Best proximity point for the proximal nonexpansive mapping on the starshaped sets

Abstract

The existence of the best proximity point for the proximal nonexpansive mapping on starshaped sets is studied. Our results are established without the assumptions of continuity, affinity or the P-property. Finally, as applications of the theorems, analogs for the nonexpansive mappings are also given.

1 Introduction

Let $T:A\rightarrow B$, where A, B are two nonempty subsets of a metric space $(X,d)$. Note that if $A\cap B=\emptyset$, the equation $Tx=x$ might have no solution. Under this circumstance it is meaningful to find a point $x\in A$ such that $d(x, Tx)$ is minimum. Essentially, if $d(x, Tx)= \operatorname {dist}(A, B)=\min\{ d(x,y): x\in A, y\in B\}$, $d(x, Tx)$ is the global minimum value $\operatorname {dist}(A, B)$ and hence x is an approximate solution of the equation $Tx=x$ with the least possible error. Such a solution is known as a best proximity point of the mapping T. A point $x\in A$ is called the best proximity point of T if

$$d(x, Tx)= \operatorname {dist}(A, B)=\min\bigl\{ d(x,y): x\in A, y\in B\bigr\} . $$

It is easy to see that if $A\cap B\neq\emptyset$, the best proximity point just is the fixed point of T.

We can find an early classical work in Ky Fan [1], and afterward, there have been many interesting results such as in Reich [2], Prolla [3], Sehgal and Singh [4, 5], Vetrivel and Veeramani [6], Sadiq and Veeramani [7], Kirk, Reich and Veeramani [8], Eldred, Kirk and Veeramani [9], Eldred and Veeramani [10], and many others.

Recently, M. Gabeleh introduced a new notion which is called the proximal nonexpansive mapping in [11].

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X,d)$. A mapping $T:A\rightarrow B$ is said to be proximal nonexpansive if

$$\left. \begin{array}{l} d(u_{1}, Tx_{1})=\operatorname {dist}(A, B)\\ d(u_{2}, Tx_{2})=\operatorname {dist}(A, B) \end{array} \right\} \quad \Rightarrow\quad d(u_{1},u_{2})\leq d (x_{1}, x_{2}) $$

for all $u_{1}, u_{2}, x_{1}, x_{2}\in A$.

There are two proximity point theorems for the proximal nonexpansive mapping, proved in [11].

The first theorem is as follows.

Theorem 1.1

[11]

Let $(A,B)$ be a pair of nonempty, closed, and convex subsets of a Banach space X such that A is compact, B is bounded and $A_{0}$ is nonempty. Assume that $T:A\rightarrow B$ and $g:A\rightarrow A$ satisfy the following conditions:

• (a) T is continuous affine proximal nonexpansive.

• (b) $T(A_{0})$ is contained in $B_{0}$.

• (c) g is an isometry.

• (d) $A_{0}$ is contained in $g(A_{0})$.

Then there exists a unique element $x^{*}\in A_{0}$ such that

$$\bigl\Vert gx^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

In this theorem, T is assumed to be continuous affine. To remove this assumption, the second theorem is given to replace it with another assumption: that the pair $(A,B)$ has the P-property.

Theorem 1.2

[11]

Let $(A,B)$ be a pair of nonempty, closed, and convex subsets of a Banach space X such that A is compact, Suppose that $A_{0}$ is nonempty and $(A,B)$ has the P-property. Assume that $T:A\rightarrow B$ and $g:A\rightarrow A$ satisfy the following conditions:

• (a) T is a proximal nonexpansive.

• (b) $T(A_{0})$ is contained in $B_{0}$.

• (c) g is an isometry.

• (d) $A_{0}$ is contained in $g(A_{0})$.

Then there exists a unique element $x^{*}\in A_{0}$ such that

$$\bigl\Vert gx^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

In this paper we focus on the sufficient conditions to ensure the existence of the proximity point for the proximal nonexpansive mapping. In our results, sets are not necessarily to be convex or to satisfy the P-property, and the mapping is not necessarily to be continuous or to be affine. The two cases that the sets are compact or weakly compact are considered, respectively. Finally, as applications of the theorems, analogs for the nonexpansive mappings are also given.

2 Preliminaries

Unless otherwise specified, we assume throughout this section that A and B are nonempty subsets of a metric space $(X, d)$. Further, we record in this section the following notations and notions that will be used in the subsequent sections:

$$\begin{aligned}& \operatorname {dist}(A, B)=\inf\bigl\{ d(x, y): \forall x\in A \mbox{ and } \forall y\in B\bigr\} , \\& A_{0}=\bigl\{ x\in A:\Vert x-y\Vert =\operatorname {dist}(A,B)\mbox{ for some } y\in B\bigr\} , \\& B_{0}=\bigl\{ y\in B:\Vert x-y\Vert =\operatorname {dist}(A,B)\mbox{ for some } x\in A\bigr\} . \end{aligned}$$

Definition 2.1

[12]

Let $(A,B)$ be a pair of nonempty subsets of a complete metric space $(X,d)$. A mapping $T:A\rightarrow B$ is said to be a proximal contraction if there exists a non-negative real number $\alpha<1$ such that, for all $u_{1},u_{2},x_{1},x_{2}\in A$,

$$\left. \begin{array}{l} d(u_{1}, Tx_{1})=\operatorname {dist}(A, B)\\ d(u_{2}, Tx_{2})=\operatorname {dist}(A, B) \end{array} \right\} \quad \Rightarrow\quad d(u_{1},u_{2})\leq\alpha d (x_{1}, x_{2}). $$

It is easy to observe that a proximal contraction for a self-mapping reduces to a contraction.

