Proof. (1): By von Neumann’s theorem, $T^{*}T: D(T^{*}T) \to H$ is a densely defined, self-adjoint, and positive operator. Thus $|T| = (T^{*}T)^{1/2}$ is well-defined, self-adjoint, and positive via the spectral theorem.

(2): By definition, $D(T)$ is isomorphic to the graph of $T$, which is a Hilbert space given that $T$ is closed.

Let $\xi \in D(T)$ with $\xi \perp D(T^{*}T)$, then for each $\eta \in D(T^{*}T)$,

By (2) of Lemma 43.8.8, $(I+T^{*}T): D(T^{*}T) \to H$ is surjective, so $\xi = 0$, and $D(T^{*}T)$ is dense in $D(T)$.

(3): By (3) of Definition 43.8.6, $|T|$ is self-adjoint, and in particular closed. Since $D(T)$ is isomorphic to the graph of $|T|$, $D(|T|)$ is a Hilbert space. Similarly, for any $\xi \in D(|T|)$ with $\xi \perp D(T^{*}T)$ and $\eta \in D(T^{*}T)$, by the Borel functional calculus^[1],

By (2) of Lemma 43.8.8, $\xi = 0$, and $D(T^{*}T)$ is dense in $D(|T|)$.

(4): For any $\xi \in D(T^{*}T)$, $\norm{\xi}_{D(T)}= \norm{\xi}_{D(|T|)}$. By (2) and (3), $D(T)$ and $D(|T|)$ are both the completions of $D(T^{*}T)$ under the same norm, so, $D(T) = D(|T|)$.

(5): For each $\xi \in D(T^{*}T)$,

By the density provided in (2) and (3), $\normn{T\xi}_{H}^{2} = \norm{|T|\xi}_{H}^{2}$ for all $\xi \in D(T)$.$\square$