FIXED POINT RESULTS FOR GERAGHTY CONTRACTION TYPE MAPPINGS IN b−METRIC SPACES

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract. In this paper, we introduce some new results on the fixed point and common fixed points of Geraghty contraction mappings in b−metric b−complete spaces. Moreover, we give a representative example to illustrate the compatibility of our results.


INTRODUCTION
The famous extensions of the concept of metric spaces have been done by Czerwik [1] where he introduced and studied the concepts of b−metric spaces. Bakhtin [2] uses b-metric spaces as a generalization of metric spaces for find fixed point. After that, several papers have been published on the theory of the fixed point in this space. For additional works and results in bmetric spaces, we encourage readers to refer to the reference ( [3,4,5,6,7,8,9,10,11,12,13]).
In this section, we recall some basic known definitions, notations and results in b−metric spaces which will be used in the sequel. Throughout this article, N, R, R + denote the set of natural numbers, the set of real numbers and the set of positive real numbers, respectively. Definition 1.1. [1]. Let Xbe a nonempty set and s ≥ 1 be a given real number. A function d : X × X → [0, ∞) is said to be a b−metric on X if the following conditions hold: (ii) d(x, y) = d(y, x) for all x, y ∈ X; (iii) d(x, y) ≤ s(d(x, z) + d(z, y)) for all x, y, z ∈ X.
In this case, the pair (X, d) is called a b−metric space.
It is worth mentioning that the class of b−metric spaces is effectively larger than that of the ordinary metric spaces. The following example illustrates the above fact. Example 1. [16]. Let (X, d) be a metric space and let β > 1, λ ≥ 0 and µ > 0. For x, y ∈ X, set ρ(x, y) = λ d(x, y) + µd(x, y) β . Then (X, ρ)is a b-metric space with the parameter s = 2 β −1 and not a metric space on X.
Definition 1.2. [17]. Let (X, d) be a b−metric space, x ∈ X and (x n ) be a sequence in X. Then  (iii) (X, d) is complete if and only if every Cauchy sequence in X is convergent. Remark 1.1. [17]. In a b−metric space (X, d), the following assertions hold: (i) A convergent sequence has a unique limit.
(ii) Each convergent sequence is Cauchy.
(iii) In general, a b−metric is not continuous.
Theorem 1.1. [18]. Let (X, d) be a complete metric space. Let f : X → X be given mapping satisfying: where α ∈ A . Then f has a unique fixed point.
Suppose that there exists β ∈ B such that: where α ∈ B. Then f has a unique fixed point. At

MAIN RESULTS
Theorem 2.1. Let (X, d) be a b−complete b−metric space with parameter s ≥ 1. Let f : X → X be a self-mapping satisfying: and β ∈ B. Then f has a unique fixed point.
Proof. Let x 0 ∈ X be arbitrary and {x n } such that x n = f x n−1 = f n x 0 , ∀ n ∈ N.
If there exists n ∈ N such that x n+1 = x n , then x n is a fixed point of f and the proof is finished.
Otherwise, we have d(x n+1 , x n ) > 0 for all n ∈ N. Using (2.1), we obtain which is a contradiction. Hence, max{d(x n−1 , Since {d(x n−1 , x n )} is a decreasing sequence of non-negative reals. Hence, there exists ρ ≥ 0 such that lim n→∞ d(x n−1 , x n ) = ρ. We will prove that ρ = 0. Suppose on contrary that ρ > 0. Then, Then, From β ∈ B, then lim sup n→∞ β (L (x n−1 , x n )) = 0. Hence, lim n→∞ d(x n−1 , x n ) = 0, which is a contradiction, that is, ρ = 0. Now, we prove that the sequence {x n } is a b−Cauchy sequence. Suppose the contrary. Then there exists ε > 0 for which we can find subsequences {x m(k) } and {x n(k) } of {x n } such that n(k) is the smallest index for which n(k) > m(k) > k and This means that Using (2.5) and the triangular inequality, we get Then, we get From the definition of L (x, y) and the above limits, Using (2.7) and (2.1), we get The completeness of X implies that there exists θ ∈ X such that x n → θ . Next, We will show that θ is a fixed point of f . Using b−triangular inequality and (2.1), we get Taking n → ∞ in the above inequality, we obtain Using (2.8), we get Finally, suppose that the set of fixed point of f is well ordered. Assume on contrary, that θ and Φ are two fixed points of f such that θ = Φ. Using (2.1), we get It is easy to check that (X, d) is a b−metric space with constant s = Proof. Let x 0 be arbitrary. Define the sequence {x n } in X by x 2n+1 = f x 2n and x 2n+2 = gx 2n+1 for all n = 0, 1, . . . . Using (2.11), for all n = 0, 1, 2, . . . , we get a contradiction. So, we have L (x 2n , x 2n+1 ) = d(x 2n , x 2n+1 ). Using (2.12), we get Hence, we have d(x n , x n+1 ) ≤ d(x n−1 , x n ). Therefore {d(x n , x n+1 )} is a nonincreasing sequence, hence there exists ρ ≥ 0 such that d(x n , x n+1 ) → ρ as n → ∞. We will show that ρ = 0. Suppose on the contrary that ρ > 0. Taking n → ∞ in (2.13), we get This is, lim n→∞ d(x 2n , x 2n+1 ) = 0. Now, we prove that the sequence {x 2n } is a b−Cauchy sequence.
Suppose the contrary. Then there exists ε > 0 for which we can find subsequences {x 2m(k) } and {x 2n(k) } of {x 2n } such that n(k) is the smallest index for which n(k) > m(k) > k and This means that Using (2.11) and (2.15), we get Taking k → ∞, we get Using the b−triangular inequality, we get Taking k → ∞ in (2.18), we get Using (2.15) and the b−triangular inequality, we get Taking k → ∞ in (2.21) and using (2.22), we obtain lim sup k→∞ d(x 2n(k) , x 2m(k) ) = 0.
This contradicts (2.15). This implies that {x 2n } is a b−Cauchy sequence and hence there exists θ ∈ X such that lim n→∞ x n = θ . If f is continuous, we get Using (2.11), we obtain From β ∈ B we conclude that Hence, gθ = θ . If g is continuous, then, by a similar argument to that of above, one can show that f , g have a common fixed point. Now, we prove the uniqueness of the common fixed point.
Let y = f y = gy, is another common fixed point for f and g. Using (2.11), we obtain Hence, θ = y and the common fixed point f and g is unique. Proof. Taking f = g in Theorem 2.2, we get the following result. Then, the conditions of Corollary 2.1 are satisfied.