Cyclic generalized contractions and fixed point results with applications to an integral equation

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.

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., [2-5]). 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.

Definition 1.1 (See [6,7])

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

(ii) .

Following the paper in [6], a number of fixed point theorems on a cyclic representation of Y with respect to a self-mapping T have appeared (see, e.g., [3,7-15]).

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.

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 .

Define and .

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;

(b) for any , (with ),

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 generalized-contraction 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

for all . Now, from

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 [1,16-21].

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;

(b′) for any, (with),

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;

(b′) for any, (with),

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;

(b′) for any, (with),

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;

(b) for any, (with),

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.

Now, let with and , we have

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 .

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.

Let , such that

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

Define the mapping by

We shall prove that

Let , that is,

Using condition (29), since for all , we obtain that

The above inequality with condition (27) implies that

for all . Then we have .

Similarly, let , that is,

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.

Now, let , that is, for all ,

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). □

The authors declare that they have no competing interests.

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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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