Consider a $C^2$ function $f \colon \mathbb{R}^n \to \mathbb{R}$ and suppose that $\nabla f(0) = 0$ and $\nabla^2 f(0) = 0$. Suppose also that $0$ is an isolated critical point (that is, there is a neighborhood of $0$ in which $0$ is the only critical point).
Given a point $x_0 \in \mathbb{R}^n$ let $x(t)$ denote the gradient flow starting from $x_0$ at time $t$.
My question is: given a neighborhood $\mathcal{U}$ of $0$, does there exist an initialization of gradient flow $x_0 \in \mathcal{U} \setminus \{0\}$ such that $x(t) \to 0$ as $t \to \infty$ or $t \to -\infty$?
I guess that such a question must have an answer somewhere. I have been looking into the dynamical systems literature (stable/unstable/center manifolds) but I haven't found much. If the assertion is false can we construct a counter-example somehow? Would it become true if we assume that $f$ is more regular?
Thanks for your time reading this.
Edit: I should emphasize that I am not assuming that $0$ is a local minimum/maximum for $f$.
Welcome Sap!
Sure it's not a problem. Think about how a gradient flow of a smooth function always goes to the local minimum. Just consider a Taylor expansion of the function around 0. First term is $f(0)$ (a constant), the next two terms are zero (by your problem statement), and then the coefficient of the next term is proportional to $\nabla^3 f(0)$ which will dominate the closer $x$ is to $0$. So checking if this is positive (or positive definite for $n > 1$) will tell you if the steady-state at $x=0$ is locally stable or not.
The reason we usually start by checking $\nabla^2 f$ around a steady state is for the same reason above: lower order terms will dominate near the steady-state. Your $\nabla^2 f$ just happens to be zero in this case so you are checking the next order term.