Let X be complete, F and g be continuous and compatible.
T and G are continuous and compatible and (M x,y,cdot ):mathbb{R}^rightarrowmathbb{I}) is continuous or.
Suppose F(X × X) ⊆ g(X), g is continuous and compatible with F and also suppose either.
Suppose that either F and g are continuous and compatible and (M x,y,cdot):mathbb{R}^rightarrowmathbb{I}) is continuous or condition (C6) holds.
They also proved the existence of a random coincidence point for a pair of reciprocally continuous and compatible single-valued and multivalued operators.
However, even though S and T are continuous, weakly compatible, and have a unique point of coincidence and therefore a unique common fixed point, S and T may have more than one coincident point.
f is g-nondecreasing with respect to ⪯ and f ( X ) ⊆ g ( X ) ; there exists x 0 ∈ X such that g x 0 ⪯ f x 0 ; f and g are continuous and compatible and ( X, d ) is complete, or.
(f(X) subseteq g(X)); there exists (x_0 in X) such that (alpha (gx_0, fx_0, fx_0 ge 1); f and g are continuous and compatible and ((X, G)) is complete, or.
Further, since f and g are continuous and compatible, we get that lim n → ∞ f g x n = f z, lim n → ∞ g f x n = g z. and lim n → ∞ d ( f g x n, g f x n ) = 0. We will show that f z = g z.
