Abstract
We set up a new variant of cyclic generalized contractive mappings for a map in a metric space and present existence and uniqueness results of fixed points for such mappings. Our results generalize or improve many existing fixed point theorems in the literature. To illustrate our results, we give some examples. At the same time as applications of the presented theorems, we prove an existence theorem for solutions of a class of nonlinear integral equations.
MSC: 47H10, 54H25.
Keywords:
fixed point; cyclic generalized contraction; integral equation1 Introduction and preliminaries
All the way through this paper, by , we designate the set of all real nonnegative numbers, while ℕ is the set of all natural numbers.
The celebrated Banach’s [1] contraction mapping principle is one of the cornerstones in the development of nonlinear analysis. This principle has been extended and improved in many ways over the years (see, e.g., [25]). Fixed point theorems have applications not only in various branches of mathematics but also in economics, chemistry, biology, computer science, engineering, and other fields. In particular, such theorems are used to demonstrate the existence and uniqueness of a solution of differential equations, integral equations, functional equations, partial differential equations, and others. Owing to the magnitude, generalizations of the Banach fixed point theorem have been explored heavily by many authors. This celebrated theorem can be stated as follows.
Theorem 1.1 ([1])
Letbe a complete metric space andTbe a mapping ofXinto itself satisfying
wherekis a constant in. ThenThas a unique fixed point.
Inequality (1) implies the continuity of T. A natural question is whether we can find contractive conditions which will imply the existence of a fixed point in a complete metric space but will not imply continuity.
On the other hand, cyclic representations and cyclic contractions were introduced by Kirk et al.[6]. A mapping is called cyclic if and , where A, B are nonempty subsets of a metric space . Moreover, T is called a cyclic contraction if there exists such that for all and . Notice that although a contraction is continuous, a cyclic contraction need not to be. This is one of the important gains of this theorem.
Let be a metric space. Let p be a positive integer, be nonempty subsets of X, , and . Then Y is said to be a cyclic representation of Y with respect to T if
(i) , are nonempty closed sets, and
Following the paper in [6], a number of fixed point theorems on a cyclic representation of Y with respect to a selfmapping T have appeared (see, e.g., [3,715]).
In this paper, we introduce a new class of cyclic generalized contractive mappings, and then investigate the existence and uniqueness of fixed points for such mappings. Our main result generalizes and improves many existing theorems in the literature. We supply appropriate examples to make obvious the validity of the propositions of our results. To end with, as applications of the presented theorems, we achieve fixed point results for a generalized contraction of integral type and we prove an existence theorem for solutions of a system of integral equations.
2 Main results
In this section, we introduce two new notions of a cyclic contraction and establish new results for such mappings.
In the sequel, we fixed the set of functions by such that
(i) ℱ is nondecreasing, continuous, and for every ;
(ii) ψ is nondecreasing, right continuous, and for every .
We state the notion of a cyclic generalized contraction as follows.
Definition 2.1 Let be a metric space. Let p be a positive integer, be nonempty subsets of X and . An operator is said to be a cyclic generalized contraction for some , , and if
(a) is a cyclic representation of Y with respect to T;
where
and
Our first main result is the following.
Theorem 2.1Letbe a complete metric space, , be nonempty closed subsets ofX, and. Supposeis a cyclic generalizedcontraction mapping for someand. ThenThas a unique fixed point. Moreover, the fixed point ofTbelongs to.
Proof Let (such a point exists since ). Define the sequence in X by
We shall prove that
If, for some k, we have , then (2) follows immediately. So, we can suppose that for all n. From the condition (a), we observe that for all n, there exists such that . Then, from the condition (b), we have
On the other hand, we have
and
Suppose that for some . Then , so
a contradiction. Hence,
and thus
Similarly, we have
Thus, from (4) and (5), we get
and the property of ψ, we obtain , and consequently (2) holds.
Now, we shall prove that is a Cauchy sequence in . Suppose, on the contrary, that is not a Cauchy sequence. Then there exists for which we can find two sequences of positive integers and such that for all positive integers k,
Further, corresponding to , we can choose in such a way that it is the smallest integer with satisfying (6). Then we have
Using (6), (7), and the triangular inequality, we get
Thus, we have
Passing to the limit as in the above inequality and using (2), we obtain
On the other hand, for all k, there exists such that . Then (for k large enough, ) and lie in different adjacently labeled sets and for certain . Using (b), we obtain
for all k. Now, we have
and
for all k. Using the triangular inequality, we get
which implies from (8) that
Using (2), we have
and
Again, using the triangular inequality, we get
Passing to the limit as in the above inequality, using (14) and (12), we get
Similarly, we have
Passing to the limit as , using (2) and (12), we obtain
Similarly, we have
Now, it follows from (12)(16) and the continuity of φ that
and
Passing to the limit as in (9), using (17), (18), (19), and the condition (ii), we obtain
which is a contradiction. Thus, we proved that is a Cauchy sequence in .
Since is complete, there exists such that
We shall prove that
From the condition (a), and since , we have . Since is closed, from (20), we get that . Again, from the condition (a), we have . Since is closed, from (20), we get that . Continuing this process, we obtain (21).
