Abstract
The aim of this paper is to introduce classes of α-admissible generalized contractive type mappings of integral type and to discuss the existence of fixed points for these mappings in complete metric spaces. Our results improve and generalize fixed point results in the literature.
MSC:46T99, 54H25, 47H10, 54E50.
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1 Introduction and preliminaries
In 2002, Branciari [1] established a fixed point theorem for a single-valued mapping satisfying a contractive inequality of integral type; we also refer the reader to [2–10]. Recently, Liu et al. [11] (see also [12, 13]) obtained fixed point theorems for general classes of contractive mappings of integral type in complete metric spaces. In this paper, using auxiliary functions, we establish some fixed point theorems for self-mappings satisfying a certain contractive inequality of integral type.
Throughout the paper , , where ℕ denotes the set of all positive integers, is a metric space and is a self-mapping. Let
= { 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 }.
In 2013, Liu et al. [11] introduced the following three contractive mappings of integral type:
where ,
where , and
where and .
The following lemmas will be used in the proof of our main results.
Lemma 1.1 [11]
Let and be a nonnegative sequence with . Then we have
Lemma 1.2 [11]
Let and be a nonnegative sequence. Then we have the following equivalence:
if and only if .
Lemma 1.3 [11]
Let . Then if and only if .
The notion of α-admissibility was defined in [14] and appreciated by several authors [15–17] (see also [18–31]).
Definition 1.1 [14]
Let and . The mapping T is said to be α-admissible if for all , we have
Definition 1.2 Let and . The mapping T is said to be weak triangular α-admissible if for all we have
2 Main results
In this section, we state and prove our main results. We start with the following general contractive inequality of integral type.
Definition 2.1 Let be a complete metric space, and be mappings. Suppose that there exist and such that
for all , where
and
and
Then f is said to be an α-admissible contractive inequality of integral type I.
Theorem 2.1 Let be a complete metric space. Suppose that is an α-admissible contractive inequality of integral type I which satisfies
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that either or ;
-
(iii)
f is continuous.
Then T has a fixed point.
Proof From (ii), there exists a point such that (due to the symmetry of the metric, the other case yields the same result). Let and we define an iterative sequence in X by for all . Note that we have
Inductively, we have
In the sequel, we use the following abbreviations:
Notice that if for some , then it is evident that is a fixed point of f. This completes the proof. Consequently, we assume that for all , that is,
We now prove that is a non-increasing sequence of real numbers, that is,
Suppose, on the contrary, that inequality (10) does not hold. Thus, there exists some such that
From (9) and (11), we get
Regarding again (9) and (11) together with the properties of ψ, we conclude that
Using equations (13), (7), (6), (8) and the fact that , we obtain immediately that
where
Thus and from (11). Hence, inequality (14) turns into
which is a contradiction. Hence (10) holds. Thus, there exists a constant such that .
Next we show that , that is,
Suppose, on the contrary, that . It follows from (6) and (7) that
where
Hence, inequality (17) becomes
Taking the upper limit in (18) and using Lemma 1.1 and noting , we get
which is a contradiction. Hence .
Next we prove that is a Cauchy sequence. Suppose, on the contrary, that is not a Cauchy sequence. Thus, 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 (20). This implies that
On the other hand, we have
In view of (21) and (22), we infer that
Using the weak triangular alpha admissible property of f, we get in view of (7)
From (6) and (24), we have
Taking the upper limit in (25) and using (23), and Lemma 1.1, we get
which is impossible. Thus is a Cauchy sequence. Now, since is complete, there exists a point such that . From the continuity of f, it follows that as . From the uniqueness of limits, we get , that is, a is a fixed point of f. This completes the proof. □
Theorem 2.2 Let be a complete metric space. Suppose that is an α-admissible contractive inequality of integral type I which satisfies
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that either or ;
-
(iii)
if is a sequence in X such that for all n and as , then for all n.
Then f has a fixed point.
