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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

Copying my pseudo-comments into the answer box:

  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.

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The correct notion of "convex" in this question should be totally convex, that is for any two points in the set ALL connecting geodesics are contained completely in the set. This is easily seen to be a necessary condition for a set to be a sublevel of a convex function. – wspin Dec 19 '12 at 17:01

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