Proposition: Purely classical return has a permanent floor

If the entire returned interface is classical, every deterministic decoder has

The full-diamond floor is $2(1-1/d)$.

Proof. A decoder of a classical channel is measure-and-prepare and hence entanglement breaking EntanglementBreaking. The Bell-input output is separable. A product vector has Bell squared overlap at most $1/d$; convexity gives the same bound for a separable state. Testing the Bell projector proves the claim.