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# 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:

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

Motivated by the fact above, in this paper, we improve the result in . 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 $\left(A,B\right)$ be a pair of nonempty subsets of a metric space $\left(X,d\right)$, and define ${A}_{0}$ and ${B}_{0}$ by
(1)
and
(2)

Then

• (Sankar Raj ) $\left(A,B\right)$ is said to have the P-property if ${A}_{0}\ne \mathrm{\varnothing }$ and the following holds:
$x,y\in {A}_{0},u,v\in {B}_{0},\phantom{\rule{1em}{0ex}}d\left(x,u\right)=d\left(y,v\right)=d\left(A,B\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}d\left(x,y\right)=d\left(u,v\right).$
• (Zhang et al.) $\left(A,B\right)$ is said to have the weak P-property if ${A}_{0}\ne \mathrm{\varnothing }$ and the following holds:
$x,y\in {A}_{0},u,v\in {B}_{0},\phantom{\rule{1em}{0ex}}d\left(x,u\right)=d\left(y,v\right)=d\left(A,B\right)\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}d\left(x,y\right)\le d\left(u,v\right).$
Proposition 2 Let $\left(A,B\right)$ be a pair of nonempty subsets of a metric space $\left(X,d\right)$, and define ${A}_{0}$ and ${B}_{0}$ by (1) and (2). Assume that ${A}_{0}\ne \mathrm{\varnothing }$. Then the following are equivalent:
1. (i)

$\left(A,B\right)$ has the weak P-property.

2. (ii)

The conjunction of the following holds:

(ii-1) For every $u\in {B}_{0}$, there exists a unique $x\in {A}_{0}$ with $d\left(x,u\right)=d\left(A,B\right)$.

(ii-2) There exists a nonexpansive mapping Q from ${B}_{0}$ into ${A}_{0}$ such that $d\left(Qu,u\right)=d\left(A,B\right)$ for every $u\in {B}_{0}$.

Proof We note that ${B}_{0}\ne \mathrm{\varnothing }$ because ${A}_{0}\ne \mathrm{\varnothing }$. First, we assume (i). Let $x,y\in {A}_{0}$ and $u\in {B}_{0}$ satisfy $d\left(x,u\right)=d\left(y,u\right)=d\left(A,B\right)$. Then from (i), we have
$d\left(x,y\right)\le d\left(u,u\right)=0,$
thus, $x=y$. So (ii-1) holds. We put $Qu=x$. Then from the definition of the weak P-property, we have $d\left(Qu,Qv\right)\le d\left(u,v\right)$ for $u,v\in {B}_{0}$, that is, Q is nonexpansive. Conversely, we assume (ii). Let $x,y\in {A}_{0}$ and $u,v\in {B}_{0}$ satisfy $d\left(x,u\right)=d\left(y,v\right)=d\left(A,B\right)$. Then from (ii-1), we have $Qu=x$ and $Qv=y$. Therefore,
$d\left(x,y\right)=d\left(Qu,Qv\right)\le d\left(u,v\right)$

holds. □

Lemma 3 Let $\left(A,B\right)$ be a pair of subsets of a metric space $\left(X,d\right)$, and define ${A}_{0}$ and ${B}_{0}$ by (1) and (2). Assume that ${A}_{0}\ne \mathrm{\varnothing }$. Let T be a mapping from A into B, and let Q be a mapping from ${B}_{0}$ into ${A}_{0}$ such that $d\left(Qu,u\right)=d\left(A,B\right)$ for every $u\in {B}_{0}$. Then the following holds:
$\left\{{u}_{n}\right\}\subset {B}_{0},\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty }}{lim}{u}_{n}=w,\phantom{\rule{2em}{0ex}}T\left(\underset{n\to \mathrm{\infty }}{lim}Q{u}_{n}\right)=w\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}w\in {B}_{0}.$
(3)
Proof Let $\left\{{u}_{n}\right\}$ be a sequence in ${B}_{0}$ such that $\left\{{u}_{n}\right\}$ converges to $w\in X$, and $T\left({lim}_{n}Q{u}_{n}\right)=w$. We put $y={lim}_{n}Q{u}_{n}$. Since $Ty=w$, we have $y\in A$ and $w\in B$. Since
$d\left(y,w\right)=\underset{n\to \mathrm{\infty }}{lim}d\left(Q{u}_{n},{u}_{n}\right)=d\left(A,B\right),$

we have $y\in {A}_{0}$ and $w\in {B}_{0}$. □

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

$\overline{Q}$ is well-defined and nonexpansive.

