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I'm asking for your help in the next problem, I can't think how to do it!

Prove that if $$\sum_{n=1}^{\infty}|(A\phi_n,\phi_n)|<\infty$$ for all orthonormal bases, then $A$ is in the trace class.

I think the fact that the hypothesis is true for all orthonormal bases is the hint, but I don't know! Thank you.

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  • $\begingroup$ What is your definition of a trace class operator? From what I know as the usual definition, this seems immediately clear. $\endgroup$ – Omnomnomnom Sep 10 '15 at 18:16
  • $\begingroup$ My definition is $A$ is in the class trace if and only if $Tr(|A|)<\infty$. Where $Tr(A)=\sum_{n=1}^{\infty}(\phi_n,A\phi_n)$ $\endgroup$ – Tom Builder Sep 10 '15 at 18:18
  • $\begingroup$ $\{\phi_n\}$ is a particular choice of orthonormal basis, right? $\endgroup$ – Omnomnomnom Sep 10 '15 at 18:20
  • $\begingroup$ Oh, wait a second... I didn't notice the $|A|$ in the definition. I see why this isn't so obvious now. $\endgroup$ – Omnomnomnom Sep 10 '15 at 18:21
  • $\begingroup$ yes, but I proved, using Parseval equation, that doesn't matter what basis you chose, the sum is the same. $\endgroup$ – Tom Builder Sep 10 '15 at 18:23
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Lets denote with $(*)$ the property that for all orthonormal basis $\phi_n$ we have $$\sum_n |(A\phi_n,\phi_n)|<\infty.$$

Assume $A$ has this property, then $A^*$ also has it and thus $A+A^*$ and $A-A^*$. By verifying that $A+A^*$ and $A-A^*$ are trace class we verify that $A$ is trace class. Now $A+A^*$ and $i(A-A^*)$ are hermitian, so if we can show that

hermitian and $(*)$ $\implies$ trace-class

we will have shown

$(*)$ $\implies$ trace-class.

For hermitian operators we can use spectral theory. Let $P_-$ denote the projection onto the negative modes of the hermitian operator we are considering and $P_+=1-P_-$ the projection onto the positive modes. With $$A_+:= P_+ A = AP_+= P_+ A P_+\qquad A_- :=-P_- A = -A P_- =- P_- AP_-$$ we have $A=A_+-A_-$, where $A_+$ and $A_-$ are positive operators. Now let $\psi_n$ be an orthonormal basis of $\mathrm{im}(P_+)$ and $\psi_n'$ and orthonormal basis of $\mathrm{im}(P_-)=\ker(P_+)$. Together they are an orthonormal basis of the Hilbert-space. We have: $$\sum_n |(A\psi_n,\psi_n)| + |(A\psi_n',\psi_n')|=\sum_n|(A_+\psi_n,\psi_n)|+|(A_+\psi_n',\psi_n')|+\sum_n|(A_-\psi_n,\psi_n)|+|(A_.\psi_n',\psi_n')| $$ by virtue of $P_+\psi_n = \psi_n$ and $P_+\psi_n' = 0$.

It follows that the positive operators $A_+$ and $A_-$ are trace-class. Then $A=A_+-A_-$ is trace-class, and we have shown the statement for an arbitrary hermitian operator. As noted above this implies it for any bounded operator.

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  • $\begingroup$ By positive (negative) modes I assume you mean the positive (negative) part of the spectrum of $A$? $\endgroup$ – Frederik vom Ende Mar 11 '18 at 17:03
  • $\begingroup$ Yes. Depending on the language you prefer to formulate the spectral theorem, $P_-$ is the image of $(-\infty,0]$ wrt the projection valued measure associated to $A$. $P_+$ is here $1-P_-$. $\endgroup$ – s.harp Mar 11 '18 at 18:04
  • $\begingroup$ Alright, I see. Thanks again for the very nice answer! $\endgroup$ – Frederik vom Ende Mar 11 '18 at 18:06

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