TITLE: How to show that $f\equiv 0$?
QUESTION [6 upvotes]: Let $f:[0;1]\to \mathbb{R}$ be a continuous function satisfying
$$f\left(\frac{x}{2}\right) + f\left(\frac{x+1}{2}\right)=3f(x).$$
How to show that $f\equiv0$?

REPLY [12 votes]: Hint:
Show that $f$ does not get maximum and minimum in the interval, which leads to that it's constant and conclude that it is the $0$ function
Update:
Basically the idea is to let $x_0$ be the point where $f$ gets its maximum.
Then we have:
$$
3f(x_0) = f(\frac{x_0}{2}) + f(\frac{x_0+1}{2}) \le 2f(x_0) \implies f(x_0) \le 0 \implies f(x) \le 0, x \in [0,1]
$$
Similarly, if $f$ gets its minimum at $x_1$ we get $f(x) \ge 0$ for $x \in [0,1]$