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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 [18] 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 [1318]). 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:

  1. (a)

    T is α-proximal admissible.

  2. (b)

    g is a τ-distance preserving with respect to p.

  3. (c)

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

  4. (d)

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

  5. (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:

  1. (a)

    T is proximally increasing.

  2. (b)

    g is a τ-distance preserving with respect to p.

  3. (c)

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

  4. (d)

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

  5. (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:

  1. (a)

    T is α-proximal admissible.

  2. (b)

    \(T(A_{0})\subseteq B_{0}\).

  3. (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:

  1. (a)

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

  2. (b)

    T is α-proximal admissible.

  3. (c)

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

    $$d(x_{1},Tx_{0})=d(A,B) \quad\textit{and} \quad \alpha(x_{0},x_{1})\geq1 . $$
  4. (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:

  1. (a)

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

  2. (b)

    T is α-proximal admissible.

  3. (c)

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

    $$d(x_{1},Tx_{0})=d(A,B) \quad\textit{and} \quad \alpha(x_{0},x_{1})\geq1 . $$
  4. (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:

  1. (a)

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

  2. (b)

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

    $$\begin{aligned} p(Tx,Ty)\leq rp(x,y), \quad \forall x,y\in A. \end{aligned}$$
  3. (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

  1. (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.

  2. (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

  1. (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.

  2. (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:

  1. (a)

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

  2. (b)

    g is a continuous τ-distance preserving with respect to p.

  3. (c)

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

  4. (d)

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

  5. (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:

  1. (a)

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

  2. (b)

    g is a continuous τ-distance preserving with respect to p.

  3. (c)

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

  4. (d)

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

  5. (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:

  1. (a)

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

  2. (b)

    g is a continuous τ-distance preserving with respect to p.

  3. (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:

  1. (a)

    T is a continuous proximal contraction of the first kind.

  2. (b)

    g is an isometry.

  3. (c)

    \(T(A_{0})\subseteq B_{0}\).

  4. (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:

  1. (a)

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

  2. (b)

    \(T(A_{0})\subseteq B_{0}\).

  3. (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:

  1. (a)

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

  2. (b)

    \(T(A_{0})\subseteq B_{0}\).

  3. (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:

  1. (a)

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

  2. (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:

  1. (a)

    T is a continuous proximal contraction of the second kind.

  2. (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.

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Omidvari, M., Vaezpour, S.M., Saadati, R. et al. Best proximity point theorems with Suzuki distances. J Inequal Appl 2015, 27 (2015). https://doi.org/10.1186/s13660-014-0538-7

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