1 Introduction and preliminaries

Let \((X,d)\) be a metric space and A, B be subsets of X. We denote by \(\operatorname{CL}(B)\), the set of all nonempty closed subsets of B. A point \(x\in A\) is called a fixed point of a mapping \(T:A\to \operatorname{CL}(B)\), if \(x\in Tx\). The multivalued map T has no fixed point if \(A\cap B=\emptyset\). In this case \(d(x,Tx)>0\) for all \(x\in A\). So, one can attempt to find the necessary condition so that the minimization problem

$$\min_{x\in A}d(x,Tx) $$

has at least one solution. A point \(x^{\ast}\in X\) is said to be a best proximity point of the mapping \(T:A\to B\) if \(d(x^{\ast},Tx^{\ast})=\operatorname{dist}(A,B)\). When \(A=B\), the best proximity point reduces to a fixed point of the mapping T. The following well-known best approximation theorem is due to Fan.

Theorem 1.1

[1]

Let A be a nonempty compact convex subset of normed linear space X and \(T:A\to X\) be a continuous function. Then there exists \(x\in A\) such that

$$\|x-Tx\|=\inf_{a\in A}\bigl\{ \Vert Tx-a\Vert \bigr\} . $$

In this paper, we discuss sufficient conditions which ensure the existence of best proximity points for multivalued non-self-mappings satisfying contraction condition on the closed ball of a complete metric space. Moreover, we discuss the stability of the best proximity points. Our results extend and generalize some results by Lim [2], and Abkar and Gabeleh [3]. Some important best proximity theorems can be found in [4–15].

Now we recollect some notions, definitions, and results, for easy reference. \(\operatorname{dist}(A,B)= \inf\{d(a,b): a\in A, b\in B\}\), \(d(x,B)=\inf\{d(x,b): b\in B\}\), \(A_{0}=\{a\in A: d(a,b)=\operatorname{dist}(A,B)\ \text{for some }b\in B\}\), \(B_{0}=\{b\in B: d(a,b)=\operatorname{dist}(A,B)\text{ for some } a\in A\}\), \(\operatorname{CB}(B)\) is the set of all nonempty closed and bounded subsets of B and \(B(x_{0},r)=\{x\in X: d(x_{0},x)\leq r\}\).

Definition 1.2

[13]

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 and only if for any \(x_{1},x_{2}\in A\) and \(y_{1},y_{2}\in B\),

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

Example 1.3

[14]

Let \(X=\{(0,1),(1,0),(0,3),(3,0)\}\), endowed with a metric \(d((x_{1},x_{2}), (y_{1},y_{2}))=|x_{1}-y_{1}|+|x_{2}-y_{2}|\). Let \(A=\{(0,1),(1,0)\}\) and \(B=\{(0,3),(3,0)\}\). Then, for

$$d\bigl((0,1),(0,3)\bigr)=\operatorname{dist}(A,B)\quad \text{and} \quad d \bigl((1,0),(3,0)\bigr)=\operatorname{dist}(A,B) $$

we have

$$d\bigl((0,1),(1,0)\bigr)< d\bigl((0,3),(3,0)\bigr). $$

Also, \(A_{0}\neq\emptyset\). Thus the pair \((A,B)\) satisfies the weak P-property.

Definition 1.4

[3]

An element \(x^{\ast}\in A\) is said to be a best proximity point of a multivalued non-self-mapping T, if \(d(x^{\ast},Tx^{\ast})=\operatorname{dist}(A,B)\).

Theorem 1.5

[3]

Let A and B be two nonempty closed subsets of a complete metric space \((X,d)\) such that \(A_{0}\) is nonempty. Let \(T: A \to \operatorname{CB}(B)\) be a mapping satisfying the following conditions:

  1. (i)

    for each \(x\in A_{0}\), we have \(Tx \subseteq B_{0}\);

  2. (ii)

    the pair \((A,B)\) satisfies the P-property;

  3. (iii)

    there exists \(\alpha\in(0,1)\) such that, for each \(x,y\in A\), we have \(H(Tx,Ty)\leq\alpha d(x,y)\).

