Let $G$ be a Lie group and $\pi$ a continuous action on $V$, a Fréchet space.

This action induces a representation of the space of compactly supported functions, $C_c(G)$, with convolution as product by: $f\in C_c(G) \mapsto \pi(f)\in End(V)$ where $\pi(f)v = \int_G f(g) \pi(g)v \; dg$.

In a similar fashion, we can consider the compactly supported bounded Borel measures: $\mu\in M_c(G) \mapsto \pi(\mu)\in End(V)$ where $\pi(\mu)v = \int_G \pi(g)v \; d\mu(g)$.

**Question:** Can we also define this representation for distributions on $G$? If so, is the homomorphism continuous?

I ask this because in the proof of Dixmier-Malliavin theorem we get some functions $f_n,g \in C^\infty_c(G)$ for which $\delta^n \ast f_n \to \delta + g$ in the sense of distributions ($\delta$ being the Dirac distribution). And then they say that $\pi(\delta^n \ast f)v \to v + \pi(g)v$ in $V$. How can I prove this?

For what I found easily accessible in the literature, $\pi$ on the measures is continuous with regard to the Banach space topology (norm given by the total variation measure). For $C^\infty_c(G)$, we also have continuity for some $L^1(G)$-norm. In symbols, $\|{\pi(\mu)v}\|_k \le \| v \|_{n(k)}\int_G |\mu|$, and likewise for $f\in C^\infty_c(G)$. But that does not seem to help me a lot with distributions... Or does it?