Definition: For integer $p\geq 1$, we say $x\in \mathbb{R}^d$ is a $p$-th order stationary point of a function $f \colon \mathbb{R}^d \to \mathbb{R}$ if there exists a $C>0$ and an $\epsilon>0$ such that $f(y) - f(x) \geq - C \|y-x\|^{p+1}$ for all $y$ such that $\|y-x\| \leq \epsilon$.
Definition: $x\in \mathbb{R}^d$ is a local minimum of a function $f \colon \mathbb{R}^d \to \mathbb{R}$ if there exists an $\epsilon>0$ such that $f(y) - f(x) \geq 0$ for all $y$ such that $\|y-x\| \leq \epsilon$.
Question: Let $d \geq 2$ and $k \geq 1$. Does there exist an integer $p = p(k, d)$ such that for every polynomial $f \colon \mathbb{R}^d \to \mathbb{R}$ of degree at most $k$, if $x$ is a $p$-th order stationary point of $f$ then $x$ is actually a local minimum of $f$?
Comments:
- Naively I expected we could take $p=k$. This is true for $k=1,2$, but not true for $k=3$. For example, consider $x = (0,0)$ and $f(x_1, x_2) = x_2^2 - x_1^2 x_2 - x_1 x_2^2$.
- For $k=3$ and $d=2$, it is not too hard to show that $p=5$ works. But the proof doesn't seem to generalize.
- I suspect $p=k$ works if we restrict to requiring that $f$ is homogeneous (around $x$) [but I care about general polynomials, not homogeneous ones].
