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.

\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
Canonical name	ProofOfInverseFunctionTheorem
Date of creation	2013-03-22 13:31:20
Last modified on	2013-03-22 13:31:20
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	6
Author	paolini (1187)
Entry type	Proof
Classification	msc 03E20