Leibniz rule for distributions

Fix $i$ and a test $\phi$. The ordinary product rule gives $a{\mathrm{\partial }}_i\phi ={\mathrm{\partial }}_i(a\phi )-({\mathrm{\partial }}_{i}a)\phi$: subtract the product at a point from the product at its coordinate increment, insert the mixed product, divide by the increment and pass to the limit. Therefore [given, F1, F2, algebra] $⟨{\mathrm{\partial }}_i(au),\phi ⟩=-⟨u,a{\mathrm{\partial }}_i\phi ⟩=⟨{\mathrm{\partial }}_{i}u,a\phi ⟩+⟨u,({\mathrm{\partial }}_{i}a)\phi ⟩\mathrm{.}$ By F2 this is the first-order formula.

From F1, evaluating two consecutive derivative operations on a test gives the sign $(-1{)}^{\mathrm{\mid }\gamma \mathrm{\mid }+1}$ times $u({\mathrm{\partial }}^{\gamma }{\mathrm{\partial }}_i\phi )$. Commutation of the smooth test partials identifies this with $({\mathrm{\partial }}^{\gamma +e_i}u)(\phi )$. Hence ${\mathrm{\partial }}_i{\mathrm{\partial }}^{\gamma }u={\mathrm{\partial }}^{\gamma +e_i}u$. The asserted formula for $\alpha =0$ is $au=au$.

Suppose the formula holds for $\alpha$. Differentiate it by ${\mathrm{\partial }}_i$ and apply step 1.1 to each smooth coefficient times its distribution. Step 2.1 yields the two terms with indices $\beta +e_i$ and $\beta$, respectively. For each resulting index $\eta$, their coefficients add to $\left(\genfrac{}{}{0px}{}{\alpha }{\eta -e_i}\right)+\left(\genfrac{}{}{0px}{}{\alpha }{\eta }\right)=\left(\genfrac{}{}{0px}{}{\alpha +e_i}{\eta }\right)$, using zero for an out-of-range binomial coefficient. The equality is Pascal's identity in coordinate $i$ with all other coordinate factors unchanged. Thus the formula holds for $\alpha +e_i$, and induction on total degree proves every case. All sums are finite. For $a=0$ or $u=0$ every term is zero, and on the empty domain the identity is between zero functionals. $\mathrm{\square }$