On Coupled Best Proximity Points in Reflexive Banach Spaces

Abstract

We investigated the existence and uniqueness of coupled best proximity points for some cyclic and semi-cyclic maps in a reflexive Banach space. We found sufficient conditions, ensuring the existence of coupled best proximity points in reflexive Banach spaces and some convexity types of conditions, ensuring uniqueness of the coupled best proximity points in strictly convex Banach spaces. We illustrate the results with examples and we present an application of one of the theorems in the modeling of duopoly markets, to have an existence of market equilibrium. We show that, in general, the iterative sequences can have chaotic behavior.

1. Introduction

The Banach contraction principle is a structural result in the fixed point theory. Cyclic contraction maps were introduced in [1]; the authors obtained existence and uniqueness of fixed points for the considered maps. It was believed that the maps introduced in [1] were generalizations of the self-map contractions; many new contractive types of conditions for cyclic maps have been introduced. Results in [2] show that these extensions (results) concerning the establishment of cyclic maps are equivalent with well known ordinary fixed point results in the literature.

For mappings, T defined on subsets of metric spaces or normed spaces, the fixed point theory is a useful tool for solving equations $Tx=x$. Because a non-self mapping $T:A\to B$ does not always have a fixed point, it is common to try to identify one, x–closest to $Tx$. In this context, the best proximity point theorems are relevant. The best proximity point is a concept that has been around for a long time, introduced in [3]; it gives one possible solution of the problem in search of an element x, which is, in some sense, closest to $Tx$. In uniformly convex Banach spaces, a sufficient condition for existence and uniqueness of the best proximity points is presented [3] for 2 sets and in [4] for p sets.

A built model may be dependent on two parameters, namely $F:X\times X\to X$. In this context, the notions of coupled fixed points [5] and coupled best proximity points for an ordered pair $(F,G)$, $F:A\times A\to B$, $G:B\times B\to A$, where $A,B\subset X$ [6,7] are relevant. Deep results in the theory of coupled fixed points, for example, can be found in [2,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23].

It seems that the classical models of best proximity points reduce to models where the coupled best proximity point $(x,y)$ satisfies $x=y$. This drawback has been solved for couple (tripled) fixed (or best proximity) points in [24,25].

A disadvantage of the classical theory of best proximity points in Banach spaces is the requirement that the Banach space should be uniformly convex [3]. This restriction was overcome in [26,27], where the underlying space was just a reflexive Banach space. There are applications in the theory of market equilibrium in the non-competitive market [28,29,30], where the underlying space is a uniformly convex Banach space. The profit of the participants in a market, selling a finite number of goods, is usually the ${\parallel ·\parallel }_1$ norm, i.e., ${\sum }_{i=1}^n\left|x_i\right|$, rather than ${\left({\sum }_{i=1}^n{\left|x_i\right|}^p\right)}^{1/p}$ for $p>1$. As far as the economic models usually taking place in finite dimensional Banach spaces, it seems relevant to stick to reflexive spaces instead of uniformly convex ones.

We tried to show that existence results, concerning a coupled best proximity point in reflexive Banach spaces, can be found; but for uniqueness, some additional conditions should be imposed. We illustrate (with examples) that the iterative sequences can have a chaotic nature.

We will apply the technique from [24,25,26] to generalize the results on the coupled best proximity points in reflexive Banach spaces and to present applications in the investigation of market equilibrium for noncompetitive markets.

2. Materials and Methods

We will recall the same (well known) notions and facts, as well as some recently defined notions and results, which will be used in the sequel.

Definition 1.

Let A, B be nonempty subsets of the metric spaces $(X,\rho )$. A distance between the sets A and B we call $dist(A,B)=inf\{\rho (a,b):a\in A,b\in B\}$.

We denote that $d=\mathrm{dist}(\mathrm{A},\mathrm{B})$ and $d_x=\mathrm{dist}({\mathrm{A}}_{\mathrm{x}},{\mathrm{B}}_{\mathrm{x}})$, as far as no confusion arises, to fit the formulas into the text field.

Besides the best proximity points for cyclic maps, the best proximity points were investigated in [31] for maps later named non-cyclic [32], i.e., $F(A)\subseteq A$ and $F(B)\subseteq B$.

We will consider two models.

Definition 2.

Let $(X,\rho )$ be a metric space, $A_x,A_y,B_x,B_y\subset X$ be non-empty, proper subsets, $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$. An ordered pair $(x,y)\in A_x\times A_y$ is called a coupled best proximity point of $(F,f)$ in $A_x\times A_y$ if $\rho (x,F(x,y))=\mathrm{dist}(A_x,B_x)$ and $\rho (y,f(x,y))=\mathrm{dist}(A_y,B_y)$.

Definition 3.

Let $A_x$, $A_y$, $B_x$, and $B_y$ be non-empty subsets of a metric space $(X,\rho )$, $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$, $G:B_x\times B_y\to A_x$ and $g:B_x\times B_y\to A_y$. The ordered pair of ordered pairs $((F,f),(G,g))$ is said to be a cyclic contraction ordered pair if there are reals $\alpha ,\beta ,\gamma ,\delta \ge 0$, so that $max\{\alpha +\gamma ,\beta +\delta \}<1$ and there holds the inequality

$$\begin{array}{ccc}{S}_1\hfill & =\hfill & \rho (F(x,y),G(u,v))+\rho (f(z,w),g(t,s))\hfill \\ \hfill & \le \hfill & \alpha \rho (x,u)+\beta \rho (y,v)+\gamma \rho (z,t)+\delta \rho (w,s)\hfill \\ \hfill & \hfill & +(1-(\alpha +\gamma ))\mathrm{dist}(A_x,B_x)+(1-(\beta +\delta ))\mathrm{dist}(A_y,B_y)\hfill \end{array}$$

(1)

for any $(x,y),(z,w)\in A_x\times A_y$ and $(u,v),(t,s)\in B_x\times B_y$.

Definition 4.

Let $A_x$, $A_y$, $B_x$ and $B_y$ be nonempty subsets of X. Let $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$, $G:B_x\times B_y\to A_x$ and $g:B_x\times B_y\to A_y$. For any initial, arbitrary chosen $(\xi ,\eta )\in A_x\times A_y$, we define the sequences ${\left\{{\xi }_n\right\}}_{n=0}^{+\infty }$ and ${\left\{{\eta }_n\right\}}_{n=0}^{+\infty }$ by ${\xi }_0=\xi$, ${\eta }_0=\eta$ and

$$\begin{array}{cc}{\xi }_{2n+1}=F({\xi }_{2n},{\eta }_{2n}),\hfill & {\eta }_{2n+1}=f({\xi }_{2n},{\eta }_{2n})\hfill \\ {\xi }_{2n+2}=G({\xi }_{2n+1},{\eta }_{2n+1}),\hfill & {\eta }_{2n+2}=g({\xi }_{2n+1},{\eta }_{2n+1})\hfill \end{array}$$

for all $n\ge 0$.

The maps in Definition 5 were introduced in [28], where they were called cyclic maps, which does not seem to be very precise. We suggest ’calling’ the maps defined in Definition 5; semi-cyclic maps.

Definition 5.

Let $(X,\rho )$ be a metric space, $A,B\subset X$ be non empty, proper subsets, and $F:A\times B\to A$ and $G:A\times B\to B$. We will call the ordered pair of maps $(F,G)$ a semi-cyclic map. An ordered pair $(\xi ,\eta )\in A\times B$ is called a coupled best proximity point of the semi-cyclic map $(F,G)$ in $A\times B$ if $\rho (\xi ,G(\xi ,\eta ))=\rho (\eta ,F(\xi ,\eta ))=d$, where $d=\mathrm{dist}(\mathrm{A},\mathrm{B})$.

We give two definitions of the best proximity points. There will be no confusion because the best proximity point notion initiated in [3] depends on the map (or the ordered pair of maps in the research), rather than just on the sets. In Definition 2, the maps $(F,f)$ generating the best proximity points $(x,y)$ are defined with the help of $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$; in Definition 5 maps $(F,G)$, generating the best proximity points $(x,y)$ are defined with the help of $F:A\times B\to A$, $G:A\times B\to B$.

Definition 6.

Let $A,B$ be subsets of a metric space $(X,\rho )$. An ordered pair of maps $(F,G)$ $F:A\times B\to A$, $G:A\times B\to B$ will be called a semi-cyclic contraction if there exists $\alpha \in [0,1/2)$, such that for any $(x,y),(u,v)\in A\times B$ there holds the inequality

$$\rho (F(x,y),G(u,v))-\mathrm{dist}(\mathrm{A},\mathrm{B})\le \alpha (\rho (x,v)+\rho (y,u)-2\mathrm{dist}(\mathrm{A},\mathrm{B})).$$

(2)

Definition 7.

Let A, B be nonempty subsets of the metric spaces $(X,\rho )$ and $(F,G)$ be a semi-cyclic map, (i.e., $F:A\times B\to A$ and $G:A\times B\to B$). For any initial, arbitrary chosen $(\xi ,\eta )\in A\times B$, we define the sequences ${\left\{{\xi }_n\right\}}_{n=0}^{+\infty }$ and ${\left\{{\eta }_n\right\}}_{n=0}^{+\infty }$ by ${\xi }_0=\xi$, ${\eta }_0=\eta$ and ${\xi }_{n+1}=F({\xi }_n,{\eta }_n)$, ${\eta }_{n+1}=G({\xi }_n,{\eta }_n)$ for all $n\ge 0$.

As far as the two models are easily distinguished, the first one consists of four maps and the second one of two; there will be no misunderstanding. When we consider the sequences ${\left\{{\xi }_n\right\}}_{n=0}^{+\infty }$ and ${\left\{{\eta }_n\right\}}_{n=0}^{+\infty }$ we will assume that they are the sequences defined in Definition 4 or Definition 7.

