Best proximity point theorems via fixed point theorems for multivalued mappings

Abstract

It is well known that the concept of a best proximity point includes that of a fixed point as a special case. In this paper, we show that the best proximity point theorems of Basha and Shahzad (Fixed Point Theory Appl. 2012:42, 2012) and of Fernández-León (J. Nonlinear Convex Anal. 15(2):313-324, 2014) can be regarded as a fixed point theorem for multivalued mappings which is modified as regards the results of Mizoguchi and Takahashi (J. Math. Anal. Appl. 141(1):177-188, 1989) and of Kada et al. (Math. Jpn. 44(2):381-391, 1996).

1 Introduction

Let X be any nonempty set and $T:X\to X$ be a given mapping. A point $x\in X$ such that $x=Tx$ is called a fixed point of T. Many problems can be reformulated to the problem of finding a fixed point of a certain mapping. If T is not a self-mapping, it is plausible that the equation $x=Tx$ has no solution. In this situation, we may find an element $x\in X$ which is close to Tx in some sense.

Now, we suppose that X is equipped with a metric d, that is, $(X,d)$ is a metric space. For two subsets A and B of X and $T:A\to B$, we are interested in finding an element $x\in A$ such that

$$d(x,Tx)=\inf\bigl\{ d(a,b):a\in A,b\in B\bigr\} =:d(A,B). $$

Such an element x is called a best proximity point of T. It follows immediately that the problem of finding a best proximity point is more general than that of finding a fixed point. In fact, if $A=B$, then $d(A,B)=0$ and hence a best proximity point of T becomes a fixed point of T. In this setting, we recall the following notions:

$$\begin{aligned} &A_{0} :=\bigl\{ a\in A: d(a,b)=d(A,B)\mbox{ for some }b\in B\bigr\} \\ &B_{0} :=\bigl\{ b\in B: d(a,b)=d(A,B)\mbox{ for some }a\in A\bigr\} . \end{aligned}$$

Basha [5] proposed the following result for the existence of a best proximity point of a non-self-mapping.

Theorem 1

([5], Theorem 3.1)

Let $(X,d)$ be a complete metric space and $A,B$ be two subsets of X such that $A_{0}\neq \varnothing$ (and hence $B_{0}\neq\varnothing$). Suppose that $T:A\to B$ is a mapping such that $T(A_{0})\subset B_{0}$. We make the following assumptions:

• A and B are closed;

• B is approximatively compact with respect to A;

• T is a proximal contraction, that is, there exists $\alpha\in [0,1)$ such that, for all $u,v,x,y\in A$,

  $$d(u,Tx)=d(A,B)=d(v,Ty) $$

  implies

  $$d(u,Tx)+d(Tx,Ty)+d(Ty,v)\leq\alpha d(x,y). $$

Then the following hold:

• (a) there exists a unique element $x\in A$ such that $d(x,Tx)=d(A,B)$;

• (b) if $\{x_{n}\}$ is a sequence in $A_{0}$ satisfying $d(x_{n+1},Tx_{n})=d(A,B)$ for all $n\geq0$, then $\lim_{n\to\infty}x_{n}=x$.

It is clear that Theorem 1 extends Banach’s contraction principle in the setting that $A=B=X$. By the way, there are plenty of papers which had generalized this result (for example, see [1, 2, 6]).

Basha and Shahzad [1] introduced the following two concepts of contractiveness for non-self-mappings.

Definition 2

([1])

Let $(X,d)$ be a metric space. Let A and B be nonempty subsets of X. We say that $T:A\to B$ is

• (a) a generalized proximal contraction of the first kind if there exist non-negative numbers α, β, γ with $\alpha+2\beta+2\gamma<1$ such that the condition

  $$d(u,Tx)=d(A,B)=d(v,Ty) $$

  implies

  $$d(u,v)\leq\alpha d(x,y)+\beta d(x,u)+\beta d(y,v)+\gamma d(x,v)+\gamma d(y,u); $$

• (b) a generalized proximal contraction of the second kind if there exist non-negative numbers α, β, γ with $\alpha+2\beta+2\gamma<1$ such that the condition

  $$d(u,Tx)=d(A,B)=d(v,Ty) $$

  implies

  $$\begin{aligned} d(Tu,Tv)\leq{}&\alpha d(Tx,Ty)+\beta d(Tx,Tu)+\beta d(Ty,Tv)\\ &{}+\gamma d(Tx,Tv)+\gamma d(Ty,Tu). \end{aligned}$$

Remark 3

Every proximal contraction is a generalized proximal contraction of the first kind.