Proof Following the proof in Theorem 2.1, we see that is a Cauchy sequence in the complete metric space . Then there exists such that as . On the other hand, from inequality (7) and hypothesis (iii), we have
Let us suppose that . In view of the above inequality and (6), we obtain that
for all . On the other hand, we have
Taking the upper limit in (27), in view of Lemmas 1.1 and 1.2 and , we infer from (28) that
which is a contradiction. Thus, we have . □
Now, we present another contractive inequality of integral type.
Definition 2.2 Let be a complete metric space and be a self-mapping. Suppose that there exist and such that
for all , where
and
and
Then f is said to be an α-admissible contractive inequality of integral type II.
We omit the proof of the following two theorems since they mimic the proof of Theorem 2.1 and Theorem 2.2.
Theorem 2.3 Let be a complete metric space. Suppose that is an α-admissible contractive inequality of integral type II which satisfies
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that either or ;
-
(iii)
f is continuous.
Then T has a fixed point.
Theorem 2.4 Let be a complete metric space. Suppose that is an α-admissible contractive inequality of integral type II which satisfies
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that either or ;
-
(iii)
if is a sequence in X such that for all n and as , then for all n.
Then T has a fixed point.
The following condition provides the uniqueness of fixed points of the maps considered in Theorem 2.3 and Theorem 2.4. Consider
() for all , there exists such that and , where denotes the set of fixed points of f.
Theorem 2.5 If the condition () is added to the hypotheses of Theorem 2.3 (respectively, Theorem 2.4), then the fixed point u of T is unique.
Proof From (), we have
Define the sequence in X by for all and . Using the weak triangular α-admissible property of f, we infer that
for all . Using inequality (6), we get
Using standard techniques, we derive that and hence the sequence converges to some . If , then the proof is complete. Indeed, we get that and analogously, as and from the uniqueness of limits, we derive that . Suppose, on the contrary, . By letting , we derive from the above inequality that
which is a contradiction. □
Now we introduce a third type of contractive inequality of integral type.
Definition 2.3 Let be a complete metric space and be a self-mapping. Suppose that there exist such that
Then f is said to be an α-admissible contractive inequality of integral type .
Theorem 2.6 Let be a complete metric space. Suppose that is an α-admissible contractive inequality of integral type which satisfies
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that either or ;
-
(iii)
f is continuous.
Then T has a fixed point.
Proof Following the lines in the proof of Theorem 2.1, we conclude the result. □
Theorem 2.7 Let be a complete metric space. Suppose that is an α-admissible contractive inequality of integral type which satisfies
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that either or ;
-
(iii)
if is a sequence in X such that for all n and as , then for all n.
Then f has a fixed point.
Proof The reasoning in Theorem 2.2 establishes the result. □
Example 2.4 Suppose that with the usual metric. We consider a mapping defined by . Define the mapping by for all . Hence, f is weak triangular α-admissible. Define by . Let us define and by and respectively for all .
Clearly, in view of the definitions of α and f, we infer that f is an α-admissible contractive inequality of integral type . There exists such that . In fact, for , we obtain
Clearly, f is continuous. Now, all the hypotheses of Theorem 2.6 are satisfied. Thus f has a fixed point in X. In this case, 0 is a fixed point of f.
Example 2.5 Suppose that with the usual metric induced by ℝ. It is a complete metric space, since X is a closed subset of ℝ. We consider a mapping defined by
Define the mapping by for all . It is clear that f is weak triangular α-admissible. Thus, the condition (i) of Theorem 2.7 is satisfied.
Now, consider the following auxiliary function φ defined as
Then, for any , we have for and for . Consequently, we have .
Clearly, in view of the definitions of α and f, we infer that f is an α-admissible contractive inequality of integral type for and , for all , where and .
There exists such that . In fact, for example, for , we obtain . Hence, the condition (ii) of Theorem 2.7 is fulfilled.
Let be a sequence in X such that for all n and as for some . From the definition of α, for all n, we have for all. So, the last condition of Theorem 2.7 is satisfied. As a result, due to Theorem 2.7, the mapping f has a fixed point. Notice that is a fixed point of f.
Theorem 2.8 If the condition () is added to the hypotheses of Theorem 2.6 (respectively, Theorem 2.7), then the fixed point u of T is unique.
3 Consequences in metric spaces
We get the following result by letting in Theorem 2.5.
