Let $f : \Bbb{R}^n \to \Bbb{R}$ be a $C^1$ function such that $$\lim\limits_{\|x\|\to\infty}f(x)=0$$ Show that there is a point $x_0 \in \Bbb{R}^n$ such that $Df(x_0)=0$.
I'm struggling to finish the proof. My idea was to distinguish some case. Let for example suppose that there is $z\in \Bbb{R}^n$ such that $f(z)>0$. Let $\epsilon>0$. By limit hypothesis at infinity we have the existence of $c>0$ s.t $\|x\|\ge c \implies |f(x)|<f(z)$. The idea now is to look what's going on in the set $\overline{B(0,c)}$. As $f$ is continuous and $\overline{B(0,c)}$ is a compact set, then $f$ has it's max and min there. But, I'm struggling to show that max cannot occur on the board of my close ball. If someone could help I would appreciate it. Thank you
As GEdgar commented, the maximum can't be attained in the boundary due to construction: the maximum of $f$ is not less than $f(z)$, but on the boundary of the $c$-ball the values of $f$ are strictly less than $f(z)$.
As a remark, in a slightly different way one can argue as follows also:
We may assume that $f$ is not constantly zero. Via the inverse of stereographic projection (which is a diffeomorphism), adding a new point and extending $f$ to be $0$ at the added new point $\infty$ we have a continuous function $F$ on the sphere $S^n$ that coincides with $f\circ\sigma$ everywhere except the added point, where $\sigma$ is the stereographic projection. $F$ attains its maximum and minimum, say at $p$ and $q$, respectively. They can't both be equal to $\infty$ for then $f=0$, so at least one of them is in the domain of $\sigma$, say $p$. Then $F(p)=f\circ\sigma(p)$, and $F'(p)=0$.