Norm of the Resolvent of a Self-Adjoint Operator

Question

Let $\mathcal H$ be a Hilbert space and $\mathcal L$ is a self- adjoint operator with a discrete spectrum $\{\lambda_{j}\}$.

I read that it is well known that for, $\lambda \notin \sigma(\mathcal L)$,

$$\left|(L- \lambda I)^{-1}\right| \leq \sup_{j}\frac{1}{|\lambda - \lambda_{j}|}.$$

How to show this well-known result?

Answer 1

That follows from the continuous functional calculus of self adjoint operator. Which states that there is an isometric homomorphism

$$\Phi: C(\sigma(\mathcal L)) \to B(\mathcal H)$$

from the space of continuous functions on $\sigma(\mathcal L)$ to $B(H)$, given by $\Phi(f) = f(\mathcal L)$ which extends the map $p(x) \mapsto p(\mathcal L)$ on polynomails. Now if $\lambda \notin \sigma(\mathcal L)$, then $f(x) = \frac{1}{x-\lambda}$ is continuous on $\sigma(\mathcal L)$. As

$$I= \Phi (1) = \Phi \bigg((x-\lambda) \frac{1}{x-\lambda} \bigg) = (\mathcal L-\lambda I) f(\mathcal L), $$ We see that $f(\mathcal L) = (\mathcal L - \lambda I)^{-1}$ and thus

$$||(\mathcal L - \lambda I)^{-1} || = ||f||_0 = \sup_j \frac{1}{|\lambda - \lambda_j|}$$

So actually we have equality.