[18]

Let X be a strictly convex Banach space, A a nonempty, compact, and convex subset of X, and B a nonempty closed subset of X. Let $\{ x_{n}\}$ be a sequence in A and $y\in B$. If $\|x_{n}-y\| \rightarrow \operatorname{dist}(A,B)$, then $x_{n}\rightarrow P_{A}(y)$.

Definition 2.6

[9]

Let A, B be nonempty convex subsets of a Banach space X. A mapping $f: A\cup B\to A\cup B$ is called affine if $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$ for all $x, y\in A$ or $x, y\in B$ and $\alpha, \beta\in[0, 1]$ with $\alpha+\beta=1$.

Lemma 2.7

[19]

If A is a nonempty, compact, and convex subset of a Banach space, and $T:A\rightarrow2^{A}$ can be expressed as a composition of finitely many upper semicontinuous multivalued mappings with nonempty, compact, and convex values, then T has a fixed point.

Let A, B be nonempty subsets of a Banach space X. $f: A\cup B\rightarrow A\cup B$ a relatively nonexpansive mapping such that $f(A) \subseteq A$, $f(B)\subseteq B$, $T:A\rightarrow \operatorname{KC}(B)$, where $\operatorname{KC}(B)$ is the set of all nonempty, compact, and convex subsets of B. The mapping f and T are said to commute if for each $x\in A$, $f(T(x)) \subseteq T(f(x))$. Define

$$\begin{aligned}& A_{0} = \bigl\{ x\in A: \Vert x-y\Vert =\operatorname{dist}(A,B) \mbox{ for some } y\in B\bigr\} , \\& B_{0} = \bigl\{ y\in B: \Vert x-y\Vert =\operatorname{dist}(A,B) \mbox{ for some } x\in A\bigr\} . \end{aligned}$$

Remark 2.8

Note that if A and B are nonempty, compact, and convex sets, then $A_{0}$ and $B_{0}$ are nonempty, compact, and convex sets with $\operatorname{dist}(A_{0},B_{0})=\operatorname{dist}(A,B)$. For details see [6] and [8]. Also, $f(A_{0})\subseteq A_{0}$ and $f(B_{0})\subseteq B_{0}$ [10].

Remark 2.9

[3, 9]

Let A be a nonempty subset of a normed space X. The metric projection operator is defined by $P_{A}(x)=\{y\in A: \|x-y\|=\operatorname{dist}(x,A)\}$ for each $x\in X$. If A is a nonempty, compact, and convex subset of a Banach space X, then $P_{A}$ is upper semicontinuous with nonempty, compact, and convex values. Observe that when A is a nonempty, compact, and convex subset of a strictly convex Banach space X, $P_{A}$ is a single-valued mapping from X to A.

Theorem 2.10

[10]

Let A, B be nonempty, compact, and convex subsets of a strictly convex Banach space X. If $f:A\cup B:\rightarrow A\cup B$ is relatively u-continuous such that $f(A)\subseteq A$ and $f(B)\subseteq B$. Then there exists $(x_{0},y_{0})\in A\times B$ such that $f(x_{0})=x_{0}$, $f(y_{0})=y_{0}$, and $\|x_{0}-y_{0}\|=\operatorname{dist}(A,B)$.

3 Main results

The following proposition is a noncyclic version of Proposition 3.2 in [9].

Proposition 3.1

Let A, B be nonempty, compact, and convex subsets of a strictly convex Banach space X. Let $f:A\cup B\rightarrow A\cup B$ be a relatively u-continuous mapping such that $f(A) \subseteq A$ and $f(B)\subseteq B$. $P: A\cup B \rightarrow A\cup B$ is a mapping defined by

$$P(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} P_{B}(x) & \textit{if } x\in A, \\ P_{A}(x) & \textit{if } x\in B. \end{array}\displaystyle \right . $$

Then $f(P(x))=P(f(x))$ for each $x\in A_{0}\cup B_{0}$, i.e., $P_{A}(f(y))=f(P_{A}(y))$ for each $y\in B_{0}$ and $P_{B}(f(x))=f(P_{B}(x))$ for each $x\in A_{0}$.

Proof

Let $x \in A_{0}$. Then there exists $y\in B$ such that $\| x-y\|=\operatorname{dist}(A,B)$. So, $y=P_{B}(x)$ and $x=P_{A}(y)$. Then for each $\delta>0$, $\|x-y\|< \delta+\operatorname{dist}(A,B)$. Since f is relatively u-continuous, for each $\epsilon>0$ we have $\operatorname{dist}(A,B) \leq\|f(x)-f(y)\| <\epsilon+\operatorname{dist}(A,B)$. Thus, $\|f(x)-f(y)\| =\operatorname{dist}(A,B)$. So, $f(x)=P_{A}(f(y))$ and $f(y)=P_{B}(f(x))$. Since A, B are nonempty, compact, and convex subsets of a strictly convex Banach space, the metric projection is unique. Now, $x=P_{A}(y)\Longrightarrow f(x)=f(P_{A}(y))\Longrightarrow P_{A}(f(y))=f(P_{A}(y))$ for each $y\in B_{0}$. Also, $y=P_{B}(x)\Longrightarrow f(y)=f(P_{B}(x))\Longrightarrow P_{B}(f(x))=f(P_{B}(x))$ for each $x\in A_{0}$. Hence, $f(P(x))=P(f(x))$ for each $x\in A_{0}\cup B_{0}$. □

A cyclic version of the following proposition can be found in [9] (see the proof of Theorem 3.1 in [9]).

