Regularized private-information formula for quantum channel capacity

Write $p_n=p_n(\mathcal{N})$. The one-state ensemble has private information zero, so $p_n\ge 0$. For any finite ensemble, [F1]–[F3] give $I_p=I(X;B^n)-I(X;E^n)\le I(X;B^n)$, since [F3] implies $I(X;E^n)\ge 0$. By [F2], [F7], and the logarithm law [F13], $I(X;B^n)=S(B^n)-{\sum }_x{p}_{x}S(B_x)\le S(B^n)$ because each conditional output entropy is nonnegative by [F6]. The output dimension and logarithm identity in [F8], together with [F6], give $S(B^n)\le n\ln \dim B$. Therefore $0\le p_n\le n\ln \dim B$.

For positive integers $m,n$, choose finite ensembles whose private informations are within an arbitrary $\delta >0$ of $p_m$ and $p_n$. Their product ensemble is finite and, by [F4], has private information equal to the sum. Thus $p_{m+n}\ge p_m+p_n-2\delta$. Letting $\delta \downarrow 0$ proves superadditivity; define $p_0=0$. The linear bound in step 1.1 and [F5] now prove that $p_n\mathrm{/}n$ converges to its supremum.

Apply the construction to any finite ensemble of inputs for a fixed block channel ${\mathcal{N}}^{\otimes j}$. Each code uses a finite average of tensor products of those possibly mixed block inputs, permitted by the operational model. Group its blocks and pad fewer than $j$ leftover original-channel uses with a fixed state. The extra complementary state is independent of the message; product spectral eigenvalues and the logarithm law [F8] show that its entropy adds the same constant to the average and every conditional output entropy, so leakage is unchanged. The decoder discards the extra receiver output. Hence $P(\mathcal{N})\ge p_j\mathrm{/}j$ for every $j$: approximate the finite-ensemble supremum first, then take an arbitrary rate strictly below its private information. The stated Fekete argument gives the entire regularized direct bound.

For any uniform $M$-message code, let $\widehat{M}$ be Bob's decoded message and $e$ its average error. Write $h(u)=-u\ln u-(1-u)\ln (1-u)$, with continuous endpoint values. The finite Fano inequality gives $H(M\mid \widehat{M})\le h(e)+e\ln (M-1)\le h(e)+e\ln M$ for $M\ge 2$. By [F12], $I(M;\widehat{M})=H(M)-H(M\mid \widehat{M})$, and $H(M)=\ln M$ for the uniform message. The logarithm law used in the finite chain-rule proof is also supplied by [F13]. The Holevo measurement bound [F11], applied to the finite receiver ensemble, gives $I(M;\widehat{M})\le {\chi }_B$. Therefore $(1-e)\ln M\le {\chi }_B+h(e)\mathrm{.}$ The uniform input ensemble $\{M^{-1},{\rho }_{m,n}\}$ is among those defining $p_n$, and its private information is ${\chi }_B-{\mathrm{\ell }}_n$. Thus $(1-e)\ln M\le p_n+{\mathrm{\ell }}_n+h(e)\mathrm{.}$ For $M=1$, the desired asymptotic bound is trivial. For a reliable strongly secret sequence, divide by $n$; since $p_n\mathrm{/}n$ is bounded and $e_n\to 0$, the displayed inequality itself bounds $\ln M_n\mathrm{/}n$, even though no dimension-based bound on the number of repeated classical messages was assumed. The terms ${\mathrm{\ell }}_n\mathrm{/}n$ and $h(e_n)\mathrm{/}n$ vanish. This proves the converse for the stated liminf-achievable rate convention and the retained full equality $P={\lim }_n{p}_n\mathrm{/}n={\sup }_n{p}_n\mathrm{/}n$.