Open Access

A best proximity point theorem for Geraghty-contractions

Fixed Point Theory and Applications20122012:231

DOI: 10.1186/1687-1812-2012-231

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 B is said to be a k-contraction if there exists k [ 0 , 1 ) such that d ( T x , T y ) k d ( x , y ) for any x , y 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 β : [ 0 , ) [ 0 , 1 ) satisfying the following condition:
β ( t n ) 1 implies t n 0 .

Theorem 1.1 ([1])

Let ( X , d ) be a complete metric space and T : X X be an operator. Suppose that there exists β F such that for any x , y X ,
d ( T x , T y ) β ( d ( x , y ) ) d ( x , y ) .
(1)

Then T has a unique fixed point.

Since the constant functions f ( t ) = k , where k [ 0 , 1 ) , 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
d ( T x , T y ) < d ( x , y ) for any  x , y X  with  x y .

Therefore, any operator T : X 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 X be a mapping. The solutions of the equation T x = x are fixed points of T. Consequently, T ( A ) A 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 , T x ) > 0 for any x A and the mapping T : A X does not have any fixed point. In this setting, our aim is to find an element x A such that d ( x , T x ) 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 B .

A natural question is whether one can find an element x 0 A such that d ( x 0 , T x 0 ) = min { d ( x , T x ) : x A } . Since d ( x , T x ) d ( A , B ) for any x 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 φ : A R given by φ ( x ) = d ( x , T x ) .

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 ( X , d ) .

We denote by A 0 and B 0 the following sets:
https://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2012-231/MediaObjects/13663_2012_Article_332_Equc_HTML.gif

where d ( A , B ) = inf { d ( x , y ) : x A  and  y 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 B is said to be a Geraghty-contraction if there exists β F such that
d ( T x , T y ) β ( d ( x , y ) ) d ( x , y ) for any  x , y A .
Notice that since β : [ 0 , ) [ 0 , 1 ) , we have
d ( T x , T y ) β ( d ( x , y ) ) d ( x , y ) < d ( x , y ) for any  x , y A  with  x 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 . Then the pair ( A , B ) is said to have the P-property if and only if for any x 1 , x 2 A 0 and y 1 , y 2 B 0 ,
d ( x 1 , y 1 ) = d ( A , B ) d ( x 2 , y 2 ) = d ( A , B ) } 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 B be a Geraghty-contraction satisfying T ( A 0 ) B 0 . Suppose that the pair ( A , B ) has the P-property. Then there exists a unique x in A such that d ( x , T x ) = d ( A , B ) .

Proof Since A 0 is nonempty, we take x 0 A .

As T x 0 T ( A 0 ) B 0 , we can find x 1 A 0 such that d ( x 1 , T x 0 ) = d ( A , B ) . Similarly, since T x 1 T ( A 0 ) B 0 , there exists x 2 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 ) for any  n 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 ) for any  n N .
Taking into account that T is a Geraghty-contraction, for any n N , we have that
d ( x n , x n + 1 ) = d ( T x n 1 , T x n ) β ( d ( x n 1 , x n ) ) d ( x n 1 , x n ) < d ( x n 1 , x n ) .
(2)

Suppose that there exists n 0 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 N .

By (2), ( d ( x n , x n + 1 ) ) is a decreasing sequence of nonnegative real numbers, and hence there exists r 0 such that
lim n d ( x n , x n + 1 ) = r .

In the sequel, we prove that r = 0 .

Assume r > 0 , then from (2) we have
0 < d ( x n , x n + 1 ) d ( x n 1 , x n ) β ( d ( x n 1 , x n ) ) < 1 for any  n N .

The last inequality implies that lim n β ( d ( x n 1 , x n ) ) = 1 and since β F , we obtain r = 0 and this contradicts our assumption.

Therefore,
lim n d ( x n , x n + 1 ) = 0 .
(3)

Notice that since d ( x n + 1 , T x n ) = d ( A , B ) for any n N , for p , q 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
lim sup m , n d ( x n , x m ) > 0 .
By using the triangular inequality,
d ( x n , x m ) 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
d ( x n , x m ) d ( x n , x n + 1 ) + d ( T x n , T x m ) + d ( x m + 1 , x m ) d ( x n , x n + 1 ) + β ( d ( x n , x m ) ) d ( x n , x m ) + d ( x m + 1 , x m ) ,
which gives us
d ( x n , x m ) ( 1 β ( d ( x n , x m ) ) ) 1 [ d ( x n , x n + 1 ) + d ( x m + 1 , x m ) ] .
Since lim sup m , n d ( x n , x m ) > 0 and by (3), lim sup n d ( x n , x n + 1 ) = 0 , from the last inequality it follows that
lim sup m , n ( 1 β ( d ( x n , x m ) ) ) 1 = .

Therefore, lim sup m , n β ( d ( x n , x m ) ) = 1 .

Taking into account that β F , we get lim sup m , n d ( x n , x m ) = 0 and this contradicts our assumption.

Therefore, ( x n ) is a Cauchy sequence.

Since ( x n ) A and A is a closed subset of the complete metric space ( X , d ) , we can find x A such that x n x .

Since any Geraghty-contraction is a contractive mapping and hence continuous, we have T x n T x .

This implies that d ( x n + 1 , T x n ) d ( x , T x ) .

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 , T x ) = d ( A , B ) .

This means that x 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 x 2 .

This means that
d ( x i , T x i ) = d ( A , B ) 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 ) β ( d ( x 1 , x 2 ) ) 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 = R 2 with the usual metric.

