Theorem: Conditional endpoint and finite-time recovery

Suppose Definition def:four is satisfied. Then

Under the KSW input,

There is an endpoint-optimal deterministic decoder ${\mathcal{D}}_{\ast }$ such that the fixed decoder ${\mathcal{D}}_{\ast }{\mathrm{Ad}}_{a_t}$ on the raw access has error at most

If $\zeta =0$, the endpoint recovery is exact and can use (eq:exact-decoder). If ${\sigma }_{\ast }$ is pure in that exact case, the endpoint radiation channel is an isometric encoding.

Proof. Every map ${\hat{V}}_t$ is isometric because it is a composition of isometries and a unitary. The first two inequalities follow from Lemma lem:stability. The triangle inequality with (eq:endpoint-complement) gives (eq:finite-leakage).

Choose an optimal endpoint decoder, which exists by compactness. Contraction and the triangle inequality give

proving (eq:fixed-endpoint-decoder). Theorem thm:approx gives ${\epsilon }_{\mathrm{rec}}({\mathcal{N}}_{\ast })\le \sqrt{2\zeta }$. Alternatively apply that theorem directly to (eq:finite-leakage). The constant decoder gives the independent cap $q_d$. Proposition prop:access-embedding identifies the optimized raw and embedded recovery errors and proves (eq:finite-endpoint-recovery).

For $\zeta =0$, (eq:endpoint-complement) is exact channel equality. Theorem thm:exact and Corollary cor:dimension give the remaining conclusions.