Let X and be metric spaces, and let f X X where X is complete For every , the function f(x) f (x, ) is contraction mapping on X with modulus f is continuous in , that is for every 0 , lim 0 f(x) f0(x) for every x X Then f has a unique fixed point x for every and lim0 x x0 Although there are many di...

By the Banach fixed point theorem for every there exists ...

Let X and be metric spaces, and let f: X X where

Question:

Let X and Θ be metric spaces, and let f: X × Θ → X where
• X is complete
• For every θ ∊ Θ, the function fθ(x) = f (x, θ) is contraction mapping on X with modulus β
• f is continuous in θ, that is for every θ0 ∊ Θ, limθ→ θ0 fθ(x) = fθ0(x) for every x ∊ X
Then fθ has a unique fixed point xθ for every θ ∊ Θ and limθ→θ0 xθ = xθ0.
Although there are many direct methods for solving systems of linear equations, iterative methods are sometimes used in practice. The following exercise outlines one such method and devises a sufficient condition for convergence.