Let A and B be the subsets of X defined by
A = { 0 } × [ 0 , ) and B = { 1 } × [ 0 , ) .

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 B be the mapping defined as
T ( 0 , x ) = ( 1 , ln ( 1 + x ) ) for any  ( 0 , x ) A .

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

In fact, for ( 0 , x ) , ( 0 , y ) A with x y , we have
d ( T ( 0 , x ) , T ( 0 , y ) ) = d ( ( 1 , ln ( 1 + x ) ) , ( 1 , ln ( 1 + y ) ) ) = | ln ( 1 + x ) ln ( 1 + y ) | = | ln ( 1 + x 1 + y ) | .
(4)
Now, we prove that
| ln ( 1 + x 1 + y ) | ln ( 1 + | x y | ) .
(5)

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

Then, since ϕ ( t ) = ln ( 1 + t ) is strictly increasing in [ 0 , ) , we have
ln ( 1 + x 1 + y ) = ln ( 1 + y + x y 1 + y ) = ln ( 1 + x y 1 + y ) ln ( 1 + x y ) = ln ( 1 + | x y | ) .

This proves (5).

Taking into account (4) and (5), we have
d ( T ( 0 , x ) , T ( 0 , y ) ) = | ln ( 1 + x 1 + y ) | ln ( 1 + | x y | ) = ln ( 1 + | x y | ) | x y | | x y | = ϕ ( d ( ( 0 , x ) , ( 0 , y ) ) ) d ( ( 0 , x ) , ( 0 , y ) ) d ( ( 0 , x ) , ( 0 , y ) ) = β ( d ( ( 0 , x ) , ( 0 , y ) ) ) d ( ( 0 , x ) , ( 0 , y ) ) ,
(6)

where ϕ ( t ) = ln ( 1 + t ) for t 0 , and β ( t ) = ϕ ( t ) t for t > 0 and β ( 0 ) = 0 .

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

It is easily seen that β ( t ) = ln ( 1 + t ) t F by using elemental calculus.

Therefore, T is a Geraghty-contraction.

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

Indeed, if
https://static-content.springer.com/image/art%3A10.1186%2F1687-1812-2012-231/MediaObjects/13663_2012_Article_332_Equy_HTML.gif
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 ) 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 B as it is proved with the following example.

Example 4.2 Consider X = R 2 with the usual metric and the subsets of X given by
A = { 0 } × [ 0 , ) and B = { 1 } × [ 0 , π 2 ) .

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

Note that A 0 = 0 × [ 0 , π 2 ) and B 0 = B .

We consider the mapping T : A B defined as
T ( 0 , x ) = ( 1 , arctan x ) for any  ( 0 , x ) A .

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

In fact, for ( 0 , x ) , ( 0 , y ) A with x y , we have
d ( T ( 0 , x ) , T ( 0 , y ) ) = d ( ( 1 , arctan x ) , ( 1 , arctan y ) ) = | arctan x arctan y | .
(7)
In what follows, we need to prove that
| arctan x arctan y | arctan | x y | .
(8)

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

Put arctan x = α and arctan y = β (notice that α > β since the function ϕ ( t ) = arctan t for t 0 is strictly increasing).

Taking into account that
tan ( α β ) = tan α tan β 1 + tan α tan β
and since α , β [ 0 , π 2 ) , we have that tan α , tan β [ 0 , ) , and consequently, from the last inequality it follows that
tan ( α β ) tan α tan β .
Applying ϕ (notice that ϕ ( t ) = arctan t ) to the last inequality and taking into account the increasing character of ϕ, we have
α β arctan ( tan α tan β ) ,
or equivalently,
arctan x arctan y = α β arctan ( x y ) ,

and this proves (8).

By (7) and (8), we get
d ( T ( 0 , x ) , T ( 0 , y ) ) = | arctan x arctan y | arctan | x y | = arctan | x y | | x y | | x y | = β ( d ( 0 , x ) , d ( 0 , y ) ) d ( ( 0 , x ) , ( 0 , y ) ) ,
(9)

where β ( t ) = arctan t t for t > 0 and β ( 0 ) = 0 . Obviously, the inequality (9) is satisfied for ( 0 , x ) , ( 0 , y ) A with x = y .

Now, we prove that β F .

In fact, if β ( t n ) = arctan t n t n 1 , then the sequence ( t n ) is a bounded sequence since in the contrary case, t n and thus β ( t n ) 0 . Suppose that t n 0 . This means that there exists ϵ > 0 such that, for each n N , there exists p n n with t p n 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 a . From β ( t n ) 1 , we obtain arctan t k n t k n arctan a a = 1 and, as the unique solution of arctan x = x is x 0 = 0 , we obtain a = 0 .

Thus, t k n 0 and this contradicts the fact that t k n ϵ for any n N .

Therefore, t n 0 and this proves that β 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 ) A is a best proximity point for T since
d ( ( 0 , 0 ) , T ( 0 , 0 ) ) = d ( ( 0 , 0 ) , ( 1 , arctan 0 ) ) = d ( ( 0 , 0 ) , ( 1 , 0 ) ) = 1 = d ( A , B ) .

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

Indeed, if ( 0 , x ) A is a best proximity point for T, then
1 = d ( A , B ) = d ( ( 0 , x ) , T ( 0 , x ) ) = d ( ( 0 , x ) , ( 1 , arctan x ) ) = 1 + ( x arctan x ) 2 ,
and this gives us
x = arctan x .

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 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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© Caballero et al.; licensee Springer. 2012

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