# A best proximity point theorem for Geraghty-contractions

- J Caballero
^{1}, - J Harjani
^{1}and - K Sadarangani
^{1}Email author

**2012**:231

**DOI: **10.1186/1687-1812-2012-231

© Caballero et al.; licensee Springer. 2012

**Received: **16 May 2012

**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 $(X,d)$.

An operator $T:A\to B$ is said to be a k-contraction if there exists $k\in [0,1)$ such that $d(Tx,Ty)\le kd(x,y)$ 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].

**Theorem 1.1** ([1])

*Let*$(X,d)$

*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*$x,y\in X$,

*Then* *T* *has a unique fixed point*.

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

**Remark 1.1**Since the functions belonging to ℱ 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 $(X,d)$ and $T:A\to X$ be a mapping. The solutions of the equation $Tx=x$ are fixed points of *T*. Consequently, $T(A)\cap A\ne \mathrm{\varnothing}$ is a necessary condition for the existence of a fixed point for the operator *T*. If this necessary condition does not hold, then $d(x,Tx)>0$ 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 $d(x,Tx)$ 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 $d({x}_{0},T{x}_{0})=min\{d(x,Tx):x\in A\}$. Since $d(x,Tx)\ge d(A,B)$ for any $x\in A$, the optimal solution to this problem will be the one for which the value $d(A,B)$ is attained by the real valued function $\phi :A\to \mathbb{R}$ given by $\phi (x)=d(x,Tx)$.

Some results about best proximity points can be found in [2–9].

## 2 Notations and basic facts

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

where $d(A,B)=inf\{d(x,y):x\in A\text{and}y\in B\}$.

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 $(X,d)$. A mapping $T:A\to B$ is said to be a Geraghty-contraction if there exists $\beta \in \mathcal{F}$ such that

Therefore, every Geraghty-contraction is a contractive mapping.

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

**Definition 2.2** ([10])

*P*-property if and only if for any ${x}_{1},{x}_{2}\in {A}_{0}$ and ${y}_{1},{y}_{2}\in {B}_{0}$,

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

In [10], the author proves that any pair $(A,B)$ 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* $(A,B)$ *be a pair of nonempty closed subsets of a complete metric space* $(X,d)$ *such that* ${A}_{0}$ *is nonempty*. *Let* $T:A\to B$ *be a Geraghty*-*contraction satisfying* $T({A}_{0})\subseteq {B}_{0}$. *Suppose that the pair* $(A,B)$ *has the* *P*-*property*. *Then there exists a unique* ${x}^{\ast}$ *in* *A* *such that* $d({x}^{\ast},T{x}^{\ast})=d(A,B)$.

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

*P*-property, we have that

*T*is a Geraghty-contraction, for any $n\in \mathbb{N}$, we have that

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

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

and this is the desired result.

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

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

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

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

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

Therefore, ${lim\hspace{0.17em}sup}_{m,n\to \mathrm{\infty}}\beta (d({x}_{n},{x}_{m}))=1$.

Taking into account that $\beta \in \mathcal{F}$, we get ${lim\hspace{0.17em}sup}_{m,n\to \mathrm{\infty}}d({x}_{n},{x}_{m})=0$ and this contradicts our assumption.

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

Since $({x}_{n})\subset A$ and *A* is a closed subset of the complete metric space $(X,d)$, 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 $d({x}_{n+1},T{x}_{n})\to d({x}^{\ast},T{x}^{\ast})$.

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 ${x}_{1}\ne {x}_{2}$.

*P*-property, we have

*T*is a Geraghty-contraction, we have

which is a contradiction.

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.

*A*and

*B*be the subsets of

*X*defined by

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

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

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

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

This proves (5).

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

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

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

Therefore, *T* is a Geraghty-contraction.

Notice that the pair $(A,B)$ satisfies the *P*-property.

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

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

The condition *A* and *B* are nonempty closed subsets of the metric space $(X,d)$ 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

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

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

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

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 $\varphi (t)=arctant$ for $t\ge 0$ is strictly increasing).

*ϕ*(notice that $\varphi (t)=arctant$) to the last inequality and taking into account the increasing character of

*ϕ*, we have

and this proves (8).

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

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

In fact, if $\beta ({t}_{n})=\frac{arctan{t}_{n}}{{t}_{n}}\to 1$, then the sequence $({t}_{n})$ is a bounded sequence since in the contrary case, ${t}_{n}\to \mathrm{\infty}$ and thus $\beta ({t}_{n})\to 0$. Suppose that ${t}_{n}\nrightarrow 0$. This means that there exists $\u03f5>0$ such that, for each $n\in \mathbb{N}$, there exists ${p}_{n}\ge n$ with ${t}_{{p}_{n}}\ge \u03f5$. The bounded character of $({t}_{n})$ gives us the existence of a subsequence $({t}_{{k}_{n}})$ of $({t}_{{p}_{n}})$ with $({t}_{{k}_{n}})$ convergent. Suppose that ${t}_{{k}_{n}}\to a$. From $\beta ({t}_{n})\to 1$, 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 ${t}_{{k}_{n}}\ge \u03f5$ 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 $(A,B)$ has the *P*-property.

*T*since

Moreover, $(0,0)$ is the unique best proximity point for *T*.

*T*, then

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 $(0,0)$.

Notice that in this case *B* is not closed.

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

**Corollary 4.1** *Let* $(X,d)$ *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

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