3
$\begingroup$

I can prove the converse of it, but I cannot do this one. Here is the problem:

Prove that the implicit function theorem implies the inverse function theorem.

$\endgroup$
  • $\begingroup$ Why not post what problems your having or what you've tried? It really helps to know what an OP is struggling with exactly. Sometimes you might be misunderstanding something or not fully understanding something and answerers on here can bridge the gap or provide valuable insight. Like a guy said in a movie once "You gotta milk it" :) Otherwise, you could just find the proof on Google.. $\endgroup$ – user70962 Jun 9 '13 at 8:14
  • $\begingroup$ Oh... I missed that point. thanks for your advice! I will soon! $\endgroup$ – syko Jun 9 '13 at 8:30
7
$\begingroup$

For $f : \mathbb{R}^n \to \mathbb{R}^n$, consider $F : \mathbb{R}^n\times\mathbb{R}^n \to \mathbb{R}^n$ given by $F({\bf x}, {\bf y}) = f({\bf y}) - {\bf x}$.

$\endgroup$
  • $\begingroup$ Is it enough to prove for the certain form for the function like F(x,y)=f(y)-x=0 ? $\endgroup$ – syko Jun 9 '13 at 9:43
  • $\begingroup$ And.. the general conditions of the implicit function Thm is... F:Rn x Rm -> Rm . Is it okay to prove just for the F : Rn x Rn -> Rn? $\endgroup$ – syko Jun 9 '13 at 9:50
  • $\begingroup$ Your post suggests you want to prove the inverse function theorem, given the implicit function theorem. Apply the implicit function theorem to the function I've suggested. $\endgroup$ – Michael Albanese Jun 9 '13 at 9:56
  • $\begingroup$ Wow!! It's solved! Thanks you!! It was very helpful. Thnaks you! $\endgroup$ – syko Jun 9 '13 at 10:04
  • $\begingroup$ I thought you should have written $F(x,y)=y-f(x)$ $\endgroup$ – user344374 Dec 27 '17 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.