Hong Kong competition problems, 2012

A possible function is $f(x) = 0$ for any $x \in [0, 1)$ and $f(x) = 3(4^k - 1)$ for any $x \in [2^k - 1, 2^{k+1} - 1)$ where $k \in \mathbb{Z}^+$.
It suffices to check the above function satisfies the conditions. It is clear that $f(x) \ge 0$ for any $x \ge 0$. If $x \in [0, 1)$, then $2x + 1 \in [1, 3)$. Therefore, we have
$$f(2x + 1) = 3(4^1 - 1) = 9 = 4(0) + 9 = 4f(x) + 9.$$
If $x \in [2^k - 1, 2^{k+1} - 1)$ where $k \in \mathbb{Z}^+$, then $2x + 1 \in [2^{k+1} - 1, 2^{k+2} - 1)$. Therefore, we have
$$f(2x + 1) = 3(4^{k+1} - 1) = 4 \cdot 3(4^k - 1) + 9 = 4f(x) + 9.$$
This completes the proof.