I'm sorry and a little ashamed to ask this very simple question. The problem is that I'm not very familiar with functional equations (I just know that they can be tricky). The question is: which are the possible solutions of the following functional equation? $$[f(x)]^{2}=f(2x)$$ I assume that the solutions (in the real field $f:\mathbb{R}\to\mathbb{R}$) are just the constant functions: $$f(x)=1, \quad f(x)=0$$ Are there other solutions? How can one prove that these are the only possible solutions?
Thanks.
Let $f(x)=a^{g(x)}$ , where $a\in\mathbb{R}^+$ and $a\neq1$ ,
Then $a^{2g(x)}=a^{g(2x)}$
$g(2x)=2g(x)$
Let $x=2^t$ ,
Then $g(2^{t+1})=2g(2^t)$
$g(2^t)=\Theta(t)2^t$ , where $\Theta(t)$ is an arbitrary periodic function with unit period
$g(x)=\Theta(\log_2x)x$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
$f(x)=a^{\Theta(\log_2x)x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period, $a\in\mathbb{R}^+$ and $a\neq1$