Let $f$ be a function $(0;+\infty)\to\mathbb{R}$ with following property:
$f(ab) = af(b) + bf(a)$.
What can $f$ be?
It can be seen that functions $\delta_p(x) = px\space ln(x)$ work.
Now, let $f_0$ be a solution and $f_0(x_0)=y_0$, $x_0\neq1$. Because $\delta$ is continuous by $p$, there exists such $p_0$ that $\delta_{p_0}(x_0)=y_0$.
It can be proven that
$\forall r\in\mathbb{Q}\space f_0(x_0^r)=\delta_{p_0}(x_0^r)$.
In other words, any solution must be a union of several $\delta_{p_S}$ functions restricted to subsets $Q_S\subset\mathbb{R}^+$, which are equivalence classes of relation $a\sim b$ if $a = b^q, q\in\mathbb{Q}$.
The question is, can a solution include two or more different $\delta_p$, i.e. can it be non-continuous, i.e. can it be not of form $\delta_p$?
Best Answer
Substitute $f(x)=xg(\ln x)$ (so $g(y)=e^{-y}f(e^y)$). Then $g\colon \mathbb R\to\mathbb R$ satisfies the Cauchy's equation $g(a+b)=g(a)+g(b)$, whose solutions are described in a known way (and may well be discontinous).