Theorem 43.8.9 (von Neumann).label Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then $T^{*}T: D(T^{*}T) \to H$ is a densely defined, self-adjoint, and positive operator.
Proof. For each $\xi, \eta \in D(T^{*}T)$, by definition of the adjoint,
(Densely Defined): Let $\eta \in D(T^{*}T)^{\perp}$. By (2) of Lemma 43.8.8, there exists $\xi \in D(T^{*}T)$ such that $\eta = \xi + T^{*}T\xi$. In which case,
so $T\xi = \xi = 0$, and $\eta = 0$ as well.
($T^{*}T \subset (T^{*}T)^{*}$): For each $\xi, \eta \in D(T^{*}T)$, $\dpn{T^*T\xi, \eta}{H}= \dpn{\xi, T^*T\eta}{H}$, so $T^{*}T \subset (T^{*}T)^{*}$.
($T^{*}T \supset (T^{*}T)^{*}$): Let $\xi \in D((T^{*}T)^{*})$, then by (2) of Lemma 43.8.8, there exists $\eta \in D(T^{*}T)$ such that $\xi + (T^{*}T)^{*}\xi = \eta + T^{*}T\eta$. In which case, for each $\theta \in D(T^{*}T)$,
As the above holds for all $\theta \in D(T^{*}T)$, (2) of Lemma 43.8.8 implies that $\dpn{\xi, \theta}{H}= \dpn{\eta, \theta}{H}$ for all $\theta \in H$. Therefore $\xi = \eta \in D(T^{*}T)$, and $T^{*}T \supset (T^{*}T)^{*}$.$\square$
Post a Comment