Abstract
Keywords:
cyclic representation; fixed point1 Introduction and preliminaries
Fixed point theory is one of the most efficient tools in nonlinear functional analysis to solve the nonlinear differential and integral equations. The existence/uniqueness of a solution of differential/integral equations turns into the existence/uniqueness of a (common) fixed point of the operators which are obtained after suitable substitutions and elementary calculations; see, e.g., [114].
In this paper, we first obtain some fixed point theorems to solve the integral equation mentioned above. For the sake of completeness, we recollect some basic definitions and elementary results. Let X be a nonempty set and T be a selfmapping on X. Then, the set of all fixed points of T on X is denoted by . Let Ψ be the set of all functions satisfying the following conditions:
(1) ψ is continuous,
Cyclic mapping and cyclic contraction were introduced by KirkSrinavasanVeeramani to improve the wellknown Banach fixed point theorem. Later, various types of cyclic contraction have been investigated by a number of authors; see, e.g., [6,1517].
Definition 1.1[18]
Suppose that is a metric space and T is a selfmapping on X. Let m be a natural number and , , be nonempty sets. Then is called a cyclic representation of X with respect to T if
Definition 1.2[17]
Let , and be two functions. We say that T is radmissible if
Definition 1.3 Let be a metric space and be a selfmapping, where is a cyclic representation of Y with respect to T. Let be two functions. An operator is called:
• a cyclic weak rCcontractive mapping of the first kind if
• a cyclic weak rCcontractive mapping of the second kind if
2 Auxiliary fixed point results
We state the main result of this section as follows.
Theorem 2.1Letbe a complete metric space, , be nonempty closed subsets ofand. Suppose thatis a cyclic weakrCcontractive mapping of the first kind such that
(ii) there existssuch thatand;
(iii) ifis a sequence inYsuch thatandfor allandas, thenand.
ThenThas a fixed point. Moreover, if, , , for all, thenThas a unique fixed point.
Proof Let there exist such that and . Since T is radmissible, then and . Again, since T is radmissible, then and . By continuing this process, we get
On the other hand, since , there exists some such that . Now implies that . Thus there exists in such that . Similarly, , where . Hence, for , there exists such that and . In case for some , then it is clear that is a fixed point of T. Now assume that for all n. Hence, we have for all n. Set . We shall show that the sequence is nonincreasing. Due to (2) with and , we get
which implies
and so for all . Then there exist such that . Suppose, on the contrary, that . Also, taking limit as in (5), we deduce
that is,
Taking limit as in (5) and using (6), we get
Consequently, we have , which yields . Hence
We shall show that is a Cauchy sequence. To reach this goal, first we prove the following claim:
(K) For every , there exists such that if with , then .
Suppose, to the contrary, that there exists such that for any , we can find with satisfying
Now, we take . Then, corresponding to , one can choose in such a way that it is the smallest integer with satisfying and . Therefore, . By using the triangular inequality,
Letting in the last inequality, keeping (7) in mind, we derive that
Again,
Taking (7) and (9) into account, we get
Also we have the following inequalities:
and
Letting in (11) and (12), we derive that
Again, we have
and
Letting in (14) and (15), we conclude that
Since and lie in different adjacently labeled sets and for certain , using the fact that T is a cyclic weak rCcontractive mapping of the first kind, we have
which implies
Letting in the inequality above and keeping the expressions (7), (9), (10), (13), (16) in mind, we conclude that
Thus, we have , which yields that . Hence, (K) is satisfied.
We shall show that the sequence is Cauchy. Fix . By the claim, we find such that if with , then
Since , we also find such that
for any . Suppose that and . Then, there exists such that . Therefore, for . So, we have, for ,
By (18) and (19) and from the last inequality, we get
This proves that is a Cauchy sequence. Since Y is closed in , then is also complete, there exists such that in . In what follows, we prove that x is a fixed point of T. In fact, since and, as is a cyclic representation of Y with respect to T, the sequence has infinite terms in each for . Suppose that , and we take a subsequence of with . Now from (iii) we have and . By using the contractive condition, we can obtain
which implies
Passing to the limit as in the last inequality, we get
which implies , i.e., . Finally, to prove the uniqueness of the fixed point, suppose that such that , , , , where . The cyclic character of T and the fact that are fixed points of T imply that . Suppose that . That is, . Using the contractive condition, we obtain
which implies
Then and so , i.e., , which is a contradiction. This finishes the proof. □
Example 2.2 Let with the metric for all . Suppose and and . Define and by
Also, define by . Clearly, , and and . Let , then . On the other hand, for all , i.e., . Similarly, implies . Therefore, T is an radmissible mapping. Let be a sequence in X such that , and as . Then . So, , i.e., and .
