Show bounded and convex function on $\mathbb R$ is constant How can we show that a bounded and convex function on $\mathbb R$ is constant? Derivatives are of no use since the function does not have to differentiable. I saw an answer here I think a while ago but did not understand it at all.

Since derivatives are useless, we would have to use the definition and somehow show that the function lies between two values which are equal to each other. But I am unable to progress any further.
Suppose that $f$ is a convex function that is not constant. Then there must be $a$ and $b$ with $f(a)&lt;f(b)$. Without loss of generality, assume that $a&lt;b$.

Then for any $c&gt;b$ we must have, due to convexity,
$$ f(c) \ge f(a) + (c-a)\frac{f(b)-f(a)}{b-a} $$
which grows without bounds as $c\to \infty$. So $f$ cannot be bounded.
I don't think that any of this is true: consider the function f(x) = \exp(-x) on [0,\inf). This function is strictly convex and bounded but not constant.