Proposition 8

Let $u$ be an operator with dense domain from E into F. Then $u^*$ has dense domain if and only if $u$ is closable. When this is so, the closure $\overline{u}$ of $u$ is equal to $u^{**}$, and the adjoint of $\overline{u}$ is equal to $u^*$.

By Prop. 7, the partial operator $u^*$ is closed. Suppose that the domain $D^*$ of $u^*$ is dense in F. Let $u^{**}$ be the adjoint of $u^*$; it is a closed partial operator from E into F. Let us prove that $u\subset u^{**}$, which will imply that $u$ is closable. Let $x\in$ dom($u$). By definition of $u^*$, the linear forms on $D^*$ given by $y\mapsto \langle x|u^*(y)\rangle$ and $y\mapsto \langle u(x)|y\rangle$ are equal; hence $x\in$ dom($u^{**}$) and $u^{**}(x) =u(x)$, whence the assertion.

Conversely, suppose that $u$ is closable; one has $\Gamma_{\overline{u}}=\overline{\Gamma}_u$ (Prop. 1 of IV, p. 228). Let $y\in F$ be a vector orthogonal to dom($u^*$). The element $(y,0)$ of $F\oplus E$ then belongs to the orthogonal complement of the graph of $u^*$. But, by Prop. 7, a), one has

It follows therefore that $(0, y)\in \Gamma_{\overline{u}}$, whence $y=\overline{u}(0) = 0$. Since the orthogonal complement of dom($u^*$) is reduced to 0, the space dom($u^*$) is dense in F.

Finally, Prop. 7, applied to $u^*$, implies that

hence $u^{**}=\overline{u}$, then $\overline{u}^*= (u^*)^{**}=\overline{u^*}=u^*$ since $u^*$ is closed.