Solving $x\partial_x f = 0$ over 'normal' functions is the same as solving $\partial_x f = 0$, i.e. one gets $f(x)=c_1$ as the complete answer. But over distributions (if my calculations are correct), $$ f(x) = (c_2-c_1)H(x)+c_1 $$ is the complete solution (with H being the Heaviside step function). For another comparison point, $x^2\partial_x f=0$ has solution $$ f(x) = (c_2-c_1)H(x)+c_1+c_3\delta(x) $$ (with $\delta$ the Dirac $\delta$ function/distribution).

My main question: are my computations correct? Are these in fact the most general solutions? [I have 3 different arguments showing that these are indeed solutions, although I am not sure that any of these constitute proper proofs, the last time I did anything with distributions was almost 20 years ago].

Motivation: What I am actually trying to do is to get a differential equation for the density function for the Pareto Distribution (where 'distribution' here is used in a different sense). The only remaining problem is to properly take care of the 'jump' at $x_m$. The above should give me what I am missing to get there.

derivativeof a solution... – Mariano Suárez-Alvarez♦ Apr 20 '11 at 2:45