Best proximity point results in partially ordered metric spaces via simulation functions

Abstract

We obtain sufficient conditions for the existence and uniqueness of best proximity points for a new class of non-self mappings involving simulation functions in a metric space endowed with a partial order. Some interesting consequences including fixed point results via simulation functions are presented.

1 Introduction

Recently, in [1] the authors introduced the class of simulation functions as follows.

Definition 1.1

We say that $\xi:[0,\infty)\times[0,\infty)\to\mathbb{R}$ is a simulation function if it satisfies the following conditions:

• (i) $\xi(0,0)=0$;

• (ii) $\xi(t,s)< s-t$, for every $t,s>0$;

• (iii) if $\{a_{n}\}$ and $\{b_{n}\}$ are two sequences in $(0,\infty )$, then

  $$\lim_{n\to\infty} a_{n}=\lim_{n\to\infty} b_{n}>0\quad \Longrightarrow\quad \limsup_{n\to\infty} \xi(a_{n},b_{n})< 0. $$

Various examples of simulation functions were presented in [1]. The class of such functions will be denoted by $\mathcal{Z}$.

Definition 1.2

([1])

Let $T: X\to X$ be a given operator, where X is a nonempty set equipped with a metric d. We say that T is a $\mathcal {Z}$-contraction with respect to $\xi\in\mathcal{Z}$ if

$$\xi\bigl(d(Tx,Ty),d(x,y)\bigr)\geq0,\quad \mbox{for all } x,y\in X. $$

In [1], the authors established the following fixed point theorem that generalizes many previous results from the literature including the Banach fixed point theorem.

Theorem 1.3

([1])

Let $T: X\to X$ be a given map, where X is a nonempty set equipped with a metric d such that $(X,d)$ is complete. Suppose that T is a $\mathcal{Z}$-contraction with respect to $\xi\in\mathcal{Z}$. Then T has a unique fixed point. Moreover, for any $x\in X$, the sequence $\{T^{n}x\}$ converges to this fixed point.

For other results via simulation functions, we refer to [2–7].

Let $(X,d)$ be a metric space. Consider a mapping $T: A\to B$, where A and B are nonempty subsets of X. If $d(x,Tx)>0$ for every $x\in A$, then the set of fixed points of T is empty. In this case, we are interested in finding a point $x\in A$ such that $d(x,Tx)$ is minimum in some sense.

Definition 1.4

We say that $z\in A$ is a best proximity point of T if

$$d(z,Tz)=d(A,B):=\inf\bigl\{ d(x,y): x\in A, y\in B\bigr\} . $$

Observe that if $d(A,B)=0$, then a best proximity point of T is a fixed point of T.

The study of the existence of best proximity points is an interesting field of optimization and it attracted recently the attention of several researchers (see [1, 8–23] and the references therein).

In the sequel, we will use the following notations. Set

$$A_{0}=\bigl\{ x \in A : d(x,y)=d(A,B), \mbox{for some }y \in B\bigr\} $$

and

$$B_{0}=\bigl\{ y\in B: d(x,y)=d(A,B), \mbox{for some }x \in A\bigr\} . $$

We refer to [19] for sufficient conditions that guarantee that $A_{0}$ and $B_{0}$ are nonempty.

Now, we endow the set X with a partial order ⪯. Let us introduce the following class of mappings. For a given simulation function $\xi\in\mathcal{Z}$, we denote by $\mathcal{T}_{\xi}$ the set of mappings $T: A\to B$ satisfying the following conditions:

• (C1) for every $x_{1},x_{2},y_{1},y_{2}\in A$, we have

  $$y_{1}\preceq y_{2},\quad d(x_{1},Ty_{1})=d(x_{2},Ty_{2})=d(A,B) \quad \Longrightarrow\quad x_{1}\preceq x_{2}; $$

• (C2) for every $x,y,u_{1},u_{2}\in A$, we have

  $$x\preceq y, x\neq y,\quad d(u_{1},Tx)=d(u_{2},Ty)=d(A,B) \quad \Longrightarrow\quad \xi \bigl(d(u_{1},u_{2}),m(x,y) \bigr)\geq0, $$

  where

  $$m(x,y)=\max \biggl\{ \frac{d(x,u_{1})d(y,u_{2})}{d(x,y)},d(x,y) \biggr\} . $$

Our aim in this paper is to study the existence and uniqueness of best proximity points for non-self mappings $T: A\to B$ that belong to the class $\mathcal{T}_{\xi}$, for some simulation function $\xi\in\mathcal{Z}$.

