Some Remarks on Fixed Point Theorems for Interpolative Kannan Contraction

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 
 
 b
 
 -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.


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 E be a metric space. A self-mapping on E is said to be a Kannan contraction if there exists λ ∈ ½0 , 1/2½ such that for all x, y ∈ E: Kannan obtained the following theorem.
Theorem 2 (see [2]). If ðE, dÞ is a complete metric space, then every Kannan contraction on E 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 ðE, dÞ be a metric space. A selfmapping T : E → E is said to be an interpolative Kannantype contraction if there are two constants λ ∈ ½0, 1½ and α ∈ 0, 1½ such that And they gave the following theorem.
Theorem 6 (see [5]). Let ðX, dÞ be a complete metric space and let T : X → X be an ðλ, α, βÞ -interpolative Kannan contraction. Then, T has a fixed point in X: 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 ðλ, α, βÞ-interpolative Kannan contraction on complete metric spaces and complete b-metric spaces.
Let us recall some basic results of b-metric spaces: Definition 7 (see [8,9]). Let X be a nonempty set and s ≥ 1 be a given real number. A function d : Definition 8 (see [10]). Let ðX, dÞ be a b-metric space Definition 9 (see [11]). Let ðX, dÞ be a b -metric space and g, h : X → X: An element x in X is called a coincidence point of g and h if gx = hx Lemma 10 (see [12]). Let fx n g be a sequence in a b -metric space ðX, d, sÞ with s ≥ 1 such that for some λ ∈ ½0, 1½, and each n ∈ ℕ. Then, fx n g is a b-Cauchy sequence in ðX, d, sÞ.
Lemma 11 (see Lemma 2.1 in [11]). Let ðX, dÞ be a b-metric space with s ≥ 1 and assume that fx n g and fy n g are b -convergent to x, y, respectively, then we have

