Convolution with a test function is smooth

Put $S=\mathrm{supp}\phi$. For compact $H\subseteq V$, all $y$-supports of $F(x,y)=\phi (x-y)$, $x\in H$, lie in $H-S$. This is compact as the continuous image of the compact product $H\times S$, and it lies in $\mathrm{\Omega }$ by the definition of $V$. The function $F$ is jointly smooth, so F2 gives smoothness of $u\ast \phi$ and ${\mathrm{\partial }}_x^{\alpha }(u\ast \phi )(x)=u(({\mathrm{\partial }}^{\alpha }\phi )(x-\cdot ))$.

Differentiating the reflected test in $y$ gives ${\mathrm{\partial }}_y^{\alpha }\phi (x-y)=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}({\mathrm{\partial }}^{\alpha }\phi )(x-y)$. F3 therefore gives $({\mathrm{\partial }}^{\alpha }u)\ast \phi (x)=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}u({\mathrm{\partial }}_y^{\alpha }\phi (x-\cdot ))=u(({\mathrm{\partial }}^{\alpha }\phi )(x-\cdot ))$, since the two signs multiply to one. Together with step 1.1 this proves both equalities.

The support of ${\mathrm{\partial }}^{\alpha }\phi$ is contained in $S$, so its safe domain contains $V$, justifying the stated restriction. If $\phi =0$, the safe domain is all of ${\mathbb{R}}^n$ and each expression is zero. If $V$ is empty the smoothness and equalities are vacuous on that open set; $\alpha =0$ gives the defining convolution identity. No choice is used. $\mathrm{\square }$