Proposition: Recorded and conditional stability

Suppose $\widetilde{{\mathcal{M}}_C}$ is a channel with the same classical record and system interface, with classical output blocks, and

For every reference-entangled input, write its two outputs as

Then

Every subsequent complete channel preserves the same error upper bound. If $p_k\ge p_{\min }>0$ and $\epsilon <p_{\min }$, then ${\tilde{p}}_k>0$ and

Proof. The diamond bound gives trace distance at most $\epsilon$ for the two complete outputs. Trace norm is additive on their orthogonal classical blocks, proving the first inequality in (eq:record-block-bound). Taking the classical marginal is a channel and proves the second. Channel postprocessing contracts trace distance, including with a reference.

Fix $k$, abbreviate

The classical marginal bound implies $\left|p-q\right|\le \epsilon$; thus $q>0$ under the stated lower bound. The sum of the norms of all other output blocks is at least $\left|p-q\right|$, because their total trace difference is $q-p$. Therefore

Using positivity and $\mathrm{Tr}\tilde{\sigma }=q$,

State trace distance is also at most one.