A best proximity point theorem for Geraghty-contractions

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).

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].

First, we introduce the class ℱ of those functions $\beta :[0,\mathrm{\infty })\to [0,1)$ satisfying the following condition:

$$\beta (t_n)\to 1\text{implies}{t}_n\to 0.$$

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 $x,y\in X$,

$$d(Tx,Ty)\le \beta (d(x,y))\cdot d(x,y).$$

(1)

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

$$d(Tx,Ty)<d(x,y)\text{for any }x,y\in X\text{ with }x\ne y.$$

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 $(X,d)$.

We denote by $A_0$ and $B_0$ the following sets:

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

$$d(Tx,Ty)\le \beta (d(x,y))\cdot d(x,y)\text{for any }x,y\in A.$$

Notice that since $\beta :[0,\mathrm{\infty })\to [0,1)$, we have

$$d(Tx,Ty)\le \beta (d(x,y))\cdot d(x,y)<d(x,y)\text{for any }x,y\in A\text{ with }x\ne y.$$

Therefore, every Geraghty-contraction is a contractive mapping.

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

Definition 2.2 ([10])

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X,d)$ with $A_0\ne \mathrm{\varnothing }$. Then the pair $(A,B)$ is said to have the P-property if and only if for any $x_1,x_2\in A_0$ and $y_1,y_2\in B_0$,

$$\begin{array}{lcr}d(x_1,y_1)& =& d(A,B)\\ d(x_2,y_2)& =& d(A,B)\end{array}\}\Rightarrow d(x_1,x_2)=d(y_1,y_2).$$

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 $x_0\in A$.

As $T{x}_0\in T(A_0)\subseteq B_0$, we can find $x_1\in A_0$ such that $d(x_1,T{x}_0)=d(A,B)$. Similarly, since $T{x}_1\in T(A_0)\subseteq B_0$, there exists $x_2\in A_0$ such that $d(x_2,T{x}_1)=d(A,B)$. Repeating this process, we can get a sequence $(x_n)$ in $A_0$ satisfying

$$d(x_{n+1},T{x}_n)=d(A,B)\text{for any }n\in \mathbb{N}.$$

Since $(A,B)$ has the P-property, we have that

$$d(x_n,x_{n+1})=d(T{x}_{n-1},T{x}_n)\text{for any }n\in \mathbb{N}.$$

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

$$d(x_n,x_{n+1})=d(T{x}_{n-1},T{x}_n)\le \beta (d(x_{n-1},x_n))\cdot d(x_{n-1},x_n)<d(x_{n-1},x_n).$$

(2)

Suppose that there exists $n_0\in \mathbb{N}$ such that $d(x_{n_0},x_{n_0+1})=0$.

In this case,

$$0=d(x_{n_0},x_{n_0+1})=d(T{x}_{n_0-1},T{x}_{n_0}),$$

and consequently, $T{x}_{n_0-1}=T{x}_{n_0}$.

Therefore,

$$d(A,B)=d(x_{n_0},T{x}_{n_0-1})=d(x_{n_0},T{x}_{n_0})$$

and this is the desired result.

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

By (2), $(d(x_n,x_{n+1}))$ is a decreasing sequence of nonnegative real numbers, and hence there exists $r\ge 0$ such that

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,x_{n+1})=r.$$

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

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

$$0<\frac{d(x_n,x_{n+1})}{d(x_{n-1},x_n)}\le \beta (d(x_{n-1},x_n))<1\text{for any }n\in \mathbb{N}.$$

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.

Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}d(x_n,x_{n+1})=0.$$

(3)

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.

In the contrary case, we have that

$$\underset{m,n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}d(x_n,x_m)>0.$$

By using the triangular inequality,

$$d(x_n,x_m)\le d(x_n,x_{n+1})+d(x_{n+1},x_{m+1})+d(x_{m+1},x_m).$$

By (2) and since $d(x_{n+1},x_{m+1})=d(T{x}_n,T{x}_m)$, by the above mentioned comment, we have

$$\begin{array}{rl}d(x_n,x_m)& \le d(x_n,x_{n+1})+d(T{x}_n,T{x}_m)+d(x_{m+1},x_m)\\ \le d(x_n,x_{n+1})+\beta (d(x_n,x_m))\cdot d(x_n,x_m)+d(x_{m+1},x_m),\end{array}$$

which gives us

$$d(x_n,x_m)\le {(1-\beta (d(x_n,x_m)))}^{-1}[d(x_n,x_{n+1})+d(x_{m+1},x_m)].$$

Since ${lim\hspace{0.17em}sup}_{m,n\to \mathrm{\infty }}d(x_n,x_m)>0$ and by (3), ${lim\hspace{0.17em}sup}_{n\to \mathrm{\infty }}d(x_n,x_{n+1})=0$, from the last inequality it follows that

$$\underset{m,n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}{(1-\beta (d(x_n,x_m)))}^{-1}=\mathrm{\infty }.$$

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 $d(x_{n+1},T{x}_n)\to d(x^{\ast },T{x}^{\ast })$.

