Lemma: Constant-channel recovery error for one qubit

If a one-qubit input channel is constant, its optimal recovery error is $3/4$.

Proof. Any decoded channel is a constant preparation $X\mapsto \mathrm{tr}(X)\sigma$. On a normalized logical Bell state with reference, its output is $\sigma \otimes I_R/2$. Its Bell overlap is $1/4$. The Bell-projector test therefore gives trace distance at least $3/4$.

The bound is attained by the completely depolarizing channel

The triangle inequality gives

The Bell-state lower bound makes this an equality.