In this paper, we show that the problem of finding a best proximity point recently established by Fernández-León [2] and Basha and Shahzad [1] reduces to a problem of finding a fixed point of a multivalued mapping. Recall that $x\in X$ is a fixed point of a multivalued mapping $T:X\to2^{X}\setminus\{\varnothing\}$ if $x\in Tx$. There are many conditions guaranteeing the existence of a fixed point of a multivalued mapping. Two of the classical works in this research are due to Nadler [7] and Caristi [8]. The interested reader is referred to [9], Chapter 5, for more discussion.

2 Main results

By studying the works of [4] and [3], we obtain the following fixed point theorem for a multivalued mapping.

Theorem 4

Let $(X,d)$ be a complete metric space. Let Y be a nonempty subset of X and let $F:Y\to(-\infty,\infty]$ be a proper function which is bounded below. Let $S:Y\to2^{Y}\setminus\{\varnothing\}$ be a multivalued mapping such that for each $x\in Y$ there exists $y\in Sx$ satisfying

$$ F(y)+d(x,y)\leq F(x).$$

(2.1)

Assume that for $z\in X$

$$ \inf\bigl\{ d(x,z)+d(x,Sx):x\in Y\bigr\} =0\quad\Longrightarrow\quad z\in Sz \cap Y.$$

(2.2)

Then there exists $w\in Y$ such that $w\in Sw$.

Proof

Let $x_{0}$ be an element in Y such that $F(x_{0})<\infty$. By the condition (2.1), there is an $x_{1}\in Sx_{0}$ such that $F(x_{1})+d(x_{0},x_{1})\leq F(x_{0})$. By induction, we have a sequence $\{ x_{n}\}$ in Y such that

$$x_{n+1}\in Sx_{n} \quad\mbox{and}\quad F(x_{n+1})+d(x_{n},x_{n+1}) \leq F(x_{n}) \quad\mbox{for all } n\geq0. $$

So $\{F(x_{n})\}$ is a decreasing sequence. Since F is bounded below, $\lim_{n\to\infty}F(x_{n})= \alpha$ for some $\alpha\in\mathbb{R}$. Let $m\geq0$. We have

$$\begin{aligned} \sum_{n=0}^{n=m} d(x_{n},x_{n+1})& \leq\sum_{n=0}^{n=m}\bigl(F(x_{n})-F(x_{n+1}) \bigr) \\ &=F(x_{0})-F(x_{m+1}) \\ &\leq F(x_{0})-\alpha. \end{aligned}$$

Then $\sum_{n=0}^{\infty}d(x_{n},x_{n+1})=\lim_{m\to\infty} \sum_{n=0}^{n=m} d(x_{n},x_{n+1})<\infty$ and hence $\{x_{n}\}$ is a Cauchy sequence. So $\lim_{n\to\infty}x_{n}= w$ for some $w\in X$. Note that

$$\lim_{n\to\infty}d(x_{n},w)= 0 \quad\mbox{and}\quad \lim _{n\to\infty }d(x_{n},Sx_{n})\leq\lim _{n\to\infty}d(x_{n},x_{n+1})= 0. $$

By the condition (2.2), we have $w\in Sw\cap Y$. □

2.1 Results for a generalized proximal contraction of the first kind

We show that the following result of Fernández-León [2] is a consequence of our Theorem 4.

Theorem 5

([2], Proposition 3.5)

Let $(X,d)$ be a complete metric space. Let A and B be nonempty subsets of X such that $A_{0}$ is nonempty. Let $T:A\to B$ be a mapping such that $T(A_{0})\subset B_{0}$. Let us assume the following conditions:

• $A_{0}$ is closed;

• T is a generalized proximal contraction of the first kind.

Then the following hold:

• (a) there exists a unique element x in A such that $d(x,Tx)=d(A,B)$;

• (b) if $\{x_{n}\}$ is a sequence in $A_{0}$ satisfying $d(x_{n+1},Tx_{n})=d(A,B)$ for each $n\geq0$, then $\lim_{n\to\infty }x_{n}= x$.

