# Relative entropy inequality for many copies of a channel

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 $$\mathcal{H}_{A^n}$$.

• 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 $\rho, \sigma$. – Rammus Nov 23 '20 at 21:40