Best proximity point theorems with Suzuki distances

Abstract

In this paper, we define the weak P-property and the α-ψ-proximal contraction by p in which p is a τ-distance on a metric space. Then, we prove some best proximity point theorems in a complete metric space X with generalized distance. Also we define two kinds of α-p-proximal contractions and prove some best proximity point theorems.

1 Introduction

Let us assume that A and B are two nonempty subsets of a metric space $(X,d)$ and $T:A\longrightarrow B$. Clearly $T(A)\cap A\neq \emptyset$ is a necessary condition for the existence of a fixed point of T. The idea of the best proximity point theory is to determine an approximate solution x such that the error of equation $d(x,Tx)=0$ is minimum. A solution x for the equation $d(x,Tx)=d(A,B)$ is called a best proximity point of T. The existence and convergence of best proximity points have been generalized by several authors [1–8] in many directions. Also, Akbar and Gabeleh [9, 10], Sadiq Basha [11] and Pragadeeswarar and Marudai [12] extended the best proximity points theorems in partially ordered metric spaces (see also [13–18]). On the other hand, Suzuki [19] introduced the concept of τ-distance on a metric space and proved some fixed point theorems for various contractive mappings by τ-distance. In this paper, by using the concept of τ-distance, we prove some best proximity point theorems.

2 Preliminaries

Let A, B be two nonempty subsets of a metric space $(X,d)$. The following notations will be used throughout this paper:

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

We recall that $x\in A$ is a best proximity point of the mapping $T:A\longrightarrow B$ if $d(x,Tx)=d(A,B)$. It can be observed that a best proximity point reduces to a fixed point if the underlying mapping is a self-mapping.

Definition 2.1

([20])

Let $(A, B)$ be a pair of nonempty subsets of a metric space X with $A\neq\emptyset$. Then the pair $(A,B)$ is said to have the P-property if and only if

$$\left.\begin{array}{r@{}} d(x_{1}, y_{1}) = d(A, B),\\ d(x_{2}, y_{2}) = d(A, B) \end{array} \right\} \quad\Longrightarrow\quad d(x_{1},x_{2})=d(y_{1},y_{2}), $$

where $x_{1}, x_{2}\in A_{0}$ and $y_{1}, y_{2}\in B_{0}$.

It is clear that, for any nonempty subset A of X, the pair $(A,A)$ has the P-property.

Definition 2.2

([5])

A is said to be approximately compact with respect to B if every sequence $\{x_{n}\}$ of A, satisfying the condition that $d(y,x_{n})\longrightarrow d(y,A)$ for some y in B, has a convergent subsequence.

Remark 2.3

([5])

Every set is approximately compact with respect to itself.

Samet et al. [21] introduced a class of contractive mappings called α-ψ-contractive mappings. Let Ψ be the family of nondecreasing functions $\psi:[0,\infty )\longrightarrow[0,\infty)$ such that $\sum_{n=1}^{\infty}\psi ^{n}(t)<\infty$ for all $t>0$, where $\psi^{n}(t)$ is the nth iterate of ψ.

Lemma 2.4

([21])

For every function $\psi:[0,\infty)\longrightarrow[0,\infty)$, the following holds:

• if ψ is nondecreasing, then, for each $t > 0$, $\lim_{n\rightarrow\infty}\psi^{n}(t)=0$ implies $\psi(t) < t$.

Definition 2.5

([1])

Let $T: A\longrightarrow B$ and $\alpha:A\times A\longrightarrow [0,\infty)$. We say that T is α-proximal admissible if

$$\left.\begin{array}{r@{}} \alpha(x_{1},x_{2})\geq1, \\ d(u_{1},Tx_{1}) = d(A, B),\\ d(u_{2},Tx_{2}) = d(A, B) \end{array} \right\} \quad\Longrightarrow\quad \alpha(u_{1},u_{2})\geq1 $$

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

Remark 2.6

Let ‘⪯’ be a partially ordered relation on A and $\alpha :A\times A\longrightarrow[0,\infty)$ be defined by

$$\alpha(x,y) = \left \{ \begin{array}{@{}l@{\quad}l} 1, & x\preceq y,\\ 0, & \mbox{otherwise}. \end{array} \right . $$

If T is α-proximal admissible, then T is said to be proximally increasing. In other words, T is proximally increasing if it satisfies the condition that

$$\left.\begin{array}{r@{}} x_{1}\preceq x_{2},\\ d(u_{1},Tx_{1}) = d(A, B),\\ d(u_{2},Tx_{2}) = d(A, B) \end{array} \right\} \quad\Longrightarrow\quad u_{1} \preceq u_{2} $$

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

Definition 2.7

([19])

Let X be a metric space with metric d. A function $p:X\times X\longrightarrow[0,\infty)$ is called τ-distance on X if there exists a function $\eta:X\times [0,\infty )\longrightarrow[0,\infty)$ such that the following are satisfied:

($\tau_{1}$):

$p(x,z)\leq p(x,y)+p(y,z)$ for all $x,y,z\in{X}$;

($\tau_{2}$):

$\eta(x,0)=0$ and $\eta(x,t)\geq t$ for all $x\in{X}$ and $t\in [0,\infty)$, and η is concave and continuous in its second variable;

($\tau_{3}$):

$\lim_{n} x_{n}=x$ and $\lim_{n}\sup\{\eta(z_{n},(z_{n},x_{m})):m\geq n\}=0$ imply $p(w,x)\leq\liminf_{n} p(w,x_{n})$ for all $w\in{X}$;

($\tau_{4}$):

$\lim_{n} \sup\{p(x_{n},y_{m}):m\geq n\}=0$ and $\lim_{n} \eta(x_{n},t_{n})=0$ imply $\lim_{n} \eta(y_{n},t_{n})=0$

($\tau_{5}$):

$\lim_{n} \eta(z_{n},p(z_{n},x_{n}))=0$ and $\lim_{n} \eta(z_{n},p(z_{n},y_{n}))=0$ imply $\lim_{n} d(x_{n},y_{n})=0$.

