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I know that if I have two operators $A$ and $B$ and one is bounded and the other is trace class, then $$ \mathrm{Tr}(AB) = \mathrm{Tr}(BA). $$ Another case where this works is when $A$ and $B$ are both Hilbert-Schmidt operators.

But I heard that it is actually sufficient to have $\mathrm{Tr}(|AB|)<\infty$ and $\mathrm{Tr}(|BA|)<\infty$. Has anyone a reference about that? Are there other cases where the cyclicity of the trace still works, or at least where it works "in a certain sense"? In particular I am interested in the case where $A$ is a really nice operator and $B$ is unbounded.

For example, say $B=x$ is the unbounded operator of multiplication by $x\in\Bbb R$ and $A$ is a compact positive operator acting on $L^2$ functions $\varphi$ through the formula $$A\varphi(x) = \sum_j \lambda_j\, \psi_j(x) \int_{\Bbb R} \psi_j\,\varphi$$ with $\sum_j \lambda_j\int_{\Bbb R} |\psi_j(x)|^2\,(1+|x|)\,\mathrm d x< C$. Then $$ \mathrm{Tr}(AB) = \mathrm{Tr}(BA) = \sum_j \lambda_j \int_{\Bbb R} |\psi_j(x)|^2\,x\,\mathrm d x $$


Remark: Another quite borderline case where I know how to do the proof is if $A$ and $B$ are positive operators and $\sqrt A\sqrt B$ is an Hilbert-Schmidt operator and one defines $\mathrm{Tr}(AB) := \mathrm{Tr}(\sqrt B\,A\,\sqrt B) = \|\sqrt A\sqrt B\|_2^2$. Then by invariance of the Hilbert-Schmidt norm by taking the adjoint, $$ \mathrm{Tr}(AB) = \|\sqrt A\sqrt B\|_2^2 = \|\sqrt B\sqrt A\|_2^2 = \mathrm{Tr}(BA). $$

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The case of $\mathrm{Tr}(\vert AB\vert)<\infty$ and $\mathrm{Tr}(\vert BA\vert)<\infty$ for bounded $A,B$ is addressed in Simon's book Trace Ideals and Their Applications, Corollary 3.8.

The idea is to show that $AB$ and $BA$ have the same nonzero eigenvalues including multiplicity and then applying Lidskii's theorem. The former is done in Deift's paper Applications of a commutation formula (this is referenced in Simon); it also treats a very specific case with unbounded operators, namely when $B=A^*$ (but in this case $A^*\!A$ looks to be unbounded, hence not trace class).

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