Common best proximity points theorem for four mappings in metric-type spaces

Abstract

In this article, we first give an existence and uniqueness common best proximity points theorem for four mappings in a metric-type space $(X, D, K)$ such that D is not necessarily continuous. An example is also given to support our main result. We also discuss the unique common fixed point existence result of four mappings defined on such a metric space.

1 Introduction and preliminary

Fixed point theory is essential for solving various equations of the form $T x = x$ for self-mappings T defined on subsets of metric spaces or normed linear spaces. Given non-void subsets A and B of a metric space and a non-self-mapping $T : A \rightarrow B$, the equation $T x = x$ does not necessarily have a solution, which is known as a fixed point of the mapping T. However, in such conditions, it may be considered to determine an element x for which the error $d(x,T x)$ is minimum, in which case x and Tx are in close proximity to each other. It is remarked that best proximity point theorems are relevant to this end. A best proximity point theorem provides sufficient conditions that confirm the existence of an optimal solution to the problem of globally minimizing the error $d(x,T x)$, and hence the existence of a complete approximate solution to the equation $T x = x$. In fact, with respect to the fact that $d(x,T x) \geq d(A,B)$ for all x, a best proximity point theorem requires the global minimum of the error $d(x,T x)$ to be the least possible value $d(A,B)$. Eventually, a best proximity point theorem offers sufficient conditions for the existence of an element x, called a best proximity point of the mapping T, satisfying the condition that $d(x,T x) = d(A,B)$. Moreover, it is interesting to observe that best proximity theorems also appear as a natural generalization of fixed point theorems, for a best proximity point reduces to a fixed point if the mapping under consideration is a self-mapping.

A study of several variants of contractions for the existence of a best proximity point can be found in [1–7]. Best proximity point theorems for multivalued mappings are available in [8–14]. Eldred et al. [15] have established a best proximity point theorem for relatively non-expansive mappings. Further, Anuradha and Veeramani have investigated best proximity point theorems for proximal pointwise contraction mappings [16].

On the other hand, Khamsi and Hussain [17] generalized the definition of a metric and defined the metric-type as follows.

Definition 1.1

[17]

Let X be a non-empty set, $K \geq1$ be a real number, and let the function $D : X \times X \rightarrow\mathbb{R}$ satisfy the following properties:

• (i) $D(x,y) = 0$ if and only if $x = y$;

• (ii) $D(x, y)= D(y, x)$ for all $x, y \in X$;

• (iii) $D(x,z)\leq K(D(x,y) + D(y, z))$ for all $x, y \in X$.

Then $(X, D,K)$ is called a metric-type space.

Obviously, for $K = 1$, a metric-type space is simply a metric space.

Afterward, other authors proved fixed point theorems in metric-type space [18–20].

Given two non-empty subsets A and B of a metric-type space $(X,D,K)$, the following notions and notations are used in the sequel.

$$\begin{aligned}& D(A,B)= \inf \bigl\{ d(x,y): x \in A , y\in B\bigr\} ; \\& A_{0}=\bigl\{ x \in A : D(x,y)= D(A,B) \mbox{ for some } y \in B\bigr\} ; \\& B_{0}=\bigl\{ y\in B : D(x,y)= D(A,B) \mbox{ for some } x \in A\bigr\} . \end{aligned}$$

This study focuses upon resolving a more general problem as regards the existence of common best proximity points for pairs of non-self-mappings in metric-type space. As a result, the finding of this study verifies a common global minimal solution to the problem of minimizing the real valued multi-objective functions $x \rightarrow d(x, Sx)$ and $x \rightarrow d(x, Tx)$, which in turn gives rise to a common optimal approximate solution of the fixed point equations $Sx = x$ and $Tx = x$, where D is a metric-type space and the non-self-mappings $S : A \rightarrow B$ and $T : A \rightarrow B$ satisfy a contraction-like condition. Our best proximity point theorem generalizes a result due to Sadiq Basha [21]. Further, a common fixed point theorem for commuting self-mappings is a special case of our common best proximity point theorem. Now, we review some definitions used throughout the paper.

