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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.2012-06-07
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    Yes, you are right! Thank you.2012-06-07
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    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).2012-06-07
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    ... So you need nonpositive curvature. If $M$ is CAT(0), then the distance function to $C$ is convex; see p.178 of Bridson-Haefliger.2012-06-07

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