Let $H$ be a Hilbert space, $(e_n)$ a complete orthonormal sequence, and $\lambda_n$ a bounded sequence of complex numbers. Let $A$ be defined such that the $(e_n)$ are the eigenvectors of $A$ and the $(\lambda_n)$ are its eigenvalues. That is, let
$$A(\sum x_n e_n) = \sum \lambda_n x_n e_n$$
whenever the sum converges. Then I know how to prove that, if $(\lambda_n) \rightarrow 0$ then $A$ is compact---I just need to pick an appropriate sequence of pairwise orthogonal projections and apply a theorem.
My question is the converse, proving that if $A$ is compact, then $(\lambda_n)$ must converge to 0. By assuming $A$ is compact I get the following:
- A is bounded and therefore continuous
- For any bounded sequence $(z_n)$ in $H$, $(Az_n)$ has a convergent subsequence
but I don't know if these two observations are enough or if I need some other piece to get the ball rolling. Furthermore, do I need to tack on any additional conditions to get self-adjointness, or can I prove that as-is?
If the $e_n$ are orthonormal, and $\lambda_{n_k}$ is a sequence with $|\lambda_{n_k}| \ge \delta >0 $, then $A e_{n_k}$ is a sequence such that the distance between any two elements is $\ge \sqrt{2} \delta$, hence cannot have a convergent subsequence.