how can we solve this equation without mellin transform?

Question

given the equation (functional equation)

$$ f(x)+f(2x)+f(3x)+.... =g(x) $$

we can use the Mobius tranform to obtain

$$ f(x)=\sum_{n=1}^{\infty}g(nx)\mu(n) $$

however, what can we do with the equation

$$ f(x)-f(2x)+f(3x)-f(4x)+... =g(x) $$ ?

how can we invert this equation to get the solution $ f(x)$ ??

Answer 1

I suppose that $g$ is given and you are looking for $f$. We can remark that $$\begin{split}g(x)+2g(2x)& =f(x)-f(2x)+f(3x)-f(4x)+\cdots+2f(2x)-2f(4x)+\cdots\\ &=f(x)+f(2x)+f(3x)-3f(4x)+\cdots\end{split}$$ and this gives the hint to consider $$g(x)+2g(2x)+4g(4x)+\cdots=f(x)+f(2x)+f(3x)+\cdots$$ and you're back to the previous problem.