Proposition 26

Let $u$ be a closed symmetric operator on a complex Hilbert space E. Let $(E_+,E_-)$ be the deficiency pair of $u$. The spaces $E_+,E_-$ and dom($u$) are closed mutually orthogonal subspaces of $E_{u^*}$ whose Hilbert sum is equal to $E_{u^*}$.

Since $u$ is symmetric, we have $u\subset u^*$, hence the space dom($u$) is closed in $E_{u^*}$ (lemma 7). The subspaces $E_+$ and $E_-$ are closed in E and contained in dom($u^*$), hence closed in $E_{u^*}$.

Let $x\in E_+$. We have $u^*(x) =ix$, whence $\langle x|u(y)\rangle =-i\langle x|y\rangle$ for every $y\in$ dom($u$). Consequently, for every $y\in$ dom($u$), we have

since $u$ is symmetric. The spaces $E_+$ and dom($u$) are therefore orthogonal in $E_{u^*}$. Analogously, $E_-$ and dom($u$) are orthogonal in $E_{u^*}$.

Let $x\in E_+$ and $y\in E_-$. Then

therefore $E_+$ and $E_-$ are orthogonal in $E_{u^*}$.

Let $x\in E_{u^*}$ be orthogonal to the closed subspace dom($u$)$\oplus E_+\oplus E_-$. For every $y\in$ dom($u$), we have $u^*(y) =u(y)$ since $u\subset u^*$, whence

This implies that $z=u^*(x)$ belongs to the domain of $u^*$ and satisfies $u^*(z) =-x$. Let $x_-=z-ix$. This is an element of dom($u^*$) satisfying $u^*(x_-) =-x-iz=-ix_-$. Hence $x_-$ belongs to $E_-$. But, for every $w\in E_-$, we have

Taking $w=x_-$, we deduce that $x_-= 0$, that is to say $z=ix$; since $z=u^*(x)$, we therefore have $x\in E_+$, whence $x$ = 0 since $x$ is orthogonal to $E_+$ in $E_{u^*}$. This completes the proof that the direct sum dom($u$)$\oplus E_+\oplus E_-$ is equal to $E_{u^*}$.