Remark 2.8

($\tau_{2}$) can be replaced by the following $(\tau_{2})'$.

$(\tau_{2})'$ :

$\inf\{\eta(x,t):t>0\}=0$ for all $x\in{X}$, and η is nondecreasing in its second variable.

Remark 2.9

If $(X,d)$ is a metric space, then the metric d is a τ-distance on X.

In the following examples, we define $\eta:X \times[0,\infty )\longrightarrow[0,\infty)$ by $\eta(x,t)= t$ for all $x\in{X}$, $t\in [0,\infty )$. It is easy to see that p is a τ-distance on a metric space X.

Example 2.10

Let $(X,d)$ be a metric space and c be a positive real number. Then $p:X\times X\longrightarrow[0,\infty)$ by $p(x,y)=c$ for $x,y\in X$ is a τ-distance on X.

Example 2.11

Let $(X,\|\cdot\|)$ be a normed space. $p:X\times X\longrightarrow[0,\infty)$ by $p(x,y)=\| x\| +\| y\|$ for $x,y\in X$ is a τ-distance on X.

Example 2.12

Let $(X,\|\cdot\|)$ be a normed space. $p:X\times X\longrightarrow[0,\infty)$ by $p(x,y)=\| y\|$ for $x,y\in X$ is a τ-distance on X.

Definition 2.13

Let $(X,d)$ be a metric space and p be a τ-distance on X. A sequence $\{x_{n}\}$ in X is called p-Cauchy if there exists a function $\eta:X \times[0,\infty)\longrightarrow[0,\infty)$ satisfying ($\tau_{2}$)-($\tau_{5}$) and a sequence $z_{n}$ in X such that $\lim_{n}\sup \{\eta(z_{n},(z_{n},x_{m})):m\geq n\}=0$.

The following lemmas are essential for the next sections.

Lemma 2.14

([19])

Let $(X,d)$ be a metric space and p be a τ-distance on X. If $\{x_{n}\}$ is a p-Cauchy sequence, then it is a Cauchy sequence. Moreover, if $\{y_{n}\}$ is a sequence satisfying $\lim_{n}\sup\{p(x_{n},y_{m}):m\geq n=0\}$, then $\{y_{n}\}$ is also a p-Cauchy sequence and $\lim_{n} d(x_{n},y_{n})=0$.

Lemma 2.15

([19])

Let $(X,d)$ be a metric space and p be a τ-distance on X. If $\{x_{n}\}$ in X satisfies $\lim_{n} p(z,x_{n})=0$ for some $z\in X$, then $\{x_{n}\}$ is a p-Cauchy sequence. Moreover, if $\{y_{n}\}$ in X also satisfies $\lim_{n}p(z,y_{n})=0$, then $\lim_{n} d(x_{n},y_{n})=0$. In particular, for $x,y,z\in X$, $p(z,x)=0$ and $p(z,y)=0$ imply $x=y$.

Lemma 2.16

([19])

Let $(X,d)$ be a metric space and p be a τ-distance on X. If a sequence $\{x_{n}\}$ in X satisfies $\lim_{n}\sup\{p(x_{n},x_{m}):m\geq n\} =0$, then $\{x_{n}\}$ is a p-Cauchy sequence. Moreover, if $\{y_{n}\}$ in X satisfies $\lim_{n} p(x_{n},y_{n})=0$, then $\{y_{n}\}$ is also a p-Cauchy sequence and $\lim_{n} d(x_{n},y_{n})=0$.

The next result is an immediate consequence of Lemma 2.14 and Lemma 2.16.

Corollary 2.17

Let $(X,d)$ be a metric space and p be a τ-distance on X. If a sequence $\{x_{n}\}$ in X satisfies $\lim_{n}\sup\{p(x_{n},x_{m}):m\geq n\} =0$, then $\{x_{n}\}$ is a Cauchy sequence.

3 Some best proximity point theorems

Now, we define the weak P-property with respect to a τ-distance as follows.

Definition 3.1

Let $(A, B)$ be a pair of nonempty subsets of a metric space $(X,d)$ with $A_{0}\neq\emptyset$. Also let p be a τ-distance on X. Then the pair $(A, B)$ is said to have the weak P-property with respect to p if and only if

$$\left.\begin{array}{r@{}} d(x_{1}, y_{1}) = d(A, B),\\ d(x_{2}, y_{2}) = d(A, B) \end{array} \right\} \quad\Longrightarrow\quad p(x_{1},x_{2})\leq p(y_{1},y_{2}), $$

where $x_{1}, x_{2}\in A_{0}$ and $y_{1}, y_{2}\in B_{0}$.

