Open Access

The existence of best proximity points with the weak P-property

Fixed Point Theory and Applications20132013:259

https://doi.org/10.1186/1687-1812-2013-259

Received: 24 June 2013

Accepted: 18 September 2013

Published: 7 November 2013

Abstract

We improve some existence theorem of best proximity points with the weak P-property, which has been recently proved by Zhang et al.

MSC:54H25, 54E50.

Keywords

best proximity point the weak P-property the Banach contraction principle Kannan’s fixed point theorem completion

1 Introduction

Let ( A , B ) be a pair of nonempty subsets of a metric space ( X , d ) , and let T be a mapping from A into B. Then x A is called a best proximity point if d ( x , T x ) = d ( A , B ) , where d ( A , B ) = inf { d ( x , y ) : x A , y B } . We have proved many existence theorems of best proximity points. See, for example, [16]. Very recently, Caballero et al. [7] proved a new type of existence theorem, and Zhang et al. [8] generalized the theorem. The theorem proved in [8] is Theorem 8 with an additional assumption of the completeness of B. The essence of the result in [7] becomes very clear in [8], however, we have not learned the essence completely.

Motivated by the fact above, in this paper, we improve the result in [8]. Also, in order to consider the discontinuous case, we give a Kannan version.

2 Preliminaries

In this section, we give some preliminaries.

Definition 1 Let ( A , B ) be a pair of nonempty subsets of a metric space ( X , d ) , and define A 0 and B 0 by
A 0 = { x A :  there exists  u B  such that  d ( x , u ) = d ( A , B ) }
(1)
and
B 0 = { u B :  there exists  x A  such that  d ( x , u ) = d ( A , B ) } .
(2)

Then

  • (Sankar Raj [9]) ( A , B ) is said to have the P-property if A 0 and the following holds:
    x , y A 0 , u , v B 0 , d ( x , u ) = d ( y , v ) = d ( A , B ) d ( x , y ) = d ( u , v ) .
  • (Zhang et al.[8]) ( A , B ) is said to have the weak P-property if A 0 and the following holds:
    x , y A 0 , u , v B 0 , d ( x , u ) = d ( y , v ) = d ( A , B ) d ( x , y ) d ( u , v ) .
Proposition 2 Let ( A , B ) be a pair of nonempty subsets of a metric space ( X , d ) , and define A 0 and B 0 by (1) and (2). Assume that A 0 . Then the following are equivalent:
  1. (i)

    ( A , B ) has the weak P-property.

     
  2. (ii)

    The conjunction of the following holds:

     

(ii-1) For every u B 0 , there exists a unique x A 0 with d ( x , u ) = d ( A , B ) .

(ii-2) There exists a nonexpansive mapping Q from B 0 into A 0 such that d ( Q u , u ) = d ( A , B ) for every u B 0 .

Proof We note that B 0 because A 0 . First, we assume (i). Let x , y A 0 and u B 0 satisfy d ( x , u ) = d ( y , u ) = d ( A , B ) . Then from (i), we have
d ( x , y ) d ( u , u ) = 0 ,
thus, x = y . So (ii-1) holds. We put Q u = x . Then from the definition of the weak P-property, we have d ( Q u , Q v ) d ( u , v ) for u , v B 0 , that is, Q is nonexpansive. Conversely, we assume (ii). Let x , y A 0 and u , v B 0 satisfy d ( x , u ) = d ( y , v ) = d ( A , B ) . Then from (ii-1), we have Q u = x and Q v = y . Therefore,
d ( x , y ) = d ( Q u , Q v ) d ( u , v )

holds. □

Lemma 3 Let ( A , B ) be a pair of subsets of a metric space ( X , d ) , and define A 0 and B 0 by (1) and (2). Assume that A 0 . Let T be a mapping from A into B, and let Q be a mapping from B 0 into A 0 such that d ( Q u , u ) = d ( A , B ) for every u B 0 . Then the following holds:
{ u n } B 0 , lim n u n = w , T ( lim n Q u n ) = w w B 0 .
(3)
Proof Let { u n } be a sequence in B 0 such that { u n } converges to w X , and T ( lim n Q u n ) = w . We put y = lim n Q u n . Since T y = w , we have y A and w B . Since
d ( y , w ) = lim n d ( Q u n , u n ) = d ( A , B ) ,

we have y A 0 and w B 0 . □

Lemma 4 Let ( X , d ) be a metric space, let A, A 0 , B 0 be nonempty subsets such that A is complete and A 0 A . Let T be a mapping from A into X such that T ( A 0 ) B 0 , and let Q be a nonexpansive mapping from B 0 into A 0 . Let Q ¯ be the mapping whose graph Gr ( Q ¯ ) is the completion of Gr ( Q ) . Assume (3). Then the following hold:
  1. (i)

    Q ¯ is well-defined and nonexpansive.

