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

This action induces a representation of the compact-supported function $C_c(G)$ (with convolution as product) by

$f\in C_c(G) \mapsto \pi(f)\in End(V)$ where $\pi(f)v = \int_G f(g) \pi(g)v \; dg$.

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

Now my question is: can we define it too to distributions in $G$? If so, is the homomorphism continuous? I ask this because, in the proof of Dixmier-Malliavin theorem, we get some functions $f_n,g \in C^\infty_c(G)$ which $\delta^n * f_n \to \delta + g$ in the sense of distributions ($\delta$ being the Dirac distribution), and then they say that $\pi(\delta^n * f)v \to v + \pi(g)v$ 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), and for $C^\infty_c(G)$, continuity with some $L^1(G)$-norm [That is, $\|{\pi(\mu)v}\|_k \le \| v \|_{n(k)}\int_G |\mu|$, and likewise for $f\in C^\infty_c(G)$.], and that does not seem to help me a lot with distributions... or does?