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This one seems intuitively obvious to me but I don't know how to prove it. Suppose you have a compact manifold $M$ with a function $f$ defined on it. Given two points $x$ and $y$ on the manifold, let $\gamma_{xy}: [0,1]\rightarrow M$ be the geodesic between $x$ and $y$. Then we say a function is convex if, $\forall x, y\in M, \lambda \in [0,1],\,\,\,\, f(\gamma_{xy}(\lambda)) \le (1-\lambda)f(x) + \lambda f(y)$

I can't think of any function on a compact manifold for which this is actually true, and I suspect that there are no convex functions on compact manifolds. Can anyone point me to the relevant proof?

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It's hard to attain maximum when you are convex. –  user53153 Mar 8 '13 at 3:54
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Hint: Every non-constant smooth function on a compact manifold has two critical points.

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What about continuous but non-smooth functions? –  David Pfau Mar 8 '13 at 4:03
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@DavidPfau Without smoothness, argue as follows: let $s=\max_M f$ and $N=f^{-1}(s)$. On one hand, $N$ is closed. On another hand, every geodesic passing through a point of $N$ must be contained in $N$. Hence $N$ is also open. –  user53153 Mar 8 '13 at 4:32
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