In this paper, new classes of rational Geraghty contractive mappings in the setup of b-metric spaces are introduced. Moreover, the existence of some fixed point for such mappings in ordered b-metric spaces are investigated. Also, some examples are provided to illustrate the results presented herein. Finally, an application of the main result is given.
MSC:47H10, 54H25.
Hinweise
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.
1 Introduction
Using different forms of contractive conditions in various generalized metric spaces, there is a large number of extensions of the Banach contraction principle [1]. Some of such generalizations are obtained via rational contractive conditions. Recently, Azam et al. [2] established some fixed point results for a pair of rational contractive mappings in complex valued metric spaces. Also, in [3], Nashine et al. proved some common fixed point theorems for a pair of mappings satisfying certain rational contractions in the framework of complex valued metric spaces. In [4], the authors proved some unique fixed point results for an operator T satisfying certain rational contractive condition in a partially ordered metric space. In fact, their results generalize the main result of Jaggi [5].
Ran and Reurings started the studying of fixed point results on partially ordered sets in [6], where they gave many useful results in matrix equations. Recently, many researchers have focused on different contractive conditions in complete metric spaces endowed with a partial order and obtained many fixed point results in such spaces. For more details on fixed point results in ordered metric spaces we refer the reader to [7, 8] and [9].
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Czerwik in [10] introduced the concept of a b-metric space. Since then, several papers dealt with fixed point theory for single-valued and multi-valued operators in b-metric spaces (see, e.g., [11‐16] and [17, 18]).
Definition 1 Let X be a (nonempty) set and be a given real number. A function is a b-metric if the following conditions are satisfied:
(b1) iff ,
(b2) ,
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(b3)
for all .
In this case, the pair is called a b-metric space.
Letbe a complete metric space, and letbe a self-map. Suppose that there existssuch that
holds for all . Thenfhas a unique fixed pointand for eachthe Picard sequenceconverges toz.
Amini-Harandi and Emami [23] generalized the result of Geraghty to the framework of a partially ordered complete metric space as follows.
Theorem 2Letbe a complete partially ordered metric space. Letbe an increasing self-map such that there existswith . Suppose that there existssuch that
holds for allwith . Assume that eitherfis continuous orXis such that if an increasing sequenceinXconverges to , thenfor alln. Thenfhas a fixed point inX. Moreover, if for eachthere existscomparable withxandy, then the fixed point offis unique.
In [24], some fixed point theorems for mappings satisfying Geraghty-type contractive conditions are proved in various generalized metric spaces. As in [24], we will consider the class ℱ of functions such that
Let , and letbe a complete metric type space. Suppose that a mappingsatisfies the condition
for alland some . Thenfhas a unique fixed point , and for eachthe Picard sequenceconverges tozin .
Also, by unification of the recent results obtained by Zabihi and Razani [25] we have the following result.
Theorem 4Letbe a partially ordered set and suppose that there exists ab-metricdonXsuch thatis ab-completeb-metric space (with parameter ). Letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Suppose there existssuch that
(1.1)
for all comparable elements , where ,
and
Iffis continuous, or, wheneveris a nondecreasing sequence inXsuch that , one hasfor all , thenfhas a fixed point. Moreover, the set of fixed points offis well ordered if and only iffhas one and only one fixed point.
The aim of this paper is to present some fixed point theorems for rational Geraghty contractive mappings in partially ordered b-metric spaces. Our results extend some existing results in the literature.
2 Main results
Let ℱ denotes the class of all functions satisfying the following condition:
Definition 3 Let be a b-metric space. A mapping is called a rational Geraghty contraction of type I if there exists such that
(2.1)
for all comparable elements , where
Theorem 5Letbe a partially ordered set and suppose there exists ab-metricdonXsuch thatis ab-completeb-metric space (with parameter ). Letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Supposefis a rational Geraghty contraction of type I. If
(I)
fis continuous, or,
(II)
wheneveris a nondecreasing sequence inXsuch that , one hasfor all ,
thenfhas a fixed point.
Moreover, the set of fixed points offis well ordered if and only iffhas one and only one fixed point.
Proof Let for all . Since and f is increasing, we obtain by induction that
We do the proof in the following steps.
Step I: We show that . Since for each , then by (2.1)
(2.2)
where
If , then from (2.2),
(2.3)
which is a contradiction.
Hence, , so from (2.2),
(2.4)
Since is a decreasing sequence, then there exists such that . We prove . Suppose on contrary that . Then, letting , from (2.4) we have
which implies that . Now, as we conclude that , which yields , a contradiction. Hence, . That is,
(2.5)
Step II: Now, we prove that the sequence is a b-Cauchy sequence. Suppose the contrary, i.e., is not a b-Cauchy sequence. Then there exists for which we can find two subsequences and of such that is the smallest index for which
(2.6)
This means that
(2.7)
From (2.5) and using the triangular inequality, we get
By taking the upper limit as , we get
(2.8)
The definition of and (2.8) imply
Now, from (2.1) and the above inequalities, we have
which implies that . Now, as we conclude that , which yields . Consequently,
which is a contradiction to (2.6). Therefore, is a b-Cauchy sequence. b-Completeness of X shows that b-converges to a point .