Proof

For each $x\in A_{0}$, we let

$$Sx=\bigl\{ y:y\in A_{0}\mbox{ and }d(y,Tx)=d(A,B)\bigr\} . $$

It follows that $S:A_{0}\to2^{A_{0}}\setminus\{\varnothing\}$.

Since T is a generalized proximal contraction of the first kind, there are $\alpha, \beta, \gamma\geq0$ with $\alpha+2\beta+2\gamma <1$ such that $d(u,Tx)=d(A,B)=d(v,Ty)$ implies

$$d(u,v)\leq\alpha d(x,y)+\beta d(x,u)+\beta d(y,v)+\gamma d(x,v)+\gamma d(y,u) $$

for all $u,v,x,y\in A$. Put $c=\frac{\alpha+\beta+\gamma}{1-\beta-\gamma }$ and $b=\frac{c+1}{2}$. Then $0\leq c< b<1$.

Claim that, for all $x,y,z\in A_{0}$, if $y\in Sx$ and $z\in Sy$, then $d(z,y)\leq cd(y,x)$. To see this, let x, y, z be elements in $A_{0}$ such that $y\in Sx$ and $z\in Sy$. Then

$$d(y,Tx)=d(A,B)=d(z,Ty). $$

Since T is a generalized proximal contraction of the first kind,

$$\begin{aligned} d(z,y)&\leq\alpha d(y,x)+\beta d(y,z)+\beta d(x,y)+\gamma d(y,y)+\gamma d(x,z)\\ &\leq\alpha d(y,x)+\beta d(y,z)+\beta d(x,y)+\gamma d(x,y)+\gamma d(y,z). \end{aligned}$$

Hence

$$d(z,y)\leq cd(y,x). $$

So we have the claim.

Next, we show that the condition (2.1) in Theorem 4 holds. Let $x\in A_{0}$. Since $0< b<1$, we can choose $y\in Sx$ so that

$$ bd(x,y)\leq d(x,Sx). $$

(2.3)

Let $z\in Sy$, then we obtain by the claim

$$ d(y,Sy)\leq d(z,y)\leq cd(y,x). $$

(2.4)

Using (2.3) and (2.4), we obtain

$$d(y,Sy)+bd(x,y)\leq cd(x,y)+d(x,Sx). $$

Then

$$\frac{1}{b-c}d(y,Sy)+d(y,x)\leq\frac{1}{b-c}d(x,Sx). $$

Let $F:A_{0}\to[0,\infty)$ be defined by $F(x)=\frac{1}{b-c}d(x,Sx)$ for each $x\in A_{0}$. So F satisfies the condition (2.1) in Theorem 4.

We show that the condition (2.2) in Theorem 4 holds. Let $\{x_{n}\}$ be a sequence in $A_{0}$ and $z\in X$ satisfying

$$\lim_{n\to\infty}d(x_{n},z)= 0 \quad\mbox{and}\quad \lim _{n\to\infty}d(x_{n},Sx_{n})= 0. $$

Since $A_{0}$ is closed, we have $z\in A_{0}$ and $Tz\in T(A_{0})\subset B_{0}$. Then there exists $u\in A_{0}$ such that

$$ d(u,Tz)=d(A,B). $$

(2.5)

We choose a sequence $\{u_{n}\}$ in $A_{0}$ so that $u_{n}\in Sx_{n}$ and

$$d(x_{n},u_{n})< d(x_{n},Sx_{n})+ \frac{1}{n} $$

for each $n\geq1$. Hence, $\lim_{n\to\infty}d(x_{n},u_{n})= 0$. Since $u_{n}\in Sx_{n}$ for each $n\geq0$,

$$ d(u_{n},Tx_{n})=d(A,B). $$

(2.6)

Since $\lim_{n\to\infty}d(x_{n},z)= 0$ and $\lim_{n\to\infty}d(x_{n},u_{n})= 0$, we get $\lim_{n\to\infty}u_{n}= z$. Using (2.5), (2.6), and the fact that T is a generalized proximal contraction of the first kind, we have, for each $n\geq0$,

$$d(u,u_{n})\leq\alpha d(z,x_{n})+\beta d(z,u)+\beta d(x_{n},u_{n})+\gamma d(z,u_{n})+\gamma d(x_{n},u). $$

