Self adjoint operator $A$ satisfies $\|A(A-i)^{-1}\|\leq 1$?

Question

Let $H$ be a seperable Hilbert space and $A=A^*:D\subset H \rightarrow H$ a possibly unbounded linear densly defined selfadjoint operator.

I try to understand a proof in a script, where a crusial step in the proof is that $\|A(A-i)^{-1}\|\leq 1$.

I don't know how to proceed and show this.

Any Ideas?

So far I have $A(A-i)^{-1}=I+i(A-i)^{-1}$ so the operator is indeed continous. But why is its norm less then $1$?

Answer 1

For $x\in D(A)$ we have $$\|(A+iI)x)|^2=\|Ax\|^2+\|x\|^2+i\langle x,Ax\rangle -i\langle Ax,x\rangle\\ =\|Ax\|^2+\|x\|^2$$ Therefore $\|Ax\|\le \|(A+iI)x\|.$ Let $y=(A+iI)x,$ for $x\in D(A).$ Then $$\|A(A+iI)^{-1}y\|=\|Ax\|\le \|(A+iI)x\|=\|y\|$$ The inequality holds for every $y\in \mathcal{H}$ as $A+iI$ is a bijection from $D(A)$ onto $\mathcal{H}.$