A classical kernel does not specify a quantum extension

Here is an explicit non-identification, used only as a comparison. For a qubit let

and

where $X$ is the Pauli bit flip. The two CPTP maps ${\mathcal{B}}_{\epsilon }$ and ${\mathcal{B}}_{\epsilon }\circ \Delta$ agree on every diagonal input. They therefore implement the same classical binary symmetric transition on computational-basis data.

Nevertheless,

The output trace distance is $1/2$. The second extension is measure-and-prepare; the classical transition table alone did not select that extension.

This is the kind of distinction between records and full quantum outputs emphasized by Cheung's readout and residual comparisons CheungVIIIA,CheungVIIIB,CheungEn07A. The standard measure-and-prepare characterization is given in HSR.

Similarly, $H(t)$ is a classical affinity: for the two Bernoulli laws

one has

The fact that classical affinities admit diagonal-matrix representations does not prove an inequality for arbitrary noncommuting density operators, entangled inputs, or collective POVMs.