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?



Property between trace class and Hilbert-Schmidt
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.


