$f$ convex, $\lim_{x\to\infty}\frac{f(x)}{x}=0$, then $f$ is constant Let $f$ be a convex function of $\Bbb R$ and suppose $\lim\limits_{x\to\pm\infty}\frac{f(x)}{x}=0$.
How we can prove that $f$ is constant function?
 A: This answer  is based on the answers here:
Show bounded and convex function on $\mathbb R$ is constant
So assume $f$ is not constant on $\mathbb R$. Then there are numbers $a<b$ such that $f(a) \ne f(b)$.
Suppose $f(a)<f(b)$. Then take $c>\max(b,0)$. The number $b$ is a convex combination of $a$ and $c$:
$$
b = a \frac{c-b}{c-a} + c \frac{b-a}{c-a} 
$$
Convexity of $f$ implies
$$
f(b) \le \frac{c-b}{c-a} f(a) + \frac{b-a}{c-a} f(c),
$$
or equivalently
$$
\frac{c-a}{b-a}f(b) \le \frac{c-b}{b-a} f(a)  + f(c),
$$
dividing by $c$ yields
$$
\frac{c-a}c \frac{f(b)}{b-a} \le \frac{c-b}c\frac{f(a)}{b-a} + \frac{ f(c)}{c}.
$$
Passing to the limit $c\to\infty$ yields $f(b)\le f(a)$, a contradiction.
Hence it follows $f(b)<f(a)$. Define the function $\tilde f(x):=f(a+b-x)$. Then $\tilde f$ is convex, satisfies $\lim_{x\to\pm\infty}\frac{\tilde f(x)}x=0$.
Moreover, $\tilde f(a)  =f(b)<f(a)=\tilde f(b)$, which is a contradiction by the first part of the proof above.
It follows that $f$ is constant.

Note, how the above proof uses that both limits $\lim_{x\to+\infty} \frac{f(x)}x$ and $\lim_{x\to-\infty} \frac{f(x)}x$ are zero.
