Let $T:H\to H$ be a compact operator on a complex Hilbert space. Assume that $$ \sup_{(e_j)}\sum_j\left|\langle Te_j,e_j\rangle\right|<\infty, $$ where the supremum extends over all orthonormal bases of $H$. Does it follow that $T$ is a trace class operator?
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$\begingroup$ Not only is it true, but you don't need the hypothesis that $T$ is compact---see math.stackexchange.com/questions/2036398. $\endgroup$ – Jonathan Gleason Dec 27 '16 at 5:44
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Yes. Break the operator into it's real and imaginary parts: $A=\frac{1}{2}(T + T^\dagger)$ and $B=\frac{1}{2i}(T - T^\dagger)$. These are also compact and satisfy the same estimate. The estimate, applied to their respective eigenbases, shows that $A$ and $B$ are trace class. Thus $T=A + i B$ is also trace class.
