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Proof Of Inverse Function Theorem

Since $\det Df(a)\neq 0$ the Jacobian matrix is invertible: let $A=(Df(a))^{-1}$ be its inverse. Choose 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

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.

is the only point in

such that

Hence given any $y\in B_r(f(a))$ we can find $x\in B$ which solves . 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 and . Clearly $f\colon U \to V$ is one to one and the inverse of is . We have to prove that is a neighbourhood of . However since is continuous in 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 . 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 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

is differentiable in

that is we know that for all

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

$\displaystyle \frac{\vert h(v(\epsilon))\vert}{\epsilon} \le \frac{2\Vert A\Ver... ...t h(v(\epsilon))\vert} {v(\epsilon)} \to 0 \qquad\mathrm{when}\ \epsilon\to 0.$

$\displaystyle \lim_{\epsilon\to 0} \frac {g(y+\epsilon)-g(y)}{\epsilon} =\lim_{... ...(x)^{-1}\cdot \frac{\epsilon w - h(v(\epsilon))}{\epsilon} = Df(x)^{-1}\cdot w$

that is

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

1	Point
1	Differntiable Function
1	Neighborhood
1	Mapping
1	Something Related To Injective Function
1	Multivalued Function
1	Equation
2	Ball
2	Classes Of Ordinals And Enumerating Functions
5	Jacobian Matrix
6	Mean Value Theorem
9	Banach Fixed Point Theorem