Absolut convergence implies being in trace class I'm asking for your help in the next problem, I can't think how to do it! 
Prove that if $$\sum_{n=1}^{\infty}|(A\phi_n,\phi_n)|<\infty$$ for all orthonormal bases, then $A$ is in the trace class.
I think the fact that the hypothesis is true for all orthonormal bases is the hint, but I don't know!  Thank you.
 A: Lets denote with $(*)$ the property that 
for all orthonormal basis $\phi_n$ we have
$$\sum_n |(A\phi_n,\phi_n)|<\infty.$$
Assume $A$ has this property, then $A^*$ also has it and thus $A+A^*$ and $A-A^*$. By verifying that $A+A^*$ and $A-A^*$ are trace class we verify that $A$ is trace class. Now $A+A^*$ and $i(A-A^*)$ are hermitian, so if we can show that

hermitian and $(*)$ $\implies$ trace-class

we will have shown

$(*)$ $\implies$ trace-class.

For hermitian operators we can use spectral theory. Let $P_-$ denote the projection onto the negative modes of the hermitian operator we are considering and $P_+=1-P_-$ the projection onto the positive modes. With
$$A_+:= P_+ A = AP_+= P_+ A P_+\qquad A_- :=-P_- A = -A P_- =- P_- AP_-$$
we have
$A=A_+-A_-$, where $A_+$ and $A_-$ are positive operators. Now let $\psi_n$ be an orthonormal basis of $\mathrm{im}(P_+)$ and $\psi_n'$ and orthonormal basis of $\mathrm{im}(P_-)=\ker(P_+)$. Together they are an orthonormal basis of the Hilbert-space. We have:
$$\sum_n |(A\psi_n,\psi_n)| + |(A\psi_n',\psi_n')|=\sum_n|(A_+\psi_n,\psi_n)|+|(A_+\psi_n',\psi_n')|+\sum_n|(A_-\psi_n,\psi_n)|+|(A_.\psi_n',\psi_n')| $$
by virtue of $P_+\psi_n = \psi_n$ and $P_+\psi_n' = 0$.
It follows that the positive operators $A_+$ and $A_-$ are trace-class. Then $A=A_+-A_-$ is trace-class, and we have shown the statement for an arbitrary hermitian operator. As noted above this implies it for any bounded operator.
