Open Access

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

Fixed Point Theory and Applications20152015:232

https://doi.org/10.1186/s13663-015-0484-1

Received: 10 August 2015

Accepted: 7 December 2015

Published: 23 December 2015

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.

Keywords

best proximity point fixed point simulation function

MSC

90C26 47H10 06A06

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:
  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 [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 [27].

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 [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:
  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 \(\xi\in\mathcal{Z}\).

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. (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 [23].

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.

Declarations

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).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Department of Mathematics, College of Science, King Saud University

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© Samet 2015