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 mapping.
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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
| \(\ds \map d {\map g s, \map g t}\) | \(=\) | \(\ds \map d {\map f {\map g s, s}, \map f {\map g t, t} }\) | 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:
| \(\ds \map d {\map g s, \map g t}\) | \(\le\) | \(\ds \dfrac 1 {1 - K} \map d {\map f {\map g t, s}, \map f {\map g t, t} }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \dfrac L {1 - K} \map d {s, t}\) | $f$ is Lipschitz continuous at $\tuple {\map g t, t}$ |
Thus $g$ is Lipschitz continuous at $t$.
$\blacksquare$