Applcation of the inverse function theorem

Kate2010
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Homework Statement



Let f \in C1(Rn) be a function such that f(0) = 0 and \delta1f(0) is nonzero. (\delta1 means partial derivative with respect to x1)
Show that there exist neighbourhoods U and V of x=0\in Rn and a diffeomorphism g:U->V such that f(g(x)) = x1 for all x = (x1,...,xn) \in U.

Homework Equations



Inverse function theorem:
Let S\subseteqRn be open, f \inC1(S,Rn) and x0\inS. If Df(x0) is invertible then there exists an open neighbourhood U of x0 such that f(U) is open and f: U -> f(U) is a diffeomorphism. Also Df-1(f(x0)) = (D(f(x0))-1.

The Attempt at a Solution



I have a hint that says to apply the inverse function theorem to some function F:Rn -> Rn.
But, I really have no idea what this should be. I think it should involve f but I'm not sure how. Also I noticed g(x) = f-1(x1) if f-1 exists but then I'm not sure this makes sense as f is defined on Rn not R. So any help with finding this F would be great!
 
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Hi Kate2010! :smile:

You're correct that you can't apply the theorem directly to f:\mathbb{R}^n\rightarrow \mathbb{R}. But maybe you can extend f so that the codomain is good. I mean, your function F should be of the form

F(x_1,...,x_n)=(f(x_1,...,x_n),?(x_1,...,x_n),...,?(x_1,...,x_n))

where ? are the functions you still need to find. Don't look too far, the ?-functions are very easy (but make sure that the condition DF(0) is invertible).

You should use [ itex] and [ /itex] to make LaTeX. The [ tex]-brackets automatically start a new line.
 
Ok, thanks!
How about F(x1,...,xn) = (f(x1,...,xn), x2,..., xn)? Then DF(0) is a matrix that is upper triangular, with 1s on the diagonal except for the first diagonal which we know is non-zero, then the det is non-zero, so it is invertible. Also we know that F(0,...,0) = (0,...,0). So by the inverse function theorem F is a diffeomorphism.

But I'm still not sure about how this helps me with g? If we look at the first component, the inverse of F will map f(x1,...,xn) to x1, so I can kind of see that it is true, but I don't know how to write this properly.
 
What if we just choose g=F-1, won't that work?
 
But then doesn't that give g(f(x)) = x1 instead of f(g(x)) = x1?
 
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