Then there exists an element \(x^{\ast}\in A_{0}\) such that \(d(x^{\ast},Tx^{\ast})=\operatorname{dist}(A,B)\).

2 Best proximity theorems

We start this section by introducing the following definition.

Definition 2.1

Let A and B be nonempty subsets of a metric space \((X,d)\), \(x_{0}\in A_{0}\), and \(B(x_{0},r)\) is a closed ball in X. A mapping \(T:A\to \operatorname{CL}(B)\) is said to be a proximal contraction on \(B(x_{0},r)\), if there exists \(\alpha\in(0,1)\) such that

$$ H(Tx,Ty)\leq\alpha d(x,y)\quad \text{for each } x,y\in B(x_{0},r)\cap A. $$
(2.1)

Lemma 2.2

[16]

Let \((X,d)\) be a metric space, \(B\in \operatorname{CL}(X)\), and \(q>1\). Then, for each \(x\in X\), there exists an element \(b\in B\) such that

$$ d(x,b)\leq q d(x,B). $$
(2.2)

Now we are in a position to state and prove our first result.

Theorem 2.3

Let A and B be nonempty closed subsets of a complete metric space \((X,d)\). Assume that \(A_{0}\) is nonempty and \(T: A \to \operatorname{CL}(B)\) is a mapping satisfying the following conditions:

  1. (i)

    for each \(x\in A_{0}\), we have \(Tx \subseteq B_{0}\);

  2. (ii)

    the pair \((A,B)\) satisfies weak P-property;

  3. (iii)

    there exists \(x_{0}\in A_{0}\) such that T is a proximal contraction on the closed ball \(B(x_{0},r)\) and \(d(x_{0},Tx_{0})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r\).

Then T has a best proximity point in \(B(x_{0},r)\cap A_{0}\).

Proof

By hypothesis (iii), we have \(x_{0}\in A_{0}\) such that T is a proximal contraction on the closed ball \(B(x_{0},r)\) and \(d(x_{0},Tx_{0})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r\). As \(x_{0}\in A_{0}\). By (i), we have \(y_{0}\in Tx_{0}\subseteq B_{0}\). Then there exists \(x_{1}\in A_{0}\) such that

$$ d(x_{1},y_{0})=\operatorname{dist}(A,B). $$
(2.3)

By using the triangular inequality, hypothesis (iii) and (2.3), we have

$$ d(x_{0},x_{1})\leq d(x_{0},Tx_{0})+d(Tx_{0},x_{1}) \leq d(x_{0},Tx_{0})+d(y_{0},x_{1}) \leq(1-\sqrt{\alpha})r. $$
(2.4)

Since \(x_{1}\in A_{0} \subseteq A\), \(x_{1}\in B(x_{0},r)\cap A\). From (2.1), we have

$$ d(y_{0},Tx_{1})\leq H(Tx_{0},Tx_{1}) \leq\alpha d(x_{0},x_{1}). $$
(2.5)

As \(\alpha>0\), by Lemma 2.2, we have \(y_{1}\in Tx_{1}\) such that

$$ d(y_{0},y_{1})\leq\frac{1}{\sqrt{\alpha}}d(y_{0},Tx_{1}) \leq\sqrt{\alpha} d(x_{0},x_{1}). $$
(2.6)

Since \(Tx_{1}\subseteq B_{0}\), there exists \(x_{2}\in A_{0}\) such that

$$ d(x_{2},y_{1})=\operatorname{dist}(A,B), $$
(2.7)

as \((A,B)\) satisfies the weak P-property. From (2.3) and (2.7), we have

$$ d(x_{1},x_{2})\leq d(y_{0},y_{1}). $$
(2.8)

From (2.6) and (2.8), we have

$$ d(x_{1},x_{2})\leq\sqrt{\alpha} d(x_{0},x_{1}). $$
(2.9)

Considering the triangular inequality, (2.4), and (2.9), we have

$$\begin{aligned} d(x_{0},x_{2}) \leq& d(x_{0},x_{1})+d(x_{1},x_{2}) \\ \leq& d(x_{0},x_{1})+\sqrt{\alpha} d(x_{0},x_{1}) \\ \leq&(1-\alpha)r< r. \end{aligned}$$

