Abstract
We present some fixed point results for nondecreasing and weakly increasing operators in a partially ordered metric space using implicit relations. Also we give an existence theorem for common solution of two integral equations.
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1. Introduction
Existence of fixed points in partially ordered sets has been considered recently in [1], and some generalizations of the result of [1] are given in [2–6]. Also, in [1] some applications to matrix equations are presented, in [3, 4] some applications to periodic boundary value problem and to some particular problems are, respectively, given. Later, in [6] O'Regan and Petruşel gave some existence results for Fredholm and Volterra type integral equations. In some of the above works, the fixed point results are given for nondecreasing mappings.
We can order the purposes of the paper as follows.
First, we give a slight generalization of some of the results of the above papers using an implicit relation in the following way.
In [1, 3], the authors used the following contractive condition in their result, there exists such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ1_HTML.gif)
Afterwards, in [2], the authors used the nonlinear contractive condition, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ2_HTML.gif)
where is anondecreasing function with
for
, instead of (1.1). Also in [2], the authors proved a fixed point theorem using generalized nonlinear contractive condition, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ3_HTML.gif)
for , where
is as above. In the Section 3, we generalized the above contractive conditions using the implicit relation technique in such a way that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ4_HTML.gif)
for , where
is a function as given in Section 2. We can obtain various contractive conditions from (1.4). For example, if we choose
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ5_HTML.gif)
in (1.4), then, we have (1.3). Similarly we can have the contractive conditions in [7–9] from (1.4).
In some of the above mentioned theorems, the fixed point results are given for nondecreasing mappings. Also in these theorems the following condition is used:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ6_HTML.gif)
In Section 4, we give some examples such that two weakly increasing mappings need not be nondecreasing. Therefore, we give a common fixed point theorem for two weakly increasing operators in partially ordered metric spaces using implicit relation technique. Also we did not use the condition (1.6) in this theorem. At the end, to see the applicability of our result, we give an existence theorem for common solution of two integral equations using a result of the Section 4.
2. Implicit Relation
Implicit relations on metric spaces have been used in many articles. See for examples, [10–15].
Let denote the nonnegative real numbers, and let
be the set of all continuous functions
satisfying the following conditions:
is nonincreasing in variables
;
there exists a right continuous function ,
for
such that for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ7_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ8_HTML.gif)
implies ;
, for all
.
Example 2.1.
where
,
,
.
Let and
. If
then
which implies
, a contradiction. Thus
and
. Similarly, let
and
then
If
, then
Thus
is satisfied with
. Also
, for all
. Therefore,
.
Example 2.2.
, where
.
Let and
. If
then
which is a contradiction. Thus
and
Similarly, let
and
then we have
If
then
Thus
is satisfied with
Also
for all
Therefore,
.
Example 2.3.
where
is right continuous and
,
for
Let and
If
then
which is a contradiction. Thus
and
Similarly, let
and
then we have
If
, then
Thus
is satisfied with
Also
, for all
Therefore,
.
Example 2.4.
, where
,
,
and
Let and
Then
Similarly, let
and
then we have
If
then
Thus
is satisfied with
. Also
, for all
. Therefore,
.
3. Fixed Point Theorem for Nondecreasing Mappings
We need the following lemma for the proof of our theorems.
Lemma 3.1 (See [16]).
Let be a right continuous function such that
for every
, then
, where
denotes the
-times repeated composition of
with itself.
Theorem 3.2.
Let be a partially ordered set and suppose that there is a metric
on
such that
is a complete metric space. Suppose
is a nondecreasing mapping such that for all
with
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ9_HTML.gif)
where . Also
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ10_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ11_HTML.gif)
hold. If there exists an with
, then
has a fixed point.
Proof.