As $n\to\infty$, we get

$$d(u,z)\leq(\beta+\gamma)d(z,u). $$

So $z=u$ and hence $d(z,Tz)=d(A,B)$, that is, $z\in Sz$. Therefore, the condition (2.2) in Theorem 4 holds. Using Theorem 4, there exists $w\in A_{0}$ such that $w\in Sw$, that is,

$$d(w,Tw)=d(A,B). $$

To see the uniqueness, we assume that $d(\widehat{w},T\widehat {w})=d(A,B)$ for some $\widehat{w}\in A$. Since T is a generalized proximal contraction of the first kind, we have

$$d(w,\widehat{w})\leq(\alpha+2\gamma) d(w,\widehat{w}). $$

That is, $w=\widehat{w}$. So we have (a).

To see (b), let $\{x_{n}\}$ be a sequence in $A_{0}$ such that

$$d(x_{n+1},Tx_{n})=d(A,B)\quad\mbox{for all }n\geq0. $$

Thus $x_{n+1}\in Sx_{n}$. By the claim, we get, for each $n\geq0$,

$$d(x_{n+2},x_{n+1})\leq cd(x_{n+1},x_{n}). $$

So $\{x_{n}\}$ is a Cauchy sequence and hence $\lim_{n\to\infty}x_{n}= x$ for some $x\in A_{0}$. Since T is a generalized proximal contraction of the first kind, we have

$$d(x_{n+1},w)\leq\alpha d(x_{n},w)+\beta d(x_{n},x_{n+1})+ \beta d(w,w)+\gamma d(x_{n},w)+\gamma d(w,x_{n+1}) $$

for each $n\geq0$. As $n\to\infty$, we get $d(x,w)\leq(\alpha+2\gamma)d(x,w)$. That is, $x=w$. Hence, $\lim_{n\to\infty}x_{n}= w$. So we have (b). □

2.2 Results for a generalized proximal contraction of the second kind

The following result of Fernández-León [2] is also a consequence of our Theorem 4.

Theorem 6

([2], Proposition 3.10)

Let $(X,d)$ be a complete metric space. Let A and B be nonempty subsets of X such that $A_{0}$ is nonempty. Let $T:A\to B$ be a mapping such that $T(A_{0})\subset B_{0}$. Let us assume the following conditions:

• $T(A_{0})$ is closed;

• T is a generalized proximal contraction of the second kind.

Then the following hold:

• (a) there exists $x\in A$ such that $d(x,Tx)=d(A,B)$;

• (b) if there is $\widehat{x}\in A$ such that $d(\widehat {x},T\widehat{x})=d(A,B)$, then $T\widehat{x}=Tx$;

• (c) if $\{x_{n}\}$ is a sequence in $A_{0}$ satisfying $d(x_{n+1},Tx_{n})=d(A,B)$ for each $n\geq0$, then $\lim_{n\to\infty }Tx_{n}= Tx$.

Proof

For each $x\in T(A_{0})$, we let

$$Sx=\bigl\{ y:y=Tu \mbox{ where } u\in A_{0}\mbox{ and }d(u,x)=d(A,B) \bigr\} . $$

It follows that $S:T(A_{0})\to2^{T(A_{0})}\setminus\{\varnothing\}$. Since T is a generalized proximal contraction of the second kind, there are $\alpha, \beta, \gamma\geq0$ with $\alpha+2\beta+2\gamma <1$ such that $d(u,Tx)=d(A,B)=d(v,Ty) \mbox{ implies}$

$$\begin{aligned} d(Tu,Tv)\leq{}&\alpha d(Tx,Ty)+\beta d(Tx,Tu)+\beta d(Ty,Tv)\\ &{}+\gamma d(Tx,Tv)+\gamma d(Ty,Tu) \end{aligned}$$

for all $u,v,x,y\in A$. Put $c=\frac{\alpha+\beta+\gamma}{1-\beta-\gamma }$ and $b=\frac{c+1}{2}$. Then $0\leq c< b<1$.

