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*
\(f\in\mathfrak{F}\).