It is clear that, for any nonempty subset A of X, the pair $(A,A)$ has the weak P-property with respect to p.

Remark 3.2

([22])

If $p=d$, then $(A,B)$ is said to have the weak P-property where $A_{0}\neq\emptyset$.

It is easy to see that if $(A,B)$ has the P-property, then $(A,B)$ has the weak P-property.

Example 3.3

Let $X=\mathbf{R}^{2}$ with the usual metric and $p_{1}$, $p_{2}$ be two τ-distances defined in Example 2.11 and Example 2.12, respectively. Consider the following:

$$\begin{aligned}& A= \bigl\{ (a,b)\in\mathbf{R}^{2}\mid a=0,2\leq b\leq3 \bigr\} , \\& B= \bigl\{ (a,b)\in\mathbf{R}^{2}\mid a=1,b\leq1 \bigr\} \cup \bigl\{ (a,b) \in \mathbf{R}^{2}\mid a=1,b\geq4 \bigr\} . \end{aligned}$$

Then $(A,B)$ has the weak P-property with respect to $p_{1}$ but has not the weak P-property with respect to $p_{2}$.

By the definition of A and B, we obtain

$$d \bigl((0,2),(1,1) \bigr)=d \bigl((0,3),(1,4) \bigr)=d(A, B)=\sqrt{2}, $$

where $(0,2),(0,3)\in A$ and $(1,1),(1,4)\in B$. We have

$$\begin{aligned}& p_{1} \bigl((0,2),(0,3) \bigr)=5\quad \mbox{and} \quad p_{1} \bigl((1,1),(1,4) \bigr)=\sqrt {2}+\sqrt{17}, \\& p_{1} \bigl((0,3),(0,2) \bigr)=5 \quad\mbox{and} \quad p_{1} \bigl((1,4),(1,1) \bigr)=\sqrt {17}+\sqrt{2}. \end{aligned}$$

Therefore $(A,B)$ has the weak P-property with respect to $p_{1}$. On the other hand, we have

$$p_{2} \bigl((0,3),(0,2) \bigr)=2 \quad\mbox{and}\quad p_{2} \bigl((1,4),(1,1) \bigr)=\sqrt{2}. $$

This implies that $(A,B)$ has not the weak P-property with respect to $p_{2}$.

Definition 3.4

Let $(X,d)$ be a metric space and let p be a τ-distance on X. A mapping $T:A\longrightarrow B$ is said to be an α-ψ-proximal contraction with respect to p if

$$\alpha(x,y)p(Tx,Ty)\leq\psi \bigl(p(x,y) \bigr) \quad\mbox{for all }x,y\in A, $$

where $\alpha:A\times A\longrightarrow[0,\infty)$ and $\psi\in \Psi$.

Remark 3.5

([1])

If $p=d$, then T is said to be an α-ψ-proximal contraction.

Example 3.6

Let $(X,d)$ be a metric space and A, B be two subsets of X. Let p be the τ-distance defined in Example 2.10. Consider the following:

$$\begin{aligned}& \psi(t)=\frac{t}{2} \quad\mbox{for all } t\geq0 ,\\& \alpha_{1}(x,y)=k_{1},\quad \mbox{where } k_{1}\in\mathbf{R}, 0\leq k_{1}\leq \frac{1}{2},\\& \alpha_{2}(x,y)=k_{2},\quad \mbox{where }k_{2}\in\mathbf{R}, k_{2}> \frac{1}{2}. \end{aligned}$$

Then $T:A\longrightarrow B$ is an $\alpha_{1}$-ψ-proximal contraction with respect to p, but it is not an $\alpha_{2}$-ψ-proximal contraction with respect to p.

Definition 3.7

$g:A\longrightarrow A$ is said to be a τ-distance preserving with respect to p if

$$p(gx_{1},gx_{2})=p(x_{1},x_{2}) $$

for all $x_{1}$ and $x_{2}$ in A.

We first prove the following lemma. Then we state our results.

Lemma 3.8

Let A and B be nonempty, closed subsets of a metric space $(X,d)$ such that $A_{0}$ is nonempty. Let p be a τ-distance on X and $\alpha:A\times A\longrightarrow[0,\infty)$. Suppose that $T:A\longrightarrow B$ and $g:A\longrightarrow A$ satisfy the following conditions:

• (a) T is α-proximal admissible.

• (b) g is a τ-distance preserving with respect to p.

• (c) $\alpha(gu,gv)\geq1$ implies that $\alpha(u,v)\geq1$ for all $u,v\in A$.

• (d) $T(A_{0})\subseteq B_{0}$ and $A_{0}\subseteq g(A_{0})$.

