On a generalization of a relatively nonexpansive mapping and best proximity pair

Abstract

Let A and B be two nonempty subsets of a normed space X, and let $T: A \cup B \to A \cup B$ be a cyclic (resp., noncyclic) mapping. The objective of this paper is to establish weak conditions on T that ensure its relative nonexpansiveness.

The idea is to recover the results mentioned in two papers by Matkowski (Banach J. Math. Anal. 2:237–244, 2007; J. Fixed Point Theory Appl. 24:70, 2022), by replacing the nonexpansive mapping $f: C \to C$ with a cyclic (resp., noncyclic) relatively nonexpansive mapping to obtain the best proximity pair. Additionally, we provide an application to a functional equation.

1 Introduction and preliminaries

Let A and B be two nonempty subsets of a normed space $(X,\|\cdot \|)$. A self-mapping $T : A \cup B \rightarrow A \cup B$ is said to be cyclic (resp., noncyclic) if $T(A) \subseteq B$ and $T (B) \subseteq A$ (resp., $T(A)\subseteq A$ and $T (B)\subseteq B$). For such a mapping, we can consider the minimization problem of finding a best proximity pair of the mapping T, that is, a pair $(p,q) \in A \times B$ such that

$$\begin{aligned}& \bigl\Vert p-T(p) \bigr\Vert = \bigl\Vert q-T(q) \bigr\Vert = \operatorname{dist}(A,B)\\& \quad \bigl(\text{resp., } T(p) = p, T(q) = q, \text{ and } \Vert p-q \Vert = \operatorname{dist}(A,B) \bigr), \end{aligned}$$

where $\operatorname{dist}(A,B)=\inf \{d(x,y) : (x,y) \in A \times B\}$.

A cyclic (resp., noncyclic) mapping $T: A \cup B \rightarrow A \cup B$ is said to be relatively nonexpansive if $\| T(x) - T(y)\| \leqslant \|x - y\|$ for all $x \in A$ and $y \in B$ (notice that in general a relatively nonexpansive mapping need not be continuous).

Recall that a real normed vector space $(X,\|\cdot \|)$ is called uniformly convex (see Clarkson [4]) if for every $\varepsilon \in (0,2]$, there is $\delta > 0$ such that for any two vectors $x,y \in X$ with $\|x\| = \|y\|= 1$, the condition $\|x - y\| \geqslant \varepsilon$ implies that

$$ \biggl\Vert \frac{x+y}{2} \biggr\Vert \leqslant 1 - \delta . $$

The existence of a best proximity pair was first considered and studied in 2005 by Eldred et al. [5]:

• if $(A,B)$ is a nonempty closed bounded convex pair of a uniformly convex Banach space X, then every cyclic relatively nonexpansive mapping defined on $A \cup B$ has a best proximity pair.

• if $(A,B)$ is a nonempty closed bounded convex pair of a uniformly convex Banach space X, then every noncyclic relatively nonexpansive mapping defined on $A \cup B$ has a best proximity pair.

The relevance of best proximity points is that they provide optimal solutions for the problem of best approximation between two sets. Some references concerning best proximity points are given in [3, 6, 8, 11–14].

Let us recall the definitions of the lower and upper bounds of a function $f: [0,+\infty ) \rightarrow [0,+\infty )$ at a point $t_{0}$:

$$ \liminf_{t\to t_{0}^{+}} f(t) = \sup_{\eta > 0}\Bigl( \inf_{ t_{0}< t < t_{0}+ \eta }f(t)\Bigr) \quad \text{and}\quad \limsup _{t\to t_{0}^{+}} f(t) = \inf_{\eta > 0}\Bigl(\sup _{ t_{0}< t < t_{0}+\eta }f(t)\Bigr). $$

Remark 1

If ${ \liminf_{t\to 0^{+}}\frac{f(t+a)}{t+a}= \ell }$ with $a,\ell \in [0,+\infty )$, then for all $\varepsilon >0$ and $\eta >0$, there exists $t_{\varepsilon}\in ]0,\eta [$ such that

$$ f(t_{\varepsilon}+a) < (\ell +\varepsilon ) (t_{\varepsilon}+a) \quad \text{and}\quad \lim_{\varepsilon \to 0^{+}}t_{\varepsilon}=0. $$

For the reader’s convenience, we recall the main results in [10].

Theorem 1

([10], Theorem 1)

Let X be a uniformly convex Banach space, let C be a nonempty bounded convex closed subset of X, and let T be a self-mapping of C. If there is a function $\beta : [0,+\infty )\rightarrow [0,+\infty )$ such that

$$\begin{aligned}& \bigl\Vert T(x) - T(y) \bigr\Vert \leqslant \beta \bigl( \Vert x-y \Vert \bigr), \quad x, y \in C, x \neq y, \end{aligned}$$

(1)

$$\begin{aligned}& \limsup_{t\to 0^{+}} \frac{\beta (t)}{t} < +\infty , \quad \textit{and}\quad \liminf_{t\to 0^{+}} \frac{\beta (t)}{t}=1, \end{aligned}$$

(2)

then T has a fixed point in C.

Proposition 2

([10], Proposition 1)

Let X be a uniformly convex Banach space, and let C be a nonempty bounded convex closed subset of X. Suppose that $T : C \rightarrow C$ is continuous. If there exist two positive sequences $(t_{n})_{n\geqslant 0}$ and $(c_{n})_{n\geqslant 0}$, ${ \lim_{n\to +\infty} t_{n} = 0}$, ${ \lim_{n\to +\infty} c_{n} = 1}$, such that for every $n \in {\mathbb{N}}$ and for all $x, y \in C$,

$$ \Vert x - y \Vert = t_{n} \quad \Rightarrow\quad \bigl\Vert T(x) - T(y) \bigr\Vert \leqslant c_{n} t_{n} , $$

(3)

then T has a fixed point.

The idea of this paper is to present weak conditions under which a cyclic (resp., noncyclic) mapping $T: A \cup B \rightarrow A \cup B$, where A and B are two subsets of a normed space X, is relatively nonexpansive. As a result, we establish the existence of the best proximity pair for this mapping.

On the other hand, Let ${\mathcal {B}} _{L,d}$ be the family consisting of functions $\beta : [0,+\infty )\rightarrow [0,+\infty )$ that satisfy the following conditions:

(i) β increases on $[0,+\infty )$,

(ii) $\beta (kt)\leqslant k\beta (t)$ for all $t\in [d,+\infty )$ and $k\in {\mathbb{N}}\setminus \{0\}$,

(iii) ${ \liminf_{t\to 0^{+}}\frac{\beta (t+d)}{t+d}= L}$.

