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:

  1. (a)

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

  2. (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

  1. (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

    $$d(u,v)\leq\alpha d(x,y)+\beta d(x,u)+\beta d(y,v)+\gamma d(x,v)+\gamma d(y,u); $$
  2. (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:

  1. (a)

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

  2. (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:

  1. (a)

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

  2. (b)

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

  3. (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:

  1. (a)

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

  2. (b)

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

  3. (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

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