1. Introduction and Preliminaries
The
order nonlinear difference equation is of the form:
where
T is a continuous function from
into
. A point
is an equilibrium point of (
1) if
. The existence of the equilibrium point of a certain difference equation is of interest and has been extensively discussed in the literature; see for example Pre
i
[
1]. On the other hand, Equation (
1) appears in many iteration methods, for example the variational iteration method and the homotopy perturbation method [
2,
3].
In the literature of fixed point theory, the result of Pre
i
[
1] is considered as one of the most important extensions of the Banach contraction principle for the operators defined on product spaces. This famous extension [
1] was stated as: Let
be a complete metric space,
k be a positive integer, and
be a mapping such that:
for every
, where
are nonnegative constants such that
. Then, there exists a unique point
such that
. Moreover, if
are arbitrary points in
X and for each
, we have:
then the sequence
is convergent and
.
Later on, this result was further extended by Ćirić and Pre
i
[
4] as: Let
be a complete metric space,
k be a positive integer, and
be a mapping such that:
for every
, where
. Then, there exists a point
such that
. Moreover, if
are arbitrary points in
X and for each
, we have:
then the sequence
is convergent and
.
Note that, if the operator
satisfies (
2), then it is considered as a Pre
i
-type operator [
4], and the fixed point of such operators is an element
x of
X such that
. Furthermore, Equation (
3) can be considered as the representation of the
order nonlinear difference equation. Thus, the fixed points of
T are the equilibrium points of difference Equation (
3). Therefore, the above stated results are taken as tools to ensure the existence and uniqueness of an equilibrium point of a
order nonlinear difference equation. To study some other forms of Pre
i
’s result, we refer to the works of: Berinde and Păcurar [
5], Khan et al. [
6], Păcurar [
7], and Shukla et al. [
8,
9].
The purpose of this paper is to study the existence of an approximate solution of the equation , where . This equation may have a solution if A and B have some common elements, but when A and B have no common element, then the above equation has no solution; hence, in this case, we can only discuss the approximate solution of the equation. The approximate solution of the equation with the error term equal to is called the best proximity point of .
The study of approximate solutions of
was inspired by the classical result of approximation theory given by Fan [
10] as: Let
A be a nonempty compact convex subset of normed linear space
X and
be a continuous function. Then, there exists
such that:
In the literature, we have seen that the existence of the best proximity points has been investigated by several researchers by using different techniques, for example: Jleli and Samet [
11] used
-
-proximal contraction to studied the best proximity points of single-valued mappings; Abkar and Gbeleh [
12] used asymptotic cyclic contraction in their results; Abkar and Gbeleh [
13] also proved the existence of best proximity points for multivalued nonself mappings satisfying contraction and nonexpansive condition along with
P-property; Alghamdi et al. [
14] studied the best proximity point theorems in geodesic metric spaces; Choudhury et al. [
15] used the structure of partially-ordered metric spaces to discuss best proximity and couple best proximity points; Bari et al. [
16] used cyclic Meir-Keeler contraction in their discussion; Eldred and Veeramani [
17] used cyclic proximal contraction to discuss the existence of best proximity point in metric space, and they further provided an algorithm to calculate a best proximity point over the structure of a uniformly-convex Banach space; Jacob et al. [
18] gave hybrid algorithms for nonself nonexpansive mappings and provided an iterative sequence of the algorithm, which converges to the proximity point of the mapping; Markin and Shahzad [
19] studied the best proximity points of relatively
u-continuous mappings; Sadiq Basha et al. [
20] discussed the existence of best proximity points of two mappings satisfying the min-max condition; Shatanawi and Pitea [
21] used the notions of
P-property and weak P-property in their best proximity theorems; Vetro [
22] gave the existence and convergence theorems for best proximity points of the mappings satisfying the
p-cyclic
-contraction.
We will use the following notions and definition in this article: Let be a metric space and be nonempty subsets of X, then , , , and .
The following definition was introduced by Basha and Shahzad [
23].
