Proposition 2

The mapping $u\mapsto b(u)$ defines a bijection of the set of closed partial operators with dense domain on E onto the set Ω(E). The inverse bijection is given by $v\mapsto B(v)$.

The relation $1_E-b(u)^*b(u) = W(u)$ (formula (3)) implies that $b(u)$ belongs to Ω(E) since $W(u)$ is positive and injective. Since moreover $b(u) =u\circ W(u)^{1/2}$ (formula (2)), one obtains

Conversely, let $v\in \Omega (E)$. Put $w= (1_E-v^*v)^{1/2}$ and let $u= B(v) =v\circ w^{-1}$. The domain of $u$ is the image of $w$, which is dense in E since $w$ is hermitian and injective (EVT, V, p. 41, prop. 4 (i)). For every $x\in$ dom($u$), one has

since $(1_E-v^*v)\circ w^{-1}$ is the restriction of $w$ to the domain of $w^{-1}$. Since $w$ is self-adjoint, one deduces that

and the partial operator $B(v) =u=v\circ w^{-1}$ is therefore closed by Lemma 2 of IV, p. 231.

Finally let us prove that $b(B(v)) =v$. Since $u=v\circ w^{-1}$, one has $v=u\circ w$, and it is therefore enough to verify that $w= W(u)^{1/2}$ (formula (2)), or even that $1_E-v^*v= W(u)$.

The endomorphism $v^*$ belongs to Ω(E) by Lemma 1. Put $w'= (1_E-vv^*)^{1/2}$. It is a positive injective endomorphism of E, and one has $v\circ w=w'\circ v$ (prop. 11 of I, p. 113). The graph of $u$ is the set of elements of the form $(w(x), v(x))$ for $x\in E$. By prop. 7 of IV, p. 236, the graph of $u^*$ is $s(\Gamma_u)^{\circ}$. Since, for all $x$ and $y\in E$, one has

the graph of $u^*$ contains elements of the form $(w'(x), v^*(x))$ for $x\in E$, and one then has $u^*(w'(x)) =v^*(x)$. In particular, the domain of $u^*$ contains the image of $w'$.

Let $x\in E$ and put $y=w^2(x) = (1_E-v^*v)(x)$, so that $x=y+v^*v(x)$. One has $y\in$ dom($u$) and $u(y) =v(w^{-1}(y)) =v(w(x)) =w'(v(x))$. In particular, $u(y)\in$ Im($w$)$\subset$ dom($u^*$), and $u^*(u(y)) =v^*(v(x))$. Hence there follows $x=y+v^*v(x) =y+u^*u(y)$, whence $y= W(u)(x)$, that is to say

Thus it has been proved that $1_E-v^*v= W(u)$, as desired.