Let $\rho'_{n,j}$ be the $j$th one-system marginal of $\rho'_n$, and put $\varepsilon_n=\|\rho'_n-\rho^{\otimes n}\|_1$. Contractivity of <trace distance> under <partial trace> gives
$$
\|\rho'_{n,j}-\rho\|_1\leq\varepsilon_n
$$
uniformly in $j$. Because the one-system state space is a <compact space>, continuity of $f(\,\cdot\,\|\sigma)$ implies <uniform continuity>. Hence there is a function $\delta(\varepsilon)\to0$ such that
$$
f(\rho'_{n,j}\|\sigma)
\geq f(\rho\|\sigma)-\delta(\varepsilon_n)
$$
for every $j$.

Multipartite superadditivity and additivity now imply
$$
\begin{aligned}
f(\rho'_n\|\sigma^{\otimes n})
&\geq\sum_{j=1}^nf(\rho'_{n,j}\|\sigma)\\
&\geq n f(\rho\|\sigma)-n\delta(\varepsilon_n),\\
f(\rho^{\otimes n}\|\sigma^{\otimes n})
&=n f(\rho\|\sigma).
\end{aligned}
$$
Therefore
$$
\frac1n\left(
f(\rho'_n\|\sigma^{\otimes n})
-f(\rho^{\otimes n}\|\sigma^{\otimes n})\right)
\geq-\delta(\varepsilon_n)\longrightarrow0,
$$
which proves the required <lower asymptotic semicontinuity of quantum relative entropy> argument.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 323 / 4 / iii / Solution

If $\lVert\rho'_n-\rho^{\otimes n}\rVert_1\to0$, then
$$
\liminf_{n\to\infty}\frac1n
\left(D(\rho'_n\|\sigma^{\otimes n})-D(\rho^{\otimes n}\|\sigma^{\otimes n})\right)\geq0.
$$
In finite dimension this follows from <multipartite superadditivity of quantum relative entropy>, <additivity of quantum relative entropy>, contraction of <trace distance> under <partial trace>, and <uniform continuity> on the compact state space.

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