Definition 2.2

Let A be a nonempty subset of a normed space X. A mapping $T: A\rightarrow X$ is called nonexpansive if $\Vert Tx-Ty\Vert \leq \Vert x-y\Vert$ for all $x, y\in A$.

Definition 2.3

[11]

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X,d)$. A mapping $T:A\rightarrow B$ is said to be proximal nonexpansive if

$$\left. \begin{array}{l} d(u_{1}, Tx_{1})=\operatorname {dist}(A, B)\\ d(u_{2}, Tx_{2})=\operatorname {dist}(A, B) \end{array} \right\} \quad \Rightarrow\quad d(u_{1},u_{2})\leq d (x_{1}, x_{2}) $$

for all $u_{1}, u_{2}, x_{1}, x_{2}\in A$.

A nonempty subset A of a linear space X is called a p-starshaped set if there exists a point $p\in A$ such that $\alpha p+(1-\alpha)x\in A$, $\forall x\in A$, $\alpha\in[0,1]$, and p is called the center of A.

Each convex set C is a p-starshaped set for each $p\in C$.

It is easy to see that in a normed space $(X, \Vert \cdot \Vert )$, if A is a p-starshaped set and B is a q-starshaped set and $\Vert p-q\Vert =\operatorname {dist}(A, B)$, $A_{0}$ is a p-starshaped set, and $B_{0}$ is a q-starshaped set, respectively. If both of A and B are closed and $A_{0}$ is nonempty, $A_{0}$ is closed.

Definition 2.4

[13]

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X,d)$ with $A_{0}\neq\emptyset$. The pair $(A,B)$ is said to have the P-property if for every $x_{1}, x_{2}\in A_{0}$ and every $y_{1}, y_{2}\in B_{0}$

$$\left. \begin{array}{l} d(x_{1},y_{1})=\operatorname {dist}(A, B)\\ d(x_{2},y_{2})=\operatorname {dist}(A, B) \end{array} \right\} \quad \Rightarrow\quad d(x_{1},x_{2})=d(y_{1},y_{2}). $$

By using the P-property, some best proximity point results were proved for various classes of non-self-mappings. But in [14] the authors have shown that some recent results with the P-property concerning the existence of best proximity points can be obtained from the same results in fixed point theory.

Definition 2.5

[15]

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X,d)$. The pair $(A, B)$ is said to be a semi-sharp proximinal pair if for each x in A (respectively, in B) there exists at most one $x^{*}$ in B (respectively, in A) such that $d(x, x^{*})=\operatorname {dist}(A, B)$.

Notice that if $(A,B)$ is a semi-sharp proximinal pair, the pair $(B, A)$ may be not be.

It is easy to see that if $(A, B)$ has the P-property, then both of $(A, B)$ and $(B, A)$ are semi-sharp proximinal pairs. Obviously a semi-sharp proximinal pair $(A, B)$ is not necessarily to have the P-property.

Next let us introduce a new notion which is important in dealing with weak convergence. In the sequel, let us write ‘⇀’ to denote ‘weak convergence’.

Definition 2.6

Let A and B be two nonempty subsets of a Banach space X. The pair $(A, B)$ is said to have the H-property if for any sequences $\{x_{n}\} \subseteq A$ and $\{y_{n}\}\subseteq B$, $x_{n}\rightharpoonup x_{0}\in A$, $y_{n}\rightharpoonup y_{0}\in B$, and $\Vert x_{n}-y_{n}\Vert \rightarrow \operatorname {dist}(A, B)$ imply that $x_{n}-y_{n}\rightarrow x_{0}-y_{0}$.

Remark 1

In the definition, since $x_{n}-y_{n}\rightharpoonup x_{0}-y_{0}$ and $\Vert x_{n}-y_{n}\Vert \rightarrow \operatorname {dist}(A, B)$, it follows that

$$\operatorname {dist}(A, B)\leq \Vert x_{0}-y_{0}\Vert \leq\lim _{n\rightarrow\infty }\inf \Vert x_{n}-y_{n}\Vert = \operatorname {dist}(A, B). $$

This implies that $\Vert x_{0}-y_{0}\Vert =\operatorname {dist}(A, B)$.

