Functional equations: $f(x^3) = \frac{1}{3}f(x)$ and $g(x/b) = b^{a}g(x)$ (uniqueness)

Question

Here are two functional equations, both of which come up in the theory of 2nd order phase transitions in statistical physics: $$f(x^3) = \frac{1}{3}f(x)$$ and $$g(x/b) = b^{a}g(x)\rm{,}$$ where $b$ is arbitrary positive real number and $a$ is a fixed real number. For both of these it is pretty easy to write down a solution by "inspection". For the first $$f(x)=\frac{C}{\log x}$$ works, while for the second $$g(x)=\frac{D}{x^a}$$ works, where $C$ and $D$ are constants. But how would one prove that these are the only possible solutions (if indeed they are)?

Answer 1

For the second one, note that if there is a real number $a$ such that $$g(\tfrac{x}{b})=b^ag(x),$$ for all $x,b\in\Bbb{R}$, then in particular for $x=1$ and $b=y^{-1}$ you have $$g(y)=y^{-a}g(1),$$ so for the constand $D:=g(1)$ you indeed have $$g(x)=\frac{D}{x^a},$$ for all $x\in\Bbb{R}$.

For the first one, there are many more solutions, unless you assume $f$ to be continuous. As it stands, you can partition $\Bbb{R}$ into subsets of the form $$P_x:=\{x^{3^n}:\ n\in\Bbb{Z}\}.$$ Then by the functional equation, given the value of $f(x)$ for some $x$, the value of $f(y)$ is determined for all $y\in P_x$. But the values on different sets are independent. For example, $$f(x)=\begin{cases}\frac{1}{\log(x)}&\text{ if } x=2^{3^n} \text{ for some }n\in\Bbb{Z}\\ 0&\text{ otherwise}\end{cases}$$