Problem, fixed model, and precise scope

The sharp dressing considered here is

The minus sign agrees with the first geometric coefficient ${{\mathcal{B}}_f}^{(1)}$. All classical configurations are smooth and unrestricted by their equations of motion.

The preceding obstruction is not interpreted as closing the interacting product problem. Here we ask a narrower question: does admitting the actual compact distributional kernel of (eq:B), with a specified normal topology, permit a continuous extension of the fixed free Wick product?

The answer proved below is negative for a named mandatory graph. The continuity assumptions are part of the theorem. An abstract algebra with a different topology, a smaller collection of bulk observables, or additional renormalized boundary operations is not excluded.