Let $A:D(A)\longrightarrow H$ be an unbounded self-adjoint (or normal) operator on a Hilbert space $H$. Then we know that $\sigma(A) \neq \emptyset$ and $$\|(\lambda-A)^{-1}\|=\frac{1}{d(\lambda,\sigma(A))}, \quad \forall \lambda \in \rho(A),$$ where $d(\lambda,\sigma(A))=\min_{\mu \in \sigma(A)} |\lambda-\mu|>0$. Do we have a similar formula for $$\|A(\lambda-A)^{-1}\|= ?$$ I point out that $A(\lambda-A)^{-1}$ is a bounded operator since $A(\lambda-A)^{-1}x=-x+\lambda(\lambda-A)^{-1}x$ for any $x \in H$. I have the basic estimate $$\|A(\lambda-A)^{-1}\| \leq 1+\frac{|\lambda|}{d(\lambda,\sigma(A))}.$$ Is it sharp ?

The following is exact: $$\|A(\lambda I-A)^{-1}\|=\sup_{\mu\in\sigma(A)}\left|\frac{\mu}{\lambda-\mu}\right| % \\ = \sup_{\mu\in\sigma(A)}\left|-1+\frac{\lambda}{\lambda-\mu}\right|.$$ If $\sigma(A)=\mathbb{R}$ and $\lambda=i$, then the above gives $$\|A(\lambda I-A)^{-1}\| = \sup_{\mu\in\mathbb{R}}\frac{|\mu|}{\sqrt{\mu^2+1}} =1.$$ while your expression gives $$1+\frac{1}{1}=2.$$
• Thanks for you answer. Do you use the spectral theorem for self-adjoint operators to prove it ? If so, do you have a more elementary proof (similar to the one to establish the expression for $\|(\lambda-A)^{-1}\|$) ? – perturbation Jan 10 '18 at 22:20
Recall that $A(\lambda-A)^{-1}=-Id+\lambda(\lambda-A)^{-1}$. Since $A$ is normal, the bounded operator $B=-Id+\lambda(\lambda-A)^{-1}$ is normal. Therefore, we have the know fact that $$\|B\|=\sup_{\eta \in \sigma(B)} |\eta|.$$ Since $\sigma(B)=\sigma(-Id+\lambda(\lambda-A)^{-1})=f(\sigma(A))$ with $f(\mu)=-1+\frac{\lambda}{\lambda-\mu}$, we have $$\begin{array}{rl} \|A(\lambda-A)^{-1}\| &=\sup_{\eta \in \sigma(A(\lambda-A)^{-1})} |\eta| \\ &=\sup_{\eta \in f(\sigma(A))} |\eta| \\ &=\sup_{\mu \in \sigma(A)} |f(\mu)| \\ &=\sup_{\mu \in \sigma(A)} |-1+\frac{\lambda}{\lambda-\mu}| \\ &=\sup_{\mu \in \sigma(A)} |\frac{\mu}{\lambda-\mu}| \end{array}.$$