Resolvent of an unbounded self-adjoint operator affiliated to a semi-finite algebra could be unbounded?

Question

Let $\mathcal{M}$ be a semi-finite von Neumann algebra and $H$ is an unbounded self-adjoint operator affiliated to $\mathcal{M}$. Then is it the true fact that $(H-iI)^{-1}\in\mathcal{M}$ ?

Answer 1

Because $H$ is selfadjoint, its spectrum is real. Then spectrum of $(H-iI)^{-1}$ consists of elements of the form $$\frac1{\lambda-i},\qquad\qquad\lambda\in\sigma(H)$$ Since $\lambda\in\mathbb R$, $$ |\lambda-i|^2=\lambda^2+1\geq1. $$ Thus $$ \frac1{|\lambda-i|}\leq1,\qquad\qquad\lambda\in\sigma(H), $$ which implies that $(H-iI)^{-1}$, being normal, is bounded and $\|(H-iI)^{-1}\|\leq1$.