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

Solution 1:

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 $\mathbb{R}$.

Solution 2:

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 $2$