2 Main results

Our first main result is the following.

Theorem 2.1

Let $T\in\mathcal{T}_{\xi}$, for some $\xi\in\mathcal{Z}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) A is closed with respect to the metric d;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) T is continuous.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Proof

By condition (4), we have

$$d(x_{1},Tx_{0})=d(A,B), $$

for some $x_{0},x_{1}\in A_{0}$ such that $x_{0}\preceq x_{1}$. Condition (3) implies that $Tx_{1}\in B_{0}$, which yields

$$d(x_{2},Tx_{1})=d(A,B), $$

for some $x_{2}\in A_{0}$. Since $x_{0}\preceq x_{1}$, condition (C1) implies that $x_{1}\preceq x_{2}$. Continuing this process, by induction, we can construct a sequence $\{x_{n}\}\subset A_{0}$ such that

$$ d(x_{n+1},Tx_{n})=d(A,B),\quad n=0,1,2,\ldots $$

(2.1)

and

$$x_{0}\preceq x_{1}\preceq x_{2}\preceq\cdots \preceq x_{n}\preceq x_{n+1}\preceq\cdots. $$

Suppose that for some $p=0,1,2,\ldots$ , we have $x_{p+1}=x_{p}$. In this case, we get $d(x_{p},Tx_{p})=d(A,B)$, that is, $x_{p}$ is a best proximity point of T. So, without restriction of the generality, we may suppose that

$$x_{n}\neq x_{n+1},\quad n=0,1,2,\ldots. $$

Since

$$x_{n}\preceq x_{n+1}, x_{n}\neq x_{n+1},\quad d(x_{n},Tx_{n-1})=d(x_{n+1},Tx_{n})=d(A,B), \quad n=1,2,3,\ldots, $$

it follows from condition (C2) that

$$\xi\bigl(d(x_{n},x_{n+1}),m(x_{n-1},x_{n}) \bigr)\geq0,\quad n=1,2,3,\ldots, $$

where

$$\begin{aligned} m(x_{n-1},x_{n}) =&\max \biggl\{ \frac {d(x_{n-1},x_{n})d(x_{n},x_{n+1})}{d(x_{n-1},x_{n})},d(x_{n-1},x_{n}) \biggr\} \\ =& \max \bigl\{ d(x_{n},x_{n+1}),d(x_{n-1},x_{n}) \bigr\} . \end{aligned}$$

Suppose that for some $n_{0}=1,2,3,\ldots$ , we have

$$\max \bigl\{ d(x_{n_{0}},x_{n_{0}+1}),d(x_{n_{0}-1},x_{n_{0}}) \bigr\} = d(x_{n_{0}},x_{n_{0}+1}). $$

In this case, we obtain

$$0\leq\xi\bigl(d(x_{n_{0}},x_{n_{0}+1}),d(x_{n_{0}},x_{n_{0}+1}) \bigr). $$

On the other hand, since $d(x_{n_{0}},x_{n_{0}+1})>0$, using the property (ii) of a simulation function, we obtain

$$\xi\bigl(d(x_{n_{0}},x_{n_{0}+1}),d(x_{n_{0}},x_{n_{0}+1}) \bigr)< 0, $$

which is a contradiction. As a consequence,

$$ \max \bigl\{ d(x_{n},x_{n+1}),d(x_{n-1},x_{n}) \bigr\} = d(x_{n-1},x_{n}),\quad n=1,2,3,\ldots. $$

