Corollary: No algebra extension under the stated requirements

There is no extension satisfying Assumption ass:extension.

Proof. Let

By assumption $X_{\epsilon }\to X$ and $Y_{\epsilon }\to Y$, so both convergent families are bounded.

Hypocontinuity implies separate continuity, hence $X_{\epsilon }{\star }_H{W}_g\to X{\star }_H{W}_g$. It also gives joint sequential continuity along convergent sequences: subtract the limit product, split the difference into two terms, and use continuity uniformly over the bounded family in the other factor. Therefore

Applying the continuous coefficient map gives convergence of ${\mathcal{G}}_{\epsilon }(f,g,f^{\prime })$, contradicting Theorem thm:log and Lemma lem:separated.