Let us recall some facts about strict convexity in Banach spaces and reflexivity of Banach spaces.

Definition 8

([33], p. 42). A Banach space $(X,\parallel ·\parallel )$ is said to be strictly convex if $x=y$, provided $x,y\in X$ are such that $\parallel x\parallel =\parallel y\parallel =1$ and $\parallel x+y\parallel =2$.

Lemma 1

([34]). Let $(X,\parallel ·\parallel )$ be a strictly convex Banach space, $A,B\in X$ be closed and proper subsets of X, such that $\mathrm{dist}(A,B)>0$ and A be convex. If $x,z\in A$ and $y\in B$ be such that $\parallel y-x\parallel =\parallel y-z\parallel =\mathrm{dist}(A,B)$, then $x=z$.

Definition 9

([35], p. 125). Let $(X,\parallel ·\parallel )$ be a normed space. The canonical embedding $\pi :X\to X^{**}$ is defined for $x\in X$ by $\pi (x):f\to f(x)$ for all $f\in X^{*}$.

Definition 10

([35], p. 126). A Banach space $(X,\parallel ·\parallel )$ is called reflexive whenever the mapping $\pi :X\to X^{**}$ maps X onto $X^{**}$.

Lemma 2

([35], p. 176). Let ${\left\{x_n\right\}}_{n=1}^{+\infty }$ be a bounded sequence in a reflexive Banach space. Then there exists a weakly convergent subsequence ${\left\{x_{n_k}\right\}}_{k=1}^{+\infty }$ of ${\left\{x_n\right\}}_{n=1}^{+\infty }$.

Lemma 3

([35], p. 176). A Banach space X is reflexive if and only if any $f\in X^{*}$ attains its norm, i.e., for any $x\in X$ there is $f\in X^{*}$, so that $f(x)=\parallel x\parallel$.

3. Auxiliary Results

We will prove some auxiliary results to make the readings of the proofs of theorems in the next section easier. Let us recall that we will use the notation $d=\mathrm{dist}(\mathrm{A},\mathrm{B})$ just to fit formulas into the text field.

Lemma 4.

Let $(X,\rho )$ be a metric space, $A,B\subset X$ be nonempty, proper subsets of X and $F:A\times B\to A$, $G:A\times B\to B$ be a semi-cyclic contraction. Then

$$\rho (x_{n+1},y_{m+1})+\rho (x_{m+1},y_{n+1})-2d\le {(2\alpha )}^n(\rho (x_1,y_{m-n+1})+\rho (x_{m-n+1},y_1)-2d).$$

holds for any $n,m\in \mathbb{N}$, $m\ge n$.

Proof. 

By applying n–times (2), we have the inequality

$$\begin{array}{ccc}{S}_2\hfill & =\hfill & \rho (x_{n+1},y_{m+1})+\rho (x_{m+1},y_{n+1})-2d\hfill \\ & =\hfill & \rho (F(x_n,y_n),G(x_m,y_m))-d+\rho (F(x_m,y_m),G(x_n,y_n))-d\hfill \\ & \le \hfill & 2\alpha (\rho (x_n,y_m)+\rho (x_m,y_n)-2d)\hfill \\ & \cdots \hfill & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \hfill \\ & \le \hfill & {(2\alpha )}^n(\rho (x_1,y_{m-n+1})+\rho (x_{m-n+1},y_1)-2d).\hfill \end{array}$$

□

If $m=n$, we have the inequality

$${\displaystyle \rho (x_{n+1},y_{n+1})-d=\frac{\rho (x_{n+1},y_{n+1})+\rho (x_{n+1},y_{n+1})-2d}{2}\le {(2\alpha )}^n(\rho (x_1,y_1)-d).}$$

(3)

If $m=n+1$, we have the inequality

$$\rho (x_{n+1},y_{n+2})+\rho (x_{n+2},y_{n+1})-2d\le {(2\alpha )}^n(\rho (x_1,y_2)+\rho (x_2,y_1)-2d).$$

(4)

Lemma 5.

Let $(X,\rho )$ be a metric space, $A,B\subset X$ be nonempty, proper subsets of X and $F:A\times B\to A$, $G:A\times B\to B$ be a semi-cyclic contraction. Then

$$\underset{n\to +\infty }{lim}\rho (x_n,y_n)=\underset{n\to +\infty }{lim}\rho (x_{n+1},y_n)=\underset{n\to +\infty }{lim}\rho (x_n,y_{n+1})=\mathrm{dist}(A,B).$$

Proof. 

From (3), we have

$$\underset{n\to +\infty }{lim}\rho (x_{n+1},y_{n+1})-\mathrm{dist}(A,B)\le \underset{n\to +\infty }{lim}{(2\alpha )}^n(\rho (x_1,y_1)-\mathrm{dist}(A,B))=0.$$

From (4), we have

$$\underset{n\to +\infty }{lim}(\rho (x_{n+1},y_{n+2})+\rho (x_{n+2},y_{n+1})-2\mathrm{dist}(A,B))=0$$

and from $\rho (x_{n+1},y_{n+2})\ge d$, $\rho (x_{n+2},y_{n+1})\ge d$ it follows that

$$\underset{n\to +\infty }{lim}\rho (x_{n+1},y_n)=\underset{n\to +\infty }{lim}\rho (x_n,y_{n+1})=\mathrm{dist}(A,B).$$

□

It is easy to observe that

$$\begin{array}{ccc}\rho (x_2,y_n)-d\hfill & =\hfill & \rho (F(x_1,y_1),G(x_{n-1},y_{n-1}))-d\hfill \\ & \le \hfill & \alpha (\rho ((x_1,y_{n-1})+\rho ((y_1,x_{n-1})-2d)\hfill \end{array}$$

and

$$\begin{array}{ccc}\rho (x_n,y_2)-d\hfill & =\hfill & \rho (F(x_{n-1},y_{n-1}),G(x_1,y_1))-d\hfill \\ & \le \hfill & \alpha (\rho ((x_1,y_{n-1})+\rho ((y_1,x_{n-1})-2d).\hfill \end{array}$$

Therefore, it holds

$$\rho (x_2,y_n)+\rho (x_n,y_2)-2d\le 2\alpha (\rho ((x_1,y_{n-1})+\rho ((y_1,x_{n-1})-2d).$$

(5)

Lemma 6.

Let $(X,\rho )$ be a metric space, $A,B\subset X$ be nonempty, proper subsets of X and $F:A\times B\to A$, $G:A\times B\to B$ be a semi-cyclic contraction. Then

$$\rho (x_1,y_n)+\rho (y_1,x_n)-2d\le \frac{1}{1-2\alpha }(\rho (x_1,x_2)+\rho (y_1,y_2))+2{(2\alpha )}^{n-1}(\rho (x_1,y_1)-d).$$

(6)

Proof. 

Using the triangular inequality and applying (5) we have the chain of inequalities

$$\begin{array}{ccc}{S}_3\hfill & =\hfill & \rho (x_1,y_n)+\rho (y_1,x_n)-2d\hfill \\ & \le \hfill & \rho (x_1,x_2)+\rho (x_2,y_n)+\rho (y_1,y_2)+\rho (y_2,x_n)-2d\hfill \\ \hfill & =\hfill & \rho (x_1,x_2)+\rho (y_1,y_2)+\rho (x_2,y_n)+\rho (y_2,x_n)-2d\hfill \\ & \le \hfill & \rho (x_1,x_2)+\rho (y_1,y_2)+2\alpha (\rho (x_1,y_{n-1})+\rho (y_1,x_{n-1})-2d)\hfill \\ & \le \hfill & (1+2\alpha )(\rho (x_1,x_2)+\rho (y_1,y_2))+{(2\alpha )}^2(\rho (x_1,y_{n-2})+\rho (y_1,x_{n-2})-2d)\hfill \\ & \le \hfill & (1+2\alpha +\cdots +{(2\alpha )}^{n-2})(\rho (x_1,x_2)+\rho (y_1,y_2))+2{(2\alpha )}^{n-1}(\rho (x_1,y_1)-d)\hfill \\ & \le \hfill & \frac{1}{1-2\alpha }(\rho (x_1,x_2)+\rho (y_1,y_2))+2{(2\alpha )}^{n-1}(\rho (x_1,y_1)-d).\hfill \end{array}$$

□

Lemma 7.

Let $(X,\rho )$ be a metric space, $A,B\subset X$ be nonempty, proper subsets of X and $F:A\times B\to A$, $G:A\times B\to B$ be a semi-cyclic contraction. Then the sequences ${\left\{x_n\right\}}_{n=1}^{+\infty }$ and ${\left\{y_n\right\}}_{n=1}^{+\infty }$ are bounded sequences.

Proof. 

From Lemma 6, it follows that

$$max\{\rho (x_1,y_n),\rho (y_1,x_n)\}\le \frac{\rho (x_1,x_2)+\rho (y_1,x_2)}{1-\alpha }+{(2\alpha )}^{n-1}(\rho (x_1,y_1)-d)+2d$$

and, therefore, the sequences ${\left\{x_n\right\}}_{n=1}^{+\infty }$ and ${\left\{y_n\right\}}_{n=1}^{+\infty }$ are bounded sequences. □

If $u=x$ and $v=y$ we have the inequality

$$\rho (F(x,y),G(x,y))-d\le 2\alpha (\rho (x,y)-d).$$

If $\mathrm{dist}(A,B)=0$, then the ordered pair $(F,G)$ satisfies the inequality

$$\rho (F(x,y),G(u,v))\le \alpha (\rho (x,v)+\rho (y,u)).$$

It is easy to observe the all lemmas hold true and for $\mathrm{dist}(A,B)=0$.