Claim 1: for each $u,v,x,y\in T(A_{0})$ if $u\in Sx$ and $v\in Sy$, then

$$\begin{aligned} d(u,v)\leq{}&\alpha d(x,y)+\beta d(x,u)+\beta d(y,v)\\ &{} +\gamma d(x,v)+\gamma d(y,u). \end{aligned}$$

To see this claim, let u, v, x, y be elements in $T(A_{0})$ such that $u\in Sx$ and $v\in Sy$. So $u=T\widehat{u}$, $v=T\widehat{v}$, $x=T\widehat{x}$ and $y=T\widehat{y}$ for some $\widehat{u},\widehat {v},\widehat{x},\widehat{y}\in A_{0}$ with

$$d(\widehat{u},T\widehat{x})=d(A,B)=d(\widehat{v},T\widehat{y}). $$

Since T is a generalized proximal contraction of the second kind,

$$\begin{aligned} d(T\widehat{u},T\widehat{v})\leq{}&\alpha d(T\widehat {x},T\widehat{y})+\beta d(T \widehat{x},T\widehat{u})+\beta d(T\widehat {y},T\widehat{v})\\ &{}+\gamma d(T\widehat{x},T\widehat{v})+\gamma d(T\widehat {y},T\widehat{u}). \end{aligned}$$

That is,

$$d(u,v)\leq\alpha d(x,y)+\beta d(x,u)+\beta d(y,v)+\gamma d(x,v)+\gamma d(y,u). $$

So we have Claim 1.

Claim 2: for each $x,y,z\in T(A_{0})$ if $y\in Sx$ and $z\in Sy$, then $d(z,y)\leq cd(x,y)$. To see this, let x, y, z be elements in $T(A_{0})$ such that $y\in Sx$ and $z\in Sy$. Using Claim 1, we have

$$\begin{aligned} d(z,y)&\leq\alpha d(y,x)+\beta d(y,z)+\beta d(x,y)+\gamma d(y,y)+\gamma d(x,z)\\ &\leq\alpha d(y,x)+\beta d(y,z)+\beta d(x,y)+\gamma d(x,y)+\gamma d(y,z). \end{aligned}$$

So $d(z,y)\leq cd(x,y)$. That is, Claim 2 holds.

Now, we show that the condition (2.1) in Theorem 4 holds. Let $x\in T(A_{0})$. Since $0< b<1$, there exists $y\in Sx$ such that

$$ bd(x,y)\leq d(x,Sx). $$

(2.7)

Let $z\in Sy$, then we obtain by Claim 2

$$ d(y,Sy)\leq d(z,y)\leq cd(x,y). $$

(2.8)

Using (2.7) and (2.8), we get

$$d(y,Sy)+bd(x,y)\leq cd(x,y)+d(x,Sx). $$

Then

$$\frac{1}{b-c}d(y,Sy)+d(x,y)\leq\frac{1}{b-c}d(x,Sx). $$

Let $F:T(A_{0})\to[0,\infty)$ be defined by $F(x)=\frac{1}{b-c}d(x,Sx)$ for each $x\in T(A_{0})$. So F satisfies the condition (2.1) in Theorem 4.

Next, we show that the condition (2.2) in Theorem 4 holds. Let $z\in X$ and let $\{x_{n}\}$ be a sequence in $T(A_{0})$ such that

$$\lim_{n\to\infty}d(x_{n},z)= 0 \quad\mbox{and}\quad \lim _{n\to\infty}d(x_{n},Sx_{n})= 0. $$

Since $T(A_{0})$ is closed, $z\in T(A_{0})$ and hence we can let

$$ \widehat{z}\in Sz. $$

(2.9)

We show that $\widehat{z}=z$. Since $\lim_{n\to\infty}d(x_{n},Sx_{n})= 0$, we can choose a sequence $\{y_{n}\}$ in $T(A_{0})$ so that

$$ y_{n}\in Sx_{n} $$

(2.10)

for each $n\geq0$ and

$$\lim_{n\to\infty}d(x_{n},y_{n})= 0. $$

Since $\lim_{n\to\infty}d(x_{n},z)= 0$ and $\lim_{n\to\infty}d(x_{n},y_{n})= 0$, we obtain $\lim_{n\to\infty}y_{n}= z$. Using (2.9), (2.10), and Claim 1,

$$d(\widehat{z},y_{n})\leq\alpha d(z,x_{n})+\beta d(z, \widehat{z})+\beta d(x_{n},y_{n})+\gamma d(z,y_{n})+\gamma d(x_{n},\widehat{z}). $$

As $n\to\infty$, we get $d(\widehat{z},z)\leq(\beta+\gamma )d(z,\widehat{z})$, that is, $\widehat{z}=z$. Hence, the condition (2.2) in Theorem 4 holds. Using Theorem 4, there exists $w\in T(A_{0})$ such that $w\in Sw$, that is, there exists $w^{*}\in A_{0}$ such that $w=Tw^{*}$ and

$$d\bigl(w^{*},Tw^{*}\bigr)=d(A,B). $$

So we have (a).

To see (b), let v be an element in A such that $d(v,Tv)=d(A,B)$. Since T is a generalized proximal contraction of the second kind,

$$\begin{aligned} d\bigl(Tv,Tw^{*}\bigr)\leq{}&\alpha d\bigl(Tv,Tw^{*}\bigr)+\beta d(Tv,Tv)+\beta d \bigl(Tw^{*},Tw^{*}\bigr)\\ &{}+\gamma d\bigl(Tv,Tw^{*}\bigr)+\gamma d\bigl(Tw^{*},Tv\bigr). \end{aligned}$$

Then $d(Tv,Tw^{*})\leq(\alpha+2\gamma)d(Tv,Tw^{*})$, which implies that $Tv=Tw^{*}$. So we have (b).

We show that (c) holds. Let $\{x_{n}\}$ be a sequence in $A_{0}$ such that $d(x_{n+1},Tx_{n})=d(A,B)$ for all $n\geq0$. So we get $Tx_{n+1}\in STx_{n}$. By using Claim 2, we have, for all $n\geq0$,

$$d(Tx_{n+2},Tx_{n+1})\leq cd(Tx_{n+1},Tx_{n}). $$

Thus $\{Tx_{n}\}$ is a Cauchy sequence and hence $\lim_{n\to\infty}Tx_{n}= s$ for some $s\in X$. Since $w\in Sw$ and $Tx_{n+1}\in STx_{n}$, we have

$$\begin{aligned} d(w,Tx_{n+1})\leq{}&\alpha d(w,Tx_{n})+\beta d(w,w)+\beta d(Tx_{n},Tx_{n+1})\\ &{}+\gamma d(w,Tx_{n+1})+\gamma d(Tx_{n},w). \end{aligned}$$

As $n\to\infty$, we have $d(w,s)\leq(\alpha+2\gamma)d(w,s)$. So $w=s$. That is, $\lim_{n\to\infty}Tx_{n}= w$. So (c) holds. □

Remark 7

The conclusion (c) of Theorem 6 is not mentioned in [2], Proposition 3.10.

Definition 8

([1])

Let $(X,d)$ be a metric space. Let A and B be nonempty subsets of X. The set B is said to be approximatively compact with respect to A if every sequence $\{y_{n}\}$ of B satisfying the condition that $\lim_{n\to\infty}d(x,y_{n})=d(x,B)$ for some x in A has a convergent subsequence.

We show that the following theorem of Basha and Shahzad [1] is also a consequence of our Theorem 4.

Theorem 9

([1], Theorem 3.4)

Let $(X,d)$ be a complete metric space. Let A and B be nonempty subsets of X such that $A_{0}$ is nonempty. Let $T:A\to B$ be a mapping such that $T(A_{0})\subset B_{0}$. Let us assume the following conditions:

• A, B are closed;

• A is approximatively compact with respect to B;

• T is continuous;

• T is a generalized proximal contraction of the second kind.

Then the following hold:

• (a) there is an element x in A such that $d(x,Tx)=d(A,B)$;

• (b) if there exists $\widehat{x}\in A$ such that $d(\widehat {x},T\widehat{x})=d(A,B)$, then $T\widehat{x}=Tx$;

• (c) if $\{x_{n}\}$ is a sequence in $A_{0}$ satisfying $d(x_{n+1},Tx_{n})=d(A,B)$ for each $n\geq0$, then $\lim_{n\to\infty }Tx_{n}= Tx$.

Proof

We define the mappings $S:T(A_{0})\to2^{T(A_{0})}\setminus\{ \varnothing\}$ and $F:T(A_{0})\to[0,\infty)$ as the ones in the proof of Theorem 6. It follows that the condition (2.1) in Theorem 4 holds.

