TITLE: Selfadjoint compact operator with finite trace
QUESTION [16 upvotes]: I have a compact selfadjoint operator $T$ on a separable Hilbert space. For some fixed orthonormal basis, the operator's diagonal is in $\ell^1(\mathbb{N})$. 

Can we conclude that $T$ is trace class?

REPLY [2 votes]: Disclaimer: Non-Compact Operators!
Given the Hilbert space $\ell^2(\mathbb{N})$.
Consider sum of shifts:*
$$A_\pm:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A_\pm:=R\pm L$$
They have finite trace:
$$\sum_n\langle A_\pm\delta_n,\delta_n\rangle=\sum_n0=0$$
But for the shifts:
$$\sum_n\langle|R|\delta_n,\delta_n\rangle=\sum_n1=\infty$$
$$\sum_n\langle|L|\delta_n,\delta_n\rangle=\sum_n1=\infty$$
Thus for the sum:
$$\operatorname{Tr}A_\pm<\infty\implies\operatorname{Tr}A_\mp<\infty$$
Concluding counterexample.
*Shifts: Right & Left