Open Access

# A best proximity point theorem for Geraghty-contractions

Fixed Point Theory and Applications20122012:231

DOI: 10.1186/1687-1812-2012-231

Accepted: 10 December 2012

Published: 27 December 2012

## Abstract

The purpose of this paper is to provide sufficient conditions for the existence of a unique best proximity point for Geraghty-contractions.

Our paper provides an extension of a result due to Geraghty (Proc. Am. Math. Soc. 40:604-608, 1973).

### Keywords

fixed point Geraghty-contraction P-property best proximity point

## 1 Introduction

Let A and B be nonempty subsets of a metric space $\left(X,d\right)$.

An operator $T:A\to B$ is said to be a k-contraction if there exists $k\in \left[0,1\right)$ such that $d\left(Tx,Ty\right)\le kd\left(x,y\right)$ for any $x,y\in A$. Banach’s contraction principle states that when A is a complete subset of X and T is a k-contraction which maps A into itself, then T has a unique fixed point in A.

A huge number of generalizations of this principle appear in the literature. Particularly, the following generalization of Banach’s contraction principle is due to Geraghty [1].

First, we introduce the class of those functions $\beta :\left[0,\mathrm{\infty }\right)\to \left[0,1\right)$ satisfying the following condition:
$\beta \left({t}_{n}\right)\to 1\phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}{t}_{n}\to 0.$

Theorem 1.1 ([1])

Let $\left(X,d\right)$ be a complete metric space and $T:X\to X$ be an operator. Suppose that there exists $\beta \in \mathcal{F}$ such that for any $x,y\in X$,
$d\left(Tx,Ty\right)\le \beta \left(d\left(x,y\right)\right)\cdot d\left(x,y\right).$
(1)

Then T has a unique fixed point.

Since the constant functions $f\left(t\right)=k$, where $k\in \left[0,1\right)$, belong to , Theorem 1.1 extends Banach’s contraction principle.

Remark 1.1 Since the functions belonging to are strictly smaller than one, condition (1) implies that

Therefore, any operator $T:X\to X$ satisfying (1) is a continuous operator.

The aim of this paper is to give a generalization of Theorem 1.1 by considering a non-self map T.

First, we present a brief discussion about a best proximity point.

Let A be a nonempty subset of a metric space $\left(X,d\right)$ and $T:A\to X$ be a mapping. The solutions of the equation $Tx=x$ are fixed points of T. Consequently, $T\left(A\right)\cap A\ne \mathrm{\varnothing }$ is a necessary condition for the existence of a fixed point for the operator T. If this necessary condition does not hold, then $d\left(x,Tx\right)>0$ for any $x\in A$ and the mapping $T:A\to X$ does not have any fixed point. In this setting, our aim is to find an element $x\in A$ such that $d\left(x,Tx\right)$ is minimum in some sense. The best approximation theory and best proximity point analysis have been developed in this direction.

In our context, we consider two nonempty subsets A and B of a complete metric space and a mapping $T:A\to B$.

A natural question is whether one can find an element ${x}_{0}\in A$ such that $d\left({x}_{0},T{x}_{0}\right)=min\left\{d\left(x,Tx\right):x\in A\right\}$. Since $d\left(x,Tx\right)\ge d\left(A,B\right)$ for any $x\in A$, the optimal solution to this problem will be the one for which the value $d\left(A,B\right)$ is attained by the real valued function $\phi :A\to \mathbb{R}$ given by $\phi \left(x\right)=d\left(x,Tx\right)$.

Some results about best proximity points can be found in [29].

## 2 Notations and basic facts

Let A and B be two nonempty subsets of a metric space $\left(X,d\right)$.

We denote by ${A}_{0}$ and ${B}_{0}$ the following sets:

where .

In [8], the authors present sufficient conditions which determine when the sets ${A}_{0}$ and ${B}_{0}$ are nonempty.

Now, we present the following definition.

