Solving $(f(x))^2 = f(\sqrt{2}x)$
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.
Solution 1:
Hints :
- assume $f(x) > 0$ for all $x$
- write $g = \log f$
- apply the equality in $g$ twice to get a term $g(2x)$
- take the second derivative of the equality in $g$ and get $g(2x) = g(x)$
- conclude that $g''$ is constant
Additional notes :
- because $f(0)^2 =f(0)$, $f(0) \in \{0,1\}$
- if $\exists x \mid f(x)=0$, then $f(0)=0$ (because $f(x/\sqrt2) = \ldots = f(x/\sqrt2^n) = 0$ and $f$ is continuous in $0$
- as pointed here, if $\exists a \mid f(a)>0$, $f(a/\sqrt{2}^k) = f(a)^{\frac1{2^k}}$ and $f(0)=1$ by continuity of $f$ in $0$
So either:
- $f(0) = 1$, and then $f$ is strictly positive and $f(x) = e^{\lambda x^2}$,
- or $f(0)=0$ and $f = 0$.
PS: as Yves' excellent post shows, relaxing the $\mathcal C^{\infty}$ assumption, even only in $0$, generates a wide class of additional solutions.
PPS: I've opened a new question to see what happens if we relax some of these conditions here: $f(\alpha x) = f(x)^{\beta}$ under different constraints
Solution 2:
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 $C=-1,A=1$:
