Lemma: Constant-channel error and universal upper bounds

Put $a_d=1-d^{-2}$. Then

Every channel from $C$ has recovery error at most $a_d$ and leakage distance from replacement channels at most $a_d$.

Proof. For any replacement state, the Bell-input output is $I_Q/d\otimes \sigma$. Its overlap with the target Bell projector is $d^{-2}$, giving the lower bound $a_d$.

The uniform twirl over the $d^2$ Weyl unitaries is ${\mathcal{T}}_{I_C/d}$. Their Bell outputs are mutually orthogonal. The triangle inequality and that Bell input give the matching distance $a_d$.

A decoder which ignores its input and prepares $I_C/d$ gives the recovery upper bound. For a channel $\Phi$,

Postprocessing contraction gives the leakage upper bound.