Is this proof of Schur's lemma (for a densely defined closed operator) mistaken? How to fix it? I'm trying to understand a version of Schur's lemma for a densely defined closed operator.  It is on page 17 of the book Nonabelian Harmonic Analysis by Howe and Tan. The confusing parts are underlined in red.

First of all, since the image of $P_U|_{\Gamma (T)}$ is the domain of $T$, which is a dense subspace of $U$, it seems like the image of $U_1$ should be $U$, not just $(\ker T)^\perp$.  Secondly, what does $\Gamma(T) \cap V =\{0\}$ have to do with T being closed?  This should be a consequence of the fact that $T$ is a function.  
So it seems like what actually needs to be shown is that $U_1$ carries $(\ker P_V)^\perp$ isometrically onto $(\ker T)^\perp$.  $U_1$ looks like an isometry, so that's okay.  Also, $P_U|_{\Gamma(T)}$ clearly carries $\ker P_V|_{\Gamma(T)}$ onto $\ker T$, so if $P_1:=   (P_U|_{\Gamma(T)}^\ast \circ P_U|_{\Gamma(T)})^{1/2}$ preserved $\ker P_V$ maybe I would be in business, but it's only evident to me that $P_U|_{\Gamma(T)}^\ast \circ P_U|_{\Gamma(T)}$ does so.
 A: I think I may have found a solution.  Any feedback would be appreciated.  Write $P_U$ (resp $P_V$) for $P_U|_{\Gamma(T)}$ (resp $P_V|_{\Gamma(T)}$).
We want to show that $U_1$ maps $(\ker P_V)^\perp$ isometrically onto $(\ker T)^\perp$.  Since $U_1$ is a Hilbert space isomorphism, it suffices to show that $U_1$ maps $\ker P_V$ onto $\ker T$.  
Take $(u, 0) \in \ker P_V$.  It is easy to seee that $P_U^\ast P_U (u,0)=(u,0)$. Therefore, since $P_1 = (P_U^\ast P_U)^{1/2}$ is the limit of polynomials in  $P_U^\ast P_U$ in the strong operator topology, and since $P_1$ is an involution of $\ker P_V$ (since its square is $P_U^\ast P_U$), we see that $P_1$ restricts to a vector space isomorphism of $\ker P_V$.   Therefore, $(u,0) = P_1 (w,0)$ for some other $(w,0) \in \ker P_V$.  Since $(w,0) \in \Gamma(T)$, this means $w \in \ker T$, and we have 
$$U(u,0)=UP_1(w,0)=P_U(w,0)=w\in \ker T$$
Converseley, if $a\in \ker T$, then $(a,0) \in \ker P_V$, so that $P_1(a,0)=(b,0)$ for some $(b,0)\in \ker P_V$.  We therefore have
$$a = P_U(a,0)=UP_1(a,0)=U(b,0).  \square$$ 