Taking into account that the sequence $(d(x_{n+1},T{x}_n))$ is a constant sequence with value $d(A,B)$, we deduce

$$d(x^{\ast },T{x}^{\ast })=d(A,B).$$

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

$$d(x_i,T{x}_i)=d(A,B)\text{for }i=1,2.$$

Using the P-property, we have

$$d(x_1,x_2)=d(T{x}_1,T{x}_2).$$

Using the fact that T is a Geraghty-contraction, we have

$$d(x_1,x_2)=d(T{x}_1,T{x}_2)\le \beta (d(x_1,x_2))\cdot d(x_1,x_2)<d(x_1,x_2),$$

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.

Let A and B be the subsets of X defined by

$$A=\{0\}\times [0,\mathrm{\infty })\text{and}B=\{1\}\times [0,\mathrm{\infty }).$$

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

Moreover, it is easily seen that $A_0=A$ and $B_0=B$.

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

$$T(0,x)=(1,ln(1+x))\text{for any }(0,x)\in A.$$

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

In fact, for $(0,x),(0,y)\in A$ with $x\ne y$, we have

$$\begin{array}{rl}d(T(0,x),T(0,y))& =d((1,ln(1+x)),(1,ln(1+y)))\\ =|ln(1+x)-ln(1+y)|\\ =|ln\left(\frac{1+x}{1+y}\right)|.\end{array}$$

(4)

Now, we prove that

$$|ln\left(\frac{1+x}{1+y}\right)|\le ln(1+|x-y|).$$

(5)

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

Then, since $\varphi (t)=ln(1+t)$ is strictly increasing in $[0,\mathrm{\infty })$, we have

$$ln\left(\frac{1+x}{1+y}\right)=ln\left(\frac{1+y+x-y}{1+y}\right)=ln(1+\frac{x-y}{1+y})\le ln(1+x-y)=ln(1+|x-y|).$$

This proves (5).

Taking into account (4) and (5), we have

$$\begin{array}{rl}d(T(0,x),T(0,y))& =|ln\left(\frac{1+x}{1+y}\right)|\\ \le ln(1+|x-y|)\\ =\frac{ln(1+|x-y|)}{|x-y|}\cdot |x-y|\\ =\frac{\varphi (d((0,x),(0,y)))}{d((0,x),(0,y))}\cdot d((0,x),(0,y))\\ =\beta \left(d((0,x),(0,y))\right)\cdot d((0,x),(0,y)),\end{array}$$

(6)

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.

Indeed, if

then $x_1=y_1$ and $x_2=y_2$ and consequently,

$$d((0,x_1),(0,x_2))=|x_1-x_2|=|y_1-y_2|=d((1,y_1),(1,y_2)).$$

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

$$A=\{0\}\times [0,\mathrm{\infty })\text{and}B=\{1\}\times [0,\frac{\pi }{2}).$$

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 $B_0=B$.

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

$$T(0,x)=(1,arctanx)\text{for any }(0,x)\in A.$$

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

In fact, for $(0,x),(0,y)\in A$ with $x\ne y$, we have

$$d(T(0,x),T(0,y))=d((1,arctanx),(1,arctany))=|arctanx-arctany|.$$

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

Taking into account that

$$tan(\alpha -\beta )=\frac{tan\alpha -tan\beta }{1+tan\alpha \cdot tan\beta }$$

and since $\alpha ,\beta \in [0,\frac{\pi }{2})$, we have that $tan\alpha ,tan\beta \in [0,\mathrm{\infty })$, and consequently, from the last inequality it follows that

$$tan(\alpha -\beta )\le tan\alpha -tan\beta .$$

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

$$\alpha -\beta \le arctan(tan\alpha -tan\beta ),$$

or equivalently,

$$arctanx-arctany=\alpha -\beta \le arctan(x-y),$$

and this proves (8).

By (7) and (8), we get

$$\begin{array}{rl}d(T(0,x),T(0,y))& =|arctanx-arctany|\\ \le arctan|x-y|\\ =\frac{arctan|x-y|}{|x-y|}\cdot |x-y|\\ =\beta (d(0,x),d(0,y))\cdot d((0,x),(0,y)),\end{array}$$

(9)

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 $ϵ>0$ such that, for each $n\in \mathbb{N}$, there exists $p_n\ge n$ with $t_{p_n}\ge ϵ$. The bounded character of $(t_n)$ 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 $t_{k_n}\ge ϵ$ 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 $(A,B)$ has the P-property.

On the other hand, the point $(0,0)\in A$ is a best proximity point for T since

$$d((0,0),T(0,0))=d((0,0),(1,arctan0))=d((0,0),(1,0))=1=d(A,B).$$

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

Indeed, if $(0,x)\in A$ is a best proximity point for T, then

$$1=d(A,B)=d((0,x),T(0,x))=d((0,x),(1,arctanx))=\sqrt{1+{(x-arctanx)}^2},$$

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 $(0,0)$.

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].