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 adominated 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$ ?