Editing Inverse function theorem (section)

=== A proof using successive approximation ===

To prove existence, it can be assumed after an affine transformation that <math>f(0)=0</math> and <math>f^\prime(0)=I</math>, so that <math> a=b=0</math>.

By the [[Mean value theorem#Mean value theorem for vector-valued functions|mean value theorem for vector-valued functions]], for a differentiable function <math>u:[0,1]\to\mathbb R^m</math>, <math display="inline">\|u(1)-u(0)\|\le \sup_{0\le t\le 1} \|u^\prime(t)\|</math>. Setting <math>u(t)=f(x+t(x^\prime -x)) - x-t(x^\prime-x)</math>, it follows that

:<math>\|f(x) - f(x^\prime) - x + x^\prime\| \le \|x -x^\prime\|\,\sup_{0\le t \le 1} \|f^\prime(x+t(x^\prime -x))-I\|.</math>

Now choose <math>\delta>0</math> so that <math display="inline">\|f'(x) - I\| < {1\over 2}</math> for <math>\|x\|< \delta</math>. Suppose that <math>\|y\|<\delta/2</math> and define <math>x_n</math> inductively by <math>x_0=0</math> and <math> x_{n+1}=x_n + y - f(x_n)</math>. The assumptions show that if <math> \|x\|, \,\, \|x^\prime\| < \delta</math> then

:<math>\|f(x)-f(x^\prime) - x + x^\prime\| \le \|x-x^\prime\|/2</math>.

In particular <math>f(x)=f(x^\prime)</math> implies <math>x=x^\prime</math>. In the inductive scheme <math>\|x_n\| <\delta</math>
and <math>\|x_{n+1} - x_n\| < \delta/2^n</math>. Thus <math>(x_n)</math> is a [[Cauchy sequence]] tending to <math>x</math>. By construction <math>f(x)=y</math> as required.

To check that <math>g=f^{-1}</math> is C<sup>1</sup>, write <math>g(y+k) = x+h</math> so that
<math>f(x+h)=f(x)+k</math>. By the inequalities above, <math>\|h-k\| <\|h\|/2</math> so that <math>\|h\|/2<\|k\| < 2\|h\|</math>.
On the other hand, if <math>A=f^\prime(x)</math>, then <math>\|A-I\|<1/2</math>. Using the [[geometric series]] for <math>B=I-A</math>, it follows that <math>\|A^{-1}\| < 2</math>. But then

:<math> {\|g(y+k) -g(y) - f^\prime(g(y))^{-1}k \| \over \|k\|} 
= {\|h -f^\prime(x)^{-1}[f(x+h)-f(x)]\| \over \|k\|} 
\le 4 {\|f(x+h) - f(x) -f^\prime(x)h\|\over \|h\|} </math>

tends to 0 as <math>k</math> and <math>h</math> tend to 0, proving that <math>g</math> is C<sup>1</sup> with <math>g^\prime(y)=f^\prime(g(y))^{-1}</math>.

The proof above is presented for a finite-dimensional space, but applies equally well for [[Banach space]]s.  If an invertible function <math>f</math> is C<sup>k</sup> with <math>k>1</math>, then so too is its inverse. This follows by induction using the fact that the map <math>F(A)=A^{-1}</math> on operators is C<sup>k</sup> for any <math>k</math> (in the finite-dimensional case this is an elementary fact because the inverse of a matrix is given as the [[adjugate matrix]] divided by its [[determinant]]).
<ref name="Hörmander" /><ref>{{cite book|title=Calcul Differentiel|language=fr|first=Henri|last= Cartan|author-link= Henri Cartan|publisher=[[Éditions Hermann|Hermann]]|year= 1971|isbn=978-0-395-12033-0 |pages=55–61}}</ref> The method of proof here can be found in the books of [[Henri Cartan]], [[Jean Dieudonné]], [[Serge Lang]], [[Roger Godement]] and [[Lars Hörmander]].