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

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 $C=f^{-1}([a, b])$?

• – Giuseppe Negro Mar 9 '17 at 14:55
• It seems it's a wrong link! The indicator function is not continuous so it cannot be convex as well. – 123... Mar 9 '17 at 15:12
• Of course, $1-\mathbf{1}_C$ is quasi-convex (but not convex, so it does not answer the question). Actually I don't know a proof with a convex function in the main role. I believe it's true. I have thought about Krein-Milman theorem: $C$ is a convex hull of its extreme points. – szw1710 Mar 9 '17 at 15:21

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 $$C = f^{-1}(\{0\}).$$
• Thank you very much for the solution. What about an arbitrary convex set?, i.e., for a given convex set $C\subseteq\mathbb{R}^n$, is there a convex function $f$ and convex set $T\subset\mathbb{R}$ such that $C=f^{-1}(T)$? – 123... Mar 9 '17 at 16:30
• That may not be possible. Consider the case when $C$ is the open unit ball along with some portion (but not all) of the boundary of the unit ball. – Umberto P. Mar 9 '17 at 20:32