Relation between the distance from a point to the resolvent set of a self-adjoint operator and the norm of a related operator

Question

$H$ is a Hilbert space and $B(H)$ is the Banach space of its bounded linear operator on $H$. Assume $A\in$ $B(H)$, $A$ is self-adjoint. Suppose $\lambda_{0}$ is in the resolvent set of $A$. Show that: $||(A-\lambda_{0}I)^{-1}||=\frac{1}{d(\lambda_{0},\sigma(A))}$

Now I can use the knowledge of functional calculus(see $(A-\lambda_{0}I)^{-1}$ as the function of $A$) to show that $||(A-\lambda_{0}I)^{-1}||\leq \frac{1}{d(\lambda_{0},\sigma(A))}$. But I don't know how to proceed.

Answer 1

For any continuous function $F$ on the spectrum of $A$, $\|F(A)\|=\sup_{\lambda\in\sigma(A)}|F(\lambda)|$. Let $F(\lambda)=\frac{1}{\lambda-\lambda_0}$. Then $F(A)=(A-\lambda_0 I)^{-1}$ and $$ \|(A-\lambda_0 I)^{-1}\| = \sup_{\lambda\in\sigma(A)}\left|\frac{1}{\lambda-\lambda_0}\right|=\frac{1}{\inf_{\lambda\in\sigma(A)}|\lambda-\lambda_0|}=\frac{1}{\mbox{dist}(\lambda_0,\sigma(A))}. $$