1. Introduction and Preliminaries
The basic concepts of proximity spaces were initially developed by Frigyes Riesz [
1] in 1908, and later on this theory was revived and axiomatized by Efremovich [
2] in 1934, which was published in 1951. Over the years, a lot of studies on proximity spaces have been carried out [
3,
4,
5,
6]. Smirnov [
5] established the link between the proximity relation and topological spaces. In addition, he was the first to introduce the relationship between proximities and uniformities.
Let be a relation on a set . Then, the pair is said to a proximity space if the following hold: for all , where the power set of
- (Pr1)
implies ;
- (Pr2)
implies ;
- (Pr3)
iff or ;
- (Pr4)
implies ;
- (Pr5)
For all , or implies ;
For all
and
, we will use the notation
and
instead of
and
, respectively. A proximity space
is called separated if
implies
for all
. The properties of proximity spaces are generalizations of uniform properties and continuity properties of metric and topology, respectively. Every proximity relation on a nonempty set
induces a topology
via the Kuratowski closure operator. The Kuratowski closure operator using proximity relation can be defined by
:
for all
. In this case, the topology
is always completely regular, further, it is Tychonoff if
is separated. If
is a topological space and
is a proximity on
such that
, then
and
are said to be compatible. Every completely regular topology on a nonempty set
has a compatible proximity. Additionally, if a sequence
converges to a point
with respect to induced topology
, then we obtain
. Furthermore, every uniform space
has an associated proximity structure defined by for all
,
if
for all
. For further information, see [
7,
8].
Now we state some examples of proximity spaces.
Example 1. Let be a metric space. Consider the relation δ on aswhere Then δ is a proximity on Ω.
In addition, the metric topology and δ are compatible. Example 2. Let be a (both normal and ) topological space. Consider the relation δ on as Then δ is a proximity on Ω. In addition, τ and δ are compatible.
Example 3. Let be a completely regular topological space. Consider the relation δ on as if and only if there does not exist a continuous function such that and Then δ is a proximity on Ω. In addition, τ and δ are compatible.
Considering that fixed-point theory on proximity space will bring a new direction and interesting results, Kostić [
9] defined the concept of
-distance and
-distance inspired by [
10] (see [
11] for more information about
-distance) in proximity space as follows and obtained the proximity space version of Banach fixed-point theorem.
Definition 1. Let be proximity space and be a function. If ω satisfies
- (w1)
, and implies for all and ,
then ω is said to be a ω-distance on Ω,
where In addition, a ω-distance on a proximity space is called -distance if the following axioms hold:
- (w2)
for all ,
- (w3)
ω is lower semicontinuous in both variables with respect to , i.e., for all , we haveandwhere is a base of neighborhoods of the point .
Remark 1. It is clear that if ω is lower semicontinuous in both variables with respect to , then we have and for every sequence converging to ξ with respect to .
Example 4. Let be endowed with the usual metric and the proximity δ defined in Example 1. Define byandthen both and are -distance on Example 5. Let be endowed with the lower limit topology and the proximity δ defined in Example 2. It is well-known that is not metrizable but normal topological space. Define bythen ω is -distance on Example 6. Let be endowed with the metric ρ defined byand the proximity δ defined in Example 1. Define bythen ω is -distance on Proof. Let
,
and
Let
. Then by (
1) there exist
and
such that
Thus we have
and hence
. Therefore, (
w1) holds, and so
is a
-distance on
. It is clear that (
w2) holds. Now, let
, then we have
that is,
is lower semicontinuous in the first variable. On the other hand, let
and
, then there exists
such that
, that is
. Hence, we have
and so we obtain
that is,
is lower semicontinuous in second variable. □
Example 7. Let be endowed with the metric ρ defined byand the proximity δ defined in Example 1. Define byandthen and are -distance on Lemma 1 ([
9,
10]).
Let be a proximity space with ω-distance ω. Then, the following properties hold:- (i)
If is separated, then and implies .
