In this paper: Bishop, R.L.; O'Neill, B., Manifolds of negative curvature, Trans. Am. Math. Soc. 145, 1-49 (1969). ZBL0191.52002, it has been proved that there is no non-trivial smooth convex function on complete Riemannian manifold with finite volume. My question is that is there any similar analogous for quasi-convex function?

  • $\begingroup$ What is a quasiconvex function? $\endgroup$ May 26, 2017 at 13:57
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    $\begingroup$ A function $f:M\rightarrow R$ is quasiconvex if for any $x,y\in M$ and for any geodesic $\gamma:[0,1]\rightarrow M$ connecting $x$ and $y$, $f(\gamma(t))\leq max(f(x),f(y)),\ \forall t\in[0,1]$. $\endgroup$
    – MAS
    May 26, 2017 at 15:11
  • $\begingroup$ I see. Then if you assume in addition that curvature of your manifold is negative then quasiconvex functions are again constant. In general, I am not sure. $\endgroup$ May 26, 2017 at 16:20
  • $\begingroup$ @chandan mondal : Do you have an example s.t. non-constant function $f$ on a Riemannian manifold of an infinite volume is not convex but quasi-convex ? $\endgroup$
    – HK Lee
    May 27, 2017 at 6:35
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    $\begingroup$ @HKLee: $f(x)=x^3$, $x\in {\mathbb R}$. $\endgroup$ May 27, 2017 at 12:13

1 Answer 1


Counterexample : Consider a flat manifold $M=S^1\times \mathbb{R}$. For $(s,t)\in M$, define $f(s,t)=0$ for $t\leq 0$, $f(s,t)=t$ for $t\in (0,1)$ and $f(s,t)=1$ for $t\geq 1$.


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