Why is $B_n$ in the domain of $H$ on a Hilbert Space? Let $H:D(H)\subset \mathcal{H}\to\mathcal{H}$ be an unbounded self-adjoint operator.
Let $B_n:= in(H+in)^{-1}$.
For $\phi\in\mathcal{H}$ it is given in my notes that $B_n\phi\in D(H)$, where $D(H)$ is the domain of $H$. Why is this? I lack the background in Functional Analysis to make the immediate connection. 

Since the operator is self-adjoint its spectrum is contained in the real line, this means that $-in$ is not in its spectrum. By definition this means that there is a bounded operator $R(H, -in)$ (formally $R(H,-in) = (H+in)^{-1}$) such that its range is in $D(H)$ that satisfies
$$(H + in)R(H, -in) = Id\quad R(H,-in)(H + in) \subseteq Id,$$
here the inclusion means that this operator defined a priori in $D(H)$ is a restriction o the identity.