     
  2. (ii)

    Q ¯ w = z is equivalent to that there exists a sequence { u n } in B 0 such that lim n u n = w and lim n Q u n = z .

     
  3. (iii)

    The domain of Q ¯ is B 0 ¯ , where B 0 ¯ is the completion of B 0 .

     
  4. (iv)

    The range of Q ¯ is a subset of A 0 ¯ , where A 0 ¯ is the completion of A 0 .

     
  5. (v)

    T Q ¯ w = w implies T Q w = w .

     
  6. (vi)

    Q ¯ T z = z implies Q T z = z .

     
  7. (vii)

    The range of Q ¯ is a subset of A.

     

Proof We consider that the whole space is the completion of X. Since Q is Lipschitz continuous, Q ¯ is well-defined. The rest of (i) and (ii)-(iv) are obvious. By using (3), we can easily prove (v) and (vi). From the completeness of A, we obtain (vii). □

3 Fixed point theorems

In this section, we give fixed point theorems, which are used in the proofs of the main results.

Theorem 5 Let ( X , d ) be a metric space, let A, A 0 , B 0 be nonempty subsets such that A is complete and A 0 A . Let T be a contraction from A into X such that T ( A 0 ) B 0 , and let Q be a nonexpansive mapping from B 0 into A 0 . Assume (3). Then Q T has a unique fixed point in A 0 .

Proof We consider that the whole space is the completion of X. Define a nonexpansive mapping Q ¯ as in Lemma 4. Since T is continuous, T ( A 0 ¯ ) is a subset of B 0 ¯ . Let S be the restriction of T to A 0 ¯ . Then Q ¯ S is a contraction on A 0 ¯ . So the Banach contraction principle yields that there exists a unique fixed point z of Q ¯ S in A 0 ¯ . Since Q ¯ T z = z , by Lemma 4(vi), z is a fixed point of Q T . □

Remark

  • If X = A = A 0 = B 0 and Q is the identity mapping on B 0 , then Theorem 5 becomes the Banach contraction principle [10].

  • We can prove Theorem 5 with the mapping T Q ¯ as in the proof of Theorem 7.

We prove generalizations of Kannan’s fixed point theorem [11].

Theorem 6 Let ( X , d ) be a metric space, let Y be a complete subset of X, and let T be a mapping from Y into X. Assume that the following hold:
  1. (i)

    There exists α [ 0 , 1 / 2 ) such that d ( T x , T y ) α d ( x , T x ) + α d ( y , T y ) for all x , y Y .

     
  2. (ii)

    There exists a nonempty subset Z of Y such that T ( Z ) Z .

     

Then there exists a unique fixed point z, and for every x Z , { T n x } converges to z.

Proof Fix x Z . Then from the proof in Kannan [11], we obtain that { T n x } converges to a fixed point, and the fixed point is unique. □

Remark If X = Y = Z , then Theorem 6 becomes Kannan’s fixed point theorem [11].

Using Theorem 6, we obtain the following.

Theorem 7 Let ( X , d ) be a metric space, let A, A 0 , B 0 be nonempty subsets such that A is complete and A 0 A . Let T be a mapping from A into X such that T ( A 0 ) B 0 , and let Q be a nonexpansive mapping from B 0 into A 0 . Assume that (3) and the following hold:

  • There exist α [ 0 , 1 / 2 ) and μ [ 0 , ) such that
    d ( T x , T y ) α ( d ( x , T x ) μ ) + α ( d ( y , T y ) μ )

for x , y A and d ( Q u , u ) μ for all u B 0 .

Then T Q has a unique fixed point in B 0 .

Proof We consider that the whole space is the completion of X. Define a nonexpansive mapping Q ¯ as in Lemma 4. From the continuity of d, d ( Q ¯ u , u ) μ for u B 0 ¯ . For u , v B 0 ¯ , we have
d ( T Q ¯ u , T Q ¯ v ) α ( d ( Q ¯ u , T Q ¯ u ) μ ) + α ( d ( Q ¯ v , T Q ¯ v ) μ ) α ( d ( Q ¯ u , u ) + d ( u , T Q ¯ u ) μ ) + α ( d ( Q ¯ v , v ) + d ( v , T Q ¯ v ) μ ) α d ( u , T Q ¯ u ) + α d ( v , T Q ¯ v ) .

Hence T Q ¯ is a Kannan mapping from B 0 ¯ into X. T Q ¯ ( B 0 ) = T Q ( B 0 ) B 0 is obvious. So by Theorem 6, there exists a unique fixed point w of T Q ¯ in B 0 ¯ . By Lemma 4(v), w B 0 and w is a fixed point of T Q . □

Remark

  • Since T is not necessarily continuous, the range of T Q ¯ is not necessarily included by B 0 ¯ . Because of the same reason, we cannot prove Theorem 7 with the mapping Q ¯ T .

  • It is interesting that we do not need the completeness of any set related to B 0 directly. Of course, we need the completeness of A.

4 Main results

In this section, we give the main results.

Theorem 8 (Zhang et al. [8])

Let ( A , B ) be a pair of subsets of a metric space ( X , d ) , and define A 0 and B 0 by (1) and (2). Let T be a contraction from A into B. Assume that the following hold:
  1. (i)

    ( A , B ) has the weak P-property.

     
  2. (ii)

    A is complete.

     
  3. (iii)

    T ( A 0 ) B 0 .

     

Then there exists a unique z A such that d ( z , T z ) = d ( A , B ) .

Proof By Proposition 2(ii-2), there exists a nonexpansive mapping Q from B 0 into A 0 such that d ( Q u , u ) = d ( A , B ) for every u B 0 . Then by Lemma 3, all the assumptions in Theorem 5 hold. So there exists a unique fixed point z of Q T in A 0 . This implies that d ( z , T z ) = d ( A , B ) . Let x A satisfy d ( x , T x ) = d ( A , B ) . Then from Proposition 2(ii-1), x A 0 , T x B 0 and Q T x = x hold. Since Q T has a unique fixed point, we obtain x = z . Hence z is unique. □

Remark

  • If we weaken (i) to the conjunction of A 0 and (ii-2) in Proposition 2, we obtain only the existence of best proximity points.

  • In [8], we assume the completeness of B.

  • Exactly speaking, in [8], we obtained a theorem connected with Geraghty’s fixed point theorem [12]. However, in this paper, the difference between the two fixed point theorems is not essential. This means that we can easily modify Theorem 8 to be connected with Geraghty’s theorem.

Theorem 9 Let ( A , B ) be a pair of subsets of a metric space ( X , d ) , and define A 0 and B 0 by (1) and (2). Let T be a mapping from A into B. Assume that (i)-(iii) in Theorem 8 and the following hold:
  1. (iv)
    There exists α [ 0 , 1 / 2 ) such that
    d ( T x , T y ) α ( d ( x , T x ) d ( A , B ) ) + α ( d ( y , T y ) d ( A , B ) )
     

for x , y A .

Then there exists a unique z A such that d ( z , T z ) = d ( A , B ) .

Proof By Proposition 2(ii-2), there exists a nonexpansive mapping Q from B 0 into A 0 such that d ( Q u , u ) = d ( A , B ) for every u B 0 . Then by Theorem 7, there exists a unique fixed point w of T Q in B 0 . This implies that d ( z , T z ) = d ( A , B ) , where z = Q w . Let x A satisfy d ( x , T x ) = d ( A , B ) . Then from Proposition 2(ii-1), x A 0 , T x B 0 and Q T x = x hold. Since T Q T x = T x , we have T x = w , and hence x = Q T x = Q w = z . Therefore, z is unique. □

Remark If we weaken (i) to the conjunction of A 0 and (ii-2) in Proposition 2, we obtain only the existence of best proximity points.

5 Additional result

In this section, we give a proposition similar to Proposition 2.

Proposition 10 Let ( A , B ) be a pair of nonempty subsets of a metric space ( X , d ) , and define A 0 and B 0 by (1) and (2). Assume that A 0 . Then the following are equivalent:
  1. (i)

    ( A , B ) has the P-property.

     
  2. (ii)

    The conjunction of the following holds:

     

(ii-1) For every u B 0 , there exists a unique x A 0 with d ( x , u ) = d ( A , B ) .

(ii-2) There exists an isometry Q from B 0 onto A 0 such that d ( Q u , u ) = d ( A , B ) for every u B 0 .

Proof We note B 0 . First, we assume (i). Let x , y A 0 and u B 0 satisfy d ( x , u ) = d ( y , u ) = d ( A , B ) . Then from (i), we have d ( x , y ) = d ( u , u ) = 0 , thus, x = y . So (ii-1) holds. We put Q u = x . Then it is obvious that Q is isometric. For every x A 0 , there exists u B 0 with d ( x , u ) = d ( A , B ) . From (ii-1), Q u = x obviously holds, and hence Q is surjective. Conversely, we assume (ii). Let x , y A 0 and u , v B 0 satisfy d ( x , u ) = d ( y , v ) = d ( A , B ) . Then we have Q u = x and Q v = y . Therefore, d ( x , y ) = d ( Q u , Q v ) = d ( u , v ) holds. □

Declarations

Acknowledgements

The author is supported in part by the Grant-in-Aid for Scientific Research from the Japan Society for the Promotion of Science.

Authors’ Affiliations

(1)
Department of Basic Sciences, Kyushu Institute of Technology

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© Suzuki; licensee Springer. 2013

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