Then we establish the following result: Any cyclic (resp., noncyclic) mapping $T :A\cup B \rightarrow A\cup B$ such that

$$ \bigl\Vert T(u)-T(v) \bigr\Vert \leqslant \beta \bigl( \Vert u-v \Vert \bigr) $$

(4)

for all $(u,v)\in A\times B$ is relatively $(L,d)$-mapping (see Sect. 2.2.), where $d \in\, ]0,+\infty )$ and $L\geqslant \frac{1}{2}$.

Note that the set ${\mathcal {B}}_{L,d}$ is not empty. For example, the function β defined as $\beta (t) = \frac{2t}{t+1}$ for $t \in [0,+\infty )$ satisfies all three conditions (i), (ii), and (iii) with $L=d=1$.

We denote by ${\mathcal {B}} _{L,0}$ ($d=0$) the family of functions $\beta : [0,+\infty )\rightarrow [0, + \infty )$ that satisfy conditions (i) and (iii).

The paper is organized as follows. Our main results are presented in Sect. 2. Theorem 3 is a modification of Theorem 1 by Matkowski [10], in which the hypothesis ${\limsup_{t\to 0^{+}} \frac{\beta (t)}{t} < +\infty }$ is replaced with $\beta \in {\mathcal {B}}_{1,0}$. In Lemma 4, we show that any cyclic (resp., noncyclic) mapping $T : A \cup B \rightarrow A \cup B$ satisfying

$$ \bigl\Vert T\circ g(u) - T\circ h(v) \bigr\Vert \leq c_{n}\bigl(t_{n}+ \Vert u-v \Vert \bigr) $$

(5)

for all $(u,v) \in (\operatorname{cov}(A\cup B))^{2}$ such that $d\leq \|u - v\| \leq 3d$, where $d=\operatorname{dist}(A,B) >0$, and $\operatorname{cov}(A\cup B)$ is the convex hull of two parts A and B, is relatively nonexpansive. Using this lemma, under certain conditions on the parts A and B of a uniformly convex space X, we present Theorem 5 on the existence of the best proximity pair. Corollary 6 describes two cases $d=0$ and $d>0$ for a mapping $T: A \cup B \rightarrow A \cup B$ satisfying

$$ \bigl( d\leq \Vert u - v \Vert \leq t_{n} +3d \Rightarrow \bigl\Vert T\circ g(u) - T \circ h(v) \bigr\Vert \leq c_{n}(t_{n}+d) \bigr), $$

(6)

$(u,v) \in (\operatorname{cov}(A\cup B))^{2}$, where $d=\operatorname{dist}(A,B)$.

Proposition 7 in Sect. 2.2 says that any cyclic (resp., noncyclic) mapping $T :A\cup B \rightarrow A\cup B$ that satisfies the condition

$$ \bigl\Vert T(u)-T(v) \bigr\Vert \leq \beta \bigl( \Vert u-v \Vert \bigr) $$

(7)

for all $(u,v)\in A\times B$ is relatively $(L,d)$-mapping on $A\cup B$ (see Definition 2).

In Sect. 2.3, we use Proposition 7 to get a result on the existence of the best proximity pair of a functional equation in $L^{2}({\mathcal {U}})$, where ${\mathcal {U}}$ is a nonempty open subset of $\mathbb{R}^{m}$.

2 Mains results

The following result is a useful reformulation of Theorem 1 in [10].

Theorem 3

Let A be a nonempty bounded closed convex subset in a uniformly convex Banach space X. Let $\beta \in {\mathcal {B}}_{1,0}$, and let $T :A \rightarrow A$ be a mapping satisfying the inequality

$$ \bigl\Vert T(x)-T(y) \bigr\Vert \leq \beta \bigl( \Vert x-y \Vert \bigr) $$

(8)

for all $(x,y)\in A^{2}$ such that $x\neq y$. Then there exists $x^{*} \in A$ such that $Tx^{*}=x^{*}$.

Proof

Take $(x,y)\in A$ such that $x\neq y$. For $\varepsilon >0$, as ${\liminf_{t\to 0^{+}}\frac{\beta (t)}{t}= 1 }$, there exists $t_{\varepsilon}>0$ such that

$$\begin{aligned}& \beta (t_{\varepsilon}) < (1 +\varepsilon )t_{\varepsilon} \end{aligned}$$

(9)

$$\begin{aligned}& \quad \text{and}\quad \lim_{\varepsilon \to 0^{+}}t_{\varepsilon}=0. \end{aligned}$$

(10)

Let $n_{\varepsilon}\in \mathbb{N}$ be such that

$$ n_{\varepsilon}t_{\varepsilon}\leq \Vert x-y \Vert < (n_{\varepsilon}+1)t_{ \varepsilon}, $$

(11)

Put

$$ z_{k}= \biggl(1- \frac{k}{n_{\varepsilon}+1}\biggr)\cdot x + \frac{k}{n_{\varepsilon}+1}\cdot y \quad \text{for } k=0,1,\dots ,n_{ \varepsilon}+1. $$

By the convexity of C, $z_{k}\in C$ for all $k\in \{0,1,\dots ,n_{\varepsilon} \}$; moreover,

$$ \Vert z_{k}-z_{k+1} \Vert = \frac{ \Vert x-y \Vert }{n_{\varepsilon}+1} < t_{\varepsilon} . $$

(12)

Applying the triangle inequality, condition (8), inequalities (9), (11), and (12), and the monotony of β, we get

$$\begin{aligned} \bigl\Vert T(x)-T(y) \bigr\Vert & \leq \sum_{j=0}^{n_{\varepsilon}} \bigl\Vert T(z_{j}) -T(z_{j+1}) \bigr\Vert \\ & \leq \sum_{j=0}^{n_{\varepsilon}} \beta \bigl( \Vert z_{j} -z_{j+1} \Vert \bigr) \\ & \leq \sum_{j=0}^{n_{\varepsilon}} \beta (t_{\varepsilon}) \\ &\leq (n_{\varepsilon}+1) (1+\varepsilon )t_{\varepsilon } \\ &\leq \bigl(t_{\varepsilon}+ \Vert x-y \Vert \bigr) (1+\varepsilon ) . \end{aligned}$$

Letting ε tend to 0⁺ and using (10), we obtain

$$ \Vert Tx-Ty \Vert \leq \Vert x-y \Vert . $$

The mapping T is relatively nonexpansive in a bounded closed convex nonempty subset of the uniformly convex Banach space X; so the result follows from the original version of the Browder–Göhde–Kirk theorem (Browder [1], Göhde [7], Kirk [8]). □

2.1 Some auxiliary results on relatively nonexpansive and best proximity pairs

We denote by $\operatorname{cov}(A\cup B)$ the convex hull of two parts A and B of a normed vector space and $d=\operatorname{dist}(A,B)$.