Step III: u is a fixed point of f.
First, let f be continuous, so we have
Now, let (II) holds. Using the assumption on X we have . Now, we show that . By Lemma 1
where
Therefore, from the above relations, we deduce that , so .
Finally, suppose that the set of fixed point of f is well ordered. Assume to the contrary that u and v are two fixed points of f such that . Then by (2.1),
(2.9)
because
So we get , a contradiction. Hence , and f has a unique fixed point. Conversely, if f has a unique fixed point, then the set of fixed points of f is a singleton, and so it is well ordered. □
Definition 4 Let be a b-metric space. A mapping is called a rational Geraghty contraction of type II if there exists such that
(2.10)
for all comparable elements , where
Theorem 6Letbe a partially ordered set and suppose that there exists ab-metricdonXsuch thatis ab-completeb-metric space. Letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Supposefis a rational Geraghty contractive mapping of type II. If
(I)
fis continuous, or,
(II)
wheneveris a nondecreasing sequence inXsuch that , one hasfor all ,
thenfhas a fixed point.
Moreover, the set of fixed points offis well ordered if and only iffhas one and only one fixed point.
Proof Set . Since and f is increasing, we obtain by induction that
We do the proof in the following steps.
Step I: We show that . Since for each , then by (2.10)
(2.11)
because
Therefore, is decreasing. Then there exists such that . We will prove that . Suppose to the contrary that . Then, letting , from (2.11)
which implies that . Hence, , a contradiction. So,
(2.12)
holds true.
Step II: Now, we prove that the sequence is a b-Cauchy sequence. Suppose the contrary, i.e., is not a b-Cauchy sequence. Then there exists for which we can find two subsequences and of such that is the smallest index for which
(2.13)
This means that
(2.14)
As in the proof of Theorem 5, we have
(2.15)
From the definition of and the above limits,
Now, from (2.10) and the above inequalities, we have
which implies that . Now, as we conclude that is a b-Cauchy sequence. b-Completeness of X shows that b-converges to a point .
Step III: u is a fixed point of f.
First, let f be continuous, so we have
Now, let (II) hold. Using the assumption on X we have . Now, we show that . By Lemma 1
because
Therefore, , so . □
Definition 5 Let be a b-metric space. A mapping is called a rational Geraghty contraction of type III if there exists such that
(2.16)
for all comparable elements , where
Theorem 7Letbe a partially ordered set and suppose that there exists ab-metricdonXsuch thatis ab-completeb-metric space. Letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Supposefis a rational Geraghty contractive mapping of type III. If
(I)
fis continuous, or,
(II)
wheneveris a nondecreasing sequence inXsuch that , one hasfor all ,
thenfhas a fixed point.
Moreover, the set of fixed points offis well ordered if and only iffhas one and only one fixed point.
Proof Set .
Step I: We show that . Since for each , then by (2.16)
(2.17)
because
Therefore, is decreasing. Similar to what we have done in Theorems 5 and 6, we have
(2.18)
Step II: Now, we prove that the sequence is a b-Cauchy sequence. Suppose the contrary, i.e., is not a b-Cauchy sequence. Then there exists for which we can find two subsequences and of such that is the smallest index for which
(2.19)
This means that
(2.20)
From (2.18) and using the triangular inequality, we get
By taking the upper limit as , we get
(2.21)
Using the triangular inequality, we have
Taking the upper limit as in the above inequality and using (2.20) we get
(2.22)
Again, using the triangular inequality, we have
Taking the upper limit as in the above inequality and using (2.20) we get
(2.23)
From the definition of and the above limits,
Now, from (2.16) and the above inequalities, we have
which implies that . Now, as we conclude that is a b-Cauchy sequence. b-Completeness of X shows that b-converges to a point .
Step III: u is a fixed point of f.
When f is continuous, the proof is straightforward.
Now, let (II) hold. By Lemma 1
where
Therefore, from the above relations, we deduce that , so . □
If in the above theorems we take , where , then we have the following corollary.
Corollary 1Letbe a partially ordered set and suppose that there exists ab-metricdonXsuch thatis ab-completeb-metric space, and letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Suppose that
for all comparable elements , where
or
or
Iffis continuous, or, for any nondecreasing sequenceinXsuch thatone hasfor all , thenfhas a fixed point.
Corollary 2Letbe a partially ordered set and suppose that there exists ab-metricdonXsuch thatis ab-completeb-metric space, and letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Suppose
or
or
for all comparable elements , whereand .
Iffis continuous, or, for any nondecreasing sequenceinXsuch thatone hasfor all , thenfhas a fixed point.
