Proposition 7

Let $u$ be a partial operator with dense domain from E into F.

a) The graph of $u^*$ is equal to $s(\Gamma^{\circ}_u) =s(\Gamma_u)^{\circ}$;

b) The partial operator $u^*$ is closed;

c) The kernel of $u^*$ is the orthogonal of the image of $u$.

Let us put $W =s(\Gamma_u)^{\circ}$. Since the linear mapping $s$ is unitary, we have $W =s(\Gamma^{\circ}_u)$.

We have $(y, x)\in W$ if and only if

for all $x'\in$ dom($u$), that is to say, if

for all $x'\in$ dom($u$). When $y\in$ dom($u^*$) and $x=u^*(y)$, this property holds (cf. formula (1), p. 236). Conversely, if this condition is satisfied, it follows that $|\langle y|u(x')\rangle |\leqslant \|x\| \|x'\|$ for all $x'\in$ dom($u$), whence it follows that $y$ belong to dom($u^*$); one then has $u^*(y) =x$. Hence $W = \Gamma_{u^*}$.

The operator $u^*$ is closed, because the space $s(\Gamma_u)^{\circ}$ is closed in $F\oplus E$.

Let us prove assertion c). If $y$ is orthogonal to the image of $u$, the linear form $\lambda_y:x\mapsto \langle y|u(x)\rangle$ on D is zero, hence $y\in$ dom($u^*$) and $u^*(y) = 0$. Conversely, let $y\in$ dom($u^*$). Then $u^*(y) = 0$ if and only if $y$ is orthogonal to $u(x)$ for all $x\in D$ (formula (1), p. 236).