Lemma 8

([24]). Let $(X,\rho )$ be a metric space, $A_x,A_y,B_x,B_y\subset X$ be nonempty, proper subsets of X, $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$, $G:B_x\times B_y\to A_x$ and $g:B_x\times B_y\to A_y$. Let $((F,f),(G,g))$ be a cyclic contraction ordered pairs. Then there holds ${lim}_{n\to +\infty }\rho (x_n,x_{n+1})=d_x$ and ${lim}_{n\to +\infty }\rho (y_n,y_{n+1})=d_y$ for arbitrary chosen $(x,y)\in A_x\times A_y$.

Lemma 9

([24]). Let $(X,\rho )$ be a metric space, $A_x,A_y,B_x,B_y\subset X$ be nonempty, proper subsets of X, $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$, $G:B_x\times B_y\to A_x$ and $g:B_x\times B_y\to A_y$. Let $((F,f),(G,g))$ be a cyclic contraction ordered pairs. For any arbitrary chosen $(x,y)\in A_x\times A_y$ the sequences ${\left\{x_{2n}\right\}}_{n=0}^{+\infty }$, ${\left\{x_{2n+1}\right\}}_{n=0}^{+\infty }$, ${\left\{y_{2n}\right\}}_{n=0}^{+\infty }$ and ${\left\{y_{2n+1}\right\}}_{n=0}^{+\infty }$ are bounded.

4. Results

We will give a variant of well known notions as weakly continuous functions and a function that satisfies the proximal property. Since the considered model in the paper differs from classical models we cannot refer to classical definitions.

Definition 11.

Let $(X,\parallel ·\parallel )$ be a Banach space and $U,V\subset X$ be subsets. We say that a map $F:U\times V\to X$ is weakly continuous if for any weakly convergent sequences ${\left\{u_n\right\}}_{n=0}^{+\infty }$ and ${\left\{v_n\right\}}_{n=0}^{+\infty }$, $u_n\in U$ and $v_n\in V$ there holds

$$w\underset{n\to \infty }{lim}F(u_n,v_n)=F(u,v),$$

where $u=w{lim}_{n\to +\infty }{u}_n$ and $v=w{lim}_{n\to +\infty }{v}_n$.

Definition 12.

Let $(X,\parallel ·\parallel )$ be a Banach space and $A_x,A_y,B_x,B_y\subset X$ be subsets and $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$. We say that the pair of maps $(F,f)$ satisfies the proximal property if for any weakly convergent sequences ${\left\{u_n\right\}}_{n=0}^{+\infty }$ and ${\left\{v_n\right\}}_{n=0}^{+\infty }$, $u_n\in A_x$ and $v_n\in A_y$ such that $u=w{lim}_{n\to +\infty }{u}_n$, $v=w{lim}_{n\to +\infty }{v}_n$ whenever there hold

$$\underset{n\to +\infty }{lim}\parallel u_n-F(u_n,v_n)\parallel =\mathrm{dist}(A_x,B_x)$$

and

$$\underset{n\to +\infty }{lim}\parallel v_n-f(u_n,v_n)\parallel =\mathrm{dist}(A_y,B_y),$$

then hold $\parallel u-F(u,v)\parallel =\mathrm{dist}(A_x,B_x)$ and $\parallel v-f(u,v)\parallel =\mathrm{dist}(A_y,B_y)$.

Theorem 1.

Let $(X,\parallel ·\parallel )$ be a reflexive Banach space, $A_x,A_y,B_x,B_y\subset X$ be nonempty, proper, weakly closed sets of X, $F:A_x\times A_y\to B_x$, $f:A_x\times A_y\to B_y$, $G:B_x\times B_y\to A_x$ and $g:B_x\times B_y\to A_y$. Let $((F,f),(G,g))$ be a cyclic contraction of ordered pairs. Let there hold one of the following

1.

F and f be weakly continuous on $A_x\times A_y$ and G and g be weakly continuous on $B_x\times B_y$

2.

$(F,f)$ satisfies the proximal property.

Then $(F,f)$ has a coupled best proximity point $({\xi }_x,{\xi }_y)\in A_x\times A_y$ and $(G,g)$ has a coupled best proximity point $({\eta }_x,{\eta }_y)\in B_x\times B_y$, (i.e., $\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel =d_x$, $\parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel =d_y$ and $\parallel {\eta }_x-G({\eta }_x,{\eta }_y)\parallel =d_x$, $\parallel {\eta }_y-g({\eta }_x,{\eta }_y)\parallel =d_y$).

If, in addition $(X,\parallel ·\parallel )$ is a strictly convex Banach space and $A_x,A_y,B_x,B_y\subset X$ are convex subsets, then

$${\xi }_x=G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y)),{\xi }_y=g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y)),$$

and

$${\eta }_x=F(G({\eta }_x,{\eta }_y),g({\eta }_x,{\eta }_y)),{\eta }_y=f(F({\eta }_x,{\eta }_y),f({\eta }_x,{\eta }_y)).$$

If $A_x=A_y$, $B_x=B_y$, $f(x,y)=F(y,x)$ and $g(x,y)=G(y,x)$, $(X,\parallel ·\parallel )$ is a strictly convex Banach space then the coupled fixed point $({\xi }_x,{\xi }_y)\in A_x\times A_x$ satisfies ${\xi }_x={\xi }_y$.

Proof. 

To start the iterative process, we chose an arbitrary initial point $(x,y)\in A_x\times A_y$. It follows from Lemma 9 that the sequences ${\left\{x_{2n}\right\}}_{n=0}^{+\infty }$, ${\left\{x_{2n+1}\right\}}_{n=0}^{+\infty }$, ${\left\{y_{2n}\right\}}_{n=0}^{+\infty }$ and ${\left\{y_{2n+1}\right\}}_{n=0}^{+\infty }$ are bounded sequences. From Lemma 2, it follows that we can choose a subsequence of naturals ${\left\{n_j\right\}}_{j=1}^{+\infty }$, such that the sequences ${\left\{x_{2n_j}\right\}}_{j=0}^{+\infty }$, ${\left\{x_{2n_j+1}\right\}}_{n=0}^{+\infty }$, ${\left\{y_{2n_j}\right\}}_{n=0}^{+\infty }$ and ${\left\{y_{2n_j+1}\right\}}_{n=0}^{+\infty }$ are weakly convergent. From the assumption that the sets $A_x$, $A_y$, $B_x$, and $B_y$ are weakly closed, it follows that $w{lim}_{j\to +\infty }{x}_{2n_j}={\xi }_x\in A_x$, $w{lim}_{j\to +\infty }{x}_{2n_j+1}={\eta }_x\in B_x$, $w{lim}_{j\to +\infty }{y}_{2n_j}={\xi }_y\in A_y$ and $w{lim}_{j\to +\infty }{y}_{2n_j+1}={\eta }_y\in B_y$.

(1) Let $F,f$, and $G,g$ be weakly continuous on $A_x\times A_y$ and $B_x\times B_y$, respectively. Then there holds

$$w\underset{j\to +\infty }{lim}G(x_{2n_j+1},y_{2n_j+1})=G({\eta }_x,{\eta }_y),w\underset{j\to +\infty }{lim}g(x_{2n_j+1},y_{2n_j+1})=g({\eta }_x,{\eta }_y).$$

and

$$w\underset{j\to +\infty }{lim}F(x_{2n_j},y_{2n_j})=F({\xi }_x,{\xi }_y),w\underset{j\to +\infty }{lim}f(x_{2n_j},y_{2n_j})=f({\xi }_x,{\xi }_y).$$

Consequently,

$${\xi }_x-{\eta }_x=w\underset{j\to +\infty }{lim}(x_{2n_j}-x_{2n_j+1})=w\underset{j\to +\infty }{lim}(x_{2n_j}-F(x_{2n_j},y_{2n_j}))={\xi }_x-F({\xi }_x,{\eta }_x).$$

From the reflexivity of $(X,\parallel ·\parallel )$, it follows the existence of a bounded linear functional $h\in S_{X^{*}}$, such that

$$h({\xi }_x-F({\xi }_x,{\xi }_y))=\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel .$$

Using that $h\in S_{X^{*}}$ and that ${\xi }_x$ and ${\xi }_y$ are a weak limit of the sequences ${\left\{x_{2n_j}\right\}}_{j=1}^{+\infty }$ and ${\left\{x_{2n_j}\right\}}_{j=1}^{+\infty }$, respectively, we have

$$\left|h(x_{2n_j}-F(x_{2n_j},y_{2n_j}))\right|\le \parallel h\parallel .\parallel x_{2n_j}-F(x_{2n_j},y_{2n_j})\parallel =\parallel x_{2n_j}-F(x_{2n_j},y_{2n_j})\parallel$$

(7)

and

$$\underset{j\to +\infty }{lim}h(x_{2n_j}-F(x_{2n_j},y_{2n_j}))=h({\xi }_x-F({\xi }_x,{\xi }_y))=\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel .$$

(8)

Using consecutively (8), (7) and Lemma 8 (${lim}_{n\to +\infty }\parallel x_n-x_{n+1}\parallel =\mathrm{dist}(A_x,B_x)=d_x$) we obtain the inequality

$$\begin{array}{ccc}{d}_x\le \parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel \hfill & =\hfill & |h({\xi }_x-F({\xi }_x,{\xi }_y))|=\underset{j\to +\infty }{lim}\left|h(x_{2n_j}-F(x_{2n_j},y_{2n_j}))\right|\hfill \\ & \le \hfill & \underset{j\to +\infty }{lim}\parallel x_{2n_j}-F(x_{2n_j},y_{2n_j})\parallel \hfill \\ & =\hfill & \underset{j\to +\infty }{lim}\parallel x_{2n_j}-x_{2n_j+1}\parallel =\mathrm{dist}(A_x,B_x).\hfill \end{array}$$

Therefore, $\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel =\mathrm{dist}(A_x,B_x)$.