Proposition 3.2

Let A, B be nonempty, compact, and convex subsets of a strictly convex Banach space X. Let $f:A\cup B \rightarrow A\cup B$ be a relatively u-continuous mapping such that $f(A) \subseteq A$ and $f(B)\subseteq B$. Then f is continuous on $A_{0}$ and $B_{0}$.

Proof

Let $x_{0}\in A_{0}$ and $\{x_{n}\} \subseteq A_{0}$ such that $x_{n}\rightarrow x_{0}$. We want to show that $f(x_{n})\rightarrow f(x_{0})$. Using the triangle inequality, we obtain

$$\begin{aligned} \begin{aligned} \bigl\Vert x_{n}-P_{B}(x_{0})\bigr\Vert & \leq \Vert x_{n}-x_{0}\Vert +\bigl\Vert x_{0}-P_{B}(x_{0})\bigr\Vert \\ &=\Vert x_{n}-x_{0}\Vert +\operatorname{dist}(A,B) \\ &\rightarrow \operatorname{dist}(A,B) . \end{aligned} \end{aligned}$$

Then for each $\delta>0$ there exists $N_{0}\in\mathbb{N}$ such that for each $n\geq N_{0}$, we have $| \|x_{n}-P_{B}(x_{0})\|-\operatorname{dist}(A,B)|<\delta$. So, $n\geq N_{0} \Longrightarrow \|x_{n}-P_{B}(x_{0})\|<\delta+\operatorname{dist}(A,B)$. By relative u-continuity of f, $\|f(x_{n})-f(P_{B}(x_{0}))\|<\epsilon+\operatorname{dist}(A,B)$ for each $n\geq N_{0}$. Since $\{f(x_{n})\} \subseteq A$ and $P_{B}(f(x_{0})) \in B$, Proposition 2.5 gives

$$f(x_{n})\rightarrow P_{A}\bigl(f\bigl(P_{B}(x_{0})\bigr)\bigr)=f\bigl(P_{A}\bigl(P_{B}(x_{0})\bigr) \bigr)=f(x_{0}). $$

Hence, $f(x_{n})\rightarrow f(x_{0})$. Since $x_{0}\in A_{0}$ was arbitrary, f is continuous on $A_{0}$. Similarly, f is continuous on $B_{0}$. Therefore, f is continuous on $A_{0}\cup B_{0}$. □

Theorem 3.3

Let A, B be nonempty, compact, and convex subsets in a strictly convex Banach space X. Suppose $f:A\cup B\rightarrow A\cup B$ is an affine relatively u-continuous mapping with $f(A) \subseteq A$, $f(B) \subseteq B$. Then there exists $(x_{0},y_{0}) \in A\times B$ such that $f(x_{0})=x_{0}$, $f(y_{0})=y_{0}$ and $\| x_{0}-y_{0} \| =\operatorname{dist}(A,B)$.

In addition, if $T:A\rightarrow \operatorname{KC}(B)$ is an upper semicontinuous multivalued mapping, f and T commute, and $T (x) \cap B_{0} \neq\emptyset$ for each $x\in A_{0}$, then there exists $a\in A$ such that $f(a)=a$ and $\operatorname{dist}(a,T(a))=\operatorname{dist}(A,B)$.

Proof

For $u \in A_{0}$, there is a $v\in B$ such that $\| u - v \| = \operatorname{dist}(A,B)$. Then by the relative u-continuity of f, $\|f(u)- f(v) \| = \operatorname{dist}(A,B)$, implying that $f(u) \in A_{0}$. Therefore, the compact convex set $A_{0}$ is invariant under the continuous mapping f, and the Schauder fixed point theorem implies the existence of a fixed point $x_{0} = f(x_{0})\in A_{0}$. Let $y_{0}$ be the unique closest point to $x_{0}$ in B. Then by the relative u-continuity of f and the uniqueness of the closest point projection onto B, $y_{0} = f(y_{0})$ and $\|x_{0} - y_{0} \| = \operatorname{dist}(A,B)$.

