Does their exist a real continuous function other than $f(x)=0$ such that $f(2x) = -2f(x)$?

Question

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?

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 03 '17 at 12:22

score 1 · Answer 1 · answered Oct 03 '17 at 15:14

1

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

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

answered Oct 03 '17 at 15:14

Tanner Swett

10,624

Does their exist a real continuous function other than $f(x)=0$ such that $f(2x) = -2f(x)$?

1 Answers1