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Let $G$ be a Lie group and $(\pi,V)$ be a continuous representation of $G$ on $V$ a Fréchet space. Let $dx$ denote the Haar measure on $G$.

The representation $\pi$ induces a representation of $M_c(G)$ - the space of bounded Borel measures on $G$ of compact support, defined as

$\pi(\mu)v = \int_G \pi(x)v \, d\mu(x). $

This is a representation, since $\pi(\mu_1)\pi(\mu_2)=\pi(\mu_1*\mu_2)$. Furthermore, it is continuous, since for a seminorm $\rho$, we have $\|\pi(\mu)v\|_\rho = \Big(\int_G |d\mu|\Big) \, \|v\|_\rho. $

I would like to know if continuity holds if, instead of the mentioned measures, we work with $C^\infty_c(G)$ - the space of smooth functions with compact support, given a Frechet topology. If known, also for distributions, Schwartz space, and tempered distributions would be nice :)

In particular, I wanted to prove that if we pick a Dirac sequence [i.e. $f_n\in C^\infty_c(G)$ sequence of $L^1$-normalized positive functions with support tending to $1_G$.], how can I prove that $\| \pi(f_n)v - v \| \to 0$. The measure approach doesn't prove it, apparently.

Thank you.

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    As for the Dirac sequence, the limit follows from the continuity of $\pi$. It is strongly continuous, i.e. for any $v$ the map $x\mapsto \pi(x)v$ is continuous and hence compactly bounded for any continuous seminorm. So if we take any $\epsilon$ there is a compact neighborhood of the identity such that \|\pi(x)v - v\|_\rho < \epsilon. Hence, for all $f_n$ supported on such neighborhood, we will have \|\pi(f_n)v-v\|_\rho<\epsilon. I keep the question, as I have not been able to follow up with other general continuity issues. Cheers.2012-07-03

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