Corollary 3Letbe an orderedb-completeb-metric space, and letbe an increasing mapping with respect to ⪯ such that there exists an elementwithand
for all comparable elements , where
or
or
for some positive integerm.
Ifis continuous, or, for any nondecreasing sequenceinXsuch thatone hasfor all , thenfhas a fixed point.
Let Ψ be the family of all nondecreasing functions such that
for all .
Lemma 2If , then the following are satisfied.
(a)
for all ;
(b)
.
As an example , for all , where , and , for all , are in Ψ.
Theorem 8Letbe a partially ordered set and suppose that there exists ab-metricdonXsuch thatis ab-completeb-metric space, and letbe an increasing mapping with respect to ⪯ such that there exists an elementwith . Suppose that
(2.24)
where
for all comparable elements . Iffis continuous, thenfhas a fixed point. Moreover, the set of fixed points offis well ordered if and only iffhas one and only one fixed point.
Proof Since and f is increasing, we obtain by induction that
Putting , we have
If there exists such that then and so we have nothing to prove. Hence, we assume that , for all .
In the following steps, we will complete the proof.
Step I: We will prove that
Using condition (2.24), we obtain
because
If , then
(2.25)
which is a contradiction. Hence, , so from (2.25),
(2.26)
Hence,
By induction,
(2.27)
As , we conclude that
(2.28)
Step II: Now, we prove that the sequence is a b-Cauchy sequence. Suppose the contrary, i.e., is not a b-Cauchy sequence. Then there exists for which we can find two subsequences and of such that is the smallest index for which
(2.29)
This means that
(2.30)
From (2.29) and using the triangular inequality, we get
Taking the upper limit as , we get
(2.31)
From the definition of and the above limits,
Now, from (2.24) and the above inequalities, we have
which is a contradiction. Consequently, is a b-Cauchy sequence. b-Completeness of X shows that b-converges to a point .
Step III: Now we show that u is a fixed point of f,
as f is continuous. □
Theorem 9Under the same hypotheses as Theorem 8, without the continuity assumption off, assume that wheneveris a nondecreasing sequence inXsuch that , for all . Thenfhas a fixed point.
Proof By repeating the proof of Theorem 8, we construct an increasing sequence in X such that . Using the assumption on X we have . Now we show that . By (2.24) we have
(2.32)
where
Letting ,
(2.33)
Again, taking the upper limit as in (2.32) and using Lemma 1 and (2.33),
So we get , i.e., . □
Remark 1 In Theorems 8 and 9, we can replace by the following:
or
Example 2 Let and define the partial order ⪯ on X by
Consider the function given as
which is increasing with respect to ⪯. Let . Hence, , so . Define first the b-metric d on X by , , , and . Then is a b-complete b-metric space with . Let is given by
and . Then
This is because
Also,
because
Also,
Hence, f satisfies all the assumptions of Theorem 5 and thus it has a fixed point (which is ).
Example 3 Let be equipped with the usual order and b-complete b-metric given by with . Consider the mapping defined by and the function β given by . It is easy to see that f is an increasing function and . For all comparable elements , by the mean value theorem, we have
So, from Theorem 5, f has a fixed point.
Example 4 Let be equipped with the usual order and b-complete b-metric d be given by with . Consider the mapping defined by and the function given by , . It is easy to see that f is increasing and . For all comparable elements , using the mean value problem, we have
so, using Theorem 8, f has a fixed point.
3 Application
In this section, we present an application where Theorem 8 can be applied. This application is inspired by [9] (also, see [26] and [27]).
Let be the set of all real continuous functions on . We first endow X with the b-metric
for all where . Clearly, is a complete b-metric space with parameter . Secondly, can also be equipped with a partial order given by
Moreover, as in [9] it is proved that is regular, that is, whenever in X is an increasing sequence such that as , we have for all .
Consider the first-order periodic boundary value problem
(3.1)
where with and is a continuous function.
A lower solution for (3.1) is a function such that
(3.2)
where .
Assume that there exists such that for all we have
(3.3)
Then the existence of a lower solution for (3.1) provides the existence of an unique solution of (3.1).
Problem (3.1) can be rewritten as
Consider
where .
Using the variation of parameters formula, we get
(3.4)
which yields
Since , we get
or
Substituting the value of in (3.4) we arrive at
where
Now define the operator by
The mapping S is nondecreasing [26]. Note that if is a fixed point of S then is a solution of (3.1).
Let . Then we have
or, equivalently,
which shows that
where
or
or
Finally, let α be a lower solution for (3.1). In [26] it was shown that .
Hence, the hypotheses of Theorem 8 are satisfied with . Therefore, there exists a fixed point such that .
Remark 2 In the above theorem, we can replace (3.3) by the following inequality:
(3.5)
for all .
Acknowledgements
The authors are grateful to the referees for valuable remarks that helped them to improve the exposition in the paper.
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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.