Statement
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be an function that is concave up and increasing. If $\displaystyle \lim_{x\to \infty}\frac{f(x)}{x}=0$, then $f$ is constant.
It'll be easy if we assume $f$ is second differentiable, the conditions just mean $f''(x)\geq 0$ and $f'(x)\geq 0$. (If $f'(x)=a>0$ at $x=b$ then for $x>b$ we have $f(x)>a(x-b)+f(b)$ by MVT and clearly in this case the limit cannot be $0$ and therefore $f'(x)$ has to be $0$ everywhere).
Can this statement be proved without assuming $f$ differentiable? Here concave up is defined by for any $x,y\in \mathbb{R}$, $t\in [0,1]$, $f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$. Increasing is defined by $y>x\Rightarrow f(y)\geq f(x)$.
Context of this problem: If the given statement is proved, then the solution in the thread How to prove Liouville's theorem for subharmonic functions given by Martin R can be extended to prove a stronger version of the result (see my comment on that answer).
Yes. For $x < y < z$ you have $$ f(y) \le \frac{z-y}{z-x} \, f(x) + \frac{y-x}{z-x}\, f(z) = \, \frac{z-y}{z-x} \, f(x) + (y-x)\frac{z}{z-x} \, \frac{f(z)}{z} . $$ For fixed $x < y$ and $z \to \infty$ it follows from $f(z)/z \to 0$ that $$ f(y) \le f(x) $$ so that $f$ is decreasing. If $f$ is both increasing and decreasing then it is constant.
Remark: The condition $\lim_{x\to \infty}\frac{f(x)}{x}=0$ can be relaxed to $\liminf_{x\to \infty}\frac{f(x)}{x}=0$ and the same conclusion holds.