Deficiency indices theorem (Von Neumann theory) Let $T: Dom(T) \rightarrow \scr H$ be a symmetric operator.
Prove the following:

$T$ admits self-adjoint extensions iff $d_+ = d_-$,
where $d_\pm = dim \ Ker(T^\dagger \pm i \mathbb{I})$

The proof I'm studying uses the Cayley transforms, but at some point I get stuck. Here's part of the proof.

*

*Let $S = S^\dagger $ be a self-adjoint extension of $T$. Its Cayley transform $U = (S-i\mathbb{I})(S+ i \mathbb{I})^{-1}$ is unitary (this fact was proven in a previous theorem) and extends the Cayley Transform of $T$, which we'll call $V$. So, in symbols:
$T \subset S \implies V \subset U$ (this implication is not proven; I don't find it obvious but ok).
The restriction $\left. U \right |_{DomV}$ is injective, because $U$ is injective by hypothesis, and maps $Ran (T+ i \mathbb{I}) $ to $Ran (T- i \mathbb{I}) $. Here’s the problem:
$$\psi \in \{ Ran (T+ i \mathbb{I})  \}^{\perp} \iff U\psi \in \{ Ran (T-i \mathbb{I})  \} ^{\perp}$$
and the only reason given for it is “$U$ is unitary”. Uhm, why?
 A: Use tha fact that $U$ bijectively maps $Ran(T+iI)$ to $Ran(T-iI)$ by construction.
Now, $$\psi \in Ran(T+iI)^\perp$$ means $$\langle \psi, (T+iI) \phi\rangle=0\quad  \mbox{ for every $\phi \in D(T)$.}$$ Since $U$ is bijective isometric, that is equivalent to $$\langle U\psi, U(T+iI) \phi\rangle=0\quad  \mbox{ for every $\phi \in D(T)$,}$$  namely $$\langle U\psi, (T-iI) \phi\rangle=0\quad  \mbox{ for every $\phi \in D(T)$.}$$
In other words,
$$\psi \in Ran(T+iI)^\perp$$
is equivalent to
$$U\psi \in Ran(T-iI)^\perp\:.$$
A: For all $\ x\in Dom(T)$
\begin{align}
\langle\psi,(T+i\mathbb{I})x\rangle&=\langle\psi,(S+i\mathbb{I})x\rangle\\
&=\langle U\psi,U(S+i\mathbb{I})x\rangle\ \ \ \ (U\ \text{ being unitary)}\\
&=\langle U\psi,(S-i\mathbb{I})(S+i\mathbb{I})^{-1}(S+i\mathbb{I})x\rangle\\
&=\langle U\psi,(S-i\mathbb{I})x\rangle\\
&=\langle U\psi,(T-i\mathbb{I})x\rangle\ .
\end{align}
Therefore $\ \langle\psi,(T+i\mathbb{I})x\rangle=0\ $ for all $\ x\in Dom(T)\ $ if and only if $\ \langle U\psi,(T-i\mathbb{I})x\rangle\ $ for all $\ x\in Dom(T)\ $.
