3
$\begingroup$

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?

Thanks in advance.

$\endgroup$
1
$\begingroup$

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 $\|f\| = (\sum_n |c_n|^2)^{1/2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.