Definition 2.1 Let A, B be two nonempty subsets of a metric space $\left(X,d\right)$. A mapping $T:A\to B$ is said to be a Geraghty-contraction if there exists $\beta \in \mathcal{F}$ such that
Notice that since $\beta :\left[0,\mathrm{\infty }\right)\to \left[0,1\right)$, we have

Therefore, every Geraghty-contraction is a contractive mapping.

In [10], the author introduces the following definition.

Definition 2.2 ([10])

Let $\left(A,B\right)$ be a pair of nonempty subsets of a metric space $\left(X,d\right)$ with ${A}_{0}\ne \mathrm{\varnothing }$. Then the pair $\left(A,B\right)$ is said to have the P-property if and only if for any ${x}_{1},{x}_{2}\in {A}_{0}$ and ${y}_{1},{y}_{2}\in {B}_{0}$,
$\begin{array}{lcr}d\left({x}_{1},{y}_{1}\right)& =& d\left(A,B\right)\\ d\left({x}_{2},{y}_{2}\right)& =& d\left(A,B\right)\end{array}\right\}\phantom{\rule{1em}{0ex}}⇒\phantom{\rule{1em}{0ex}}d\left({x}_{1},{x}_{2}\right)=d\left({y}_{1},{y}_{2}\right).$

It is easily seen that for any nonempty subset A of $\left(X,d\right)$, the pair $\left(A,A\right)$ has the P-property.

In [10], the author proves that any pair $\left(A,B\right)$ of nonempty closed convex subsets of a real Hilbert space H satisfies the P-property.

## 3 Main results

We start this section presenting our main result.

Theorem 3.1 Let $\left(A,B\right)$ be a pair of nonempty closed subsets of a complete metric space $\left(X,d\right)$ such that ${A}_{0}$ is nonempty. Let $T:A\to B$ be a Geraghty-contraction satisfying $T\left({A}_{0}\right)\subseteq {B}_{0}$. Suppose that the pair $\left(A,B\right)$ has the P-property. Then there exists a unique ${x}^{\ast }$ in A such that $d\left({x}^{\ast },T{x}^{\ast }\right)=d\left(A,B\right)$.

Proof Since ${A}_{0}$ is nonempty, we take ${x}_{0}\in A$.

As $T{x}_{0}\in T\left({A}_{0}\right)\subseteq {B}_{0}$, we can find ${x}_{1}\in {A}_{0}$ such that $d\left({x}_{1},T{x}_{0}\right)=d\left(A,B\right)$. Similarly, since $T{x}_{1}\in T\left({A}_{0}\right)\subseteq {B}_{0}$, there exists ${x}_{2}\in {A}_{0}$ such that $d\left({x}_{2},T{x}_{1}\right)=d\left(A,B\right)$. Repeating this process, we can get a sequence $\left({x}_{n}\right)$ in ${A}_{0}$ satisfying
Since $\left(A,B\right)$ has the P-property, we have that
Taking into account that T is a Geraghty-contraction, for any $n\in \mathbb{N}$, we have that
$d\left({x}_{n},{x}_{n+1}\right)=d\left(T{x}_{n-1},T{x}_{n}\right)\le \beta \left(d\left({x}_{n-1},{x}_{n}\right)\right)\cdot d\left({x}_{n-1},{x}_{n}\right)
(2)

Suppose that there exists ${n}_{0}\in \mathbb{N}$ such that $d\left({x}_{{n}_{0}},{x}_{{n}_{0}+1}\right)=0$.

In this case,
$0=d\left({x}_{{n}_{0}},{x}_{{n}_{0}+1}\right)=d\left(T{x}_{{n}_{0}-1},T{x}_{{n}_{0}}\right),$

and consequently, $T{x}_{{n}_{0}-1}=T{x}_{{n}_{0}}$.

