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Best proximity point results in partially ordered metric spaces via simulation functions
Fixed Point Theory and Applications volume 2015, Article number: 232 (2015)
Abstract
We obtain sufficient conditions for the existence and uniqueness of best proximity points for a new class of nonself 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 [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)
\(\xi(t,s)< st\), 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
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
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
and
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 nonself 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)
\((X,d)\) 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
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
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
and
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
Since
it follows from condition (C2) that
where
Suppose that for some \(n_{0}=1,2,3,\ldots \) , we have
In this case, we obtain
On the other hand, since \(d(x_{n_{0}},x_{n_{0}+1})>0\), using the property (ii) of a simulation function, we obtain
which is a contradiction. As a consequence,
Thus, we obtain
From (2.2), we deduce that the sequence \(\{r_{n}\}\) defined by
is decreasing, which yields
where \(r\in[0,\infty)\). Suppose that \(r>0\). Using (2.3) and the property (iii) of a simulation function, we deduce that
which is a contradiction. As consequence, we have
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
Using the triangle inequality, we have
Thus we have
Letting \(k\to\infty\) and using (2.4), we obtain
Again, the triangle inequality yields
Letting \(k\to\infty\), using (2.4) and (2.5), we obtain
Similarly, we have
Letting \(k\to\infty\), using (2.4) and (2.6), we obtain
Observe that for k large enough, we have
and
Then condition (C2) yields
On the other hand, for all k, we have
Passing \(k\to\infty\) and using (2.4) and (2.7), we get
Using (2.5), (2.9), (2.8) and the condition (iii) of a simulation function, we have
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
The continuity of T (from condition (5)) yields
Since \(d(x_{n+1},Tx_{n})=d(A,B)\) for all \(n=0,1,2,\ldots \) , we obtain
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:
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
From condition (3), we have \(Tz\in B_{0}\), which yields
for some \(y_{1}\in A_{0}\). On the other hand, the regularity condition (5) implies that
Since for all n,
condition (C1) yields
On the other hand, we know that \(z=\sup\{x_{n}\}\), which implies that
Thus we have
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
Set \(y_{0}=z\) and continuing this process, we can build a sequence \(\{ y_{n}\}\subset A_{0}\) such that
and
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
for some \(y\in A_{0}\). The regularity assumption (5) implies that \(y=\sup \{y_{n}\}\). So, we have
We claim that \(z=y\). In order to prove our claim, suppose that \(d(z,y)>0\). Set
We consider two cases.
Case 1. If \(I=\infty\).
In this case, there is a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that
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
From condition (C2), for n large enough, we obtain
where
Observe that
From the property (iii) of simulation functions, we obtain
which is a contradiction. As consequence, we have \(z=y\).
Since \(z=y\), we obtain
which implies that
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
We endow X with the partial order ⪯ defined by
Let d be the Euclidean metric on \(\mathbb{R}^{3}\). Then \((X,d)\) is a complete metric space. Set
In this case, we have
Let \(T: A\to B\) be the mapping defined by
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
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,
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
where
Thus we have
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
Since \(T(A_{0})\subseteq B_{0}\), we can build a sequence \(\{w_{n}\}\subset A_{0}\) such that
with \(w_{0}=w\). From condition (C1), we get
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
If \(w_{n}\neq z_{1}\) for every n, from condition (C2), we have
where
Thus we have
On the other hand, from the property (ii) of a simulation function, we have
We deduce that the sequence \(\{s_{n}\}\) defined by
converges to some \(s\geq0\). But the property (ii) of a simulation function gives us that \(s=0\). Thus, in all cases, we have
Analogously, we can prove that
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 \(\xi(t,s)=kst\), 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
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 semicontinuous 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
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
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
Then T has a unique fixed point.
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Acknowledgements
The author extends his sincere appreciation to the Deanship of Scientific Research at King Saud University for its funding this Prolific Research group (PRG143610).
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 90C26
 47H10
 06A06
Keywords
 best proximity point
 fixed point
 simulation function