Inverse Function Theorem

Inverse Function Theorem

A smooth (or $C^k$ for $k\ge 1$) map $f:U\to {\mathbb{R}}^n$ where $U\subset {\mathbb{R}}^n$ for which ${\text{D}}_{x}f$ is invertible is a local smooth (or $C^k$ for $k\ge 1$) diffeomorphism about $x$.

Proof To simplify the algebraic exposition, we define a normalized function $F:U\to {\mathbb{R}}^n$ by composing $f$ with the linear inversion of its derivative at $x_0$: $F:={\left(Df{|}_{x_0}\right)}^{-1}\circ f.$ By the chain rule, the derivative of $F$ at $x_0$ satisfies: $DF{|}_{x_0}={\left(Df{|}_{x_0}\right)}^{-1}\circ Df{|}_{x_0}=\text{Id}.$Consequently, proving that $f$ is a local diffeomorphism at $x_0$ is completely equivalent to demonstrating that $F$ is a local diffeomorphism at $x_0$. (As invertible linear map is always a diffeomorphism.) Step 1: Surjectivity (Existence via Contraction Mapping). We aim to show that $F$ is locally surjective near $F(x_0)$. Since $F$ is $C^1$, the derivative map $x\mapsto DF{|}_x$ is continuous. Because $DF{|}_{x_0}=\text{Id}$, we can choose a sufficiently small radius $\delta >0$ such that the closed ball ${\bar{B}}_{\delta }(x_0)\subseteq U$, and for all $x\in {\bar{B}}_{\delta }(x_0)$, the operator norm satisfies: $\Vert \text{Id}-DF{|}_x\Vert \le \frac{1}{3}.$For a fixed target vector $y\in B_{\delta /3}(F(x_0))$, we define an auxiliary setup function $G_y:U\to {\mathbb{R}}^n$ by: $G_y(x)=y-F(x)+x.$Evidently, finding a solution to $F(x)=y$ is mathematically equivalent to finding a fixed point of $G_y$, since $G_y(x)=x⟺F(x)=y$. We first check that $G_y$ maps the closed ball ${\bar{B}}_{\delta }(x_0)$ into itself. For any $x\in {\bar{B}}_{\delta }(x_0)$, we add and subtract $F(x_0)$ to execute the following triangle inequality bound: \|G_y(x) - x_0\| = \|y - F(x) + x - x_0\| \\ = \|(y - F(x_0)) + (F(x_0) - F(x) + x - x_0)\| $$$$\le \|y - F(x_0)\| + \|F(x_0) - F(x) + (\text{Id} \cdot x - \text{Id} \cdot x_0)\|. Let $\hat{F}(x)=F(x)-x$. Then $D\hat{F}{|}_x=DF{|}_x-\text{Id}$. By the Mean Value theorem and the bound $\Vert \text{Id}-DF{|}_x\Vert \le \frac{1}{3}$, we have:$\Vert F(x_0)-F(x)+x-x_0\Vert =\Vert \hat{F}(x)-\hat{F}(x_0)\Vert \le {\sup }_{c\in [x_0,x]}\Vert DF{|}_c-\text{Id}\Vert \cdot \Vert x-x_0\Vert \le \frac{1}{3}\delta .$ Since $y\in B_{\delta /3}(F(x_0))$, we obtain: $\Vert G_y(x)-x_0\Vert \le \frac{\delta }{3}+\frac{\delta }{3}=\frac{2\delta }{3}\le \delta .$ Thus, $G_y\left({\bar{B}}_{\delta }(x_0)\right)\subseteq {\bar{B}}_{\delta }(x_0)$. Next, we establish that $G_y$ is a strict contraction on this ball. For any $x_1,x_2\in {\bar{B}}_{\delta }(x_0)$, applying the Mean Value theorem yields: $\Vert G_y(x_2)-G_y(x_1)\Vert =\Vert (x_2-x_1)-(F(x_2)-F(x_1))\Vert \le \frac{1}{3}\Vert x_2-x_1\Vert .