From the reflexivity of $(X,\parallel ·\parallel )$, it follows the existence of a bounded linear functional $s\in S_{X^{*}}$, such that

$$s({\xi }_y-f({\xi }_x,{\xi }_y))=\parallel {\xi }_y-F({\xi }_x,{\xi }_y)\parallel .$$

By similar arguments from the inequalities

$$\left|s(y_{2n_j}-f(x_{2n_j},y_{2n_j}))\right|\le \parallel s\parallel .\parallel y_{2n_j}-f(x_{2n_j},y_{2n_j})\parallel =\parallel y_{2n_j}-f(x_{2n_j},y_{2n_j})\parallel ,$$

$$\underset{j\to +\infty }{lim}s(y_{2n_j}-f(x_{2n_j},y_{2n_j}))=s({\xi }_y-f({\xi }_x,{\xi }_y))=\parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel$$

and Lemma 8 (${lim}_{n\to +\infty }\parallel y_n-y_{n+1}\parallel =\mathrm{dist}(A_y,B_y)=d_y$), we obtain the inequality

$$\begin{array}{ccc}{d}_y\le \parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel \hfill & =\hfill & |s({\xi }_y-f({\xi }_x,{\xi }_y))|=\underset{j\to +\infty }{lim}\left|s(y_{2n_j}-f(x_{2n_j},y_{2n_j}))\right|\hfill \\ & \le \hfill & \underset{j\to +\infty }{lim}\parallel y_{2n_j}-f(x_{2n_j},y_{2n_j})\parallel \hfill \\ & =\hfill & \underset{j\to +\infty }{lim}\parallel y_{2n_j}-y_{2n_j+1}\parallel =\mathrm{dist}(A_y,B_y).\hfill \end{array}$$

Therefore, $\parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel =\mathrm{dist}(A_y,B_y)$.

Thus $({\xi }_x,{\xi }_y)$ is a coupled best proximity point of $(F,f)$.

(2) Let $(F,f)$ satisfy the proximal property. From Lemma 5, we have

$$\underset{j\to +\infty }{lim}\parallel x_{2n_j}-F(x_{2n_j},y_{2n_j})\parallel =\mathrm{dist}(A_x,B_x)$$

and

$$\underset{j\to +\infty }{lim}\parallel y_{2n_j}-f(x_{2n_j},y_{2n_j})\parallel =\mathrm{dist}(A_y,B_y).$$

By the assumption that $(F,f)$ satisfies the proximal property, it follows that

$$\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel =\mathrm{dist}(A_x,B_x)$$

and

$$\parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel =\mathrm{dist}(A_y,B_y)$$

and thus $({\xi }_x,{\xi }_y)$ is a best proximity point of $(F,f)$.

Let us assume in addition that $(X,\parallel ·\parallel )$ be a strictly convex Banach space. From the inequalities, assuming that $({\xi }_x,{\xi }_y)\in A_x\times A_y$ be a coupled best proximity point of the ordered pair of maps $(F,f)$, we have

$$\begin{array}{ccc}{S}_4\hfill & =\hfill & \parallel G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-F({\xi }_x,{\xi }_y)\parallel +\parallel g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-f({\xi }_x,{\xi }_y)\parallel \hfill \\ \hfill & \le \hfill & \alpha \parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel +\beta \parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel +\gamma \parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel \hfill \\ \hfill & \hfill & +\delta \parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel +(1-(\alpha +\gamma ))d_x+(1-(\beta +\delta ))d_y=d_x+d_y.\hfill \end{array}$$

(9)

From the inequalities,

$$d_x\le \parallel G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-F({\xi }_x,{\xi }_y)\parallel ,d_y\le \parallel g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-f({\xi }_x,{\xi }_y)\parallel$$

and (9) it follows that $\parallel G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-F({\xi }_x,{\xi }_y)\parallel =d_x$, $\parallel g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-f({\xi }_x,{\xi }_y)\parallel =d_y$. From the assumption that $({\xi }_x,{\xi }_y)$ is a coupled best proximity pair of $(F,f)$ i.e., $\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel =d_x$, $\parallel {\xi }_y-f({\xi }_x,{\xi }_y)\parallel =d_y$, the strict convexity of $(X,\parallel ·\parallel )$ and Lemma 1, it follows that $G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))={\xi }_x$, $g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))={\xi }_y$.

The proof that $g(F({\xi }_x,{\xi }_y),f({\xi }_y,{\xi }_x))={\eta }_x$ can be done in a similar fashion.

Let $(X,\parallel ·\parallel )$ be a strictly convex Banach space, $({\xi }_x,{\xi }_y)$ be a coupled best proximity point of $(F,f)$, where $f(x,y)=F(y,x)$ and let us denote $A=A_x=A_y$ and $B=B_x=B_y$. We have just proven that ${\xi }_x=G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y)$ and ${\xi }_y=g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))$. Therefore, we have

$$\begin{array}{ccc}{S}_5\hfill & =\hfill & \parallel {\xi }_y-F({\xi }_x,{\xi }_y)\parallel +\parallel {\xi }_x-f({\xi }_x,{\xi }_y)\parallel \hfill \\ & =\hfill & \parallel g(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-f({\xi }_y,{\xi }_x)\parallel +\parallel G(F({\xi }_x,{\xi }_y),f({\xi }_x,{\xi }_y))-F({\xi }_y,{\xi }_x)\parallel \hfill \\ & \le \hfill & (\alpha +\gamma )\parallel {\xi }_y-F({\xi }_x,{\xi }_y)\parallel +(\beta +\delta )\parallel ({\xi }_x-f({\xi }_x,{\xi }_y)\parallel +(2-(\alpha +\gamma +\beta +\delta ))d.\hfill \end{array}$$

Consequently,

$$(1-(\alpha +\gamma ))(\parallel \eta -F({\xi }_x,{\xi }_y)\parallel -\mathrm{dist}(A,B))+(1-(\beta +\delta ))(\parallel \xi -f({\xi }_x,{\xi }_y)\parallel -d)\le 0.$$

Thus, we have

$$\parallel {\xi }_y-F({\xi }_x,{\xi }_y)\parallel =\parallel {\xi }_x-F({\xi }_y,{\xi }_x)\parallel =\mathrm{dist}(A,B).$$

Using the strict convexity of $(X,\parallel ·\parallel )$, the convexity of the sets A and B, the equalities $\parallel {\xi }_x-F({\xi }_x,{\xi }_y)\parallel =\parallel {\xi }_y-F({\xi }_y,{\xi }_x)\parallel =\mathrm{dist}(A,B)$ we have $\xi =\eta$. Indeed let us denote $z=F({\xi }_x,{\xi }_y)\in B$. Let us consider the ball $B_z\left[d\right]=\{u\in X:\parallel u-z\parallel \le d\}$. From the strict convexity of $\parallel ·\parallel$ it follows that $∥\frac{{\xi }_x+{\xi }_y}{2}-z∥<d$. From the convexity of the set A we have $\frac{{\xi }_x+{\xi }_y}{2}\in A$. □

Definition 13.

Let $(X,\parallel ·\parallel )$ be a Banach space and $U,V\subset X$ be subsets and $F:U\times V\to U$, $G:U\times V\to V$. We say that the pair of maps $(F,G)$ satisfies the proximal property if for any weakly convergent sequences ${\left\{u_n\right\}}_{n=0}^{+\infty }$ and ${\left\{v_n\right\}}_{n=0}^{+\infty }$, $u_n\in U$ and $v_n\in V$ such that $u=w{lim}_{n\to +\infty }{u}_n$, $v=w{lim}_{n\to +\infty }{v}_n$ whenever it holds

$$\underset{n\to +\infty }{lim}\parallel u_n-G(u_n,v_n)\parallel =\mathrm{dist}(U,V)$$

and

$$\underset{n\to +\infty }{lim}\parallel v_n-F(u_n,v_n)\parallel =\mathrm{dist}(U,V)$$

there hold $\parallel u-G(u,v)\parallel =\parallel v-F(u,v)\parallel =\mathrm{dist}(U,V)$.

Theorem 2.

Let $(X,\parallel ·\parallel )$ be a reflexive Banach space and $A,B\subset X$, be nonempty, proper, weakly closed sets, and $F:A\times B\to A$, $G:A\times B\to B$ be a semi-cyclic contraction. Let there hold one of the following

1.

F and G are weakly continuous on $A\times B$;

2.

$(F,G)$ satisfy the proximal property.

Then there exists $(\xi ,\eta )\in A\times B$, which is a best proximity point of $(F,G)$.

If in addition $(X,\parallel ·\parallel )$ is a strictly convex Banach space and $A,B\subset X$ are convex subsets, then $\xi =F(F(\xi ,\eta ),G(\xi ,\eta ))$, $\eta =G(F(\xi ,\eta ),G(\xi ,\eta ))$ and $\xi =F(\xi ,\eta )$, $\eta =G(\xi ,\eta )$.

Proof. 

For an arbitrary $(x,y)\in A\times B$, we consider the sequence

$$(x_n,y_n)=(F(x_{n-1},y_{n-1}),G(x_{n-1},y_{n-1})).$$

By Lemma 7 the sequences ${\left\{x_n\right\}}_{n=1}^{+\infty }$ and ${\left\{y_n\right\}}_{n=1}^{+\infty }$ are bounded sequences. From the assumption that the sets A and B are weakly closed, it follows that we can choose a subsequence of naturals ${\left\{n_j\right\}}_{j=1}^{+\infty }$, such that the sequences ${\left\{x_{n_j}\right\}}_{j=1}^{+\infty }$ and ${\left\{y_{n_j}\right\}}_{j=1}^{+\infty }$ are weakly convergent. Let us denote $w{lim}_{j\to +\infty }{x}_{n_j}=\xi$ and $w{lim}_{j\to +\infty }{y}_{n_j}=\eta$.

(1) Let F and G be weakly continuous on $A\times B$. Then there holds

$$\xi =w\underset{j\to +\infty }{lim}{x}_{n_j},w\underset{j\to +\infty }{lim}F(x_{n_j},y_{n_j})=F(\xi ,\eta )$$

and

$$\eta =w\underset{j\to +\infty }{lim}{y}_{n_j},w\underset{j\to +\infty }{lim}G(x_{n_j},y_{n_j})=G(\xi ,\eta ).$$

Consequently,

$$\xi -\eta =w\underset{j\to +\infty }{lim}(x_{n_j}-y_{n_j}),$$

$$w\underset{j\to +\infty }{lim}(x_{n_j}-G(x_{n_j},y_{n_j}))=\xi -G(\xi ,\eta )$$

and

$$w\underset{j\to +\infty }{lim}(y_{n_j}-F(x_{n_j},y_{n_j}))=\eta -F(\xi ,\eta ).$$

From the reflexivity of $(X,\parallel ·\parallel )$, it follows the existence of a bounded linear functional $f\in S_{X^{*}}$, such that

$$f(\xi -G(\xi ,\eta ))=\parallel \xi -G(\xi ,\eta )\parallel .$$

From the inequalities

$$\left|f(x_{n_j}-G(x_{n_j},y_{n_j}))\right|\le \parallel f\parallel .\parallel x_{n_j}-G(x_{n_j},y_{n_j})\parallel =\parallel x_{n_j}-G(x_{n_j},y_{n_j})\parallel ,$$

$$\underset{j\to +\infty }{lim}f(x_{n_j}-G(x_{n_j},y_{n_j}))=f(\xi -G(\xi ,\eta ))=\parallel \xi -G(\xi ,\eta )\parallel$$

and Lemma 5, we obtain the inequality

$$\begin{array}{ccc}\parallel \xi -G(\xi ,\eta )\parallel \hfill & =\hfill & \left|f(\xi -G(\xi ,\eta ))\right|=\underset{j\to +\infty }{lim}\left|f(x_{n_j}-G(x_{n_j},y_{n_j}))\right|\hfill \\ & \le \hfill & \underset{j\to +\infty }{lim}\parallel x_{n_j}-G(x_{n_j},y_{n_j})\parallel \hfill \\ & \le \hfill & \underset{j\to +\infty }{lim}\parallel x_{n_j}-y_{n_j+1}\parallel =\mathrm{dist}(A,B).\hfill \end{array}$$

Therefore, $\parallel \xi -G(\xi ,\eta )\parallel =\mathrm{dist}(A,B)$.

From the reflexivity of $(X,\parallel ·\parallel )$, it follows the existence of a bounded linear functional $g\in S_{X^{*}}$, such that

$$g(\eta -F(\xi ,\eta ))=\parallel \eta -F(\xi ,\eta )\parallel .$$

From the inequalities

$$\left|g(y_{n_j}-F(x_{n_j},y_{n_j}))\right|\le \parallel g\parallel .\parallel y_{n_j}-F(x_{n_j},y_{n_j})\parallel =\parallel y_{n_j}-F(x_{n_j},y_{n_j})\parallel ,$$

$$\underset{j\to +\infty }{lim}g(y_{n_j}-F(x_{n_j},y_{n_j}))=g(\eta -G(\xi ,\eta ))=\parallel \eta -F(\xi ,\eta )\parallel$$

and Lemma 5, we obtain the inequality

$$\begin{array}{ccc}\parallel \eta -F(\xi ,\eta )\parallel \hfill & =\hfill & \left|f(\eta -F(\xi ,\eta ))\right|=\underset{j\to +\infty }{lim}\left|f(y_{n_j}-F(x_{n_j},y_{n_j}))\right|\hfill \\ & \le \hfill & \underset{j\to +\infty }{lim}\parallel y_{n_j}-F(x_{n_j},y_{n_j})\parallel \hfill \\ & =\hfill & \underset{j\to +\infty }{lim}\parallel y_{n_j}-x_{n_j+1}\parallel =\mathrm{dist}(A,B).\hfill \end{array}$$

Therefore, $\parallel \eta -F(\xi ,\eta )\parallel =\mathrm{dist}(A,B)$.

Thus, $(\xi ,\eta )$ is a coupled best proximity point of $(F,G)$.

(2) Let $(F,G)$ satisfy the proximal property. From Lemma 5, we have

$$\underset{j\to +\infty }{lim}\parallel x_{n_j}-G(x_{n_j},y_{n_j})\parallel =\underset{j\to +\infty }{lim}\parallel y_{n_j}-F(x_{n_j},y_{n_j})\parallel =\mathrm{dist}(A,B).$$

By the assumption that $(F,G)$ satisfies the proximal property, it follows that

$$\parallel \xi -G(\xi ,\eta )\parallel =\parallel \eta -F(\xi ,\eta )\parallel =\mathrm{dist}(A,B)$$

and, thus, $(\xi ,\eta )$ is a best proximity point of $(F,G)$.

Let us assume in addition that $(X,\parallel ·\parallel )$ is a strictly convex Banach space. From the inequalities, assuming that $(\xi ,\eta )\in A\times B$ is a coupled best proximity point of the ordered pair of maps $(F,G)$, we have

$$\parallel F(F(\xi ,\eta ),G(\xi ,\eta ))-G(\xi ,\eta )\parallel -d\le \alpha (\parallel F(\xi ,\eta )-\eta \parallel +\parallel G(\xi ,\eta )-\xi \parallel -2d)=0$$

and

$$\parallel G(F(\xi ,\eta ),G(\xi ,\eta ))-F(\xi ,\eta )\parallel -d\le \alpha (\parallel F(\xi ,\eta )-\eta \parallel +\parallel G(\xi ,\eta )-\xi \parallel -2d)=0.$$

From the assumption that $(X,\parallel ·\parallel )$ is a strictly convex Banach space, it follows that

$$G(F(\xi ,\eta ),G(\xi ,\eta ))=\eta ,F(F(\xi ,\eta ),G(\xi ,\eta ))=\xi .$$

From the inequalities

$$\begin{array}{ccc}0\hfill & \le \hfill & \parallel \xi -\eta \parallel -d=\parallel F(F(\xi ,\eta ),G(\xi ,\eta ))-G(F(\xi ,\eta ),G(\xi ,\eta ))\parallel -d\hfill \\ & \le \hfill & 2\alpha (\parallel F(\xi ,\eta )-G(\xi ,\eta )\parallel -d)\hfill \\ & \le \hfill & 2\alpha (2\alpha (\parallel \xi -\eta \parallel -d)=0,\hfill \end{array}$$

$\parallel \xi -G(\xi ,\eta )\parallel =\parallel \eta -F(\xi ,\eta )\parallel =d$, the strict convexity of $(X,\parallel ·\parallel )$ and the convexity of A and B we have—by similar arguments to that at the end of the proof of Theorem 1—that $\xi =F(\xi ,\eta )$ and $\eta =G(\xi ,\eta )$. □

Remark 1.

We can replace the additional condition $(X,\parallel ·\parallel )$ to be strictly convex with the condition that the sets $A_x,A_y,B_x,B_x$ and $A,B$ be strictly convex sets in Theorem 1 and Theorem 2, respectively.

Remark 2.

Following [26], if we remove the assumption on the maps to be weakly continuous or to satisfy the proximal property, we can have $(\xi ,\eta )\in A\times B(A_x\times B_x)$ in Theorem 2 (Theorem 1), so that $\parallel \xi -\eta \parallel =d(=d_x)$. It follows directly from $\parallel \xi -\eta \parallel \le {lim}_{j\to +\infty }\parallel x_{n_j}-y_{n_j}\parallel$, whenever ${\left\{x_{n_j}\right\}}_{j=1}^{+\infty }$ and ${\left\{y_{n_j}\right\}}_{j=1}^{+\infty }$ are weakly convergent to ξ and η, respectively.

5. Examples and Applications

We will present some examples and we will construct a model of a duopoly market to have the existence of equilibrium productions.

5.1. Examples

Example 1.

Let us consider the space $X=({\mathbb{R}}^2,\parallel ·{\parallel }_{\infty })$, where ${\parallel (x,y)\parallel }_{\infty }=max\left\{\right|x|,|y\left|\right\}$. It is a reflexive Banach space. Let $A_x,A_y,B_x,B_y\subset X$ be defined by

$$A_x=[6,+\infty )\times [15,25],A_y=[6,+\infty )\times [0,1],$$

$$B_x=(-\infty ,-6]\times [15,25],B_y=(-\infty ,-6]\times [-1,0].$$

Let us consider the function $p(x,y)=-\frac{x}{6}-\frac{y}{6}$.

