Proposition 25

Let $u$ be a closed symmetric operator on a complex Hilbert space E. We have

Ker($u^*-i$)$^{\circ}=$ Im($u+i$),

Ker($u^*+i$)$^{\circ}=$ Im($u-i$).

By prop. 7, c) of IV, p. 236, it is enough to prove that the image of $u+i$ (resp. of $u-i$) is closed in E. Since $u$ is symmetric, we have $\langle (u+i)(x)|(u+i)(y)\rangle = (x|y)_u$ for all $x$ and $y$ in dom($u$). The mapping $x\mapsto u(x) +ix$ of $E_u$ into E is therefore isometric. Its image is closed in E (lemma 8 of I, p. 107). Since this is also the image of $u+i$, the latter is closed in E. Analogously, the image of $u-i$ is closed in E.