# Norm of an operator and eigenvalues

I have $K\colon L^2(0,T) \to L^2(0,T)$ a Hilbert-Schmidt integral operator (and so $K$ is linear, bounded, compact and self-adjoint) and I have obtained its eigenvalues and eigenvectors. From them, I got $\{e_n\mid n\ge 0\}$ a Hilbert base of eigenvectors, and using those, I have calculated (supposing $\lambda$ is not an eigenvalue) $(K-\lambda I)^{-1}$ as a series, obtaining that $$(K-\lambda I)^{-1}(f) = \sum_{n=0}^\infty \frac{\langle f,e_n\rangle}{\lambda_n-\lambda}e_n.$$ where $\lambda_n$ is the eigenvalue corresponding to $e_n$.

Nos I have to obtain $\|(K-\lambda I)^{-1}\|$, and I think it's going to be the supreme of the module of the eigenvalues of the operator, that is $$\|(K-\lambda I)^{-1}\|=\sup_{n\ge 0} \frac{1}{|\lambda_n-\lambda|}$$ but so far I failed to prove it. Is there an easy way to do it?

Obviously, the norm is at least what you anticipate by setting $f = e_n$ where $n$ almost realizes the supremum of $|\lambda_n - \lambda|^{-1}$.
Conversely, write $f$ as $f = \sum_n c_n e_n$. Then $$\|(K - \lambda I)^{-1} f\|^2 = \sum_n |c_n|^2 |\lambda_n - \lambda|^{-2} \leq (\sup_n |\lambda_n - \lambda|^{-2}) (\sum_n |c_n|^2).$$ Taking the square root shows that the norm of the operator is at most the supremum of $|\lambda_n - \lambda|^{-1}$, because $\|f\| = (\sum_n |c_n|^2)^{1/2}$.