Lemma 4

Let $(A,B)$ be a nonempty pair in a normed space $(X,\|\cdot \|)$. Let $g: \operatorname{cov}(A\cup B)\rightarrow A$ and $h: \operatorname{cov}(A\cup B)\rightarrow B$ be two mappings such that $g_{|A}=Id_{A}$ and $h_{|B}=Id_{B}$. Let $T :A\cup B \rightarrow A\cup B$ be a cyclic (resp., noncyclic) mapping, and let $(t_{n})_{n}$ and $(c_{n})_{n}$ be two positive sequences, ${\lim_{n\to +\infty} t_{n} = 0}$, ${\lim_{n\to +\infty} c_{n} = 1 }$, such that for every $n \in \mathbb{N}$ and for all $(u,v) \in (\operatorname{cov}(A\cup B))^{2}$ such that $d\leq \|u - v\| \leq 3d$, where $d=\operatorname{dist}(A,B) >0$,

$$ \bigl\Vert T\circ g(u) - T\circ h(v) \bigr\Vert \leq c_{n}\bigl(t_{n}+ \Vert u-v \Vert \bigr). $$

(13)

Then, for all $(x,y)\in A\times B$,

$$ \bigl\Vert T(x) - T(y) \bigr\Vert \leq \Vert x-y \Vert . $$

Proof

Let $(u,v)\in (\operatorname{cov}(A\cup B))^{2}$ be such that $d \leq \|u-v\|\leq 3d$. Then for all $n \in \mathbb{N}$,

$$ \bigl\Vert T\circ g(u) - T\circ h(v) \bigr\Vert \leq c_{n} \bigl(t_{n}+ \Vert u-v \Vert \bigr). $$

Taking the limit as n goes to +∞, we have

$$ \bigl\Vert T\circ g(u) - T\circ h(v) \bigr\Vert \leq \Vert u - v \Vert ; $$

(14)

in particular, if $(x,y)\in A \times B$ and $\|x-y\| \leq 3d$, then we have

$$ \bigl\Vert T(x) - T(y) \bigr\Vert \leq \Vert x - y \Vert . $$

Now let $(x,y)\in A\times B$ be such that $\|x-y\|> 3d$; in this case, $\operatorname{diam}(A,B) > d$. There is $p \in \mathbb{N}$ such that

$$ 2p+1 < \frac{ \Vert x-y \Vert }{3d} \leq 2p+3. $$

For $k=0,1,\dots ,2p+3$, let $x_{k}=x+\frac{k}{2p+3}(y-x)$. We have $x_{0}=x$, $x_{2p+3}=y$, $x_{k} \in \operatorname{cov}(A\cup B)$ for every k in $\{0,1,\dots ,2p+2\}$, and

$$ \Vert x_{k+1}-x_{k} \Vert = \frac{ \Vert x-y \Vert }{2p+3} \in \,]d,3d]. $$

Applying the triangle inequality and (14), we have

$$\begin{aligned} \bigl\Vert T(x)-T(y) \bigr\Vert \leq{}& \sum_{k=0}^{p} \bigl\Vert T\circ g(x_{2k})-T\circ h(x_{2k+1}) \bigr\Vert \\ & {}+ \sum_{k=0}^{p} \bigl\Vert T\circ h(x_{2k+1})-T\circ g(x_{2k+2}) \bigr\Vert \\ & {}+ \bigl\Vert T\circ g(x_{2p+2})-T\circ h(x_{2p+3}) \bigr\Vert \\ \leq{}& \sum_{k=0}^{p} \bigl( \Vert x_{2k} -x_{2k+1} \Vert + \Vert x_{2k+1}-x_{2k+2} \Vert \bigr)+ \Vert x_{2p+2}-x_{2p+3} \Vert \\ ={}& \Vert x-y \Vert . \end{aligned}$$

This finishes the proof. □

Theorem 5

Let $(A,B)$ be a nonempty closed bounded convex pair in a uniformly convex Banach space X. Let $g: \operatorname{cov}(A\cup B)\rightarrow A$ and $h: \operatorname{cov}(A\cup B)\rightarrow B$ be mappings such that $g_{|A}=Id_{A}$ and $h_{|B}=Id_{B}$. Let $T :A\cup B \rightarrow A\cup B$ be a cyclic (resp., noncyclic) mapping, and let $(t_{n})_{n}$ and $(c_{n})_{n}$ be positive sequences, ${\lim_{n\to +\infty} t_{n} = 0}$, ${\lim_{n\to +\infty} c_{n} = 1}$, such that for every $n \in \mathbb{N}$ and for all $(u,v) \in (\operatorname{cov}(A\cup B))^{2}$ such that $d\leq \|u - v\| \leq 3d$, where $d=\operatorname{dist}(A,B) >0$,

$$ \bigl\Vert T\circ g(u) - T\circ h(v) \bigr\Vert \leq c_{n} \bigl(t_{n}+ \Vert u-v \Vert \bigr) . $$

Then there exists $(x^{*},y^{*})\in A\times B$ such that

$$\begin{aligned}& \bigl\Vert x^{*}-Tx^{*} \bigr\Vert = \operatorname{dist}(A,B)= \bigl\Vert y^{*}-Ty^{*} \bigr\Vert \end{aligned}$$

(15)

$$\begin{aligned}& \quad \bigl(\textit{resp., }Tx^{*}=x^{*}, Ty^{*}=y^{*}, \textit{ and } \bigl\Vert x^{*}-y^{*} \bigr\Vert = \operatorname{dist}(A,B) \bigr). \end{aligned}$$

(16)

Proof

According to Lemma 4, the mapping T is cyclic (resp., noncyclic) relatively nonexpansive in $A\cup B$, where $(A,B)$ is a nonempty closed bounded convex pair of the uniformly convex Banach space X; so the result follows from the paper of Eldred et al. [5]. □

Corollary 6

Let $(A,B)$ be a nonempty closed bounded convex pair in a uniformly convex Banach space X. Let $g: \operatorname{cov}(A\cup B)\rightarrow A$ and $h: \operatorname{cov}(A\cup B)\rightarrow B$ be mappings such that $g_{|A}=Id_{A}$ and $h_{|B}=Id_{B}$. Let $T :A\cup B \rightarrow A\cup B$ be a cyclic (resp., noncyclic) mapping, and let $(t_{n})_{n}$ and $(c_{n})_{n}$ be strictly positive sequences, ${\lim_{n\to +\infty} t_{n} = 0}$, ${\lim_{n\to +\infty} c_{n} = 1 }$, such that for every $n \in \mathbb{N}$ and for all $(u,v) \in (\operatorname{cov}(A\cup B))^{2}$,

$$ \bigl( d\leq \Vert u - v \Vert \leq t_{n} +3d \Rightarrow \bigl\Vert T\circ g(u) - T \circ h(v) \bigr\Vert \leq c_{n}(t_{n}+d) \bigr), $$

(17)

where $d=\operatorname{dist}(A,B)$. Then there exists $(x^{*},y^{*})\in A\times B$ such that

$$\begin{aligned}& \bigl\Vert x^{*}-Tx^{*} \bigr\Vert = \operatorname{dist}(A,B)= \bigl\Vert y^{*}-Ty^{*} \bigr\Vert \\& \quad \bigl(\textit{resp., }Tx^{*}=x^{*}, Ty^{*}=y^{*}, \textit{ and } \bigl\Vert x^{*}-y^{*} \bigr\Vert = \operatorname{dist}(A,B) \bigr). \end{aligned}$$

Proof

We distinguish two cases $d>0$ and $d=0$.