Next, we show that the condition (2.2) in Theorem 4 holds. Let $\{x_{n}\}$ be a sequence in $T(A_{0})$ and let $z\in X$. Assume that

$$ \lim_{n\to\infty}d(x_{n},z)= 0 \quad\mbox{and}\quad \lim_{n\to\infty}d(x_{n},Sx_{n})= 0. $$

(2.11)

Since $x_{n}\in T(A_{0})\subset T(A)\subset B$ and B is closed, $z\in B$. We choose a sequence $\{y_{n}\}$ in $T(A_{0})$ so that $y_{n}\in Sx_{n}$ for each $n\geq0$ and

$$ \lim_{n\to\infty}d(x_{n},y_{n})= 0. $$

(2.12)

Since $y_{n}\in Sx_{n}$ for each $n\geq0$, we write $y_{n}=Tu_{n}$ for some $u_{n}\in A_{0}$ with

$$d(u_{n},x_{n})=d(A,B). $$

We have

$$d(A,B)\leq d(u_{n},z)\leq d(u_{n},x_{n})+d(x_{n},z)=d(A,B)+d(x_{n},z). $$

So $\lim_{n\to\infty}d(u_{n},z)= d(A,B)$. Since A is approximatively compact with respect to B, there is a subsequence $\{u_{n_{k}}\}$ of $\{ u_{n}\}$ such that $u_{n_{k}}\to u$ for some $u\in A$. Since T is continuous, we get $Tu_{n_{k}}\to Tu$. Using (2.11) and (2.12), we get $y_{n}\to z$ and hence

$$Tu=\lim_{k\to\infty}Tu_{n_{k}}=\lim_{n\to\infty}Tu_{n}= \lim_{n\to\infty}y_{n}=z. $$

Therefore,

$$d(u,Tu)=d(u,z)=\lim_{k\to\infty}d(u_{n_{k}},x_{n_{k}})=d(A,B). $$

That is, $Tu\in STu$ or $z\in Sz\cap T(A_{0})$. Therefore, the condition (2.2) in Theorem 4 holds. Using Theorem 4, there is $w\in T(A_{0})$ such that $w\in Sw$, that is, $w=T\widehat{w}$ for some $\widehat{w}\in A_{0}$ with $d(\widehat{w},T\widehat{w})=d(A,B)$. So (a) holds. The rest of the conclusions follow from Theorem 6. □

For a generalized proximal contraction of the first kind, the closedness of $A_{0}$ is more general than the condition that B is approximatively compact with respect to A (see Proposition 3.3 of [2]). Hence Proposition 3.5 of [2] (see our Theorem 5) is a generalized version of Theorem 3.1 of [1]. However, this is not the case for a generalized proximal contraction of the second kind. The following example is applicable in Theorem 9 but not in Theorem 6. That is, there is a continuous generalized proximal contraction of the second kind $T:A\to B$ such that $T(A_{0})$ is not closed but A is approximatively compact with respect to B.

Example 10

We consider the 2-dimensional Euclidean metric space $\mathbb{R}^{2}$. Let $A=\{(a,0):a\geq0\}$ and $B=\{(b,1):b\geq0\}$. We have $A_{0}=A$ and $B_{0}=B$. Let $T:A\to B$ be a mapping defined by, for each $(a,0)\in A$,

$$T(a,0)= \bigl(f(a),1 \bigr), $$

where

$$f(a)=\frac{1}{2}-\frac{1}{a+2}. $$

It is clear that, for each $a,b\geq0$,

$$\bigl|f(a)-f(b)\bigr|\leq\frac{1}{4} |a-b|. $$

Note that $T(A_{0})=T(A)=\{(x,1):x\in[0,\frac{1}{2})\}$ is not closed. It is clear that A is approximatively compact with respect to B and T is continuous. We show that T is a generalized proximal contraction of the second kind. In fact, let u, v, x, y be elements in A such that $d(u,Tx)=d(v,Ty)=d(A,B)$. We write $x=(a_{1},0)$ and $y=(a_{2},0)$ for some $a_{1},a_{2}\geq0$. So $u=(f(a_{1}),1)$ and $v=(f(a_{2}),1)$. We obtain

$$d(Tu,Tv)=\bigl|f^{2}(a_{1})-f^{2}(a_{2})\bigr| \leq\frac{1}{4} \bigl|f(a_{1})-f(a_{2})\bigr|= \frac{1}{4}d(Tx,Ty). $$