Proposition: One-use information transfer under implementation error

For two qubit-output channels on the same input interface, if $d_{\diamond }(\mathcal{N},\hat{\mathcal{N}})\le \delta$, then

In particular, a certified implementation allowance for the frozen channel gives

Proof. For qubit states, trace distance $\delta$ bounds the difference of their larger eigenvalues by $\delta$, as follows from the Bloch-vector formula and the reverse triangle inequality.

The binary entropy modulus is

For fixed $|s-t|$, this follows by differentiating the entropy difference and using the decreasing derivative of binary entropy. For a distance upper bound greater than $1/2$, use the trivial entropy bound one. Thus output entropy differences are at most $b(\delta )$.

Apply this once to the ensemble-average output and once to the average of the conditional output entropies. Every fixed-ensemble Holevo value changes by at most $2b(\delta )$. Take suprema over the same ensemble class. Both channel Holevo informations are between zero and one.