TITLE: If the diagonal of a positive operator is compact, is the operator itself compact?
QUESTION [8 upvotes]: Let $H$ be a separable Hilbert space with a fixed orthonormal basis $\{e_n\}_n$.  For a bounded operator $T$ on $H$, the diagonal of $T$ is the unique operator $D_T$ on $H$ which is diagonal with respect to the above basis, and whose diagonal entries are given by $d_n=\langle T(e_n),e_n\rangle$.  It is well know that if $T$ is positive and $D_T=0$, then necessarily $T=0$.

Question: If $T$ is positive and $D_T$ is compact, is $T$ necessarily compact?

REPLY [5 votes]: If T also is sparse, i.e.  for some number N then for each i
<Te_i, e_j> is non-zero for at most  N values of j’s,
(and for each j it is non-zero for at most  N values of i’s) then the claim is true.
I have a proof, but I guess this is well known. My own literature search has been unsuccessful, so anyone having a reference will make me happy.