Therefore,
$d\left(A,B\right)=d\left({x}_{{n}_{0}},T{x}_{{n}_{0}-1}\right)=d\left({x}_{{n}_{0}},T{x}_{{n}_{0}}\right)$

and this is the desired result.

In the contrary case, suppose that $d\left({x}_{n},{x}_{n+1}\right)>0$ for any $n\in \mathbb{N}$.

By (2), $\left(d\left({x}_{n},{x}_{n+1}\right)\right)$ is a decreasing sequence of nonnegative real numbers, and hence there exists $r\ge 0$ such that
$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{n+1}\right)=r.$

In the sequel, we prove that $r=0$.

Assume $r>0$, then from (2) we have

The last inequality implies that ${lim}_{n\to \mathrm{\infty }}\beta \left(d\left({x}_{n-1},{x}_{n}\right)\right)=1$ and since $\beta \in \mathcal{F}$, we obtain $r=0$ and this contradicts our assumption.

Therefore,
$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{n+1}\right)=0.$
(3)

Notice that since $d\left({x}_{n+1},T{x}_{n}\right)=d\left(A,B\right)$ for any $n\in \mathbb{N}$, for $p,q\in \mathbb{N}$ fixed, we have $d\left({x}_{p},T{x}_{p-1}\right)=d\left({x}_{q},T{x}_{q-1}\right)=d\left(A,B\right)$, and since $\left(A,B\right)$ satisfies the P-property, $d\left({x}_{p},{x}_{q}\right)=d\left(T{x}_{p-1},T{x}_{q-1}\right)$.

In what follows, we prove that $\left({x}_{n}\right)$ is a Cauchy sequence.

In the contrary case, we have that
$\underset{m,n\to \mathrm{\infty }}{lim sup}d\left({x}_{n},{x}_{m}\right)>0.$
By using the triangular inequality,
$d\left({x}_{n},{x}_{m}\right)\le d\left({x}_{n},{x}_{n+1}\right)+d\left({x}_{n+1},{x}_{m+1}\right)+d\left({x}_{m+1},{x}_{m}\right).$
By (2) and since $d\left({x}_{n+1},{x}_{m+1}\right)=d\left(T{x}_{n},T{x}_{m}\right)$, by the above mentioned comment, we have
$\begin{array}{rl}d\left({x}_{n},{x}_{m}\right)& \le d\left({x}_{n},{x}_{n+1}\right)+d\left(T{x}_{n},T{x}_{m}\right)+d\left({x}_{m+1},{x}_{m}\right)\\ \le d\left({x}_{n},{x}_{n+1}\right)+\beta \left(d\left({x}_{n},{x}_{m}\right)\right)\cdot d\left({x}_{n},{x}_{m}\right)+d\left({x}_{m+1},{x}_{m}\right),\end{array}$
which gives us
$d\left({x}_{n},{x}_{m}\right)\le {\left(1-\beta \left(d\left({x}_{n},{x}_{m}\right)\right)\right)}^{-1}\left[d\left({x}_{n},{x}_{n+1}\right)+d\left({x}_{m+1},{x}_{m}\right)\right].$
Since ${lim sup}_{m,n\to \mathrm{\infty }}d\left({x}_{n},{x}_{m}\right)>0$ and by (3), ${lim sup}_{n\to \mathrm{\infty }}d\left({x}_{n},{x}_{n+1}\right)=0$, from the last inequality it follows that
$\underset{m,n\to \mathrm{\infty }}{lim sup}{\left(1-\beta \left(d\left({x}_{n},{x}_{m}\right)\right)\right)}^{-1}=\mathrm{\infty }.$

Therefore, ${lim sup}_{m,n\to \mathrm{\infty }}\beta \left(d\left({x}_{n},{x}_{m}\right)\right)=1$.

Taking into account that $\beta \in \mathcal{F}$, we get ${lim sup}_{m,n\to \mathrm{\infty }}d\left({x}_{n},{x}_{m}\right)=0$ and this contradicts our assumption.