Case 1: $d>0$.

Let $(u,v) \in (\operatorname{cov}(A\cup B))^{2}$ be such that $d\leq \|u - v\| \leq 3d$, so for each $n\in \mathbb{N}$, $d\leq \|u - v\| \leq t_{n} +3d$, and according to implication (17), we get

$$ \bigl\Vert T\circ g(u) - T\circ h(v) \bigr\Vert \leq c_{n}(t_{n}+d)\quad \text{for all } n \in \mathbb{N}. $$

We thus obtain the result according to Theorem 5.

Case 2: $d=0$.

In this case, we claim that $A\cap B\neq \emptyset$. Indeed, since $\operatorname{dist}(A,B) = 0$, there exists a sequence $((x_{m},y_{m}))_{m\geq 0}$ in $A\times B$ such that ${\lim_{m\to +\infty} d(x_{m},y_{m})=0}$. Since the space X is a uniformly convex Banach space, it is therefore reflexive, and since A and B are closed and bounded, the sequence $((x_{m},y_{m} ) )_{m \geq 0}$ admits a subsequence $((x_{\phi (m)},y_{\phi (m)}))_{m\geq 0}$ that converges weakly to $(a,b)\in A\times B$. By the weak lower semicontinuity of the norm $\|\cdot \|$ we have

$$ \Vert a-b \Vert \leq \lim_{m\to +\infty} \Vert x_{\phi (m)}-y_{\phi (m)} \Vert = 0. $$

Thus $a=b$, which shows that $A\cap B \neq \emptyset$.

Take $x, y\in A\cap B$ such that $x\neq y$ and $n\in \mathbb{N}$. There is a unique $p_{n}\in \mathbb{N}$ such that

$$ p_{n}\leq \frac{ \Vert x-y \Vert }{t_{n}} < p_{n}+1. $$

(18)

Put

$$ z_{k}= \biggl(1- \frac{k}{p_{n}+1}\biggr)\cdot x + \frac{k}{p_{n}+1}\cdot y \quad \text{for } k=0,1,\dots ,p_{n}+1. $$

Then $z_{k}\in A\cap B$ for all $k\in \{0,1,\dots ,p_{n}+1 \}$, because $A\cap B$ is convex; moreover,

$$ \Vert z_{k}-z_{k+1} \Vert = \frac{ \Vert x-y \Vert }{p_{n}+1}< t_{n} \quad \text{for all }k \in \{0,1,\dots ,p_{n} \}. $$

(19)

Applying implication (17) and inequalities (18) and (19), we obtain

$$\begin{aligned} \Vert Tx-Ty \Vert & \leq \sum_{j=0}^{p_{n}} \bigl\Vert T(z_{k}) -T(z_{k+1}) \bigr\Vert \\ & \leq \sum_{j=0}^{p_{n}} c_{n}t_{n} \\ & \leq (p_{n}+1) c_{n}t_{n} \\ & \leq c_{n} \Vert x-y \Vert +c_{n}t_{n}. \end{aligned}$$

Letting n tend to +∞ in the previous inequality, since ${\lim_{n\to +\infty}c_{n}=1}$ and ${\lim_{n\to +\infty}t_{n}=0}$, we obtain

$$ \Vert Tx-Ty \Vert \leq \Vert x-y \Vert . $$

In this case, $A \cap B \neq \emptyset$, the restriction of T to $A \cap B$ is nonexpansive, and the result follows from the Browder–Göhde–Kirk result. □

Remark 2

Under the hypotheses of the corollary, if we take $A=B$, then $\operatorname{cov}(A\cup A)=A$, $d=\operatorname{dist}(A,A)=0$, and $h=g=Id_{ A}$, and there is a fixed point of T. The difference between this corollary (for the case $d=0$) and Proposition 1 in [10] is that the corollary uses the implication

$$ \bigl( \Vert u - v \Vert \leq t_{n} \Rightarrow \bigl\Vert T(u) - T(v) \bigr\Vert \leq c_{n}t_{n} \bigr), $$

(20)

whereas Matkowski’s proposition uses (3) and the continuity of T.

We will provide an example of Corollary 6, which justifies that for elements u and v in $\operatorname{cov}(A\cup B)$, where $|u-v|>3d$, we can infer the following:

$$ \bigl( \vert u-v \vert \leq t_{n}+3d \Rightarrow \bigl\vert T \circ g(u)-T\circ h(v) \bigr\vert \leq c_{n}(t_{n}+d) \bigr). $$

(21)

Furthermore, we can obtain the result of this example using Theorem 5 in a straightforward manner.

Example 1

Let $A= [-6,-1]$ and $B= [0,1]$ be two parts of $\mathbb{R}$. We denote by $pr_{A}$ and $pr_{B}$ the projections on A and B, respectively. Let T be the mapping defined on $A\cup B$ by

$$ T(x)= \textstyle\begin{cases} \frac{1}{n} &\text{if } x\in ]\frac{1}{n+1},\frac{1}{n}] \text{ and } n \in \mathbb{N}\setminus \{0\}, \\ 0& \text{if } x= 0, \\ -1 &\text{if } x\in A.\end{cases} $$

It is clear that $(A,B)$ is a nonempty bounded closed convex pair in a uniformly convex Banach space $\mathbb{R}$, $\operatorname{cov}(A\cup B)= [-6,1]$, $d=\operatorname{dist}(A,B)= 1$, $\operatorname{diam}(A,B)= 7$, and $TA \subset A$, $TB \subset B$. Consider the sequences $(t_{n})_{\geq 1}$ and $(c_{n})_{\geq 1}$ defined by $t_{n}= \frac{1}{n}$ and $c_{n}= 1+\frac{1}{n}$ for $n\in \mathbb{N}\setminus \{0\}$.