Now, we will prove that there exists $a\in A$ such that $\operatorname{dist}(a , T(a))=\operatorname{dist}(A,B)$. Define $\operatorname{Fix}(f)=\{ x\in A\cup B: f(x)=x\}$, $\operatorname{Fix}_{A}(f)=\operatorname{Fix}(f) \cap A_{0}$ and $\operatorname{Fix}_{B}(f)=\operatorname{Fix}(f) \cap B_{0}$. Clearly, $\operatorname{Fix}_{A}(f)$ and $\operatorname{Fix}_{B}(f)$ are nonempty, because $x_{0} \in \operatorname{Fix}_{A}(f)$ and $y_{0} \in \operatorname{Fix}_{B}(f)$. The set $\operatorname{Fix}_{A}(f)$ is closed. Indeed, let $\{x_{n}\} \subseteq \operatorname{Fix}_{A}(f)$ such that $x_{n}\rightarrow x_{0}$. Since $\{x_{n}\} \subseteq A_{0}$ and $A_{0}$ is closed by Remark 2.8, we have $x_{0}\in A_{0} \subseteq A$. Using Proposition 3.2, $f(x_{n}) \rightarrow f(x_{0})$. But $f(x_{n})=x_{n}$ for each n. So $x_{n} \rightarrow f(x_{0})$. Consequently $x_{0}=f(x_{0})$. Thus $x_{0} \in \operatorname{Fix}_{A}(f)$. Therefore, $\operatorname{Fix}_{A}(f)$ is closed. Similarly, $\operatorname{Fix}_{B}(f)$ is closed. So, $\operatorname{Fix}_{A}(f)$ and $\operatorname{Fix}_{B}(f)$ are compact sets as they are closed subsets of the compact sets $A_{0}$, $B_{0}$. In addition, $\operatorname{Fix}_{A}(f)$ is a convex set. Indeed, let $x,y \in \operatorname{Fix}_{A}(f)$ and $\alpha ,\beta\in[0,1]$ with $\alpha+\beta=1$. Since f is affine, $f(\alpha x+\beta y)= \alpha f(x)+ \beta f(y)= \alpha x+\beta y$, i.e., $\alpha x+\beta y \in \operatorname{Fix}(f)$. Also, $\alpha x+\beta y \in A_{0}$ as $A_{0}$ is convex and $x,y\in A_{0}$. Consequently, $\alpha x+\beta y \in \operatorname{Fix}(f)\cap A_{0}= \operatorname{Fix}_{A}(f)$. Similarly, $\operatorname{Fix}_{B}(f)$ is a convex set.

Assume $x \in \operatorname{Fix}_{A}(f)$ and choose $v \in T(x)$. Since f and T commute, $f(v) \in T(f(x)) = T(x)$, which implies that $T(x)$ is invariant under f. Then the invariance of $B_{0}$ under f shows that the compact convex set $T(x) \cap B_{0}$ is invariant under f. Since f is continuous on $B_{0}$, by the Schauder fixed point theorem f has a fixed point in $T(x) \cap B_{0}$, implying that $T(x) \cap \operatorname{Fix}_{B} (f) \neq\emptyset$ for each $x\in \operatorname{Fix}_{A}(f)$.

Now, define $F: \operatorname{Fix}_{A}(f)\rightarrow 2^{\operatorname{Fix}_{B}(f)}$ by $F(x)=T(x) \cap \operatorname{Fix}_{B}(f)$ for each $x\in \operatorname{Fix}_{A}(f)$. Then F is an upper semicontinuous multivalued mapping with nonempty, compact, and convex values. Note that $P_{A}:\operatorname{Fix}_{B}(f) \rightarrow \operatorname{Fix}_{A}(f)$. To see this, let $x\in \operatorname{Fix}_{B}(f) \subseteq B_{0}$. Then there exists $y\in A$ such that $\|x-y\|=\operatorname{dist}(A,B)$. So, $y=P_{A}(x)$ and $x=P_{B}(y)$. For each $\delta>0$, we have $\|x-y\|<\delta+\operatorname{dist}(A,B)$. Using the relative u-continuity for any f, $\operatorname{dist}(A,B)\leq \|f(x)-f(y)\|<\epsilon+\operatorname{dist}(A,B)$ for each $\epsilon>0$. Thus, $\|f(x)-f(y)\|=\operatorname{dist}(A,B)$. This implies that $f(y)=P_{A}(f(x))$ and $f(x)=P_{B}(f(y))$. Since $x\in \operatorname{Fix}_{B}(f)$ and $y=P_{A}(x)$, we have $f(y)=f(P_{A}(x))=P_{A}(f(x))=P_{A}(x)$ and so $P_{A}(x) \in \operatorname{Fix}_{A}(f) \subseteq A$. Note that $P_{A} \circ F: \operatorname{Fix}_{A}(f) \rightarrow 2^{\operatorname{Fix}_{A}(f)}$. By Lemma 2.7, there exists $a\in \operatorname{Fix}_{A}(f) \subseteq A_{0}$ such that $a\in(P_{A} \circ F)(a)$, i.e., $a=f(a)$ and $a\in P_{A}(F(a))$. So, there exists $b \in F(a) = T(a) \cap \operatorname{Fix}_{B}(f) \subseteq B_{0}$ such that $a= P_{A}(b) \subseteq \operatorname{Fix}_{A}(f)$. As $a=P_{A}(b)$, $\|a-b\|=\operatorname{dist}(b,A)$. Since $b\in F(a)=T(a) \cap \operatorname{Fix}_{B}(f) \subseteq B_{0}$, then $b\in T(a)$ and $b\in B_{0}$. Since $b\in B_{0}$, there exists $a' \in A$ such that $\|a'-b\|=\operatorname{dist}(A,B)$. Since $a\in A$ and $T(a) \subseteq B$, we have

$$\begin{aligned} \operatorname{dist}(A,B) \leq& \operatorname{dist}\bigl(a, T(a)\bigr) \\ \leq&\|a-b\| \\ =&\operatorname{dist}(b,A) \\ \leq&\bigl\Vert b-a'\bigr\Vert \\ =&\operatorname{dist}(A,B) . \end{aligned}$$

