# References for this statement on convex functions?

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.

• You need the zero set to be non-empty: think of $f(x)=e^{-x_1}$. Jun 11 '19 at 19:51
• What do you mean by $x \to +\infty$ if $x \in \mathbb R^n$? Maybe you mean $|x| \to +\infty$? Jun 11 '19 at 20:22
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 $$\|x\| \ge R$$.
• Thank you but, I do not see why if we look at the restriction of $f$ to a line through $0$, $f$ is positive ? Jun 11 '19 at 22:28
• If $g$ is a convex function on $\mathbb R$, you can't have $g(0)=0$, $g(R) < 0$ and $g(-R) < 0$. Jun 12 '19 at 0:59