(2.2)

Thus, we obtain

$$ \xi\bigl(d(x_{n},x_{n+1}),d(x_{n-1},x_{n}) \bigr)\geq0,\quad n=1,2,3,\ldots. $$

(2.3)

From (2.2), we deduce that the sequence $\{r_{n}\}$ defined by

$$r_{n}=d(x_{n},x_{n+1}),\quad n=0,1,2,\ldots $$

is decreasing, which yields

$$\lim_{n\to\infty}r_{n}=r, $$

where $r\in[0,\infty)$. Suppose that $r>0$. Using (2.3) and the property (iii) of a simulation function, we deduce that

$$0\leq\limsup_{n\to\infty} \xi\bigl(d(x_{n},x_{n+1}),d(x_{n-1},x_{n}) \bigr)< 0, $$

which is a contradiction. As consequence, we have

$$ \lim_{n\to\infty} d(x_{n},x_{n+1})=0. $$

(2.4)

Let us prove now that $\{x_{n}\}$ is a Cauchy sequence. We argue by contradiction by supposing that $\{x_{n}\}$ is not a Cauchy sequence. In this case, there is some $\varepsilon>0$ for which there are subsequences $\{x_{m(k)}\}$ and $\{x_{n(k)}\}$ of $\{x_{n}\}$ such that

$$n(k)>m(k)>k,\quad d(x_{m(k)},x_{n(k)})\geq\varepsilon,\qquad d(x_{m(k)},x_{n(k)-1})< \varepsilon. $$

Using the triangle inequality, we have

$$ \varepsilon\leq d(x_{m(k)},x_{n(k)}) \leq d(x_{m(k)},x_{n(k)-1})+d(x_{n(k)-1},x_{n(k)}) < \varepsilon+d(x_{n(k)-1},x_{n(k)}). $$

Thus we have

$$\varepsilon\leq d(x_{m(k)},x_{n(k)})< \varepsilon +d(x_{n(k)-1},x_{n(k)}),\quad \mbox{for all } k. $$

Letting $k\to\infty$ and using (2.4), we obtain

$$ \lim_{n\to\infty} d(x_{m(k)},x_{n(k)})= \varepsilon. $$

(2.5)

Again, the triangle inequality yields

$$\bigl\vert d(x_{n(k)-1},x_{m(k)})-d(x_{m(k)},x_{n(k)}) \bigr\vert \leq d(x_{n(k)-1},x_{n(k)}),\quad \mbox{for all } k. $$

Letting $k\to\infty$, using (2.4) and (2.5), we obtain

$$ \lim_{n\to\infty} d(x_{n(k)-1},x_{m(k)})= \varepsilon. $$

(2.6)

Similarly, we have

$$\bigl\vert d(x_{n(k)-1},x_{m(k)-1})-d(x_{n(k)-1},x_{m(k)}) \bigr\vert \leq d(x_{m(k)-1},x_{m(k)}),\quad \mbox{for all } k. $$

Letting $k\to\infty$, using (2.4) and (2.6), we obtain

$$ \lim_{n\to\infty} d(x_{n(k)-1},x_{m(k)-1})= \varepsilon. $$

(2.7)

Observe that for k large enough, we have

$$x_{m(k)-1}\preceq x_{n(k)-1},\qquad x_{m(k)-1}\neq x_{n(k)-1} $$

and

$$d(x_{m(k)},Tx_{m(k)-1})= d(x_{n(k)},Tx_{n(k)-1})=d(A,B). $$

Then condition (C2) yields

$$ \xi\bigl(d(x_{m(k)},x_{n(k)}),m(x_{m(k)-1},x_{n(k)-1}) \bigr)\geq0, \quad \mbox{for all } k. $$

(2.8)

On the other hand, for all k, we have

$$m(x_{m(k)-1},x_{n(k)-1})=\max \biggl\{ \frac {d(x_{m(k)-1},x_{m(k)})d(x_{n(k)-1},x_{n(k)})}{d(x_{m(k)-1},x_{n(k)-1})},d(x_{m(k)-1},x_{n(k)-1}) \biggr\} . $$

