The purpose of this paper is to present a fixed point theorem for generalized contractions in partially ordered complete metric spaces. We also present an application to firstorder ordinary differential equations.
1. Introduction
Existence of fixed point in partially ordered sets has been considered recently in [1–17]. Tarski's theorem is used in [9] to show the existence of solutions for fuzzy equations and in [11] to prove existence theorems for fuzzy differential equations. In [2, 6, 7, 10, 13] some applications to ordinary differential equations and to matrix equations are presented. In [3–5, 17] some fixed point theorems are proved for a mixed monotone mapping in a metric space endowed with partial order and the authors apply their results to problems of existence and uniqueness of solutions for some boundary value problems.
In the context of ordered metric spaces, the usual contraction is weakened but at the expense that the operator is monotone. The main tool in the proof of the results in this context combines the ideas in the contraction principle with those in the monotone iterative technique [18].
Let denote the class of the class of the functions which satisfies the condition
In [19] the following generalization of Banach's contraction principle appears.
Theorem 1.1.
Let be a complete metric space and let be a mapping satisfying
where . Then has a unique fixed point and converges to for each .
Recently, in [2] the authors prove a version of Theorem 1.1 in the context of ordered complete metric spaces. More precisely, they prove the following result.
Theorem 1.2.
Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a nondecreasing mapping such that
where . Assume that either is continuous or satisfies the following condition:
Besides, suppose that for each there exists which is comparable to and . If there exists with , then has a unique fixed point.
The purpose of this paper is to generalize Theorem 1.2 with the help of the altering functions.
We recall the definition of such functions.
Definition 1.3.
An altering function is a function which satisfies the following.
(a) is continuous and nondecreasing.
(b) if and only if .
Altering functions have been used in metric fixed point theory in recent papers [20–22].
In [7] the authors use these functions and they prove some fixed point theorems in ordered metric spaces.
2. Fixed Point Theorems
Definition 2.1.
If is a partially ordered set and , we say that is monotone nondecreasing if for ,
This definition coincides with the notion of a nondecreasing function in the case and represents the usual total order in .
In the sequel, we prove the main result of the paper.
Theorem 2.2.
Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Let be a continuous and nondecreasing mapping such that
where is an altering function and .
If there exist with , then has a fixed point.
Proof.
If , then the proof is finished. Suppose that . Since and is a nondecreasing mapping, we obtain by induction that
Put . Taking into account that and since for each then, by (2.2), we get
Using the fact that is nondecreasing, we have
If there exists such that , then and is a fixed point and the proof is finished. In another case, suppose that for all . Then, taking into account (2.5), the sequence is decreasing and bounded below, so
Assume that .
Then, from (2.4), we have
Letting in the last inequality and by the fact that is an altering function, we get
and, consequently, Since this implies that = and this contradicts our assumption that Hence,
In what follows, we will show that is a Cauchy sequence.
Suppose that is not a Cauchy sequence. Then, there exists for which we can find subsequences and of with such that
Further, corresponding to , we can choose in such a way that it is the smallest integer with and satisfying (2.10), then
Using (2.10), (2.11), and the triangular inequality, we have
Letting and using (2.9), we get
Again, the triangular inequality gives us
Letting in the above two inequalities and using (2.9) and (2.13), we have
As and , by (2.2), we obtain
Taking into account (2.13) and (2.15) and the fact that is continuous and letting in (2.16), we get
As is an altering function, , the last inequality gives us
Since , this means that
This fact and (2.15) give us which is a contradiction.
This shows that is a Cauchy sequence.
Since is a complete metric space, there exists such that Moreover, the continuity of implies that
and this proves that is a fixed point.
In what follows, we prove that Theorem 2.2 is still valid for not necessarily continuous, assuming the following hypothesis in (which appears in [10, Theorem ]):
Theorem 2.3.
Let be a partially ordered set and suppose that there exists a metric in such that is a complete metric space. Assume that satisfies (2.21). Let be a nondecreasing mapping such that
where is an altering function and . If there exists with , then has a fixed point.
Proof.
Following the proof of Theorem 2.2, we only have to check that . As is a nondecreasing sequence in and then, by (2.21), we have for all , and, consequently,
Letting and using the continuity of , we have
or, equivalently,
As is an altering function, this gives us and, thus,
Now, we present an example where it can be appreciated that the hypotheses in Theorems 2.2 and 2.3 do not guarantee uniqueness of the fixed point. This example appears in [10].
Let and consider the usual order
is a partially ordered set whose different elements are not comparable. Besides, is a complete metric space considering as the Euclidean distance. The identity map is trivially continuous and nondecreasing and condition (2.2) of Theorem 2.2 is satisfied since elements in are only comparable to themselves. Moreover, and has two fixed points in .
