Here is the statement :
$\forall n \ge 2$, if $f : \mathbb{R}^n \to \mathbb{R}$ is a continuous convex function whose the non empty set $f^{-1}\{(0) \}$ is compact then $\lim \limits_{\Vert x \Vert \to +\infty } f(x)=+\infty.$
It seems similar to Whitney's statement on the set of zeroes of a continuous function.
So, it would be great if someone had references about this result.
Thanks in advance !
For convenience, translate so $f(0) = 0$. Take $R$ large enough that $f^{-1}(\{0\})$ is contained in the open ball $B$ of radius $R$ about $0$. Since $f$ is continuous and the complement of this ball is connected, $f$ is either always positive or always negative outside $B$. By looking at the restriction of $f$ to a line through $0$, we see that it must be positive. Moreover, if $m = \inf_{\|x\| = R} f(x)$, we find that for every $s$ on the sphere $S = \{x: \|x\| = R\}$ and $t > 1$, since $s = (1-\frac1t) 0 + \frac{1}{t} (ts)$, $ f(s) \le \frac{1}{t} f(ts)$, i.e. $f(ts) \ge t f(s)$, and thus taking $t = \|x\|/R$ and $s = R x/\|x\|$, $ f(x) \ge m \|x\|/R $ for $\|x\| \ge R$.