Let $(u,v)\in (\operatorname{cov}(A\cup B))^{2}$ and $n\in \mathbb{N}\setminus \{0\}$ be such that $1 \leq |u-v| \leq t_{n}+3$ (we can have elements u and v such that $3d=3 < |u-v| \leq t_{n}+3$; for example, $x= t_{n}$ and $y= -3$). We have

$$ \bigl\vert T\circ pr_{A}(u)-T\circ pr_{B}(v) \bigr\vert = \textstyle\begin{cases} \vert T(u)-T(0) \vert &\text{if } (u,v) \in A\times (\operatorname{cov}(A\cup B)\setminus B), \\ \vert T(u)-T(v) \vert &\text{if } (u,v) \in A\times B \text{ or } (u,v) \in B \times A, \\ \vert T(-1)-T(0) \vert &\text{if } u \in (\operatorname{cov}(A\cup B)\setminus A) \\ &\text{and } v \in (\operatorname{cov}(A\cup B)\setminus B), \\ \vert T(-1)-T(v) \vert &\text{if } u \in (\operatorname{cov}(A\cup B)\setminus A) \\ & \text{and } v \in B.\end{cases} $$

In the case where $v\in B\setminus \{0\}= \,]0,1]$ and $u\in A$, there exists $m\in \mathbb{N} \setminus \{0\}$ such that $v\in\, ]t_{m+1} ,t_{m}]$,

$$ 1 \leq v-u \leq t_{n}+3 \quad \text{if and only if}\quad 0< v \leq t_{n}+3+u. $$

We must have $-3 \leq u \leq -1$. In particular, for the element $u=-3$, since $1 \leq v-u \leq t_{n}+3$, we have $v \leq t_{n}$, so $t_{m} \leq t_{n}$, and, consequently,

$$ \bigl\vert T\circ pr_{A}(u)-T\circ pr_{B}(v) \bigr\vert = t_{m}+1 \leq t_{n}+ 1 \leq c_{n}(t_{n}+ 1). $$

This inequality is also true for the other cases,

$$ \bigl\vert T\circ pr_{A}(u)-T\circ pr_{B}(v) \bigr\vert = 1 < t_{n}+1 \leq c_{n}(t_{n}+ 1). $$

Hence

$$ \bigl( 1 \leq \vert u-v \vert \leq t_{n}+3 \Rightarrow \bigl\vert T\circ pr_{A}(u)-T \circ pr_{B}(v) \bigr\vert \leq c_{n}(t_{n}+1) \bigr). $$

Since ${\lim_{n\to +\infty}t_{n}= 0 }$ and ${\lim_{n\to +\infty}c_{n}= 1 }$, by Corollary 6 there exists a pair $(x^{*},y^{*})$ in $A \times B$ such that

$$ Tx^{*} =x^{*}, \qquad Ty^{*}=y^{*}, \quad \text{and} \quad \bigl\vert x^{*}-y^{*} \bigr\vert = \operatorname{dist}(A,B) $$

with $x^{*}= -1$ and $y^{*}=0$.

2.2 Some auxiliary results on relatively $(L,d)$-mappings

Definition 1

Let $(A,B)$ be a nonempty pair in a normed space $(X,\|\cdot \|)$, $d:=\operatorname{dist}(A,B)$, and $L \geq \frac{1}{2}$. A mapping $T :A\cup B \rightarrow A\cup B$ is said to be a relatively $(L,d)$-mapping on $A\cup B$ if for all $(x,y)\in A\times B$,

$$ \bigl\Vert T(x) - T(y) \bigr\Vert \leq {L}\bigl( \Vert x-y \Vert +d\bigr). $$

(22)

Proposition 7

Let $(A,B)$ be a nonempty pair in a normed space X. Let $\beta \in {\mathcal {B}}_{L,d}$ with $d=\operatorname{dist}(A,B)>0$ and $L \geq \frac{1}{2}$. Let $T :A\cup B \rightarrow A\cup B$ be a cyclic (resp., noncyclic) mapping satisfying the following condition:

$$ \bigl\Vert T(u)-T(v) \bigr\Vert \leq \beta \bigl( \Vert u-v \Vert \bigr) $$

(23)

for all $(u,v)\in A\times B$. Then T is a relatively $(L,d)$-mapping on $A\cup B$.

Proof

• As ${\liminf_{t\to 0^{+}}\frac{\beta (t+d)}{t+d}= L }$, for every $\varepsilon >0$, there exists $t_{\varepsilon} >0$ such that

  $$\begin{aligned}& \beta (t_{\varepsilon}+d) < (L +\varepsilon ) (t_{\varepsilon}+d) \end{aligned}$$

  (24)
  $$\begin{aligned}& \quad \text{and}\quad \lim_{\varepsilon \to 0^{+}}t_{\varepsilon}=0. \end{aligned}$$

  (25)

  By the monotony of β and inequalities(24) and (25) we have

  $$ \beta (d)\leq \liminf_{\varepsilon \rightarrow 0^{+}}\beta (t_{ \varepsilon}+d) \leq \lim_{\varepsilon \rightarrow 0^{+}}(L + \varepsilon ) (t_{\varepsilon}+d) =Ld . $$

  Thus $\beta (d)\leq Ld$. Moreover, by (ii), $\beta (nd)\leq ndL$ for all $n\in \mathbb{N}\setminus \{0\}$.

• Let $t\in [d,+\infty [$ . Then there is an integer $n \geq 1$ such that

  $$ nd \leq t < (n+1)d \leq t+d, $$

  and then

  $$ \beta (t)\leq \beta \bigl((n+1)d\bigr) \leq (n+1)\beta (d) \leq (n+1)Ld \leq L(t+d). $$

  (26)

• Let $(u,v)\in A\times B$. Then $\|u-v\| \geq d$, and from (26) we have

  $$ \bigl\Vert T(u)-T(v) \bigr\Vert \leq \beta \bigl( \Vert u-v \Vert \bigr) \leq L\bigl( \Vert u-v \Vert +d\bigr). $$

This finishes the proof. □

We give the following simple example of a relatively $(L,d)$-mapping T that is not relatively nonexpansive.

Example 2

Consider the space $\ell ^{p}(\mathbb{R})$, $1\leq p < \infty$, endowed with the norm $\|\cdot \| := \|\cdot \|_{p}$. Let

$$ A= \Biggl\{ x=\lambda .e_{1}+\mu .e_{2}+ \sum _{n=3}^{+\infty} \frac{1}{2^{n}}.e_{n} \in \ell ^{p}(\mathbb{R}) : 0\leq \lambda \leq 2 \text{ and } 1 \leq \mu \leq 2\Biggr\} $$

and

$$ B= \Biggl\{ y= \lambda '.e_{1}+\mu '.e_{2}+ \sum_{n=3}^{+\infty} \frac{1}{2^{n}}.e_{n} \in \ell ^{p}(\mathbb{R}) : 0\leq \lambda ' \leq 2\text{ and } -2\leq \mu ' \leq -1\Biggr\} , $$

where $e_{n}$ the sequence consisting of 1s at the nth place and 0s elsewhere.

