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 ?

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
$$
              1+\frac{1}{1}=2.
$$
