Lemma 43.8.8.label Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, and
be the graph of $T$, then:
- (1)
$\Gamma(T)^{\perp} = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$.
- (2)
If $T$ is closed, then the mapping $I + T^{*}T: D(T^{*}T) \to H$ is surjective.
Proof. (1): Let $\Gamma(T) = \bracsn{(\xi, T\xi)| \xi \in D(T)}$ be the graph of $T$, then for any $(\xi, \eta) \in H \oplus H$, the following are equivalent:
- (i)
$(\xi, \eta) \perp \Gamma(T)$.
- (ii)
For each $\theta \in D(T)$, $\dpn{\eta, T\theta}{H}= -\dpn{\xi, \theta}{H}$.
- (iii)
$\eta \in D(T^{*})$, with $\dpn{T^*\eta, \theta}{H}= -\dpn{\xi, \theta}{H}$ for all $\theta \in H$.
- (iv)
$\eta \in D(T^{*})$ with $\xi = -T^{*}\eta$.
so $\Gamma(T)^{\perp} = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$.
(2): Given that $T$ is closed, $H \oplus H = \Gamma(T) \oplus \Gamma(T)^{\perp}$. In particular, for each $\theta \in H$, there exists $\xi \in D(T)$ and $\eta \in D(T^{*})$ such that
In which case, $T\xi = -\eta \in D(T^{*})$, so $\xi \in D(T^{*}T)$ with
so $I + T^{*}T: D(T^{*}T) \to H$ is surjective.$\square$
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