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 $g \left({t}\right) \in M$ such that $f(g \left({t}\right), t) = g \left({t}\right)$, and the mapping $g: N \to M$ is continuous.

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

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

We show that $g$ is continuous.

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

Let $s, t \in N$.

Then

and thus:
 * $d \left({g \left({s}\right), g \left({t}\right)}\right) \le \dfrac 1 {1 - K} d \left({f \left({g \left({t}\right), s}\right), f \left({g \left({t}\right), t}\right)}\right)$

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

Also see

 * Implicit Function Theorem for Lipschitz Contractions