Thus, $\operatorname{dist}(a ,T(a))=\operatorname{dist}(A,B)$. □

Remark 3.4

The condition $T(x) \cap B_{0} \neq \emptyset$ for each $x\in A_{0}$ is necessary in Theorem 3.3. For example, in the real space if $A=[1,5]\times[-5,5]$, $B=[-1,\frac{-1}{25}]\times[-5,5]$. Define

$$f:A\cup B \rightarrow A\cup B \quad \mbox{by}\quad f(x,y)=\biggl(x,\frac {y+1}{2} \biggr) $$

and

$$T:A\rightarrow \operatorname{KC}(B) \quad \mbox{by}\quad T(x,y)=\biggl[-1, \frac{-1}{x^{2}}\biggr]\times\{y\}. $$

Clearly, T is upper semicontinuous and f is affine and relatively u-continuous. Also, $f(A)\subseteq A$ and $f(B)\subseteq B$. There are fixed points of f, $x_{0}=(1,1)\in A$, $y_{0}=(\frac{-1}{25},1)\in B$ such that $\|x_{0}-y_{0}\|=\operatorname{dist}(A,B)=1.04$. In addition, f and T commute. Suppose that there exists $a\in \operatorname{Fix}(f)\cap A$ such that $\operatorname{dist}(a,T(a))=1.04$. Then $a=(z,1)$, for some $1\leq z\leq5$. So,

$$\operatorname{dist}\bigl(a,T(a)\bigr)=\operatorname{dist}\biggl((z,1),\biggl[-1, \frac{-1}{z^{2}}\biggr]\times\{1\}\biggr)=\biggl\Vert (z,1)-\biggl( \frac{-1}{z^{2}},1\biggr)\biggr\Vert =1.04. $$

Consequently, $z^{3}-1.04 z^{2}+1=0$. So, $z_{1}=0.893939214944 + 0.7334769205376 i$, $z_{2}= 0.893939214944 - 0.7334769205376 i$, which are not real numbers, and $z_{3}= -0.747878429888$, which does not belong to $[1,5]$. Note that $A_{0}=\{1\}\times[-5,5]$, $B_{0}=\{\frac{-1}{25}\}\times[-5,5]$. For $x=(1,y)\in A_{0}$, we have $T(x)=T(1,y)=\{(-1,y)\}$. So, $T(x) \cap B_{0} = \{(-1,y)\} \cap \{(\frac{-1}{25},y): -5\leq y\leq5\}=\emptyset$.

Corollary 3.5

Let A, B be nonempty, compact, and convex sets in a strictly convex Banach space X. If $T:A\rightarrow \operatorname{KC}(B)$ is an upper semicontinuous multivalued mapping and $T(x)\cap B_{0}\neq\emptyset$ for each $x\in A_{0}$, then there exists $a\in A$ such that $\operatorname{dist}(a,T(a))=\operatorname{dist}(A,B)$.

Proof

Taking $f=I$ (the identity mapping on $A\cup B$) in Theorem 3.3, we obtain the desired result. □

Corollary 3.6

Let A be a nonempty, compact, and convex set in a strictly convex Banach space. Suppose $f:A\rightarrow A$ is an affine continuous mapping. If $T:A\rightarrow \operatorname{KC}(A)$ is an upper semicontinuous multivalued mapping and f, T commute, then there exists $a\in A$ such that $a \in \operatorname{Fix}(f)\cap \operatorname{Fix}(T)$.

Proof

Since any continuous mapping on a compact set is relatively u-continuous on that set, taking $A=B$ in Theorem 3.3, we see that there exists a∈A such that $f(a)=a$ and $\operatorname{dist}(a , T(a))=\operatorname{dist}(A,A)=0$, i.e., $a\in T(a)$. So, $f(a)=a\in T(a)$. Therefore, $a \in \operatorname{Fix}(f)\cap \operatorname{Fix}(T)$. □

Theorem 3.7

Let X be a strictly convex Banach space. Let A, B be nonempty, compact, and convex subsets of X and let $f, g: A \cup B \to A \cup B$ be commuting, affine, and relatively u-continuous mappings such that $f(A) \subseteq A$, $f(B) \subseteq B$ and $g(A) \subseteq A$, $g(B) \subseteq B$. Then there exist points $x_{0} \in A$ and $y_{0} \in B$ such that $x_{0} = f(x_{0}) = g(x_{0})$, $y_{0} = f(y_{0}) =g(y_{0})$ and $\|x_{0} - y_{0}\| = \operatorname{dist}(A, B)$.