• (e) There exist $x_{0},x_{1}\in A$ such that

  $$d(gx_{1},Tx_{0})=d(A,B) \quad \textit{and} \quad \alpha(x_{0},x_{1})\geq1 . $$

Then there exists a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$d(gx_{n+1},Tx_{n})=d(A,B) \quad\textit{and} \quad \alpha(x_{n},x_{n+1})\geq 1 \quad\textit{for all }n\in\mathbf{N}\cup \{0\}. $$

Proof

By condition (e) there exist $x_{0},x_{1}\in A$ such that

$$\begin{aligned} d(gx_{1},Tx_{0})=d(A,B) \quad\mbox{and} \quad \alpha(x_{0},x_{1})\geq1. \end{aligned}$$

(1)

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

$$\begin{aligned} d(gx_{2}, Tx_{1})=d(A,B). \end{aligned}$$

(2)

T is α-proximal admissible, therefore by (1) and (2) we have

$$\alpha(gx_{1},gx_{2})\geq1 . $$

By condition (c) we obtain

$$\alpha(x_{1},x_{2})\geq1 . $$

Continuing this process, we can find a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$\begin{aligned} d(gx_{n+1},Tx_{n})=d(A,B) \quad\mbox{and}\quad \alpha(x_{n},x_{n+1})\geq 1 \quad \mbox{for all }n\in\mathbf{N} \cup\{0\}. \end{aligned}$$

(3)

This completes the proof of the lemma. □

The following result is a special case of Lemma 3.8 obtained by setting α defined in Remark 2.6.

Corollary 3.9

Let A and B be nonempty, closed subsets of a metric space $(X,d)$ such that $A_{0}$ is nonempty. Let ‘⪯’ be a partially ordered relation on A and p be a τ-distance on X. Suppose that $T:A\longrightarrow B$ and $g:A\longrightarrow A$ satisfy the following conditions:

• (a) T is proximally increasing.

• (b) g is a τ-distance preserving with respect to p.

• (c) $gu\preceq gv$ implies that $u\preceq v$ for all $u,v\in A$.

• (d) $T(A_{0})\subseteq B_{0}$ and $A_{0}\subseteq g(A_{0})$.

• (e) There exist $x_{0},x_{1}\in A$ such that

  $$d(gx_{1},Tx_{0})=d(A,B) \quad\textit{and}\quad x_{0}\preceq x_{1}. $$

Then there exists a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$d(gx_{n+1},Tx_{n})=d(A,B) \quad \textit{and}\quad x_{n}\preceq x_{n+1} \quad \textit{for all }n\in \mathbf{N}\cup \{0\}. $$

The following result is a spacial case of Lemma 3.8 if g is the identity map.

Corollary 3.10

Let A and B be nonempty, closed subsets of a metric space $(X,d)$ such that $A_{0}$ is nonempty and $\alpha:A\times A\longrightarrow [0,\infty)$. Suppose that $T:A\longrightarrow B$ satisfies the following conditions:

• (a) T is α-proximal admissible.

• (b) $T(A_{0})\subseteq B_{0}$.

• (c) There exist $x_{0},x_{1}\in A$ such that

  $$d(x_{1},Tx_{0})=d(A,B) \quad \textit{and}\quad \alpha(x_{0},x_{1})\geq1. $$

Then there exists a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$d(x_{n+1},Tx_{n})=d(A,B) \quad\textit{and} \quad \alpha(x_{n},x_{n+1})\geq 1 \quad\textit{for all }n\in\mathbf{N} \cup\{0\}. $$

Theorem 3.11

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that $A_{0}$ is nonempty. Let $\alpha:A\times A\longrightarrow[0,\infty)$ and $\psi\in\Psi$. Also suppose that p is a τ-distance on X and $T:A\longrightarrow B$ satisfies the following conditions:

• (a) $T(A_{0})\subseteq B_{0}$ and $(A,B)$ has the weak P-property with respect to p.

• (b) T is α-proximal admissible.

• (c) There exist $x_{0},x_{1}\in A$ such that

  $$d(x_{1},Tx_{0})=d(A,B) \quad\textit{and} \quad \alpha(x_{0},x_{1})\geq1 . $$

• (d) T is a continuous α-ψ-proximal contraction with respect to p.

Then T has a best proximity point in A.

Proof

By Corollary 3.10 there exists a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$\begin{aligned} d(x_{n+1},Tx_{n})=d(A,B) \quad\mbox{and}\quad \alpha(x_{n},x_{n+1})\geq 1 \quad \mbox{for all }n\in\mathbf{N} \cup\{0\}. \end{aligned}$$

(4)

$(A,B)$ satisfies the weak P-property with respect to p, therefore by (4) we obtain that

$$\begin{aligned} p(x_{n},x_{n+1})\leq p(Tx_{n-1},Tx_{n}) \quad \mbox{for all }n\in\mathbf{N}. \end{aligned}$$

(5)

Also, by the definition of T, we have

$$\alpha(x_{n-1},x_{n})p(Tx_{n-1},Tx_{n}) \leq\psi \bigl(p(x_{n-1},x_{n}) \bigr) \quad\mbox{for all }n \in\mathbf{N}. $$

On the other hand, we have $\alpha(x_{n-1},x_{n})\geq1$ for all $n\in \mathbf{N}$, which implies that

$$\begin{aligned} p(Tx_{n-1},Tx_{n})\leq\psi \bigl(p(x_{n-1},x_{n}) \bigr) \quad \mbox{for all }n\in \mathbf{N}. \end{aligned}$$

(6)

From (5) and (6), we get that

$$\begin{aligned} p(x_{n},x_{n+1})\leq\psi \bigl(p(x_{n-1},x_{n}) \bigr) \quad \mbox{for all }n\in \mathbf{N}. \end{aligned}$$

(7)

If there exists $n_{0}\in\mathbf{N}$ such that $p(x_{n_{0}},x_{n_{0}-1})=0$, then, by the definition of ψ, we obtain that $\psi(p(x_{n_{0}-1},x_{n_{0}}))=0$. Therefore by (7) we have $p(x_{n},x_{n+1})=0$ for all $n>n_{0}$. Thus by Lemma 3.8 the sequence $\{x_{n}\}$ is Cauchy.

