TITLE: Solving $(f(x))^2 = f(\sqrt{2}x)$
QUESTION [10 upvotes]: I would like to know how to solve this equation :
$$f(x)^2 = f(\sqrt{2}x)$$
We assume that $f : \mathbb R \to \mathbb R$ is $\mathcal C^{2}$.
The answer should be $f(x)=e^{-x^{2}/2}$, but I don't know how to show this.

REPLY [6 votes]: Setting $x=2^{t/2}$ and taking the logarithm twice,
$$(f(x))^2=f(\sqrt2x)$$
becomes
$$\log_2(\log_2(f(2^{t/2})))+1=\log_2(\log_2(f(2^{(t+1)/2})))$$
or
$$h(t)+1=h(t+1).$$
An obvious solution is $h(t)=t+c$, or $\log_2(\log_2(f(2^{t/2})))=t+c=2\log_2(x)+c$, $$f(x)=2^{Cx^2}.$$
More solutions are found by adding smooth periodic functions of period $1$, like
$$h(t)=t+A\sin(2\pi t)+c,$$
that yield
$$f(x)=2^{Cx^22^{A\sin(4\pi\log_2(x))}}.$$
Example with $C=-1,A=1$: