Suppose that $A$ is an operator on a dense domain $D(A)\subset L^2$ with compact resolvent, and with quadratic form $q(f,g):=\langle f,Ag\rangle$.
Let $(r_n)_{n\in\mathbb N}$ be a sequence of quadratic forms on $D(A)$ that are uniformly form-bounded by $A$, in the sense that there exists $0<\alpha<1$ and $\beta>0$ independent of $n$ such that $$|r_n(f,f)|\leq \alpha\cdot q(f,f)+\beta\cdot\|f\|_2,\qquad f\in D(A).$$ Since $\alpha<1$ the operators $A_n$ defined by the form $q+r_n$ all have compact resolvent.
Suppose that the $r_n$ have a limit $r_\infty$, in the sense that $$\lim_{n\to\infty}r_n(f,f)\to r_\infty(f,f)$$ for every $f\in D(A)$. Define the operator $A_\infty$ through the form $q+r_\infty$. Clearly $$|r_\infty(f,f)|\leq \alpha\cdot q(f,f)+\beta\cdot\|f\|_2,\qquad f\in D(A),$$ and thus $A_\infty$ has compact resolvent as well.
Question. Does the uniform form-bound of the $r_n$ give rise to a dominated convergence-type result for the spectrum of $A_n$, that is, the eigenvalues of $A_n$ converge to that of $A_\infty$, and the eigenfunctions converge in $L^2$?