If , then the proof is finished; so suppose
. Now let
for
. Notice that, since
and
is nondecreasing, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ12_HTML.gif)
Now since , we can use the inequality (3.1) for these points, then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ13_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ14_HTML.gif)
Now using , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ15_HTML.gif)
and from there exists a right continuous function
,
,
, for
, such that for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ16_HTML.gif)
If we continue this procedure, we can have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ17_HTML.gif)
and so from Lemma 3.1,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ18_HTML.gif)
Next we show that is a Cauchy sequence. Suppose it is not true. Then we can find a
and two sequence of integers
with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ19_HTML.gif)
We may also assume
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ20_HTML.gif)
by choosing to be the smallest number exceeding
for which (3.11) holds. Now (3.9), (3.11), and (3.12) imply
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ21_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ22_HTML.gif)
Also since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ23_HTML.gif)
we have from (3.9) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ24_HTML.gif)
On the other hand, since , we can use the condition (3.1) for these points. Therefore, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ25_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ26_HTML.gif)
Now letting and using (3.14), we have, by continuity of
that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ27_HTML.gif)
From , we have
. Therefore, letting
in (3.16), we have
This is a contradiction since
for
Thus
is a Cauchy sequence in
so there exists an
with
.
If (3.2) holds, then clearly . Now suppose (3.3) holds. Suppose
. Now since
, then from (3.3),
for all
. Using the inequality (3.1), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ28_HTML.gif)
so letting from the last inequality, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ29_HTML.gif)
which is a contradiction to . Thus
and so
.
Remark 3.3.
Note that if we take that
there exists a nondecreasing function with
for each
such that for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ30_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ31_HTML.gif)
implies ,
Instead of in Theorem 3.2, again we can have the same result.
If we combine Theorem 3.2 with Example 2.1, we obtain the following result.
Corollary 3.4.
Let be a partially ordered set and suppose that there is a metric
on
such that
is a complete metric space. Suppose
is a nondecreasing mapping such that for all
with
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ32_HTML.gif)
where ,
,
. Also
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ33_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ34_HTML.gif)
hold. If there exists an with
, then
has a fixed point.
Remark 3.5.
Theorem of [2] follows from Example 2.3, Remark 3.3, and Theorem 3.2.
Remark 3.6.
We can have some new results from other examples and Theorem 3.2.
Remark 3.7.
In Theorem [1], it is proved that if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ35_HTML.gif)
then for every ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ36_HTML.gif)
where is the fixed point of
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ37_HTML.gif)
and hence has a unique fixed point. If condition (3.27) fails, it is possible to find examples of functions
with more than one fixed point. There exist some examples to illustrate this fact in [3].
4. Fixed Point Theorem for Weakly Increasing Mappings
Now we give a fixed point theorem for two weakly increasing mappings in ordered metric spaces using an implicit relation. Before this, we will define an implicit relation for the contractive condition of the theorem.
Let be the set of all continuous functions
satisfying
and the following conditions:
there exists a right continuous function ,
for
such that for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ38_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ39_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ40_HTML.gif)
implies ;
and
, for all
.
We can easily show that, all functions in the Examples in Section 2 are in .
Definition 4.1 (See [17, 18]).
Let be a partially ordered set. Two mappings
are said to be weakly increasing if
and
for all
.
Note that, two weakly increasing mappings need not be nondecreasing.
Example 4.2.
Let endowed with usual ordering. Let
defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ41_HTML.gif)
then it is obvious that and
for all
. Thus
and
are weakly increasing mappings. Note that both
and
are not nondecreasing.
Example 4.3.
Let be endowed with the coordinate ordering, that is,
and
. Let
be defined by
and
, then
and
. Thus
and
are weakly increasing mappings.
Example 4.4.
Let be endowed with the lexicographical ordering, that is,
or
if
then
. Let
be defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ42_HTML.gif)
then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ43_HTML.gif)
Thus and
are weakly increasing mappings. Note that
but
, then
is not nondecreasing. Similarly,
is not nondecreasing.
Theorem 4.5.
Let be a partially ordered set and suppose that there is a metric
on
such that
is a complete metric space. Suppose
are two weakly increasing mappings such that for all comparable
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ44_HTML.gif)
where . Also
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ45_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ46_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ47_HTML.gif)
hold, then and
have a common fixed point.
Remark 4.6.
Note that, in this theorem we remove the condition "there exists an with
'' of Theorem 3.2. Again we can consider the result of Remark 3.7 for this theorem.
Proof of Theorem 4.5.
