Does their exist a real continuous function other than $f(x)=0$ such that $f(2x) = -2f(x)$? I have a gut feeling it doesn't exist but I'm not sure how to prove/disprove it.
My attempt: Suppose there exists $a \in \mathbb{R}\setminus\left\{0\right\}$ such that $f(a) \neq 0$ . Define $x_n = \frac{a}{2^n}$ 
$f(x_{n+1}) = \frac{-1}{2} f(x_n)$ and inductively $f(x_n) = (\frac{-1}{2})^n f(a)$
What can I do from here? 

(Rewriting achille hui's comment as an answer.)
Yes, there are other functions satisfying that equation. One such function is  $f(x) = x \sin(\pi \log_2(x))$.
