# Is every closed convex subset a sublevel set?

Is it true that on every Riemannian manifold $M$ (whether compact or merely complete), every closed convex set C in M is the sublevel set $f((-\infty,t])$ of some convex function $f : M \rightarrow \mathbb{R}$? Thank you every much!

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Given the (riemannian-geometry) tag, I suppose you mean a Riemannian manifold? Certainly, on a general manifold, convexity doesn't make any sense. – Harald Hanche-Olsen Jun 7 '12 at 14:58
Yes, you are right! Thank you. – Peter Hu Jun 7 '12 at 15:02
The equator of $\mathbb S^2$ is convex (unless I'm using a wrong definition of convexity), but is not a sublevel set for a convex function (otherwise the function would have a maximum somewhere on the sphere). – user31373 Jun 7 '12 at 15:35
... So you need nonpositive curvature. If $M$ is CAT(0), then the distance function to $C$ is convex; see p.178 of Bridson-Haefliger. – user31373 Jun 7 '12 at 15:43

1. The equator of $\mathbb S^2$ is convex but is not a sublevel set of any convex function: otherwise the function would have a maximum somewhere on the sphere.
2. But if $M$ is a $CAT(0)$ space, then the distance function to any closed convex subset of $M$ is convex. See p.178 of Metric spaces of nonpositive curvature by Bridson and Haefliger.