For all $(x,y)\in A\times B$,

$$ \Vert x-y \Vert _{p} = \bigl( \bigl(\lambda -\lambda '\bigr)^{p}+ \bigl(\mu -\mu ' \bigr)^{p} \bigr)^{\frac{1}{p}} \geq 2, $$

where $2= \| a-b\|_{p}$ with ${a= e_{2}+ \sum_{n=3}^{+\infty}\frac{1}{2^{n}}\cdot e_{n} \in A }$ and ${b=- e_{2}+ \sum_{n=3}^{+\infty}\frac{1}{2^{n}}\cdot e_{n} \in B}$. Hence $d=\operatorname{dist}(A,B)=2$.

Consider the mapping $T: A\cup B \rightarrow A\cup B$ defined by

$$\begin{aligned}& T(x)= {2.e_{1}}+\frac{\mu +1}{2}\cdot e_{2}+\sum _{n=3}^{+\infty} \frac{1}{2^{n}}\cdot e_{n} \quad \text{if } x\in A,\\& T(y)= \frac{\mu '-1}{2}\cdot e_{2}+\sum _{n=3}^{+\infty} \frac{1}{2^{n}}\cdot e_{n} \quad \text{if } y\in B. \end{aligned}$$

We have $T(A) \subset A$ and $T(B)\subset B$.

• Letting $x\in A$ and $y\in B$,

  $$\begin{aligned} \bigl\Vert T(x)-T(y) \bigr\Vert _{p} & = \biggl( 2^{p} + \biggl(\frac{\mu -\mu '}{2}+ 1\biggr)^{p} \biggr)^{\frac{1}{p}} \\ & \leq \bigl( 2^{p} + \bigl(\mu -\mu ' \bigr)^{p} \bigr)^{\frac{1}{p}} \\ & \leq \bigl( 2^{p} + \Vert x-y \Vert ^{p} \bigr)^{\frac{1}{p}}. \end{aligned}$$

  Hence

  $$ \bigl\Vert T(x)-T(y) \bigr\Vert _{p} \leq \beta \bigl( \Vert x-y \Vert _{p}\bigr) $$

  with $\beta (t)= ( 2^{p} + t^{p} )^{\frac{1}{p}}$ for $t\in [0,+\infty [$ . We have

  $$ \liminf_{t\to 0^{+}}\frac{\beta (t+2)}{t+2}= \lim _{t\to 0^{+}} \frac{ ( 2^{p} + (t+2)^{p} )^{\frac{1}{p}}}{t+2} = 2^{ \frac{1}{p}} > \frac{1}{2}. $$

  Moreover, β satisfies conditions $(i)$ and $(ii)$, and so $\beta \in {\mathcal {B}}_{2^{\frac{1}{p}},2}$.

  By Proposition 7, for all $(x,y)\in A\times B$,

  $$ \bigl\Vert T(x)-T(y) \bigr\Vert _{p} \leq 2^{\frac{1}{p}} \bigl( \Vert x-y \Vert _{p}+\operatorname{dist}(A,B)\bigr). $$

  Then T is a noncyclic relatively $(2^{\frac{1}{p}},2)$-mapping on $A\cup B$.

• The mapping T from the previous example is not relatively nonexpansive on $A\cup B/$ For example, taking ${x= e_{1}+ e_{2}+\sum_{n=3}^{+\infty}\frac{1}{2^{n}} \cdot e_{n}}$ and ${y= -e_{2}+\sum_{n=3}^{+\infty}\frac{1}{2^{n}}\cdot e_{n}}$, we get

  $$\begin{aligned} 2&=\operatorname{dist}(A,B)< \Vert x-y \Vert _{p} = \bigl(1+2^{p}\bigr)^{\frac{1}{p}} \\ & < \bigl\Vert T(x)-T(y) \bigr\Vert _{p} = \bigl(2^{p}+2^{p} \bigr)^{\frac{1}{p}}=2^{1+\frac{1}{p}}. \end{aligned}$$

2.3 An application to functional equations

Let ${\mathcal {U}}$ be a nonempty open subset of $\mathbb{R}^{m}$ such that $\mu ({\mathcal {U}})=1$, where $m\in \mathbb{N}\setminus \{0\}$, and μ is the Lebesgue measure on $\mathbb{R}^{m}$. We denote by $L^{2}({\mathcal {U}})$ the space of measurable functions $f: {\mathcal {U}} \rightarrow \mathbb{R}$ for which $|f|^{2}$ is integrable with respect to μ. We equip $L^{2}({\mathcal {U}})$ with the norm

$$ \Vert f \Vert _{2}= \biggl( \int _{{\mathcal {U}}} \vert f \vert ^{2}\,d\mu \biggr)^{ \frac{1}{2}}. $$

It is known that $(L^{2}({\mathcal {U}}),\|\cdot\|_{2})$ is a uniformly convex Banach space (see Clarkson [4]).

We assume the following conditions:

$(H_{1})$:

Let ${\mathcal {M}}: {\mathcal {U}} \rightarrow {\mathcal {U}}$ be a locally Lipschitzian homeomorphic mapping, and let $\gamma : {\mathcal {U}}\times {\mathcal {U}} \rightarrow [0,+\infty )$ be a measurable Lebesgue function such that

$$ \bigl(\gamma \bigl(x,{\mathcal {M}}(x)\bigr)\bigr)^{2} = \bigl\vert J_{{\mathcal {M}}}(x) \bigr\vert \text{ a.e. in }{\mathcal {U}} \text{ and } \mu \bigl({\mathcal {M}}({ \mathcal {U}})\bigr)=1, $$

(27)

where for ${\mathcal {M}}=({\mathcal {M}}_{1},\dots ,{\mathcal {M}}_{m})$ and $x=(x_{1},\dots ,x_{m})$.

$$ J_{{\mathcal {M}}}(x): = \frac{ \partial ({\mathcal {M}}_{1},\dots ,{\mathcal {M}}_{m})}{\partial (x_{1},\dots ,x_{m})} $$

is the Jacobian of ${\mathcal {M}}$.

