3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Yes: en.wikipedia.org/wiki/… $\endgroup$ – Giuseppe Negro Mar 9 '17 at 14:55
  • $\begingroup$ It seems it's a wrong link! The indicator function is not continuous so it cannot be convex as well. $\endgroup$ – 123... Mar 9 '17 at 15:12
  • $\begingroup$ 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. $\endgroup$ – szw1710 Mar 9 '17 at 15:21
2
$\begingroup$

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 $$C = f^{-1}(\{0\}).$$

$\endgroup$
  • $\begingroup$ 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 $C=f^{-1}(T)$? $\endgroup$ – 123... Mar 9 '17 at 16:30
  • $\begingroup$ 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. $\endgroup$ – Umberto P. Mar 9 '17 at 20:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.