Proposition 1

We have the formulas

Let $x\in E$. Put $y= W(u)(x)\in$ dom($u^*u$). We have $y\in$ dom($u$) and $u(y)\in$ dom($u^*$). For every element $(y_1, u(y_1))$ of $\Gamma_u$, where $y_1\in$ dom($u$), we compute

This means that $p^*_1(x) = (y, u(y))$, whence $p_1\circ p^*_1(x) =y= W(u)(x)$. Consequently, we have $p_1\circ p^*_1= W(u)$, hence $|p^*_1|= W(u)^{1/2}$, that is, formula (1). In particular, since Im($p_1$) $=$ Im($|p^*_1|$) by the Cor. to Prop. 11 of I, p. 140, it follows that dom($u$) $=$ Im($p_1$) $=$ Im(W($u$)$^{1/2}$). The partial operator $u\circ W(u)^{1/2}$ therefore has the space E as domain. The relation $p_1\circ j= (p^*_1)^*\circ j=|p^*_1|= W(u)^{1/2}$ (Prop. 11, a) of I, p. 140) then implies that

which is formula (2).

On has Ker($j$) $=$ Ker($|p^*_1|$) (prop. 10, b) of I, p. 139) and formula (1) therefore implies that the partial isometry $j: E\rightarrow \Gamma_u$ is injective, hence isometric, whence $j^*\circ j= 1_E$. Writing $j= (p_1\circ j, b(u))$, we find

whence formula (3).

Let $x\in$ dom($u$) and put $y= W(u)(x)$. We have $y\in$ dom($u$) and formula $y+u^*u(y) =x$ implies that $u^*u(y)\in$ dom($u$). We then have

which means that the reduction of $u\circ W(u)$ to the domain of $u$ is equal to $W(u^*)\circ u$. Since the image of $|p^*_1|$ is equal to that of $p_1$ (cor. of prop. 11 of I, p. 140) and hence to the domain of $u$, it follows that

(formula (2)). Moreover, $|p^*_1|= W(u)^{1/2}$ commutes with $W(u)$, hence we obtain

This relation implies that $b(u)\circ f(W(u)) =f(W(u^*))\circ b(u)$ for every function $f\in \mathscr{C}(\mathbf{R}_+)$ (prop. 11 of I, p. 113), hence in particular that $b(u)\circ W(u)^{1/2}= W(u^*)^{1/2}\circ b(u)$. This proves formula (4) and concludes the proof.