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This would be a basic problem in representation theory.

Let $G$ be a unimodular real Lie group, $(\pi,V)$ a smooth representation of $G$ in a Frechet space $V$. Let $f$ be a smooth function on $G$. Now define the operator $\pi(f)$ as $$\pi(f)v=\int_G f(g)\pi(g)vdg$$ for any $v\in V$.

Now the question is that in order to show $\pi(f)$ is an operator on $V$, i.e. $\pi(f)v\in V$ for any $v$, it suffices to check $$|\pi(f)v|_{\mu}<\infty$$

for some particular seminorm $||_{\mu}$, or to check that is finite for ALL seminorms on $V$ ?

In particular, if $(\pi,V)$ is the smooth vectors in a unitary representation, then for all smooth function $f$, which is also in $L^1(G)$, $\pi(f)$ is an operator on $V$, and in fact continuous,right?

Many thanks.

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    $\begingroup$ There are two things. First you need to show that the integral for $\pi(f)v$ converges, usually in the weak sense (i.e. $\int_G f(g)h(\pi(g)v)dg$ exists in $\mathbb{C}$ for every $h\in V'$, the topological dual of $V$). Then you need to show that $\pi(f)$ is continuous as an operator $V\rightarrow V$, i.e. for every seminorm $\mu$ on $V$, there is a constant $C>0$ and seminorms $\nu_1,...,\nu_k$ such that $|\pi(f)v|_\mu \leq C.\max_{1\leq i\leq k} |v|_{\nu_i}$. $\endgroup$
    – Alain Valette
    Commented Jan 15, 2012 at 16:43
  • $\begingroup$ Thanks for your comment. Here I'm more interested in if $\pi(f)$ converges to certain vector in $V$. Do we need the continuity prior to this question? $\endgroup$
    – user1832
    Commented Jan 15, 2012 at 17:43
  • $\begingroup$ I extended my response to include more general $f$'s. I don't think the integral makes sense for every $f\in L^1(G)$, one needs some "decay at infinity", see below. $\endgroup$
    – GH from MO
    Commented Jan 16, 2012 at 1:52
  • $\begingroup$ @unknown: I don't understand what you mean by "$\pi(f)$ converges to certain vector in $V$", as this mixes objects of different nature. Now remember that we can always add the zero seminorm to the family of seminorms defining the topology of $V$. For this one, the condition $|\pi(f)v|_\mu <\infty$ is trivial, so you can get no satisfaction with just one seminorm. $\endgroup$
    – Alain Valette
    Commented Jan 16, 2012 at 10:58
  • $\begingroup$ Alain: Thanks for your comment. It should be $\pi(f)v$ converges to a vector in V for any $v\in V$. Sorry for the confusion. $\endgroup$
    – user1832
    Commented Jan 16, 2012 at 11:57

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I would think the answer is yes and follows by applying "[1.0.1] Theorem" on page 2 here for the finite Borel measures $\Re^+f(g)dg$, $\Re^-f(g)dg$, $\Im^+f(g)dg$, $\Im^-f(g)dg$ on $G$, where $\pm$ stands for positive and negative part. Note that Frechet spaces satisfy the conditions there.

EDIT. As the OP pointed out, the result is stated for compactly supported functions. By an approximation argument and the last line of the quoted theorem, an extension seems possible to functions $f:G\to\mathbb{C}$ satisfying $\int_G |f(g)|\ |\pi(g)v|_\mu\ dg < \infty$ for each seminorm $|\cdot|_\mu$ that participates in the definition of the topology of $V$.

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  • $\begingroup$ Thanks. But it seems in that theorem it requires f to be compactly-supported... $\endgroup$
    – user1832
    Commented Jan 16, 2012 at 0:57
  • $\begingroup$ Thanks a lot. So it looks like the question for the particular case when $\pi$ is from a unitary representation I asked is fine. $\endgroup$
    – user1832
    Commented Jan 16, 2012 at 2:33

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