Definition 1 ([
23])
. Let A and B be nonempty subsets of metric space . Then, B is said to be approximatively compact with respect to A, if each sequence in B with , for some x in A, has a convergent subsequence. 2. Main Results
Throughout the article, we assume that is a directed graph defined on a metric space such that the set of its vertices and the set of its edges contain all loops, but it has no parallel edge. Further, we say that for , we have a path from x to y, denoted by , if we have with and satisfying for each .
Definition 2. Let A and B be nonempty subsets of metric space endowed with the above mentioned graph G. A mapping is said to be path admissible, if:where . Here, by , we mean, for the above-mentioned , we have and . We now state and prove the first result of the article:
Theorem 1. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each in A with , that is, , and , we have:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
B is approximately compact with respect to A;
- (vi)
if is a sequence in X such that for each and as , then for each and .
Then, T has a best proximity point, that is there exists satisfying .
Proof. Hypothesis (ii) implies that we have
satisfying
and
, that is,
. By using Hypothesis (iv), we have
, and by the definition of
, we have
satisfying
. Since the mapping
T is path admissible, hence we have
. Thus,
. By considering the same arguments further, we construct a sequence
in
satisfying:
and:
From (
4), we have:
For convenience, we take
for each
. By using induction, we can get:
where
and
. Clearly,
and
. We obtain:
Thus,
for each
. By using the triangle inequality, for each
, we have:
Note that
. Therefore,
is a Cauchy sequence in a closed subset
A of the complete metric space
X. Then, there is a point
in
A such that
. Furthermore,
Therefore,
as
. Since
B is approximatively compact with respect to
A, the sequence
has a subsequence
, which converges to a point
in
B. This implies that:
Hence,
. As we know
, we have
satisfying
. By Hypothesis (vi), we have
for each
. Thus, we get
, that is
, for each
. Hence, from (
4), we get:
Applying the limit when
n tends to infinity in the above inequality, we get
, that is
. Furthermore, note that
, and there is
satisfying
. By Hypothesis (vi), we further have
. Hence, we have
,
, and
, that is,
and
. Thus, from (
4), we get:
By taking limit as
n tends to infinity in the above inequality, we get
, that is
. Thus, we have
. □
Example 1. Let be endowed with a metric for each and a graph G be defined as and . Take and . Define: Then, for each with and , we have:where . Consider such that and , then , since and . Thus, T is path admissible. We also have , , such that and . Moreover, B is approximately compact with respect to A and for each sequence in X such that for each and as , then for each and . Hence, all the conditions of Theorem 1, are satisfied. Therefore, T has a best proximity point. Theorem 2. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each with , that is , and , we have:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
B is approximately compact with respect to A;
- (vi)
if is a sequence in X such that for each and as , then for each and .
Then, T has a best proximity point, that is there exists satisfying .
Proof. Following the proof of the above theorem, we will construct a sequence
in
satisfying:
and:
From (
8), we have:
Following the above inequality and the proof of Theorem 1, we conclude that
is a Cauchy sequence in
A such that
and
. As
, we have
satisfying
. By Hypothesis (vi), we have
for each
. Thus, we get
, that is
, for each
. Hence, from (
8), we get:
Applying the limit when
n tends to infinity in the above inequality, we get
, that is
. Furthermore, note that
, and there is
satisfying
. By Hypothesis (vi), we further have
. Hence, we have
,
,
, that is
and
. Thus, from (
8), we get:
By taking the limit as
n tends to infinity in the above inequality, we get
, that is
. Thus, we have
. □
Example 2. Let be endowed with a metric for each and a graph G be defined as and . Take and . Define: Then, for each with and , we have:where . The rest of the conditions of Theorem 2 are obviously fulfilled. Thus, T has a best proximity point. Remark 1. Note that Theorem 1 is not applicable on the above example. To see this, use , and in (4). Theorem 3. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each in A with , that is , and , we have:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
A is approximately compact with respect to B;
- (vi)
if and are sequences in X such that and , then .
Then, T has a best proximity point, that is there exists satisfying .
