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This question already has an answer here:

Does there exist a continuous function $f$ such that $f(x^2)=2f(x)$ and $f(0)=0$? How do you solve this? I understand that this is nothing like a normal equation, because you can't solve for $x$, or $f(x)$, because of that $f(x^2)

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marked as duplicate by Martin R, Dietrich Burde, Elliot G, Leucippus, achille hui Jun 1 at 19:03

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    $\begingroup$ What about $f(x)=0$ for all $x$? $\endgroup$ – lulu Jun 1 at 18:34
  • $\begingroup$ Are there any nontrivial solutions? $\endgroup$ – Math Jun 1 at 18:35
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    $\begingroup$ Is there any restriction on the domain of $x$? $\endgroup$ – Nick Guerrero Jun 1 at 18:38
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    $\begingroup$ Possible duplicate of $f(x^2) = 2f(x)$ and $f(x)$ continuous – found with Approach0 $\endgroup$ – Martin R Jun 1 at 18:41
  • $\begingroup$ $f(x) = \max( 0, \log x^2)$ $\endgroup$ – achille hui Jun 1 at 18:52
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Since $f(x)=\frac12 f(x^2)$ is even, we just need to check $x>0$. Clearly $f(1)=0$. If $x\in(0,\,1)$, $$0=f(0)=f\left(\lim_{n\to\infty}x^{2^n}\right)=\lim_{n\to\infty}f\left(x^{2^n}\right)=\lim_{n\to\infty}(2^nf(x))\implies f(x)=0.$$But we may obtain nonzero $f(x)$ if $x>1$. A comment by @achillehui notes that $f(x)=\max(0,\,\log x^2)$ works. We can generalize this a little, e.g. we may multiply by a constant, or (I think) multiply $f$ for $x\ge2$ by $1+\{\log_2(\log_2 x^2)\}$ where $\{y\}$ is the fractional part of $y$.

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For $f:\mathbb{R}^+\to\mathbb{R}$ we have a cool function that works, the $\log$ function. This works with any base, is nontrivial, and satisfies your things. To extend this to negatives, set $f(-x)=f(x)$. I'm sorry about $0$, but let's just keep it out of the domain.

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  • $\begingroup$ That's discontinuous at $x=0$. $\endgroup$ – J.G. Jun 1 at 18:52

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