Now, let $p(x_{n-1},x_{n})\neq0$ for all $n\in\mathbf{N}$. By the monotony of ψ and using induction in (7), we obtain

$$ p(x_{n},x_{n+1})\leq\psi^{n} \bigl(p(x_{0},x_{1}) \bigr) \quad\mbox{for all }n\in \mathbf{N}. $$

(8)

By the definition of ψ, we have $\sum_{k=1}^{\infty}\psi ^{k}(p(x_{0},x_{1}))<\infty$. So, for all $\varepsilon>0$, there exists some positive integer $h=h(\varepsilon)$ such that

$$\sum_{k\geq h}^{\infty}\psi^{k} \bigl(p(x_{0},x_{1}) \bigr)<\varepsilon. $$

Now let $m>n>h$. By the triangle inequality and (8), we have

$$\begin{aligned} p(x_{n},x_{m})\leq\sum_{k=n}^{m-1}p(x_{k},x_{k+1}) \leq\sum_{k=n}^{m-1}\psi^{k} \bigl(p(x_{0},x_{1}) \bigr)\leq\sum _{k\geq h} \psi ^{k} \bigl(p(x_{0},x_{1}) \bigr)<\varepsilon. \end{aligned}$$

This implies that

$$\begin{aligned} \lim_{n}\sup\bigl\{ p(x_{n},x_{m}):m \geq n\bigr\} =0. \end{aligned}$$

By Corollary 2.17 $\{x_{n}\}$ is a Cauchy sequence in A. Since X is a complete metric space and A is a closed subset of X, there exists $x\in A$ such that $\lim_{n\rightarrow\infty}x_{n}=x$.

T is continuous, therefore, by letting $n\longrightarrow\infty$ in (4), we obtain

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

This completes the proof of the theorem. □

The following result is the special case of Theorem 3.11 obtained by setting $p=d$.

Corollary 3.12

([1])

Let A and B be nonempty closed subsets of a complete metric space $(X,d)$ such that $A_{0}$ is nonempty. Let $\alpha:A\times A\longrightarrow[0,\infty)$ and $\psi\in\Psi$. Suppose that $T:A\longrightarrow B$ is a nonself mapping satisfying the following conditions:

• (a) $T(A_{0})\subseteq B_{0}$ and $(A,B)$ has the P-property.

• (b) T is α-proximal admissible.

• (c) There exist $x_{0},x_{1}\in A$ such that

  $$d(x_{1},Tx_{0})=d(A,B) \quad\textit{and} \quad \alpha(x_{0},x_{1})\geq1 . $$

• (d) T is a continuous α-ψ-proximal contraction.

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

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

Theorem 3.13

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that $A_{0}$ is nonempty. Also suppose that p is a τ-distance on X and $T:A\longrightarrow B$ satisfies the following conditions:

• (a) $T(A_{0})\subseteq B_{0}$ and $(A,B)$ has the weak P-property with respect to p.

• (b) There exists $r\in[0,1)$ such that

  $$\begin{aligned} p(Tx,Ty)\leq rp(x,y), \quad \forall x,y\in A. \end{aligned}$$

• (c) T is continuous.

Then T has a best proximity point in A. Moreover, if $d(x,Tx)=d(x^{*},Tx^{*})=d(A,B)$ for some $x,x^{*}\in A$, then $p(x,x^{*})=0$.

Proof

Define $\alpha:A\times A\longrightarrow[0,\infty)$ and $\psi :[0,\infty)\longrightarrow[0,\infty)$ by $\alpha(x,y) = 1$ for all $x,y\in A$ and $\psi(t)=t$ for all $t\geq0$. Therefore by Theorem 3.11, T has a best proximity point in A. Now let x, $x^{*}$ be best proximity points in A. Therefore we have

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

The pair $(A,B)$ has the weak P-property with respect to p, hence by the definition of T we obtain that

$$p\bigl(x,x^{*}\bigr)\leq p\bigl(Tx,Tx^{*}\bigr)\leq rp \bigl(x,x^{*}\bigr). $$

Hence $p(x,x^{*})=0$ and this completes the proof of the theorem. □

The next result is an immediate consequence of Theorem 3.13 by taking $A=B$ and $p=d$.

Corollary 3.14

(Banach’s contraction principle)

Let $(X,d)$ be a complete metric space and A be a nonempty closed subset of X. Let $T:A\longrightarrow A$ be a contractive self-map. Then T has a unique fixed point $x^{*}$ in A.

4 α-p-Proximal contractions

Definition 4.1

Let A, B be subsets of a metric space $(X,d)$ and p be a τ-distance on X. A mapping $T:A\longrightarrow B$ is said to be an α-p-proximal contraction of the first kind if there exists $r\in [0,1)$ such that

$$\left.\begin{array}{r@{}} \alpha(x_{1},x_{2})\geq1,\\ d(u_{1},Tx_{1}) = d(A, B),\\ d(u_{2},Tx_{2}) = d(A, B) \end{array} \right\} \quad\Longrightarrow\quad p(u_{1},u_{2})\leq rp(x_{1},x_{2}), $$

where $\alpha:A\times A\longrightarrow[0,\infty)$ and $u_{1}, u_{2},x_{1}, x_{2}\in A$.

Also if T is an α-p-proximal contraction of the first kind, then

• (i) T is said to be an ordered p-proximal contraction of the first kind if ‘⪯’ is a partially ordered relation on A and α is defined in Remark 2.6.

• (ii) T is said to be p-proximal contraction of the first kind if $\alpha(x,y)=1$ for all $x,y\in A$.

Remark 4.2

([11])

If T is an ordered p-proximal contraction of the first kind and $p=d$, then T is said to be an ordered proximal contraction of the first kind.

Remark 4.3

If T is a p-proximal contraction of the first kind and $p=d$, then T is said to be a proximal contraction of the first kind (see [5]).

Definition 4.4

Let A, B be subsets of a metric space $(X,d)$ and p be a τ-distance on X. A mapping $T:A\longrightarrow B$ is said to be an α-p-proximal contraction of the second kind if there exists $r\in [0,1)$ such that

$$\left.\begin{array}{r@{}} \alpha(x_{1},x_{2})\geq1,\\ d(u_{1},Tx_{1}) = d(A, B),\\ d(u_{2},Tx_{2}) = d(A, B) \end{array} \right\}\quad \Longrightarrow\quad p(Tu_{1},Tu_{2})\leq rp(Tx_{1},Tx_{2}), $$

where $\alpha:A\times A\longrightarrow[0,\infty)$ and $u_{1}, u_{2},x_{1}, x_{2}\in A$.

Also if T is an α-p-proximal contraction of the second kind, then

• (i) T is said to be an ordered p-proximal contraction of the second kind if ‘⪯’ is a partially ordered relation on A and α is defined in Remark 2.6.

• (ii) T is said to be a p-proximal contraction of the second kind if $\alpha(x,y)=1$ for all $x,y\in A$.

Remark 4.5

If T is an ordered p-proximal contraction of the second kind and $p=d$, then T is said to be an ordered proximal contraction of the second kind.

Remark 4.6

If T is a p-proximal contraction of the second kind and $p=d$, then T is said to be a proximal contraction of the second kind.

Example 4.7

Let $X=\mathbf{R}$ with the usual metric and p be the τ-distance defined in Example 2.11. Given $A=[-3,-2]\cup[2,3]$, $B=[-1,1]$ and $T:A\longrightarrow B$ by

$$T(x) = \left \{ \begin{array}{@{}l@{\quad}l} x+2, & -3\leq x \leq-2,\\ x-2, & 2\leq x\leq3, \end{array} \right . $$

then T is a p-proximal contraction of the first and second kind.

It is easy to see that

$$d \bigl(-2,T(-3) \bigr)=d \bigl(2,T(3) \bigr)=d(A,B)=1. $$

If $r\in[\frac{2}{3},1)$, then we have

$$\begin{aligned}& p(-2,2)\leq rp(-3,3), \\& p(2,-2)\leq rp(3,-3). \end{aligned}$$

Hence T is a p-proximal contraction of the first kind. Also,

$$\begin{aligned}& p \bigl(T(-2),T(2) \bigr)\leq rp \bigl(T(-3),T(3) \bigr), \\& p \bigl(T(2),T(-2) \bigr)\leq rp \bigl(T(3),T(-3) \bigr) \end{aligned}$$

for all $r\in[0,1)$. This implies that T is a p-proximal contraction of the second kind.

Example 4.8

Let $X=\mathbf{R}$ with the usual metric and p be the τ-distance defined in Example 2.12. Let ‘⪯’ be the usual partially ordered relation in R. Given $A=\{-2\}\cup[2,3]$, $B=[-1,1]$ and $T:A\longrightarrow B$ by

$$T(x) = \left \{ \begin{array}{@{}l@{\quad}l} -1, & x =-2,\\ x-2, & 2\leq x\leq3, \end{array} \right . $$

then T is an ordered p-proximal contraction of the first and second kind, but it is not a p-proximal contraction of the first and second kind.

It is easy to see that

$$d \bigl(-2,T(-2) \bigr)=d \bigl(2,T(3) \bigr)=d(A,B)=1 \quad\mbox{and}\quad {-}2 \preceq3 . $$

If $r\in[\frac{2}{3},1)$, then we have

$$p(-2,2)\leq rp(-2,3) . $$

$p(2,-2)\nleq rp(3,-2)$, but it is not necessary because $3\npreceq -2$. Hence T is an ordered p-proximal contraction of the first kind. But T is not a p-proximal contraction of the first kind because $p(2,-2)\nleq rp(3,-2)$ for all $r\in[0,1)$. Also,

$$p \bigl(T(-2),T(2) \bigr)\leq rp \bigl(T(-2),T(3) \bigr) $$

for all $r\in[0,1)$. Notice that $p (T(2),T(-2) )\nleq rp (T(3),T(-2) )$, but it is not necessary because $3\npreceq-2$. This implies that T is an ordered p-proximal contraction of the second kind. But T is not a p-proximal contraction of the second kind because $p (T(2),T(-2) )\nleq rp (T(3),T(-2) )$ for all $r\in[0,1)$.

Theorem 4.9

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that $A_{0}$ is nonempty. Let p be a w-distance on X and $\alpha:A\times A\longrightarrow[0,\infty)$. Suppose that $T:A\longrightarrow B$ and $g:A\longrightarrow A$ satisfy the following conditions:

• (a) T is an α-proximal admissible and continuous α-p-proximal contraction of the first kind.