In what follows, we give a sufficient condition for the uniqueness of the fixed point in Theorems 2.2 and 2.3. This condition appears in [16] and says that
In [10] it is proved that condition (2.27) is equivalent to
Theorem 2.4.
Adding condition (2.28) to the hypotheses of Theorem 2.2 (resp., Theorem 2.3), we obtain uniqueness of the fixed point of .
Proof.
Suppose that there exist which are fixed points of and . We distinguish two cases.
Case 1.
If and are comparable, then and are comparable for Using the contractive condition appearing in Theorem 2.2 (or Theorem 2.3) and the fact that , we get
which is a contradiction.
Case 2.
Using condition (2.28), there exists comparable to and . Monotonicity of implies that is comparable to and to , for Moreover, as , we get
Since is nondecreasing the above inequality gives us
Thus,
Assume that .
Taking into account that is an altering function and letting in (2.30), we obtain
and this implies that .
Since then we get
and, consequently, , which is a contradiction.
Hence, .
Analogously, it can be proved that
Finally, as
and taking limit, we obtain .
This finishes the proof.
Remark 2.5.
Under the assumptions of Theorem 2.4, it can be proved that for every , , where is the fixed point (i.e., the operator is Picard).
In fact, for and comparable to then using the same argument that is in Case 1 of Theorem 2.4 can prove that and, hence, .
If is not comparable to , we take that is comparable to and . Using a similar argument that is in Case 2 of Theorem 2.4, we obtain
Finally,
and taking limit as , we obtain or, equivalently, = .
Remark 2.6.
Notice that if is totally ordered, condition (2.28) is obviously satisfied.
Remark 2.7.
Considering the identity mapping in Theorem 2.4, we obtain Theorem 1.2, being the main result of [2].
3. Application to Ordinary Differential Equations
In this section we present an example where our results can be applied.
This example is inspired by [10].
We study the existence of solution for the following firstorder periodic problem
where and is a continuous function.
Previously, we considered the space () of continuous functions defined on . Obviously, this space with the metric given by
is a complete metric space. can also be equipped with a partial order given by
Clearly, satisfies condition (2.28), since for the function .
Moreover, in [10] it is proved that with the abovementioned metric satisfies condition (2.21).
Now, let denote the class of functions satisfying the following.
(i) is nondecreasing.
(ii), .
(iii),
where is the class of functions defined in Section 1.
Examples of such functions are , with , , and .
Recall now the following definition
Definition 3.1.
A lower solution for (3.1) is a function such that
Now, we present the following theorem about the existence of solution for problem (3.1) in presence of a lower solution.
Theorem 3.2.
Consider problem (3.1) with continuous and suppose that there exist with
such that for with
where . Then the existence of a lower solution for (3.1) provides the existence of a unique solution of (3.1).
Proof.
Problem (3.1) can be written as
This problem is equivalent to the integral equation
where is the Green function given by
Define by
Notice that if is a fixed point of , then is a solution of (3.1).
In the sequel, we check that hypotheses in Theorem 2.4 are satisfied.
The mapping is nondecreasing since, by hypothesis, for
and this implies, taking into account that for , that
Besides, for , we have
Using the CauchySchwarz inequality in the last integral, we get
The first integral gives us
As is nondecreasing, the second integral in (3.14) can be estimated by
Taking into account (3.14), (3.15), and (3.16), from (3.13) we get
Since , the last inequality gives us
or, equivalently,
This implies that
Putting , which is an altering function, and because , we have
This proves that the operator satisfies condition (2.2) of Theorem 2.2.
Finally, letting be a lower solution for (3.1), we claim that
In fact,
Multiplying by ,
and this gives us
As , the last inequality implies that
and so
This and (3.24) give us
and, consequently,
Finally, Theorem 2.4 gives that has a unique fixed point.
Remark 3.3.
Notice that if , then . In fact, as , then is nondecreasing and, consequently, is also nondecreasing.
Moreover, as , then , and, thus, .
Finally, as , and as , then it is easily seen that .
Example 3.4.
Consider given by
It is easily seen that . Taking into account Remark 3.3, .
Now, we consider problem (3.1) with continuous and suppose that there exist with
such that for with
where is the function above mentioned.
This example can be treated by our Theorem 3.2 but it cannot be covered by the results of [6] because is not increasing.
Acknowledgments
This research was partially supported by "Ministerio de Educación y Ciencia", Project MTM 2007/65706. This work is dedicated to Professor W. Takahashi on the occasion of his retirement.
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