proof of inverse function theorem

Since detDf(a)0 the Jacobian matrix Df(a) is invertible: let A=(Df(a))-1 be its inversePlanetmathPlanetmathPlanetmath. Choose r>0 and ρ>0 such that


Let yBr(f(a)) and consider the mapping


If xB we have


Let us verify that Ty is a contraction mapping. Given x1,x2B, by the Mean-value Theorem on n we have


Also notice that Ty(B)B. In fact, given xB,


So Ty:BB is a contraction mapping and hence by the contraction principle there exists one and only one solution to the equation


i.e. x is the only point in B such that f(x)=y.

Hence given any yBr(f(a)) we can find xB which solves f(x)=y. Let us call g:Br(f(a))B the mapping which gives this solution, i.e.


Let V=Br(f(a)) and U=g(V). Clearly f:UV 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 continuousMathworldPlanetmathPlanetmath in a we know that there exists a ball Bδ(a) such that f(Bδ(a))Br(y0) and hence we have Bδ(a)U.

We now want to study the differentiability of g. Let yV be any point, take wn and ϵ>0 so small that y+ϵwV. Let x=g(y) and define v(ϵ)=g(y+ϵw)-g(y).

First of all notice that being


we have


and hence


On the other hand we know that f is differentiableMathworldPlanetmathPlanetmath in x that is we know that for all v it holds


with limv0h(v)/|v|=0. So we get

|h(v(ϵ))|ϵ2A|w||h(v(ϵ))|v(ϵ)0  whenϵ0.



that is

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