Passing $k\to\infty$ and using (2.4) and (2.7), we get

$$ \lim_{k\to\infty}m(x_{m(k)-1},x_{n(k)-1})= \varepsilon. $$

(2.9)

Using (2.5), (2.9), (2.8) and the condition (iii) of a simulation function, we have

$$0\leq\limsup_{k\to\infty} \xi \bigl(d(x_{m(k)},x_{n(k)}),m(x_{m(k)-1},x_{n(k)-1}) \bigr)< 0, $$

which is a contradiction. As consequence, the sequence $\{x_{n}\}$ is Cauchy. Since A is a closed subset of the complete metric space $(X,d)$ (from conditions (1) and (2)), there is some $z\in A$ such that

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

The continuity of T (from condition (5)) yields

$$\lim_{n\to\infty}d(Tx_{n},Tz)=0. $$

Since $d(x_{n+1},Tx_{n})=d(A,B)$ for all $n=0,1,2,\ldots$ , we obtain

$$d(A,B)=\lim_{n\to\infty}d(x_{n+1},Tx_{n})=d(z,Tz), $$

that is, $z\in A$ is a best proximity point of T. This ends the proof. □

Next, we obtain a best proximity point result for mappings $T\in \mathcal{T}_{\xi}$ that are not necessarily continuous.

We say that the set A is $(d,\preceq)$-regular if it satisfies the following property:

$$\{a_{n}\} \subset A \mbox{ is nondecreasing w.r.t.}\preceq\quad \mbox{and}\quad \lim_{n\to\infty}d(a_{n},a)=0\quad \Longrightarrow \quad a=\sup\{a_{n}\}. $$

Theorem 2.2

Let $T\in\mathcal{T}_{\xi}$, for some $\xi\in\mathcal{Z}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) $A_{0}$ is closed;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) A is $(d,\preceq)$-regular.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Proof

Let us consider the sequence $\{x_{n}\}\subset A_{0}$ defined by (2.1). Following the proof of Theorem 2.1, we know that $\{x_{n}\}$ is a Cauchy sequence. Since $A_{0}$ is closed, there is some $z\in A_{0}$ such that

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

From condition (3), we have $Tz\in B_{0}$, which yields

$$d(y_{1},Tz)=d(A,B), $$

for some $y_{1}\in A_{0}$. On the other hand, the regularity condition (5) implies that

$$x_{n}\preceq z,\quad \mbox{for all } n. $$

Since for all n,

$$x_{n}\preceq z,\quad d(x_{n+1},Tx_{n})=d(y_{1},Tz)=d(A,B), $$

condition (C1) yields

$$x_{n+1}\preceq y_{1}, \quad \mbox{for all } n. $$

On the other hand, we know that $z=\sup\{x_{n}\}$, which implies that

$$z\preceq y_{1}. $$

Thus we have

$$d(y_{1},Tz)=d(A,B),\quad z\preceq y_{1}. $$

Again, since $Ty_{1}\in B_{0}$, there is some $y_{2}\in A_{0}$ such that $d(y_{2},Ty_{1})=d(A,B)$. Condition (C1) yields $y_{1}\preceq y_{2}$. Thus we have

$$d(y_{2},Ty_{1})=d(A,B),\quad y_{1}\preceq y_{2}. $$

Set $y_{0}=z$ and continuing this process, we can build a sequence $\{ y_{n}\}\subset A_{0}$ such that

$$d(y_{n+1},Ty_{n})=d(A,B),\quad n=0,1,2,\ldots $$

and

$$y_{0}\preceq y_{1}\preceq y_{2}\preceq\cdots \preceq y_{n}\preceq y_{n+1}\preceq\cdots. $$

Following similar arguments as in the proof of Theorem 2.1, we can prove that $\{y_{n}\}$ is a Cauchy sequence in the closed subset $A_{0}$ of the complete metric space $(X,d)$, which yields

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

for some $y\in A_{0}$. The regularity assumption (5) implies that $y=\sup \{y_{n}\}$. So, we have

$$x_{n}\preceq z=y_{0}\preceq y_{1}\preceq\cdots \preceq y_{n}\preceq y,\quad \mbox{for all } n. $$

We claim that $z=y$. In order to prove our claim, suppose that $d(z,y)>0$. Set

$$I=\{n: x_{n}=z\}. $$

We consider two cases.