Remark 2

If $B=A$, $\operatorname {dist}(A, B)=0$. For any $\{x_{n}\}\subseteq A$, $\{y_{n}\} \subseteq B$ satisfying $\Vert x_{n}-y_{n}\Vert \rightarrow \operatorname {dist}(A, B)=0$ must have $x_{n}-y_{n}\rightarrow0$. So for any nonempty subset A, the pair $(A, A)$ must have the H-property.

Recall that a Banach space X is said to have the H-property if for any sequence $\{x_{n}\}\subset X$, $x_{n}\rightharpoonup x_{0}$, and $\Vert x_{n}\Vert \rightarrow \Vert x_{0}\Vert$ imply that $x_{n}\rightarrow x_{0}$.

Let X be a locally uniformly convex space, that is, given $\varepsilon>0$ and an element x with $\Vert x\Vert =1$, there exists $\delta(\varepsilon, x)>0$ such that

$$\frac{\Vert x+y\Vert }{2}\leq1-\delta(\varepsilon, x), \quad \mbox{whenever } \Vert x-y \Vert \geq\varepsilon, \Vert y\Vert =1. $$

We have the following.

• (i) Uniform convexity implies locally uniform convexity, and locally uniform convexity implies strict convexity. But a locally uniformly convex space is not necessarily a reflexive space.

• (ii) Locally uniform convexity is different from uniform convexity or strict convexity; see [16].

• (iii) X is a locally uniformly convex space if and only if

  $$\forall x, x_{n}\in S(X),\quad x_{n}\rightarrow x\quad \mbox{whenever } \Vert x_{n}+x\Vert \rightarrow2. $$

It is well known that if X is a locally uniformly convex space, X has the H-property. In fact, $\forall x_{n}$, $x\in X$, if $x_{n}\rightharpoonup x$ and $\Vert x_{n}\Vert \rightarrow \Vert x\Vert$, let us show $x_{n}\rightarrow x$. Without loss of generality, we may assume that $\Vert x_{n}\Vert =\Vert x\Vert =1$ and show $\Vert x_{n}+x\Vert \rightarrow2$ by use of (iii). It is easy to see that $\lim_{n\rightarrow\infty}\sup \Vert x_{n}+x\Vert \leq2$ since ∀n, $\Vert x_{n}+x\Vert \leq \Vert x_{n}\Vert +\Vert x\Vert =2$. We also have $\lim_{n\rightarrow\infty}\sup \Vert x_{n}+x\Vert \geq2$ since $x_{n}+x\rightharpoonup2x$ and $2=\Vert 2x\Vert \leq\lim_{n\rightarrow \infty}\inf \Vert x_{n}+x\Vert$. Hence $\Vert x_{n}+x\Vert \rightarrow2$ and thus $x_{n}\rightarrow x$.

It is not difficult to see that if X has the H-property, for any sets $A, B\subset X$, the pair $(A, B)$ satisfies the H-property.

Definition 2.7

[17]

A Banach space X satisfies the Opial condition for the weak topology if $x_{n}\rightharpoonup x\in X$ implies that $\lim_{n}\inf \Vert x_{n}-x\Vert <\lim_{n}\inf \Vert x_{n}-y\Vert$ for all $y\neq x$.

All Hilbert spaces, all finite dimensional Banach spaces, and $l_{p}$ ($1< p<\infty$) have the Opial property.

Definition 2.8

A mapping $T: A\rightarrow X$ is called demiclosed if for any sequence $\{x_{n}\}\subseteq A$ which converges weakly to $x_{0}\in A$, the strong convergence of the sequence $\{Tx_{n}\}$ to $y_{0}$ in X implies that $Tx_{0} = y_{0}$.

It can be shown that in a Banach space X with the Opial property, $(I-T)$ must be demiclosed if T is a nonexpansive mapping (see Lemma 2 in [17]).

Definition 2.9

[18]

Let A and B be nonempty subsets of a normed space X, $T: A\cup B\rightarrow A\cup B$, $T(A)\subseteq B$ and $T(B)\subseteq A$. We say that T satisfies the proximal property if $x_{n}\rightharpoonup x\in A\cup B$ and $\Vert x_{n}-Tx_{n}\Vert \rightarrow \operatorname {dist}(A, B)$ imply that $\Vert x-Tx\Vert =\operatorname {dist}(A, B)$.

If $\operatorname {dist}(A, B)=0$, the proximal property reduces to the usual demiclosedness property of $I-T$ at 0.

3 Proximity point for the proximal nonexpansive

First let us prove the following lemma, which plays an important role in our main results.

Lemma 3.1

Let $(A,B)$ be a pair of nonempty closed subsets of a complete metric space $(X,d)$ and $A_{0}$ be nonempty. Assume that $T:A\rightarrow B$ satisfies the following conditions:

• (a) T is a proximal contraction;

• (b) $T(A_{0})\subseteq B_{0}$.

