Convolution of distributions when one has compact support

Let $u,v\in {\mathcal{D}}^{\mathrm{\prime }}({\mathbb{R}}^n)$, $n\ge 1$, with at least one compact support. Write $S=\mathrm{supp}u$, $T=\mathrm{supp}v$ and $a(x,y)=x+y$. For $\psi \in \mathcal{D}({\mathbb{R}}^n)$ put $E_{\psi }=(S\times T)\cap a^{-1}(\mathrm{supp}\psi )$. For any $\chi \in \mathcal{D}({\mathbb{R}}^{2n})$ equal to one on a neighborhood of $E_{\psi }$, set the candidate value $C_{u,v,\psi }(\chi )=(u\otimes v)(\chi \psi \circ a)\mathrm{.}$

The tensor and its support are supplied by Tensor product distributions and iterated pairings and Support of a distribution. The set $E_{\psi }$ is compact: if $S$ is compact, it is a closed subset of the compact set $S\times (\mathrm{supp}\psi -S)$; interchange factors if $T$ is compact. A cutoff therefore exists by Test function cutoffs and euclidean localization. The lemma Convolution of distributions is well defined under the support hypothesis ↗ proves that the candidate value is independent of $\chi$ and is a continuous linear functional of $\psi$; define $(u\ast v)(\psi )$ to be this common value. If both supports are compact it agrees with the smooth extension pairing of Compactly supported distributions extend to smooth functions. Empty support gives the zero candidate, and $\chi =0$ is allowed when $E_{\psi }$ is empty. No infinite selection is part of the definition.