Case 1. If $|I|=\infty$.

In this case, there is a subsequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that

$$x_{n_{k}}=z,\quad \mbox{for all } k, $$

which implies that z is a best proximity point. So, this case is trivial.

Case 2. If $|I|<\infty$.

In this case, for n large enough, we have

$$x_{n}\neq z, \quad x_{n}\preceq z\preceq y_{n}, \quad \mbox{for all } n. $$

From condition (C2), for n large enough, we obtain

$$\xi\bigl(d(x_{n+1},y_{n+1}),m(x_{n},y_{n}) \bigr)\geq0, $$

where

$$m(x_{n},y_{n})=\max \biggl\{ \frac {d(x_{n},x_{n+1})d(y_{n},y_{n+1})}{d(x_{n},y_{n})},d(x_{n},y_{n}) \biggr\} . $$

Observe that

$$\lim_{n\to\infty}d(x_{n+1},y_{n+1})=\lim _{n\to\infty}m(x_{n},y_{n})=d(z,y)>0. $$

From the property (iii) of simulation functions, we obtain

$$0\leq\limsup_{n\to\infty} \xi\bigl(d(x_{n+1},y_{n+1}),m(x_{n},y_{n}) \bigr) < 0, $$

which is a contradiction. As consequence, we have $z=y$.

Since $z=y$, we obtain

$$x_{n}\preceq z=y_{0}\preceq y_{1}\preceq\cdots \preceq y_{n}\preceq y=z, \quad \mbox{for all } n, $$

which implies that

$$y_{n}=z,\quad \mbox{for all } n. $$

Since $d(y_{n+1},Ty_{n})=d(A,B)$, we have $d(z,Tz)=d(A,b)$, that is, z is a best proximity point of T. This completes the proof. □

Note that the assumptions in Theorems 2.1 and 2.2 do not guarantee the uniqueness of the best proximity point. The next example shows this fact.

Example 2.3

Let X be the subset of $\mathbb{R}^{3}$ given by

$$X=\bigl\{ (0,0,1),(1,0,0),(0,0,-1),(-1,0,0)\bigr\} . $$

We endow X with the partial order ⪯ defined by

$$(x,y,z)\preceq\bigl(x',y',z'\bigr)\quad \Longleftrightarrow\quad x\leq x',y\leq y', z\leq z'. $$

Let d be the Euclidean metric on $\mathbb{R}^{3}$. Then $(X,d)$ is a complete metric space. Set

$$A=\bigl\{ (0,0,1),(1,0,0)\bigr\} \quad \textit{and}\quad B=\bigl\{ (0,0,-1),(-1,0,0) \bigr\} . $$

In this case, we have

$$d(A,B)=\sqrt{2},\qquad A_{0}=A,\qquad B_{0}=B. $$

Let $T: A\to B$ be the mapping defined by

$$T(x,y,z)=(-z,-y,-x), \quad (x,y,z)\in A. $$

Then T is continuous and $T\in\mathcal{T}_{\xi}$ for every $\xi\in \mathcal{Z}$. Moreover, it can be shown that all the other conditions of Theorems 2.1 and 2.2 are satisfied. However, $z_{1}=(0,0,1)$ and $z_{2}=(1,0,0)$ are two best proximity points of T.

In the next theorem, we give a sufficient condition for the uniqueness of the best proximity point.

Theorem 2.4

In addition to the assumptions of Theorem  2.1 (resp. Theorem  2.2), suppose that

$$\textit{for every } (x,y)\in A_{0}\times A_{0}, \textit{there is some } w\in A_{0} \textit{ such that } x\preceq w, y \preceq w. $$

Then T has a unique best proximity point.

