Any convex set is pre-image of a convex set under a convex function

Question

Let $C\subseteq R^n$ be a compact convex set. Is there a convex function $f : R^n\to R$ and real numbers $a\leq b$, such that $C=f^{-1}([a, b])$?

Answer 1

The distance function $$f(x) = \newcommand{\dist}{\mathrm{dist}}\dist(x,C) = \inf_{c \in C} |x-c|$$ should do the job.

Select two points $x,y \in \mathbb R^n$ and let $0 \le t \le 1$. Select two points $c,d \in C$. You have $tc + (1-t) d \in C$ by convexity so that $$\dist(tx + (1-t)y,C) \le |(tx + (1-t)y) - (tc + (1-t)d)| \le t|x-c| + (1-t)|y-d|.$$

Take the appropriate infima over $c$ and $d$ individually to find $$\dist(tx + (1-t)y,C) \le t \dist(x,C) + (1-t) \dist(y,C).$$

Thus $$f(tx + (1-t)y) \le tf(x) + (1-t) f(y)$$ meaning $f$ is convex. Since $C$ is closed you have $f(x) = 0$ if and only if $x \in C$ so that $$C = f^{-1}(\{0\}).$$