Proposition 12

Let $u$ be a closed operator with dense domain from E into F. The partial operator $u^*\circ u$ on E is self-adjoint and positive. Its domain is a core for $u$.

Let us denote by $v$ the partial operator $1_E+u^*\circ u$. Its domain is dom($u^*\circ u$), which is contained in dom($u$). For all $x\in$ dom($u$) and $y\in$ dom($v$), one has $u(y)\in$ dom($u^*$) and

Formula (4) implies that the partial operator $v$ is injective. Moreover, by prop. 7 of IV, p. 236, a), one has $F\oplus E = \Gamma_{u^*}\oplus s(\Gamma_u)$. Let $x\in E$. There exist $y'\in$ dom($u^*$) and $x'\in$ dom($u$) such that

Hence one has $y'=u(x')$, whence $x'\in$ dom($u^*\circ u$) $=$ dom($v$) and

The partial operator $v$ on E is therefore surjective, and induces a bijective linear mapping of dom($v$) onto E.

Let $x\in E$ be orthogonal to dom($v$). Write $x=v(x')$ where $x'\in$ dom($v$). By formula (4), one obtains

whence $x'= 0$, and then $x= 0$. The domain of $v$ is therefore dense in E.

The partial operator $v$ has dense domain; it is bijective and formula (3) shows that it is symmetric. One concludes that $v$ is self-adjoint by applying the corollary to proposition 10. Consequently, $u^*\circ u=v-1_E$ is self-adjoint. Moreover, the formula

for all $x\in$ dom($u^*\circ u$) implies that $u^*\circ u$ is positive.

Finally, let $y\in E_u$ be orthogonal to dom($u^*\circ u$). There exists an element $x$ in the domain of $u^*\circ u$ such that $y=v(x) =x+ (u^*\circ u)(x)$. One then has $0 = (x|y)_u=\langle y|y\rangle$, whence $y= 0$.