Proof

From Theorem 2.1 (resp. Theorem 2.2), the set of best proximity points of T is not empty. Suppose that $z_{1},z_{2}\in A_{0}$ are two distinct best proximity points of T, that is,

$$d(z_{1},Tz_{1})=d(z_{2},Tz_{2})=d(A,B), \qquad d(z_{1},z_{2})>0. $$

We consider two cases.

Case 1. If $z_{1}$ and $z_{2}$ are comparable.

We may assume that $z_{1}\preceq z_{2}$. From condition (C2), we have

$$\xi\bigl(d(z_{1},z_{2}),m(z_{1},z_{2}) \bigr)\geq0, $$

where

$$m(z_{1},z_{2})=\max \biggl\{ \frac{d(z_{1},z_{1})d(z_{2},z_{2})}{ d(z_{1},z_{2})},d(z_{1},z_{2}) \biggr\} =d(z_{1},z_{2}). $$

Thus we have

$$\xi\bigl(d(z_{1},z_{2}),d(z_{1},z_{2}) \bigr)\geq0, $$

which is a contradiction with the property (ii) of a simulation function.

Case 2. If $z_{1}$ and $z_{2}$ are not comparable.

In this case, there is some $w\in A_{0}$ such that

$$z_{1}\preceq w, \qquad z_{2}\preceq w,\quad w\notin \{z_{1},z_{2}\}. $$

Since $T(A_{0})\subseteq B_{0}$, we can build a sequence $\{w_{n}\}\subset A_{0}$ such that

$$d(w_{n+1},Tw_{n})=d(A,B),\quad n=0,1,2,\ldots $$

with $w_{0}=w$. From condition (C1), we get

$$z_{1}\preceq w_{n}, \quad n=0,1,2,\ldots. $$

If for some k, we have $z_{1}=w_{k}$, using condition (C1), we have $w_{k+1}\preceq z_{1}$, which yields $w_{k+1}=z_{1}$. Arguing similarly, we obtain $w_{n}=z_{1}$ for every $n\geq k$. Thus we have

$$\lim_{n\to\infty} d(w_{n},z_{1})=0. $$

If $w_{n}\neq z_{1}$ for every n, from condition (C2), we have

$$\xi\bigl(d(z_{1},w_{n+1}),m(z_{1},w_{n}) \bigr)\geq0,\quad n=0,1,2,\ldots, $$

where

$$m(z_{1},w_{n})=\max \biggl\{ \frac{d(z_{1},z_{1})d(w_{n},w_{n+1})}{d(z_{1},w_{n})}, d(z_{1},w_{n}) \biggr\} =d(z_{1},w_{n}). $$

Thus we have

$$\xi\bigl(d(z_{1},w_{n+1}),d(z_{1},w_{n}) \bigr)\geq0, \quad n=0,1,2,\ldots. $$

On the other hand, from the property (ii) of a simulation function, we have

$$0\leq\xi\bigl(d(z_{1},w_{n+1}),d(z_{1},w_{n}) \bigr)< d(z_{1},w_{n})-d(z_{1},w_{n+1}), \quad n=0,1,2,\ldots. $$

We deduce that the sequence $\{s_{n}\}$ defined by

$$s_{n}=d(z_{1},w_{n}),\quad n=0,1,2,\ldots $$

converges to some $s\geq0$. But the property (ii) of a simulation function gives us that $s=0$. Thus, in all cases, we have

$$\lim_{n\to\infty}d(w_{n},z_{1})=0. $$

Analogously, we can prove that

$$\lim_{n\to\infty}d(w_{n},z_{2})=0. $$

Finally, the uniqueness of the limit yields the desired result. □

In the following corollaries we deduce some known and some new results in best proximity point theory via various choices of simulation functions.

