I know that any local minimum $x_0$ of a (twice continuously differentiable) function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ has positive semi-definite Hessian $H(x_0) \succeq 0$. Can we say moreover that there exists a neighbourhood $U$ of $x_0$ such that $$ H(x) \succeq 0 $$ for all $x \in U$? I don't see any reason for this being true since positive semi-definiteness is not a continuous property (while positive definiteness is), but can't find a counter-example.

Edit: Follow-up posted here.

  • $\begingroup$ I would think of an example of the form $f(x)=x^2\sin^2(1/x)$ or something like that. This has a global minimum at $0$, and I expect the second derivative to oscillate wildly in any neighborhood. $\endgroup$ – Giuseppe Negro Jul 11 '18 at 15:04
  • $\begingroup$ The second derivative does not exist at the origin. I added this condition in the question for clarity. $\endgroup$ – Nao Jul 11 '18 at 15:10
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    $\begingroup$ Yeah, of course. I mean, I would consider a modification of that example, it is just a conjecture and not an answer. Try putting $x^4$ instead of $x^2$ $\endgroup$ – Giuseppe Negro Jul 11 '18 at 15:24
  • $\begingroup$ Your intuition was right Giuseppe...! $\endgroup$ – Nao Jul 11 '18 at 16:22
  • $\begingroup$ Thanks, but I didn't do anything, all credit goes to Will Jagy. $\endgroup$ – Giuseppe Negro Jul 11 '18 at 16:31

No. Take$$\begin{array}{rccc}f\colon&\mathbb{R}&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}x^4\left(2+\sin\left(\frac1x\right)\right)&\text{ if }x\neq0\\0&\text{ otherwise.}\end{cases}\end{array}$$Then $f$ is twice differentiable, it has a local minimum at $0$ and $f''(0)=0$. But there are critical points $x_0$ of $f$ as close as you wish from $0$ such that $f''(x_0)<0$.

  • $\begingroup$ Thank you José, this works. Do you know whether a counter-example exists for analytic functions? I will post this as a separate question tomorrow. $\endgroup$ – Nao Jul 11 '18 at 16:23
  • $\begingroup$ I don't think so. Do that: post it as a separate question. $\endgroup$ – José Carlos Santos Jul 11 '18 at 16:26
  • $\begingroup$ Will do as soon as I'm home, and thanks again. It would be interesting to see if this can be proved for analytic functions, because I'm happy to assume this in my work. $\endgroup$ – Nao Jul 11 '18 at 16:27
  • $\begingroup$ Follow-up question posted here. Any help appreciated. $\endgroup$ – Nao Jul 12 '18 at 13:18

$$ e^{ \left(\frac{-1}{x^2} \right)} \; \sin^2 \left(\frac{1}{x^2} \right) $$

is $C^\infty$



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