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

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.

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 $x\in A$ 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.

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

$$A_0=\{x\in A:\text{ there exists }u\in B\text{ such that }d(x,u)=d(A,B)\}$$

(1)

and

$$B_0=\{u\in B:\text{ there exists }x\in A\text{ such that }d(x,u)=d(A,B)\}.$$

(2)

Then

• (Sankar Raj [9]) $(A,B)$ is said to have the P-property if $A_0\ne \mathrm{\varnothing }$ and the following holds:

  $$x,y\in A_0,u,v\in B_0,d(x,u)=d(y,v)=d(A,B)⟹d(x,y)=d(u,v).$$

• (Zhang et al.[8]) $(A,B)$ is said to have the weak P-property if $A_0\ne \mathrm{\varnothing }$ and the following holds:

  $$x,y\in A_0,u,v\in B_0,d(x,u)=d(y,v)=d(A,B)⟹d(x,y)\le d(u,v).$$

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 $A_0\ne \mathrm{\varnothing }$. Then the following are equivalent:

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

• (ii) The conjunction of the following holds:

(ii-1) For every $u\in B_0$, there exists a unique $x\in A_0$ with $d(x,u)=d(A,B)$.

(ii-2) There exists a nonexpansive mapping Q from $B_0$ into $A_0$ such that $d(Qu,u)=d(A,B)$ for every $u\in B_0$.

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 $d(x,u)=d(y,u)=d(A,B)$. Then from (i), we have

$$d(x,y)\le d(u,u)=0,$$

thus, $x=y$. So (ii-1) holds. We put $Qu=x$. Then from the definition of the weak P-property, we have $d(Qu,Qv)\le d(u,v)$ 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 $d(x,u)=d(y,v)=d(A,B)$. Then from (ii-1), we have $Qu=x$ and $Qv=y$. Therefore,

$$d(x,y)=d(Qu,Qv)\le d(u,v)$$

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 $A_0\ne \mathrm{\varnothing }$. Let T be a mapping from A into B, and let Q be a mapping from $B_0$ into $A_0$ such that $d(Qu,u)=d(A,B)$ for every $u\in B_0$. Then the following holds:

$$\{u_n\}\subset B_0,\underset{n\to \mathrm{\infty }}{lim}{u}_n=w,T\left(\underset{n\to \mathrm{\infty }}{lim}Q{u}_n\right)=w⟹w\in B_0.$$

(3)

Proof Let $\{u_n\}$ be a sequence in $B_0$ such that $\{u_n\}$ converges to $w\in X$, and $T({lim}_{n}Q{u}_n)=w$. We put $y={lim}_{n}Q{u}_n$. Since $Ty=w$, we have $y\in A$ and $w\in B$. Since

$$d(y,w)=\underset{n\to \mathrm{\infty }}{lim}d(Q{u}_n,u_n)=d(A,B),$$

we have $y\in A_0$ and $w\in B_0$. □

Lemma 4 Let $(X,d)$ be a metric space, let A, $A_0$, $B_0$ be nonempty subsets such that A is complete and $A_0\subset A$. 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 $A_0$. Let $\overline{Q}$ be the mapping whose graph $Gr(\overline{Q})$ is the completion of $Gr(Q)$. Assume (3). Then the following hold:

• (i) $\overline{Q}$ is well-defined 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 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 $A_0\subset A$. 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].

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:

• (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 $B_0$.

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, $d(\overline{Q}u,u)\le \mu$ for $u\in \overline{B_0}$. For $u,v\in \overline{B_0}$, we have

$$\begin{array}{r}d(T\circ \overline{Q}u,T\circ \overline{Q}v)\\ \le \alpha (d(\overline{Q}u,T\circ \overline{Q}u)-\mu )+\alpha (d(\overline{Q}v,T\circ \overline{Q}v)-\mu )\\ \le \alpha (d(\overline{Q}u,u)+d(u,T\circ \overline{Q}u)-\mu )+\alpha (d(\overline{Q}v,v)+d(v,T\circ \overline{Q}v)-\mu )\\ \le \alpha d(u,T\circ \overline{Q}u)+\alpha d(v,T\circ \overline{Q}v).\end{array}$$

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])

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:

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

• (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(ii-2), 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(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 [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:

• (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(ii-2), 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(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 $(A,B)$ 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 $A_0\ne \mathrm{\varnothing }$. Then the following are equivalent:

• (i) $(A,B)$ has the P-property.

• (ii) The conjunction of the following holds:

(ii-1) For every $u\in B_0$, there exists a unique $x\in A_0$ with $d(x,u)=d(A,B)$.

(ii-2) 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 (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 $d(x,u)=d(A,B)$. 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 $d(x,u)=d(y,v)=d(A,B)$. Then we have $Qu=x$ and $Qv=y$. Therefore, $d(x,y)=d(Qu,Qv)=d(u,v)$ holds. □