Proposition 12

Let $u$ be a closed operator with dense domain from E into a complex Hilbert space F.

a) The domain of the positive self-adjoint partial operator $|u|$ coincides with the domain of $u$;

b) There exists a unique partially isometric linear mapping $j$ of E into F such that $u=j|u|$ and Ker($j$) $=$ Ker($u$);

c) Let $u_1$ be a positive self-adjoint operator on E and $j_1$ a partially isometric linear mapping of E into F such that $u=j_1u_1$ and Ker($j_1$) $=$ Ker($u_1$). Then $u_1=|u|$ and $j_1=j$.

The domain of $u^*u$ is contained in dom($u$) and in dom($|u|$). It is dense in the Hilbert spaces $E_u($loc. cit.) and $E_{|u|}$ (Cor. 1, e) of IV, p. 273) and, for every $x\in$ dom($u^*u$), one has

Consequently, the Hilbert spaces $E_u$ and $E_{|u|}$ are equal (Lemma 8), hence dom($u$) $=$ dom($|u|$) and $\|u(x)\|$ = $\||u|(x)\|$ for every $x\in$ dom($u$).

Formula (16) implies that Ker($u$) $=$ Ker($|u|$) and that there exists a unique isometric linear mapping $v$ of Im($|u|$) onto Im($u$) satisfying $v(|u|(x)) =u(x)$ for every $x\in$ dom($|u|$). Since $|u|$ is self-adjoint, one has Im($|u|$)$^{\circ}=$ Ker($|u|$) (Prop. 7, c) of IV, p. 236). There therefore exists a unique partial isometry $j$ of E into F which extends $v$ and vanishes on Ker($|u|$) $=$ Ker($u$). Since E = Ker($u$)$\oplus$ Im($|u|$), this mapping is the unique partially isometric mapping such that $u=j|u|$ and Ker($j$) $=$ Ker($u$).

Let us prove c). We have $u_1j_1^*j_1u_1\subset (j_1u_1)^*j_1u_1=u^*u$. The linear mapping $j_1^*j_1$ is the orthogonal projector with kernel Ker($j_1$) $=$ Ker($u_1$) (TVS, V, p. 41, Prop. 5 (ii)) and hence with image Ker($u_1$)$^{\circ}=$ Im($u_1$) (Prop. 7, c) of IV, p. 236). Consequently, $u^2_1\subset u^*u$, whence $u^2_1=u^*u$ since these two operators are self-adjoint. Thus, it follows that $u_1= (u^*u)^{1/2}$, and the assertion of uniqueness in b) finally proves that $j_1=j$.