Proof

For $u \in A_{0}$, there is a $v \in B$ such that $\| u - v \| = \operatorname{dist}(A,B)$. Then by the relative u-continuity of f, $\|f(u)- f(v) \| = \operatorname{dist}(A,B)$, implying that $f(u) \in A_{0}$. Therefore, the compact convex set $A_{0}$ is invariant under the continuous mapping f, and the Schauder fixed point theorem implies the existence of a fixed point $x = f(x)\in A_{0}$. The set of fixed points of f in $A_{0}$ (denoted by $\operatorname{Fix}_{A}(f)$) is closed and convex since f is continuous and affine. If $x \in \operatorname{Fix}_{A}(f)$, commutativity of f and g implies $f(g(x)) = g(f(x)) =g(x)$. Therefore, $\operatorname{Fix}_{A}(f)$ is invariant under g, and since g is continuous it has a fixed point in $\operatorname{Fix}_{A}(f)$. Let $x_{0}$ be a common fixed point of f and g in $A_{0}$, that is, $x_{0} = f(x_{0}) = g(x_{0})$, and let $y_{0}$ be the unique closest point to $x_{0}$ in B. Then by the relative u-continuity of f and g and the uniqueness of the closest point projection onto B, $y_{0} = f(y_{0}) = g(y_{0})$ and $\| x_{0} - y_{0} \| = \operatorname{dist}(A,B)$. □

The previous theorem can be extended to an arbitrary family of commuting affine and noncyclic mappings. The proof depends on the following common fixed point result for commuting affine u-continuous mappings in strictly convex Banach spaces. The proof of this result is adapted from Przebieracz ([20], Theorem 1.1) and is included for convenience of the reader.

Lemma 3.8

(Markov-Kakutani theorem)

Let X be a strictly convex Banach space. Let A, B be nonempty, compact, and convex subsets of X and let $\mathfrak{F}$ be a family of commuting affine and relatively u-continuous mappings on $A\cup B$ such that $f(A) \subseteq A$ and $f(B) \subseteq B$. Then there is an $x_{0}\in A_{0}$ such that $f(x_{0}) = x_{0}$ for every $f \in \mathfrak{F}$. There is a $y_{0}\in B_{0}$ such that $f(y_{0}) = y_{0}$ for every $f \in\mathfrak{F}$.

Proof

Notice that the mappings in the family $\mathfrak{F}$ are continuous on $A_{0}\cup B_{0}$. Let $\operatorname{Fix}(f) = \{x \in A\cup B: f(x) = x\}$, $\operatorname{Fix}_{A}(f)=\operatorname{Fix}(f) \cap A_{0}$, $f\in\mathfrak{F}$. As shown in the proof of Theorem 3.7, $\operatorname{Fix}_{A}(f) \ne\emptyset$ and $\operatorname{Fix}_{A}(f)$ is convex and compact. To prove that $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)\neq\emptyset$, consider any finite collection from $\mathfrak{F}$, say $f_{1},\ldots, f_{n}$. Assume that

$$C = \bigcap_{1\leq i\leq n} \operatorname{Fix}_{A}(f_{i}) \neq\emptyset. $$

For each $x\in C$ and $k \in\{1,\ldots, n\}$, $f_{k} f_{n+1}(x) = f_{n+1} f_{k} (x) = f_{n+1}(x)$, which implies that $f_{n+1}(x)\in C$. Therefore, the compact convex set C is invariant under $f_{n+1}$, implying that $\operatorname{Fix}_{A}(f_{n+1}) \cap C \ne\emptyset$ since $f_{n+1}$ is continuous on $A_{0}$. Since every finite collection of the sets $\operatorname{Fix}_{A}(f)$, $f\in\mathfrak{F}$, has a nonempty intersection, we have $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)\neq\emptyset$. Similarly, $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)\neq\emptyset$. □

Theorem 3.9

Let X be a strictly convex Banach space. Let A, B be nonempty, compact, and convex subsets of X and let $\mathfrak{F}$ be a family of commuting affine and relatively u-continuous mappings on $A\cup B$ such that $f(A) \subseteq A$ and $f(B) \subseteq B$. Then there exist points $x_{0} \in A$ and $y_{0} \in B$ such that $x_{0} = f(x_{0})$ and $y_{0} = f(y_{0})$, for all $f\in\mathfrak{F}$ where $\|x_{0} - y_{0}\| = \operatorname{dist}(A, B)$.

Proof

By Lemma 3.8 the mappings in the family $\mathfrak{F}$ have a common fixed point $x_{0} \in A$, that is, $f(x_{0}) = x_{0}$ for $f\in \mathfrak{F}$. Let $y_{0}\in B$ be the unique closest point to $x_{0}$ in B. Then, for any $f\in\mathfrak{F}$, $\|f(x_{0})- y_{0}\| = \operatorname{dist}(A,B)$, but by the relative u-continuity of f, $\|f(x_{0})-f(y_{0})\| = \operatorname{dist}(A,B)$. By the uniqueness of the closest point, $y_{0} = f(y_{0})$ for $f\in\mathfrak{F}$. □

Theorem 3.10

Let A, B be nonempty, compact, and convex subsets of a strictly convex Banach space X and let $\mathfrak{F}$ be a family of commuting, affine and relatively u-continuous mappings on $A\cup B$ with $f(A)\subseteq A$, $f(B)\subseteq B$ for each $f\in\mathfrak{F}$. Let $T:A \rightarrow \operatorname{KC}(B)$ be an upper semicontinuous mapping such that $T(x)\cap B_{0} \neq\emptyset$ for each $x\in A_{0}$. If $\mathfrak{F}$ and T commute, then there exists a point $a\in A$ such that $f(a)=a$ for each $f\in\mathfrak{F}$ and $\operatorname{dist}(a,T(a))=\operatorname{dist}(A,B)$.

