Additivity of optimized entanglement-assisted information

Strong subadditivity [F4] with the middle register one-dimensional gives subadditivity $S(XY)\le S(X)+S(Y)$ for every bipartite state: tensoring with the unique one-dimensional state leaves all nontrivial spectra unchanged. The proof of this consequence uses the natural-log entropy convention and its product-spectrum calculation, with [F7] supplying the logarithm product law. Thus $I(X:Y)\ge 0$ by [F2].

Fix an arbitrary state ${\rho }_{A_1{A}_2}$, and set ${\rho }_i$ to its marginals, ${\omega }_{B_1{B}_2}=({\mathcal{N}}_1\otimes {\mathcal{N}}_2)(\rho )$, and ${\eta }_{E_1{E}_2}=({\mathcal{N}}_1^c\otimes {\mathcal{N}}_2^c)(\rho )$ using product Stinespring dilations. Expanding each entropy by [F1] and each mutual information by [F2] yields $f_{{\mathcal{N}}_1\otimes {\mathcal{N}}_2}(\rho )-f_{{\mathcal{N}}_1}({\rho }_1)-f_{{\mathcal{N}}_2}({\rho }_2)=-I(A_1:A_2{)}_{\rho }-I(B_1:B_2{)}_{\omega }+I(E_1:E_2{)}_{\eta }$. By [F3] applied to the local complementary channels, $I(E_1:E_2{)}_{\eta }\le I(A_1:A_2{)}_{\rho }$. Step 1.1 gives $I(B_1:B_2{)}_{\omega }\ge 0$, so the displayed difference is nonpositive. Therefore $F({\mathcal{N}}_1\otimes {\mathcal{N}}_2)\le F({\mathcal{N}}_1)+F({\mathcal{N}}_2)$.

For product inputs ${\rho }_1\otimes {\rho }_2$, diagonalize each density operator using [F5]. The positive eigenvalues of the product are ${\lambda }_i{\mu }_j$, so expanding $-{\sum }_{i,j}{\lambda }_i{\mu }_j\ln ({\lambda }_i{\mu }_j)$ and applying [F7] proves entropy additivity and hence $f_{{\mathcal{N}}_1\otimes {\mathcal{N}}_2}({\rho }_1\otimes {\rho }_2)=f_{{\mathcal{N}}_1}({\rho }_1)+f_{{\mathcal{N}}_2}({\rho }_2)$. For any density operator, [F5] and [F8] give nonnegative eigenvalues summing to one, so every positive eigenvalue lies in $(0,1]$. Monotonicity and $\ln 1=0$ in [F7] therefore make each term $-{\lambda }_j\ln {\lambda }_j$ nonnegative. Together with [F6], this bounds each of $S({\rho }_i)$, $S({\mathcal{N}}_i({\rho }_i))$, and $S({\mathcal{N}}_i^c({\rho }_i))$ between zero and the logarithm of its fixed register dimension, so $f_{{\mathcal{N}}_i}$ is bounded above and below and its supremum is finite. For any $\delta >0$, choose ${\rho }_i$ with $f_{{\mathcal{N}}_i}({\rho }_i)>F({\mathcal{N}}_i)-\delta$. The product input gives $F({\mathcal{N}}_1\otimes {\mathcal{N}}_2)\ge F({\mathcal{N}}_1)+F({\mathcal{N}}_2)-2\delta$. Letting $\delta \downarrow 0$ proves the reverse inequality.