Lemma 6.2.30C82C2C21EE7

Proof. If $y_ j\in \operatorname{Dom}T^*$, $y_ j\to y$ and $T^*y_ j\to w$, then for every $x\in \operatorname{Dom}T$

hence $y\in \operatorname{Dom}T^*$ and $T^*y=w$, so $T^*$ is closed. A vector $y\in H_1$ is orthogonal to $\operatorname {Ran}T$ exactly when the functional $x\mapsto \langle Tx,y\rangle $ vanishes, that is, when $y\in \operatorname{Dom}T^*$ and $T^*y=0$. This proves $\ker T^*=(\operatorname {Ran}T)^\perp $. In the same way, $(a,b)\in H_0\oplus H_1$ is orthogonal to $\operatorname {Gr}T$ if and only if $\langle Tx,b\rangle =\langle x,-a\rangle $ for all $x\in \operatorname{Dom}T$, so

Now let $T$ be closed, and let $v\in H_1$ be orthogonal to $\operatorname{Dom}T^*$. Since $\operatorname {Gr}T$ is closed, proposition 6.2.1 and the description of $(\operatorname {Gr}T)^\perp $ decompose $(0,v)$ as $(x,Tx)+(-T^*y,y)$ with $x\in \operatorname{Dom}T$ and $y\in \operatorname{Dom}T^*$. Thus $x=T^*y$ and $v=Tx+y$. Since $v$ is orthogonal to $y\in \operatorname{Dom}T^*$, we obtain

so $x=y=0$ and $v=0$. Hence $\operatorname{Dom}T^*$ is dense, and $T^{**}$ is defined. Let $J\colon H_1\oplus H_0\to H_0\oplus H_1$ and $J'\colon H_0\oplus H_1\to H_1\oplus H_0$ be the unitary maps $J(b,a)=(-a,b)$ and $J'(a,b)=(-b,a)$. The description of the orthogonal complement of a graph, applied to $T$ and to the densely defined operator $T^*$, reads $(\operatorname {Gr}T)^\perp =J(\operatorname {Gr}T^*)$ and $(\operatorname {Gr}T^*)^\perp =J'(\operatorname {Gr}T^{**})$. For a closed subspace $M$ we have $(M^\perp )^\perp =M$ by proposition 6.2.1, and unitary maps carry orthogonal complements to orthogonal complements. Therefore

because $JJ'=-\mathrm{id}$ and graphs are linear subspaces. Thus $T^{**}=T$. $\square$