Follow-up to $f(x)^2 = f(\sqrt2 x)$

Question

This is a follow-up to: Solving $(f(x))^2 = f(\sqrt{2}x)$ .

So $f : \Bbb R \to \Bbb R$ is $\mathcal C^2$ and verifies $\forall x,\, f(x)^2 = f(\sqrt2 x)$.

We already know that $f(0) \in \{0,1\}$ and that if $f(0)=1$, $f$ is strictly positive and $f(x) = e^{\lambda x^2}$.

The question is: what if $f(0) = 0$? Specifically:

• are there solutions other than $0$?
• are there solutions where $f(0) = 0$ and $f(x\neq0)\neq0$?
• what would be generic descriptions of solutions with $f(0) = 0$?

Answer 1

Assume that $f(a)=A\ne0$ for some $a\in{\mathbb R}$. Then $A=f^2\bigl({a\over\sqrt{2}}\bigr)>0$, and $$f\left({a\over2^{n/2}}\right)=A^{1/2^n}\qquad(n\geq1)\ .$$ Taking the limit as $n\to\infty$ we conclude that $f(0)=1$. You already have found out that in this case necessarily $f(x)=e^{\lambda x^2}$ for some $\lambda\in{\mathbb R}$.

This shows that $f(0)=0$ leads to the sole solution $f(x)\equiv0$.