Then there exists a unique $x^{*}\in A_{0}$ such that

$$ d\bigl(x^{*},Tx^{*}\bigr)=\operatorname {dist}(A,B). $$

Proof

Since $A_{0}$ is nonempty and $T(A_{0})\subseteq B_{0}$, choose $x_{0}\in A_{0}$, there exists $x_{1}\in A_{0}$ such that $d(x_{1},Tx_{0})=\operatorname {dist}(A,B)$. Since $Tx_{1}\in B_{0}$, there exists $x_{2}\in A_{0}$ such that $d(x_{2},Tx_{1})=\operatorname {dist}(A,B)$. Continuing this process, we obtain a sequence $\{x_{n}\}\subseteq A_{0}$ such that

$$d(x_{n+1},Tx_{n})=\operatorname {dist}(A,B),\quad \mbox{for all } n\in N. $$

Next, let us show $\{x_{n}\}$ is a Cauchy sequence and its limit just is the unique best proximity point of T.

As T is a proximal contraction and, for all $n\in N$,

$$\begin{aligned}& d(x_{n},Tx_{n-1})=\operatorname {dist}(A,B), \\& d(x_{n+1},Tx_{n})=\operatorname {dist}(A,B), \end{aligned}$$

we have

$$d(x_{n+1},x_{n})\leq\alpha d(x_{n},x_{n-1})\quad (0< \alpha<1). $$

Further, for all $p\in N$,

$$\begin{aligned} d(x_{n+p},x_{n})&\leq d(x_{n+p},x_{n+p-1})+d(x_{n+p-1},x_{n+p-2})+ \cdots +d(x_{n+1},x_{n}) \\ &\leq\alpha^{n+p-1}d(x_{1},x_{0})+ \alpha^{n+p-2}d(x_{1},x_{0})+\cdots + \alpha^{n}d(x_{1},x_{0}) \\ &\leq\frac{\alpha^{n}}{1-\alpha}d(x_{1},x_{0}). \end{aligned} $$

Therefore, $\{x_{n}\}$ is a Cauchy sequence and converges to some $x^{*}$ in $A_{0}$ since $A_{0}$ is closed. With use of the assumption $T(A_{0})\subseteq B_{0}$ again, $Tx^{*}\in B_{0}$. Then there exists an element $u\in A_{0}$ such that

$$ d\bigl(u,Tx^{*}\bigr)=\operatorname {dist}(A,B). $$

As we know, for all $n\in N$,

$$ d(x_{n+1},Tx_{n})=\operatorname {dist}(A,B). $$

Hence for all $n\in N$,

$$ d(u,x_{n+1})\leq\alpha d\bigl(x^{*},x_{n}\bigr). $$

Let $n\rightarrow\infty$, then $d(u,x_{n+1})\rightarrow0$ since $d(x^{*},x_{n})\rightarrow0$. Therefore, $x_{n} \rightarrow u$ and thus $u=x^{*}$. So

$$ d\bigl(x^{*},Tx^{*}\bigr)=\operatorname {dist}(A,B). $$

Suppose that there is another element $x^{**}$ such that

$$ d\bigl(x^{**},Tx^{**}\bigr)=\operatorname {dist}(A,B). $$

Since T is a proximal contraction, we have

$$ d\bigl(x^{*},x^{**}\bigr)\leq\alpha d\bigl(x^{*},x^{**}\bigr), $$

which implies that $x^{*}$ and $x^{**}$ are identical. □

Let T satisfy (a) and (b) in the above lemma and $g: A\rightarrow A$ be an isometry satisfying $A_{0}\subseteq g(A_{0})$. Denote $G=g(A)\subseteq A$ and $G_{0}=\{z\in G:d(z,y)=\operatorname {dist}(A,B) \mbox{ for some }y\in B\}$, then $G_{0}=A_{0}$ and $Tg^{-1}: G\rightarrow B$ is a proximal contraction. So we have the following corollary.

Corollary 3.2

Let $(A,B)$ be a pair of nonempty closed subsets of a complete metric space $(X,d)$ and $A_{0}$ be nonempty. Assume that $T:A\rightarrow B$ and $g: A\rightarrow A$ satisfy the following conditions:

• (a) T is a proximal contraction.

• (b) $T(A_{0})\subseteq B_{0}$.

• (c) g is an isometry.

• (d) $A_{0}$ is contained in $g(A_{0})$.

Then there exists a unique $x^{*}\in A_{0}$ such that

$$d(gx, Tx)=d(A, B). $$

Let g in the above corollary be the identity, we return to Lemma 3.1. So Lemma 3.1 and Corollary 3.2 are equivalent.

If we contrast Corollary 3.2 with Theorem 3.1 in [12] and Theorem 3.3 in [19], we may find that they have the same assertion, but in Theorem 3.1 in [12] the set B was restricted to be approximatively compact with respect to A, and in Theorem 3.3 in [19] the mapping T was restricted to be continuous.

Now let us present the first main theorem in this section.

Theorem 3.3

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set, and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Suppose A is compact, and $(A,B)$ is a semi-sharp proximinal pair. Assume that $T:A\rightarrow B$ satisfies the following conditions:

• (a) T is a proximal nonexpansive.

• (b) $T(A_{0})\subseteq B_{0}$.

