= Lower asymptotic semicontinuity of quantum relative entropy
Let $\rho,\sigma$ be finite-dimensional <density operators> with $D(\rho\|\sigma)<\infty$. If $\|\rho'_n-\rho^{\otimes n}\|_1\to0$, then
$$
\liminf_{n\to\infty}\frac1n
\left[D(\rho'_n\|\sigma^{\otimes n})-nD(\rho\|\sigma)\right]\geq0.
$$
The finite-reference condition is essential to make the subtraction meaningful; one cannot subtract two infinite <quantum relative entropies>.
Put $\varepsilon_n=\|\rho'_n-\rho^{\otimes n}\|_1/2$. Each one-site marginal $\rho'_{n,i}$ has <trace distance> at most $\varepsilon_n$ from $\rho$, by contraction under <partial trace>. If $\rho'_n$ lies outside the support of $\sigma^{\otimes n}$, its <quantum relative entropy> is infinite and the lower bound is automatic. Otherwise restrict to the <support of a positive operator> of $\sigma$. There $\log\sigma$ is bounded, and <entropy continuity from Jordan decomposition> gives a uniform $c_n\to0$ such that
$$
D(\rho'_{n,i}\|\sigma)\geq D(\rho\|\sigma)-c_n.
$$
<Multipartite superadditivity of quantum relative entropy> then yields
$$
D(\rho'_n\|\sigma^{\otimes n})\geq\sum_{i=1}^nD(\rho'_{n,i}\|\sigma)
\geq nD(\rho\|\sigma)-nc_n,
$$
which proves the claim. Compactness alone does not make relative <entropy> continuous across a change of reference support.
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