1. Introduction and Preliminaries
Banach Contraction Principle is the most important result in metric fixed point theory. This result was due to Banach [
1] in 1922. Banach contraction principle has been generalized by many researchers. Among the first generalizations in the setting of ordered metric spaces was proved by Ran-Reurings [
2] in 2004. Many papers have been reported in ordered metric spaces (see [
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14,
15]).
In 2012, Wardowski [
16] generalized the Banach Contraction Principle by introducing a new type of contractions, called
F-contractions. This concept attracted many researchers to contribute in this field. Many papers are reported on the existence of fixed points using
F-contractions in different spaces (see [
17,
18,
19,
20,
21,
22,
23,
24,
25,
26]). For instance, let
be the set of all functions
satisfying the following conditions:
F is strictly increasing, i.e., for all with , then .
For each sequence
of positive numbers,
There exists such that .
Consider , , and . We have that for all .
Definition 1. [16]. Let be a metric space and be a self-mapping. Then, T is said to be an F-contraction if for , there exists such that If we take
, the previous inequality becomes
In addition, for
such that
, the inequality
holds. Hence,
T is a contraction mapping where the Lipshitiz constant is
. Thus, every contraction is also an
F-contraction, but the converse is not true in general as it is proved in Example 2.5 of [
16].
Definition 2. [27]. A sequence in a partially ordered set is said to be increasing or ascending if for , . It is said strictly increasing if and . We denote it as . In most fixed point results (including the ones dealing with
F-contractions of Wardowski [
16]), the completeness hypothesis is essential to ensure the existence of a fixed point. Note that this hypothesis is strong and it would be interesting to obtain fixed point results without the set being complete. The aim of this paper goes in this direction, that is, we have strong results for weaker hypotheses. More precisely, our motivation is based on a very recent paper [
28], where the authors introduced the concept of
t-property (for partially ordered metric spaces) to ovoid the completeness hypothesis, that is, the metric space may be incomplete.
Definition 3. [28]. Let be any ordered metric space. X has the t-property if every strictly increasing Cauchy sequence in X has a strict upper bound in X, i.e., there exists such that . We present the following examples illustrating Definition 3.
Example 1. [28]. Let be equipped with the natural ordering ≤ and the usual metric. Then, X has the t-property. Example 2. [28]. Let . We define ⪯ in X by iff and . Let d be the Euclidean metric on X. Then, has the t-property. Example 3. [28]. Let be equipped with the metric d defined as . Then, is not a complete metric space. For , iff for each . Obviously, has t-property. In the following example, the increasing Cauchy sequence does not have any strict upper bound.
Example 4. [28]. Let us consider . Endow X with the Euclidean metric on . Define ⪯ in X by if , and . Consider in X such that and is strictly increasing in . We have that for all . In addition, is a strictly increasing Cauchy sequence in X, but it does not have any strict upper bound in X. In this paper, we prove some fixed point results for -contraction mappings (introduced in Definition 4) without requiring that the metric space is complete, but using the concept of the t-property. We give some examples to illustrate our obtained results.
2. Main Results
Definition 4. Let be an ordered metric space and be a self-mapping. T is said an -contraction if for , there exists such that for all with , and , we have In the following example, the considered mapping T is not an F-contraction, but it is an -contraction.
Example 5. Let be endowed by the usual metric of and the natural ordering ≤. Define by For and , we get . Let . By , we have Thus, T is not an F-contraction. Now, we show that T is an -contraction. Clearly, for . Set . We show that Equation (1) is satisfied. Let such that and . Then, and . Further, and . In addition, Thus, . This shows that T is an -contraction.
Example 6. Let and . Endow with the usual metric of and the natural ordering ≤. Define by Let be defined by . Clearly, . It can be easily proved that T is an -contraction.
Our first fixed point result is:
Theorem 1. Let be an ordered metric space having t-property. Let be an -contraction. Suppose that T is non-decreasing and there exists such that . Then, T has a fixed point in X.
Proof. By assumption, we have
such that
. If
, the proof is completed. Otherwise, choose
such that
. By monotonicity of
T, we have
, that is,
. If
, the proof is completed. Otherwise, choose
such that
. Again, by monotonicity of
T, we have
. Continuing this process, we get a strictly increasing sequence
in
X such that
. As
, by Equation (
1), we have
Again, since
, by Equation (
1), we have
From Equations (
2) and (
3), we get
Continuing in this process, we get
Denote
for
. From Equation (
4), we obtain
We get
. Using property
,
By
, there exists
such that
By Equation (
5), we have for all
nThus, there exists
such that
for all
, that is,
for all
. Now, we show that
is a Cauchy sequence. Let
with
. Using Equation (
6), one writes
Taking
, we get
. Thus,
is a strictly increasing Cauchy sequence in
X, which has
t-property. Therefore, there exists
such that
. If
, the proof is completed. Otherwise, by Equation (
1), we have
. Using Equation (
4), we get
At the limit, . By , we have . Thus, u is a fixed point of T in X. □
Now, we report some examples to illustrate our obtained result. The first example clarifies Theorem 1 where the mapping T is not an F-contraction.
Example 7. Let and . Take , so . Endow X with the usual metric on and the natural ordering ≤. Clearly, is not complete but has the t-property. Define by Obviously, T is non-decreasing. Now, it remains to prove that T satisfies Equation (1). Letting with , and , we have . Then, , and . If we take and . Then, , i.e., T is an -contraction. Hence, all the conditions of Theorem 1 are satisfied. B is the set of fixed points of T. Example 8. Let be endowed with the Euclidean metric. Consider: iff and . Then, is an ordered metric space having t-property. Take . Let be defined by . Clearly, for all such that , we have Define by Clearly, T is non-decreasing. We show that T satisfies Equation (1). Let such that , and . Then, and In addition,and Since , we have . Take and . We have Hence, for all with , and , we have Thus, all the conditions of Theorem 1 are satisfied. Any element of is a fixed point of T.
The following example clarifies Theorem 1, where the space is not complete.
Example 9. Let be equipped with the metric d defined as . For , iff for each . Note that is an ordered metric space having t-property, but it is not complete. Let be a subset of X. Define byfor . Clearly, T is non-decreasing. We prove that T is an -contraction. Let with , and . Then, and for each . We haveand Consider and . We have Thus, for all with , and , we have Hence, all the conditions of Theorem 1 are satisfied. is the set of fixed points of T.
Definition 5. A function is said to be a sublinear altering distance function, if it satisfies the following:
- 1.
ψ is monotonic increasing and continuous.
- 2.
iff .
- 3.
, for any .
Example 10. The map defined by () is a sublinear altering function.
Example 11. Let us define defined by . Then, ψ is a sublinear altering function.
Definition 6. Let be an ordered metric space and be a self-mapping. T is said an -contraction, if for , there exists such that for all with , and , we havewhere ψ is a sublinear altering function, is such that iff and for each . Our second fixed point result is:
Theorem 2. Let be an ordered metric space having t-property. Let be an -contraction. Suppose that T is non-decreasing and there exists such that . Then, T has a fixed point in X.
Proof. Let
be such that
. If
, the proof is completed. Otherwise, choose
such that
. Proceeding similarly as Theorem 1, we get a strictly increasing sequence
in X such that
. As
, by Equation (
7),
By a property of
ϕ, we get
Since
, by Equation (
7), we have
Continuing in the same way, we get
Denote
for
. By (
8), we obtain
We get
. By
, we have
Using
, there exists
such that
By Equation (
9), we have for all
nLetting
in Equation (
12) and using Equations (
10) and (
11), we get
Hence, there exists
such that
for all
, i.e.,
Now, we show that
is a Cauchy sequence. Using the triangular inequality, properties of
ψ and Equation (
13), we have
Letting
, we get
. By properties of
ψ, we get
. Thus,
is a strictly increasing Cauchy sequence in
X, which has the
t-property, so there exists
such that
. If
, the proof is completed. Otherwise, by Equation (
7), we have
Hence, . By , we have . This implies that , . Hence, u is a fixed point of T in X. □
Example 12. Let and . Take , then . Endow X with the usual metric on and the natural ordering ≤. Clearly, has the t-property, but it is not complete. Define by Then, T is non-decreasing. Further, define by and . We prove that Equation (7) is satisfied. Take and . Let such that , and . Then, , , and . In addition, Thus, all the conditions of Theorem 2 hold, so there exists a fixed point of T in X.