$ Since $\frac{1}{3}<1$, $G_y$ is a strict contraction mapping. By the Banach Fixed-Point Theorem, there exists a unique $x\in {\bar{B}}_{\delta }(x_0)$ such that $G_y(x)=x$, which implies $F(x)=y$. Local surjectivity is thus established. Step 2: Injectivity (Uniqueness via Contradiction). We now verify that $F$ is locally injective near $x_0$. Suppose for contradiction that for every integer $N\gg 1$, $F$ fails to be injective on $B_{1/N}(x_0)$. Then there exist distinct points $x_N^{(1)}\ne x_N^{(2)}\in B_{1/N}(x_0)$ such that: $F(x_N^{(1)})=F(x_N^{(2)})=y_N.$ As $N\to \infty$, both sequences naturally converge to the center point: ${\lim }_{N\to \infty }{x}_N^{(1)}={\lim }_{N\to \infty }{x}_N^{(2)}=x_0$. For a sufficiently large $N$ such that $1/N<\delta /3$, the target $y_N$ falls well within $B_{\delta /3}(F(x_0))$. Returning to our auxiliary mapping $G_{y_N}(x)=y_N-F(x)+x$, we observe that: $G_{y_N}(x_N^{(1)})=y_N-F(x_N^{(1)})+x_N^{(1)}=y_N-y_N+x_N^{(1)}=x_N^{(1)},$ and similarly $G_{y_N}(x_N^{(2)})=x_N^{(2)}$. This implies that $x_N^{(1)}$ and $x_N^{(2)}$ are two distinct fixed points of $G_{y_N}$ inside ${\bar{B}}_{\delta }(x_0)$, which directly violates the uniqueness guarantee of the Banach Fixed-Point Theorem established in Step 1. Hence, $F$ must be locally injective. We can now define our open domains as $U_{x_0}=B_{\delta }(x_0)\cap F^{-1}\left(B_{\delta /3}(F(x_0))\right)$ and $V_{f(x_0)}=F(U_{x_0})$, making $F{|}_{U_{x_0}}:U_{x_0}\to V_{f(x_0)}$ a strict bijection. Step 3: Differentiability of the Inverse. It remains to prove that the inverse mapping $F^{-1}:V_{f(x_0)}\to U_{x_0}$ is $C^1$. Let $y,y_0\in V_{f(x_0)}$ and set $x=F^{-1}(y)$, $x_0=F^{-1}(y_0)$. Since $DF{|}_{x_0}=\text{Id}$, the first-order Taylor expansion of $F$ at $x_0$ reads:$y-y_0=F(x)-F(x_0)=\text{Id}\cdot (x-x_0)+o(\Vert x-x_0\Vert )=(x-x_0)+o(\Vert x-x_0\Vert ).$ By employing the contraction property from Step 1, it follows via the triangle inequality that for a sufficiently small neighborhood, we have the Lipschitz continuity bound: $\Vert x-x_0\Vert \le 4\Vert y-y_0\Vert .$ This relation guarantees that any remainder term in terms of $x$ safely translates into a remainder term in terms of $y$; that is, $o(\Vert x-x_0\Vert )=o(\Vert y-y_0\Vert )$. Substituting this back into the Taylor expansion gives: $y-y_0=F^{-1}(y)-F^{-1}(y_0)+o(\Vert y-y_0\Vert ),$ which simplifies directly to the definition of differentiability for $F^{-1}$:$F^{-1}(y)-F^{-1}(y_0)=\text{Id}\cdot (y-y_0)+o(\Vert y-y_0\Vert ).$This proves that $F^{-1}$ is differentiable at $y_0$ with $D{F}^{-1}{|}_{y_0}=\text{Id}=(DF{|}_{x_0}{)}^{-1}$. Reverting the normalization step back to our original function $f$, we conclude that $f^{-1}$ is differentiable everywhere on its local image with a derivative matrix given by $D(f^{-1}){|}_{f(x)}=(Df{|}_x{)}^{-1}$. Since matrix inversion is a smooth operation on the general linear group, the continuity of $Df$ ensures that $D(f^{-1})$ is continuous, completing the proof that $f$ is a local $C^1$ diffeomorphism.