ProofOfInverseFunctionTheorem 2003-03-17 09:20:50 2004-03-01 04:57:18 Proof inverse function theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\R}{\mathbb R} Since $\det Df(a)\neq 0$ the Jacobian matrix $Df(a)$ is invertible: let $A=(Df(a))^{-1}$ be its inverse. Choose $r>0$ and $\rho>0$ such that \[ B=\overline{B_\rho(a)} \subset E, \] \[ \Vert Df(x) - Df(a)\Vert \le \frac{1}{2n\Vert A \Vert}\quad \forall x \in B, \] \[ r\le \frac{\rho}{2\Vert A\Vert}. \] Let $y\in B_r(f(a))$ and consider the mapping \[ T_y \colon B \to \R^n \] \[ T_y(x) = x + A\cdot(y-f(x)). \] If $x\in B$ we have \[ \Vert D T_y(x)\Vert = \Vert 1- A\cdot Df(x)\Vert \le \Vert A\Vert \cdot \Vert Df(a) - Df(x)\Vert \le \frac{1}{2n}. \] Let us verify that $T_y$ is a contraction mapping. Given $x_1,x_2\in B$, by the Mean-value Theorem on $\R^n$ we have \[ \vert T_y(x_1) - T_y(x_2)\vert \le \sup_{x\in[x_1,x_2]} n \Vert DT_y(x)\Vert \cdot \vert x_1 - x_2\vert \le \frac 1 2 \vert x_1 -x_2\vert. \] Also notice that $T_y(B)\subset B$. In fact, given $x\in B$, \[ \vert T_y(x)-a\vert \le \vert T_y(x) - T_y(a)\vert + \vert T_y(a)-a\vert \le \frac 1 2 \vert x -a\vert + \vert A\cdot (y-f(a))\vert \le \frac \rho 2 + \Vert A\Vert r \le \rho. \] So $T_y\colon B\to B$ is a contraction mapping and hence by the contraction principle there exists one and only one solution to the equation \[ T_y(x)=x, \] i.e. $x$ is the only point in $B$ such that $f(x)=y$. Hence given any $y\in B_r(f(a))$ we can find $x\in B$ which solves $f(x)=y$. Let us call $g\colon B_r(f(a))\to B$ the mapping which gives this solution, i.e. \[ f(g(y))=y. \] Let $V=B_r(f(a))$ and $U=g(V)$. Clearly $f\colon U \to V$ is one to one and the inverse of $f$ is $g$. We have to prove that $U$ is a neighbourhood of $a$. However since $f$ is continuous in $a$ we know that there exists a ball $B_\delta(a)$ such that $f(B_\delta(a))\subset B_r(y_0)$ and hence we have $B_\delta(a)\subset U$. We now want to study the differentiability of $g$. Let $y\in V$ be any point, take $w\in \R^n$ and $\epsilon>0$ so small that $y+\epsilon w\in V$. Let $x=g(y)$ and define $v(\epsilon)=g(y+\epsilon w)-g(y)$. First of all notice that being \[ \vert T_y(x+v(\epsilon)) - T_y(x)\vert \le \frac 1 2 \vert v(\epsilon)\vert \] we have \[ \frac 1 2 \vert v(\epsilon) \ge \vert v(\epsilon)-\epsilon A\cdot w\vert \ge \vert v(\epsilon)\vert - \epsilon\Vert A\Vert\cdot \vert w\vert \] and hence \[ \vert v(\epsilon)\vert \le 2\epsilon \Vert A\Vert\cdot\vert w\vert. \] On the other hand we know that $f$ is differentiable in $x$ that is we know that for all $v$ it holds \[ f(x+v)-f(x) = Df(x)\cdot v + h(v) \] with $\lim_{v\to 0} h(v)/\vert v\vert = 0$. So we get \[ \frac{\vert h(v(\epsilon))\vert}{\epsilon} \le \frac{2\Vert A\Vert\cdot \vert w\vert\cdot \vert h(v(\epsilon))\vert} {v(\epsilon)} \to 0 \qquad\mathrm{when}\ \epsilon\to 0. \] So \[ \lim_{\epsilon\to 0} \frac {g(y+\epsilon)-g(y)}{\epsilon} =\lim_{\epsilon\to 0} \frac {v(\epsilon)}{\epsilon} = \lim_{\epsilon\to 0} Df(x)^{-1}\cdot \frac{\epsilon w - h(v(\epsilon))}{\epsilon} = Df(x)^{-1}\cdot w \] that is \[ Dg(y)=Df(x)^{-1}. \]