Regularized coherent-information formula for quantum capacity

Set $q_0=0$. A unit vector in the nonzero space $A^{\otimes n}$ defines a pure input state by [F7]. Its Stinespring output on receiver and environment is pure; its two marginals have equal entropy by the Schmidt decomposition in [F5], so the coherent-information/complement identity in [F5] gives zero. For every input state, [F1] gives $I_c=S(B^n)-S(R{B}^n)\le S(B^n)$ because entropy is nonnegative by [F3]. The state $B^n$ lies on a space of dimension $(\dim B{)}^n$ by [F6], so [F3] gives $S(B^n)\le n\ln \dim B$. Thus $0\le q_n\le n\ln \dim B$.

For positive $m,n$, choose states within $\delta >0$ of the respective suprema $q_m$ and $q_n$. Their product is an admissible input for ${\mathcal{N}}^{\otimes (m+n)}$, and [F2] makes its coherent information the sum of the two values. Letting $\delta \downarrow 0$ gives $q_{m+n}\ge q_m+q_n$. The linear bound in step 1.1 and [F4] show that $q_n\mathrm{/}n$ converges to ${\sup }_{n\ge 1}{q}_n\mathrm{/}n$.

For any state $\rho$, apply the exact type reduction to ${\rho }^{\otimes m}$ and choose a type with $I_c({\tau }_t,{\mathcal{N}}^{\otimes m})\ge m{I}_c(\rho ,\mathcal{N})-d\ln (m+1)\mathrm{.}$ Restrict the input of the $m$-use channel to that type subspace and apply the flat-input direct construction. Concatenating the type embedding with its isometric code preserves the operational encoder condition. Thus $Q(\mathcal{N})\ge I_c(\rho ,\mathcal{N})-\frac{d\ln (m+1)}{m}\mathrm{.}$ Here and below a negative right side is covered by rate zero. To pass from codes of length divisible by $m$ to all lengths $n$, take $k=\lfloor n\mathrm{/}m\rfloor$, use the $k$-block code, and pad the remaining fewer than $m$ inputs with a fixed pure state; the decoder discards their outputs. Fidelity is unchanged and the rate loss tends to zero. Since $\ln (m+1)\mathrm{/}m=((m+1)\mathrm{/}m)\ln (m+1)\mathrm{/}(m+1)\to 0$ by [F15], letting $m\to \mathrm{\infty }$ gives $Q(\mathcal{N})\ge I_c(\rho ,\mathcal{N})$. Applying this result to any fixed block channel ${\mathcal{N}}^{\otimes j}$ and padding again gives $Q(\mathcal{N})\ge q_j\mathrm{/}j$ for every $j$, and therefore the full regularized direct bound.

For a length-$n$ isometric code of dimension $K$, let $\rho =W(I_K\mathrm{/}K)W^{\ast }$, let ${\zeta }_{RM}$ be its decoded entangled state, and put $t=T(\zeta ,{\mathrm{\Phi }}_K)$. Write $g(t)=(1+t)\ln (1+t)-t\ln t$, with $g(0)=0$. Fuchs–van de Graaf gives $t\le \sqrt{1-F_n}$, because $F_n$ is the squared overlap with the pure target. Partial-trace contraction gives $T({\zeta }_M,I_K\mathrm{/}K)\le t$. Quantitative continuity on dimensions $K$ and $K^2$ yields $S(M{)}_{\zeta }\ge \ln K-t\ln K-g(t),S(RM{)}_{\zeta }\le 2t\ln K+g(t)\mathrm{.}$ Thus decoded coherent information is at least $(1-3t)\ln K-2g(t)$. By [F14] and [F1], $I(R:B)=S(R)+I_c$ for the pre-decoding purification and analogously after decoding. The reference entropy is unchanged, and data processing [F13] applied to $I(R;B)$ gives $S(B)-S(RB)\ge S(M)-S(RM)$. Before decoding, the purification is the encoded half-maximally-entangled state, so its coherent information equals $I_c(\rho ,{\mathcal{N}}^{\otimes n})\le q_n$. Consequently $q_n\ge (1-3t)\ln K-2g(t)\mathrm{.}$ For a reliable code sequence $t\to 0$, and $K\le (\dim A{)}^n$ because the encoder is isometric. Divide by $n$ and use this uniform dimension bound to conclude that the limsup code rate is at most $\lim q_n\mathrm{/}n$. Dimension-one codes need no continuity estimate. Combining the converse, the direct bound for every fixed block, and the stated proved superadditive limit gives exactly the retained full quantum-capacity statement.