By construction, we have \(x_{2}\in A_{0} \subseteq A\). Thus \(x_{2}\in B(x_{0},r)\cap A\). Again from (2.1), we have

$$ d(y_{1},Tx_{2})\leq H(Tx_{1},Tx_{2}) \leq\alpha d(x_{1},x_{2}). $$
(2.10)

By using Lemma 2.2, we have \(y_{2}\in Tx_{2}\) such that

$$ d(y_{1},y_{2})\leq\frac{1}{\sqrt{\alpha}}d(y_{1},Tx_{2}) \leq\sqrt{\alpha} d(x_{1},x_{2}). $$
(2.11)

Since \(Tx_{2}\subseteq B_{0}\), there exists \(x_{3}\in A_{0}\) such that

$$ d(x_{3},y_{2})=\operatorname{dist}(A,B), $$
(2.12)

as \((A,B)\) satisfies the weak P-property. From (2.7) and (2.12), we have

$$ d(x_{2},x_{3})\leq d(y_{1},y_{2}). $$
(2.13)

From (2.11) and (2.13), we have

$$ d(x_{2},x_{3})\leq\sqrt{\alpha} d(x_{1},x_{2}) \leq\alpha d(x_{0},x_{1}). $$
(2.14)

By considering the triangular inequality, (2.9), and (2.14), we have

$$\begin{aligned} d(x_{0},x_{3}) \leq& d(x_{0},x_{1})+d(x_{1},x_{2})+d(x_{2},x_{3}) \\ \leq& \bigl[1+\sqrt{\alpha}+(\sqrt{\alpha})^{2}\bigr]d(x_{0},x_{1}) \\ \leq& \bigl[1+\sqrt{\alpha}+(\sqrt{\alpha})^{2}\bigr](1-\sqrt{ \alpha})r< r, \end{aligned}$$

as \(x_{3}\in A_{0} \subseteq A\). Thus, \(x_{3}\in B(x_{0},r)\cap A\). Continuing in the same way, we get two sequences \(\{x_{n}\} \subseteq A_{0}\) with \(x_{n}\in B(x_{0},r)\cap A\) and \(\{y_{n}\}\subseteq B_{0}\) with \(y_{n}\in Tx_{n}\) such that

$$ d(x_{n},y_{n-1})=\operatorname{dist}(A,B)\quad \text{for each } n\in \mathbb{N} . $$
(2.15)

Moreover,

$$ d(x_{n},x_{n+1})\leq d(y_{n-1},y_{n}) \leq(\sqrt{\alpha})^{n} d(x_{0},x_{1})\quad \text{for each } n\in\mathbb{N}. $$
(2.16)

For \(n>m\), we have

$$ d(x_{n},x_{m})\leq\sum_{i=n}^{m-1}d(x_{i},x_{i+1}) \leq \sum_{i=n}^{m-1}(\sqrt{ \alpha})^{i}d(x_{0},x_{1})< \sum _{i=n}^{\infty}(\sqrt{\alpha})^{i}d(x_{0},x_{1})< \infty. $$
(2.17)

Hence \(\{x_{n}\}\) is a Cauchy sequence in \(B(x_{0},r)\cap A \subseteq A\). A similar reasoning shows that \(\{y_{n}\}\) is a Cauchy sequence in B. Since \(B(x_{0},r)\cap A\) is closed in A, and A, B are closed subsets of a complete metric space, there exist \(x^{\ast} \in B(x_{0},r)\cap A\) and \(y^{\ast}\in B\) such that \(x_{n} \to x^{\ast}\) and \(y_{n} \to y^{\ast}\). By (2.15), we conclude that \(d(x^{\ast},y^{\ast})=\operatorname{dist}(A,B)\) as \(n\to\infty\). Clearly, \(y^{\ast}\in Tx^{\ast}\), since \(\lim_{n \to \infty}d(y_{n},Tx^{\ast})\leq\lim_{n\to \infty}H(Tx_{n},Tx^{\ast})=0\). Hence \(\operatorname{dist}(A,B)\leq d(x^{\ast},Tx^{\ast})\leq d(x^{\ast},y^{\ast})=\operatorname{dist}(A,B)\). Therefore, \(x^{\ast}\) is a best proximity point of the mapping T. □