Then there exists an element $x^{*}\in A_{0}$ such that

$$ \bigl\Vert x^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

Proof

For each positive integer $k\geq1$, define $T_{k}:A_{0}\rightarrow B_{0}$ by

$$ T_{k}x=\biggl(1-\frac{1}{k}\biggr)Tx+\frac{1}{k}q,\quad \forall x\in A_{0}. $$

Then $T_{k}(A_{0})\subseteq B_{0}$, since $T(A_{0})\subseteq B_{0}$ and $B_{0}$ is a q-starshaped set.

Next, let us show that for each k, all $x_{1},x_{2},u_{1},u_{2}\in A_{0}$ assumed as follows:

$$\begin{aligned}& \Vert u_{1}-T_{k}x_{1}\Vert = \operatorname {dist}(A_{0}, B_{0}), \\& \Vert u_{2}-T_{k}x_{2}\Vert = \operatorname {dist}(A_{0}, B_{0}), \end{aligned}$$

must satisfy

$$\begin{aligned} \Vert u_{1}-u_{2}\Vert \leq\biggl(1-\frac{1}{k} \biggr)\Vert x_{1}-x_{2}\Vert , \end{aligned}$$

that is, for each k, $T_{k}$ is a proximal contraction with $\alpha =1-\frac{1}{k}<1$.

Let $s_{1}, s_{2}\in A_{0}$ satisfy

$$\begin{aligned}& \Vert s_{1}-Tx_{1}\Vert =\operatorname {dist}(A_{0}, B_{0}), \\& \Vert s_{2}-Tx_{2}\Vert =\operatorname {dist}(A_{0} ,B_{0}), \end{aligned}$$

then $\Vert s_{1}-s_{2}\Vert \leq \Vert x_{1}-x_{2}\Vert$, since T is a proximal nonexpansive mapping.

Now write

$$\begin{aligned}& u_{1}'=\biggl(1-\frac{1}{k}\biggr)s_{1}+ \frac{1}{k}p\in A_{0}, \\& u_{2}'=\biggl(1-\frac{1}{k}\biggr)s_{2}+ \frac{1}{k}p\in A_{0}, \end{aligned}$$

we have

$$\begin{aligned} \operatorname {dist}(A_{0}, B_{0})&\leq\bigl\Vert u_{1}'-T_{k}x_{1} \bigr\Vert \\ &=\biggl\Vert \biggl(1-\frac{1}{k}\biggr)s_{1}+ \frac{1}{k}p-\biggl(1-\frac{1}{k}\biggr)Tx_{1}- \frac {1}{k}q\biggr\Vert \\ &\leq\biggl(1-\frac{1}{k}\biggr)\Vert s_{1}-Tx_{1} \Vert +\frac{1}{k}\Vert p-q\Vert \\ &=\operatorname {dist}(A_{0}, B_{0}). \end{aligned}$$

Hence

$$ \bigl\Vert u_{1}'-T_{k}x_{1}\bigr\Vert =\operatorname {dist}(A_{0}, B_{0}). $$

And thus $u_{1}'=u_{1}$ since $\Vert u_{1}-T_{k}x_{1}\Vert =\operatorname {dist}(A_{0}, B_{0})$ and $(A_{0}, B_{0})$ is a semi-sharp proximinal pair.

By the same method we also have $u_{2}'=u_{2}$.

Therefore,

$$\begin{aligned} \Vert u_{1}-u_{2}\Vert &=\bigl\Vert u_{1}'-u_{2}'\bigr\Vert \\ &=\biggl\Vert \biggl(1-\frac{1}{k}\biggr) (s_{1}-s_{2}) \biggr\Vert \\ &\leq\biggl(1-\frac{1}{k}\biggr)\Vert x_{1}-x_{2} \Vert . \end{aligned}$$

So, for each k, $T_{k}$ is a proximal contraction.

Invoking Lemma 3.1, for each $k\geq1$, there exists a unique $u_{k}\in A_{0}$ such that

$$ \Vert u_{k}-T_{k}u_{k}\Vert = \operatorname {dist}(A_{0}, B_{0}). $$

Since $A_{0}$ is compact and $\{u_{k}\}\subseteq A_{0}$, without loss of generality, we may assume that $u_{k}$ is a convergent sequence and $u_{k}\rightarrow x^{*}\in A_{0}$.

Next, let us show $x^{*}$ is the proximity point of T to finish the proof.

