0
$\begingroup$

Consider a self-adjoint operator $T$ on a Hilbert space $\mathcal{H}$, with domain $\mathcal{D}(T)$ and suppose that $A$ is a symmetric operator satisfying: $\mathcal{D}(T) \subset \mathcal{D}(A)$ and the inequality:

$$||Au|| \leq b||Tu|| + a||u||$$ with $b<1$ it is a well known theorem of Kato that $S=T+A$ is also self-adjoint on $\mathcal{D}(T)$. Furthermore, if $T$ has compact resolvent and if there is $z \in \rho(T)$ such that:

$$a||R(z,T)|| +b||TR(z,T)|| <1 $$ then $S+A$ also has compact resolvent. In particular there is a complete orthonormal system of eigenvectors.

I was wondering if there are there any "easy" results that allow to discuss the dimension of the new spectral subspaces of the perturbed eigenvalues? Say, if all of the eigenvalues of $T$ are simple i.e. the spectral subspace associated to any eigenvalue of $T$ is of dimension 1, are there sufficient conditions for this also to be the case for the eigenvalues of $S=T+A$?

$\endgroup$

Your Answer

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

Browse other questions tagged or ask your own question.