Abstract
Three fixed point theorems for three general classes of contractive mappings of integral type in complete metric spaces are proved. Three examples are included.
MSC:54H25.
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1 Introduction
Branciari [1] was the first to study the existence of fixed points for the contractive mapping of integral type. He established a nice integral version of the Banach contraction principle and proved the following fixed point theorem.
Theorem 1.1 Let f be a mapping from a complete metric space into itself satisfying
where is a constant and . Then f has a unique fixed point such that for each .
Afterwards, many authors continued the study of Branciari and obtained many fixed point theorems for several classes of contractive mappings of integral type; see, e.g., [1–8] and the references therein. In particular, in 2011, Liu et al. [5] extended the result of Branciari [1] and deduced the following fixed point theorems.
Theorem 1.2 Let f be a mapping from a complete metric space into itself satisfying
where and is a function with
Then f has a unique fixed point such that for each .
Theorem 1.3 Let f be a mapping from a complete metric space into itself satisfying
where and are two functions with
Then f has a unique fixed point such that for each .
In 2008, Dutta and Choudhuty [9] proved the following result.
Theorem 1.4 Let f be a mapping from a complete metric space into itself satisfying
where are both continuous and monotone nondecreasing functions with if and only if . Then f has a unique fixed point such that for each .
However, to the best of our knowledge, no one studied the following contractive mappings of integral type:
where ;
where ;
where and .
It is clear that the above contractive mappings of integral type include these mappings in Theorems 1.1-1.4 as special cases. The purpose of this paper is to investigate the existence of fixed points for contractive mappings (1.1)-(1.3) of integral type. Under certain conditions, we prove the existence, uniqueness and iterative approximations of fixed points for contractive mappings (1.1)-(1.3) of integral type in complete metric spaces. Three examples with uncountably many points are constructed.
2 Preliminaries
Throughout this paper, we assume that , , ℕ denotes the set of all positive integers, is a metric space, is a self-mapping and
= { is Lebesgue integrable, summable on each compact subset of and for each };
= { satisfies that for each };
= { is nondecreasing continuous and };
= { satisfies that };
= { satisfies that for each };
= { satisfy that , and for each }.
The following lemmas play important roles in this paper.
Lemma 2.1 ([5])
Let and be a nonnegative sequence with . Then
Lemma 2.2 ([5])
Let and be a nonnegative sequence. Then
if and only if .
Lemma 2.3 Let . Then if and only if .
Proof Let . Put for each . It is easy to see that , which together with ensures that
Conversely, suppose that for some . Set for each . It is clear that , which together with guarantees that
This completes the proof. □
3 Main results
In this section we show the existence, uniqueness and iterative approximations of fixed points for contractive mappings (1.1)-(1.3) of integral type, respectively.
Theorem 3.1 Let f be a mapping from a complete metric space into itself satisfying (1.1). Then f has a unique fixed point such that for each .
Proof Let x be an arbitrary point in X. Firstly, we show that
Suppose that (3.1) does not hold. It follows that there exists some satisfying
Note that (3.2) and imply that
Using (1.1), (3.2) and , we conclude immediately that
which yields that
and
Combining (3.5) and Lemma 2.3, we get that
which together with and (3.4) means that
that is,
which contradicts (3.3). Hence (3.1) holds.
Secondly, we show that
In view of (3.1), we deduce that the nonnegative sequence is nonincreasing, which means that there exists a constant c with . Suppose that . It follows from (1.1) that
Taking upper limit in (3.7) and using Lemma 2.1 and , we conclude that
which is a contradiction. Hence .
Thirdly, we show that is a Cauchy sequence. Suppose that is not a Cauchy sequence, which means that there is a constant such that for each positive integer k, there are positive integers and with satisfying
For each positive integer k, let denote the least integer exceeding and satisfying (3.8). It follows that
Note that
In light of (3.9) and (3.10), we get that
In view of (1.1), we deduce that
Taking upper limit in (3.12) and using (3.11), and Lemma 2.1, we deduce that
which is impossible. Thus is a Cauchy sequence.
Since is complete, it follows that there exists a point satisfying . By virtue of (1.1), we infer that
which together with and Lemmas 2.1 and 2.2 gives that
which together with yields that
that is, .
Finally, we show that a is a unique fixed point of f in X. Suppose that f has another fixed point . It follows from (1.1) and that
which is a contradiction. This completes the proof. □
Theorem 3.2 Let f be a mapping from a complete metric space into itself satisfying (1.2). Then f has a unique fixed point such that for each .