Therefore, $\left({x}_{n}\right)$ is a Cauchy sequence.

Since $\left({x}_{n}\right)\subset A$ and A is a closed subset of the complete metric space $\left(X,d\right)$, we can find ${x}^{\ast }\in A$ such that ${x}_{n}\to {x}^{\ast }$.

Since any Geraghty-contraction is a contractive mapping and hence continuous, we have $T{x}_{n}\to T{x}^{\ast }$.

This implies that $d\left({x}_{n+1},T{x}_{n}\right)\to d\left({x}^{\ast },T{x}^{\ast }\right)$.

Taking into account that the sequence $\left(d\left({x}_{n+1},T{x}_{n}\right)\right)$ is a constant sequence with value $d\left(A,B\right)$, we deduce
$d\left({x}^{\ast },T{x}^{\ast }\right)=d\left(A,B\right).$

This means that ${x}^{\ast }$ is a best proximity point of T.

This proves the part of existence of our theorem.

For the uniqueness, suppose that ${x}_{1}$ and ${x}_{2}$ are two best proximity points of T with ${x}_{1}\ne {x}_{2}$.

This means that
Using the P-property, we have
$d\left({x}_{1},{x}_{2}\right)=d\left(T{x}_{1},T{x}_{2}\right).$
Using the fact that T is a Geraghty-contraction, we have
$d\left({x}_{1},{x}_{2}\right)=d\left(T{x}_{1},T{x}_{2}\right)\le \beta \left(d\left({x}_{1},{x}_{2}\right)\right)\cdot d\left({x}_{1},{x}_{2}\right)

Therefore, ${x}_{1}={x}_{2}$.

This finishes the proof. □

## 4 Examples

In order to illustrate our results, we present some examples.

Example 4.1 Consider $X={\mathbb{R}}^{2}$ with the usual metric.

Let A and B be the subsets of X defined by
$A=\left\{0\right\}×\left[0,\mathrm{\infty }\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\left\{1\right\}×\left[0,\mathrm{\infty }\right).$

Obviously, $d\left(A,B\right)=1$ and A, B are nonempty closed subsets of X.

Moreover, it is easily seen that ${A}_{0}=A$ and ${B}_{0}=B$.

Let $T:A\to B$ be the mapping defined as

In the sequel, we check that T is a Geraghty-contraction.

In fact, for $\left(0,x\right),\left(0,y\right)\in A$ with $x\ne y$, we have
$\begin{array}{rl}d\left(T\left(0,x\right),T\left(0,y\right)\right)& =d\left(\left(1,ln\left(1+x\right)\right),\left(1,ln\left(1+y\right)\right)\right)\\ =|ln\left(1+x\right)-ln\left(1+y\right)|\\ =|ln\left(\frac{1+x}{1+y}\right)|.\end{array}$
(4)
Now, we prove that
$|ln\left(\frac{1+x}{1+y}\right)|\le ln\left(1+|x-y|\right).$
(5)

Suppose that $x>y$ (the same reasoning works for $y>x$).

Then, since $\varphi \left(t\right)=ln\left(1+t\right)$ is strictly increasing in $\left[0,\mathrm{\infty }\right)$, we have
$ln\left(\frac{1+x}{1+y}\right)=ln\left(\frac{1+y+x-y}{1+y}\right)=ln\left(1+\frac{x-y}{1+y}\right)\le ln\left(1+x-y\right)=ln\left(1+|x-y|\right).$

This proves (5).

