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Given a convex set $U \subset \mathbb{R}^{n}$, can one always construct a convex function $f: \mathbb{R}^{n} \to \mathbb{R}$ such that its $0$-sublevel set $\{ u \in \mathbb{R}^{n} | f(u) \leq 0 \}$ is the convex set $U$?

Jinrae Kim
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    Ah, I think the distance to the convex set $U$ is a convex function which renders that the $0$-sublevel set is $U$...! – Jinrae Kim Apr 27 '23 at 02:56
  • You need to assume that $U$ is closed. Such a convex function will be continuous... – daw Apr 27 '23 at 06:32
  • @daw Does the closedness affect when $ f(u) =0 $? Or, is that assumption necessary for more general reasons? – Jinrae Kim Apr 28 '23 at 07:06
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    If $f$ is as in the question, then $f$ is continuous, and the sublevel sets are closed. – daw Apr 28 '23 at 08:04
  • @daw If the closedness of $U$ is assumed, is the distance function enough to construct such a convex function? – Jinrae Kim Apr 29 '23 at 08:56

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