- (ii)
If and as , then subsequently converges to ξ with respect to .
In [
9], Kostić introduced the Banach contraction principle in proximity space using
-distance
as follows:
Theorem 1. Let be a separated proximity space with -distance ω. Suppose that the mapping satisfies the following:
- (i)
There exists such thatfor all ; - (ii)
For all , any iterative sequence has a convergent subsequence with respect to .
Then, there exists , such that and .
In this study, we will use some auxiliary functions in the contraction inequality to generalize Kostić’s theorem. Thus, we will obtain equivalents in proximity space of the famous Boyd–Wong [
12] and Matkowski [
13] fixed-point theorems known in metric fixed-point theory. In addition, for the equivalent of Edelstein’s theorem [
14], in this space, we will consider that the space is compact according to the topology induced by proximity.
Let be a function. Next, we will consider the below properties for :
- ()
is non-decreasing;
- ()
For all , ;
- ()
For all ,
- ()
is upper semicontinuous from the right.
We will symbolize the set of all functions with properties (–) and (–), by and respectively. It is easy to see that both and are nonempty and also every function in satisfies the property (). Some examples of the functions belonging both and are , where and .
2. Main Result
Here, we present our main theorems.
Theorem 2. Let be a separated proximity space with -distance ω and be a mapping satisfying the followings:
- (i)
There exists , satisfyingfor all ; - (ii)
For all , any iterative sequence has a convergent subsequence with respect to .
Then there exists , such that and .
Proof. Let
be arbitrary. Consider the corresponding Picard sequence
constructed by
Since
is non-decreasing, and by (
i), we have
On the other hand, since
, we have
for all
. Hence by (
2), there exists a natural number
N satisfying
for all
.
Since
is a
-distance and
is non-decreasing, then we have
Continuing this process, we obtain
for all
and similarly by (
3) we obtain
for all
. Thus, for all
, we have
By (
ii), there exists a subsequence
of the sequence
which is convergent with respect to
to some
, and by
, Remark 1 and Inequality (
4), we have
and symmetrically, we obtain
for all
. By Inequality (
6), we have
for all
. Finally, by Inequalities (
5) and (
7), we have
for all
. Since, by (
),
as
, it follows that
. On the other hand, by
and Remark 1, we have
and consequently,
. Thus, Lemma 1 (
i), we have
. To check the uniqueness, let
be a different fixed point of
T. Then,
, because
and
. By (
i), we obtain
a contradiction. Hence, the fixed point of
T is unique. □
Remark 2. Note that for , , Theorem 2 reduces to Theorem 1.
Example 8. Let be the proximity space and ω be the -distance given in Example 5. Clearly is separated proximity space. Define a mapping by and consider the function as . In this case, we havefor all . In addition, for all , is convergent with respect to . Hence, all conditions of Theorem 2 hold. Therefore, T has a unique fixed point. Note that, sincewe can not find a constant satisfying Hence, Theorem 1 cannot be applied.
Theorem 3. Let be a separated proximity space with -distance ω and be a mapping satisfying the followings:
- (i)
There exists such that and ϕ is right continuous at 0 satisfying for all ;
- (ii)
For all , any iterative sequence has a convergent subsequence with respect to .
Then, there exists such that and .
Proof. Let
be arbitrary. Consider the corresponding Picard sequence
constructed by
for all
. For simplicity, call
. If there exists
such that
, then by triangular inequality we have
and so by Lemma 1 (
i) we have
. This shows that
is a fixed point of
T. Now assume that any of consecutive terms of the
are not zero. In this case, there exists a subsequence
of
such that
for all
. By (
i), we have
Since the index sequence
is strictly increasing, we have
and so
. Therefore, from (
8), we have
It follows that
is a decreasing sequence which is also bounded below. Thus, there exists
such that
. If
, then by (
9),
and
, we have
a contradiction. Therefore,
, and consequently,
Similarly, it can be obtained that the limit of all subsequences of
whose all terms are greater than zero, is zero. Hence we have
Now, if we call
, similarly we can obtain
Now, we claim that for every
, there exists a natural number
N satisfying
for all
. Assume the contrary, then there exist
,
with
,
such that
. Here, we can choose
as the smallest positive integer that satisfies
. Therefore, we obtain
, and since
is a
-distance, we obtain
By Equation (
11), it follows that
. On other hand,
Taking limit
and by
, we have
, a contradiction. Thus, for every
, there exists
, such that
for all
. Similarly, we obtain
for all
.
