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Let $\pi$ be a continuous irreducible representation of $G:=\mathrm{SL}(2,\mathbb{R})$ in a Banach space $H$, and $\pi^1$ the representation of $\mathcal{C}_c(G)$ induced by $\pi$. Suppose $\mathcal{S}$ is an $\mathrm{L}^1$-dense subspace of $\mathcal{C}_c(G)$. Why does $\mathcal{S}$-invariance imply $\mathcal{C}_c(G)$-invariance for every closed subspace of $H$? In his book on $\mathrm{SL}(2,\mathbb{R})$, Serge Lang makes use of this in the proof of the fact that $K$-finite vectors are dense, $K$ denoting $\mathrm{SO}(2)$ as usual (Theorem 3, p. 24). The answer is clear when $\pi$ is a bounded (in particular, when $H$ is a Hilbert space and $\pi$ is unitary), since in this case $\pi^1$ has a continuous extension to $\mathrm{L}^1(G)$.

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With due respect to Lang's memory, whom I deeply admire: in 1981-82, when I was a 2nd year graduate student, my supervisor at Paris gave me this book $SL(2,\mathbb{R})$ and commented: "If you want to learn representation theory, find all the mistakes in Lang's book!" That was an extremely good advice. –  Alain Valette Apr 21 '13 at 19:30
    
Let me record here that $K$-finite vectors are dense in Hilbert representations—neither irreducibility nor boundedness is required. Indeed, one can construct a new inner product on $H$ which induces the same topology and with respect to which $\pi|_K$ is unitary. It follows then by Peter-Weyl theorem that $K$-finite vectors make up a dense subspace. –  Murat Güngör Jun 7 '13 at 11:19
    
Let me also record that in Hilbert representations $\mathcal{S}$-invariance implies $\mathcal{C}_c(G)$-invariance for every closed subspace if $\mathcal{S}$ is taken to be $\sum S_{n,m}$ (as in Theorem 3): For any open set $U\subseteq G$ with compact closure, $\pi^1:\{f\in\mathrm{L}^1(G)\mid\mathrm{supp}f\subseteq U\}\to\mathrm{End}(H)$ makes sense and is continuous. Consequently, for each $f\in\mathcal{C}_c(G)$ one can find by using the last assertion of Lemma 1 on p. 19 a sequence $(f_n)$ in $\mathcal{S}$ such that $\pi^1(f_n)\to\pi^1(f)$ in $\mathrm{End}(H)$. The result follows. –  Murat Güngör Jul 14 '13 at 14:11
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