FIXED POINT POINTS OF RATIONAL TYPE CONTRACTIONS IN MULTIPLICATIVE METRIC SPACES

In this paper, we prove common fixed point results for mappings satisfying some contractive condition in multiplicative metric spaces. We also give examples in support of our results.


Introduction and Preliminaries
It is well know that the set of positive real numbers R + is not complete according to the usual metric.To overcome this problem, in 2008, Bashirov et al. [2] introduced the concept of multiplicative metric spaces as follows: Definition 1.1.Let X be a nonempty set.A multiplicative metric is a mapping d : X × X → R + satisfying the following conditions: (i) d(x, y) ≥ 1 for all x, y ∈ X and d(x, y) = 1 if and only if x = y; (ii) d(x, y) = d(y, x) for all x, y ∈ X; (iii) d(x, y) ≤ d(x, z) • d(z, y) for all x, y, z ∈ X (multiplicative triangle inequality).
Then the mapping d together with X, that is, (X, d) is a multiplicative metric space.
Example 1.2.( [6]) Let R n + be the collection of all n-tuples of positive real numbers.Let d * : R n + × R n + → R be defined as follows: , where x = (x 1 , . . ., x n ), y = (y 1 , . . ., y n ) ∈ R n + and | • | * : R + → R + is defined by Then it is obvious that all conditions of a multiplicative metric are satisfied.Therefore (R n + , d * ) is a multiplicative metric space.Example 1.3.( [7]) Let d : R × R → [1, ∞) be defined as d(x, y) = a |x−y| for all x, y ∈ R, where a > 1.Then d is a multiplicative metric and (R, d) is a multiplicative metric space.We may call it usual multiplicative metric spaces.
Remark 1.4.We note that the Example 1.2 is valid for positive real numbers and Example 1.3 is valid for all real numbers.Example 1.5.( [7]) Let (X, d) be a metric space.Define a mapping d a on X by for all x, y ∈ X, where a > 1.Then d a is a multiplicative metric and (X, d a ) is known as the discrete multiplicative metric space.
Remark 1.7.( [7]) We note that multiplicative metric and metric spaces are independent.
Indeed, the mapping d * defined in Example 1.2 is multiplicative metric but not metric as it does not satisfy triangular inequality.Consider On the other hand the usual metric on R is not multiplicative metric as it doesnt satisfy multiplicative triangular inequality, since One can refer to [3,6] for detailed multiplicative metric topology.Definition 1.8.Let (X, d) be a multiplicative metric space.Then a sequence {x n } in X said to be (1) a multiplicative convergent to x if for every multiplicative open ball (2) a multiplicative Cauchy sequence if for all ǫ > 1, there exists (3) We call a multiplicative metric space complete if every multiplicative Cauchy sequence in it is multiplicative convergent to x ∈ X.
Remark 1.9.The set of positive real numbers R + is not complete according to the usual metric.Let X = R + and the sequence {x n } = { 1 n }.It is obvious {x n } is a Cauchy sequence in X with respect to usual metric and X is not a complete metric space, since 0 / ∈ R + .In case of a multiplicative metric space, we take a sequence {x n } = {a is a complete multiplicative metric space.
In 2012, Özavsar and C ¸evikel [6] gave the concept of multiplicative contraction mappings and proved some fixed point theorems of such mappings in a multiplicative metric space.Definition 1.10.Let f be a mapping of a multiplicative metric space (X, d) into itself.Then f is said to be a multiplicative contraction if there exists a real constant λ ∈ [0, 1) such that Also they proved the Banach Contraction Principle in the setting of multiplicative metric spaces as follows: Theorem 1.11.Let f be a multiplicative contraction mapping of a complete multiplicative metric space (X, d) into itself.Then f has a unique fixed point.
In 1996, Jungck [4] introduce the notion of weakly compatible mappings in a metric space.Now, we introduce the notions in multiplicative metric spaces Definition 1.12.Let f and g be two mappings of a multiplicative metric space (X, d) into itself.Then f and g are said to be weakly compatible if they commute at coincidence points, that is, if f t = gt for some t ∈ X implies that f gt = gf t.
Khan et al. [5] initiated the use of the control function as follows: is called an alternating distance function if the following properties are satisfied: (1) φ is increasing and continuous, (2) φ(t) = 1 if and only if t = 1.
In our results we will use the following class of functions.

