Additive entanglement-assisted classical capacity formula

For any purification ${\psi }_{RA}$, its reference marginal is ${\rho }_R$ with the same positive eigenvalues as ${\rho }_A$, so $S(R)=S(\rho )$ by [F1]. Substituting [F1] into [F2] gives $I(R;B)=S(\rho )+I_c(\rho ,\mathcal{N})$. Thus the two maximized information expressions in the Statement agree, and [F3] supplies their maximizers.

By [F4], the optimized one-use information is additive across every finite tensor power; hence its regularized information value is the same one-use maximum. This proves additivity for the information expression itself.

For any input state $\rho$ and positive integer $m$, exact spectral type reduction gives a flat state ${\tau }_t$ on an $m$-use input subspace such that $f_{{\mathcal{N}}^{\otimes m}}({\tau }_t)\ge m{f}_{\mathcal{N}}(\rho )-2d\ln (m+1),d=\dim A\mathrm{.}$ Apply the flat-state construction to that fixed block channel. For fixed $m$, both its Weyl alphabet and its receiver output dimension are finite, so the HSW theorem applies before any limit in $m$ is taken. The shared resource is the flat type-subspace maximally entangled state for each outer block. Padding fewer than $m$ unused original-channel inputs with a fixed state gives codes for all sufficiently large lengths and no asymptotic rate loss. Therefore $C_E(\mathcal{N})\ge f_{\mathcal{N}}(\rho )-\frac{2d\ln (m+1)}{m}\mathrm{.}$ Since $\ln (m+1)\mathrm{/}m\to 0$ by [F13], letting $m\to \mathrm{\infty }$ and maximizing over $\rho$ proves the general direct bound with precisely the stated unlimited finite-shared-entanglement model. No unrestricted quantum-input HSW single-letter theorem and no inference from Schumacher compression alone is used.

Consider any length-$n$ stated EA code with uniform message register $M$, receiver's prior share $S$, channel input $A^n$, and receiver output $B^{n}S$. Before transmission its joint state is ${\omega }_{M{A}^{n}S}=\frac{1}{K}{\sum }_m\mathrm{\mid }m⟩⟨m\mathrm{\mid }\otimes ({\mathcal{E}}_m\otimes {\mathrm{id}}_S)(\sigma )\mathrm{.}$ The marginal $MS$ is the product of the uniform message and the original share: each encoder is trace-preserving and hence cannot change $S$. In particular $I(M:S)=0$. Let ${\bar{\rho }}_{A^n}$ be its average input. By the extension lemma, the $MS{A}^n$ state is obtained by a channel on the reference of a purification of ${\bar{\rho }}_{A^n}$. After transmission, quantum-mutual-information data processing gives $I(MS:B^n)\le I(R:B^n{)}_{(\mathrm{id}\otimes {\mathcal{N}}^{\otimes n})({\psi }_{\bar{\rho }})}\mathrm{.}$ Expanding the entropy definitions gives the exact identity $I(M:B^{n}S)=I(MS:B^n)-I(S:B^n)+I(M:S)\mathrm{.}$ The last term is zero, and [F14] gives $I(S:B^n)\ge 0$. Therefore $I(M:B^{n}S)\le I(R:B^n)\le F({\mathcal{N}}^{\otimes n})=nF(\mathcal{N})\mathrm{.}$ The equality uses the stated optimized-information additivity lemma, whose proof follows local-channel mutual-information data processing and product-state entropy additivity. Write $h(u)=-u\ln u-(1-u)\ln (1-u)$ with continuous endpoint values. By [F11], $I(M;\widehat{M})=H(M)-H(M\mid \widehat{M})$; [F15] identifies $I(M:B^{n}S)$ with the Holevo quantity of the cq output state, and the Holevo measurement bound [F10] gives $I(M;\widehat{M})\le I(M:B^{n}S)$. Fano then gives $(1-e_n)\ln K_n\le nF(\mathcal{N})+h(e_n)\mathrm{.}$ Singleton message sets are trivial. Divide by $n$ and let average error tend to zero to bound every achievable rate by $F(\mathcal{N})$, without restricting the shared state, its dimension, or the message-dependent CPTP encoders. Together with the general direct bound, purification identity and attained maximum, this proves the entire retained operational formula. Finally applying the identity $C_E=F$ to both channels and their product, and using the stated proved additivity of $F$, proves the capacity additivity assertion as well.