$f$ convex, $\lim_{x\to\infty}\frac{f(x)}{x}=0$, then $f$ is constant

Question

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?

Answer 1

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.