Proof

By Lemma 3.8, $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)$ and $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)$ are nonempty.

As in the proof of Theorem 3.3, $T(x)$ is invariant under each $f\in\mathfrak{F}$, for $x \in \operatorname{Fix}_{A}(f)$. Since $\bigcap_{f\in \mathfrak{F}} \operatorname{Fix}_{A}(f)\ne\emptyset$, for $x\in\bigcap_{f\in \mathfrak{F}} \operatorname{Fix}_{A}(f)$, $T(x)$ is invariant under $\mathfrak{F}$. Also, $B_{0}$ is invariant under $\mathfrak{F}$. Therefore as in the proof of Theorem 3.3, since $T(x)\cap B_{0}$ is a compact convex set, $T(x)\cap(\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f))\ne\emptyset$. By the proof of Theorem 3.3, $\operatorname{Fix}_{A}(f)$ and $\operatorname{Fix}_{B}(f)$ are compact and convex sets for $f \in \mathfrak{F}$. Therefore, $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)$ and $\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)$ are compact and convex.

Now define $F: \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)\rightarrow 2^{\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)}$ by $F(x)=T(x) \cap (\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f))$ for each $x\in \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)$. Clearly, F is an upper semicontinuous multivalued mapping with compact convex values. Now, $P_{A}:\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)\rightarrow \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)$. To see this, let $x\in \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)$. Then $x\in B_{0}$ and $f(x)=x$ for each $f\in\mathfrak{F}$. So, there exists $y\in A$ such that $\|x-y\|=\operatorname{dist}(A,B)$. This implies $x=P_{B}(y)$ and $y=P_{A}(x)$. For each $\delta>0$, we have $\|x-y\|<\delta+\operatorname{dist}(A,B)$. Using the relative u-continuity for any $f\in\mathfrak{F}$, $\operatorname{dist}(A,B)\leq \|f(x)-f(y)\|<\epsilon+\operatorname{dist}(A,B)$ for each $\epsilon>0$. Thus, $\|f(x)-f(y)\|=\operatorname{dist}(A,B)$. Therefore, $f(y)=P_{A}(f(x))$ and $f(x)=P_{B}(f(y))$ for each $f \in\mathfrak{F}$. Now, $y=P_{A}(x)\Longrightarrow f(y)=f(P_{A}(x)) \Longrightarrow P_{A}(x)= f(P_{A}(x))$ for each $f \in\mathfrak{F}$. Hence, $P_{A}(x)\in \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)$ for each $x\in \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{B}(f)$. Note that $P_{A} \circ F: \bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)\rightarrow 2^{\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)}$. By Lemma 2.7, $P_{A}\circ F$ has a fixed point. So, there exists $a \in\bigcap_{f\in\mathfrak{F}} \operatorname{Fix}_{A}(f)$ such that $a\in(P_{A}\circ F)(a)$. So, $f(a)=a$ for each $f\in\mathfrak{F}$ and $a\in P_{A}(F(a))$, i.e., there exists $b\in F(a)$ such that $a=P_{A}(b)$. Since $b\in F(a)$, $b\in T(a)\cap(\bigcap_{f\in \mathfrak{F}} \operatorname{Fix}_{B}(f))$. So, $b\in T(a)$, $b\in B_{0}$, and $f(b)=b$ for each $f\in\mathfrak{F}$. $a=P_{A}(b)$ implies $\|a-b\|=\operatorname{dist}(b,A)$. Since $b\in B_{0}$, there exists $a' \in A$ such that $\|a'-b\|=\operatorname{dist}(A,B)$. Since $a\in A$ and $T(a)\subseteq B$, we have

$$\begin{aligned} \operatorname{dist}(A,B) \leq &\operatorname{dist}\bigl(a, T(a)\bigr) \\ \leq&\|a-b\| \\ =&\operatorname{dist}(b,A) \\ \leq&\bigl\Vert b-a'\bigr\Vert \\ =&\operatorname{dist}(A,B) . \end{aligned}$$

Thus, $\operatorname{dist}(a ,T(a))=\operatorname{dist}(A,B)$. □

Corollary 3.11

Let A be a nonempty, compact, and convex subset of a strictly convex Banach space X and let $\mathfrak{F}$ be a family of commuting, affine and continuous self-mappings of A. Let $T:A \rightarrow \operatorname{KC}(A)$ be an upper semicontinuous mapping. If $\mathfrak{F}$ and T commute, then there exists a point $a\in A$ such that $a=f(a)\in T(a)$ for each $f\in\mathfrak{F}$.

4 Examples

Examples 4.1 to 4.4 are related to Theorem 3.3. On other hand, the last two examples are related to Theorem 3.7 (and Theorem 3.10).