2. (ii)

$\overline{Q}w=z$ is equivalent to that there exists a sequence $\left\{{u}_{n}\right\}$ in ${B}_{0}$ such that ${lim}_{n}{u}_{n}=w$ and ${lim}_{n}Q{u}_{n}=z$.

3. (iii)

The domain of $\overline{Q}$ is $\overline{{B}_{0}}$, where $\overline{{B}_{0}}$ is the completion of ${B}_{0}$.

4. (iv)

The range of $\overline{Q}$ is a subset of $\overline{{A}_{0}}$, where $\overline{{A}_{0}}$ is the completion of ${A}_{0}$.

5. (v)

$T\circ \overline{Q}w=w$ implies $T\circ Qw=w$.

6. (vi)

$\overline{Q}\circ Tz=z$ implies $Q\circ Tz=z$.

7. (vii)

The range of $\overline{Q}$ is a subset of A.

Proof We consider that the whole space is the completion of X. Since Q is Lipschitz continuous, $\overline{Q}$ 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 $\left(X,d\right)$ be a metric space, let A, ${A}_{0}$, ${B}_{0}$ be nonempty subsets such that A is complete and ${A}_{0}\subset A$. Let T be a contraction from A into X such that $T\left({A}_{0}\right)\subset {B}_{0}$, and let Q be a nonexpansive mapping from ${B}_{0}$ into ${A}_{0}$. Assume (3). Then $Q\circ T$ has a unique fixed point in ${A}_{0}$.

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

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 .

• We can prove Theorem 5 with the mapping $T\circ \overline{Q}$ as in the proof of Theorem 7.

We prove generalizations of Kannan’s fixed point theorem .

Theorem 6 Let $\left(X,d\right)$ 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 $\alpha \in \left[0,1/2\right)$ such that $d\left(Tx,Ty\right)\le \alpha d\left(x,Tx\right)+\alpha d\left(y,Ty\right)$ for all $x,y\in Y$.

2. (ii)

There exists a nonempty subset Z of Y such that $T\left(Z\right)\subset Z$.

Then there exists a unique fixed point z, and for every $x\in Z$, $\left\{{T}^{n}x\right\}$ converges to z.

Proof Fix $x\in Z$. Then from the proof in Kannan , we obtain that $\left\{{T}^{n}x\right\}$ 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 .

Using Theorem 6, we obtain the following.

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

• There exist $\alpha \in \left[0,1/2\right)$ and $\mu \in \left[0,\mathrm{\infty }\right)$ such that
$d\left(Tx,Ty\right)\le \alpha \left(d\left(x,Tx\right)-\mu \right)+\alpha \left(d\left(y,Ty\right)-\mu \right)$

for $x,y\in A$ and $d\left(Qu,u\right)\le \mu$ for all $u\in {B}_{0}$.

Then $T\circ Q$ has a unique fixed point in ${B}_{0}$.

Proof We consider that the whole space is the completion of X. Define a nonexpansive mapping $\overline{Q}$ as in Lemma 4. From the continuity of d, $d\left(\overline{Q}u,u\right)\le \mu$ for $u\in \overline{{B}_{0}}$. For $u,v\in \overline{{B}_{0}}$, we have
$\begin{array}{r}d\left(T\circ \overline{Q}u,T\circ \overline{Q}v\right)\\ \phantom{\rule{1em}{0ex}}\le \alpha \left(d\left(\overline{Q}u,T\circ \overline{Q}u\right)-\mu \right)+\alpha \left(d\left(\overline{Q}v,T\circ \overline{Q}v\right)-\mu \right)\\ \phantom{\rule{1em}{0ex}}\le \alpha \left(d\left(\overline{Q}u,u\right)+d\left(u,T\circ \overline{Q}u\right)-\mu \right)+\alpha \left(d\left(\overline{Q}v,v\right)+d\left(v,T\circ \overline{Q}v\right)-\mu \right)\\ \phantom{\rule{1em}{0ex}}\le \alpha d\left(u,T\circ \overline{Q}u\right)+\alpha d\left(v,T\circ \overline{Q}v\right).\end{array}$

