Let $\mathcal H$ be a Hilbert space and $f: \mathcal H \to \mathbb R$ a $ C^2$ function. Suppose $z \in \mathcal H$ is a critical point of $f$, and the Hessian of $f$ is strictly positive at $z$.
Can we prove that $z$ is a local minimum point of $f$?
No, a counterexample exists on this site but I could not find it.
Take $H=l^2$, and set $$ f(x) = \sum_k \frac1k x_k^2 - 2x_k^3. $$ Then $f'(0)=0$, $f''(0)(d,d)>0$ for all $d\ne 0$, but $0$ is not a local minimum, as $f(e_n/n) <0$. (Here $e_n$ is the sequence of unit vectors)