Singular support of a distribution is {0}.

Question

If T is a Distribution function with support {$0$} then T can be written as a linear combination of $\delta$ and its derivatives.

Can anything be said about a distribution with singular support {$0$}?

Answer 1

If $u \in \mathcal{D}'(\Omega)$ is a distribuction, the singular support $\mathrm{singsupp}(u)$ of $u$ is the set of all points $x \in \Omega$ such that does not exist open neighborhood $U\subset \Omega$ where $u_{|U}$ is a smooth function \begin{align*} \displaystyle \mathrm{singsupp}(u) = \lbrace x \in \Omega : \nexists U \subset \Omega : u_{|U} \in \mathcal{E}(U) \rbrace \end{align*} in other words if $U_{max}$ is the maximum open in $\Omega$where the distribuction $u \in \mathcal{E}(U_{\max})$ then \begin{align*} \displaystyle \mathrm{singsupp}(u) = \Omega \setminus U_{max} \end{align*} Note that since $u=0$ in the largest open $G \subset \Omega$,then $\mathrm{singsupp}(u) \subset \mathrm{supp}(u)$.

So in general I think you can not apply the theorem, there is little information.