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Let $H$ be a separable Hilbert space and $\{e_n\}_{n=1}^\infty$ some ON-basis for $H$. Suppose $A:H\to H$ is a bounded linear operator that fulfills the following summability condition $$\sum_{n=1}^\infty\sum_{m=1}^\infty \left| \langle Ae_n,e_m \rangle\right|<\infty .$$ Here we just assume the summability for some specific ON-basis, it might not be true for others.

  1. Does the above condition imply that $A$ is compact?

  2. Is there any standard term for this condition (or perhaps some equivalent condition)?

Accto3
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1 Answers1

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Yes. Your condition implies that $$\sum_{n\ge1} \sum_{m \ge 1} |(Ae_n \mid e_m ) |^2 < \infty$$ and the latter is equivalent to $$\sum_{n \ge 1} \|Ae_n\|^2 < \infty \tag{1} $$ by Parseval's inequality. Operators that satisfy (1) are called Hilbert Schmidt and they are compact. Moreover, the Hilbert-Schmidt norm $\|A\|^2_{HS} : = \sum_{n \ge 1} \|Ae_n\|^2$ is independent of the choice of ON basis $(e_n)$.

Evangelopoulos Foivos
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    Should'nt the condition for that instead be

    $$\sum_{n=1}^\infty\sum_{m=1}^\infty \left| \langle Ae_n,e_m \rangle\right|^2<\infty .$$

    Otherwise I do not see how you utilize Parseval

     – Accto3 Oct 23 '24 at 07:42
  • @Accto3 W.l.o.g we may suppose $|A|<1$. In that case $| \langle Ae_n, e_m \rangle|<1$ so $| \langle Ae_n, e_m \rangle |^{2}\le | \langle Ae_n, e_m \rangle |$ – Kavi Rama Murthy Oct 23 '24 at 07:45
  • Yes, but how do you show the reverse implication? The claim was that the condition was equivalent to the HS-property. Not that it just implies the HS property. – Accto3 Oct 23 '24 at 07:49
  • @Accto3 You are right, only one implication holds. My bad. – Evangelopoulos Foivos Oct 23 '24 at 07:53
  • Also, while the HS condition is indeed basis independent I don't see how that has any bearing on whether or not the provided condition is basis independent. But thank you for settling the compactness issue. – Accto3 Oct 23 '24 at 08:02