We denote by $\mathcal{F}$ the set of mappings $T: A\to B$ satisfying the following conditions:

• (F1) for every $x_{1},x_{2},y_{1},y_{2}\in A$, we have

  $$y_{1}\preceq y_{2},\quad d(x_{1},Ty_{1})=d(x_{2},Ty_{2})=d(A,B) \quad \Longrightarrow\quad x_{1}\preceq x_{2}; $$

• (F2) for every $x,y,u_{1},u_{2}\in A$, we have

  $$\begin{aligned}& x\preceq y, x\neq y,\quad d(u_{1},Tx)=d(u_{2},Ty)=d(A,B) \\& \quad \Longrightarrow\quad d(u_{1},u_{2})\leq k \max \biggl\{ \frac {d(x,u_{1})d(y,u_{2})}{d(x,y)},d(x,y) \biggr\} , \end{aligned}$$

  for some constant $k\in(0,1)$.

Take $\xi(t,s)=ks-t$, for $t,s\geq0$, we deduce from Theorems 2.1, 2.2 and 2.4 the following results.

Corollary 2.5

Let $T\in\mathcal{F}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) A is closed with respect to the metric d;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) T is continuous.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Corollary 2.6

Let $T\in\mathcal{F}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) $A_{0}$ is closed;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) A is $(d,\preceq)$-regular.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Corollary 2.7

In addition to the assumptions of Corollary  2.5 (resp. Corollary  2.6), suppose that

$$\textit{for every } (x,y)\in A_{0}\times A_{0}, \textit{there is some } w\in A_{0} \textit{ such that } x\preceq w, y \preceq w. $$

Then T has a unique best proximity point.

We denote by $\mathcal{G}$ the set of mappings $T: A\to B$ satisfying the following conditions:

• (G1) for every $x_{1},x_{2},y_{1},y_{2}\in A$, we have

  $$y_{1}\preceq y_{2},\quad d(x_{1},Ty_{1})=d(x_{2},Ty_{2})=d(A,B) \quad \Longrightarrow \quad x_{1}\preceq x_{2}; $$

• (G2) for every $x,y,u_{1},u_{2}\in A$, we have

  $$\begin{aligned}& x\preceq y, x\neq y,\quad d(u_{1},Tx)=d(u_{2},Ty)=d(A,B) \\& \quad \Longrightarrow \quad d(u_{1},u_{2})\leq\max \biggl\{ \frac {d(x,u_{1})d(y,u_{2})}{d(x,y)},d(x,y) \biggr\} \\& \hphantom{\quad \Longrightarrow \quad d(u_{1},u_{2})\leq{}}{}-\varphi \biggl(\max \biggl\{ \frac {d(x,u_{1})d(y,u_{2})}{d(x,y)},d(x,y) \biggr\} \biggr), \end{aligned}$$

  where $\varphi:[0,\infty)\to[0,\infty)$ is lower semi-continuous function and $\varphi^{-1}(\{0\})=\{0\}$.

Take $\xi(t,s)=s-\varphi(s)-t$, for $t,s\geq0$, we deduce from Theorems 2.1, 2.2 and 2.4 the following results obtained in [23].

Corollary 2.8

Let $T\in\mathcal{G}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) A is closed with respect to the metric d;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) T is continuous.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Corollary 2.9

Let $T\in\mathcal{G}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) $A_{0}$ is closed;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) A is $(d,\preceq)$-regular.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Corollary 2.10

In addition to the assumptions of Corollary  2.8 (resp. Corollary  2.9), suppose that

$$\textit{for every } (x,y)\in A_{0}\times A_{0}, \textit{there is some } w\in A_{0} \textit{ such that } x\preceq w, y \preceq w. $$

Then T has a unique best proximity point.

