Solve the functional equation $2f(x)=f(ax)$ for some $a$.

I am trying to solve the following functional equation, and could use some help. $$2f(x)=f(ax)$$

For some $a\in\mathbb{R}$. By repeated adding $2f(x)$ together we notice that $$2nf(x)=f(a^nx).$$

Also $$2mf(a^{-m}x)=f(x)\Rightarrow \frac{1}{2m}f(x)=f(a^{-m}x).$$

Putting it all together I have for all $m,n\in\mathbb{N}$, $$\frac{n}{m}f(x)=\frac{2n}{2m}f(x)=f(a^{n-m}x)$$

I am unsure of how to proceed from here.

Answer

I give only some hints for the case $a>0$, $a\not =1$, and $f$ defined on $I=]0,+\infty[$.

Put $\displaystyle g(x)=\exp(-x\log 2)f(\exp(x\log a))$. Then show that $f(au)=2f(u)$ for all $u>0$ is equivalent to $g(x+1)=g(x)$ for all $x\in \mathbb{R}$. Hence we get that the solutions are the functions $f$ of the form $\displaystyle f(x)=g(\frac{\log x}{\log a})\exp(\frac{x\log2}{\log a})$ for any $g$, periodic of period $1$ on $\mathbb{R}$.