Abstract
In this paper, we use interpolation to obtain fixedpoint and common fixed point results for a new type of Kannan contraction mappings in complete metric and -metric spaces. Our results extend and improve some results on fixed point theory in the literature. We also give some examples to illustrate the given results.
1. Introduction and Preliminaries
It is well known that fixed point theory played a central role in various scientific fields. The well-known result in this area is undoubtedly the famous Banach contraction principle (see [1]) which motivated researchers to find other forms of contractions. In this line, we cite the well-known Kannan contraction that does not require continuous mapping.
Definition 1 (see [2]). Let be a metric space. A self-mapping on is said to be a Kannan contraction if there exists such that for all
Kannan obtained the following theorem.
Theorem 2 (see [2]). If is a complete metric space, then every Kannan contraction on has a unique fixed point.
In 2018, Karapinar published a new type of contraction obtained from the definition of the Kannan contraction by interpolation as follows.
Definition 3 (see [3]). Let be a metric space. A self-mapping is said to be an interpolative Kannan-type contraction if there are two constants and such that for all with and
Karapinar obtained the following result.
Theorem 4 (see [3]). Let be a complete metric space and be an interpolative Kannan-type contraction mapping. Then, has a fixed point.
This theorem has been generalized by Noorwali [4] who proposed a common fixed point for two maps, and in 2019, Gabba and Karapinar defined the -interpolative Kannan contraction as follows.
Definition 5 (see [5]). A self-mapping on a metric space is called -interpolative Kannan contraction if with and such that for all with and
And they gave the following theorem.
Theorem 6 (see [5]). Let be a complete metric space and let be an -interpolative Kannan contraction. Then, has a fixed point in
The interpolative method has been used by several researchers to obtain generalizations of other forms of contractions (see [6, 7]).
In this paper, we discuss the results of [4], and we give some generalizations for existence of fixed points for -interpolative Kannan contraction on complete metric spaces and complete -metric spaces.
Let us recall some basic results of -metric spaces:
Definition 7 (see [8, 9]). Let be a nonempty set and be a given real number. A function is a -metric if for all , the following conditions are satisfied:
(b1) if and only if
(b2)
(b3)
The is called a -metric space.
Definition 8 (see [10]). Let be a -metric space (a) is -convergent in if there exists such that as . In this case, we write (b) is a -Cauchy sequence in if as (c) is -complete if every -Cauchy sequence in -converges
Definition 9 (see [11]). Let be a -metric space and An element in is called a coincidence point of and if
Lemma 10 (see [12]). Let be a sequence in a -metric space with such that for some , and each . Then, is a -Cauchy sequence in .
Lemma 11 (see Lemma 2.1 in [11]). Let be a -metric space with and assume that and are -convergent to , respectively, then we have In particular, if , then we have . Moreover, for each , we have
2. Discussion and Results
Using the -interpolative Kannan contraction defined as above, we give our first main result.
Theorem 12. Let be a complete metric space and is self-mapping on such that for all with and , and where and such that If there exists such that then has a fixed point in
Proof. For the case see Karapinar et al. [3]. Assume and define a sequence by and for all integer and assume that for all
We have
Since we obtain by (8)
Assume that there exists a real such that we obtain
which gives
It follows that
where for all with and
Since we have It follows that
which is convergent, and consequently, is a Cauchy sequence in Thus, converges to some Assume that we obtain, by (1):
If we obtain , which is a contradiction. Then,
Example 13. Let be set endowed with the metric defined by and define the self-mapping on by For , and we have for all Moreover, has a fixed point in
In [4], Noorwali gave the following result.
Theorem 14. Let be a complete metric space and be self-mappings. Assume that there are two constants such that the condition is satisfied for all such that and . Then, and have a unique common fixed point.
The uniqueness is not true in the case moreover, in the proof, the author considers the sequence defined by , and In the case where there is no three consecutive identical terms in the sequence the author uses the inequality
We note that the above inequality is not valuable if or Moreover, the following example shows that the theorem is not true in this form.
Example 15. If is the identity map on , it is clear that the result is not valid for any mapping without a fixed point.
Example 16. Let be endowed with the metric defined by the following Table 1 of values.
And we define two self-mappings on by the following matrix of values:
For and , we obtain for all with and , but and have no common fixed point in
As an alternative of this theorem, we give the following result.
