Probabilistic G-Metric space and some fixed point results
Journal of Nonlinear Analysis and Application
In this note we introduce the notions of generalized probabilistic metric spaces and generalized Menger probabilistic metric spaces. After making our elementary observations and proving some basic properties of these spaces, we are going to prove some fixed point result in these spaces. In the following two examples, we construct two probabilistic G-metric space using a PM-space and a G-metric space, respectively. Commutativity of G m is trivial by commutativity of F. For proving G 5 , let p,
... ving G 5 , let p, q, r, s ∈ X. We have Also if a proper probabilistic G-metric is of the form (2.1), then F p,q,r is a G-metric. Suppose (X, G, τ) is a proper probabilistic G-metric space and there exists a function F : X × X × X → R + , such that G p,q,r = ε F p,q,r then (X, F) is a G-metric space. which implies that (X, G) is a G-metric space. In the following proposition, it is proved that we may construct a probabilistic G-metric space using a pseudo probabilistic G-metric space. To do this, we introduce the following relation: Let (X, G, τ) be a probabilistic pseudo G-metric space. For p, q ∈ X, we say p ∼ q if and only if G(p, p, q) = ε 0 and G(p, q, q) = ε 0 . This relation is an equivalence relation. Indeed if p ∼ q and q ∼ r, then G(p, p, q) = ε 0 , G(p, q, q) = ε 0 and G(q, q, r) = ε 0 , G(r, r, q) = ε 0 But G is a probabilistic pseudo G-metric, so G(p, p, r) = G(r, p, p) ≥ τ(G(r, q, q), G(q, p, p)) = τ(ε 0 , ε 0 ) = ε 0 , which implies that G(p, p, r) ≥ ε 0 . Now maximality of ε 0 implies that G(p, p, r) = ε 0 . Similarly G(p, r, r) = ε 0 . This prove that ∼ is transitive. The other properties of ∼ to be an equivalence relation is trivial. Proposition 2.1. Let (X, G, τ) be a probabilistic pseudo G-metric space, for every p ∈ S, let p ⋆ denote the equivalence class of p and let X ⋆ denotes the set of these equivalence classes. Then the expression is a probabilistic G-metric space, the quotient space of (X, G, τ). Suppose t 1 > G ⋆ (p, s, s) and t 2 > G ⋆ s,q,r are upper bounds for B and C, respectively. Then G(p, s, s)(t 1 ) > 1 − t 1 and G s,q,r (t 2 ) > 1 − t 2 . Therefore G p,q,r (t 1 + t 2 ) ≥ T L (G p,s,s (t 1 ), G s,q,r (t 2 )) ≥ G p,s,s (t 1 ) + G s,q,r (t 2 ) − 1 > 1 − (t 1 + t 2 ). Thus t 1 + t 2 is an upper bound for A.