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.
 A: 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 $\mathbb{R}$. 
A: Case $1$: $a=0$
Then $2f(x)=f(0)$
$f(x)=0$
Case $2$: $a=1$
Then $2f(x)=f(x)$
$f(x)=0$
Case $3$: $a>0$ and $a\neq1$
Then $f(ax)=2f(x)$
$f(aa^x)=2f(a^x)$
$f(a^{x+1})=2f(a^x)$
$f(a^x)=\Theta(x)2^x$, where $\Theta(x)$ is an arbitrary periodic function with unit period
$f(x)=\Theta(\log_ax)x^{\log_a2}$, where $\Theta(x)$ is an arbitrary periodic function with unit period
Case $4$: $a=-1$
Then $2f(x)=f(-x)$
Let $f(x)=2^{g(x)}$ ,
Then $2^{g(x)+1}=2^{g(-x)}$
$g(x)+1=g(-x)$
$g(-x)-g(x)=1$
Case $5$: $a<0$ and $a\neq-1$
Then $f(ax)=2f(x)$
$f(a^2x)=2f(ax)=4f(x)$
$f((-a)^2x)=4f(x)$
$f((-a)^2(-a)^x)=4f((-a)^x)$
$f((-a)^{x+2})=4f((-a)^x)$
$f((-a)^x)=\Theta(x)2^x$, where $\Theta(x)$ is an arbitrary periodic function with period $2$
$f(x)=\Theta(\log_{-a}x)x^{\log_{-a}2}$, where $\Theta(x)$ is an arbitrary periodic function with period $2$