Definition 1.2

An element $x\in A$ is said to be a common best proximity point of the non-self-mappings $f_{1}, f_{2},\ldots,f_{n} : A \rightarrow B$ if it satisfies the condition that

$$D(x,f_{1}x) = D(x,f_{2}x) = \cdots= D(x,f_{n}x) = D(A,B). $$

Definition 1.3

The mappings $S:A\rightarrow B$ and $T:A\rightarrow B$ are said to be commute proximally if they satisfy the condition that

$$\bigl[D(u,Sx) = D(v,Tx) = D(A,B)\bigr]\quad\Rightarrow\quad Sv=Tu. $$

Definition 1.4

If $A_{0}\ne\varnothing$ then the pair $(A,B)$ is said to have P-property if and only if for any $x_{1}, x_{2}\in A_{0}$ and $y_{2}, y_{2}\in B_{0}$

$$ \left \{ \begin{array}{@{}l} D(x_{1},y_{1})=D(A,B),\\ D(x_{2},y_{2})=D(A,B) \end{array} \right . \quad\Longrightarrow\quad D(x_{1},x_{2})=D(y_{1},y_{2}). $$

2 Main result

We begin our study with the following definition.

Definition 2.1

Let A and B be two non-empty subsets of a metric-type space $(X,D,K)$. Non-self-mappings $f,g,S,T: A\rightarrow B$ are said to satisfy a K-contractive condition if there exists a non-negative number $\alpha< {\frac{1}{K}}$ such that for each $x,y \in A$

$$D(fx,gy) \leq{\alpha} \max \biggl\{ D(Sx,Ty) , D(fx,Sx) , D(Ty,gy), \frac {1}{2K}\bigl[D(Sx, gy)+D(fx, Ty)\bigr]\biggr\} . $$

Theorem 2.2

Let A and B be non-empty subsets of a complete metric-type space $(X,D,K)$. Moreover, assume that $A_{0}$ and $B_{0}$ are non-empty and $A_{0}$ is closed. Let the non-self-mappings $f,g,S,T: A\rightarrow B$ satisfy the following conditions:

• (i) $\{f,S\}$ and $\{g,T\}$ commute proximally;

• (ii) the pair $(A,B)$ has the P-property;

• (iii) f, g, S and T are continuous;

• (iv) f, g, S, and T satisfy the K-contractive condition;

• (v) $f(A_{0})\subseteq T(A_{0})$, $g(A_{0})\subseteq S(A_{0})$ and $g(A_{0})\subseteq B_{0}$, $f(A_{0})\subseteq B_{0}$.

Then f, g, S, and T have a unique common best proximity point.

Proof

Fix $x_{0}$ in $A_{0}$, since $f(A_{0})\subseteq T(A_{0})$, then there exists an element $x_{1}$ in $A_{0}$ such that $f(x_{0})=T(x_{1})$. Similarly, a point $x_{2} \in A_{0}$ can be chosen such that $g(x_{1}) = S(x_{2})$. Continuing this process, we obtain a sequence $\{x_{n}\}\in A_{0}$ such that $f(x_{2n}) = T(x_{2n+1})$ and $g(x_{2n+1}) = S(x_{2n+2})$.

Since $f(A_{0})\subseteq B_{0}$ and $g(A_{0})\subseteq B_{0}$, there exists $\{u_{n}\} \in A_{0}$ such that

$$ D\bigl(u_{2n},f(x_{2n})\bigr) = D(A,B) \quad \mbox{and} \quad D\bigl(u_{2n+1}, g(x_{2n+1})\bigr)= D(A,B). $$

(1)

Since the pair $(A,B)$ has the P-property, by (1) we have

$$\begin{aligned} D(u_{2n},u_{2n+1}) =&D(fx_{2n},gx_{2n+1}) \\ \leq&\alpha \max\biggl\{ D(Sx_{2n},Tx_{2n+1}) , D(fx_{2n},Sx_{2n}) , D(Tx_{2n+1},gx_{2n+1}), \\ &{}\frac{1}{2K}\bigl[D(Sx_{2n}, gx_{2n+1})+D(fx_{2n}, Tx_{2n+1})\bigr]\biggr\} \\ \leq&\alpha \max\biggl\{ D(u_{2n-1},u_{2n}) , D(u_{2n},u_{2n-1}) , D(u_{2n},u_{2n+1}),\\ &{} \frac{1}{2K}\bigl[D(u_{2n-1}, u_{2n+1})+D(u_{2n}, u_{2n})\bigr]\biggr\} , \end{aligned}$$

thus (note that $\frac{1}{2K}D(u_{2n-1}, u_{2n+1}) \leq\frac {1}{2}[D(u_{2n-1},u_{2n})+D(u_{2n},u_{2n+1})]$ and $\alpha< 1$)

$$ D(u_{2n},u_{2n+1})\leq{\alpha}D(u_{2n-1},u_{2n}). $$

(2)

Similarly

$$\begin{aligned} D(u_{2n+1},u_{2n+2}) =&D(fx_{2n+2},gx_{2n+1}) \\ \leq&\alpha \max\biggl\{ D(Sx_{2n+2},Tx_{2n+1}) , D(fx_{2n+2},Sx_{2n+2}) , D(Tx_{2n+1},gx_{2n+1}), \\ &{}\frac{1}{2K}\bigl[D(Sx_{2n+2}, gx_{2n+1})+D(fx_{2n+2}, Tx_{2n+1})\bigr]\biggr\} \\ \leq&\alpha \max\biggl\{ D(u_{2n+1},u_{2n}) , D(u_{2n+2},u_{2n+1}) , D(u_{2n},u_{2n+1}),\\ &{} \frac{1}{2K}\bigl[D(u_{2n+1}, u_{2n+1})+D(u_{2n+2}, u_{2n})\bigr]\biggr\} , \end{aligned}$$

thus (note that $\frac{1}{2K}D(u_{2n+2}, u_{2n}) \leq\frac {1}{2}[D(u_{2n+2},u_{2n+1})+D(u_{2n+1},u_{2n})]$ and $\alpha< 1$)

$$ D(u_{2n+1},u_{2n+2})\leq{\alpha}D(u_{2n},u_{2n+1}). $$

(3)

Therefore, by (2) and (3) we have

$$ D(u_{n},u_{n+1})\leq\alpha D(u_{n-1},u_{n}), $$

and then

$$ D(u_{n},u_{n+1})\leq\alpha^{n} D(u_{0}, u_{1}). $$

(4)

Let $m, n \in\mathbb{N}$ and $m < n$; we have

$$\begin{aligned} D(u_{m}, u_{n}) \leq& K\bigl[D(u_{m}, u_{m+1})+ D(u_{m+1}, u_{n})\bigr] \\ \leq&K D(u_{m}, u_{m+1}) + K^{2} \bigl[D(u_{m+1}, u_{m+2}) + D(u_{m+2}, u_{n}) \bigr]\\ \leq& \cdots \\ \leq& K D(u_{m}, u_{m+1}) + K^{2}D(u_{m+1}, u_{m+2})+ \cdots\\ &{}+ K^{n-m-1}\bigl[D(u_{n-2}, u_{n-1}) + D(u_{n-1}, u_{n})\bigr] \\ \leq& K D(u_{m}, u_{m+1}) + K^{2}D(u_{m+1}, u_{m+2} )+\cdots\\ &{}+ K^{n-m-1}D(u_{n-2}, u_{n-1})+ K^{n-m}D(u_{n-1}, u_{n}). \end{aligned}$$

Now (4) and $K \alpha< 1$ imply that

$$\begin{aligned} D(u_{m},u_{n}) \leq& \bigl(K \alpha^{m} + K^{2} \alpha^{m+1}+\cdots + K^{n-m}\alpha^{n-1} \bigr)D(u_{0} , u_{1}) \\ \leq& K {\alpha}^{m} \bigl( 1+ K \alpha+ \cdots+ {(K \alpha)}^{n-m-1}\bigr)D(u_{0} , u_{1}) \\ \leq& \frac{K {\alpha}^{m}}{1-K\alpha}D(u_{0},u_{1})\rightarrow0 \quad \mbox{when } m \rightarrow\infty; \end{aligned}$$

then $\{u_{n}\}$ is a Cauchy sequence.

Since $\{u_{n}\}\subset A_{0}$ and $A_{0}$ is a closed subset of the complete metric-type space $(X,D,K)$, we can find $u \in A_{0}$ such that $\lim_{n \rightarrow\infty }u_{n}=u$.

By (1) and because of the fact $\{f,S\}$ and $\{g,T\}$ commute proximally, $fu_{2n-1} = Su_{2n}$ and $gu_{2n} = Tu_{2n+1}$. Therefore, the continuity of f, g, S, and T and $n \rightarrow\infty$ ascertain that $fu=gu=Tu=Su$.

Since $f(A_{0})\subseteq B_{0}$, there exists $x \in A_{0}$ such that

$$D(A,B)=D(x, fu)=D(x,gu)=D(x,Su)=D(x,Tu). $$

As $\{f,S\}$ and $\{g,T\}$ commute proximally, $fx=gx=Sx=Tx$. Since $f(A_{0})\subseteq B_{0}$, there exists $z \in A_{0}$ such that

$$D(A,B)=D(z,fx)=D(z,gx)=D(z,Sx)=D(z,Tx). $$

Because the pair $(A,B)$ has the P-property

$$\begin{aligned} D(x,z) =&D(fu,gx)\\ \leq&\alpha \max\biggl\{ D(Su,Tx),D(fu,Su),D(Tx,gx), \frac {1}{2K}\bigl[D(Su,gx)+D(fu,Tx)\bigr]\biggr\} \\ \leq& \alpha \max\biggl\{ D(x,z),D(x,x),D(z,z),\frac {1}{2K} \bigl[D(x,z)+d(x,z)\bigr]\biggr\} \\ \leq& \alpha D(x,z), \end{aligned}$$

which implies that $x=z$. Thus, it follows that

$$ D(A,B)=D(x,fx)=(x,gx)=(x,Tx)=(x,Sx), $$

(5)

then x is a common best proximity point of the mappings f, g, S, and T.

Suppose that y is another common best proximity point of the mappings f, g, S, and T, so that

$$ D(A,B)=D(y,fy)=(y,gy)=(y,Ty)=(y,Sy). $$

(6)

As the pair $(A,B)$ has the P-property, from (5) and (6), we have

$$D(x,y)\leq\alpha D(x,y), $$

which implies that $x = y$. □

Now we illustrate our common best proximity point theorem by the following example.

Example 2.3

Let $X = [0,1]\times[0,1]$. Suppose that $D(x,y) = d^{2}(x,y)$ for all $x,y \in X$, where d is the Euclidean metric. Then $(X,D,K)$ is a complete metric-type space with $K=2$. Let

$$A:=\bigl\{ (0,x): 0 \leq x \leq1\bigr\} , \qquad B := \bigl\{ (1,y) : 0 \leq y \leq 1\bigr\} . $$

Then $D( A,B)=1$, $A_{0} = A$, and $B_{0} =B$. Let f, g, S, and T be defined as $f(0,y)=(1,\frac{y}{8})$, $g(0,y) = (1,\frac{y}{32})$, $S(0,y) = (1,y)$, and $T(0,y) = (1,\frac{y}{4})$. Then for all x and $y \in X$ we have

$$D(fx,gy)= \biggl(\frac{x}{8}-\frac{y}{32}\biggr)^{2} = \frac{1}{64}D(Sx , Ty). $$

Now, all the required hypotheses of Theorem 2.2 are satisfied. Clearly $(0,0)$ is unique common best proximity point of f, g, S, and T.

By Theorem 2.2 we also obtain the following common fixed point theorem in metric-type space.

Theorem 2.4

Let $(X,D,K)$ be a complete metric-type space. Let $f,g,S,T : X \rightarrow X$ be given continuous mappings satisfying the K-contractive condition such that S and T commute with f and g, respectively. Further let $f(X)\subseteq T(X)$, $g(X)\subseteq S(x)$. Then f, g, S, and T have a unique common fixed point.

Proof

We take the same sequence $\{u_{n}\}$ and u as in the proof of Theorem 2.2. Due to the fact that S and T commute with f and g, respectively, we have

$$fu_{2n-1} = Su_{2n}, \qquad gu_{2n}=Tu_{2n+1}. $$

By continuity of f, g, S, T, and $n \rightarrow\infty$ we have

$$\begin{aligned} fu = Su, \qquad gu = Tu. \end{aligned}$$

(7)

Since $f,g,S,T : X \rightarrow X$ satisfy the K-contractive condition, and by (7),

$$\begin{aligned} D(fu,gu) \leq& \alpha \max \biggl\{ D(Su,Tu),D(fu,Su),D(Tu,gu),\frac {1}{2K} \bigl[D(Su,gu)+D(fu,Tu)\bigr]\biggr\} \\ \leq& \alpha \max \biggl\{ D(fu,gu),D(fu,fu),D(gu,gu),\frac {1}{2K} \bigl[(fu,gu)+(fu,gu)\bigr]\biggr\} , \end{aligned}$$

we have $D(fu,gu) \leq\alpha D(fu,gu)$. Therefore $fu = gu$, and by (7), $fu = gu = Su = Tu$.

We set $w = fu = gu = Su = Tu$. Because of the fact that T commutes with g we obtain

$$gw = gTu = Tgu = Tw, $$

and

$$\begin{aligned} D(w,gw) =& D(fu ,gw)\\ \leq&\alpha \max \biggl\{ D(Su,Tw),D(fu,Su),D(Tw,gw), \frac {1}{2K}\bigl[D(Su,gw)+D(fu,Tw)\bigr]\biggr\} \\ \leq&\alpha \max \biggl\{ D(w,gw),D(w,w),D(gw,gw),\frac {1}{2K} \bigl[(w,gw)+(w,gw)\bigr]\biggr\} . \end{aligned}$$

Therefore, $D(w,gw) \leq\alpha D(w,gw)$ and consequently

$$\begin{aligned} w = gw = Tw. \end{aligned}$$

(8)

Similarly, we can show that

$$\begin{aligned} w = fw = Sw. \end{aligned}$$

(9)

Hence, by (8) and (9) we deduce that $w = fw = gw = Sw = Tw$. Therefore, w is a common fixed point of f, g, S, and T.

Assume to the contrary that $p = fp = gp = Sp = Tp$ and $q = fq = gq = Sq = Tq$ but $p \neq q$.

We have

$$\begin{aligned} D(p,q) &= D(fp ,gq)\\ &\leq\alpha \max \biggl\{ D(Sp,Tq),D(fp,Sp),D(Tq,gq), \frac {1}{2K}\bigl[D(Sp,gq)+D(fp,Tq)\bigr]\biggr\} \\ &\leq\alpha \max \biggl\{ D(p,q),D(p,p),D(q,q),\frac{1}{2K} \bigl[(p,q)+(p,q)\bigr]\biggr\} . \end{aligned}$$

Consequently $D(p,q) \leq\alpha D(p,q)$ and $\alpha< 1$; then $D(p,q) = 0$, a contradiction. Therefore, f, g, S, and T have a unique fixed point. □