How do I find a function $f$ such that $f(x^2)=2f(x)$?

Question

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)

Answer 1

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.

Answer 2

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 $y$.