In what follows, $G$ is a compact Lie group and $\Phi$ is a unitary representation of $G$ on a Hilbert space $V$ iff $\Phi$ is a homomorphism of $G$ to the group of unitary operators on $V$ and for all $v$ in $V$ the map $g\mapsto\Phi(g)v$ is continuous. I was toying around with the proof of statement 1.12.d, page 17 in Anthony Knapp - Representation Theory of Semisimple Groups (2001):
Let $\Phi$ be a unitary representation of $G$ on a Hilbert space $V$. Then $V$ is the orthogonal sum of finite-dimensional, irreducible invariant subspaces.
In order to prove to prove the statement, the author uses the obvious approach: using Zorn's Lemma, he determines a maximal subspace $W\subseteq V$ which decomposes as an orthogonal, finite-dimensional sum and then goes ahead to prove that $W^{\perp}=\{0\}$.
In this context he uses the homomorphism $L^{1}(G)\to L(V)$, where $L(V)$ is (if I recall correctly) the space of bounded linear operators on $V$, given by: $$\langle f(v),w\rangle_{V}=\int_{G}f(g)\langle \Phi(g)v,w\rangle_{V}\operatorname{d}\mu_{G}(g)$$ where $\mu_{G}$ is a Haar measure on $G$.
I have found that if $B_{n}$ is a collection of compact neighbourhoods of $1\in G$ such that $B_{n+1}\subseteq B_{n}$ for all $n\in\mathbb{N}$ and $\cap_{n\in\mathbb{N}}B_{n}=\{1\}$, then $\frac{1}{\mu_{G}(B_{n})}\mathbf{1}_{B_{n}}(v)\to v$ when $n\to\infty$. I wanted to construct such sets from $\Phi$ and I approached it as follows: let $v\in V\setminus\{0\}$, $N_{0}$ a compact neighbourhood of $1\in G$ and for $n\geq 1$ let: $$N_{n}:=N_{0}\cap\left\{g\in G;\lVert\Phi(g)v-v\rVert_{V}\leq\frac{1}{n}\right\}$$
Now $\cap_{n\geq\mathbb{N}}B_{n}=\{g\in N_{0};\Phi(g)v=v\}$, which is not exactly what I wanted. I know, that I can choose neighbourhoods $U_{n}$ of the identity such that $\hat{N}_{n}=N_{n}\cap U_{n}$ would be appropriate. Still I was interested, whether I could drop choosing the $U_{n}$, i.e. when can I be sure that I can choose a compact neighbourhood $N_{0}$ such that: $$\{1\}=\cap_{n\geq 1}\left(N_{0}\cap\left\{g\in G;\lVert \Phi(g)v-v\rVert_{V}\leq\frac{1}{n}\right\}\right)$$ It is clear that this is equivalent to $G_{v}:=\{g\in G;\Phi(g)v=v\}$ being discrete, so I have to find out when the latter is the case. In the context I am interested in, there are two equivalent formulations of what I wanted to show:
Let $\Phi$ be a unitary representation of a compact Lie group $G$ on an infinite dimensional Hilbert space $V$ containing no finite dimensional subrepresentations (assuming its existence contrary to Peter-Weyl). Let $v\in V\setminus\{0\}$. Then $G_{v}:=\{g\in G;\Phi(g)v=v\}$ is discrete.
Let $\Phi$ be a unitary representation of a compact Lie group $G$ on an infinite dimensional Hilbert space $V$. Let $v\in V\setminus\{0\}$ and assume that $g_{n}\to 1$ such that $\Phi(g_{n})v=v$ for all $n\in\mathbb{N}$. Then it follows (without using Peter-Weyl) that $V$ contains a finite dimensional subrepresentation.
Does anybody have an idea on whether this could be true? In the second formulation I started toying around with the closure of the group generated by the $(g_{n})_{n\in\mathbb{N}}$ being compact subgroup necessarily of positive dimension and then wanted to use the index but I might not know enough about Lie groups at that point. Is there no relation at all or do you know a hint? Of course it might be useful to use facts about general Lie groups acting on infinite dimensional Hilbert spaces but I really do not know where to start.