Example 2.4

Let \(X=\mathbb{R}^{2}\) be endowed with the metric \(d((x_{1},y_{1}),(x_{2},y_{2}))=|x_{1}-x_{2}|+{|y_{1}-y_{2}|}\). Suppose that \(A=\{(1,x): x\in\mathbb{R}\}\) and \(B=\{(0,x): x\in\mathbb{R}\}\). Define \(T:A\to \operatorname{CL}(B)\) by

$$T(1,x)=\left \{ \textstyle\begin{array}{l@{\quad}l} \{(0,0)\} & \mbox{if } x\leq0, \\ \{(0,0),(0,x/2)\} & \mbox{if } 0\leq x\leq10, \\ \{(0,x)\} & \mbox{if } x>10. \end{array}\displaystyle \right . $$

Let us consider a ball \(B(x_{0},r)\) with \(x_{0}=(1,0.1)\) and \(r=7.5\). Then it is easy to see that T is a proximal contraction on the closed ball \(B((1,0.1),7.5)\) with \(\alpha=\frac{1}{2}\). Also, we have \(d(x_{0},Tx_{0})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r\). Furthermore, \(A_{0}=A\), \(B_{0}=B\); for each \(x\in A_{0}\) we have \(Tx \subseteq B_{0}\) and the pair \((A,B)\) satisfies the weak P-property. Therefore, all the conditions of Theorem 2.3 hold and T has a best proximity point.

Corollary 2.5

Let A and B be nonempty closed subsets of a complete metric space \((X,d)\). Assume that \(A_{0}\) is nonempty and \(T: A \to B\) is a mapping satisfying the following conditions:

  1. (i)

    for each \(x\in A_{0}\), we have \(Tx \in B_{0}\);

  2. (ii)

    the pair \((A,B)\) satisfies the weak P-property;

  3. (iii)

    there exists \(x_{0}\in A_{0}\) such that T is a proximal contraction on the closed ball \(B(x_{0},r)\), that is,

    $$ d(Tx,Ty)\leq\alpha d(x,y)\quad \textit{for each }x,y\in B(x_{0},r) \cap A, $$
    (2.18)

    and \(d(x_{0},Tx_{0})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r\).

Then T has a best proximity point in \(B(x_{0},r)\cap A_{0}\).

If we assume that \(X=A=B\), then Theorem 2.3 reduces to the following fixed point theorem.

Corollary 2.6

Let \((X,d)\) be a complete metric space and \(T: X \to \operatorname{CL}(X)\) be a mapping. Assume that there exist \(x_{0}\in X\) and \(\alpha\in(0,1)\) satisfying

$$ H(Tx,Ty)\leq\alpha d(x,y)\quad \textit{for each } x,y\in B(x_{0},r) $$

and \(d(x_{0},Tx_{0})\leq(1-\sqrt{\alpha})r\). Then T has a fixed point.

3 Stability of best proximity points

Stability of fixed point sets of multivalued mappings was initially investigated by Markin [15] and Nadler [16] with some strong conditions. Lim [2] proved the stability theorem for fixed point sets of multivalued contraction mappings by relaxing the condition assumed by Markin [15]. Abkar and Gabeleh [3] discussed the stability of best proximity point sets of non-self-multivalued mappings. In this section, we extend and generalize the stability theorems due to Abkar and Gabeleh [3], and Lim [2].

In this section, by \(B_{T_{1}}\) and \(B_{T_{2}}\) we denote the sets of best proximity points of \(T_{1}\) and \(T_{2}\), respectively.

