eu SOME COMMON RANDOM FIXED POINT THEOREMS OF COMPATIBLE MAPPINGS OF VARIOUS TYPES FOR RANDOM MAPPINGS IN MULTIPLICATIVE METRIC SPACES

In this paper first, we introduce the notions of compatible mappings of various types in framework of random fixed points in multiplicative metric spaces and using these notions we prove common random fixed point theorems for these mappings. AMS Subject Classification: 47H10, 54H25


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. [1] studied the multiplicative calculus and defined a new distance so called multiplicative distance.By using this idea, Özavsar and C ¸evikel [3] introduced the concept of multiplicative metric spaces and studied some topological properties in such spaces.Definition 1.1.( [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.( [3]) 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.Remark 1.3.( [6]) We note that multiplicative metrics and metric spaces are independent.
One can refer to [3] for detailed multiplicative metric topology.Definition 1.4.Let (X, d) be a multiplicative metric space.Then a sequence {x n } in X is said to be (1) a multiplicative convergent to x if for every multiplicative open ball B ǫ (x) = {y | d(x, y) < ǫ}, ǫ > 1, there exists N ∈ N such that x n ∈ B ǫ (x) for all n ≥ N, that is, d(x n , x) → 1 as n → ∞.
(2) a multiplicative Cauchy sequence if for all ǫ > 1, there exists N ∈ N such that d(x n , x m ) < ǫ for all m, n ≥ N , that is, d(x n , x m ) → 1 as n, m → ∞.
(3) We call a multiplicative metric space complete if every multiplicative Cauchy sequence in it is multiplicative convergent to x ∈ X.
In 2012, Özavsar and C ¸evikel [3] gave the concept of multiplicative contraction mapping and proved some fixed point theorem of such mappings on a complete multiplicative metric spaces.Definition 1.5.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 number λ ∈ [0, 1) such that d(f x, f y) ≤ d λ (x, y) for all x, y ∈ X.

Relationships and Properties of Compatible Mappings and its
Variants for Random Mappings Now we introduce the following concepts.
Definition 2.1.Let (X, d) be a multiplicative metric space and F : R × X → X be a mapping, where X is a nonempty set.Then a mapping g : R → X is said to be a random fixed point of the mapping 6 and a mapping g : R → X defined by g(t) = t for every t ∈ R. Then T has a unique random fixed point in X Next we introduce the notions of compatible mappings and its variants for random mappings in multiplicative metric spaces as follows: Definition 2.3.Let (X, d) be a multiplicative metric space.Two mappings A and B : R × X → X are called (1)  as derived.
Proposition 2.7.Let (X, d) be a multiplicative metric space and A and B : R × X → X be jointly continuous.If A and B are compatible of type (B), then they are compatible of type (A).
Proof.Let {g n } be a sequence in X such that lim n→∞ A(t, g n (t)) = lim n→∞ B(t, g n (t)) = g(t), t ∈ R.
Since A and B are jointly continuous, we have Therefore, A and B compatible of type (A).This completes the proof.
Proposition 2.8.Let (X, d) be a multiplicative metric space and A and B : R × X → X be jointly continuous.If A and B are compatible of type (B), then they are compatible.
Since A and B are jointly continuous, we have By multiplicative triangle inequality, we have Letting n → ∞, we have Therefore, A and B are compatible.This completes the proof.
Proposition 2.9.Let (X, d) be a multiplicative metric space and A and B : R × X → X be jointly continuous.If A and B are compatible, then they are compatible of type (B).
Proof.Since A and B are compatible so there exists {g n } be a sequence in Since A and B are jointly continuous, we have Proof.(a) One can easily prove it using Propositions 2.8 and 2.9.(b) One can easily prove it using Propositions 2.6 and 2.7.Proposition 2.11.Let (X, d) be a multiplicative metric space and A and B : R × X → X be mappings.Assume that A and B are compatible mappings of type (B).If A(t, g(t)) = B(t, g(t)) for some g(t) ∈ X, then Proof.Suppose that {g n } is a sequence in X defined by g n (t) = g(t), n = 1, 2, . . ., for some t ∈ R and A(t, g(t)) = B(t, g(t)) = z(t), say.Then we have A(t, g n (t)), B(t, g n (t)) → A(t, g(t)) as n → ∞.Since A and B are compatible of type (B), we have Hence we have A(t, B(t, g(t)) = B(t, B(t, g(t)).Therefore, since A(t, g(t)) = B(t, g(t)), we have This completes the proof.
From Proposition 2.11, we have Proposition 2.12.Let (X, d) be a multiplicative metric space and A and B : R×X → X be mappings.Assume that lim n→∞ Therefore, lim n→∞ B(t, B(t, g n (t))) = A(t, g(t)).This completes the proof.
(b) The proof of lim n→∞ A(t, A(t, g n (t))) = B(t, g(t)) follows by similar arguments as in (a).
(c) Suppose that A and B are jointly continuous.Since B(t, g n (t)) → g(t) as n → ∞ and A is jointly continuous, by (a), B(t, B(t, g n (t))) → A(t, g(t)) as n → ∞.On the other hand, B is also jointly continuous and B(t, B(t, g n (t))) → B(t, g(t)).Thus, we have A(t, g(t)) = B(t, g(t)) by the uniqueness of limit and so by Proposition 2.11, A(t, B(t, g(t))) = B(t, A(t, g(t))).This completes the proof.
Remark 2.13.In Proposition 2.11, assume that A and B be compatible mappings of type (C) or of type (P ) instead of type (B), the conclusion of Proposition 2.11 still holds.
Remark 2.14.In Proposition 2.12, assume that A and B be compatible mappings of type (C) or of type (P ) instead of of type (B), the conclusion of Proposition 2.12 still holds.

