Flat-input coherent information is achievable by subspace codes

Put $s_B=S(\mathcal{N}(\tau ))$ and $s_E=S({\mathcal{N}}^c(\tau ))$. By [F11], $s_B-s_E=I_c(\tau ,\mathcal{N})$; choose $\eta >0$ with $R<s_B-s_E-2\eta$. On $n$ uses, let $\mathrm{\mid }{v}_n⟩=(I\otimes V^{\otimes n})\mathrm{\mid }{\mathrm{\Phi }}_{d^n}⟩$ in factor order $A^{\mathrm{\prime }}{B}^n{E}^n$. Let $P_n,Q_n$ be the spectral projectors for the IID receiver and environment states from Exponential averaged spectral projectors for a finite cq ensemble, and let ${ϵ}_n\le C{e}^{-cn}$ bound the sum of their failure probabilities. The commuting joint projection has acceptance at least $1-{ϵ}_n$. For $\mathrm{\mid }{\widehat{v}}_n⟩=(I_{A^{\mathrm{\prime }}}\otimes P_n\otimes Q_n)\mathrm{\mid }{v}_n⟩,$ the rank-two pure-vector calculation used in gentle measurement gives $\parallel \mathrm{\mid }{v}_n⟩⟨v_n\mathrm{\mid }-\mathrm{\mid }{\widehat{v}}_n⟩⟨{\widehat{v}}_n\mathrm{\mid }{\parallel }_1\le 2\sqrt{{ϵ}_n}\mathrm{.}$ This follows also by applying the gentle-measurement lemma to the pure state and the joint projection. Let $\omega ={\mathrm{Tr}}_{B^n}\mathrm{\mid }{\widehat{v}}_n⟩⟨{\widehat{v}}_n\mathrm{\mid }$. Its environment rank is at most $e^{n(s_E+\eta )}$. Since the projected vector is pure, $\mathrm{tr}({\omega }^2)=\mathrm{tr}({\omega }_{B^n}^2)\mathrm{.}$ Moreover $0⪯{\omega }_{B^n}⪯P_n\mathcal{N}(\tau {)}^{\otimes n}{P}_n⪯e^{-n(s_B-\eta )}{P}_n\mathrm{.}$ To verify the first inequality, take the partial trace after $Q_n$; the complementary environment projection produces another positive summand of the original receiver marginal. Hence $\mathrm{tr}({\omega }^2)\le e^{-n(s_B-\eta )}$, because its trace is at most one.

First prove $s_B\le \ln d+s_E$. The one-use Stinespring output of $\mathrm{\mid }{\mathrm{\Phi }}_d⟩$ is pure on $A^{\mathrm{\prime }}BE$, so [F5] gives $s_B=S(A^{\mathrm{\prime }}E)$. For its marginal state ${\omega }_{A^{\mathrm{\prime }}E}$, append a one-dimensional register $C$ in a pure state and order the resulting state as $A^{\mathrm{\prime }}CE$. Strong subadditivity [F10] gives $S(A^{\mathrm{\prime }}C)+S(CE)\ge S(C)+S(A^{\mathrm{\prime }}CE)$, that is $S(A^{\mathrm{\prime }})+S(E)\ge S(A^{\mathrm{\prime }}E)$; [F12] gives $S(A^{\mathrm{\prime }})=\ln d$. Hence $s_B\le \ln d+s_E$, so $I_c=s_B-s_E\le \ln d$ and $0<R<I_c$ implies $K_n=\lfloor e^{nR}\rfloor \le d^n$ by [F12]. For an isometry $W:{\mathbb{C}}^{K_n}\to A^{\otimes n}$, let ${\rho }_{R{E}^n}^W$ be the environment/reference state produced by encoding half of $\mathrm{\mid }{\mathrm{\Phi }}_{K_n}⟩$. Coefficient matrices give exactly the normalized-compression expression for the unprojected marginal. Use the finite ensemble above with tolerance $e^{-n}$. Apply its compression estimate to $\omega$, and its averaged trace-contraction estimate to the difference between projected and unprojected marginals. Their environment marginals differ by at most $2\sqrt{{ϵ}_n}$. The triangle inequality therefore yields $\mathbb{E}{\parallel {\rho }_{R{E}^n}^W-\frac{I_R}{K_n}\otimes {\mathcal{N}}^c(\tau {)}^{\otimes n}\parallel }_1\le 6\sqrt{{ϵ}_n}+2e^{-n}+\sqrt{2e^{-n(s_B-s_E-R-2\eta )}}⟶0.$ Order the finite generator list and its length-$L$ words. For each $n$, take the first word whose displayed norm is at most twice its expectation; if that expectation is zero, take the first zero-norm word. A qualifying word exists by finite averaging. This specifies the entire encoder sequence without countable choices.

The product target is purified by $\mathrm{\mid }{\mathrm{\Phi }}_{K_n}{⟩}_{RM}\otimes \mathrm{\mid }{e}_n{⟩}_{E^{n}F}$, with a sufficiently large finite $F$. The encoded Stinespring output is another purification on reference/environment versus receiver. Root-fidelity purification characterization, after adding a pure finite ancilla to the receiver if needed, supplies an isometry on the receiver's purifying space whose overlap with that target is the root fidelity of the two marginals. The fixed-purification form follows from the purification/unitary-equivalence theorem. Fuchs–van de Graaf makes this root fidelity at least one minus half the displayed norm. Trace out the extra decoder factor $F$: the resulting CPTP decoder has squared entanglement fidelity at least the square of that overlap and hence tending to one. The encoder is exactly the isometry $W$; no generating-code/transmitting-code equivalence is assumed. Fix ordered bases throughout. In the purification characterization use the polar/SVD attaining map, extending every kernel partial isometry by the ordered basis completion; in repeated eigenspaces, project the ordered coordinate vectors and Gram–Schmidt in that order. This specifies a decoder at each block length by one formula, rather than choosing a sequence of unrelated attaining maps.

If $I_c(\tau ,\mathcal{N})\le 0$, choose a unit vector of $A^{\otimes n}$ as the encoder for $K_n=1$. The decoder to $\mathbb{C}$ is $X\mapsto \mathrm{tr}(X)$, a channel by [F13] using the isometry $x\mapsto 1\otimes x$ and tracing $A^{\otimes n}$. Its entanglement fidelity is one and its rate is zero. For a subspace inclusion $J:{\mathbb{C}}^D\to A$, the restricted channel ${\mathcal{N}}_J(X)=\mathcal{N}(JX{J}^{\ast })$ has dilation $VJ$, and its flat input $I_D\mathrm{/}D$ has coherent information $I_c(J{J}^{\ast }\mathrm{/}D,\mathcal{N})$ by the fixed-dilation definition. Applying steps 1.1–3.1 to ${\mathcal{N}}_J$ and composing each encoder with $J^{\otimes n}$ proves the stated subspace case.