Proof Let x be an arbitrary point in X. Suppose that (3.2) holds for some . Using (1.2), (3.2) and , we get that
and
which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) is true. Notice that the nonnegative sequence is nonincreasing, which implies that there exists a constant with . Suppose that . In light of (1.2), we infer that
Taking upper limit in (3.13) and using Lemma 2.1 and , we know that
which is a contradiction, and hence , that is, (3.6) holds.
Now we show that is a Cauchy sequence. Suppose that is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist and with for each satisfying (3.8)-(3.11). By means of (1.2), (3.11), Lemma 2.1 and , we get that
which is a contradiction. Hence is a Cauchy sequence.
It follows from completeness of that there exists with . In view of (1.2), we have
Taking upper limit in (3.14) and making use of and Lemmas 2.1 and 2.2, we get that
which means that
that is, .
Next we prove that a is a unique fixed point of f in X. Suppose that f has another fixed point . It follows from (1.2) and that
which is a contradiction. This completes the proof. □
Theorem 3.3 Let f be a mapping from a complete metric space into itself satisfying (1.3) and
Then f has a unique fixed point such that for each .
Proof Let x be an arbitrary point in X. If there exists satisfying , it is clear that is a fixed point of f and . Now we assume that for all . Suppose that (3.2) holds for some . It follows from (1.3) that
which together with (3.2), (3.15), and implies that
which is a contradiction, and hence (3.2) does not hold. Consequently, (3.1) holds.
Next we show that . Note that the nonnegative sequence is nonincreasing, which implies that there exists a constant with . Suppose that . It follows from (1.3) that
which means that
Taking upper limit in (3.16) and using (3.15), , and Lemma 2.1, we arrive at
which is impossible. Therefore , that is, .
Next we show that is a Cauchy sequence. Suppose that is not a Cauchy sequence. As in the proof of Theorem 3.1, we conclude that there exist and with for each satisfying (3.8)-(3.11). By means of (3.12), we deduce that
Taking upper limit in (3.17) and making use of (1.3), (3.11), Lemma 2.1, and , we deduce that
which is a contradiction. Hence is a Cauchy sequence.
Completeness of implies that there exists a point such that . In view of (1.3), , and Lemma 2.1, we infer that
which together with yields that
which gives that , that is, .
Finally, we prove that a is a unique fixed point of f in X. Suppose that f has another fixed point . It follows from (1.3) and and that
which is a contradiction. This completes the proof. □
4 Three examples
Now we construct three examples to explain Theorems 3.1-3.3.
Example 4.1 Let be endowed with the Euclidean metric . Assume that and are defined by
Clearly, is a complete metric and . Let with . In order to verify (1.1), we have to consider the following four cases.
Case 1. Let . Note that
Case 2. Let and . It follows that
Case 3. Let and . It follows that
Case 4. Let and . Note that
That is, (1.1) holds. Thus Theorem 3.1 guarantees that f has a unique fixed point such that for each .
Example 4.2 Let be endowed with the Euclidean metric . Assume that and and are defined by
Obviously, . Put with . In order to verify (1.2), we have to consider three possible cases as follows.
Case 1. Let . It is clear that
Case 2. Let . It follows that
Case 3. Let and . It follows that
That is, (1.2) holds. Consequently, the conditions of Theorem 3.2 are satisfied. It follows from Theorem 3.2 that f has a unique fixed point such that for each .
Example 4.3 Let be endowed with the Euclidean metric . Assume that , and are defined by
It is easy to see that , and (3.15) holds. In order to verify (1.3), we have to consider the five possible cases below.
Case 1. Let with . Note that
Case 2. Let with . Note that
Case 3. Let and . It follows that
Case 4. Let and . Note that
Case 5. Let . Notice that . It follows that
That is, (1.3) holds. Thus all the conditions of Theorem 3.3 are satisfied. It follows from Theorem 3.3 that f has a unique fixed point such that for each .
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Acknowledgements
This research was supported by the Science Research Foundation of Educational Department of Liaoning Province (L2012380).
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Liu, Z., Li, J. & Kang, S.M. Fixed point theorems of contractive mappings of integral type. Fixed Point Theory Appl 2013, 300 (2013). https://doi.org/10.1186/1687-1812-2013-300
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DOI: https://doi.org/10.1186/1687-1812-2013-300