TITLE: Property between trace class and Hilbert-Schmidt
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: 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.