Uniform Contraction Mapping Theorem

Theorem
Let $M$ and $N$ be metric spaces.

Let $M$ be complete.

Let $f : M \times N \to M$ be a continuous uniform contraction.

Then for all $t\in N$ there exists a unique $g(t) \in M$ such that $f(g(t), t) = g(t)$, and the mapping $g : N \to M$ is continuous.

Proof
For every $t\in N$, the mapping:
 * $f_t : M \to M : x \mapsto f(x,t)$ is a contraction.

By the Banach Fixed-Point Theorem, there exists a unique $g(t) \in M$ such that $f_t(g(t)) = g(t)$.

We show that $g$ is continuous.

Let $K<1$ be a uniform Lipschitz constant for $f$.

Let $s,t\in N$.

Then

and thus:
 * $d(g(s), g(t)) \leq \dfrac1{1-K}d(f(g(t), s), f(g(t), t))$

The continuity of $g$ now follows from that of $f$ using the definition of product metric

Also see

 * Implicit Function Theorem for Lipschitz Contractions