Implicit Function Theorem for Lipschitz Contraction at Point

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

Let $M$ be complete.

Let $f: M \times N \to M$ be a 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 if $f$ is Lipschitz continuous at a point $\tuple {\map g t, t}$, then $g$ is Lipschitz continuous at $t$.

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 $\map {f_t} {\map g t} = \map g t$.

Let $f$ be Lipschitz continuous at $\tuple {\map g t, t}$.

We show that $g$ is Lipschitz continuous at $t$.

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

Let $L$ be a Lipschitz constant for $f$ at $a$.

Let $s\in N$.

Then

and thus:

Thus $g$ is Lipschitz continuous at $t$.

Also see

 * Implicit Function Theorem for Lipschitz Contractions