> [!theoremb] Inverse Function Theorem (Simplified) > > Let $\cx$ be a [[Banach Space|Banach space]], $U \subset \cx$ be an [[Open Set|open set]] containing $0$ and $f: U \to \cx$ be a [$C^p$](Space%20of%20Continuously%20Differentiable%20Functions) map with $p \ge 1$. Suppose that $Df(0) = I$ is the identity and $f(0) = 0$, then $f$ is a local $C^p$-isomorphism at $0$. > > ### Neighbourhood > > There exists $r > 0$ such that $Df(x)$ is invertible for all $x \in B(0, r)$. Moreover, if $g(x) = x - f(x)$, then $g(\ol{B(0, r)}) \subset \ol{B(0, r/2)}$. > > *Proof*. Since $Dg(0) = 0$ and $f \in C^p$, by continuity, there exists $r > 0$ such that > $ > \begin{align*} > \norm{Dg(x)} &< \frac{1}{2} \\ > \norm{Df(x) - I} &< \frac{1}{2} > \end{align*} > $ > for all $x$ with $\norm{x} \le r$. Since the space of [[Bounded Linear Map|isomorphisms]] is open, and $\norm{Df(x) - I} < \norm{I^{-1}}^{-1}$, $Df(x)$ is invertible for all $x$ with $\norm{x} < r$. > > For any $x \in \ol{B(0, r)}$, by the [[Mean Value Theorem|mean value theorem]], > $ > \begin{align*} > \norm{g(x)} = \norm{g(x) - g(0)} \le \norm{x} \cdot \sup_{t \in [0, 1]}\norm{Dg(tx)} \le \frac{1}{2}\norm{x} > \end{align*} > $ > we have $x \in \ol{B(0, r/2)}$. > > ### Contraction > > The map $g: \ol{B(0, r)} \to \ol{B(0, r)}$ is a contractor. > > *Proof*. Let $x_1, x_2 \in \ol{B(0, r)}$, then by the mean value theorem again > $ > \begin{align*} > \norm{g(x_1) - g(x_2)} &\le \norm{x_1 - x_2} \cdot \sup_{y \in \ol{B(0, r)}}\norm{Dg(y)} < \frac{1}{2}\norm{x_1 - x_2} > \end{align*} > $ > > ### Inverse > > For any $y \in \ol{B(0, r/2)}$, there exists a unique $x \in \ol{B(0, r)}$ such that $f(x) = y$. > > *Proof (see [[Fixed Point Problem]])*. Let $y \in \ol{B(0, r/2)}$ and $g_y(x) = x - f(x) + y$, then $f(x) = y$ if and only if $g_y(x) = x$. If $\norm{y} \le r/2$ and $\norm{x} \le r$, > $ > \norm{g_y(x)} = \norm{g(x) + y} \le \frac{1}{2}\norm{x} + \norm{y} \le r > $ > and $g_y$ maps $\ol{B(0, r)}$ to $\ol{B(0, r)}$. Now for any $x_1, x_2 \in \ol{B(0, r)}$, > $ > \norm{g_y(x_1) - g_y(x_2)} = \norm{g(x_1) - g(x_2)} > $ > Since $g$ is a contractor, so is $g_y$. By the [[Contractor|Banach fixed-point theorem]], $g_y$ has a unique fixed point $x \in \ol{B(0, r)}$, which is the solution to $f(x) = y$. > > ### Continuity > > Let $\phi: \ol{B(0, r/2)} \to \ol{B(0, r)}$ be the inverse of $f$, then $\phi$ is continuous. > > *Proof*. Let $x_1, x_2 \in \ol{B(0, r)}$, then > $ > x_1 - x_2 = f(x_1) + g(x_1) - f(x_2) - g(x_2) > $ > and > $ > \begin{align*} > \norm{x_1 - x_2} &\le \norm{f(x_1) - f(x_2)} + \norm{g(x_1) - g(x_2)} \\ > &\le 2\norm{f(x_1) - f(x_2)} > \end{align*} > $ > So for any $y_1, y_2 \in \ol{B(0, r/2)}$, > $ > \norm{\phi(y_1) - \phi(y_2)} \le 2\norm{y_1 - y_2} > $ > and $\phi$ is continuous. > > > ### Differentiability > > Let $\phi: \ol{B(0, r/2)} \to \ol{B(0, r)}$ be the inverse of $f$, then $\phi$ is differentiable with $D\phi(x) = Df(x)^{-1}$. > > *Proof*. Let $x_1, x_2 \in \ol{B(0, r/2)}$, and $y_1 = f(x_1)$, $y_2 = f(x_2)$, then > $ > \begin{align*} > &\norm{\phi(y_1) - \phi(y_2) - Df(x_2)^{-1}(y_1 - y_2)} \\ > &= \norm{x_1 - x_2 - Df(x_2)^{-1}(y_1 - y_2)} \\ > &\le \norm{Df(x_2)} \cdot \norm{Df(x_2)(x_1 - x_2) - (y_1 - y_2)} > \end{align*} > $ > By the differentiability of $f$, > $ > \begin{align*} > f(x_1) &= f(x_2) + Df(x_2)(x_1 - x_2) + o(x_1 - x_2) \\ > f(x_1) - f(x_2) &= Df(x_2)(x_1 - x_2) + o(x_1 - x_2) > \end{align*} > $ > so > $ > \begin{align*} > &\norm{Df(x_2)} \cdot \norm{Df(x_2)(x_1 - x_2) - (y_1 - y_2)} \\ > &= \norm{Df(x_2)} \cdot \norm{o(x_1 - x_2)} > \end{align*} > $ > is [[Little-O|little-o]] of $(x_1 - x_2)$. Since as shown before, > $ > \norm{x_1 - x_2} \le 2\norm{y_1 - y_2} > $ > meaning that > $ > \begin{align*} > \frac{\norm{o(x_1 - x_2)}}{\norm{y_1 - y_2}} &= \frac{\norm{o(x_1 - x_2)}}{\norm{x_1 - x_2}} \cdot \frac{\norm{x_1 - x_2}}{\norm{y_1 - y_2}} \\ > &\le 2\frac{\norm{o(x_1 - x_2)}}{\norm{x_1 - x_2}} > \end{align*} > $ > and the difference is little-o of $(y_1 - y_2)$ as well. > > > ### Smoothness > > Let $\phi: \ol{B(0, r/2)} \to \ol{B(0, r)}$ be the inverse of $f$, then $\phi$ is of class $C^p$. > > *Proof*. Let $\psi: \text{Laut}(\cx) \to \text{Laut}(\cx)$ with $x \mapsto x^{-1}$ be the inversion map, then since $D\phi = \psi \circ Df$ and $\psi \in C^\infty$[^1], $D\phi \in C^p$. > [!theoremb] Inverse Function Theorem > > Let $\cx$ be a [[Banach Space|Banach space]], $U \subset \cx$ be an [[Open Set|open set]] and $f: U \to \cx$ be of class $C^p$ with $p \ge 1$. If there exists $x_0 \in U$ such that $Df(x_0) \in \text{Laut}(\cx)$ is [[Space of Toplinear Isomorphisms|invertible]], then $f$ is a local $C^p$-isomorphism. > > *Proof*. Let $g(x) = Df(x_0)^{-1} \circ f(x + x_0)$, then $g: (U - x_0) \to \cx$ is of class $C^p$ with $Dg(0) = I$. By the simplified version, there exists $V, W \in \cn(x_0)$ such that $g: V \to W$ is a $C^p$-isomorphism. > > Since $f(x) = Df(x_0) \circ g(x - x_0)$, $f: (V + x_0) \to Df(x_0)(W)$ is a $C^p$-isomorphism. [^1]: See [[Banach Algebra#Smoothness of the Inverse|Smoothness of the Inverse]].