Example 4.1

Let $X=\mathbb{R}^{2}$ with the usual metric. The sets $A=\{(x,y): 0\leq x\leq4, 1\leq y \leq5\}$, $B=\{(x,0): 0\leq x \leq4\}$ are nonempty, compact, and convex with $\operatorname{dist}(A,B)=1$. Define $f:A\cup B \rightarrow A\cup B$ by $f(x,y)=(\frac{2x+1}{3},y)$ and $T:A\rightarrow \operatorname{KC}(B)$ by $T(x,y)=[x,4]\times\{0\}$. Then T is upper semicontinuous and f is relatively u-continuous and affine with $f(A)\subseteq A$ and $f(B) \subseteq B$. As $\operatorname{Fix}(f)=\{(1,y): 1 \leq y \leq5\mbox{ or } y=0 \}$, we get $x_{0}=(1,1) \in \operatorname{Fix}(f) \cap A$, $y_{0}=(1,0)\in \operatorname{Fix}(f) \cap B$ with $\|x_{0}-y_{0}\|=1$. In addition, f and T commute. Indeed, $f(T(x,y))=f([x,4]\times\{0\})=\{\frac{2z+1}{3}:z\in [x,4]\}\times\{0\}$ and $T(f(x,y))=T(\frac{2x+1}{3},y)=[\frac{2x+1}{3},4]\times\{0\}$. For $z\in[x,4]$, $\frac{2z+1}{3}\in[\frac{2x+1}{3},3] \subseteq [\frac{2x+1}{3},4]$. Thus, $f(T(x,y))\subseteq T(f(x,y))$ for each $(x,y) \in A$. Also, $T(x)\cap B_{0}\neq\emptyset$ for each $x\in A_{0}$ since $A_{0}=\{(x,1):0\leq x\leq4 \}$ and $B_{0}=B$. For $(1, 1)\in A$, we have $f(a)=a$ and $\operatorname{dist}(a, T(a))=\operatorname{dist}(A,B)=1$.

Example 4.2

Let $X=\mathbb{R}^{2}$ with the usual metric. The sets $A=\{ (0,a): 1\leq a\leq3\}$, $B=\{(x,y): 1\leq x \leq5, 1\leq y \leq5\}$ are nonempty, compact, and convex with $\operatorname{dist}(A,B)=1$. Define $f:A\cup B \rightarrow A\cup B$ by $f(x,y)=(x,\frac{y+3}{2})$ and $T:A\rightarrow \operatorname{KC}(B)$ by $T(0,a)=[1,a]\times\{3\}$. Then T is upper semicontinuous and f is relatively u-continuous and affine with $f(A)\subseteq A$ and $f(B) \subseteq B$. As $\operatorname{Fix}(f)=\{(x,3): x=0\mbox{ or } 1 \leq x \leq5 \}$, we get $x_{0}=(0,3) \in \operatorname{Fix}(f) \cap A$, $y_{0}=(1,3)\in \operatorname{Fix}(f) \cap B$ with $\|x_{0}-y_{0}\|=1$. In addition, f and T commute. Indeed, $f(T(0,a))=f([1,a]\times\{3\})=[1,a]\times\{3\}$ and $T(f(0,a))=T(0,\frac{a+3}{2})=[1,\frac{a+3}{2}]\times\{3\}$. For $a\in[1,3]$, $\frac{a+3}{2}\geq a$, i.e., $[1,a]\subseteq[1,\frac{a+3}{2}]$. Thus, $f(T(0,a))\subseteq T(f(0,a))$ for each $(0,a) \in A$. Also, $T(x)\cap B_{0}\neq \emptyset$ for each $x\in A_{0}$ since $A_{0}=A$ and $B_{0}=\{(1,y):1\leq y\leq3\}$. For $a=(0,3)\in A$, we have $f(a)=a$ and $\operatorname{dist}(a, T(a))=\operatorname{dist}(A,B)=1$.

Example 4.3

Let $X=\mathbb{R}^{2}$ with the usual metric. The sets $A=\{ (x,y): -1\leq x\leq-0.04, -5\leq y \leq 5\}$, $B=\{(x,y): 0\leq x \leq5, -5\leq y\leq5\}$ are nonempty, compact, and convex with $\operatorname{dist}(A,B)=0.04$. Define $f:A\cup B \rightarrow A\cup B$ by $f(x,y)=(x,\frac{y+1}{2})$ and $T:A\rightarrow \operatorname{KC}(B)$ by $T(x,y)=[0,x^{2}]\times\{y\}$. Then T is upper semicontinuous and f is relatively u-continuous and affine with $f(A)\subseteq A$ and $f(B) \subseteq B$. As $\operatorname{Fix}(f)=\{(x,1): -1\leq x\leq-0.04\mbox{ or } 0 \leq x \leq5 \}$, we get $x_{0}=(-0.04,1) \in \operatorname{Fix}(f) \cap A$, $y_{0}=(0,1)\in \operatorname{Fix}(f) \cap B$ with $\|x_{0}-y_{0}\|=0.04$. In addition, f and T commute. Also, $T(x)\cap B_{0}\neq\emptyset$ for each $x\in A_{0}$ since $A_{0}=\{ (-0.04,y): -5 \leq y \leq5\}$ and $B_{0}=\{(0,y):-5\leq y\leq5\}$. For $a=(-0.04,1)\in A$, we have $f(a)=a$ and $\operatorname{dist}(a, T(a))=\operatorname{dist}(A,B)=0.04$.

