Implicit => inverse function theorem ( due to exam, )

[yoyo_fen]

Implicit => inverse function theorem (urgent due to exam, please help)

Homework Statement

Prove the inverse function theorem, knowing the implicit function theorem.

Homework Equations

The statements of both theorems... Can't think of much else.

The Attempt at a Solution

Alright... I have a feeling that this is going to come up on my exam which is only two days from now, so please help me :(

I got a hint today: define g(x,y) = f(x)-y or g(x,y)=f(y)-x, the guy was not sure. I presume it should be the first because then the derivative with respect to the second variable y is 1 which is invertible and in the first case we are not sure whether it is.
I have a problem though because in the statement of the implicit function theorem I have, the function needs to be from R^n to R^(n+p) so that we can take the differential of the matrix w.r.t. the p+i variables (i goes from 1 to n). In the hint the function is from R^2 -> R.

Even if that was not a problem I don't know how to end up with the inverse function theorem.

I thank in advance for anyone who is going to help me out :)

 

[hunt_mat]

Isn't this just bookwork?

 

[yoyo_fen]

What do you mean? Even if it is... I still cannot do it. :(

 

[hunt_mat]

what does the implicit and inverse functon theorems say?

 

[yoyo_fen]

Inverse
f:R^n->R^n,
if f is C^1, f(a)=b and differential is invertible at a, then the inverse is well defined and C^1 on some open subsets U and V where a is in U and b is in V

Implicit
f:R^p+n-->R^n , f(a,b)=0, f is C^1 and the partials with respect to the variables from p onwards form and invertible matrix, then F(a)=b is well defined and C^1 on some open subsets U and V...

I know them... And I have tried... If I knew the answer I wasn't going to be asking right?

 

[hunt_mat]

write g(a,b)=b-f(a)=0, what does the implicit function tell you now?

 

[yoyo_fen]

The implicit f(a)= b is C^1 and well defined in a neighbourhood around those points...?

 

[hunt_mat]

Or that there is an F such that F(b)=a...

 

[yoyo_fen]

How? I really don't see it... Is mine wrong?

 

[hunt_mat]

Your's isn't wrong, but I used a different variable for mine.

 

[yoyo_fen]

The neighbourhoods are not the same, how did you just switch the letters a and b? And even so... Okay F(b) = b but how did f(a) become just a? What we want to show is that there is an inverse function of f in my previous post (f(a)=b), s.t. f^-1 (b)=a. This is why I don't get? How do we get that function. How are you able to write what you have written.

 

[hunt_mat]

It didn't, remember the implicit function theorem says that if f(a,b)=0 which we have then there is an F such that a=F(b) but likewise there exists a G such that b=G(a), we are just applying the implicit function to each variable.

 

[yoyo_fen]

OH SNAP!
So I was wrong after all saying this "The implicit f(a)= b is C^1 and well defined in a neighbourhood around those points...?". This follows from b-f(a)= 0 as we've defined g(a,b) and the other result follows from implicit function theorem.

THANKS MAN! YOU ARE THE BEST!

 

[hunt_mat]

There is more work to be done if you want to make it fully rigourous but what I said is the gist of the proof.

I help when I can.