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# 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

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

## Mathematics Subject Classification

03E20*no label found*

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