# Property between trace class and Hilbert-Schmidt

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?

• A diagonal operator shows that (A) is strictly stronger than Hilbert-Schmidt. – Matthew Daws Jul 19 '17 at 8:05
• That requires to show that the sum is infinite for every ONB. How do you show that? – user1688 Jul 19 '17 at 8:08
• 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. – Mikael de la Salle Jul 19 '17 at 10:12
• But for rank $1$ operators, trace norm and operator norm are equal, so the sum is convergent for the trace norm. – Mikael de la Salle Jul 19 '17 at 10:15
• 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. – 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.