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 !

  • 2
    $\begingroup$ You need the zero set to be non-empty: think of $f(x)=e^{-x_1}$. $\endgroup$
    – Mindlack
    Jun 11 '19 at 19:51
  • 1
    $\begingroup$ What do you mean by $x \to +\infty$ if $x \in \mathbb R^n$? Maybe you mean $|x| \to +\infty$? $\endgroup$ Jun 11 '19 at 20:22
  • $\begingroup$ @Mindlak Indeed my mistake ! $\endgroup$
    – Maman
    Jun 11 '19 at 20:43
  • $\begingroup$ @RobertIsrael That's it ! $\endgroup$
    – Maman
    Jun 11 '19 at 20:44

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 $\|x\| \ge R$.

  • $\begingroup$ Thank you but, I do not see why if we look at the restriction of $f$ to a line through $0$, $f$ is positive ? $\endgroup$
    – Maman
    Jun 11 '19 at 22:28
  • 1
    $\begingroup$ If $g$ is a convex function on $\mathbb R$, you can't have $g(0)=0$, $g(R) < 0$ and $g(-R) < 0$. $\endgroup$ Jun 12 '19 at 0:59

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