Proposition 43.8.10.label Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, then $T$ is closable if and only if $D(T^{*})$ is dense in $H$.
Proof. ($\Rightarrow$): Let $S \supset T$ be a closed extension of $T$, and $\Gamma(T) = \bracsn{(\xi, T\xi)| \xi \in D(T)}$ be the graph of $T$, then by Lemma 43.8.9,
\[\Gamma(T)^{\perp} = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}\]
Let $\xi \in D(T^{*})^{\perp}$, then for any $\eta \in D(T^{*})$, $(0, \xi) \perp (-T^{*}\eta, \eta)$, so $(0, \xi) \in \Gamma(T)^{\perp\perp}= \ol{\Gamma(T)}$. As such, there exists $\theta \in D(S)$ such that $(0, \xi) = (\theta, T\theta)$, so $\xi = 0$, and $D(T^{*})$ is dense.
($\Leftarrow$): If $D(T^{*})$ is dense, then $T^{**}$ is a closed extension of $T$.$\square$
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