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?

$\begingroup$ Not only is it true, but you don't need the hypothesis that $T$ is compactsee math.stackexchange.com/questions/2036398. $\endgroup$– Jonathan GleasonDec 27, 2016 at 5:44
1 Answer
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.