Assume I have a probability density function $\rho$ on $R$. (e.g $\rho \geq 0$ $\int \rho dx =1 $ $\rho \in L^1(R)$ ...). So $\rho$ is the density wrt the lebesgue measure.
Now I try to understand if $\rho$ itself has again a weak derivative.
Is this true in general?
If not: What are conditions on $\rho$ to have weak derivative?
Edit: Is absolute continuity of $\rho$ in this case equivalent to existence of a weak derivative?
(I know absolute cont ==> weak derivative.. Is the other direction also true? So is absolute continuity of $\rho$ necessary and sufficient?)
Any function has a distributional derivative - I'm guessing you want the derivative to also be an $L^1$ function, i.e. $\rho \in W^{1,1}(\mathbb R)$? In this case you're almost right - $W^{1,1}(\mathbb R)$ is the space of absolutely continuous functions that are also $L^1$ with $L^1$ derivatives. If you restrict yourself to a bounded interval (or work with locally weakly differentiable functions) then the notions coincide.