Hence $T\circ \overline{Q}$ is a Kannan mapping from $\overline{{B}_{0}}$ into X. $T\circ \overline{Q}\left({B}_{0}\right)=T\circ Q\left({B}_{0}\right)\subset {B}_{0}$ is obvious. So by Theorem 6, there exists a unique fixed point w of $T\circ \overline{Q}$ in $\overline{{B}_{0}}$. By Lemma 4(v), $w\in {B}_{0}$ and w is a fixed point of $T\circ Q$. □

Remark

• Since T is not necessarily continuous, the range of $T\circ \overline{Q}$ is not necessarily included by $\overline{{B}_{0}}$. Because of the same reason, we cannot prove Theorem 7 with the mapping $\overline{Q}\circ T$.

• 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. )

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

$\left(A,B\right)$ has the weak P-property.

2. (ii)

A is complete.

3. (iii)

$T\left({A}_{0}\right)\subset {B}_{0}$.

Then there exists a unique $z\in A$ such that $d\left(z,Tz\right)=d\left(A,B\right)$.

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

Remark

• If we weaken (i) to the conjunction of ${A}_{0}\ne \mathrm{\varnothing }$ and (ii-2) in Proposition 2, we obtain only the existence of best proximity points.

• In , we assume the completeness of B.

• Exactly speaking, in , we obtained a theorem connected with Geraghty’s fixed point theorem . 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 $\left(A,B\right)$ be a pair of subsets of a metric space $\left(X,d\right)$, 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 $\alpha \in \left[0,1/2\right)$ such that
$d\left(Tx,Ty\right)\le \alpha \left(d\left(x,Tx\right)-d\left(A,B\right)\right)+\alpha \left(d\left(y,Ty\right)-d\left(A,B\right)\right)$

for $x,y\in A$.

Then there exists a unique $z\in A$ such that $d\left(z,Tz\right)=d\left(A,B\right)$.

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

Remark If we weaken (i) to the conjunction of ${A}_{0}\ne \mathrm{\varnothing }$ 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 $\left(A,B\right)$ be a pair of nonempty subsets of a metric space $\left(X,d\right)$, and define ${A}_{0}$ and ${B}_{0}$ by (1) and (2). Assume that ${A}_{0}\ne \mathrm{\varnothing }$. Then the following are equivalent:
1. (i)

$\left(A,B\right)$ has the P-property.

2. (ii)

The conjunction of the following holds:

(ii-1) For every $u\in {B}_{0}$, there exists a unique $x\in {A}_{0}$ with $d\left(x,u\right)=d\left(A,B\right)$.

(ii-2) There exists an isometry Q from ${B}_{0}$ onto ${A}_{0}$ such that $d\left(Qu,u\right)=d\left(A,B\right)$ for every $u\in {B}_{0}$.

Proof We note ${B}_{0}\ne \mathrm{\varnothing }$. First, we assume (i). Let $x,y\in {A}_{0}$ and $u\in {B}_{0}$ satisfy $d\left(x,u\right)=d\left(y,u\right)=d\left(A,B\right)$. Then from (i), we have $d\left(x,y\right)=d\left(u,u\right)=0$, thus, $x=y$. So (ii-1) holds. We put $Qu=x$. Then it is obvious that Q is isometric. For every $x\in {A}_{0}$, there exists $u\in {B}_{0}$ with $d\left(x,u\right)=d\left(A,B\right)$. From (ii-1), $Qu=x$ obviously holds, and hence Q is surjective. Conversely, we assume (ii). Let $x,y\in {A}_{0}$ and $u,v\in {B}_{0}$ satisfy $d\left(x,u\right)=d\left(y,v\right)=d\left(A,B\right)$. Then we have $Qu=x$ and $Qv=y$. Therefore, $d\left(x,y\right)=d\left(Qu,Qv\right)=d\left(u,v\right)$ 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, Tobata, Kitakyushu 804-8550, Japan

## References

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