Consider the following condition on a bounded operator $T$ on a Hilbert space:

$\ \ \ \ \ $(A) there exists an orthonormal basis $(e_j)$ with $\sum_j\parallel Te_j\parallel<\infty$.

We have the implications

$\ \ \ \ \ $(trace class) $\ \Rightarrow\ $ (A) $\ \Rightarrow\ $ (Hilbert-Schmidt)

But what about the converse directions? Are they both false or is one of them true?

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    $\begingroup$ A diagonal operator shows that (A) is strictly stronger than Hilbert-Schmidt. $\endgroup$ – Matthew Daws Jul 19 '17 at 8:05
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    $\begingroup$ That requires to show that the sum is infinite for every ONB. How do you show that? $\endgroup$ – user1688 Jul 19 '17 at 8:08
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    $\begingroup$ Isn't (A) equivalent to (trace class)? Indeed, if $y_j=T e_j$ satisfies $\sum_j \|y_j\| <\infty$, then $T= \sum_j \langle \cdot,e_j\rangle y_j$ is a norm-converging series of rank $1$ operators. $\endgroup$ – Mikael de la Salle Jul 19 '17 at 10:12
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    $\begingroup$ But for rank $1$ operators, trace norm and operator norm are equal, so the sum is convergent for the trace norm. $\endgroup$ – Mikael de la Salle Jul 19 '17 at 10:15
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    $\begingroup$ I also think that (A) implies (trace class). Indeed, assuming (A), we have that $\sum_j\langle|T|e_j,e_j\rangle\leq\sum_j\||T|e_j\|=\sum_j\|Te_j\|<\infty$, so $T$ is trace class. $\endgroup$ – GH from MO Jul 19 '17 at 10:38

Condition (A) is equivalent to $T$ being trace class. I will use the usual notation $|T|:=(T^\ast T)^{1/2}$.

1. Assume that condition (A) holds. Then $$\sum_j\langle|T|e_j,e_j\rangle\leq\sum_j\||T|e_j\|=\sum_j\|Te_j\|<\infty.$$ The left hand side is finite, hence $T$ is trace class by definition.

2. Assume that $T$ is trace class. Let $(e_j)$ be an orthonormal eigenbasis of $|T|$. Then $$\sum_j\|Te_j\|=\sum_j\||T|e_j\|=\sum_j\langle|T|e_j,e_j\rangle<\infty.$$ The left hand side is finite, hence condition (A) holds.

Note that this answers the question completely, since trace class is a strictly stronger property than Hilbert-Schmidt.

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