Distributional derivatives commute with test-function convolution

For $\mathrm{\Omega }={\mathbb{R}}^n$, every translated compact support $x-\mathrm{supp}\phi$ lies in $\mathrm{\Omega }$, so the safe domain is all of ${\mathbb{R}}^n$. Also ${\mathrm{\partial }}^{\alpha }\phi$ is a test function with support contained in $\mathrm{supp}\phi$. Thus all three convolutions are defined there by [F2].

By the smoothness and parameter-differentiation conclusion in [F1], differentiating the pairing in [F2] gives ${\mathrm{\partial }}_x^{\alpha }(T\ast \phi )(x)=⟨T(y),({\mathrm{\partial }}^{\alpha }\phi )(x-y)⟩=T\ast ({\mathrm{\partial }}^{\alpha }\phi )(x)$.

The signed derivative definition [F3] gives $({\mathrm{\partial }}^{\alpha }T)\ast \phi (x)=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}⟨T(y),{\mathrm{\partial }}_y^{\alpha }\phi (x-y)⟩$. Since ${\mathrm{\partial }}_y^{\alpha }\phi (x-y)=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}({\mathrm{\partial }}^{\alpha }\phi )(x-y)$, the two signs cancel and this equals the expression in step 1.2. Hence all three smooth functions agree on the safe domain.

The same calculation includes $\alpha =0$; if $\phi =0$ or $T=0$ each side is zero; for $n=1$ it is the same one-variable derivative calculation, and formally for $n=0$ the only multi-index is zero and the identity is tautological. There are no spatial boundary endpoints on ${\mathbb{R}}^n$, and no choice is used.