Resolvent estimate self-adjoint operator 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 ?
 A: 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
$$
              1+\frac{1}{1}=2.
$$
A: Actually, I have found a simple answer in the bible Kato (1966), p. 277 identities (3.31).
The proof is the same as for the bounded case, to which Kato refers (see (3.17), p. 273).
I reproduced the arguments below for the sake of completeness.
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
$$
\begin{array}{rl}
\|A(\lambda-A)^{-1}\| &=\sup_{\eta \in \sigma(A(\lambda-A)^{-1})} |\eta| \\
&=\sup_{\eta \in f(\sigma(A))} |\eta| \\
&=\sup_{\mu \in \sigma(A)} |f(\mu)| \\
&=\sup_{\mu \in \sigma(A)} |-1+\frac{\lambda}{\lambda-\mu}| \\
&=\sup_{\mu \in \sigma(A)} |\frac{\mu}{\lambda-\mu}|
\end{array}.
$$
