# 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?

• It does exist, e.g. $f(x) = x \sin(\pi \log_2(x))$. – achille hui Oct 3 '17 at 12:20
• goodness me^ thanks – Rishi Oct 3 '17 at 12:20
• @achillehui Argh, beat me to it :( – orlp Oct 3 '17 at 12:22
• To force the function to be $0$ you need to control the behavior near $0$ a bit. I think that requiring $f'(0)=0$ should be enough. – lulu Oct 3 '17 at 12:22
• @achillehui - Oh, ok, thanks. – uniquesolution Oct 3 '17 at 12:25

## 1 Answer

(Rewriting achille hui's comment as an answer.)

Yes, there are other functions satisfying that equation. One such function is $f(x) = x \sin(\pi \log_2(x))$.