Proof. Based on a similar argument to the one used in the proof of Theorem 1, we will construct a sequence
in
satisfying:
and:
From (
10), we have:
Inductively, we get:
Based on the triangle inequality, from the above inequality, for each
, we get:
This proves that
is a Cauchy sequence in the closed subset
B of a complete space
X. Then, there is a point
in
B such that
. Furthermore, we have:
Therefore,
as
. Since
A is approximatively compact with respect to
B, the sequence
has a subsequence
that converges to a point
in
A. This implies that:
and the proof is complete. □
Example 3. Let be endowed with a metric for each and a graph G be defined as and . Take and . Define: Then, for each with , we have:where . One can easily check that the remaining conditions of Theorem 3 are also satisfied. Thus, T has a best proximity point. Theorem 4. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each with , that is , and , we have:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
A is approximately compact with respect to B;
- (vi)
if and are sequences in X such that and , then .
Then, T has a best proximity point, that is there exists satisfying .
Proof. Using the hypothesis of the theorem, we will construct a sequence
in
satisfying:
and:
From (
12), we have:
otherwise, we have a contradiction. Iteratively, we get:
The rest of the proof is similar to the proof of Theorem 3. □
Theorem 5. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each in A with , that is , and , we have:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
A is approximately compact with respect to B;
- (vi)
if and are sequences in X such that and , then .
Then, T has a best proximity point, that is there exists satisfying .
Proof. This theorem can be proven in a similar way to the proof of Theorem 4. □
3. Further Extension of the Main Results
In this section, we will extend the above-mentioned results for the operators that map from into B, where k is any natural number.
Theorem 6. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each , with , that is , and , satisfies one of the following inequalities:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
B is approximately compact with respect to A;
- (vi)
if is a sequence in X such that for each and , then for each and .
Then, T has a best proximity point, that is there exists satisfying .
Proof. This theorem can be proven similarly to Theorems 1 and 2. □
Theorem 7. Let A and B be nonempty closed subsets of a complete metric space endowed with the graph G. Let be a mapping such that for each , with , that is , and , satisfies one of the following inequalities:where . Further, assume that the following conditions hold: - (i)
T is path admissible;
- (ii)
there exist satisfying and ;
- (iii)
is nonempty;
- (iv)
;
- (v)
A is approximately compact with respect to B;
- (vi)
T is continuous with respect to each coordinate.
Then, T has a best proximity point, that is there exists satisfying .
Proof. This theorem can be proven similarly to Theorems 3 and 4. □
Remark 2. Note that is path admissible, if for each , with , that is and , we have
4. Consequences
Considering in Theorems 6 and 7, then we obtain the following theorems, which ensure the existence of fixed points of the operator .
Theorem 8. Let be a complete metric space endowed with the graph G. Let be a mapping such that for each with , that is , , satisfies one of the following inequalities:where . Further, assume that the following conditions hold: - (i)
If , that is , then we have: - (ii)
there exist with and ;
- (iii)
if is a sequence in X such that for each and , then for each and .
Then, T has a fixed point in X, that is there exists with .
Theorem 9. Let be a complete metric space endowed with the graph G. Let be a mapping such that for each with , that is , , satisfies one of the following inequalities:where . Further, assume that the following conditions hold: - (i)
If , that is , then we have: - (ii)
there exist with and ;
- (iii)
T is continuous with respect to each coordinate.
Then, T has a fixed point in X, that is there exists with .
Remark 3. Note that if is an operator satisfying Theorem 8 or Theorem 9 and is a sequence in X such that for each and for each , then the sequence converges to fixed point of T.
Considering that the graph is defined as and , then Theorems 8 and 9 reduce to the following corollaries, respectively.
Corollary 1. Let be a complete metric space, and let be a mapping such that for each , one of the following inequalities is satisfied:where . Then, T has a fixed point in X, that is there exists with . Corollary 2. Let be a complete metric space, and let be a mapping such that for each , one of the following inequalities is satisfied:where . Further, assume that T is continuous with respect to each coordinate. Then, T has a fixed point in X, that is there exists with .