Proof, [Theorem XIV.1.2, Lan93]. By translation, assume without loss of generality that $x_{0} = f(x_{0}) = 0$ and $Df(x_{0}) = Df(0) = I$.

Existence and Uniqueness of Inverse: Since $f \in C^{1}$, there exists $r > 0$ such that $\norm{Df(x) - I}_{L(E; E)}< 1/2$ for all $x \in \ol{B_E(0, r)}$. In which case, by Lemma 28.2.2, $Df(x)$ is an isomorphism for all $x \in B(0, r)$. Let

For any $x, y \in \overline{B_E(0, r)}$, by the Mean Value Theorem,

In particular, for any $x \in \overline{B_E(0, r)}$, $\norm{g(x)}_{E} = \norm{g(x) - g(0)}_{E} \le \norm{x}_{E}/2$, so $g: \ol{B_E(0, r)}\to \ol{B_E(0, r/2)}$ is a contraction.

For each $y \in B(0, r/2)$, the mapping

is also a contraction. By Banach’s Fixed Point Theorem, there exists a unique $x \in B(0, r)$ such that $g_{y}(x) = x$. In which case, $f(x) = y$. Therefore $f$ restricted to $V = f^{-1}(B(0, r))$ is invertible.

Differentiability of Inverse: Let $f^{-1}: f(V) \to V$ be the local inverse of $f$ on $V$. By assumption, it is sufficient to show that $Df^{-1}(0) = I$ as well. For each $y \in \overline{B(0, r/2)}$,

where $r(x)/\norm{x}_{E} \to 0$ as $x \to 0$. In addition,

so $[f^{-1}(y) - y]/\norm{y}_{E} \to 0$ as $y \to 0$. Therefore $f^{-1}$ is differentiable at $0$ with $Df^{-1}= I$.

Smoothness of Inverse: By the above argument, the inverse is differentiable on every point in $B(0, r/2)$, and $Df^{-1}(f(x)) = [Df(x)]^{-1}$ for all $x \in V$. By Proposition 28.2.3, the inversion map $T \mapsto T^{-1}$ is smooth. Therefore if $Df \in C^{p - 1}$, then $f \in C^{p - 1}$ as well.$\square$