By a result of Shing-Tung Yau [1974, Mathematische Annalen 207: 269-270], there are no non-trivial continuous geodesic convex functions on complete manifolds with finite volume. What happened if we relaxed assumptions, for example:
Q1 if we consider weaker condition than continuity, for example, considering "lower semi-continuity"?
Q2 if we consider a generalized notion of convexity, for example, quasi-convexity?
Any result or reference is appreciated.
Actually, this statement was proved by Richard Bishop and Barrett O’Neill [see Proposition 2.2 in “Manifolds of negative curvature.” Trans. Am. Math. Soc. 145 (1969)].
For Q1, the asnwer is "yes". Let $d$ be the dimension of the manifold. Assume $f$ is not constant.
It is easy to get a converging sequence of points, say $p_n\to p_\infty$, such that $f(p_n)\to \infty$ as $n\to\infty$. Choose $d$ points $a_1,\dots, a_d$ near $p_\infty$ in nearly ortogonal directions from $p_\infty$.
Consider polygonal geodesic lines starting from $p_n$ and going in the directions opposite to $a_1,\dots, a_d$. The ends of these lines fill a cube $\square_n$ with one vertex at $p_n$. By convexity of $f$, the values in $\square_n$ are large if $n$ is large. Note that the intersection $\square$ of all $\square_n$ have nonempty interior. It follows that $f|_\square\equiv\infty$; by convexity, $f\equiv \infty$ --- a contradiction.
For Q2, the asnwer is also "yes". Moreover, it requires only minor modification of the solution of "Volume and convexity" in my PIGTIKAL.
Instead of condition on gradinet, you can get an open set $\Omega$ such that $\phi^{t}(\Omega)\cap \Omega=\varnothing$ for any $t\ge T$. Observe that $\phi^{m\cdot T}(\Omega)\cap \phi^{n\cdot T}(\Omega)=\varnothing$ for any positive integers $m\ne n$. Then you arrive at a contradition the same way.
For Q1+Q2 the answer is "no"; that is, there is a lower-semcontinuous quasiconvex function $f$ that is not constant: Choose a rotationally symmetric manifolds of finite volume. Let $z$ be the center of rotation. Note that there is no geodesic loop with base at $z$. So we may assume that $f(z)=0$ and $f(x)=1$ for $x\ne z$.
