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Let $A$ be a trace-class operator on a separable Hilbert space $H$. By definition, this means that the quantity $$\sum_i\langle|A|e_i,e_i\rangle$$ is finite, where $\{e_i\}_i$ is an orthonormal basis of $H$.

This has the consequence that the trace of $A$, $$\sum_i\langle Ae_i,e_i\rangle,$$ is well-defined in the sense that it is independent of the choice of orthonormal basis, as well as the ordering of elements within a basis.

I'm wondering if the converse is true, i.e. whether having a well-defined trace in the above sense implies that $A$ is trace class. Equivalently:

Question: Is there a bounded operator $T$ on $H$ that is not trace-class, yet $\sum_i\langle Te_i,e_i\rangle$ is well-defined independent of the choice of orthonormal basis $\{e_i\}_i$ and of reordering of basis elements?

geometricK
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1 Answers1

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Consider an operator $A$ for which $$\sum_i\langle Ae_i,e_i\rangle $$ is well-defined and finite for all orthonormal bases of $H$.

Assume that one of these traces is only conditionally convergent. Then, by the Riemann rearrangement theorem, you would be able to find a rearrangement of your basis for which the trace would diverge (got this argument from Robert Israel's answer to the following post: Which (non-trace class) operators have a well-defined trace (if any)?)

This means that all these traces need to be absolutely convergent, which is actually equivalent to $A$ being trace-class by the accepted answer of this post: Absolut convergence implies being in trace class.

In other words, if all these traces exist finite, then they are all absolutely convergent, which then implies that the operator was trace class and that they were all equal.

Bruno B
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