Diagonal non-compact operator

Question

Suppose we have an operator $I:l_2 \rightarrow l_2$ which is diagonal but not compact. Does that follow: there exists a constant $C$ such that infinite number of diagonal terms $>C$?

Answer 1

Presumably you are asking about the absolute value of the diagonal terms, rather than the diagonal terms. In this case the answer is:

Yes, since otherwise the diagonal elements would converge to zero. This would mean that $I$ is a norm limit of the finite-rank operators $$I_n = \sum_{k \le n} \lambda_k P_k,$$ where $\{P_k\}$ are the orthogonal projectors onto the elements of the canonical basis $\{\delta_k\}$ and $\{\lambda_k\}$ are the respective diagonal elements, and hence compact. (Proof: compute $\|I - I_n\| = \sup_{m > n} |\lambda_m|$.)

If you were really asking about the diagonal elements themselves, the answer is yes if and only if $I$ is bounded from below, i.e., there is a real number $c$ such that $$\langle x, Ix \rangle \geq c\|x\|^2$$ for all $x \in \ell^2$. In particular, if $I$ is bounded the claim is true.