The existence of best proximity points with the weak Pproperty
 Tomonari Suzuki^{1}Email author
https://doi.org/10.1186/168718122013259
© Suzuki; licensee Springer. 2013
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 Pproperty, which has been recently proved by Zhang et al.
MSC:54H25, 54E50.
Keywords
best proximity point the weak Pproperty the Banach contraction principle Kannan’s fixed point theorem completion1 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 $x\in A$ is called a best proximity point if $d(x,Tx)=d(A,B)$, where $d(A,B)=inf\{d(x,y):x\in A,y\in B\}$. We have proved many existence theorems of best proximity points. See, for example, [1–6]. 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.
Then

(Sankar Raj [9]) $(A,B)$ is said to have the Pproperty if ${A}_{0}\ne \mathrm{\varnothing}$ and the following holds:$x,y\in {A}_{0},u,v\in {B}_{0},\phantom{\rule{1em}{0ex}}d(x,u)=d(y,v)=d(A,B)\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(x,y)=d(u,v).$

(Zhang et al.[8]) $(A,B)$ is said to have the weak Pproperty if ${A}_{0}\ne \mathrm{\varnothing}$ and the following holds:$x,y\in {A}_{0},u,v\in {B}_{0},\phantom{\rule{1em}{0ex}}d(x,u)=d(y,v)=d(A,B)\phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}d(x,y)\le d(u,v).$
 (i)
$(A,B)$ has the weak Pproperty.
 (ii)
The conjunction of the following holds:
(ii1) For every $u\in {B}_{0}$, there exists a unique $x\in {A}_{0}$ with $d(x,u)=d(A,B)$.
(ii2) There exists a nonexpansive mapping Q from ${B}_{0}$ into ${A}_{0}$ such that $d(Qu,u)=d(A,B)$ for every $u\in {B}_{0}$.
holds. □
we have $y\in {A}_{0}$ and $w\in {B}_{0}$. □
 (i)
$\overline{Q}$ is welldefined and nonexpansive.
 (ii)
$\overline{Q}w=z$ is equivalent to that there exists a sequence $\{{u}_{n}\}$ in ${B}_{0}$ such that ${lim}_{n}{u}_{n}=w$ and ${lim}_{n}Q{u}_{n}=z$.
 (iii)
The domain of $\overline{Q}$ is $\overline{{B}_{0}}$, where $\overline{{B}_{0}}$ is the completion of ${B}_{0}$.
 (iv)
The range of $\overline{Q}$ is a subset of $\overline{{A}_{0}}$, where $\overline{{A}_{0}}$ is the completion of ${A}_{0}$.
 (v)
$T\circ \overline{Q}w=w$ implies $T\circ Qw=w$.
 (vi)
$\overline{Q}\circ Tz=z$ implies $Q\circ Tz=z$.
 (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 welldefined. 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 ${A}_{0}\subset A$. Let T be a contraction from A into X such that $T({A}_{0})\subset {B}_{0}$, 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, $T(\overline{{A}_{0}})$ 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 [10].

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 [11].
 (i)
There exists $\alpha \in [0,1/2)$ such that $d(Tx,Ty)\le \alpha d(x,Tx)+\alpha d(y,Ty)$ for all $x,y\in Y$.
 (ii)
There exists a nonempty subset Z of Y such that $T(Z)\subset Z$.
Then there exists a unique fixed point z, and for every $x\in Z$, $\{{T}^{n}x\}$ converges to z.
Proof Fix $x\in Z$. 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 ${A}_{0}\subset A$. Let T be a mapping from A into X such that $T({A}_{0})\subset {B}_{0}$, and let Q be a nonexpansive mapping from ${B}_{0}$ into ${A}_{0}$. Assume that (3) and the following hold:

There exist $\alpha \in [0,1/2)$ and $\mu \in [0,\mathrm{\infty})$ such that$d(Tx,Ty)\le \alpha (d(x,Tx)\mu )+\alpha (d(y,Ty)\mu )$
for $x,y\in A$ and $d(Qu,u)\le \mu $ for all $u\in {B}_{0}$.
Then $T\circ Q$ has a unique fixed point in ${B}_{0}$.
Hence $T\circ \overline{Q}$ is a Kannan mapping from $\overline{{B}_{0}}$ into X. $T\circ \overline{Q}({B}_{0})=T\circ Q({B}_{0})\subset {B}_{0}$ 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. [8])
 (i)
$(A,B)$ has the weak Pproperty.
 (ii)
A is complete.
 (iii)
$T({A}_{0})\subset {B}_{0}$.
Then there exists a unique $z\in A$ such that $d(z,Tz)=d(A,B)$.
Proof By Proposition 2(ii2), there exists a nonexpansive mapping Q from ${B}_{0}$ into ${A}_{0}$ such that $d(Qu,u)=d(A,B)$ 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 $d(z,Tz)=d(A,B)$. Let $x\in A$ satisfy $d(x,Tx)=d(A,B)$. Then from Proposition 2(ii1), $x\in {A}_{0}$, $Tx\in {B}_{0}$ 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 (ii2) 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.
 (iv)There exists $\alpha \in [0,1/2)$ such that$d(Tx,Ty)\le \alpha (d(x,Tx)d(A,B))+\alpha (d(y,Ty)d(A,B))$
for $x,y\in A$.
Then there exists a unique $z\in A$ such that $d(z,Tz)=d(A,B)$.
Proof By Proposition 2(ii2), there exists a nonexpansive mapping Q from ${B}_{0}$ into ${A}_{0}$ such that $d(Qu,u)=d(A,B)$ 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 $d(z,Tz)=d(A,B)$, where $z=Qw$. Let $x\in A$ satisfy $d(x,Tx)=d(A,B)$. Then from Proposition 2(ii1), $x\in {A}_{0}$, $Tx\in {B}_{0}$ 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 (ii2) 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.
 (i)
$(A,B)$ has the Pproperty.
 (ii)
The conjunction of the following holds:
(ii1) For every $u\in {B}_{0}$, there exists a unique $x\in {A}_{0}$ with $d(x,u)=d(A,B)$.
(ii2) There exists an isometry Q from ${B}_{0}$ onto ${A}_{0}$ such that $d(Qu,u)=d(A,B)$ 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 $d(x,u)=d(y,u)=d(A,B)$. Then from (i), we have $d(x,y)=d(u,u)=0$, thus, $x=y$. So (ii1) 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 $d(x,u)=d(A,B)$. From (ii1), $Qu=x$ obviously holds, and hence Q is surjective. Conversely, we assume (ii). Let $x,y\in {A}_{0}$ and $u,v\in {B}_{0}$ satisfy $d(x,u)=d(y,v)=d(A,B)$. Then we have $Qu=x$ and $Qv=y$. Therefore, $d(x,y)=d(Qu,Qv)=d(u,v)$ holds. □
Declarations
Acknowledgements
The author is supported in part by the GrantinAid for Scientific Research from the Japan Society for the Promotion of Science.
Authors’ Affiliations
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