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 $g(t) \in M$ such that $f(g(t), t) = g(t)$, and if $f$ is Lipschitz continuous at a point $(g(t),t)$, then $g$ is Lipschitz continuous at $t$.

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

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

Let $f$ be Lipschitz continuous at $(g(t),t)$.

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