Theorem 3.1 Let f be a mapping from a complete metric space into itself satisfying, for all ,
where , , and . Then f has a unique fixed point such that for each .
If we take in Theorem 3.1, we get the following result.
Theorem 3.2 Let f be a mapping from a complete metric space into itself satisfying, for all ,
where , and . Then f has a unique fixed point such that for each .
If we take in Theorem 3.2, we get the following result.
Theorem 3.3 Let f be a mapping from a complete metric space into itself satisfying, for all ,
where , and . Then f has a unique fixed point such that for each .
Remark 3.1 The following theorem is the main result of [11] that can be easily deduced by taking , for all , in Theorem 2.8. Consequently, all corollaries of the main result of [11] can be deduced evidently.
Theorem 3.4 Let f be a mapping from a complete metric space into itself satisfying, for all ,
where . Then f has a unique fixed point such that for each .
If we take in Theorem 3.4, we get the following result.
Theorem 3.5 Let f be a mapping from a complete metric space into itself satisfying, for all ,
where . Then f has a unique fixed point such that for each .
Theorem 3.6 Let f be a mapping from a complete metric space into itself. If there is satisfying the following condition for all :
then f has a unique fixed point such that for each .
4 Consequences in partially ordered metric spaces
Definition 4.1 Let be a partially ordered set and be a given mapping. We say that T is nondecreasing with respect to ⪯ if
Definition 4.2 Let be a partially ordered set. A sequence is said to be nondecreasing with respect to ⪯ if for all n.
Definition 4.3 Let be a partially ordered set and d be a metric on X. We say that is regular if for every nondecreasing sequence such that as , there exists a subsequence of such that for all k.
We have the following result.
Corollary 4.1 Let be a partially ordered set and d be a metric on X such that is complete. Let be a nondecreasing mapping with respect to ⪯ and satisfy the following inequality:
for all with , where , , and . Suppose also that the following conditions hold:
-
(i)
there exists such that ;
-
(ii)
f is continuous or is regular.
Then f has a fixed point. Moreover, if for all there exists such that and , we have uniqueness of the fixed point.
Proof Define the mapping by
Clearly, f is an α-admissible contractive inequality of integral type I. From condition (i), we have . Moreover, for all , from the monotone property of f, we have
Thus f is α-admissible. Now, if f is continuous, the existence of a fixed point follows from Theorem 2.1. Suppose now that is regular. Let be a sequence in X such that for all n and as . From the regularity hypothesis, there exists a subsequence of such that for all k. This implies from the definition of α that for all k. In this case, the existence of a fixed point follows from Theorem 2.2. To show the uniqueness, let . By hypothesis, there exists such that and , which implies from the definition of α that and . Thus we deduce the uniqueness of the fixed point by Theorem 2.5. □
The following result is an immediate consequence of Corollary 4.1.
Corollary 4.2 Let be a partially ordered set and d be a metric on X such that is complete. Let be a nondecreasing mapping with respect to ⪯. Suppose that there exists a function such that
for all with . Suppose also that the following conditions hold:
-
(i)
there exists such that ;
-
(ii)
T is continuous or is regular.
Then T has a fixed point. Moreover, if for all there exists such that and , we have uniqueness of the fixed point.
Remark 4.1 Let be nonempty closed subsets of a complete metric space and be a given mapping, where . If the following condition holds:
then T is called a cyclic mapping. Since and are closed subsets of the complete metric space , then is complete. Define the mapping by
By using the observation above, it is possible to deduce some fixed point results of a cyclic mapping that satisfies, e.g., one of the inequalities between (31)-(36), and so on. For more details on such approach, we refer, e.g., to [14, 16].
5 Further results
Theorem 5.1 Let be a complete metric space, and be self-mappings. Suppose that there exist and such that
for all , where
and
Suppose also that the following conditions hold:
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that ;
-
(iii)
f is continuous or is regular.
Then f has a unique fixed point such that for each .
Proof From (ii), there exits a point such that (due to the symmetry of the metric, the other case yields the same result). Let and consider an iterative sequence in X by for all . Note that we have
By mathematical induction, we get
Let us denote
Now, if for some , then is a fixed point of f. This completes the proof. Consequently, suppose for all , that is,
Now, we proceed to show that is a non-increasing sequence of real numbers, that is,
Suppose, on the contrary, that inequality (42) does not hold. Thus, there exists some such that
From (41) and (43), we get
Regarding again (41) and (43) together with the properties of ψ, we conclude that
Using equations (38)-(40), (45) we obtain that
where
From (43), we have . Hence, inequality (46) implies
which contradicts inequality (44). Hence, (42) holds. Thus, there exists a constant such that .
Next we show that , that is,
Suppose, on the contrary, that . It follows from (38) and (39) that
where
Hence, inequality (49) becomes
Taking the upper limit in (50) and using Lemma 1.1, we get
which is a contradiction. Hence .
Next we show that is a Cauchy sequence. Suppose, on the contrary, that is not a Cauchy sequence. Thus, 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 (52). This implies that
On the other hand, we have
In view of (53) and (54), we infer that
Using the weak triangular alpha admissible property of f, we get in view of (39)
From (38) and (56), we have, for all ,
Taking the upper limit in (57) and using (55) and Lemma 1.1, we get
which is impossible. Thus is a Cauchy sequence. Now, since is complete, there exists a point such that . From the continuity of f, it follows that as . From the uniqueness of limits, we get , that is, u is a fixed point of f. This completes the proof. □
Theorem 5.2 Let be a complete metric space and be a self-mapping. Suppose that there exist with for all , , and such that
for all , where
and
Suppose also that the following conditions hold:
-
(i)
f is weak triangular α-admissible;
-
(ii)
there exists such that ;
-
(iii)
f is continuous or is regular.
Then f has a unique fixed point such that for each .
Proof From condition (ii), there exists a point such that (due to the symmetry of the metric, the other case yields the same result). Let and let the iterative sequence in X be defined by for all . Note that we have
Using mathematical induction, we obtain
Set
If for some , , then is a fixed point of f. This completes the proof. Suppose that for all , that is,
Now, we need to show that is a non-increasing sequence of real numbers, that is,
Suppose, on the contrary, that inequality (62) does not hold. Thus, there exists some such that
From (61) and (63), we get
From equations (61) and (63) and using the properties of ψ, we get
In view of equations (58)-(60), (65) we infer that
where
From (63), we have . Hence, inequality (66) implies
which contradicts inequality (64). Hence, (62) holds. Thus, there exists a constant such that .
Next we show that , that is,
Suppose, on the contrary, that . It follows from (58) and (59) that
where
Hence, inequality (69) becomes
Taking the upper limit in (70) and using Lemma 1.1, we get
which is a contradiction. Hence .
Now, we show that is a Cauchy sequence. Suppose, on the contrary, that is not a Cauchy sequence. Thus, 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 (72). This implies that
On the other hand, we have
In view of (73) and (74), we infer that
Using the weak triangular alpha admissible property of f, we get in view of (59)
From (58) and (76), we have, for all ,
Taking the upper limit in (77) and using (75) and Lemma 1.1, we get
which is impossible. Thus is a Cauchy sequence. Now, since is complete, there exists a point such that . From the continuity of f, it follows that as . From the uniqueness of limits, we get , that is, u is a fixed point of f. This completes the proof. □
6 Conclusion
In this paper, we handle contractive mappings of integral type in a more general frame via α-admissible mappings. More precisely, we examine the contractive mapping of integral type given in [11] by using α-admissible mappings. Very recently, some new contractive mappings of integral type were introduced in [12] and [13]. We assert that our techniques are also valid to extent the results of [12] and [13] in the frame of α-admissible mappings.
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Acknowledgements
This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant No. (55-130-35-HiCi). The authors, therefore, acknowledge technical and financial support of KAU.
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Alsulami, H.H., Karapınar, E., O’Regan, D. et al. Fixed points of generalized contractive mappings of integral type. Fixed Point Theory Appl 2014, 213 (2014). https://doi.org/10.1186/1687-1812-2014-213
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DOI: https://doi.org/10.1186/1687-1812-2014-213