# 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.

• The equation $2nf(x)=f(a^nx)$ does not hold for $a=2, f(x)=x$... – gebruiker Nov 22 '14 at 17:35
• It seems like it should be $f(a^{n}x) = f(a(a^{n-1}x))=2f(a^{n-1}x)$, so by induction we find $f(a^{n}x) = 2^{n}f(x)$. – Alex Wertheim Nov 22 '14 at 17:37
• oop, you are right – kjhvzd Nov 24 '14 at 4:18

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 $\mathbb{R}$.

• That is a beautiful proof. Can you give me the intuition as to why you thought to make that particular function transformation?Also any ideas on how to extend this for $f$ defined on $R$? – kjhvzd Nov 24 '14 at 4:20

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 $2$