Taking into account (4) and (5), we have
$\begin{array}{rl}d\left(T\left(0,x\right),T\left(0,y\right)\right)& =|ln\left(\frac{1+x}{1+y}\right)|\\ \le ln\left(1+|x-y|\right)\\ =\frac{ln\left(1+|x-y|\right)}{|x-y|}\cdot |x-y|\\ =\frac{\varphi \left(d\left(\left(0,x\right),\left(0,y\right)\right)\right)}{d\left(\left(0,x\right),\left(0,y\right)\right)}\cdot d\left(\left(0,x\right),\left(0,y\right)\right)\\ =\beta \left(d\left(\left(0,x\right),\left(0,y\right)\right)\right)\cdot d\left(\left(0,x\right),\left(0,y\right)\right),\end{array}$
(6)

where $\varphi \left(t\right)=ln\left(1+t\right)$ for $t\ge 0$, and $\beta \left(t\right)=\frac{\varphi \left(t\right)}{t}$ for $t>0$ and $\beta \left(0\right)=0$.

Obviously, when $x=y$, the inequality (6) is satisfied.

It is easily seen that $\beta \left(t\right)=\frac{ln\left(1+t\right)}{t}\in \mathcal{F}$ by using elemental calculus.

Therefore, T is a Geraghty-contraction.

Notice that the pair $\left(A,B\right)$ satisfies the P-property.

Indeed, if
then ${x}_{1}={y}_{1}$ and ${x}_{2}={y}_{2}$ and consequently,
$d\left(\left(0,{x}_{1}\right),\left(0,{x}_{2}\right)\right)=|{x}_{1}-{x}_{2}|=|{y}_{1}-{y}_{2}|=d\left(\left(1,{y}_{1}\right),\left(1,{y}_{2}\right)\right).$

By Theorem 3.1, T has a unique best proximity point.

Obviously, this point is $\left(0,0\right)\in A$.

The condition A and B are nonempty closed subsets of the metric space $\left(X,d\right)$ is not a necessary condition for the existence of a unique best proximity point for a Geraghty-contraction $T:A\to B$ as it is proved with the following example.

Example 4.2 Consider $X={\mathbb{R}}^{2}$ with the usual metric and the subsets of X given by
$A=\left\{0\right\}×\left[0,\mathrm{\infty }\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\left\{1\right\}×\left[0,\frac{\pi }{2}\right).$

Obviously, $d\left(A,B\right)=1$ and B is not a closed subset of X.

Note that ${A}_{0}=0×\left[0,\frac{\pi }{2}\right)$ and ${B}_{0}=B$.

We consider the mapping $T:A\to B$ defined as

Now, we check that T is a Geraghty-contraction.

In fact, for $\left(0,x\right),\left(0,y\right)\in A$ with $x\ne y$, we have
$d\left(T\left(0,x\right),T\left(0,y\right)\right)=d\left(\left(1,arctanx\right),\left(1,arctany\right)\right)=|arctanx-arctany|.$
(7)
In what follows, we need to prove that
$|arctanx-arctany|\le arctan|x-y|.$
(8)

In fact, suppose that $x>y$ (the same argument works for $y>x$).

Put $arctanx=\alpha$ and $arctany=\beta$ (notice that $\alpha >\beta$ since the function $\varphi \left(t\right)=arctant$ for $t\ge 0$ is strictly increasing).

Taking into account that
$tan\left(\alpha -\beta \right)=\frac{tan\alpha -tan\beta }{1+tan\alpha \cdot tan\beta }$
and since $\alpha ,\beta \in \left[0,\frac{\pi }{2}\right)$, we have that $tan\alpha ,tan\beta \in \left[0,\mathrm{\infty }\right)$, and consequently, from the last inequality it follows that
$tan\left(\alpha -\beta \right)\le tan\alpha -tan\beta .$
Applying ϕ (notice that $\varphi \left(t\right)=arctant$) to the last inequality and taking into account the increasing character of ϕ, we have
$\alpha -\beta \le arctan\left(tan\alpha -tan\beta \right),$
or equivalently,
$arctanx-arctany=\alpha -\beta \le arctan\left(x-y\right),$

and this proves (8).

