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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?

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    $\begingroup$ It does exist, e.g. $f(x) = x \sin(\pi \log_2(x))$. $\endgroup$ – achille hui Oct 3 '17 at 12:20
  • $\begingroup$ goodness me^ thanks $\endgroup$ – Rishi Oct 3 '17 at 12:20
  • $\begingroup$ @achillehui Argh, beat me to it :( $\endgroup$ – orlp Oct 3 '17 at 12:22
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    $\begingroup$ 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. $\endgroup$ – lulu Oct 3 '17 at 12:22
  • $\begingroup$ @achillehui - Oh, ok, thanks. $\endgroup$ – uniquesolution Oct 3 '17 at 12:25
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(Rewriting achille hui's comment as an answer.)

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

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