I try to understand the proof of (ii) in the following lemma.
Schur's lemma. Let $G$ be a topological group. Suppose $(\pi_1,V_1)$ and $(\pi_2,V_2)$ are any two irreducible unitary representations of $G$ for some nonzero complex Hilbert spaces $V_1$, $V_2$. Let $T: V_1\to V_2$ be a bounded linear operator such that $T\pi_1(g)=\pi_2(g)T$ for all $g\in G$. Then
(i) If $T\neq 0$ then $T$ is invertible.
(ii) If $(\pi_1,V_1)=(\pi_2,V_2)$ then $T$ is scalar.
$Proof \ of \ (ii).$
Assume that $(\pi_1,V_1)=(\pi_2,V_2)$. For brevity, we put $(\pi, V):=(\pi_1,V_1)$. Suppose $T$ is not scalar. Note that for any $g\in G$ we have $(T\pi(g))^*=(\pi(g)T)^*$ which implies $\pi(g)^* T^*=T^*\pi(g)^*$. By the unitarity of $\pi$, the operator $T^*$ commutes with the action of $G$ , that is, $\pi(g)T^*=T^*\pi(g)$ for all $g\in G$. We now define the bounded linear operators $A:=\frac{T+T^*}{2}$ and $B:=\frac{T-T^*}{2i}$. Note that $A^*=A$ and $B^*=\overline{(1/(2i))}(T^*-T)=B$. So they are (bounded) self-adjoint operators. Moreover, $A$ and $B$ commute with the action of $G$. Since $T=A+iB$ and $T$ is not scalar, at least one of $A$ and $B$ must be non-scalar. Say $A$ is non-scalar. By the Spectral Theorem, applied to $A$, there is a non-scalar orthogonal projection $E$ that is a function of $A$. Moreover $E$ commutes with the action of $G$ since $E$ is a function of $A$. So $E$ must be invertible by the first part $(i)$, contradiction.
My question: Which version of Spectral theorem do we use in the above proof? I have some confusion about the statement of spectral theorem. Thanks!
A sufficient version of a spectral theorem for bounded hermitian operators $T$ for this form of Schur's lemma would be that the operator-norm closure of $\mathbb R[T]$ is isomorphic to $C^o(\sigma T)$, where $\sigma T$ is the spectrum. An immediate corollary is that if $\sigma T$ is a single point $\lambda$, then $T$ is the scalar operator by $\lambda$.
EDIT: In more detail, suppose $\sigma(T)$ contains at least two distinct points $x,y$. We can make (e.g., using Tietze-Urysohn-Brouwer extension) two continuous functions $f,g$ on $\sigma(T)$ (and in fact extending to an interval $[a,b]$ containing $\sigma(T)$), such that $f(x)=1=g(y)$, and $fg=0$ on $\sigma(T)$. The spectral theorem gives an (isometric) isomorphism of $C^o(\sigma(T))$ to the operator-norm closure of $\mathbb R[T]$. Thus, neither $f(T)$ nor $g(T)$ is the $0$ operator, since they are not the $0$ functions on $\sigma(T)$. From this point, it is relatively easy to finish the argument...