Now, we shall prove that is a fixed point of T. Indeed, from (21), since for all n there exists such that , applying (b) with and , we obtain
for all n. On the other hand, we have
and
Passing to the limit as in the above inequality and using (20), we obtain that
Passing to the limit as in (22), using (23) and (20), we get
Suppose that . In this case, we have
which implies that
a contradiction. Then we have , that is, is a fixed point of T.
Finally, we prove that is the unique fixed point of T. Assume that is another fixed point of T, that is, . From the condition (a), this implies that . Then we can apply (b) for and . We obtain
Since and are fixed points of T, we can show easily that and . If , we get
a contradiction. Then we have , that is, . Thus, we proved the uniqueness of the fixed point. □
In the following, we deduce some fixed point theorems from our main result given by Theorem 2.1.
If we take and in Theorem 2.1, then we get immediately the following fixed point theorem.
Corollary 2.1Letbe a complete metric space andsatisfy the following condition: there exist, , andsuch that
for all. ThenThas a unique fixed point.
Remark 2.1 Corollary 2.1 extends and generalizes many existing fixed point theorems in the literature [1,1621].
Corollary 2.2Letbe a complete metric space, , be nonempty closed subsets ofX, , and. Suppose that there existandsuch that
(a′) is a cyclic representation ofYwith respect toT;
ThenThas a unique fixed point. Moreover, the fixed point ofTbelongs to.
Remark 2.2 Corollary 2.2 is similar to Theorem 2.1 in [7].
Remark 2.3 Taking in Corollary 2.2 with , we obtain a generalized version of Theorem 1.3 in [6].
Corollary 2.3Letbe a complete metric space, , be nonempty closed subsets ofX, , and. Suppose that there existandsuch that
(a′) is a cyclic representation ofYwith respect toT;
ThenThas a unique fixed point. Moreover, the fixed point ofTbelongs to.
Remark 2.4 Taking in Corollary 2.3 with , we obtain a generalized version of Theorem 3 in [13].
Corollary 2.4Letbe a complete metric space, , be nonempty closed subsets ofX, , and. Suppose that there existandsuch that
(a′) is a cyclic representation ofYwith respect toT;
ThenThas a unique fixed point. Moreover, the fixed point ofTbelongs to.
Remark 2.5 Taking in Corollary 2.4 with , we obtain a generalized version of Theorem 5 in [13].
Corollary 2.5Letbe a complete metric space, , be nonempty closed subsets ofX, , and. Suppose that there existandsuch that
(a) is a cyclic representation ofYwith respect toT;
ThenThas a unique fixed point. Moreover, the fixed point ofTbelongs to.
We provide some examples to illustrate our obtained Theorem 2.1.
Example 2.1 Let with the usual metric. Suppose and and . Define such that for all . It is clear that is a cyclic representation of Y with respect to T. Let be defined by and of the form , . For all and , we have
So, T is a cyclic generalized contraction for any . Therefore, all conditions of Theorem 2.1 are satisfied (), and so T has a unique fixed point (which is ).
Example 2.2 Let with the usual metric. Suppose and and . Define the mapping by
Clearly, we have and . Moreover, and are nonempty closed subsets of X. Therefore, is a cyclic representation of Y with respect to T.
On the other hand, we have
and
Then we have
Define the function by and of the form , and . Then we have
Moreover, we can show that (24) holds if or . Similarly, we also get (24) holds for .
Now, all the conditions of Theorem 2.1 are satisfied (with ), we deduce that T has a unique fixed point .
3 An application to an integral equation
In this section, we apply the result given by Theorem 2.1 to study the existence and uniqueness of solutions to a class of nonlinear integral equations.
We consider the nonlinear integral equation
where , and are continuous functions.
Let be the set of real continuous functions on . We endow X with the standard metric
It is well known that is a complete metric space.
We suppose that for all , we have
and
We suppose that for all , is a decreasing function, that is,
We suppose that
Finally, we suppose that, for all , for all with ( and ) or ( and ),
where is a nondecreasing function that belongs to and .
Now, define the set
We have the following result.
Theorem 3.1Under the assumptions (26)(31), problem (25) has one and only one solution.
Proof Define the closed subsets of X, and , by
and
We shall prove that
Using condition (29), since for all , we obtain that
The above inequality with condition (27) implies that
Using condition (29), since for all , we obtain that
The above inequality with condition (28) implies that
for all . Then we have . Finally, we deduce that (32) holds.
This implies, from condition (26), that for all ,
Now, using conditions (30) and (31), we can write that for all , we have
This implies that
where of the form . Using the same technique, we can show that the above inequality holds also if we take .
Now, all the conditions of Theorem 2.1 are satisfied (with ), we deduce that T has a unique fixed point , that is, is the unique solution to (25). □
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.
Acknowledgements
The second author would like to thank the Research Professional Development Project under the Science Achievement Scholarship of Thailand (SAST). Moreover, the third author was supported by the Commission on Higher Education (CHE), the Thailand Research Fund (TRF) and the King Mongkut’s University of Technology Thonburi (KMUTT) (Grant No. MRG5580213).
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