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Let $f : \mathbb{R}^n \rightarrow \mathbb{R}, f \in C^2$. I have to prove that for every nondegenerate critical point of $f$, there exists a neighbourhood which does not contain any further critical points.

I know that I will have to look at the Taylor formula and I'm pretty sure that I have to use that $\det (\partial_{ij} f(x)) \neq 0$ for all $x$ which are nondegenerate, that is to say the Jacobi-matrix is invertible. This makes me feel like I should use the Inverse function theorem, but to what end?

I'd be really delighted if I could get some advice. Thanks in advance.

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You're on the right track. Apply the inverse function theorem to $({\partial f \over \partial x_1}(x),...,{\partial f \over \partial x_n}(x))$ on a neighborhood of the critical point.

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Let $x_0$ be the critical nondegenerate point. As far as I know, the inverse function theorem tells me that there exist neighbourhoods $U$ of $x_0$ and $V$ of $0$, such that the inverse function of $Df$ is continuously differentiable. But how do I proceed now? –  Huy Mar 27 '11 at 18:02
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It will say that $({\partial f \over \partial x_1}(x),...,{\partial f \over \partial x_n}(x))$ is one-to-one on a neighborhood of the critical point (since it has an inverse). So each value can be taken at most once. This includes the value $(0,...,0)$. So there can be just the one critical point. –  Zarrax Mar 27 '11 at 18:15
    
Wow, this is incredibly easy... So I actually don't even need Taylor's formula? –  Huy Mar 27 '11 at 18:28
    
no need for Taylor :) –  Zarrax Mar 27 '11 at 18:29
    
Incidentally, another way to see this is to use the fact that around a nondegenerate critical point, a smooth function $f $ looks like $\sum x_i^2 - \sum x_j^2$ (in suitable coordinates). –  Akhil Mathew Mar 27 '11 at 20:56

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