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Let $\rho ,\sigma$ be finite-dimensional density operators with $D(\rho \Vert \sigma )<\infty$. If $\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1\to 0$, thenThe finite-reference condition is essential to make the subtraction meaningful; one cannot subtract two infinite quantum relative entropies.

Put ${\epsilon }_n=\Vert {\rho }_n^{\prime }-{\rho }^{\otimes n}{\Vert }_1/2$. Each one-site marginal ${\rho }_{n,i}^{\prime }$ has trace distance at most ${\epsilon }_n$ from $\rho$, by contraction under partial trace. If ${\rho }_n^{\prime }$ 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 $c_n\to 0$ such thatMultipartite superadditivity of quantum relative entropy then yieldswhich proves the claim. Compactness alone does not make relative entropy continuous across a change of reference support.

Apply bipartite superadditivity of quantum relative entropy first to system $1$ and systems $2,\dots ,n$, and then repeat on the remaining joint state. Mathematical induction givesThis is multipartite superadditivity of quantum relative entropy. Repeated application of bipartite additivity similarly gives