Let $\mathcal{H}_A\otimes\mathcal{H}_B$ be a finite-dimensional bipartite Hilbert space, $P_A$ a positive semi-definite operator on $\mathcal{H}_A$, $P_B$ a positive semi-definite operator on $\mathcal{H}_B$, and $U_{AB}$ a unitary operator on $\mathcal{H}_A\otimes\mathcal{H}_B$. In some roundabout way involving integrals over the unitary group it seems I can prove the following inequality: $$\mathrm{tr}_A\left(\mathrm{tr}_B((P_A\otimes P_B)U_{AB})\mathrm{tr}_B((P_A\otimes P_B)U_{AB}^\dagger)\right)\leq\mathrm{tr}(P_A^2)\mathrm{tr}(P_B)^2.$$ I have been unable to find a more ``conventional'' proof of this inequality, e.g. using von Neumann's trace inequality or Holder's inequality, so I am curious if this is a known result or if someone can provide a proof which uses only linear algebra.
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You can rewrite the left-hand side $\mathrm{tr}_A\left(P_AX P_AX^*\right)$, where $X=\mathrm{tr}_B((1\otimes P_B)U_{AB})$. By Hölder's inequality this is less than $\|P_A\|_2^2 \|X\|_\infty^2$, so your inequality is just that $\|X\|_\infty \leq \mathrm{tr}(P_B)$, which is certainly "conventional". For a justification use Hölder's inequality to obtain $\mathrm{tr}_A(CX) = \mathrm{tr}((C\otimes P_B) U_{AB}) \leq \|C\otimes P_B\|_1 = \|C\|_1 \mathrm{tr}(P_B)$, so this conventional inequality follows by taking the supremum over all $C$ with $\|C\|_1 =1$.
answered Mar 17, 2022 at 22:58
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1$\begingroup$ Perfect, just what I was looking for thanks! (you have a small typo $\tau\to \mathrm{tr}$) $\endgroup$ Commented Mar 18, 2022 at 1:42
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$\begingroup$ @DanielHarlow Thanks, I corrected the typo. $\endgroup$ Commented Mar 18, 2022 at 16:33