Proposition: Zero endpoint leakage is exact constancy

For finite input and separable normal outputs, ${\delta }_{\mathrm{full}}(\mathcal{E})=0$ implies that $\mathcal{E}$ is a replacement channel.

Proof. Choose states ${\sigma }_n$ with $d_{\diamond }(\mathcal{E},{\mathcal{T}}_{{\sigma }_n})\to 0$. Evaluation on $I_C/d$ gives ${\sigma }_n\to \mathcal{E}(I_C/d)$ in trace distance. Replacement-channel distance equals the corresponding state distance. The triangle inequality gives $\mathcal{E}={\mathcal{T}}_{\mathcal{E}(I_C/d)}$.