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Suppose we have two quantum channels $\mathcal{E}_{A\rightarrow B}, \mathcal{F}_{A\rightarrow B}$ that satisfy

$$D(\mathcal{E}(\rho_A)\|\mathcal{E}(\sigma_A))\geq D(\mathcal{F}(\rho_A)\|\mathcal{F}(\sigma_A)) \quad \forall \rho_A,\sigma_A\in \mathcal{H}_A,$$

where $D(\rho\|\sigma) = Tr(\rho\log\rho) - Tr(\rho\log\sigma)$. Is it then true that

$$D\left(\mathcal{E}^{\otimes n}(\rho_{A^n})\|\mathcal{E}^{\otimes n}(\sigma_{A^n})\right)\geq D\left(\mathcal{F}^{\otimes n}(\rho_{A^n})\|\mathcal{F}^{\otimes n}(\sigma_{A^n})\right) \quad \forall \rho_{A^n},\sigma_{A^n}\in \mathcal{H}_{A^n}$$

The question is due to a claim made in https://arxiv.org/abs/1110.5746. Proposition 4 is shown to hold for $n=1$ but on Page 3 (after Eq (8)), the claim is made for all $n$. I am not sure why. I understand it for the case where $\rho_{A^n}, \sigma_{A^n}$ are also $n$-fold tensor products i.e. $\rho_A^{\otimes n}$ but the statement here is for any pair of states in $\mathcal{H}_{A^n}$.

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  • $\begingroup$ After reading the passage you are referring to I don't believe that the author is claiming that this always holds. Instead they are defining $\mathcal{E}$ to be "less divergence contracting" IF the inequality holds for all $n\geq 1$ and all operators $\rho, \sigma$. $\endgroup$ – Rammus Nov 23 '20 at 21:40

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