I'd like to have the equality condition in the Araki–Lieb–Thirring inequality $$\operatorname{Tr} [(BAB)^r]\leq \operatorname{Tr} [(B^{r}A^{r}B^{r})],$$ valid for $A,B$ semidefinite positive and $r\geq1$ (I'm interested in the case $r=2$). Any clue? I didn't find the proof of it yet, I guess it would be a good start!

As explained in comment, finding the equality condition for the inequality $$\operatorname{Tr} [X^2]\leq \operatorname{Tr} [XX^\dagger]$$ is sufficient!

It is the inequality n°13 of Lieb and Thirring - Inequalities for the moments of the Eigenvalues of the Schrödinger Hamiltonian and their relation to Sobolev inequalities.

  • $\begingroup$ At least for $r=2$ this can be proven with the inequality $$\operatorname{Tr} [X^2]\leq \operatorname{Tr} [XX^\dagger],$$ with $X=B^2A$. It is (4) of aip.scitation.org/doi/pdf/10.1063/1.1704727?class=pdf, proven for instance here: ncbi.nlm.nih.gov/pmc/articles/PMC1063049/pdf/pnas01544-0068.pdf But the equality condition is not clear for me... $\endgroup$ – MarcO Feb 15 at 9:17
  • $\begingroup$ You are asking for the equality case in the Cauchy-Schwartz inequality $\langle X, X^*\rangle \leq \|X\|^2$ for the scalar product $\langle \cdot ,\cdot\rangle = Tr(\cdot (\cdot)^*)$... so the answer is that $Tr(X^2) = Tr(X X^*)$ if and only if $X=X^*$, so (if $X=B^2 A$ for positive $A,B$) if and only if $A$ and $B$ commute. $\endgroup$ – Mikael de la Salle Feb 15 at 13:38
  • $\begingroup$ Yes, this is what found using the Schur decomposition, thanks! $\endgroup$ – MarcO Feb 18 at 9:50

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