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If $u$ is a unitary representation of a locally compact group $G$ on a Hilbert Space $\mathcal{H}$ then $\pi_u :L^1(G)\rightarrow \mathcal{B}(\mathcal{H})$ given as $$\langle \pi_u(f)\xi\vert\eta\rangle = \int f(x) \langle u(x)\xi\vert\eta\rangle dx$$ gives a non-degenerate $*$ representation of $L^1(G)$ on $\mathcal{H}$.

I have proved all parts, but I don't understand why the map $\pi_u$ is strongly continuous, i.e the map $f\mapsto \pi_u(f)\xi $ is continuous $\forall \xi \in \mathcal{H}$.
Reference: G.B.Folland's book A Course in Abstract Harmonic Analysis, page 73.

user26857
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Parish
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  • I've proved this myself and the way that I did it was by showing that $|\pi_u f| \le |f|$. The way to do this easily is by first proving it for characteristic functions (then extending by standard measure theory arguments). – Cameron Williams Mar 03 '15 at 05:07

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