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

Proof

For every $t\in N$, the mapping:

$f_t: M \to M : x \mapsto \map f {x, t}$ is a contraction.

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

We show that $g$ is continuous.

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

Let $s, t \in N$.

Then

\(\ds \map d {\map g s, \map g t}\)

\(=\)

\(\ds \map d {\map f {\map g s, s}, \map f {\map g t, t} }\)

Definition of $g$

\(\ds \)

\(\le\)

\(\ds \map d {\map f {\map g s, s}, \map f {\map g t, s} } + \map d {\map f {\map g t, s}, \map f {\map g t, t} }\)

Definition of Metric

\(\ds \)

\(\le\)

\(\ds K \cdot \map d {\map g s, \map g t} + \map d {\map f {\map g t, s}, \map f {\map g t, t} }\)

$f$ is a uniform contraction

and thus:

$\map d {\map g s, \map g t} \le \dfrac 1 {1 - K} \map d {\map f {\map g t, s}, \map f {\map g t, t} }$

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

$\blacksquare$

Also see

• Implicit Function Theorem for Lipschitz Contractions