$(H_{2})$:

$h,k: {\mathcal {U}} \times \mathbb{R} \rightarrow \mathbb{R}$ are functions such that for all $y\in \mathbb{R}$, the functions $x\mapsto h(x,y)$ and $x\mapsto k(x,y)$ are Lebesgue measurable on ${\mathcal {U}}$, and for almost all $x\in {\mathcal {U}}$ (with respect to μ), the functions $y\mapsto h(x,y)$ and $y\mapsto k(x,y)$ are continuous on $\mathbb{R}$,

$(H_{3})$:

Let $g_{1},g_{2}\in L^{2}({\mathcal {U}})$ be such that $0< g_{1}\leq g_{2}$ a.e. in ${\mathcal {U}}$ and for all $(x,y)\in {\mathcal {U}} \times \mathbb{R}$, we have the following implications:

$$\begin{aligned}& g_{1}\bigl({\mathcal {M}}(x)\bigr) \leq y\leq g_{2}\bigl({\mathcal {M}}(x)\bigr) \Rightarrow \textstyle\begin{cases} g_{1}(x) \leq \gamma (x,{\mathcal {M}}(x))h(x,y) \leq g_{2}(x), \\ g_{1}(x) \leq \gamma (x,{\mathcal {M}}(x))k(x,y) \leq g_{2}(x),\end{cases}\displaystyle \end{aligned}$$

(28)

$$\begin{aligned}& h_{1}\bigl({\mathcal {M}}(x)\bigr) \leq y\leq h_{2}\bigl({\mathcal {M}}(x)\bigr) \Rightarrow \textstyle\begin{cases} h_{1}(x) \leq \gamma (x,{\mathcal {M}}(x))h(x,y) \leq h_{2}(x), \\ h_{1}(x) \leq \gamma (x,{\mathcal {M}}(x))k(x,y) \leq h_{2}(x),\end{cases}\displaystyle \end{aligned}$$

(29)

where $h_{1}= -g_{2}$ and $h_{2}= -g_{1}$

We set

$$\begin{aligned}& A= \bigl\{ \phi \in L^{2}({\mathcal {U}}): g_{1}\leq \phi \leq g_{2} a.e. \text{ in } {\mathcal {U}}\bigr\} \\& \quad \text{and}\quad B= \bigl\{ \phi \in L^{2}({\mathcal {U}}): h_{1}\leq \phi \leq h_{2} a.e. \text{ in } { \mathcal {U}} \bigr\} . \end{aligned}$$

We have $d= \operatorname{dist}(A,B)= \|g_{1}-h_{2}\|_{2} >0$ and $A\cap B=\emptyset$.

We define the maps T and S on $A\cup B$ by

$$ \textstyle\begin{cases} T(\phi )(x)= \gamma (x,{\mathcal {M}}(x)) h(x,\phi ({\mathcal {M}}(x))), \\ S(\phi )(x)= \gamma (x,{\mathcal {M}}(x)) k(x,\phi ({\mathcal {M}}(x))), \end{cases}\displaystyle {\phi \in A\cup B \text{ and } x \in {\mathcal {U}}}. $$

(30)

Theorem 8

Assume that hypotheses $(H_{1})$–$(H_{3})$ hold. Suppose

$$ \sqrt{2}\bigl( \bigl\vert k(x,y_{1}) -k(x,y_{2}) \bigr\vert +d\bigr) \leq \bigl\vert h(x,y_{1}) -h(x,y_{2}) \bigr\vert \leq \beta \bigl( \vert y_{1}-y_{2} \vert \bigr) $$

(31)

for all $x\in {\mathcal {U}}$ and $y_{1},y_{2}\in \mathbb{R}$, where $\beta \in {\mathcal {B}}_{1,d}$, and $t\mapsto [\beta (t^{\frac{1}{2}})]^{2}$ is concave on $[0,+\infty )$. Then there exists $(\phi _{0},\psi _{0}) \in A\times B$ such that

$$ S(\phi _{0})= \phi _{0},\qquad S(\psi _{0})= \psi _{0} \quad \textit{and}\quad \operatorname{dist}(A,B)= \Vert \psi _{0} - \psi _{0}) \Vert _{2}. $$

Proof

First, we verify without difficulty that $(A,B)$ is nonempty bounded closed and convex in the Hilbert space $(L^{2}({\mathcal {U}}),\|\cdot \|_{2})$ equipped with the real scalar product

$$ \langle \phi ,\psi \rangle = \int _{{\mathcal {U}}} \phi (x)\psi (x)\,dx \quad \text{for }\phi ,\psi \in L^{2}({\mathcal {U}}). $$

Take an arbitrary $\phi \in A\cup B$. Then, in view of the Carathéodory theorem [2], conditions $(H_{1})$ and $(H_{2})$ imply that the functions $T(\phi )$ and $S(\phi )$ are Lebesgue measurable.

Note that $T(A)\subseteq A$. Indeed, for $\phi \in A$, we have $g_{1}\leq \phi \leq g_{2}$ a.e. in $\mathcal {U}$, so from implication (28) we have

$$ g_{1}\bigl({\mathcal {M}}(x)\bigr)\leq \phi \bigl({\mathcal {M}}(x) \bigr) \leq g_{2}\bigl({ \mathcal {M}}(x)\bigr) \quad \text{a.e. in } \Omega , $$

and thus, in view of condition $(H_{3})$,

$$ g_{1}(x)\leq \gamma \bigl(x,{\mathcal {M}}(x)\bigr)h\bigl(x,\phi \bigl({\mathcal {M}}(x)\bigr)\bigr) \leq g_{2}(x)\quad \text{a.e. in } { \mathcal {U}} , $$

that is, $T(\phi )\in A$.

Similarly, we justify that $T(B)\subseteq B$, $S(A)\subseteq A$, and $S(B)\subseteq B$ using condition $(H_{3})$ and implications (28) and (29).

• Step 1: Let $(\phi ,\psi )\in A\times B$. Using the assumptions $\mu ({\mathcal {M}}({\mathcal {U}}))=1$, $(\gamma (x,{\mathcal {M}}(x)))^{2} = |J_{{\mathcal {M}}}(x)|$ a.e. in $\mathcal {U}$ (27), and $\beta \in { \mathcal {B}}_{1,d}$, we obtain the following inequalities:

  $$\begin{aligned} \bigl\Vert T(\phi )-T(\psi ) \bigr\Vert _{2}^{2} & = \int _{{\mathcal {U}}} \bigl\vert T(\phi ) (x)-T( \psi ) (x) \bigr\vert ^{2}\,dx \\ & = \int _{{\mathcal {U}}}\bigl(\gamma \bigl(x,{\mathcal {M}}(x)\bigr) \bigr)^{2} \bigl\vert h(x,\phi \bigl({\mathcal {M}}(x)\bigr)-h\bigl(x, \psi \bigl({\mathcal {M}}(x)\bigr)\bigr) \bigr\vert ^{2}\,dx \\ & = \int _{{\mathcal {U}}}\bigl(\gamma (x),{\mathcal {M}}(x) \bigr)^{2} (\beta \bigl( \phi \bigl({\mathcal {M}}(x)\bigr)-\psi \bigl({ \mathcal {M}}(x)|\bigr)\bigr)^{2}\,dx \\ & = \int _{{\mathcal {U}}} \bigl\vert J_{{\mathcal {M}}}(x) \bigr\vert ( \beta \bigl( \bigl\vert \phi \bigl({\mathcal {M}}(x)\bigr)-\psi \bigl({\mathcal {M}}(x) \bigr\vert \bigr)\bigr)^{2}\,dx \\ & = \int _{{\mathcal {M}}({\mathcal {U}})} \bigl(\beta \bigl( \bigl\vert \phi (x)-\psi (x) \bigr\vert \bigr)\bigr)^{2}\,dx \\ & \leq \int _{{\mathcal {U}}} \bigl(\beta \bigl( \bigl\vert \phi (x)-\psi (x) \bigr\vert \bigr)\bigr)^{2}\,dx. \end{aligned}$$

