Let $H$ be a complex Hilbert space, $A:D(A) \subset H \rightarrow H$ be an unbounded linear operator and $K \in L(H)$ be a compact operator. Let $A_K$ be defined by $$A_K x=Ax+Kx, \quad x \in D(A_K)=D(A).$$ Assume that:
(1) There exists an orthonormal basis $\{\phi_k\}_{k \in \mathbb{N}}$ of $H$ and a sequence of distinct complex numbers $\{\lambda_k\}_{k \in \mathbb{N}}$ such that $\phi_k \in D(A)$ and $A\phi_k=\lambda_k \phi_k$ for every $k \in \mathbb{N}$.
(2) $|\lambda_k| \to +\infty$ as $k \to +\infty$.
We know that $A_K$ has a compact resolvent (since so has $A$, and $K$ is a bounded perturbation). In particular it has generalized eigenfunctions, etc.
My question is: can we prove that, asymptotically $A_K$ behaves like $A$ ? For instance, are the following properties true ?
(1') Asymptotically, $A_K$ has only eigenfunctions, and they are of multiplicity one.
(2') If $\{\mu_k\}_{k \in \mathbb{N}}$ denotes an enumeration of the eigenvalues of $A_K$, then $\mu_k \sim C \lambda_k$ as $k \to +\infty$ ($C>0$ is a constant).
Is it an open problem ? If not, I would appreciate any reference on this (articles/books). I already know the classics Kato, Dunford-Schwartz, Gohberg-Krein, etc.
Thank you !
A few remarks on the first question: Weyl's Theorem says that a compact self-adjoint perturbation of a self-adjoint operator preserves the essential spectrum. This has been generalized to normal operators, and your $A$ is normal if you give it the natural domain $D(A)=\{ \sum c_k\phi_k: \sum\lambda_k |c_k|^2<\infty \}$.
Since $\sigma_{ess}(A)=\emptyset$, this means that indeed $A+K$ will have pure point spectrum, with eigenvalues $|\lambda_n(A+K)|\to\infty$, if $K$ is normal, in addition to being compact.
In general, this will not follow: a trivial counterexample is obtained if the eigenvalues of $A$ almost come in pairs, that is, $\lambda_{2n+1}\simeq\lambda_{2n}$. Then a compact perturbation can make them equal to one another, plus you can introduce a non-zero entry in the top right corner of each such $2\times 2$ block, so that you get non-diagonalizable Jordan blocks.