Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a $C^2$ function, and the origin is a non-degenerate critical point and suppose $f(x,mx)$ is a local minimum at the origin for all $m$, then does $f$ have a local minimum at the origin?
I understand that if the function is not degenerate, then we either have a local maximum (which we can rule out, since the function has local minimums as you approach the origin via a straight line), a local minimum, or a saddle point.
I am not sure how to rule out that it's not a saddle point. In particular, I'm not sure how you apply the non-degenerate condition rigorously.
Best Answer
Suppose that $f$ has a saddle point at $(0,0)$. Let $v\in\Bbb R^2$ be an eigenvector of the Hessian of $f$ at $(0,0)$ such that the corresponding eigenvalue $\mu$ is negative; $v$ must exist, since $(0,0)$ is a non-degenerate critical point. Consider the map $\varphi\colon\Bbb R\longrightarrow\Bbb R$ defined by $\lambda\mapsto f(\lambda v)$. Then $\varphi'(0)=0$ (since $(0,0)$ is a critical point of $f$), and $\varphi''(0)<0$ (since $\mu<0$). Therefore, $0$ is a local maximum of $\varphi$.