Theorem: Finite-code endpoint continuity

If the preceding strong convergence holds for a finite code, it holds in operator norm, and $V_{\mathrm{out}}$ is an isometry. Writing

the complete channel and either reduced channel differ from their endpoint channels by half-diamond distance at most $a_t$.

In particular, on a common residual space,

Thus a proved sequence ${\delta }_{\mathrm{full}}(t)\to 0$, together with the declared asymptotic construction, implies ${\delta }_{\mathrm{end}}=0$.

Proof. Choose an orthonormal code basis. Convergence on its finitely many vectors bounds the operator norm difference by the square root of the sum of their squared differences. Preservation of inner products passes to the limit, so $V_{\mathrm{out}}$ is an isometry.

Lemma lem:isometry gives channel convergence. For every fixed replacement state, the two distances differ by at most the channel comparison allowance. Taking infima in both directions proves the last estimate.