Show bounded and convex function on $\mathbb R$ is constant
Solution 1:
Suppose that $f$ is a convex function that is not constant. Then there must be $a$ and $b$ with $f(a)<f(b)$. Without loss of generality, assume that $a<b$.
Then for any $c>b$ 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.
Solution 2:
Hint: Without loss of generality, let $f(y)<f(z)$ for $y<z$. Then $$\frac{z-y}{z-x}f(x)+\frac{y-x}{z-x}f(z) \le f(y)$$ for all $x < y$. Now let $x \to -\infty$ and use boundedness of $f$.