For each $k\geq1$, there exists $v_{k}\in A_{0}$, such that

$$\Vert v_{k}-Tu_{k}\Vert =\operatorname {dist}(A_{0}, B_{0}). $$

By the following equalities:

$$\begin{aligned} \operatorname {dist}(A_{0}, B_{0})&\leq\biggl\Vert \biggl(1- \frac{1}{k}\biggr)v_{k}+\frac{1}{k}p-T_{k}u_{k} \biggr\Vert \\ &=\biggl\Vert \biggl(1-\frac{1}{k}\biggr)v_{k}+ \frac{1}{k}p-\biggl(1-\frac{1}{k}\biggr)Tu_{k}- \frac {1}{k}q\biggr\Vert \\ &\leq\biggl(1-\frac{1}{k}\biggr)\Vert v_{k}-Tu_{k} \Vert +\frac{1}{k}\Vert p-q\Vert \\ &=\operatorname {dist}(A_{0}, B_{0}), \end{aligned}$$

we deduce that

$$\biggl\Vert \biggl(1-\frac{1}{k}\biggr)v_{k}+ \frac{1}{k}p-T_{k}u_{k}\biggr\Vert = \operatorname {dist}(A_{0}, B_{0}). $$

Since $(A_{0}, B_{0})$ is a semi-sharp proximinal pair and $\Vert u_{k}-T_{k}u_{k}\Vert =\operatorname {dist}(A_{0}, B_{0})$, we have $u_{k}=(1-\frac{1}{k})v_{k}+\frac{1}{k}p$, and thus

$$\Vert u_{k}-v_{k}\Vert =\frac{1}{k}\Vert v_{k}-p\Vert \rightarrow0\quad (k\rightarrow\infty). $$

So $\{v_{k}\}$ is also a convergent sequence and $\lim_{k\rightarrow\infty}v_{k}=\lim_{k\rightarrow\infty}u_{k}=x^{*}$.

Note that $Tx^{*}\in B_{0}$, there must exist $u\in A_{0}$ such that

$$\bigl\Vert u-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

As we know,

$$\Vert v_{k}-Tu_{k}\Vert =\operatorname {dist}(A,B) $$

and T is a proximal nonexpansive mapping, therefore

$$ \Vert v_{k}-u\Vert \leq\bigl\Vert u_{k}- x^{*}\bigr\Vert \rightarrow0\quad (k\rightarrow\infty). $$

This implies $u=\lim_{k\rightarrow\infty}v_{k}=x^{*}$ and then we deduce that

$$ \bigl\Vert x^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

The proof is finished. □

The following theorem is the equivalent of Theorem 3.3; we omit the proof.

Theorem 3.4

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set, and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Suppose A is compact, and $(A,B)$ is a semi-sharp proximinal pair. Assume that $T:A\rightarrow B$ and $g:A\rightarrow A$ satisfy the following conditions:

• (a) T is a proximal nonexpansive.

• (b) $T(A_{0})\subseteq B_{0}$.

• (c) $A_{0}\subseteq g(A_{0})$

• (d) g is an isometry.

Then there exists an element $x^{*}\in A_{0}$ such that

$$ \bigl\Vert gx^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

In Theorem 3.4, T is not necessarily continuous affine, $(A, B)$ does not have the P-property and each of A and B is not necessarily convex.

It is easy to see that Theorem 1.2 is the corollary of Theorem 3.4.

Now let us begin to consider the case that the sets are weakly compact.

Lemma 3.5

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set, and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Let $T:A\rightarrow B$ be a proximal nonexpansive and $T(A_{0})\subseteq B_{0}$. Suppose that $(A,B)$ is a semi-sharp proximinal pair and $A_{0}$ is weakly compact. Then T has at least one best proximity point in A provided that one of the following conditions is satisfied:

• (a) T is weakly continuous.

• (b) T satisfies the proximal property.

Proof

For each positive integer $k\geq1$, define $T_{k}:A_{0}\rightarrow B_{0}$ by

$$ T_{k}x=\biggl(1-\frac{1}{k}\biggr)Tx+\frac{1}{k}q,\quad \forall x\in A_{0}. $$

In the same way as the proof of Theorem 3.3, for each $k\geq1$, there exists a unique $u_{k}\in A_{0}$ such that

$$ \Vert u_{k}-T_{k}u_{k}\Vert = \operatorname {dist}(A_{0}, B_{0}). $$

Note that for each $k\geq1$,

$$\begin{aligned} \Vert u_{k}-Tu_{k}\Vert &\leq \Vert u_{k}-T_{k}u_{k}\Vert +\Vert T_{k}u_{k}-Tu_{k}\Vert \\ &=\operatorname {dist}(A,B)+\biggl\Vert \biggl(1-\frac{1}{k}\biggr)Tu_{k}+ \frac{1}{k}q-Tu_{k}\biggr\Vert \\ &=\frac{1}{k}\Vert q-Tu_{k}\Vert +\operatorname {dist}(A_{0}, B_{0}). \end{aligned}$$

By the assumption that $A_{0}$ is weakly compact, it follows that $A_{0}$ is bounded and this implies that $B_{0}$ is also bounded. So there exists a constant M such that $\Vert q-Tu_{k}\Vert \leq M$ for each $k\geq1$. Now let $k\rightarrow\infty$, we obtain

$$ \Vert u_{k}-Tu_{k}\Vert \rightarrow \operatorname {dist}(A,B). $$

Since $A_{0}$ is weakly compact, without loss of generality, we may assume that $u_{k}\rightharpoonup x^{*}\in A_{0}$.

If T is weakly continuous, $Tu_{k}\rightharpoonup Tx^{*}\in B_{0}$ and

$$ \operatorname {dist}(A,B)\leq\bigl\Vert x^{*}-Tx^{*}\bigr\Vert \leq\lim_{n} \inf \Vert u_{k}-Tu_{k}\Vert =\operatorname {dist}(A,B). $$

Therefore, we have

$$ \bigl\Vert x^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

If T satisfies the proximal property, since $u_{k}\rightharpoonup x^{*}\in A_{0}$ and $\Vert u_{k}-Tu_{k}\Vert \rightarrow \operatorname {dist}(A,B)$ we also have

$$ \bigl\Vert x^{*}-Tx^{*}\bigr\Vert =\operatorname {dist}(A,B). $$

 □

Let us present the second main result in this section.

Theorem 3.6

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Let $T:A\rightarrow B$ be a proximal nonexpansive and $T(A_{0})\subseteq B_{0}$. Suppose that $(A,B)$ is a semi-sharp proximinal pair, $(A_{0}, B_{0})$ is a weakly compact pair and satisfies the H-property. Then T has at least one best proximity point in A provided that $(I-T)$ is demiclosed.

Proof

Invoking Lemma 3.5, it suffices to prove T satisfies the proximal property.

Suppose $u_{k}\in A_{0}$ such that $u_{k}\rightharpoonup x^{*}\in A_{0}$ and $\Vert u_{k}-Tu_{k}\Vert \rightarrow \operatorname {dist}(A,B)$. Since $A_{0},B_{0}$ are weakly compact, without loss of generality, we may assume that $u_{k}\rightharpoonup x^{*}\in A_{0}$ and $Tu_{k}\rightharpoonup y^{*}\in B_{0}$. So we have $u_{k}-Tu_{k}\rightharpoonup x^{*}-y^{*}$ and $\Vert u_{k}-Tu_{k}\Vert \rightarrow \operatorname {dist}(A,B)$. This implies that $(I-T)u_{k}=u_{k}-Tu_{k}\rightarrow x^{*}-y^{*}$ by the assumption that $(A_{0}, B_{0})$ satisfies the H-property. Therefore $(I-T)x^{*}=x^{*}-y^{*}$ since $(I-T)$ is demiclosed. It is easy to see that $\Vert x^{*}-Tx^{*}\Vert =\Vert x^{*}-y^{*}\Vert =\operatorname {dist}(A,B)$. So T satisfies the proximal property. □

Recall that a locally uniformly convex Banach space X must be strictly convex and have the H-property, then each convex subset pair $(A, B)$ is a semi-sharp proximinal pair and $(A_{0}, B_{0})$ satisfies the H-property. So we have the following corollary.

Corollary 3.7

Let $(A,B)$ be a pair of nonempty, closed, and convex subsets of a locally uniformly convex Banach space X and $A_{0}\neq\emptyset$. Suppose that $T:A\rightarrow B$ is a proximal nonexpansive and $T(A_{0})\subseteq B_{0}$. Then T has at least one best proximity point in A provided that $(A_{0}, B_{0})$ is a weakly compact pair and $(I-T)$ is demiclosed.

Recently a new notion of the proximal generalized nonexpansive mapping was introduced in [20], and a few results about the proximity point for this class of mappings are given. For more information, see [20].

Let us end this section by an example.

Example

Let $(X, \Vert \cdot \Vert )=(R^{2}, l_{1})$, that is, for any $(x, y)\in R^{2}$, $\Vert (x, y)\Vert =\vert x\vert +\vert y\vert$,

$$\begin{aligned}& A=\bigl\{ (x,y):0\leq x\leq1, y=0\bigr\} , \\& B_{1}=\bigl\{ (x, y): -x+y=1, -1\leq x\leq0\bigr\} , \\& B_{2}=\bigl\{ (x,y):0\leq x\leq1, y=1\bigr\} , \\& B=B_{1}\cup B_{2}. \end{aligned}$$

Define $T: A\rightarrow B$ as follows:

$$ T \bigl((x, 0) \bigr)= \begin{cases} (-\frac{1}{2}, \frac{1}{2}), & \mbox{if } x=0;\\ (\sin x, 1),& \mbox{if } x\neq0. \end{cases} $$

We have the following assertions.

• (i) $p=(0,1)\in B$ and set B is not convex but is a p-starshaped set;

• (ii) for any $s\in B_{1}$ and $O=(0, 0)\in A$, $d(s, O)=1=\operatorname {dist}(A, B)$ and $d(s, t)>1$ for any $t\in A$, $t\neq O$;

• (iii) $A_{0}=A$, $B_{0}=B$, and $(A, B)$ is a semi-sharp proximinal pair;

• (iv) $(B, A)$ is not a semi-sharp proximinal pair;

• (v) T is not a nonexpansive mapping but is a proximal nonexpansive mapping;

• (vi) T is not continuous or affine;

• (vii) $T(A_{0})\subseteq B_{0}$ and $A_{0}$ is compact;

• (viii) $\Vert (0, 0)-T((0, 0))\Vert =\Vert (0, 0)-(-\frac{1}{2}, \frac {1}{2})\Vert =1=\operatorname {dist}(A, B)$.

4 Proximity point for nonexpansive mapping

In this section, as applications of Theorem 3.3, Lemma 3.5, and Theorem 3.6, three results to ensure the existence of the best proximity for the nonexpansive mapping are given.

In [21], Abkar and Gabeleh obtained two proximity point theorems for the nonexpansive mapping.

Theorem 4.1

[21]

Let $(A,B)$ be a pair of nonempty, closed, and convex subsets of a Banach space X such that A is compact. Suppose that $A_{0}$ is nonempty and $(A,B)$ has the P-property. Let $T: A\rightarrow B$ be a nonexpansive non-self-mapping such that $T(A_{0})\subseteq B_{0}$. Then T has at least one best proximity point in A.

Theorem 4.2

[21]

Let $(A,B)$ be a pair of nonempty, closed, and convex subsets of a Banach space X such that $A_{0}$ is nonempty. Let $T:A\rightarrow B$ be a nonexpansive mapping such that $T(A_{0})\subseteq B_{0}$. Suppose the pair $(A,B)$ has the P-property and A is weakly compact. Then T has at least one best proximity point in A provided that one of the following conditions is satisfied:

• (a) T is weakly continuous.

• (b) T satisfies the proximal property.

We can find that the ‘P-property’ appears in both of Theorems 4.1 and Theorem 4.2. Next, let us show that this assumption can be replaced by the so-called ‘weak P-property’. Besides, in both of the theorems, the pair $(A, B)$ can be assumed only as a starshaped pair instead of a convex pair.

Definition 4.3

[22]

Let $(A,B)$ be a pair of nonempty subsets of a metric space $(X, d)$ with $A_{0}\neq\emptyset$. Then the pair $(A,B)$ is said to have the weak P-property if $\forall x_{1}, x_{2}\in A_{0}$, and $y_{1}, y_{2}\in B_{0}$,

$$\left. \begin{array}{l} d(x_{1}, y_{1})=\operatorname {dist}(A, B)\\ d(x_{2}, y_{2})=\operatorname {dist}(A, B) \end{array} \right\} \quad \Rightarrow\quad d(x_{1}, x_{2})\leq d (y_{1}, y_{2}). $$

It is clear that the weak P-property is weaker than the P-property and $(A, B)$ has the P-property if and only if both $(A, B)$ and $(B, A)$ have the weak P-property. See [23] and [24]. Obviously, if a pair $(A,B)$ has the weak P-property it must be a semi-sharp proximinal pair.

If $(A, B)$ has the weak P-property and $T: A\rightarrow B$ is a nonexpansive mapping, for all $u_{1}, u_{2}, x_{1}, x_{2}\in A$

$$\left. \begin{array}{l} d(u_{1}, Tx_{1})=\operatorname {dist}(A, B)\\ d(u_{2}, Tx_{2})=\operatorname {dist}(A, B) \end{array} \right\} \quad \Rightarrow\quad d(u_{1},u_{2})\leq d (Tx_{1}, Tx_{2})\leq d(x_{1}, x_{2}). $$

That is, T is a proximal nonexpansive mapping. As consequences of Theorem 3.3, Lemma 3.5, and Theorem 3.6, we have the following theorems.

Theorem 4.4

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set, and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Suppose that $A_{0}$ is compact and $(A,B)$ has the weak P-property. Let $T: A\rightarrow B$ be a nonexpansive mapping such that $T(A_{0})\subseteq B_{0}$. Then T has at least one best proximity point in $A_{0}$.

Proof

It is the direct consequence of Theorem 3.3. □

Theorem 4.5

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set, and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Let $T:A\rightarrow B$ be a nonexpansive mapping such that $T(A_{0})\subseteq B_{0}$. Suppose the pair $(A,B)$ has the weak P-property and A is weakly compact. Then T has at least one best proximity point in A provided that one of the following conditions is satisfied:

• (a) T is weakly continuous;

• (b) T satisfies the proximal property.

Proof

It is the direct consequence of Lemma 3.5. □

Theorem 4.6

Let $(A,B)$ be a pair of nonempty, closed subsets of a Banach space X such that A is a p-starshaped set, B is a q-starshaped set, and $\Vert p-q\Vert =\operatorname {dist}(A,B)$. Let $T:A\rightarrow B$ be nonexpansive and $T(A_{0})\subseteq B_{0}$. Suppose that $(A,B)$ has the weak P-property, $(A_{0}, B_{0})$ is a weakly compact pair and satisfies the H-property. Then T has at least one best proximity point in A provided that one of the following conditions is satisfied:

• (a) $(I-T)$ is demiclosed.

• (b) X satisfies the Opial property.

Proof

If X satisfies the Opial property, then $(I-T)$ must be demiclosed since T is a nonexpansive mapping. To prove the theorem, it suffices to prove the assertion for the case (a). But this is the direct consequence of Theorem 3.6. □