By (
ii), there exists a subsequence
of sequence
which is convergent with respect to
to some
, and by
and Inequality (
12), we have
for all
and symmetrically by (
13), we obtain
for all
. Then, we have
and consequently,
. On the other hand, we have
Since and is right continuous at 0, we have Thus, by Lemma 1 (i), we have .
To check the uniqueness, let
with
be another fixed point of mapping
T. Since
it must be
. Hence by (
i), we obtain
a contradiction. Thus, the fixed point of
T is unique. □
Finally, in the theoretical part, we present a fixed-point result, taking into account the compactness of the space and a strict contraction inequality. Here, we will use the following well-known lemma.
Lemma 2 ([
15]).
Let a lower semicontinuous function, where Ω
is a compact topological space. Then, there exists an element such that Definition 2. Let be a proximity space with ω-distance ω and be a mapping. Then T is said to have 0-property if whenever for all .
Example 9. Let be endowed with the lower limit topology and the proximity δ defined in Example 2. Consider the ω-distance on Ω defined by . Then every self-mapping T of Ω has 0-property, provided that .
Theorem 4. Let be a separated proximity space with -distance ω such that Ω
is compact with respect to , and be a continuous mapping with 0-
property. Assume thatfor all with . Then, there exists such that . Proof. Define
by
and
. Since
is lower semicontinuous in both variables and
T is continuous, then
f and
g are lower semicontinuous. Since
is compact with respect to
, then from Lemma 2 there exist
, such that
and
In this case, if
, then we obtain
which is a contradiction. Hence, we obtain
. Similarly, we obtain
.
Now, if
, then we obtain
which is a contradiction. Hence, we have
. Therefore, from Lemma 1 (
i), we have
and so
. Now,
and so from Lemma 1 (
i), we have
. Additionally,
Now let
v be another fixed point of
T. Then, we have
and
. Therefore, we obtain
which is a contradiction. □
3. Application
Now, by considering Theorem 2, we give existence and uniqueness results about the solution of the second-order boundary value problem (BVP) as follows:
where
is continuous function. By taking into account some certain conditions
some existence theorems were recently presented for problem (
16) in the literature (see [
16,
17,
18]). Here, we will consider some different conditions on
F, and we provide a new theorem. We can see that the problem (
16) is equivalent to the integral equation
where
is associated Green’s function defined as
Therefore,
is a solution of (
16) if and only if it is a solution of (
17). It is clear that
Let
be the proximity space, where
and
is induced by the uniform metric
In this case,
is separated proximity space. Consider the following
-distances
on
defined by
where
is a constant.
Theorem 5. The second-order BVP given by (16) has a unique solution under the following assumptions: (a) There exists a non-decreasing function such thatfor all . (b) for all , where and Proof. Consider the operator
defined by
Then, for any
and
, we have
and then we obtain
By taking supremum over
, we have
Therefore, condition (
i) of Theorem 2 is satisfied. Now let
be an arbitrary function. Define a sequence of functions
as
. As in the proof of Theorem 2 and by (
18), we have
as
. On the other hand, since
for all
, we have
as
. That is, the sequence
is Cauchy and so has a convergent subsequence with respect to
since
is complete. Therefore, condition (
ii) of Theorem 2 is satisfied. Consequently, there exists unique
which is a fixed point of the operator
T, moreover
. Hence, the (
16) has a unique solution. □