I need a limit definition for the Hessian, does this work?

Question

Let $f: \mathbb{R}^n \to \mathbb{R}$ be of class $C^2$. Let $x$ be a non-degenerate critical point of $f$. Prove that there is an open neighborhood of $x$ which contains no other critical points of $f$.

$\textbf{Proof:}$ Suppose the contrary. Then for every $n \in \mathbb{Z}_+$ there is an $x_n \in B\left(x,\frac{1}{n}\right)$, which is a critical point of $f$ not equal to $x$. So $x_n$ converge to $x$. Now let $v_n = \frac{x_n - x}{||x_n-x||}$. Let us just assume $v_n$ converges to some $v \in \mathbb{S}^{n-1}$ (since $v_n$ is a bounded sequence, it has a convergent subsequence). So by the definition of the Frechet derivative, and since $x_n \to x$, we have that \begin{align*} 0 &= \lim_{n\to\infty} \frac{||Df(x_n) - Df(x) - D^2f(x)(x_n -x)||}{||x_n - x||}\\ &=\lim_{n\to\infty} \frac{||- D^2f(x)(x_n -x)||}{||x_n - x||}\\ &=\lim_{n\to\infty} ||D^2f(x)v_n||\\ &= D^2f(x)v, \end{align*} where we use the fact that $Df(x) = Df(x_n) = 0$ for all $n \in \mathbb{Z}_+$ ($x$ and $x_n$ are critical points). This implies that $D^2f(x)v = 0$, but since $x$ is a nondegenerate point, $D^2f(x)$ is invertible, and since $v \ne 0$, we have a contradiction. Thus, there is an open neighborhood of $x$ that contains no other critical points of $f$.

Does this approach work? I am not sure if I used a correct definition for the Hessian. Also, I did not use that the Hessian is continuous, which worries me. Regardless of whether this is correct or not, is there a direct approach, or even maybe a more elegant one?

Answer 1

That argument looks fine to me. Another way would be to use Morse's lemma to say that after a coordinate change, for some $m$ one has $f(x+h) = f(x) + \sum_{i = 1}^m h_i^2 - \sum_{i = m+1}^n h_i^2$ for small enough $|h|$. Therefore if $h_i \neq 0$, ${\partial f \over \partial x_i}(x + h)$ is nonzero in the new coordinates.

Answer 2

Your proof, as it is, is correct! You do not need continuity of the second derivative, only that it is non-degenerate. Indeed, the following is universally true:

Lemma 1: If $g: R^n \to R^m$ (not necessarily $C^1$) has injective derivative at $x_0$ (as a linear map), then there is a neighborhood of $x_0$ on which no other point maps to $g(x_0)$.

In the above, you will apply this to $g=Df$, $n=m$, as people pointed out. Indeed, your proof is basically how this lemma is proved!

Last, probably your proof will benefit from justifying the very first line of calculations, which is equivalent to saying that Hessian is the Frechet derivative of $Df$:

Lemma 2: If $f\colon R^n \to R$ is $C^2$, then $$ \lim_{h \to 0} \frac{\nabla f(x_0+h) - \nabla f(x_0) - Hf(x)h}{|h|} = 0 \, . $$