• (b) g is a continuous τ-distance preserving with respect to p.

• (c) $\alpha(gu,gv)\geq1$ implies that $\alpha(u,v)\geq1$ for all $u,v\in A$.

• (d) $T(A_{0})\subseteq B_{0}$ and $A_{0}\subseteq g(A_{0})$.

• (e) There exist $x_{0},x_{1}\in A$ such that

  $$d(gx_{1},Tx_{0})=d(A,B) \quad\textit{and} \quad \alpha(x_{0},x_{1})\geq1. $$

Then there exists an element $x\in A$ such that

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

Proof

By Lemma 3.8 there exists a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$\begin{aligned} d(gx_{n+1},Tx_{n})=d(A,B) \quad\mbox{and}\quad \alpha(x_{n},x_{n+1})\geq 1 \quad\mbox{for all }n\in\mathbf{N}\cup \{0\}. \end{aligned}$$

(9)

We will prove the convergence of a sequence $\{x_{n}\}$ in A. T is an α-p-proximal contraction of the first kind and (3) holds, hence, for any positive integer n, we have

$$p(gx_{n},gx_{n+1})\leq rp(x_{n},x_{n-1}). $$

Also g is a τ-distance preserving with respect to p, so we get that

$$p(x_{n},x_{n+1})\leq rp(x_{n},x_{n-1})\leq \cdots\leq r^{n}p(x_{0},x_{1}) $$

for every $n\in\mathbf{N}$. Hence, if $m>n$,

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

This implies that

$$\begin{aligned} \lim_{n}\sup\bigl\{ p(x_{n},x_{m}):m \geq n\bigr\} =0. \end{aligned}$$

By Corollary 2.17, $\{x_{n}\}$ is a Cauchy sequence in A. Since X is a complete metric space and A is a closed subset of X, there exists $x\in A$ such that $\lim_{n\rightarrow\infty}x_{n}=x$.

T and g are continuous, therefore by letting $n\longrightarrow \infty$ in (3), we obtain

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

This completes the proof of the theorem. □

The next result is an immediate consequence of Theorem 4.9 by setting α defined in Remark 2.6.

Corollary 4.10

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that $A_{0}$ is nonempty. Let ‘⪯’ be a partially ordered relation on A and p be a τ-distance on X. Suppose that $T:A\longrightarrow B$ and $g:A\longrightarrow A$ satisfy the following conditions:

• (a) T is a proximally increasing and continuous ordered p-proximal contraction of the first kind.

• (b) g is a continuous τ-distance preserving with respect to p.

• (c) $gu\preceq gv$ implies that $u\preceq v$ for all $u,v\in A$.

• (d) $T(A_{0})\subseteq B_{0}$ and $A_{0}\subseteq g(A_{0})$.

• (e) There exist $x_{0},x_{1}\in A$ such that

  $$d(gx_{1},Tx_{0})=d(A,B) \quad\textit{and} \quad x_{0}\preceq x_{1}. $$

Then there exists an element $x\in A$ such that

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

Theorem 4.11

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that $A_{0}$ is nonempty. Let p be a τ-distance on X. Suppose that $T:A\longrightarrow B$ and $g:A\longrightarrow A$ satisfy the following conditions:

• (a) T is a continuous p-proximal contraction of the first kind.

• (b) g is a continuous τ-distance preserving with respect to p.

• (c) $T(A_{0})\subseteq B_{0}$ and $A_{0}\subseteq g(A_{0})$.

Then there exists an element $x\in A$ such that

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

Moreover, if $d(gx,Tx)=d(gx^{*},Tx^{*})=d(A,B)$ for some $x,x^{*}\in A$, then $p(x,x^{*})=0$.

Proof

By Theorem 4.9 there exists an element $x\in A$ such that

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

Now let $x^{*}$ be in A such that

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

T is a p-proximal contraction of the first kind and g is a τ-distance preserving with respect to p, therefore

$$p\bigl(x,x^{*}\bigr)\leq rp\bigl(x,x^{*}\bigr). $$

Hence $p(x,x^{*})=0$ and this completes the proof of the theorem. □

The next result is obtained by taking $p=d$ in Theorem 4.11.

Corollary 4.12

([5])

Let X be a complete metric space. Let A and B be nonempty, closed subsets of X. Further, suppose that $A_{0}$ and $B_{0}$ are nonempty. Let $T:A\longrightarrow B$ and $g:A\longrightarrow A$ satisfy the following conditions:

• (a) T is a continuous proximal contraction of the first kind.

• (b) g is an isometry.

• (c) $T(A_{0})\subseteq B_{0}$.

• (d) $A_{0}\subseteq g(A_{0})$.

Then there exists a unique element $x\in A$ such that

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

The following result is a best proximity point theorem for nonself α-p-proximal contraction of the second kind.

Theorem 4.13

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that A is approximately compact with respect to B and $A_{0}$ is nonempty. Let p be a τ-distance on X and $\alpha :A\times A\longrightarrow[0,\infty)$. Suppose that $T:A\longrightarrow B$ satisfies the following conditions:

• (a) T is an α-proximal admissible and continuous α-p-proximal contraction of the second kind.

• (b) $T(A_{0})\subseteq B_{0}$.

