Abstract
In this paper, we prove a fixed point theorem for a contraction in generalized complete metric spaces endowed with partial order. As an application, we use the fixed point theorem to prove the Hyers-Ulam stability of the Cauchy functional equation in Banach spaces endowed with a partial order.
MSC:54H25, 47H10, 39B52.
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1 Introduction
In 1940, Ulam gave a wide ranging talk in front of the mathematics club of University of Wisconsin in which he discussed a number of important unsolved problems (see [1]). One of the problems was the question concerning the stability of homomorphisms:
Let be a group and be a metric group with a metric . Given , does there exist such that, if a mapping satisfies the inequality for all , then there exists a homomorphism with for all ?
In 1941, Hyers [2] affirmatively answered the question of Ulam for the case where and are Banach spaces. Taking this fact into account, the additive Cauchy functional equation is said to satisfy the Hyers-Ulam stability.
On the other hand, Banach’s contraction principle is one of the pivotal results of analysis. It is widely considered as the source of metric fixed point theory. Also, its significance lies in its vast applicability in a number of branches of mathematics. Many kinds of generalizations of the above principle have been a heavily investigated branch of research. In particular, Diaz and Margolis [3] presented the following definition and fixed point theorem in a ‘generalized complete metric space’.
Definition 1.1 Let X be an abstract (nonempty) set and assume that, in the Cartesian product , a distance function ( for all ) is defined and satisfies the following conditions:
-
(D1)
if and only if ;
-
(D2)
(symmetry);
-
(D3)
(triangle inequality);
-
(D4)
every d-Cauchy sequence in X is d-convergent, i.e., for a sequence in X implies the existence of an element with (the point x is unique by (D1) and (D3)).
Then we call a generalized complete metric space.
Theorem 1.2 Suppose that is a generalized complete metric space and the function is a contraction, that is, T satisfies the following condition:
(CI) There exists a constant q with such that, whenever ,
Let and consider a sequence of successive approximations with initial element . Then the following alternative holds: either
(A) for all , one has
or
(B) the sequence is d-convergent to a fixed point of T.
Recently, Nieto and Rodriguez-Lopez [4] proved a fixed point theorem in partially ordered sets as follows.
Theorem 1.3 Let be a partially ordered set. Suppose that there exists a metric d in X such that is a complete metric space. Let be a continuous and nondecreasing mapping such that there exists with
for all . If there exists with , then f has a fixed point.
In 2003, Cǎdariu and Radu [5] applied the fixed point method to investigate the Jensen functional equation (see also [6–9]) and presented a short and simple proof (different from the direct method initiated by Hyers in 1941) for the Hyers-Ulam stability of the Jensen functional equation [5] for proving properties of generalized Hyers-Ulam stability for some functional equations in a single variable [7] for the stability of some nonlinear equations [6]. Recently, Brzdek [10], Brzdek and Cieplinski [11, 12] reported some interesting results in this direction (see also [13–16]).
In this paper, we prove a fixed point theorem for self-mappings on a partially ordered set X which has a generalized metric d. Moreover, we give a generalization of the Hyers-Ulam stability of the conditional Cauchy equation as an important result of our fixed point theorem.
2 Main results
We start our work by the following fixed point theorem in generalized complete metric spaces.
Theorem 2.1 Let be a generalized complete metric space and ≤ be a partial order on X. Let be a continuous and nondecreasing mapping such that there exists with
for all with . If there exists , then the following alternative holds: either
(A) for all , one has
or
(B) the sequence of is d-convergent to a fixed point of f.
Proof Consider the sequence of real numbers. Then we consider two cases as follows:
-
(a)
If, for all , , then (A) holds;
-
(b)
If, for some integer l, , then denotes the smallest nonnegative integer such that .
We see that and, by induction, we have
for all .
Note that f is nondecreasing. Then we have
and we can write for all .
Now, by induction, we show that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1687-1812-2014-15/MediaObjects/13663_2013_Article_659_Equg_HTML.gif)
for all . For , since , we have
Supposing that (Ω) holds for some n and using that , we obtain
Thus it follows that is a Cauchy sequence in X. Indeed, let . Then we have
On the other hand, since X is a complete generalized metric space, there exists such that .