Main Results
Now we prove some fixed point results as follows: Theorem 2.1.Let f be a continuous mapping of a complete multiplicative metric space (X, d) into itself such that for all x, y ∈ X where φ ∈ Φ and ψ ∈ Ψ, and Then f has a unique fixed point.
Proof.Let x 0 ∈ X be arbitrary.Then there exists x 1 ∈ X such that If there exists some n ∈ N such that x ( n + 1) = x n .Then we have x n+1 = f x n = x n , which implies that x n is a fixed point of f .Suppose that where Therefore, we have 2), we have Since {R n } is a decreasing sequence of positive real numbers and it is bounded below, there exists r ≥ 1 such that as n → ∞.Now we shall show that r = 1.Assume that r > 1. Taking limit on both sides (2.3) and using (2.4), the property of ψ and the continuity of φ, we get , as n → ∞.
Next to show that {x n } is a multiplicative Cauchy sequence.Suppose that {x n } is not a multiplicative Cauchy sequence.Then there exists an ǫ > 1 for which we can find two sequences of positive integers {m(k)} and {n(k)} such that for all positive integers k, n(k) > m(k) ≥ k and Letting k → ∞ and using (2.5), we have and Letting k → ∞ in above inequalities and using (2.5) and (2.6), we have Letting k → ∞ in the above inequalities and using (2.5) and (2.7), we have Similarly, we have (2.10) and (2.11) Letting k → ∞ in (2.10) and (2.11), using (2.5), (2.7), (2.8) and (2.9), we have lim (2.13) From (2.1), using (2.10) and (2.11), we have ) .
Taking limit on both the sides and using (2.6), (2.12) and (2.13), the property of ψ and the continuity of φ, we have , which is a contradiction by property of ψ.Thus {x n } is a multiplicative Cauchy sequence in X.Since X is complete, there exists u ∈ X such that lim n→∞ x n = u.Then using continuity of f, we get Hence u is a fixed point of f.Finally, we shall prove the uniqueness of the fixed point of f.Suppose that u and v (u = v) be two fixed points of f.Consider where φ ∈ Φ, ψ ∈ Ψ and and Therefore, we have which implies that ψ(d(u, v)) ≤ 1, which is a contraction by definition of ψ.Hence u = v.Therefore f has a unique fixed point.This completes the proof.
Next we prove the following result without the condition of continuity of f.Theorem 2.2.Let f be a mapping of a complete multiplicative metric space (X, d) into itself such that for all x, y ∈ X where φ ∈ Φ and ψ ∈ Ψ, and Then f has a unique fixed point.
Proof.From the proof of Theorem 2.1 {x n } is a multiplicative Cauchy sequence and hence there exists u ∈ X such that lim where and Letting n → ∞ in (2.15) and (2.16),we have Again letting n → ∞ in (2.14), using (2.15), (2.16) and property of φ and ψ, we have which implies that lim n→∞ ψ(N (x n , u)) ≤ 1, which is a contradiction by property of ψ.Therefore, f u = u and hence u is a fixed point of f.
Uniqueness easily follows from Theorem 2.1.This completes the proof.
which implies that ψ(d(y n+1 , y n )) ≤ 1, which is a contradiction.Therefore, M (x n , x n+1 ) = N (x n , x n+1 ) = d(y n−1 , y n ) and we have which implies that Since φ is increasing, therefore d(y n , y n+1 ) < d(y n−1 , y n ), which implies that {d(y n , y n+1 )} is decreasing sequence and bounded below.Hence it converges to some positive number r ≥ 1.Now we prove that r = 1.Letting n → ∞ in (2.17), and using the continuity of φ, we have , which implies that lim (2.18) Next to show that {y n } is a multiplicative Cauchy sequence.Suppose that {y n } is not a multiplicative Cauchy sequence.Then there exists an ǫ > 1 for which we can find two sequences of positive integers {m(k)} and {n(k)} such that for all positive integers k, n(k) > m(k) ≥ k and d(y m(k) , y n(k) ) ≥ ǫ.Assume that n(k) is the smallest such positive integer, we get n Letting k → ∞ and using (2.18), we have and Letting k → ∞ in above inequality and using (2.18) and (2.19), we have Letting k → ∞ in the above inequalities and using (2.20), we have Similarly, we have lim where  Therefore the coincidence point of f and g is unique.
Finally suppose that f and g are weakly compatible.Since f u = gu = p, hence f gu = gf u, that is, f p = gp.Using uniqueness of the coincidence point of f and g, we have p = u.Thus u = f u = gu.Hence u is a unique common fixed point of f and g.This completes the proof.