Let us consider the maps $F:A_x\times A_y\to B_x$, $G:B_x\times B_y\to A_x$, $\phi ,\psi :[15,25]\to [15,25]$, defined by

$$F((x,y),(u,v))=\left(p(x,u)-4,\phi (y)\right),$$

$$G((x,y),(u,v))=\left(p(x,u)+4,\psi (y)\right),$$

where $|\phi (a)-\psi (b)|<8$ for any $a,b\in [15,25]$, and the maps $f:A_x\times A_y\to B_y$, $g:B_x\times B_y\to A_y$, defined by

$$f((x,y),(u,v))=\left(p(x,u)-4,-\frac{u}{2}\right),$$

$$g((x,y),(u,v))=\left(p(x,u)+4,-\frac{u}{2}\right).$$

Let us put $x=(x_1,x_2)$, $y=(y_1,y_2)$, $u=(u_1,u_2)$, $v=(v_1,v_2)$, $z=(z_1,z_2)$, $w=(w_1,w_2)$, $t=(t_1,t_2)$, $s=(s_1,s_2)$ just to fit the next formulas into the text field. Using that $F:A_x\times A_y\to B_x$, $G:B_x\times B_y\to A_x$ and $|\phi (a)-\psi (b)|<8$ it follows that

$$\begin{array}{ccc}{S}_6\hfill & =\hfill & \parallel F((x_1,x_2),(y_1,y_2))-G((u_1,u_2),(v_1,v_2)){\parallel }_{\infty }\hfill \\ & =\hfill & {∥\left(p(x_1,y_1)-p(u_1,v_1)-8,\phi (x_2)-\psi (u_2)\right)∥}_{\infty }\hfill \\ & \le \hfill & \frac{|u_1-x_1|}{6}+\frac{|v_1-y_1|}{6}+8\hfill \end{array}$$

and

$$\parallel f((z_1,z_2),(w_1,w_2))-g((t_1,t_2),(s_1,s_2)){\parallel }_{\infty }={∥\left(q(z_1,w_1)-q(t_1,s_1)-8,\frac{w_2-s_2}{2}\right)∥}_{\infty }.$$

From the inequality $\frac{|w_2-s_2|}{2}<8$ for any $w_2\in [0,1]$ and any $s_2\in [-1,0]$ we have

$$\parallel f((z_1,z_2),(w_1,w_2))-g((t_1,t_2),(s_1,s_2)){\parallel }_{\infty }\le \frac{|z_1-t_1|}{6}+\frac{|w_1-s_1|}{6}+8.$$

Let us denote $d_x=\mathrm{dist}(A_x,B_x)=12$ and $d_y=\mathrm{dist}(A_y,B_y)=12$. It is easy to check that

$$8+8=d_x\left(1-\left(\frac{1}{6}+\frac{1}{6}\right)\right)+d_y\left(1-\left(\frac{1}{6}+\frac{1}{6}\right)\right).$$

Consequently, we have

$$\begin{array}{ccc}{S}_7\hfill & =\hfill & {\parallel F(x,y)-G(u,v)\parallel }_{\infty }+{\parallel f(z,w)-g(t,s)\parallel }_{\infty }\hfill \\ & \le \hfill & \frac{|u_1-x_1|}{6}+\frac{|v_1-y_1|}{6}+8+\frac{|z_1-t_1|}{6}+\frac{|w_1-s_1|}{6}+8\hfill \\ & \le \hfill & \frac{1}{6}max\left\{\right|x_1-u_1|,|x_2-u_2\left|\right\}+\frac{1}{6}max\left\{\right|y_1-v_1|,|y_2-v_2\left|\right\}\hfill \\ & & +\frac{1}{6}max\left\{\right|z_1-t_2|,|z_1-t_2\left|\right\}+\frac{1}{6}max\left\{\right|w_1-s_1|,|w_2-s_2\left|\right\}+16\hfill \\ & \le \hfill & \frac{1}{6}\parallel x-u\parallel +\frac{1}{6}\parallel y-v\parallel +\frac{1}{6}\parallel z-t\parallel +\frac{1}{6}\parallel w-s\parallel \hfill \\ & & +d_x\left(1-\left(\frac{1}{6}+\frac{1}{6}\right)\right)+d_y\left(1-\left(\frac{1}{6}+\frac{1}{6}\right)\right).\hfill \end{array}$$

Therefore, we can apply Theorem 1. Consequently, there are $\xi \in A_x$ and $\eta \in A_y$ such that ${\parallel \xi -F(\xi ,\eta )\parallel }_{\infty }=d_x=12$ and ${\parallel \eta -f(\xi ,\eta )\parallel }_{\infty }=d_y=12$. Indeed any points $\xi =(6,\alpha )$ and $\eta =(6,\beta )$ for $\alpha \in [15,20]$ and $\beta \in [0,1]$ is a coupled best proximity point for $(F,f)$. It is easy to observe that $F(\xi ,\eta )=(-6,\phi (\alpha ))$, $f(\xi ,\eta )=(-6,\beta /2)$ and therefore ${\parallel (6,\alpha )-(-6,\phi (\alpha ))\parallel }_{\infty }=max\{12,|\alpha -\phi (\alpha )\left|\right\}=12$ and ${\parallel (6,\beta )-(-6,-\beta /2)\parallel }_{\infty }=max\{26,|3\beta /2\left|\right\}=12$.

Let us consider a particular numeric example with the function $\phi$ defined by

$$\phi (x)=\left\{\begin{array}{cc}\hfill x+1,& x\in [15,24]\hfill \\ \hfill x-9,& x\in (24,25].\hfill \end{array}\right.$$

Let us start from an initial guess $(x_0,y_0)\in A_x$, $(u_0,v_0)\in A_y$. We obtain the iterative sequences ${\left\{(x_n,y_n)\right\}}_{n=0}^{+\infty }$ and ${\left\{(u_n,v_n)\right\}}_{n=0}^{+\infty }$.

From

Table 1. Values of the iterated sequence $(x_n,y_n)$ and $(u_n,v_n)$ if started with $(8,17)$ and $(10,1)$.

n	${\mathit{x}}_{\mathit{n}}$	${\mathit{y}}_{\mathit{n}}$	${\mathit{u}}_{\mathit{n}}$	${\mathit{v}}_{\mathit{n}}$
0	8	17	10	1
1	$-7$	$17.34$	$-7$	$-0.5$
2	$6.33$	$17.974$	$6.33$	$0.25$
6	$6.0041$	$20.05$	$6.0041$	$0.0156$
7	$-6.0013$	$20.51$	$-6.0013$	$0.0078$
12	$6.000016$	$23.08$	$6.000016$	$0.00048$
13	$-6.0000056$	$24.08$	$-6.0000056$	$0.00024$
20	$6.00000000086$	$18.42$	$6.00000000086$	$0.000000095$
21	$-6.00000000028$	$19.32$	$-6.00000000028$	$0.000000047$

, we see a chaotic behavior of the iterative sequence. It is easy to see that ${lim}_{n\to +\infty }{x}_{2n}={lim}_{n\to +\infty }{u}_{2n}=6$, ${lim}_{n\to +\infty }{x}_{2n+1}={lim}_{n\to +\infty }{u}_{2n+1}=-6$, ${lim}_{n\to +\infty }{v}_n=0$ and the sequences $y_n$ oscillates between $[15,25]$.

Example 2.

Let us consider the space $(\mathbb{R},|·|)$, which is a reflexive Banach space. Let $A=[0,1]$ and $B=[2,4]$. Let us consider the maps $F:A\times B\to A$ and $G:A\times B\to B$, defined by

$$F(x,y)=\frac{1+x}{2}·\frac{y}{4}+\left(1-\frac{y}{4}\right)$$

and

$$G(x,y)=\frac{2+y}{2}x+2\left(1-x\right).$$

It can be observed that $1\ge \frac{1+x}{2}\ge F(x,y)\ge x$ and $2\le \frac{2+y}{2}\le G(x,y)\le y$ and thus $F:A\times B\to A$ and $G:A\times B\to B$. From the inequality $F(x,y)\le \frac{1+x}{2}<\frac{2+y}{2}\le G(u,v)$ we have

$$|F(x,y)-G(u,v)|<\left|\frac{2+v}{2}-\frac{1+x}{2}\right|\le \frac{1}{2}\left(|v-x|+|y-u|\right)+\left(1-2\frac{1}{2}\right)\mathrm{dist}(A,B).$$

Therefore, the ordered pair $(F,G)$ satisfies the conditions of Theorem 2 and, thus, it has coupled best proximity points, which is the solution of the system

$$\left|\begin{array}{c}\frac{1+x}{2}·\frac{y}{4}+\left(1-\frac{y}{4}\right)=y\\ \frac{2+y}{2}x+2\left(1-x\right)-y=x\end{array}\right.$$

We will present an example in the infinite dimensional reflexive Banach space. The next example is a variation of Example 2.

Example 3.

Let us consider the space ${\ell }_{\infty }$, which is a reflexive Banach space. Let $A,B\subset {\ell }_{\infty }$, defined as

$$A=\{x={\left\{x_i\right\}}_{n=1}^{+\infty }:x_i\in [0,1]\}andB=\{y={\left\{y_i\right\}}_{n=1}^{+\infty }:y_i\in [2,4]\}.$$

Let us consider the maps $\Phi :[0,1]\times [2,4]\to [0,1]$ and $\Psi :[0,1]\times [2,4]\to [2,4]$, defined by

$$\Phi (x,y)=\frac{1+x}{2}·\frac{y}{4}+\left(1-\frac{y}{4}\right)$$

and

$$\Psi (x,y)=\frac{2+y}{2}x+2\left(1-x\right).$$

Let $F:{\ell }_{\infty }\times {\ell }_{\infty }\to {\ell }_{\infty }$ and $G:{\ell }_{\infty }\times {\ell }_{\infty }\to {\ell }_{\infty }$ be defined by $F={\{\Phi (x_i,y_i)\}}_{i=1}^{+\infty }$ and $F={\{\Psi (x_i,y_i)\}}_{i=1}^{+\infty }$ for ${\left\{x_i\right\}}_{i=1}^{+\infty },{\left\{y_i\right\}}_{i=1}^{+\infty }\in {\ell }_{\infty }$.

It is easy to see that $x_i\le \Phi (x_i,y_i)\le \frac{1+x}{2}\le 1$ and $2\le \frac{2+y_i}{2}\le \Psi (x_i,y_i)\le y$, provided that $x_i\in [0,1]$ and $y_i\in [2,4]$. Thus $F:A\times B\to A$ and $G:A\times B\to B$.

From the inequality $\Phi (x,y)\le \frac{1+x}{2}<\frac{2+y}{2}\le \Psi (u,v)$ we have

$$\begin{array}{ccc}\underset{i\in \mathbb{N}}{sup}\left\{\right|\Phi (x_i,y_i)-\Psi (u_i,v_i)\left|\right\}\hfill & <\hfill & \underset{i\in \mathbb{N}}{sup}\left\{\left|\frac{2+v_i}{2}-\frac{1+x_i}{2}\right|\right\}\hfill \\ & \le \hfill & \frac{1}{2}\underset{i\in \mathbb{N}}{sup}\left\{\left(|v_i-x_i|+|y_i-u_i|\right)\right\}\hfill \\ & \le \hfill & \frac{1}{2}\left(\underset{i\in \mathbb{N}}{sup}\left\{|v_i-x_i|\right\}+\underset{i\in \mathbb{N}}{sup}\left\{|y_i-u_i|\right\}\right)\hfill \\ & \le \hfill & \frac{1}{2}\left({\parallel v-x\parallel }_{\infty }+{\parallel y-u\parallel }_{\infty }\right)+\left(1-2\frac{1}{2}\right)\mathrm{dist}(A,B)\hfill \end{array}$$

Therefore, the ordered pair $(F,G)$ satisfies the conditions of Theorem 2 and, thus, it has a coupled best proximity point.

Example 4.

Let us consider the space $({\mathbb{R}}^2,\parallel ·{\parallel }_{\infty })$, which is a reflexive Banach space. Let $A=[0,1]\times [0,1]$ and $B=[3,4]\times [0,1]$. Let $\alpha ,\beta :[0,1]\to [0,1/2]$. Let us consider the maps $F:A\times B\to A$ and $G:A\times B\to B$, defined by

$$F((x_1,x_2),(y_1,y_2))=\left(\frac{1+x_1}{2},\alpha (y_2)x_2\right)$$

and

$$G((x_1,x_2),(y_1,y_2))=\left(\frac{3+y_1}{2},\beta (x_2)y_2\right).$$

It can be observed that $F:A\times B\to A$ and $G:A\times B\to B$. Let us denote $x=(x_1,x_2),u=(u_1,u_2)\in A$ and $y=(y_1,y_2),v=(v_1,v_2)\in B$. From the inequality $\frac{1+x}{2}<\frac{3+y}{2}$ we have

$$\begin{array}{ccc}{S}_7\hfill & =\hfill & \parallel F((x_1,x_2),(y_1,y_2))-G((u_1,u_2),(v_1,v_2)){\parallel }_{\infty }\hfill \\ & <\hfill & {∥\left(\left(\frac{1+x_1}{2},\alpha (y_2)x_2\right)-\left(\frac{3+v_1}{2},\beta (u_2)v_2\right)\right)∥}_{\infty }\hfill \\ & =\hfill & {∥\left(2+\frac{x_1-v_1}{2},\alpha (y_2)x_2-\beta (u_2)v_2\right)∥}_{\infty }\hfill \\ & =\hfill & max\left\{\left|2+\frac{v_1-x_1}{2}\right|,\left|\alpha (y_2)x_2-\beta (u_2)v_2\right|\right\}=2+\left|\frac{x_1-v_1}{2}\right|\hfill \\ & \le \hfill & \frac{1}{2}max\left\{\right|x_1-v_1|,|x_2-v_2\left|\right\}+\frac{1}{2}max\left\{\right|y_1-u_1|,|y_2-u_2\left|\right\}+2\hfill \\ & =\hfill & \frac{1}{2}{(\parallel x-v\parallel }_{\infty }{+\parallel y-u\parallel }_{\infty })+(1-2\frac{1}{2})\mathrm{dist}(A,B)\hfill \end{array}$$

and

$$\begin{array}{ccc}{S}_8\hfill & =\hfill & \parallel F((x_1,x_2),(y_1,y_2))-G((u_1,u_2),(v_1,v_2)){\parallel }_1\hfill \\ & <\hfill & {∥\left(\left(\frac{1+x_1}{2},\alpha (y_2)x_2\right)-\left(\frac{3+v_1}{2},\beta (u_2)v_2\right)\right)∥}_1\hfill \\ & =\hfill & {∥\left(2+\frac{x_1-v_1}{2},\alpha (y_2)x_2-\beta (u_2)v_2\right)∥}_1\hfill \\ & =\hfill & \left|2+\frac{v_1-x_1}{2}\right|+\left|\alpha (y_2)x_2-\beta (u_2)v_2\right|=2+\left|\frac{x_1-v_1}{2}\right|+\left|\frac{x_2-v_2}{2}\right|\hfill \\ & \le \hfill & \frac{1}{2}{(\parallel x-v\parallel }_1{+\parallel y-u\parallel }_1)+(1-2\frac{1}{2})\mathrm{dist}(A,B).\hfill \end{array}$$

Therefore, the ordered pair $(F,G)$ satisfies the conditions of Theorem 2 and, thus, it has a coupled best proximity point.

In the case $({\mathbb{R}}^2,\parallel ·{\parallel }_{\infty })$, we have

$$\left|\begin{array}{c}max\left\{\left|y_1-\frac{1+x_1}{2}\right|,\alpha (y_2)x_2-y_2\right\}=y_1-\frac{1+x_1}{2}=2\\ max\left\{\left|\frac{3+y_1}{2}-x_1\right|,\beta (x_2)y_2-x_2\right\}=\frac{3+y_1}{2}-x_1=2\end{array}\right.$$

and, therefore, any points $(1,x_2)\in A$ and $(3,y_2)\in B$ are solutions of the example.

In the case $({\mathbb{R}}^2,\parallel ·{\parallel }_1)$, we have

$$\left|\begin{array}{c}\left|y_1-\frac{1+x_1}{2}\right|+|\alpha (y_2)x_2-y_2|\ge 2+|\alpha (y_2)x_2-y_2|=2\\ \left|\frac{3+y_1}{2}-x_1\right|+|\beta (x_2)y_2-x_2|\ge 2+|\beta (x_2)y_2-x_2|=2\end{array}\right.$$

and, therefore, any points $(1,x_2)\in A$ and $(3,y_2)\in B$, such that

$$\left|\begin{array}{c}\alpha (y_2)x_2=y_2\\ \beta (x_2)y_2=x_2\end{array}\right.$$

are solutions of the example.

Example 5.

Let us consider the space $({\mathbb{R}}^2,\parallel ·{\parallel }_1)$, which is a reflexive Banach space. Let $A,B\subset {\mathbb{R}}^2$ be defined as

$$A:\left|\begin{array}{c}y\ge \frac{x}{\sqrt{3}}\\ y\le \sqrt{3}x\\ y\le 1-x\end{array}\right.B:\left|\begin{array}{c}y\ge \frac{x}{\sqrt{3}}\\ y\le \sqrt{3}x\\ y\ge 4-x.\end{array}\right.$$

Then $dist(A,B)=3$. Let us consider the maps $F:A\times B\to A$ and $G:A\times B\to B$, defined by

$$F((x,y),(u,v))=\left(\frac{u}{u+v}\frac{1+x+y}{2},\frac{v}{u+v}\frac{1+x+y}{2}\right)$$

and

$$G((x,y),(u,v))=\left(\frac{x}{x+y}\frac{4+u+v}{2},\frac{y}{x+y}\frac{4+u+v}{2}\right).$$

It is easy to see that $F:A\times B\to A$ and $G:A\times B\to B$. Indeed

$$\frac{u}{u+v}\frac{1+x+y}{2}\ge \frac{v}{u+v}\frac{1+x+y}{2}$$

for any $v\ge \frac{u}{\sqrt{3}}$,

$$\frac{v}{u+v}\frac{1+x+y}{2}\ge \sqrt{3}\frac{u}{u+v}\frac{1+x+y}{2}$$

for any $v\le \sqrt{3}u$ and

$$\frac{v}{u+v}\frac{1+x+y}{2}\ge 1-\frac{u}{u+v}\frac{1+x+y}{2}$$

for any $x+y\le 1$.

Therefore, $F((x,y),(u,v))\subseteq A$. The proof that $G((x,y),(u,v))\subseteq B$ can be done in a similar fashion. Let us put $\alpha =\frac{u}{u+v}$, $\beta =\frac{w}{w+z}$, $a=\frac{1+x+y}{2}$ and $b=\frac{4+s+t}{2}$, where $(u,v)\in B$ and $(w,z)\in A$. It s easy to check that $\alpha \le \frac{1}{\sqrt{3}}$, $\beta \ge \frac{1}{1+\sqrt{3}}$ and $a\le 1$ and $b\ge 4$. Then

$$\beta b-\alpha a\le \frac{4}{1+\sqrt{3}}-\frac{1}{\sqrt{3}}<3\le b-a.$$

Thus, we can write the chain of inequalities

$$\begin{array}{ccc}{S}_9\hfill & =\hfill & {∥F((x,y),(u,v))-G((w,z),(s,t))∥}_1\hfill \\ & =\hfill & {∥\left(\frac{u}{u+v}\frac{1+x+y}{2},\frac{v}{u+v}\frac{1+x+y}{2}\right),\left(\frac{w}{w+z}\frac{4+s+t}{2},\frac{z}{w+z}\frac{4+s+t}{2}\right)∥}_1\hfill \\ & =\hfill & \left|\frac{u}{u+v}\frac{1+x+y}{2}-\frac{w}{w+z}\frac{4+s+t}{2}\right|+\left|\frac{v}{u+v}\frac{1+x+y}{2}-\frac{z}{w+z}\frac{4+s+t}{2}\right|\hfill \\ & =\hfill & \left|\alpha a-\beta b\right|+\left|(1-\alpha )a-(1-\beta )b\right|=\left|\alpha a-\beta b\right|+\left|(b-a)-(\beta b-\alpha a)\right|\hfill \\ & =\hfill & \left|\alpha a-\beta b\right|+\left|b-a\right|-\left|\beta b-\alpha a\right|\le \frac{1}{2}(|x-s+|y-t|)\hfill \\ & =\hfill & \frac{1}{2}{\parallel (x,y)-(s,t)\parallel }_1+\frac{1}{2}{\parallel (u,v)-(w,z)\parallel }_1+\left(1-2.\frac{1}{2}\right)d.\hfill \end{array}$$

Therefore, the ordered pair $(F,G)$ satisfies the conditions of Theorem 2 and, thus, it has a coupled best proximity point.

