In the first paragraph of the section2: proof of the Theorem of the paper
Yau, Shing-Tung, Non-existence of continuous convex functions on certain Riemannian manifolds, Math. Ann. 207, 269-270 (1974). ZBL0261.53036. there is a statement: If $g(t)$ is any geodesic in $M^n$ with $\lim_{t_i\rightarrow \infty}g(t_i)=g(0)$ for some sequence $\{t_i\}\rightarrow\infty$, then $f$ is constant on the ray $\{g(t):t>0\}$, $f$ is a continuous convex function on $M^n$. I can not get the proof the this statement. Please help me. Thank you
Assume that $s_i\rightarrow \infty$ and $f(g(s_i))\rightarrow \infty $.
By continuity of $f$, $f$ is bounded on $B_\epsilon (p)$ i.e. $f\leq C$ where $p=g(0)$. So there is $i_0$ s.t. $g(t_i)\in B_\epsilon (p)$ for $i\geq i_0$. Hence $$f(g(t_i)) \leq C $$
So there is large $j,\ n_1,\ n_2$ s.t. $$ i_0\leq n_1,\ t_{n_1}<j< t_{n_2} $$ Hence $$ f(g(s_j) ) \leq C$$
This is a contraction. And since $f$ is bounded above on $g$, then it is a constant.
That is, $f$ is bounded above by $C$ so that it is constant on geodesic $g$.