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[parent] proof of inverse function theorem (Proof)

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

$\displaystyle B=\overline{B_\rho(a)} \subset E, $
$\displaystyle \Vert Df(x) - Df(a)\Vert \le \frac{1}{2n\Vert A \Vert}\quad \forall x \in B, $
$\displaystyle r\le \frac{\rho}{2\Vert A\Vert}. $

Let $ y\in B_r(f(a))$ and consider the mapping

$\displaystyle T_y \colon B \to \mathbb{R}^n $
$\displaystyle T_y(x) = x + A\cdot(y-f(x)). $
If $ x\in B$ we have
$\displaystyle \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 $ \mathbb{R}^n$ we have
$\displaystyle \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$,
$\displaystyle \vert T_y(x)-a\vert \le \vert T_y(x) - T_y(a)\vert + \vert T_y(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

$\displaystyle 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.

$\displaystyle 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 \mathbb{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

$\displaystyle \vert T_y(x+v(\epsilon)) - T_y(x)\vert \le \frac 1 2 \vert v(\epsilon)\vert $
we have
$\displaystyle \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
$\displaystyle \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
$\displaystyle 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
$\displaystyle \frac{\vert h(v(\epsilon))\vert}{\epsilon} \le \frac{2\Vert A\Ver... ...t h(v(\epsilon))\vert} {v(\epsilon)} \to 0 \qquad\mathrm{when}\ \epsilon\to 0. $
So
$\displaystyle \lim_{\epsilon\to 0} \frac {g(y+\epsilon)-g(y)}{\epsilon} =\lim_{... ...(x)^{-1}\cdot \frac{\epsilon w - h(v(\epsilon))}{\epsilon} = Df(x)^{-1}\cdot w $
that is
$\displaystyle Dg(y)=Df(x)^{-1}. $



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Cross-references: differentiable, ball, continuous, neighbourhood, point, equation, solution, contraction principle, mean-value theorem, contraction mapping, mapping, inverse, invertible, Jacobian matrix

This is version 3 of proof of inverse function theorem, born on 2003-03-17, modified 2004-03-01.
Object id is 4112, canonical name is ProofOfInverseFunctionTheorem.
Accessed 8159 times total.

Classification:
AMS MSC: 03E20 (Mathematical logic and foundations :: Set theory :: Other classical set theory )
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