# proof of inverse function theorem

Since $detDf(a)\ne 0$ the Jacobian matrix $Df(a)$ is invertible:
let $A={(Df(a))}^{-1}$ be its inverse^{}.
Choose $r>0$ and $\rho >0$ such that

$$B=\overline{{B}_{\rho}(a)}\subset E,$$ |

$$\parallel Df(x)-Df(a)\parallel \le \frac{1}{2n\parallel A\parallel}\mathit{\hspace{1em}}\forall x\in B,$$ |

$$r\le \frac{\rho}{2\parallel A\parallel}.$$ |

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

$${T}_{y}:B\to {\mathbb{R}}^{n}$$ |

$${T}_{y}(x)=x+A\cdot (y-f(x)).$$ |

If $x\in B$ we have

$$\parallel D{T}_{y}(x)\parallel =\parallel 1-A\cdot Df(x)\parallel \le \parallel A\parallel \cdot \parallel Df(a)-Df(x)\parallel \le \frac{1}{2n}.$$ |

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

$$|{T}_{y}({x}_{1})-{T}_{y}({x}_{2})|\le \underset{x\in [{x}_{1},{x}_{2}]}{sup}n\parallel D{T}_{y}(x)\parallel \cdot |{x}_{1}-{x}_{2}|\le \frac{1}{2}|{x}_{1}-{x}_{2}|.$$ |

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

$$|{T}_{y}(x)-a|\le |{T}_{y}(x)-{T}_{y}(a)|+|{T}_{y}(a)-a|\le \frac{1}{2}|x-a|+|A\cdot (y-f(a))|\le \frac{\rho}{2}+\parallel A\parallel r\le \rho .$$ |

So ${T}_{y}:B\to B$ 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 $g:{B}_{r}(f(a))\to B$ the mapping which gives this solution, i.e.

$$f(g(y))=y.$$ |

Let $V={B}_{r}(f(a))$ and $U=g(V)$. Clearly $f:U\to V$ 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 $\u03f5>0$ so small that $y+\u03f5w\in V$. Let $x=g(y)$ and define $v(\u03f5)=g(y+\u03f5w)-g(y)$.

First of all notice that being

$$|{T}_{y}(x+v(\u03f5))-{T}_{y}(x)|\le \frac{1}{2}|v(\u03f5)|$$ |

we have

$$\frac{1}{2}|v(\u03f5)\ge |v(\u03f5)-\u03f5A\cdot w|\ge |v(\u03f5)|-\u03f5\parallel A\parallel \cdot |w|$$ |

and hence

$$|v(\u03f5)|\le 2\u03f5\parallel A\parallel \cdot |w|.$$ |

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

$$\frac{|h(v(\u03f5))|}{\u03f5}\le \frac{2\parallel A\parallel \cdot |w|\cdot |h(v(\u03f5))|}{v(\u03f5)}\to 0\mathit{\hspace{1em}\hspace{1em}}\mathrm{when}\u03f5\to 0.$$ |

So

$$\underset{\u03f5\to 0}{lim}\frac{g(y+\u03f5)-g(y)}{\u03f5}=\underset{\u03f5\to 0}{lim}\frac{v(\u03f5)}{\u03f5}=\underset{\u03f5\to 0}{lim}Df{(x)}^{-1}\cdot \frac{\u03f5w-h(v(\u03f5))}{\u03f5}=Df{(x)}^{-1}\cdot w$$ |

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 |