Random Fixed Points for Compatible Mappings of Various Types
Now, we give the random fixed point theorem of compatible mappings of type (A) for random mappings.Theorem 3.1.Let (X, d) be a complete multiplicative metric space and A, B, S and T : R × X → X be mappings satisfying the following conditions; S(t, X) ⊂ B(t, X) and T (t, X) ⊂ A(t, X); for all x, y ∈ X and t ∈ R, where λ ∈ (0, 1/2); (C 3 ) one of A, B, S and T is jointly continuous.Assume that the pairs A, S and B, T are compatible of type (A).Then A, B, S and T have a unique common random fixed point.
Proof.Suppose that A is jointly continuous.Since A and S be compatible of type (A), by Proposition 2.5, A and S be compatible so results easily follows from [5,Theorem 3.1].
Similarly, we can complete the proof when B or S or T is jointly continuous.This complete the proof.
Next, we give the random fixed point theorem of compatible mappings of type (B) for random mappings.Theorem 3.2.Let (X, d) be a complete multiplicative metric space and A, B, S and T : R × X → X be mappings satisfying the conditions (C 1 )-(C 3 ).
Assume that the pairs A, S and B, T are compatible of type (B).Then A, B, S and T have a unique common random fixed point.
Proof.Let g 0 : R → X be an arbitrary mapping.By (C 1 ), there exists g 1 : R → X such that B(t, g 1 (t)) = S(t, g 0 (t)), t ∈ R and for this g 1 : R → X we can choose g 2 : R → X such that T (t, g 1 (t)) = A(t, g 2 (t)), t ∈ R, and so on.By the method of induction we can define a sequence {y n (t)}, t ∈ R, of mappings as follows: Letting n → ∞, we obtain This implies that d(z(t), T (t, z(t))) = 1 and hence T (t, z(t)) = z(t).Since T (t, X) ⊂ A(t, X), so there exists a point v(t) ∈ X such that z(t) = T (t, z(t)) = A(t, v(t)).Putting x(t) = v(t) and y(t) = z(t) in (C 2 ), we have This implies that S(t, v(t)) = z(t).Since A and S are compatible (B) and Letting n → ∞, we have This implies that d(S(t, z(t)), z(t)) = 1 and hence S(t, z(t)) = z(t).Since z(t) = S(t, z(t)) ∈ S(t, X) ⊂ B(t, X), there exists w(t) ∈ X such that z(t) = B(t, w(t)), t ∈ R.
Putting x(t) = z(t) and y(t) = w(t) in (C 2 ), we have This implies that d(z(t), T (t, w(t)) = 1 and hence T (t, w(t)) = z(t).Since B and T are compatible of type (B) and B(t, w(t)) = T (t, w(t)) = z(t), by Proposition 2.11, we have Also, putting x(t) = z(t) and y(t) = z(t) in (C 2 ), we have Similarly, we can prove theorem when B is jointly continuous.Uniqueness follows easily from (C 2 ).Therefore, A, B, S and T have a unique common fixed point.This completes the proof.
Next, we give the random fixed point theorem of compatible mappings of type (C) for random mappings.Theorem 3.3.Let (X, d) be a complete multiplicative metric space and A, B, S and T : R × X → X be mappings satisfying the conditions (C 1 )-(C 3 ).
Assume that the pairs A, S and B, T are compatible of type (C).Then A, B, S and T have a unique common random fixed point.
Proof.From the proof of Theorem 3.2, {y n (t)} is a multiplicative Cauchy sequence and since X is complete so {y n (t)} converges to a point z(t) ∈ X as n → ∞.Also subsequences of {y n (t)} also converges to a point z(t) ∈ X, that is Therefore z(t) is a common random fixed point of A, B, S and T.
Similarly, we can complete the proof when A or B or T is jointly continuous.The uniqueness follows easily.Therefore A, B, S and T have a unique common random fixed point.This completes the proof.
Finally, we give the random fixed point theorem of compatible mappings of type (P ) for random mappings.Theorem 3.4.Let (X, d) be a complete multiplicative metric space and A, B, S and T : R × X → X be mappings satisfying the conditions (C 1 )-(C 3 ).
Assume that the pairs A, S and B, T are compatible of type (P ).Then A, B, S and T have a unique common random fixed point.
Proof.From the proof of Theorem 3.2, {y n (t)} is a multiplicative Cauchy sequence and since X is complete so {y n (t)} converges to a point z(t) ∈ X as n → ∞.Also subsequences of {y n (t)} also converges to a point z( This implies that d(z(t), T (t, u(t)) = 1 and hence T (t, u(t)) = z(t).Therefore, B(t, u(t)) = T (t, u(t)) = z(t).
Therefore z(t) is a common random fixed point of A, B, S and T.
Similarly, we can complete the proof when A or B or T is jointly continuous.The uniqueness follows easily.Therefore A, B, S and T have a unique common random fixed point.This completes the proof.Remark 3.5.If we take R to be a singleton set, then the results can reduce into the result of Kang et al. [2].
where {g n } is a sequence Let (X, d) be a multiplicative metric space and A and B : R×X → X be mappings.Assume that A and B are compatible mappings of type (A).If one of A and B is jointly continuous, then A and B are compatible.