Finally, we prove that is a fixed point of f, that is, . Let be a positive real number. Using the continuity of f at y, for , there exists such that implies that . Now, by the convergence of to y and , there exists such that and, for all , . Therefore, for all , we have
and hence . This completes the proof. □
Theorem 2.2 In Theorem 2.1, we can replace the following condition with the continuity of f:
If is a nondecreasing sequence and in X, then for all .
Then f has a fixed point.
Proof In Theorem 2.1, we just showed that y is a fixed point of f. Let be given. Since and is a nondecreasing sequence, we have . For any , there exists such that and, for all , . Therefore, we have
This shows that . This completes the proof. □
Remark 2.3 In Theorem 2.2, since and is nondecreasing, we have and
Therefore, we have
Thus it follows that
Theorem 2.4 If, for all , there exists z which is comparable to x and y and , , then, in Theorems 2.1 and 2.2, the uniqueness of the fixed point of f follows.
Proof If is another fixed point of f, then we prove that , where . Since there exists which is comparable to x and y, is comparable to and for all and
whenever and so we have . This completes the proof. □
3 Application
In this section, we suppose that is a partially ordered normed space with the following conditions:
-
(a)
for all , for all ;
-
(b)
for all , there exists such that z is comparable to x and y.
Also, we suppose that is a partially ordered Banach space with the condition (i) and satisfies the following:
-
(c)
for all , there exists such that z is an upper bound of ;
-
(d)
if is a nondecreasing sequence in and , then for all .
As a simple example, we can show that ℝ satisfies the conditions (a), (b), (c) and (d). Also, in this section, we consider .
Now, we prove the main result of this section as follows.
Theorem 3.1 Suppose that is a mapping satisfying
and
for all , where x is comparable to z, y is comparable to w, where is a function satisfying and the following condition:
for all , where x is comparable to y and is a constant. Then there exists a unique additive mapping such that
for all .
Proof It is clear that . Putting and in (3.2), we get
for all . Hence we have
for all . Consider and introduce the generalized metric d on X by
for all . It is easy to show that is a complete generalized metric space.
Now, we put the partial order ≤ on X as follows: for all ,
for all . Now, we define a mapping by
for all . For any with , it follows that, for all ,
It follows that
It is easy to show that J is a nondecreasing mapping.
Now, we show that J is a continuous function. To this end, let be a sequence in which converges to and let be given. Then there exist and with such that
for all and and so
for all and . By inequality (3.3) and the definition of J, we get
for all and . Hence
for all . It follows that J is continuous. On the other hand, by (3.1), we have and, by applying inequality (3.5), we see that . Applying Theorem 2.1, it follows that J has a fixed point such that . It follows that
for all . For any , it follows from (3.1) that the sequence is a nondecreasing sequence in and so, by (3.6), we have for all . In particular, . This shows that . Now, we can see that
and hence
This implies inequality (3.4).
On the other hand, by using inequality (3.3), we have
for all and , where x is comparable to y. Let be arbitrary elements. Then there exists such that z is comparable to x and y. This implies that is comparable to and for all . It follows from (3.2) that
for all . By using (3.6) and (3.7), it follows that T is a Cauchy mapping.
To prove the uniqueness property of T, suppose that is another additive function satisfying (3.4). It is clear that . Then, for any , there exists such that is an upper bound of . This shows that is a mapping which is comparable to T and . Hence we have
for all . Since , . This completes the proof. □
Corollary 3.2 Let and be a function such that and
and
for all , where x is comparable to z and y is comparable to w. Then there exists a unique additive mapping such that
for all .
Proof Set for all with , , and let in Theorem 3.1. Then we get the desired result. □
Corollary 3.3 Let and . Suppose that is a mapping such that
and
for all , where x is comparable to z and y is comparable to w. Then there exists a unique additive mapping such that
for all .
Proof Set for all , and let in Theorem 3.1. Then we get the desired result. □
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Acknowledgements
YJ Cho and C Park were supported by the Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education, Science and Technology (NRF-2012-0008170) and (NRF-2012R1A1A2004299), respectively.
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Eshaghi Gordji, M., Ramezani, M., Sajadian, F. et al. A new type fixed point theorem for a contraction on partially ordered generalized complete metric spaces with applications. Fixed Point Theory Appl 2014, 15 (2014). https://doi.org/10.1186/1687-1812-2014-15
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DOI: https://doi.org/10.1186/1687-1812-2014-15