  Since the function $t\mapsto [\beta (t^{\frac{1}{2}})]^{2}$ is concave on $[0,+\infty )$, we have

  $$\begin{aligned} \bigl\Vert T(\phi )-T(\psi ) \bigr\Vert _{2} & \leq \biggl( \int _{{\mathcal {U}}} \bigl[ \beta \bigl( \bigl\vert \phi (x)-\psi (x) \bigr\vert \bigr) \bigr]^{2}\,dx \biggr)^{\frac{1}{2}} \\ & = \biggl( \int _{{\mathcal {U}}} \bigl[\beta \bigl(\bigl( \bigl\vert \phi (x)- \psi (x) \bigr\vert ^{2}\bigr)^{ \frac{1}{2}}\bigr) \bigr]^{2}\,dx \biggr)^{\frac{1}{2}} \\ & \leq \biggl[ \biggl(\beta \biggl( \int _{{\mathcal {U}}} \bigl\vert \phi (x)-\psi (x) \bigr\vert ^{2}\,dx \biggr)^{\frac{1}{2}} \biggr)^{2} \biggr]^{\frac{1}{2}} \\ & \leq \beta \bigl( \Vert \phi -\psi \Vert _{2}\bigr), \end{aligned}$$

  Thus all the assumptions of Proposition 7 are satisfied. Consequently, for $(\phi ,\psi )\in A\times B$,

  $$ \bigl\Vert T(\phi )-T(\psi ) \bigr\Vert _{2} \leq \Vert \phi -\psi \Vert _{2} +d . $$

  (32)

• Step 2: Let $(\phi ,\psi )\in A\times B$. By inequality (31), for each $x\in {\mathcal {U}}$, we have

  $$ 2\bigl( \bigl\vert k(x,\phi \bigl({\mathcal {M}}(x)\bigr) -k\bigl(x,\psi \bigl({ \mathcal {M}}(x)\bigr) \bigr\vert +d\bigr)^{2} \leq \bigl\vert h(x, \phi \bigl({\mathcal {M}}(x)\bigr)-h\bigl(x,\psi \bigl({\mathcal {M}}(x)\bigr)\bigr) \bigr\vert ^{2}, $$

  so that

  $$\begin{aligned}& 2\bigl(\gamma \bigl(x,{\mathcal {M}}(x)\bigr)\bigr)^{2}( \bigl\vert k(x,\phi \bigl({\mathcal {M}}(x)\bigr) -k\bigl(x, \psi \bigl({\mathcal {M}}(x)\bigr) \bigr\vert ^{2} +d^{2} \bigr) \\& \quad \leq \bigl(\gamma \bigl(x,{\mathcal {M}}(x)\bigr)\bigr)^{2}( \bigl\vert h\bigl(x,\phi \bigl({\mathcal {M}}(x)\bigr)-h\bigl(x, \psi \bigl({ \mathcal {M}}(x)\bigr)\bigr) \bigr\vert ^{2}\bigr) . \end{aligned}$$

  Integrating both sides, we get

  $$\begin{aligned}& 2 \int _{{\mathcal {U}}}\bigl(\gamma \bigl(x,{\mathcal {M}}(x)\bigr) \bigr)^{2}( \bigl\vert k(x,\phi \bigl({\mathcal {M}}(x)\bigr) -k \bigl(x,\psi \bigl({\mathcal {M}}(x)\bigr) \bigr\vert ^{2} +d^{2} \bigr)\,dx \\& \quad \leq \int _{{\mathcal {U}}}\bigl(\gamma \bigl(x,{\mathcal {M}}(x)\bigr) \bigr)^{2}( \bigl\vert h\bigl(x,\phi \bigl({\mathcal {M}}(x)\bigr)-h \bigl(x,\psi \bigl({\mathcal {M}}(x)\bigr)\bigr) \bigr\vert ^{2} \bigr)\,dx , \end{aligned}$$

  whence

  $$\begin{aligned}& {2 \biggl( \int _{{\mathcal {U}}} \bigl\vert S(\phi ) (x) -S(\psi ) (x) \bigr\vert ^{2}\,dx +d^{2} \biggr) \leq \int _{{\mathcal {U}}} \bigl\vert T(\phi ) (x) -T(\psi ) (x) \bigr\vert ^{2}\,dx, }\\& \sqrt{2} \sqrt{ \int _{{\mathcal {U}}} \bigl\vert S(\phi ) (x) -S(\psi ) (x) \bigr\vert ^{2}\,dx +d^{2}} \leq \sqrt{ \int _{{\mathcal {U}}} \bigl\vert T(\phi ) (x) -T(\psi ) (x) \bigr\vert ^{2}\,dx }. \end{aligned}$$

Since

$$ {\sqrt{ \int _{{\mathcal {U}}} \bigl\vert S(\phi ) (x) -S(\psi ) (x) \bigr\vert ^{2}\,dx} +d \leq \sqrt{2} \sqrt{ \int _{{\mathcal {U}}} \bigl\vert S(\phi ) (x) -S(\psi ) (x) \bigr\vert ^{2}\,dx +d^{2}} }, $$

we have

$$ \bigl\Vert S(\phi ) -S(\psi ) \bigr\Vert _{2} +d \leq \bigl\Vert T(\phi ) -T(\psi ) \bigr\Vert _{2} $$

and, according to inequality (32),

$$ \bigl\Vert S(\phi ) -S(\psi ) \bigr\Vert _{2} +d \leq \bigl\Vert T(\phi ) -T(\psi ) \bigr\Vert _{2} \leq \Vert \phi -\psi \Vert _{2} +d . $$

Hence, for all $(\phi ,\psi )\in A\times B$,

$$ \bigl\Vert S(\phi ) -S(\psi ) \bigr\Vert _{2} \leq \Vert \phi -\psi \Vert _{2}. $$

Thus S is relatively nonexpansive on $A\cup B$. The hypotheses of the result of Eldred et al. [5] for a noncyclic mapping hold for S, so there exists $(\phi _{0},\psi _{0}) \in A\times B$ such that

$$ S(\phi _{0})= \phi _{0},\qquad S(\psi _{0})= \psi _{0}, \quad \text{and}\quad \operatorname{dist}(A,B)= \Vert \phi _{0} - \psi _{0}) \Vert _{2}, $$

and necessarily $\phi _{0}= g_{1}$ and $\psi _{0}= h_{2}$. □