Example 4.4

Let $X=\mathbb{R}^{2}$ with the usual metric. The sets $A=\{ (x,y): -3\leq x\leq3, -1\leq y\leq -0.25\}$, $B=\{(x,y): -3\leq x \leq3, 0\leq y \leq4\}$ are nonempty, compact, and convex with $\operatorname{dist}(A,B)=0.25$. Define $f:A\cup B \rightarrow A\cup B$ by $f(x,y)=(\frac{x}{2},y)$ and $T:A\rightarrow \operatorname{KC}(B)$ by $T(x,y)=\{x\}\times[0,y^{2}]$. Then T is upper semicontinuous and f is relatively u-continuous and affine with $f(A)\subseteq A$ and $f(B) \subseteq B$. As $\operatorname{Fix}(f)=\{(0,y): 0\leq y\leq4\mbox{ or } -1 \leq y \leq-0.25 \}$, we get $x_{0}=(0,-0.25) \in \operatorname{Fix}(f) \cap A$, $y_{0}=(0,0)\in \operatorname{Fix}(f) \cap B$ with $\|x_{0}-y_{0}\|=0.25$. In addition, f and T commute. Also, $T(x)\cap B_{0}\neq \emptyset$ for each $x\in A_{0}$ since $A_{0}=\{ (x,-0.25): -3 \leq x \leq3\}$ and $B_{0}=\{(x,0):-3\leq x\leq3\}$. For $a=(0,-0.25)$, we have $f(a)=a$ and $\operatorname{dist}(a, T(a))=\operatorname{dist}(A,B)=0.25$.

Example 4.5

Let $X=\mathbb{R}^{2}$ with the usual metric. The sets $A=\{ (x,y): 0\leq x \leq5 , y=-1\}$, $B=\{(x,y): -5\leq x \leq0, y =1\}$ are nonempty, compact, and convex subsets of a strictly convex Banach space with $\operatorname{dist}(A,B)=2$. Define $f,g:A\cup B \rightarrow A\cup B$ by $f(x,y)=(\frac{2 x}{5},y)$ and $g(x,y)=(\frac{x}{2},y)$. Then f, g are relatively u-continuous and affine with $f(A)\subseteq A$, $f(B) \subseteq B$, $g(A)\subseteq A$, and $g(B) \subseteq B$. Also f, g commute. Now, define $T:A\rightarrow \operatorname{KC}(B)$ by $T(x,y)=[-5,-x]\times\{y^{2}\}$. Then T is upper semicontinuous with nonempty, compact, and convex values. In addition, T commutes with f and g. Clearly, $A_{0}=\{(0,-1)\}$, $B_{0}=\{(0,1)\}$, and $(0,1)\in T(0,-1)=[-5,0]\times\{1\}$. For $a=(0,-1)\in A$ and $b=(0, 1)\in B$, we have $f(a)=g(a)=a$, $f(b)=g(b)=b$, and $\|a-b\|=\operatorname{dist}(A, B)=2$. Moreover, $\operatorname{dist}(a,T(a))=\operatorname{dist}(A,B)$.

Example 4.6

Let $X=\mathbb{R}^{2}$ with the usual metric. The sets $A=\{ (x,y): -4\leq x \leq-1 , -6\leq y\leq6 \}$, $B=\{(x,y): 0\leq x \leq4, -6\leq y\leq6\}$ are nonempty, compact, and convex subsets of a strictly convex Banach space with $\operatorname{dist}(A,B)=1$. Define $f,g:A\cup B \rightarrow A\cup B$ by $f(x,y)=(x,\frac{y}{3})$ and $g(x,y)=(x,\frac{y}{2})$. Then f, g are relatively u-continuous and affine with $f(A)\subseteq A$, $f(B) \subseteq B$, $g(A)\subseteq A$, and $g(B) \subseteq B$. Also f, g commute. Now, define $T:A\rightarrow \operatorname{KC}(B)$ by $T(x,y)=[0,-x]\times\{y\}$. Then T is upper semicontinuous with nonempty, compact, and convex values. In addition, T commutes with f and g. Clearly, $A_{0}=\{(-1,y): -6 \leq y \leq6\}$, $B_{0}=\{(0,y): -6\leq y\leq6\}$. So, $(0,y)\in T(-1,y)\cap B_{0}=([0,1]\times\{y\})\cap B_{0}$ for each $(-1,y)\in A_{0}$. For $a=(-1, 0)\in A$ and $b=(0, 0)\in B$, we have $f(a)=g(a)=a$, $f(b)=g(b)=b$, and $\|a-b\|=\operatorname{dist}(A, B)=1$. Moreover, $\operatorname{dist}(a,T(a))=\operatorname{dist}(A,B)$.