We cannot obtain uniqueness of the coupled best proximity points. Indeed for any $m\in [1/\sqrt{3},\sqrt{3}]$, there holds $x=\left(\frac{1}{1+m},\frac{m}{1+m}\right)\in A$ and $y=\left(\frac{4}{1+m},\frac{4m}{1+m}\right)\in B$. Then $F\left(x,y\right)=\left(\frac{1}{m},\frac{m}{1+m}\right)$, $G\left(x,y\right)=\left(\frac{4}{m},\frac{4m}{1+m}\right)$,

$${\parallel y-F(x,y)\parallel }_1=\mathrm{dist}(A,B),{\parallel x-G(x,y)\parallel }_1=\mathrm{dist}(A,B).$$

In this particular choice of points, we have $(x,y)$ that $x=F(x,y)$ and $y=G(,x,y)$.

If we choose $n,m\in [1/\sqrt{3},\sqrt{3}]$, there holds $x=\left(\frac{1}{1+m},\frac{m}{1+m}\right)\in A$ and $y=\left(\frac{4}{1+n},\frac{4n}{1+n}\right)\in B$. Then $F\left(x,y\right)=\left(\frac{1}{n},\frac{n}{1+n}\right)$, $G\left(x,y\right)=\left(\frac{4}{m},\frac{4m}{1+m}\right)$,

$${\parallel y-F(x,y)\parallel }_1={\mathrm{dist}(A,B),\parallel x-G(x,y)\parallel }_1=\mathrm{dist}(A,B)\parallel ,$$

(10)

but $x\ne F(x.y)$ and $y\ne G(x,y)$, provided that $n\ne m$.

Let us consider an initial guess $(x_0,y_0)=(0.5,0.4)$ and $(u_0,v_0)=(2,2.5)$. Then we have the tables of the successive iterations.

It can be seen in

Table 2. The iterated sequence values $(x_n,y_n)$ and $(u_n,v_n)$ being started with $(0.5,0.4)$ and $(2,2.5)$.

n	${\mathit{x}}_{\mathit{n}}$	${\mathit{y}}_{\mathit{n}}$	${\mathit{u}}_{\mathit{n}}$	${\mathit{v}}_{\mathit{n}}$
0	$0.5000$	$0.4000$	$2.0000$	$2.4000$
1	$0.4222$	$0.5278$	$2.3611$	$1.8889$
2	$0.5417$	$0.4333$	$1.8333$	$2.2917$
3	$0.4389$	$0.5486$	$2.2569$	$1.8056$
4	$0.5521$	$0.4417$	$1.7917$	$2.2396$
5	$0.4431$	$0.5538$	$2.2309$	$1.7847$
10	$0.5555$	$0.4444$	$1.7780$	$2.2225$
11	$0.4444$	$0.5555$	$2.2224$	$1.7779$
20	$0.5555$	$0.4444$	$1.7780$	$2.2225$
21	$0.4444$	$0.5555$	$2.2224$	$1.7779$
48	$0.5555$	$0.4444$	$1.7777$	$2.2222$
49	$0.4444$	$0.5555$	$2.2222$	$1.7777$

that the iterative sequences $(x_n,y_n)$ and $(u_n,v_n)$ have two convergent sub-sequences

$$\underset{n\to +\infty }{lim}(x_{2n},y_{2n})=(5/9,4/9),\underset{n\to +\infty }{lim}(u_{2n},v_{2n})=(16/9,20/9)$$

and

$$\underset{n\to +\infty }{lim}(x_{2n+1},y_{2n+1})=(4/9,5/9),\underset{n\to +\infty }{lim}(u_{2n+1},v_{2n+1})=(20/9,16/9),$$

which are solutions of the system (10).

5.2. Market Equilibrium in Duopoly Markets

Consider two businesses that provide identical goods or services. These could range from localized health care to simple grocery delivery in a neighborhood. The Cournot’s oligopoly is a fully rational game, in which each firm seeks to maximize its profit under the following assumptions:

• Each company must know the rival’s production before making its optimal production decision, and both firms must make their decisions at the same time.
• Each company is well-versed in the market demand function.

Let us begin with a duopoly model [36,37]—two companies competing for the same consumers and attempting to meet the demand with a total production of $Z=x+y$. The market price $P(Z)=P(x+y)$ is defined as the inverse of the demand function. Market participants have cost functions $c_1(x)$ and $c_2(y)$. Assuming that both firms are acting rationally, the profit functions of the first and second firms are ${\Pi }_1(x,y)=xP(x+y)-c_1(x)$ and ${\Pi }_2(x,y)=yP(x+y)-c_2(y)$, respectively. Each company’s goal is to maximize its profit, i.e., $max\{{\Pi }_1(x,y):x,assumingthatyisfixed\}$ and $max\{{\Pi }_2(x,y):y,assumingthatxisfixed\}$. We have equations (11), if the functions P and $c_i$, $i=1,2$ are differentiable.

$$\left|\begin{array}{c}\frac{\partial {\Pi }_1(x,y)}{\partial x}=P(x+y)+x{P}^{\prime }(x+y)-c_1^{\prime }(x)=0\hfill \\ \frac{\partial {\Pi }_2(x,y)}{\partial y}=P(x+y)+y{P}^{\prime }(x+y)-c_2^{\prime }(y)=0.\hfill \end{array}\right.$$

(11)

The solution of (11) represents the equilibrium pair of production in the duopoly market [36,37].

Equations (11) frequently have solutions in the form of $x=b_1(y)$ and $y=b_2(x)$, which are known as response functions [36].

It may prove difficult or impossible to solve (11), so it is frequently advised to seek an approximate solution. Another disadvantage of seeking an approximate solution is that it may not be stable. Fortunately, an implicit formula for the response function can be found in (11), i.e.,

$$x=\frac{c_1^{\prime }(x)-P(x+y)}{P^{\prime }(x+y)}=F(x,y)andy=\frac{c_2^{\prime }(y)-P(x+y)}{P^{\prime }(x+y)}=G(x,y).$$

It is still possible that we will end up with response functions that do not maximize profit $\Pi$. As commonly assumed, each participant’s response is determined by its own production level, as well as the level of other players. For example, if the output quantities are $(x_n,y_n)$ at a point n and the first player changes its productions to $x_{n+1}=F(x_n,y_n)$, the second player will also change its output to $y_{n+1}=G(x_n,y_n)$. If there are two productions x and y, we have an equilibrium if $\rho (x,F(x,y))=\rho (y,G(x,y))=0$ [36,37]. Unfortunately, as pointed in [28], it may happen that one of the firms has much bigger production sets than the other one and $A\cap B=\varnothing$. In this case, the equilibrium production will be reached if $\rho (x,F(x,y))=\rho (y,G(x,y))=\mathrm{dist}(A,B)$ [28], which is a coupled best proximity point for the ordered pair $(F,G)$ of maps.

The functions ${\Pi }_i$ are referred to as payoff functions. To ensure that the solutions of (11) present a maximization of the payoff functions, sufficient conditions are ${\Pi }_i$ to be concave functions [38,39,40]. We change the maximization problem into a coupled fixed point one; thus, all assumptions of concavity and differentiability can be removed. The problem of finding coupled best proximity points for an ordered pair of maps $(F,G)$ [5] is the problem of solving the equations $\rho (x,F(x,y))=\rho (y,G(x,y))=\mathrm{dist}(A,B)$. However, one important limitation may be that players cannot change the output too quickly and, thus, may not perform to maximize their profits.

The problem of finding market equilibrium by the help of best proximity points is investigated in [28] when the underlying space is a uniformly convex Banach space. Unfortunately in economics, the used metrics are usually ${\parallel ·-·\parallel }_1$ as far as the profit is ${\sum }_{i=1}^n{\parallel p_i\parallel }_1$, where $p_i$ is the profit of the i–th good of the player. Fortunately the underlying space in the economic models is finite dimensional and $({\mathbb{R}}^n,\parallel ·{\parallel }_1)$ is a reflexive space; we can apply Theorem 2.

Example 3 fits to such a model. As pointed out in Example 3, we can say that there is an equilibrium point in the market, which is not unique. For any initial start of the market the sequence of successive production will have a subsequence, convergent to the equilibrium point. The equilibrium point $((x,y),(u,v))$ will satisfy the condition $x+y=1$ and $u+v=4$.

6. Discussion

The results in Theorems 1 and 2 show that it is possible to obtain existence results about coupled best proximity points in reflexive Banach spaces. The illustrative examples show that in real models the iterative process can have chaotic nature. It will be interesting whether some additional conditions can be imposed for the iterative process to be convergent.