Theorem 17. Let be a complete metric space, be two self-mappings on . Assume that there are some such that the followings conditions hold: (i) for all with and (ii) for all with and (iii) for all with and Then, and have a common fixed point.
Proof. Let we define by induction a sequence by , and for all
We shall discuss the following cases:
(a) for all Then, by the same arguments as in the proof of Theorem 2.1 in [4], we can prove that is a Cauchy sequence in which converges to a common fixed point of and Note that this common fixed point is not necessarily unique(b)There exists such that Then, is a common fixed point of and (c)Assume that is without three consecutive identical terms, but it contains a subsequence such that for all We have the following situations(c1) Let such that if for some integer we have
It follows that
and then
And if for some integer we have
By (ii), we obtain
It follows that
From (23) and (26) and since , we obtain
(c2) Assume that and if for some we obtain
which leads to
And if for some we obtain
By (i), we obtain
which leads to
The inequalities (29) and (32) imply that
where is such that and
In view of (27) and (33), we obtain
for all
Let defined by if and if It is clear that is monotone nondecreasing and the sequences and have the same set of values.
Moreover, for all , and we have
for all It follows that is a Cauchy sequence, and then is also a Cauchy sequence. And since is complete, there exists such that
(c3) Assume that we have the following cases:
First case: there exists such that for all Thus, for all And we obtain
It follows that , which is a contradiction.
Second case: there exists a subsequence with for all we have for all Thus,
It follows that , which is a contradiction. Then,
By identical arguments, we can prove that , which ends the proof.
Remark 18. If , the conditions (i), (ii), and (iii) are identical, and we obtain Theorem 2.2 of [3].
Example 19. Let ; we define on a metric and two self-mappings and by
For and we have
(i)
First case: the inequality (39) is obvious.
Second case: and with we have
Third case: and with we have
Fourth case: ; we have
Then, equation (39) is satisfied.
(ii) To prove
we consider the following cases:
First case: the inequality (43) is obvious.
Second case: and with we have
Third case: we have and
(iii) By the same arguments, we prove
First case: the inequality (46) is obvious.
Second case: and with we have
Third case: we have and
The inequalities (39), (43), and (46) show that all conditions of Theorem 17 are satisfied and 1 is a common fixed point of and
Theorem 20. Let be a -complete -metric space, a self-mapping on , and such that for all , . Assume that there exists such that for all with and Then, has a fixed point in
Proof. Let we define a sequence by and for all And assume that otherwise, has a fixed point in We obtain for all integer The sequence is nonincreasing in then it converges to some positive real It follows that there exists such that for large integers. Thus, by Lemma 10, is a -Cauchy sequence in Let such that and assume that we have for all Then, by Lemma 11, we have which is a contradiction, and then is a fixed point of
Definition 21. Let be a -metric space and be self-mappings on . We say that the mapping is an -interpolative Kannan-type contraction if there exist two constants and such that for all with and
Theorem 22. Let be a-complete a-metric space and a -interpolative Kannan-type contraction. Assume that and is closed. If there exist such that (53) holds. Then, and have a coincidence point in .
Proof. Let since we can define inductively a sequence such that and for all integer
If there exists such that then is a coincidence point of and Assume that for all By (53), we obtain
Thus,
Lemma 10 implies that is a -Cauchy sequence, and consequently, is also a -Cauchy sequence. Let such that
And since there exists such that We claim that is a coincidence point of and For this, if we assume that we obtain
By Lemma 11, we obtain
which is a contradiction. Then,
Example 23. Let and the -metric defined on by
is a complete -metric space.
Define two self-mappings and on by for all and
is a -interpolative Kannan contraction for and For this, we discuss the following cases:
(i)If , then (ii)If with we have(iii)If and with we have(iv)If and with we obtainWe conclude that for all such that and And we remark that is a coincidence point of and
Remark 24. In Theorem 22, and need not have a common fixed point.
Example 25. Let be endowed with the metric . We define and by their matrix of values as follows: For and we have 0 and 3 are coincidence points of and , but and have no common fixed point.
Remark 26. If we define , and as in Example 23 and , we obtain for all such that and And has a fixed point.
Data Availability
No data were used to support this study.
Conflicts of Interest
The authors declare that they have no conflicts of interest regarding this article.