Theorem 8.1.3 (Banach’s Fixed Point Theorem).label Let $(X, d)$ be a metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that
\[d(f(x), f(y)) \le Cd(x, y) \quad \forall x, y \in X\]
then:
- (1)
There exists a unique $x \in X$ such that $f(x) = x$.
- (2)
For any $y \in X$, $\limv{n}f^{n}(y) = x$.
Proof. Let $x_{0} \in X$ be arbitrary, and $x_{n} = f^{n}(x_{0})$, then for ecah $n \in \natp$,
\[d(x_{n}, x_{n+1}) \le C d(x_{n-1}, x_{n}) \le C^{n} d(x_{0}, x_{1})\]
Thus $\seq{x_n}\subset X$ is Cauchy, and converges to a point $x \in X$.
(2): For any $y_{0} \in X$, let $y_{n} = f^{n}(y_{0})$, then $d(x_{n}, y_{n}) \to 0$ as $n \to \infty$, so $\limv{n}f^{n}(y_{0}) = x$.
(1): Since $f$ is Lipschitz continuous,
\[f(x) = f\braks{\limv{n}f^n(x)}= \limv{n}f^{n+1}(x) = x\]
$\square$