Corollary

We have $b(u)^*=b(u^*)$.

Let $q_1$ and $q_2$ be the projections $\Gamma_{u^*}\rightarrow E$ and let $(k,|q_1^*|)$ be the polar decomposition of $q_1^*$, so that $b(u^*) =q_2\circ k$. For all $x$ and $y$ in E, we have $j(x)\in \Gamma_u$ and $k(y)\in \Gamma_{u^*}$, and since $\Gamma_u$ is orthogonal to $s(\Gamma_{u^*})$ by prop. 7 of IV, p. 236, it follows that

Since $p_1\circ j=|p^*_1|$ and $q_1\circ k=|q^*_1|$ (prop. 11, b) of I, p. 140) we obtain

hence $|q^*_1|b(u) =b(u^*)^*|p^*_1|$. Using formulas (1) and (4), we conclude that $b(u)|p^*_1|=|q^*_1|b(u) =b(u^*)^*|p^*_1|$, and since the image of $|p^*_1|$ is the domain of $u$, which is dense in E, we deduce $b(u) =b(u^*)^*$.

We denote by Ω(E) the set of $v\in \mathscr{L}(E)$ such that $1_E-v^*v$ is positive and injective. For $v\in \Omega (E)$, the hermitian endomorphism $(1_E-v^*v)^{1/2}$ is injective, since its square is so, and we denote by $B(v)$ the partial operator $v\circ ((1_E-v^*v)^{1/2})^{-1}$.

Note that $1_E-v^*v$ is positive if and only if $\|v\|\leqslant 1$, since one has $\langle x|(1_E-v^*v)(x)\rangle =\|x\|^2- \|v(x)\|^2$ for every $x\in E$.