Selfadjoint compact operator with finite trace 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?

 A: No, we cannot conclude that the operator is trace class.
For example, let a Hilbert space have orthonormal basis $e_1,f_2,e_2,f_2,e_3,f_3,\ldots$, and $T$ interchanges $e_i,f_i$, while multiplying both by a positive real $\lambda_i$. That is, in these coordinates, the matrix of $T$ is a list of diagonal blocks, with the $i$-th diagonal block being anti-diagonal $\lambda_i,\lambda_i$.
For $\lambda_i\rightarrow 0$, the operator is compact, almost from the definition.
All the diagonal entries are $0$.
The operator is self-adjoint because the matrix is symmetric real.
However, the operator is not trace class unless $\sum_i |\lambda_i|<\infty$, which easily fails for many sequences of positive reals $\lambda_i\rightarrow 0$.
Edit: It is noteworthy that the analogous characterization (I pointedly don't say "definition") of "Hilbert-Schmidt" does not depend on choice of basis. Thus, "defining" trace-class as composition of two Hilbert-Schmidt operators is sometimes usefully more intrinsic, less basis/coordinate-dependent.
A: 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