Let and . Now, if or , then . Also, if and , then . That is, for all and all . Hence,
for all and . Then T is a cyclic weak rCcontractive mapping of the first kind. Therefore all the conditions of Theorem 2.1 hold and T has a fixed point in . Here, is a fixed point of T.
Theorem 2.3Letbe a complete metric space, , be nonempty closed subsets ofand. Suppose thatis a cyclic weakrCcontractive mapping of the second kind such that
(ii) there existssuch thatand;
(iii) ifis a sequence inYsuch thatandfor allandas, thenand.
ThenThas a fixed point. Moreover, if, , , for all, thenThas a unique fixed point.
Proof By a similar method as in the proof of Theorem 2.1, we have
We shall show that the sequence is nonincreasing. Due to (3) with and , we get
which implies
and so for all . Then there exists such that . We shall show that by the method of reductio ad absurdum. Suppose that . By letting in (22), we deduce
that is,
Taking limit as in (22) and using (23), we get
Thus, we have and hence , which is a contradiction. Consequently, we have
We shall show that is a Cauchy sequence. To reach this goal, first we prove the following claim:
(K) For every , there exists such that if with , then .
Suppose, to the contrary, that there exists such that for any we can find with satisfying
Following the related lines in Theorem 2.1, we deduce
and
Since and lie in different adjacently labeled sets and for certain , using the fact that a cyclic weak rCcontractive mapping of the second kind, we have
which implies
Letting in the inequality above and by applying (24) (26), (27), (28), (29), we deduce that
Consequently, we have , and hence . As a result, we conclude that (K) is satisfied. We assert that the sequence is Cauchy. Fix . By the claim, we find such that if with , then
Since , we also find such that
for any . Suppose that and . Then there exists such that . Therefore, for . So, we have, for , ,
By (30) and (31) and from the last inequality, we get
This proves that is a Cauchy sequence. Since Y is closed in , then is also complete, there exists such that in . In what follows, we prove that x is a fixed point of T. In fact, since and, as is a cyclic representation of Y with respect to T, the sequence has infinite terms in each for . Suppose that , and we take a subsequence of with . Now from (iii) we have and . By using the contractive condition, we can obtain
which implies
Passing to the limit as in the last inequality, we get
which implies , i.e., . Finally, to prove the uniqueness of the fixed point, suppose that such that , , , , where . The cyclic character of T and the fact that are fixed points of T imply that . Suppose that . That is, . Using the contractive condition, we obtain
which implies
Hence, we obtain , which implies , that is, a contradiction. □
3 Existence of solutions of an integral equation
For , we denote by the set of real continuous functions on . We endow X with the metric
It is evident that is a complete metric space.
Consider the integral equation
(1) and are continuous functions.
and
Let for all , be a decreasing function, that is,
Let . There exist and such that if and with ( and ) or ( and ), then for every , we have
(3) Assume that
for all , where . Suppose that
(4) If is a sequence in such that for all and as , then .
Theorem 3.1Under assumptions (1)(5), integral equation (33) has a solution in.
Proof Define the closed subsets of X, and by
and
Let us prove that
Applying condition (37), since for all , we obtain that
The above inequality with condition (35) imply that
Using condition (37), since for all , we obtain that
The above inequality with condition (36) imply that
for all . Then we have . Also, we deduce that (41) holds.
This implies from condition (34) that for all ,
Now, by conditions (39) and (38), we have, for all ,
which implies
Hence,
for all . By a similar method, we can show that the above inequality holds if . Now, all the conditions of Theorem 2.1 hold and T has a fixed point in
That is, is the solution to (33). □
Example 3.2 In this example, we denote by the set of real continuous functions on . We endow X with the metric
Consider the following continuous functions:
and
and
Clearly, . Also, if , then . On the other hand,
for all . That is, . Hence, implies .
Assume and with ( and ) or ( and ). Thus, and , which implies . That is,
and so
Assume that is a sequence in X such that for all and as . Then . So, . That is, .
Therefore, all of the conditions of Theorem 3.1 are satisfied. Then the integral equation
has a solution in . Here, is a solution.
But if we chose and , then and . That is,
Also,
and so
That is, Theorem 3.1 of [6] cannot be applied to this example.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Acknowledgements
The authors thank the anonymous referees for their remarkable comments, suggestions and ideas that helped to improve this paper. The third author (V Rakocević) is supported by Grant No. 174025 of the Ministry of Science, Technology and Development, Republic of Serbia.
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