• (c) There exist $x_{0},x_{1}\in A$ such that

  $$d(x_{1},Tx_{0})=d(A,B) \quad \textit{and}\quad \alpha(x_{0},x_{1})\geq1. $$

Then there exists an element $x\in A$ such that

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

Proof

By Corollary 3.10 there exists a sequence $\{x_{n}\}$ in $A_{0}$ such that

$$\begin{aligned} d(x_{n+1},Tx_{n})=d(A,B) \quad\mbox{and}\quad \alpha(x_{n},x_{n+1})\geq 1 \quad\mbox{for all }n\in\mathbf{N}\cup \{0\}. \end{aligned}$$

(10)

We will prove the convergence of a sequence $\{x_{n}\}$ in A. T is an α-p-proximal contraction of the second kind and (10) holds, hence, for any positive integer n, we have

$$p(Tx_{n},Tx_{n+1})\leq rp(Tx_{n-1},Tx_{n}) \leq\cdots\leq r^{n}p(Tx_{0},Tx_{1}) $$

for every $n\in\mathbf{N}$. Hence, if $m>n$,

$$\begin{aligned} p(Tx_{n},Tx_{m}) \leq& p(Tx_{n},Tx_{n+1})+ \cdots+p(Tx_{m-1},Tx_{m}) \\ \leq& r^{n}p(Tx_{0},Tx_{1})+ \cdots+r^{m-1}p(Tx_{0},Tx_{1}) \\ \leq& \frac{r^{n}}{1-r}p(Tx_{0},Tx_{1}). \end{aligned}$$

This implies that

$$\begin{aligned} \lim_{n}\sup\bigl\{ p(Tx_{n},Tx_{m}):m \geq n\bigr\} =0. \end{aligned}$$

By Corollary 2.17, $\{Tx_{n}\}$ is a Cauchy sequence in B. Since X is a complete metric space and B is a closed subset of X, there exists $y\in B$ such that $\lim_{n\rightarrow\infty}Tx_{n}=y$. By the triangle inequality, we have

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

Letting $n\longrightarrow\infty$ in the above inequality, we obtain

$$\lim_{n\rightarrow\infty}d(y,x_{n})=d(y,A). $$

Since A is approximately compact with respect to B, there exists a subsequence $\{ x_{n_{k}} \}$ of $\{ x_{n}\}$ converging to some $x\in A$. Therefore

$$d(x,y)=\lim_{k\rightarrow\infty}d(x_{n_{k}},Tx_{n_{k}-1})=d(A,B). $$

This implies that $x\in A_{0}$. T is continuous and $\{Tx_{n}\}$ is convergent to y, therefore

$$\lim_{n_{k}\rightarrow\infty}Tx_{n_{k}}=Tx=y. $$

Thus, it follows that

$$d(x,Tx)=\lim_{n_{k}\rightarrow\infty}d(x_{n_{k}},Tx_{n_{k}-1})=d(A,B). $$

This completes the proof of the theorem. □

The next result is an immediate consequence of Theorem 4.13 by setting α defined in Remark 2.6.

Corollary 4.14

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that A is approximately compact with respect to B and $A_{0}$ is nonempty. Let ‘⪯’ be a partially ordered relation on A and p be a τ-distance on X. Suppose that $T:A\longrightarrow B$ satisfies the following conditions:

• (a) T is a proximally increasing and continuous ordered p-proximal contraction of the second kind.

• (b) $T(A_{0})\subseteq B_{0}$.

• (c) There exist $x_{0},x_{1}\in A$ such that

  $$d(x_{1},Tx_{0})=d(A,B) \quad \textit{and}\quad x_{0}\preceq x_{1} . $$

Then there exists an element $x\in A$ such that

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

Theorem 4.15

Let A and B be nonempty, closed subsets of a complete metric space $(X,d)$ such that A is approximately compact with respect to B, and let p be a τ-distance on X. Further, suppose that $A_{0}$ is nonempty. Let $T:A\longrightarrow B$ satisfy the following conditions:

• (a) T is a continuous p-proximal contraction of the second kind.

• (b) $T(A_{0})\subseteq B_{0}$.

Then there exists an element $x\in A$ such that

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

Moreover, if $d(x,Tx)=d(x^{*},Tx^{*})=d(A,B)$ for some $x,x^{*}\in A$, then $p(Tx,Tx^{*})=0$.

Proof

By Theorem 4.13 there exists an element $x\in A$ such that

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

Now let $x^{*}$ be an element in A such that

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

We will show that $p(Tx,Tx^{*})=0$. T is a p-proximal contraction of the second kind, therefore

$$p\bigl(Tx,Tx^{*}\bigr)\leq rp\bigl(Tx,Tx^{*}\bigr). $$

Hence $p(Tx,Tx^{*})=0$ and this completes the proof of the theorem. □

The following result is obtained by taking $p=d$ in Theorem 4.15.

Corollary 4.16

([5])

Let A and B be nonempty, closed subsets of a complete metric space such that A is approximately compact with respect to B. Further, suppose that $A_{0}$ and $B_{0}$ are nonempty. Let $T:A\longrightarrow B$ satisfy the following conditions:

• (a) T is a continuous proximal contraction of the second kind.

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

Then there exists an element $x\in A$ such that

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

Moreover, if $x^{*}$ is another best proximity point of T, then Tx and $Tx^{*}$ are identical.