Discussion and Results
Using the ðλ, α, βÞ-interpolative Kannan contraction defined as above, we give our first main result.
Theorem 12. Let ðX, dÞ be a complete metric space and T is self-mapping on X such that for all x, y ∈ X with x ≠ Tx and y ≠ Ty, and where λ ∈ 0, 1½ and α, β ∈ 0, 1½ such that α + β ≥ 1: If there exists x ∈ X such that dðx, TxÞ ≤ 1, then T has a fixed point in X: Proof. For the case α + β = 1, see Karapinar et al. [3]. Assume α + β > 1 and define a sequence fx n g by x 0 = x and x n+1 = T x n , for all integer n and assume that x n ≠ Tx n , for all n: We have Since dðx 0 , Assume that there exists a real γðnÞ such that dðx n , x n+1 Þ ≤ λ γðnÞ ; we obtain which gives It follows that where γðn + 1Þ = 1/ð1 − βÞð1 + αγðnÞÞ, for all n ≥ 1 with γð0Þ = 0 and γð1Þ = 1: Since α/ð1 − βÞ > 1, we have lim n γðnÞ = +∞: It follows that which is convergent, and consequently, fx n g is a Cauchy sequence in ðX, dÞ: Thus, fx n g converges to some x ∈ X: Assume that x ≠ T x, we obtain, by (1): If n → +∞, we obtain dð x, T xÞ = 0, which is a contradiction. Then, T x = x: Journal of Function Spaces Example 13. Let X = fx, y, z, wg be set endowed with the metric d defined by and define the self-mapping T on X by For λ = 99/100,α = 5/8, and β = 1/2, we have for all u, v ∈ X − fxg: Moreover, T has a fixed point in X: In [4], Noorwali gave the following result.
Theorem 14. Let ðM, dÞ be a complete metric space and S, T : M → M be self-mappings. Assume that there are two constants λ ∈ ½0, 1½,α ∈ 0, 1½ such that the condition is satisfied for all p, q ∈ M such that Tp ≠ p and Sq ≠ q. Then, S and T have a unique common fixed point.
The uniqueness is not true in the case S = T ; moreover, in the proof, the author considers the sequence defined by p 0 ∈ M,p 2n+1 = Tp 2n , and p 2n+2 = Sp 2n+1 : In the case where there is no three consecutive identical terms in the sequence fp n g, the author uses the inequality We note that the above inequality is not valuable if p 2n = p 2n+1 or p 2n+1 = p 2n+2 : Moreover, the following example shows that the theorem is not true in this form.
Example 15. If S is the identity map on M, it is clear that the result is not valid for any mapping T without a fixed point.
Example 16. Let M = fp, q, r, sg be endowed with the metric defined by the following Table 1 of values.
And we define two self-mappings on M by the following matrix of values: For λ = 99/100 and α = 3/5, we obtain for all x, y ∈ M with Sx ≠ x and Ty ≠ y, but S and T have no common fixed point in M: As an alternative of this theorem, we give the following result. Proof. Let x ∈ M ; we define by induction a sequence fx n g by x 0 = x,x 2n+1 = Sx 2n , and x 2n+2 = Tx 2n+1 , for all n: We shall discuss the following cases: (a) dðx n , x n+1 Þ > 0, for all n: Then, by the same arguments as in the proof of Theorem 2.1 in [4], we can prove that fx n g n is a Cauchy sequence in M which converges to a common fixed point of S and T: Note that this common fixed point is not necessarily unique (b) There exists n 0 such that x n 0 = x n 0 +1 = x n 0 +2 : Then, x n 0 is a common fixed point of S and T (c) Assume that fx n g is without three consecutive identical terms, but it contains a subsequence fx n k g such that x n k = x n k +1 , for all k ∈ ℕ: We have the following situations (c1) Let p ∈ ℕ such that x p = x p+1 ; if p = 2n for some integer n, we have and then And if p = 2n − 1 for some integer n, we have By (ii), we obtain It follows that From (23) and (26) and since λ 1/α ≤ λ, we obtain (c2) Assume that x p−1 ≠ x p and x p ≠ x p+1 ; if p = 2n, for some n ∈ ℕ, we obtain which leads to And if p = 2n + 1, for some n ∈ ℕ, we obtain By (i), we obtain which leads to The inequalities (29) and (32) imply that where p is such that x p−1 ≠ x p and x p ≠ x p+1 : In view of (27) and (33), we obtain for all p ∈ ℕ: Let φ : ℕ → ℕ defined by φðpÞ = p if x p ≠ x p+1 and φðpÞ = p + 1 if x p = x p+1 : It is clear that φ is monotone nondecreasing and the sequences fx n g and fx φðnÞ g have the same set of values.
Moreover, x φðpÞ ≠ x φðp+1Þ , for all p ∈ ℕ * , and we have for all p ∈ ℕ: It follows that fx φðnÞ g is a Cauchy sequence, and then fx n g is also a Cauchy sequence. And since M is complete, there exists u ∈ M such that lim n dðx n , uÞ = 0: (c3) Assume that Su ≠ u ; we have the following cases: First case: there exists n 0 such that x 2n = x 2n+1 , for all n ≥ n 0 : Thus, Tx 2n−1 ≠ x 2n−1 , for all n ≥ n 0 : And we obtain It follows that lim n dðSu, x 2n Þ = dðSu, uÞ = 0, which is a contradiction.
Second case: there exists a subsequence fx 2n k g k with x 2n k ≠ x 2n k +1 , for all k ; we have Sx 2n k ≠ x 2n k , for all k: Thus, It follows that lim k dðSu, x 2n k +1 Þ = dðSu, uÞ = 0, which is a contradiction. Then, Su = u: By identical arguments, we can prove that Tu = u, which ends the proof.
Remark 18. If S = T, the conditions (i), (ii), and (iii) are identical, and we obtain Theorem 2.2 of [3]. 4 Journal of Function Spaces Example 19. Let X = ½0,+∞½; we define on X a metric d and two self-mappings S and T by The inequalities (39), (43), and (46) show that all conditions of Theorem 17 are satisfied and 1 is a common fixed point of S and T: