# proof of inverse function theorem

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

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

Let $y\in B_{r}(f(a))$ and consider the mapping

 $T_{y}\colon B\to\mathbb{R}^{n}$
 $T_{y}(x)=x+A\cdot(y-f(x)).$

If $x\in B$ we have

 $\|DT_{y}(x)\|=\|1-A\cdot Df(x)\|\leq\|A\|\cdot\|Df(a)-Df(x)\|\leq\frac{1}{2n}.$

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

 $|T_{y}(x_{1})-T_{y}(x_{2})|\leq\sup_{x\in[x_{1},x_{2}]}n\|DT_{y}(x)\|\cdot|x_{% 1}-x_{2}|\leq\frac{1}{2}|x_{1}-x_{2}|.$

Also notice that $T_{y}(B)\subset B$. In fact, given $x\in B$,

 $|T_{y}(x)-a|\leq|T_{y}(x)-T_{y}(a)|+|T_{y}(a)-a|\leq\frac{1}{2}|x-a|+|A\cdot(y% -f(a))|\leq\frac{\rho}{2}+\|A\|r\leq\rho.$

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 $w\in\mathbb{R}^{n}$ 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

 $|T_{y}(x+v(\epsilon))-T_{y}(x)|\leq\frac{1}{2}|v(\epsilon)|$

we have

 $\frac{1}{2}|v(\epsilon)\geq|v(\epsilon)-\epsilon A\cdot w|\geq|v(\epsilon)|-% \epsilon\|A\|\cdot|w|$

and hence

 $|v(\epsilon)|\leq 2\epsilon\|A\|\cdot|w|.$

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 $\lim_{v\to 0}h(v)/|v|=0$. So we get

 $\frac{|h(v(\epsilon))|}{\epsilon}\leq\frac{2\|A\|\cdot|w|\cdot|h(v(\epsilon))|% }{v(\epsilon)}\to 0\qquad\mathrm{when}\ \epsilon\to 0.$

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

 $Dg(y)=Df(x)^{-1}.$
Title proof of inverse function theorem ProofOfInverseFunctionTheorem 2013-03-22 13:31:20 2013-03-22 13:31:20 paolini (1187) paolini (1187) 6 paolini (1187) Proof msc 03E20