# 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.

## Introduction

Recently, in  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:

1. (i)

$$\xi(0,0)=0$$;

2. (ii)

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

3. (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 . The class of such functions will be denoted by $$\mathcal{Z}$$.

### Definition 1.2

()

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 , 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

()

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 .

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, 823] 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  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:

1. (C1)

for every $$x_{1},x_{2},y_{1},y_{2}\in A$$, we have

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

## 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. (1)

$$(X,d)$$ is complete;

2. (2)

A is closed with respect to the metric d;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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. (1)

$$(X,d)$$ is complete;

2. (2)

$$A_{0}$$ is closed;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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:

1. (F1)

for every $$x_{1},x_{2},y_{1},y_{2}\in A$$, we have

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

$$(X,d)$$ is complete;

2. (2)

A is closed with respect to the metric d;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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. (1)

$$(X,d)$$ is complete;

2. (2)

$$A_{0}$$ is closed;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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:

1. (G1)

for every $$x_{1},x_{2},y_{1},y_{2}\in A$$, we have

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

### Corollary 2.8

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

1. (1)

$$(X,d)$$ is complete;

2. (2)

A is closed with respect to the metric d;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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. (1)

$$(X,d)$$ is complete;

2. (2)

$$A_{0}$$ is closed;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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:

1. (H1)

for every $$x_{1},x_{2},y_{1},y_{2}\in A$$, we have

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

$$(X,d)$$ is complete;

2. (2)

A is closed with respect to the metric d;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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. (1)

$$(X,d)$$ is complete;

2. (2)

$$A_{0}$$ is closed;

3. (3)

$$T(A_{0})\subseteq B_{0}$$;

4. (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. (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:

1. (I)

for every $$x,y\in X$$, we have

$$x\preceq y \quad \Longrightarrow\quad Tx\preceq Ty;$$
2. (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. (1)

$$(X,d)$$ is complete;

2. (2)

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

3. (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. (1)

$$(X,d)$$ is complete;

2. (2)

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

3. (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.

## References

1. 1.

Khojasteh, F, Shukla, S, Radenović, S: A new approach to the study of fixed point theory for simulation functions. Filomat 29(6), 1189-1194 (2015)

2. 2.

Argoubi, H, Samet, B, Vetro, C: Nonlinear contractions involving simulation functions in a metric space with a partial order. J. Nonlinear Sci. Appl. 8, 1082-1094 (2015)

3. 3.

Du, WS, Khojasteh, F: New results and generalizations for approximate fixed point property and their applications. Abstr. Appl. Anal. 2014, Article ID 581267 (2014)

4. 4.

Du, WS, Khojasteh, F, Chiu, YN: Some generalizations of Mizoguchi-Takahashi’s fixed point theorem with new local constraints. Fixed Point Theory Appl. 2014, 31 (2014)

5. 5.

Khojasteh, F, Karapinar, E, Radenović, S: Metric space: a generalization. Math. Probl. Eng. 2013, Article ID 504609 (2013)

6. 6.

Roldán-López-de-Hierro, AF, Karapinar, E, Roldán-López-de-Hierro, C, Martínez-Moreno, J: Coincidence point theorems on metric spaces via simulation functions. J. Comput. Appl. Math. 275, 345-355 (2015)

7. 7.

Roldán-López-de-Hierro, AF, Shahzad, N: New fixed point theorem under R-contractions. Fixed Point Theory Appl. 2015, 98 (2015)

8. 8.

Abkar, A, Gabeleh, M: Best proximity points for cyclic mappings in ordered metric spaces. J. Optim. Theory Appl. 150(1), 188-193 (2011)

9. 9.

Basha, SS: Discrete optimization in partially ordered sets. J. Glob. Optim. 54(3), 511-517 (2012)

10. 10.

Bilgili, N, Karapinar, E, Sadarangani, K: A generalization for the best proximity point of Geraghty-contractions. J. Inequal. Appl. 2013, 286 (2013)

11. 11.

de la Sen, M, Agarwal, RP: Some fixed point-type results for a class of extended cyclic self-mappings with a more general contractive condition. Fixed Point Theory Appl. 2011, 59 (2011)

12. 12.

Eldred, AA, Veeramani, P: Existence and convergence of best proximity points. J. Math. Anal. Appl. 323(2), 1001-1006 (2006)

13. 13.

Jleli, M, Karapinar, E, Samet, B: A best proximity point result in modular spaces with the Fatou property. Abstr. Appl. Anal. 2013, Article ID 329451 (2013)

14. 14.

Jleli, M, Karapinar, E, Samet, B: A short note on the equivalence between best proximity points and fixed point results. J. Inequal. Appl. 2014, 246 (2014)

15. 15.

Jleli, M, Samet, B: An optimization problem involving proximal quasi-contraction mappings. Fixed Point Theory Appl. 2014, 141 (2014)

16. 16.

Karapinar, E: Fixed point theory for cyclic weak ϕ-contraction. Appl. Math. Lett. 24(6), 822-825 (2011)

17. 17.

Karapinar, E, Pragadeeswarar, V, Marudai, M: Best proximity point for generalized proximal weak contractions in complete metric space. J. Appl. Math. 2014, Article ID 150941 (2014)

18. 18.

Kim, WK, Lee, KH: Existence of best proximity pairs and equilibrium pairs. J. Math. Anal. Appl. 316(2), 433-446 (2006)

19. 19.

Kirk, WA, Reich, S, Veeramani, P: Proximinal retracts and best proximity pair theorems. Numer. Funct. Anal. Optim. 24(7-8), 851-862 (2003)

20. 20.

Nashine, HK, Kumam, P, Vetro, C: Best proximity point theorems for rational proximal contractions. Fixed Point Theory Appl. 2013, 95 (2013)

21. 21.

Raj, VS: A best proximity point theorem for weakly contractive non-self-mappings. Nonlinear Anal. 74(14), 4804-4808 (2011)

22. 22.

Srinivasan, PS, Veeramani, P: On existence of equilibrium pair for constrained generalized games. Fixed Point Theory Appl. 2004(1), 21-29 (2004)

23. 23.

Pragadeeswarar, V, Marudai, M: Best proximity points for generalized proximal weak contractions satisfying rational expression on ordered metric spaces. Abstr. Appl. Anal. 2015, Article ID 361657 (2015)

## Acknowledgements

The author extends his sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding this Prolific Research group (PRG-1436-10).

## Author information

Authors

### Corresponding author

Correspondence to Bessem Samet. 