About the linear functional equations: $f(x + a) = bf(x)$ and $f(ax) = bf(x)$, Marek Kuczma e Polyanin A.D. they got the respective solutions (http://eqworld.ipmnet.ru):
$f(x) = g(x)b^{x/a}$, where $g(x) = g(x + a)$ is an arbitrary periodic function with period $a$.
And
$f(x) = g(\log x)x^{\log b/\log a}$, where $g(x) = g(x +\log a)$ is an arbitrary periodic function with period log(a).
By the induction method I got the particular solutions: $f(x) = Cb^{x/a}$ and $f(x) = Cx^{\log b/\log a}$, where $C$ is an arbitrary constant. But I did not understand how they arrived at the generic solutions with the arbitrary periodic function "$g(x)$"?
Would anyone have a demonstration of how they arrived at the generic solutions including the arbitrary periodic functions?
I searched all over the net and found no proof and no book accessible. Thank You.
I'll explain with a very simple example.
From the equation
$$f(x+1)=f(x)$$ you will conclude $$f(x)=c.$$
But as $x$ is a continuous variable it can take fractional values. Given that e.g. $f(0)$ and $f(0.3)$ are unrelated by the equation, you might very well have $f(0)=f(1)=f(2)=\cdots=4$ while $f(0.3)=f(1.3)=f(2.3)=\cdots=-5$.
In fact, $c$ is not a constant but a function of the fractional part of $x$, or if you prefer, $f$ is an arbitrary periodic function of period $1$.
For example,
$$f(x)=e^{\sin(2\pi x)}$$ or
$$f(x)=(x-\lfloor x\rfloor)^2$$ are solutions.