By (7) and (8), we get
$\begin{array}{rl}d\left(T\left(0,x\right),T\left(0,y\right)\right)& =|arctanx-arctany|\\ \le arctan|x-y|\\ =\frac{arctan|x-y|}{|x-y|}\cdot |x-y|\\ =\beta \left(d\left(0,x\right),d\left(0,y\right)\right)\cdot d\left(\left(0,x\right),\left(0,y\right)\right),\end{array}$
(9)

where $\beta \left(t\right)=\frac{arctant}{t}$ for $t>0$ and $\beta \left(0\right)=0$. Obviously, the inequality (9) is satisfied for $\left(0,x\right),\left(0,y\right)\in A$ with $x=y$.

Now, we prove that $\beta \in \mathcal{F}$.

In fact, if $\beta \left({t}_{n}\right)=\frac{arctan{t}_{n}}{{t}_{n}}\to 1$, then the sequence $\left({t}_{n}\right)$ is a bounded sequence since in the contrary case, ${t}_{n}\to \mathrm{\infty }$ and thus $\beta \left({t}_{n}\right)\to 0$. Suppose that ${t}_{n}↛0$. This means that there exists $ϵ>0$ such that, for each $n\in \mathbb{N}$, there exists ${p}_{n}\ge n$ with ${t}_{{p}_{n}}\ge ϵ$. The bounded character of $\left({t}_{n}\right)$ gives us the existence of a subsequence $\left({t}_{{k}_{n}}\right)$ of $\left({t}_{{p}_{n}}\right)$ with $\left({t}_{{k}_{n}}\right)$ convergent. Suppose that ${t}_{{k}_{n}}\to a$. From $\beta \left({t}_{n}\right)\to 1$, we obtain $\frac{arctan{t}_{{k}_{n}}}{{t}_{{k}_{n}}}\to \frac{arctana}{a}=1$ and, as the unique solution of $arctanx=x$ is ${x}_{0}=0$, we obtain $a=0$.

Thus, ${t}_{{k}_{n}}\to 0$ and this contradicts the fact that ${t}_{{k}_{n}}\ge ϵ$ for any $n\in \mathbb{N}$.

Therefore, ${t}_{n}\to 0$ and this proves that $\beta \in \mathcal{F}$.

A similar argument to the one used in Example 4.1 proves that the pair $\left(A,B\right)$ has the P-property.

On the other hand, the point $\left(0,0\right)\in A$ is a best proximity point for T since
$d\left(\left(0,0\right),T\left(0,0\right)\right)=d\left(\left(0,0\right),\left(1,arctan0\right)\right)=d\left(\left(0,0\right),\left(1,0\right)\right)=1=d\left(A,B\right).$

Moreover, $\left(0,0\right)$ is the unique best proximity point for T.

Indeed, if $\left(0,x\right)\in A$ is a best proximity point for T, then
$1=d\left(A,B\right)=d\left(\left(0,x\right),T\left(0,x\right)\right)=d\left(\left(0,x\right),\left(1,arctanx\right)\right)=\sqrt{1+{\left(x-arctanx\right)}^{2}},$
and this gives us
$x=arctanx.$

Taking into account that the unique solution of this equation is $x=0$, we have proved that T has a unique best proximity point which is $\left(0,0\right)$.

Notice that in this case B is not closed.

Since for any nonempty subset A of X, the pair $\left(A,A\right)$ satisfies the P-property, we have the following corollary.

Corollary 4.1 Let $\left(X,d\right)$ be a complete metric space and A be a nonempty closed subset of X. Let $T:A\to A$ be a Geraghty-contraction. Then T has a unique fixed point.

Proof Using Theorem 3.1 when $A=B$, the desired result follows. □

Notice that when $A=X$, Corollary 4.1 is Theorem 1.1 due to Gerahty [1].

## Declarations

### Acknowledgements

This research was partially supported by ‘Universidad de Las Palmas de Gran Canaria’, Project ULPGC 2010-006.

## Authors’ Affiliations

(1)
Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria

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