fact that densities of selfdecomposable distributions fulfill an equality like can also be found in [@steutelvanharn Thm. V.2.16]. Here we see that actually all solutions to correspond to distributions in ${\mathrm{L}}^+$.

\[cor:densitySD\] Let $f(t)$ be a probability density with support $[a, \infty)$, $a\geq 0$, which is continuous on $(a, \infty)$. Then $f$ corresponds to a selfdecomposable distribution, if and only if $f$ fulfills $$\begin{aligned}
\label{eq:densitySD}
 (a-t) f(t) + \int_a^t  \nu ((t-s, \infty))  f(s) ds=0, \quad t\geq a,\end{aligned}$$ for some Lévy measure $\nu$ such that $\int_0^\infty \log^+(x) \nu(dx)<\infty$.

Every distribution $\mu\in{\mathrm{L}}^+$ which is non-degenerate is absolutely continuous and can be represented as $\mu=\Phi_\xi({\mathcal{L}}(L_1))$ for $\xi_t=t$ and some subordinator $L$ with ${\mathbb{E}}[\log^+(L_1)]<\infty$. Further ${\mathrm{supp}\,}(\mu)=[a, \infty)$, $a\geq 0$, implies by [@BLMranges Thm. 1(ii)] that $L$ has drift $a$. Hence by Theorem \[thm:density\](i) the density of $\mu$ fulfills .\
Conversely, if $f(t)$ is a density with support $[a, \infty)$, $a\geq 0$, which is continuous on $(a, \infty)$ and which fulfills , then by Theorem \[thm:density\] it is the density of the exponential functional $\int_{(0,\infty)} e^{-t} dL_t$ for some subordinator $L$ with Lévy measure $\nu$ and drift $a\geq 0$. Thus we conclude that $\mu\in{\mathrm{L}}^+$.

Semi-selfdecomposability
------------------------

We say a random variable $X$ (or its probability 