Theorem 3.1

Let A and B be nonempty closed subsets of a complete metric space \((X,d)\). Assume that \(A_{0}\) is nonempty and \(T_{i}: A \to \operatorname{CL}(B)\), \(i=1,2\) are mappings satisfying the following conditions:

  1. (i)

    for each \(x\in A_{0}\), we have \(T_{i}x \subseteq B_{0}\), \(i=1,2\);

  2. (ii)

    the pair \((A,B)\) satisfies the weak P-property;

  3. (iii)

    for each \(i=1,2\), there exists \(a_{i}\) such that \(T_{i}\) is proximal contraction on the closed ball \(B(a_{i},r_{i})\) with the same α as a contraction constant, that is,

    $$ H(T_{i}x,T_{i}y)\leq\alpha d(x,y)\quad \textit{for each } x,y\in B(a_{i},r_{i})\cap A, $$
    (3.1)

    and \(d(a_{i},T_{i}a_{i})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r_{i}\).

Then

$$H(B_{T_{1}},B_{T_{2}})\leq\frac{1}{1-\sqrt{\alpha}} \Bigl[\sup _{x\in A}H(T_{1}x,T_{2}x) +2 \operatorname{dist}(A,B) \Bigr]. $$

Proof

Let \(x_{0}\in B_{T_{1}}\), then we have \(y_{0}\in T_{2}x_{0}\) such that

$$d(x_{0},y_{0})\leq H(T_{1}x_{0},T_{2}x_{0})+ \operatorname{dist}(A,B). $$

Since \(y_{0}\in T_{2}x_{0}\subseteq B_{0}\), we have \(x_{1}\in A_{0}\) such that

$$ d(x_{1},y_{0})=\operatorname{dist}(A,B). $$
(3.2)

We know that \(T_{2}\) is a proximal contraction for closed ball \(B(a_{2},r_{2})\). Without loss of generality, we take \(a_{2}=x_{0}\) and \(r_{2}=r\) such that \(d(x_{0},T_{2}x_{0})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r\). Clearly, \(x_{1}\in B(x_{0},r)\cap A\), since \(x_{1}\in A_{0}\subseteq A\) and

$$ d(x_{0},x_{1})\leq d(x_{0},T_{2}x_{0})+d(T_{2}x_{0},x_{1}) \leq d(x_{0},T_{2}x_{0})+d(y_{0},x_{1}) \leq(1-\sqrt{\alpha})r. $$
(3.3)

By hypothesis (iii), we have

$$ d(y_{0},T_{2}x_{1})\leq H(T_{2}x_{0},T_{2}x_{1})\leq\alpha d(x_{0},x_{1}). $$
(3.4)

As \(\alpha>0\), by Lemma 2.2, we have \(y_{1}\in T_{2}x_{1}\) such that

$$ d(y_{0},y_{1})\leq\frac{1}{\sqrt{\alpha}}d(y_{0},T_{2}x_{1}) \leq \sqrt{\alpha} d(x_{0},x_{1}). $$
(3.5)

Since \(T_{2}x_{1}\subseteq B_{0}\), there exists \(x_{2}\in A_{0}\) such that

$$ d(x_{2},y_{1})=\operatorname{dist}(A,B), $$
(3.6)

as \((A,B)\) satisfies the weak P-property. From (3.2) and (3.6), we have

$$ d(x_{1},x_{2})\leq d(y_{0},y_{1}). $$
(3.7)

From (3.5) and (3.7), we have

$$ d(x_{1},x_{2})\leq\sqrt{\alpha} d(x_{0},x_{1}). $$
(3.8)

Considering the triangular inequality, (3.3), and (3.8), we have

$$\begin{aligned} d(x_{0},x_{2}) \leq& d(x_{0},x_{1})+d(x_{1},x_{2}) \\ \leq& d(x_{0},x_{1})+\sqrt{\alpha} d(x_{0},x_{1}) \\ \leq&(1-\alpha)r< r. \end{aligned}$$

Also, \(x_{2}\in A_{0}\subseteq A\). Thus, \(x_{2}\in B(x_{0},r)\cap A\). Continuing in the same way, we get two sequences \(\{x_{n}\} \subseteq A_{0}\) with \(x_{n}\in B(x_{0},r)\cap A\) and \(\{y_{n}\}\subseteq B_{0}\) with \(y_{n}\in T_{2}x_{n}\) such that

$$ d(x_{n},y_{n-1})=\operatorname{dist}(A,B)\quad \text{for each } n\in \mathbb{N} . $$
(3.9)

Moreover,

$$ d(x_{n},x_{n+1})\leq d(y_{n-1},y_{n}) \leq(\sqrt{\alpha})^{n} d(x_{0},x_{1})\quad \text{for each } n\in\mathbb{N}. $$
(3.10)

For \(n>m\), we have

$$ d(x_{n},x_{m})\leq\sum_{i=n}^{m-1}d(x_{i},x_{i+1}) \leq \sum_{i=n}^{m-1}(\sqrt{ \alpha})^{i}d(x_{0},x_{1})< \sum _{i=n}^{\infty}(\sqrt{\alpha})^{i}d(x_{0},x_{1})< \infty. $$
(3.11)

Hence \(\{x_{n}\}\) is a Cauchy sequence in \(B(x_{0},r)\cap A\subseteq A\). A similar reasoning shows that \(\{y_{n}\}\) is a Cauchy sequence in B. Since \(B(x_{0},r)\cap A\) is closed in A, and A, B are closed subsets of a complete metric space, there exist \(u^{\ast} \in B(x_{0},r)\cap A\) and \(v^{\ast}\in B\) such that \(x_{n} \to u^{\ast}\) and \(y_{n} \to v^{\ast}\). By (3.9), we conclude that \(d(u^{\ast},v^{\ast})=\operatorname{dist}(A,B)\) as \(n\to\infty\). Clearly, \(v^{\ast}\in T_{2}u^{\ast}\). Then we have \(\operatorname{dist}(A,B)\leq d(u^{\ast},T_{2}u^{\ast})\leq d(u^{\ast},v^{\ast})=\operatorname{dist}(A,B)\). Therefore \(u^{\ast}\) is a best proximity point of \(T_{2}\). Now, we have

$$\begin{aligned} d\bigl(x_{0},u^{\ast}\bigr) \leq& \sum _{n=0}^{\infty}d(x_{n},x_{n+1}) \\ \leq& \sum_{n=0}^{\infty}(\sqrt{ \alpha})^{n}d(x_{0},x_{1}) \\ =&\frac{1}{1-\sqrt{\alpha}}d(x_{0},x_{1}) \\ \leq& \frac{1}{1-\sqrt{\alpha}}\bigl[d(x_{0},y_{0})+d(y_{0},x_{1}) \bigr] \\ =&\frac{1}{1-\sqrt{\alpha}}\bigl[d(x_{0},y_{0})+ \operatorname{dist}(A,B)\bigr] \\ \leq&\frac{1}{1-\sqrt{\alpha}}\bigl[H(T_{1}x_{0},T_{2}x_{0})+2 \operatorname{dist}(A,B)\bigr]. \end{aligned}$$

Similarly, if \(\mathfrak{x}_{\mathfrak{0}}\in B_{T_{2}}\), then we have \(\mathfrak{u}^{\ast}\in B_{T_{1}}\) such that

$$\begin{aligned} d\bigl(\mathfrak{x}_{\mathfrak{0}}, \mathfrak{u}^{\ast}\bigr) \leq \frac{1}{1-\sqrt{\alpha}}\bigl[H(T_{1}\mathfrak{x}_{\mathfrak{0}},T_{2} \mathfrak {x}_{\mathfrak{0}})+2\operatorname{dist}(A,B)\bigr]. \end{aligned}$$

Thus, we have

$$H(B_{T_{1}},B_{T_{2}})\leq\frac{1}{1-\sqrt{\alpha}} \Bigl[\sup _{x\in A} H(T_{1}x,T_{2}x)+2 \operatorname{dist}(A,B) \Bigr]. $$

 □

Example 3.2

Let \(X=\mathbb{R}^{2}\) be endowed with the metric \(d((x_{1},y_{1}),(x_{2},y_{2}))=|x_{1}-x_{2}|+{|y_{1}-y_{2}|}\). Suppose that \(A=\{(1,x): x\in\mathbb{R}\}\) and \(B=\{(0,x): x\in\mathbb{R}\}\). Define \(T_{1},T_{2}:A\to \operatorname{CL}(B)\) by

$$T_{1}(1,x)=\left \{ \textstyle\begin{array}{l@{\quad}l} \{(0,0)\} &\mbox{if } x\leq0, \\ \{(0,0),(0,x/2)\}& \mbox{if } 0\leq x\leq10, \\ \{(0,x^{2})\} & \mbox{if } x>10 \end{array}\displaystyle \right . $$

and

$$T_{2}(1,x)=\left \{ \textstyle\begin{array}{l@{\quad}l} \{(0,1)\} & \mbox{if } x\leq1, \\ \{(0,1),(0,(x+1)/2)\} & \mbox{if } x>1. \end{array}\displaystyle \right . $$