We denote by $\mathcal{H}$ the set of mappings $T: A\to B$ satisfying the following conditions:

• (H1) for every $x_{1},x_{2},y_{1},y_{2}\in A$, we have

  $$y_{1}\preceq y_{2}, \quad d(x_{1},Ty_{1})=d(x_{2},Ty_{2})=d(A,B) \quad \Longrightarrow\quad x_{1}\preceq x_{2}; $$

• (H2) for every $x,y,u_{1},u_{2}\in A$, we have

  $$\begin{aligned}& x\preceq y, x\neq y,\quad d(u_{1},Tx)=d(u_{2},Ty)=d(A,B) \\& \quad \Longrightarrow \quad d(u_{1},u_{2})\leq\varphi \biggl(\max \biggl\{ \frac {d(x,u_{1})d(y,u_{2})}{d(x,y)},d(x,y) \biggr\} \biggr) \\& \hphantom{\quad \Longrightarrow \quad d(u_{1},u_{2})\leq{}}{}\times\max \biggl\{ \frac {d(x,u_{1})d(y,u_{2})}{d(x,y)},d(x,y) \biggr\} , \end{aligned}$$

  where $\varphi:[0,\infty)\to[0,1)$ is a function such that $\limsup_{t\to r^{+}}\varphi(t)<1$, for all $r>0$.

Take $\xi(t,s)=s\varphi(s)-t$, for $t,s\geq0$, we deduce from Theorems 2.1, 2.2 and 2.4 the following results.

Corollary 2.11

Let $T\in\mathcal{H}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) A is closed with respect to the metric d;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) T is continuous.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Corollary 2.12

Let $T\in\mathcal{H}$. Suppose that the following conditions hold:

• (1) $(X,d)$ is complete;

• (2) $A_{0}$ is closed;

• (3) $T(A_{0})\subseteq B_{0}$;

• (4) there exist $x_{0},x_{1}\in A_{0}$ such that

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

• (5) A is $(d,\preceq)$-regular.

Then T has a best proximity point, that is, there is some $z\in A$ such that $d(z,Tz)=d(A,B)$.

Corollary 2.13

In addition to the assumptions of Corollary  2.11 (resp. Corollary  2.12), suppose that

$$\textit{for every } (x,y)\in A_{0}\times A_{0}, \textit{there is some } w\in A_{0} \textit{ such that } x\preceq w, y \preceq w. $$

Then T has a unique best proximity point.

Finally, take $A=B=X$ in Theorems 2.1, 2.2 and 2.4, we obtain the following fixed point theorems.

For a given simulation function $\xi\in\mathcal{Z}$, we denote by $\mathcal{C}_{\xi}$ the class of mappings $T: X\to X$ satisfying the following conditions:

• (I) for every $x,y\in X$, we have

  $$x\preceq y \quad \Longrightarrow\quad Tx\preceq Ty; $$

• (II) for every $x,y\in X$, we have

  $$x\preceq y, x\neq y \quad \Longrightarrow\quad \xi \biggl(d(Tx,Ty),\max \biggl\{ \frac{d(x,Tx)d(y,Ty)}{d(x,y)},d(x,y) \biggr\} \biggr)\geq0. $$

Corollary 2.14

Let $T\in\mathcal{C}_{\xi}$, for some $\xi\in\mathcal{Z}$. Suppose that

• (1) $(X,d)$ is complete;

• (2) there exists some $x_{0}\in X$ such that $x_{0}\preceq Tx_{0}$;

• (3) T is continuous.

Then T has a fixed point, that is, there is some $z\in X$ such that $z=Tz$.

Corollary 2.15

Let $T\in\mathcal{C}_{\xi}$, for some $\xi\in\mathcal{Z}$. Suppose that

• (1) $(X,d)$ is complete;

• (2) there exists some $x_{0}\in X$ such that $x_{0}\preceq Tx_{0}$;

• (3) X is $(d,\preceq)$-regular.

Then T has a fixed point, that is, there is some $z\in X$ such that $z=Tz$.

Corollary 2.16

In addition to the assumptions of Corollary  2.14 (resp. Corollary  2.15), suppose that

$$\textit{for every } (x,y)\in X\times X, \textit{there is some } w\in X \textit{ such that } x\preceq w, y\preceq w. $$

Then T has a unique fixed point.