Proposition 4

(i) The initial (resp. final) subspace of $u^*$ is equal to the final (resp. initial) subspace of $u$.

(ii) Suppose that $E = F$. Let $M$ be a closed vector subspace of $E$ and $M^\circ$ its orthogonal. The relations $u(M) \subset M$ and $u^*(M^\circ) \subset M^\circ$ are equivalent.

Let $Q = \overline{u(E)}$ be the final subspace of $u$. The orthogonal $Q^\circ$ of $Q$ in $F$ consists of all vectors $y$ such that $\langle u(x)|y\rangle = 0$ for all $x \in E$; this is equivalent to : $\langle x|u^*(y)\rangle = 0$ for all $x \in E$, or to $u^*(y) = 0$. Hence we have $Q^\circ = \mathrm{Ker}\, u^*$, and $Q$ is the initial subspace of $u^*$. Since $u$ is the adjoint of $u^*$, the final subspace of $u^*$ is also the initial subspace of $u$. This proves (i).

The relation $u(M) \subset M$ implies that $u(M)$ is orthogonal to $M^\circ$, and the relation $u^*(M^\circ) \subset M^\circ$ implies that $u^*(M^\circ)$ is orthogonal to $M$. But we have $\langle u(x)|y\rangle = \langle u^*(y)|x\rangle$ for all $x \in M$ and $y \in M^\circ$; hence (ii) follows.

We remark that prop. 4 can be deduced from the general properties of the transpose (II, p. 51, cor. 2) in view of remark 3, V, p. 41.