It is easy to see that \(T_{1}\) is a proximal contraction on the closed ball \(B(x_{0}=(1,0.1),r=7.5)\) with \(\alpha=\frac{1}{2}\) and \(d(x_{0},Tx_{0})+\operatorname{dist}(A,B)\leq(1-\sqrt{\alpha})r\). Further, \(T_{2}\) is a proximal contraction on the closed ball \(B(x_{1}=(1,1.25),r_{1}=8)\) with \(\alpha=\frac{1}{2}\) and \(d(x_{1},Tx_{1})+\operatorname{dist}(A,B)\leq (1-\sqrt{\alpha})r_{1}\). Furthermore, it is easy to see that \(A_{0}=A\), \(B_{0}=B\), and for each \(x\in A_{0}\) we have \(T_{i}x \subseteq B_{0}\) for each \(i=1,2\) and the pair \((A,B)\) satisfies the weak P-property. All the conditions of Theorem 3.1 hold. Thus the conclusion holds. That is,

$$H(B_{T_{1}},B_{T_{2}})\leq \frac{1}{1-\sqrt{\alpha}}\Bigl[\sup _{x\in A}H(T_{1}x,T_{2}x)+2 \operatorname{dist}(A,B)\Bigr]. $$

If we assume that \(X=A=B\), then Theorem 3.1 reduces to the following stability result.

Corollary 3.3

Let \((X,d)\) be a complete metric space and \(T_{i}: X \to \operatorname{CL}(X)\), \(i=1,2\) be mappings. Assume that there exist \(\alpha\in(0,1)\) and \(a_{1},a_{2}\in X\) such that, for each i, we have

$$ H(T_{i}x,T_{i}y)\leq\alpha d(x,y) \quad \textit{for each } x,y\in B(a_{i},r_{i}) $$
(3.12)

and \(d(a_{i},T_{i}a_{i})\leq(1-\sqrt{\alpha})r_{i}\). Let \(F_{T_{1}}\) and \(F_{T_{2}}\) denote the sets of fixed points of \(T_{1}\) and \(T_{2}\) respectively. Then

$$H(F_{T_{1}},F_{T_{2}})\leq\frac{1}{1-\sqrt{\alpha}}\sup _{x\in A}H(T_{1}x,T_{2}x). $$

Note that in this theorem \(B(a_{i},r_{i})\) are closed balls.

Remark 3.4

If \(r_{1}\), \(r_{2}\) are sufficiently large, then \(B(a_{1},r_{1})\) and \(B(a_{2},r_{2})\) are equal to X. In this case, from Corollary 3.3, we get the following result.

Corollary 3.5

(Lim [2], Lemma 1)

Let \((X,d)\) be a complete metric space and \(T_{i}: X \to \operatorname{CL}(X)\), \(i=1,2\) be α-contractions with the same α, that is,

$$ H(T_{i}x,T_{i}y)\leq\alpha d(x,y)\quad \textit{for each } x,y\in X, $$

where \(\alpha\in(0,1)\). Then

$$H(F_{T_{1}},F_{T_{2}})\leq\frac{1}{1- {\alpha}}\sup _{x\in X}H(T_{1}x,T_{2}x). $$

Corollary 3.6

(Lim [2], Theorem 1)

Let \((X,d)\) be a complete metric space and \(T_{i}: X \to \operatorname{CL}(X)\), \(i=1,2,\ldots \) , be α-contractions with the same α. If \(\lim_{i\to\infty} H(T_{i}x,T_{0}x)=0\) uniformly for all \(x\in X\), then \(\lim_{i\to\infty} H(F_{T_{i}},F_{T_{0}})=0\).