proof of inverse function theorem

Since $\\det Df(a)\eq 0$ 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 uscall $g\\colon B_{r}(f(a))\\to B$ 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 hencewe 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 thatfor 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{\\epsilonw%-h(v(\\epsilon))}{\\epsilon}=Df(x)^{-1}\\cdot w

that is

Dg(y)=Df(x)^{-1}.