First of all we show that if or
has a fixed point, then it is a common fixed point of
and
. Indeed, let
be a fixed point of
. Now assume
. If we use the inequality (4.7), for
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ48_HTML.gif)
which is a contradiction to . Thus
and so
is a common fixed point of
and
. Similarly, if
is a fixed point of
, then it is also a fixed point of
. Now let
be an arbitrary point of
. If
, the proof is finished, so assume
. We can define a sequence
in
as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ49_HTML.gif)
Without loss of generality, we can suppose that the successive terms of are different. Otherwise, we are again finished. Note that since
and
are weakly increasing, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ50_HTML.gif)
and continuing this process, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ51_HTML.gif)
Now since and
are comparable then, we can use the inequality (4.7) for these points then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ52_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ53_HTML.gif)
Now using , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ54_HTML.gif)
and form there exists a right continuous function
for
, we have for all
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ55_HTML.gif)
Similarly, since and
are comparable, then we can use the inequality (4.7) for these points then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ56_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ57_HTML.gif)
Now again using , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ58_HTML.gif)
and form , we have for all
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ59_HTML.gif)
Therefore, from (4.18) and (4.22), we can have, for all
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ60_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ61_HTML.gif)
Thus from Lemma 3.1, we have, since ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ62_HTML.gif)
Next we show that is a Cauchy sequence. For this it is sufficient to show that
is a Cauchy sequence. Suppose it is not true. Then we can find an
such that for each even integer
, there exist even integers
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ63_HTML.gif)
We may also assumethat
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ64_HTML.gif)
by choosing to be the smallest number exceeding
for which (4.26) holds. Now (4.24), (4.26), and (4.27) imply
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ65_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ66_HTML.gif)
Also, by the triangular inequality,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ67_HTML.gif)
Therefore, we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ68_HTML.gif)
Also we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ69_HTML.gif)
On the other hand, since and
are comparable, we can use the condition (4.7) for these points. Therefore, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ70_HTML.gif)
and so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ71_HTML.gif)
Now, considering (4.29) and (4.31) and letting in the last inequality, we have, by continuity of
, that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ72_HTML.gif)
From , we have
. Therefore, letting
in (4.32), we have
. This is a contradiction since
for
. Thus
is a Cauchy sequence in
, so
is a Cauchy sequence. Therefore, there exists an
with
.
If (4.8) or (4.9) hold then clearly . Now suppose (4.10) holds. Suppose
. Now since
, then from (4.10),
for all
. Using the inequality (4.7), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ73_HTML.gif)
So letting from the last inequality, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ74_HTML.gif)
which is a contradiction to . Thus
and so
.
Remark 4.7.
We can have some new results from Theorem 4.5 with some examples for .
For example, we can have the following corollary.
Corollary 4.8.
Let be a partially ordered set and suppose that there is a metric
on
such that
is a complete metric space. Suppose
are two weakly increasing mappings such that for all comparable
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ75_HTML.gif)
where is a right continuous function such that
,
for
. Also
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ76_HTML.gif)
or
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ77_HTML.gif)
hold, then and
have a common fixed point.
Proof.
Let , then it is obvious that
. Therefore, the proof is complete from Theorem 4.5.
5. Application
Consider the integral equations
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ78_HTML.gif)
The purpose of this section is to give an existence theorem for common solution of (5.1) using Corollary 4.8. This section is related to those [19–22].
Let be a partial order relation on
.
Theorem 5.1.
Consider the integral equations (5.1).
(i) and
are continuous;
(ii)for each ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ79_HTML.gif)
(iii)there exist a continuous function and a right continuous and nondecreasing function
such that
and
for
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ80_HTML.gif)
for each and comparable
;
(iv).
Then the integral equations (5.1) have a unique common solution in
.
Proof.
Let with the usual supremum norm, that is,
, for
. Consider on
the partial order defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ81_HTML.gif)
Then is a partially ordered set. Also
is a complete metric space. Moreover, for any increasing sequence
in
converging to
, we have
for any
. Also for every
, there exists
which is comparable to
and
[6].
Define , by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ82_HTML.gif)
Now from (ii), we have, for all
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ83_HTML.gif)
Thus, we have and
for all
. This shows that
and
are weakly increasing. Also for each comparable
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F621469/MediaObjects/13663_2009_Article_1315_Equ84_HTML.gif)
Hence for each comparable
. Therefore, all conditions of Corollary 4.8 are satisfied. Thus the conclusion follows.
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Altun, I., Simsek, H. Some Fixed Point Theorems on Ordered Metric Spaces and Application. Fixed Point Theory Appl 2010, 621469 (2010). https://doi.org/10.1155/2010/621469
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DOI: https://doi.org/10.1155/2010/621469