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I have a question regarding the isolation of critical points of a function:

Suppose $f : \mathbb{R}^n \to \mathbb{R}$ is a $C^\infty$ function such that $f$ has a non - degenerate critical point at $0 \in \mathbb{R}^n$. That is, we have $\triangledown f (0) = 0$, and the Hessian $\left((\partial^2 f/ \partial x_i \partial x_j ) (0) \right)$ is invertible.

Can I deduce from this that the critical point at $0$ is an isolated critical point ? My guess is to say yes, because the fact that the Hessian is non-degenerate forces $f$ to change its value in the vicinity, and in all directions. But I am unsure, in particular with regards to the last statements ("in all directions") - though that should be encoded by the fact all eigenvalues of the Hessian are non - zero.

Is this the right way to think about the question of whether the critical point is isolated ? Thanks for your thoughts !

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Yes, this is a right way of thinking about the question. You can see this in several ways.

1: The Morse lemma: The Morse lemma declares that about any nondegenerate critical point $p$ of a smooth function $f$ there is a coordinate neighborhood $x_i$ so that in those coordinates, $f(x_i) = x_1^2 + \cdots + x_i^2 - x_{i+1}^2 - \cdots - x_n^2 + f(p)$. It's an easy exercise to compute the gradient in these coordinates, and you can immediately see that the only zero of the gradient in the neighborhood is at $p$. (In this case, $p = 0$ and the claim is that you can precompose $f$ with a diffeomorphism $\varphi$ of $\mathbb{R}^n$ so that $f\circ\varphi = \sum x_i^2 - \sum x_j^2$.)

2: Directly: Take a curve (easiest, a straight line) $\gamma$ through $0$, and for simplicity's sake, parametrize the curve so that $\gamma(0)=0$. Consider $f|_\gamma$. Then $f'(t) = (\nabla f\cdot \gamma')(t)$ and $f''(t) = (\gamma'^THess_f\gamma')(t)$. Leaving out some details you should verify as an exercise, the Hessian is invertible, so in a neighborhood of $0$, $f$ has a local extremum, hence the zero of $f'$ is isolated. This is true for any curve, thus, every direction, so $0$ is an isolated critical point. This formalizes your intuition about "every direction."

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it's an extremum, not an extremeum :-) –  user20266 May 10 '12 at 18:35
    
It's an extremuseum! Thanks :) –  Neal May 10 '12 at 20:15
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Yes. But the Morse Lemma is a nice and worthy exhibit. –  user20266 May 11 '12 at 16:01
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