Suppose $f:\mathbb{R}^2\to\mathbb{R}$ is a $\mathcal{C}^1$ proper map having a local but not global minimum, then $f$ must have another point where the gradient is $0\,$.
My attempt.
Suppose that the local minimum is achieved in $P=(x_0,y_0)$ and $f(P)=m$. Since $f$ is proper then it has to be unbounded or from below or from above or both. Since $P$ is a local minimum there must exist another point $Q$ where $f(Q)<m$ but at the same time there exist $\varepsilon>0$ such that $f(R)>f(P)$ for every $r\in B(P,\varepsilon)$. From this point I'm unable to proceed. Any ideas?
Thanks.