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Let $G=\mathrm{SL}(2,\mathbb{R})$ and $K=\mathrm{SO}(2)$. Suppose $\pi$ is a continuous irreducible representation of $G$ in a Banach space $H$. Can one always find a nonzero $v\in H$ such that $\langle \pi(k)v : k\in K\rangle$ is finite-dimensional? Serge Lang uses this implicitly in the proof of Theorem 3 on p. 24 of his book on $\mathrm{SL}(2,\mathbb{R}).$ (I know that the answer is affirmative when $H$ is a Hilbert space and $\pi$ is unitary on $K$.)

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Let me record here that $K$-finite vectors not only exist but are actually dense in Hilbert representations—neither irreducibility of $\pi$ nor unitarity of $\pi|_K$ 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 11:23

1 Answer 1

You must combine lemma 5 and lemma 1 in Serge Lang$(2,\mathbb{R})$: lemma 5 (which holds for Banach spaces) tells you how to construct $K$-finite vectors. Now you must rule out the case where $\pi(f)v=0$ for every $v\in H$ and $f\in \oplus_{m,n} S_{n,m}$ (notations as in Lang). But the latter direct sum is dense in $C_c(G)$, by lemma 1. And using an approximation of identity in $C_c(G)$, you see that $\pi(f)v=0$ for every $f\in C_c(G)$, forces $v=0$.

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Thank you for your answer, professor. The argument you give essentially reduces my question to the following one: mathoverflow.net/questions/128268/…. –  Murat Güngör Apr 21 '13 at 18:23

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