Associativity of distribution convolution under compact support

First suppose all three supports are compact. By F4 every pairing with a smooth function may use a product of cutoffs equal to one near the three supports. Expanding either bracketed convolution on a test $\psi$ then gives the same nested pairing $u_x(v_y(w_z(\psi (x+y+z))))$: for example the inner $w$ pairing is smooth, and the smooth extension of $u\ast v$ pairs it with $x+y$; inserting the three fixed cutoffs makes this exactly the tensor definition. F2's associativity identifies the bracketings. The equality holds also for all smooth $\psi$ by the same cutoffs.

In the general case let $q$ be the possibly noncompact factor and $S,T$ the two compact supports of the other factors. For a fixed compact test support $K$, the closed set $E=\mathrm{supp}q\cap (K-S-T)$ is compact. Choose $\theta =1$ near $E$ with compact support, and write $q=q_0+q_1$, where $q_0=\theta q$. The support of $q_1$ is contained in $\mathrm{supp}q$ and avoids a neighborhood of $E$, so $K$ is disjoint from $S+T+\mathrm{supp}{q}_1$. This last sum is closed: add the compact set $S+T$ to the closed support of $q_1$, using F1's compact-sum argument. Every permitted bracketing containing $q_1$ has support in that sum by F1 twice, and hence pairs to zero on tests supported in $K$. Both bracketings exist: either an inner pair is compactly supported, or the outer factor is one of the compact factors. Bilinearity therefore reduces both bracketings on this test to those with $q_0$. They agree by step 1.1. Since $K$ was arbitrary, associativity follows.

Derivatives have support contained in the original support by F3 and F5, so all asserted convolutions are permitted. For a test $\psi$ use a cutoff $\chi$ equal to one near $(\mathrm{supp}u\times \mathrm{supp}v)\cap a^{-1}(\mathrm{supp}\psi )$, where $a(x,y)=x+y$. The tensor derivative formula and F3 yield [step 2.1, F1, F2, F3, F5] $(({\mathrm{\partial }}^{\alpha }u)\ast v)(\psi )=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}(u\otimes v)({\mathrm{\partial }}_x^{\alpha }(\chi \psi \circ a))\mathrm{.}$ Every term differentiating $\chi$ is zero near the tensor support: on the relevant intersection all such derivatives vanish, and outside it all derivatives of $\psi \circ a$ vanish locally. These terms therefore pair to zero. The remaining term is $(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}(u\ast v)({\mathrm{\partial }}^{\alpha }\psi )$, giving the first identity. Differentiating in $y$ gives the other one, with the same total sign.

In the cutoff definition of ${\delta }_0\ast u$, take a product cutoff with its first factor equal to one near zero and its second equal to one near $\mathrm{supp}\psi$. Evaluating the first pairing at zero leaves exactly $u(\psi )$ by F3. The differentiated identity follows from step 3.1. Zero factors give zero throughout; order zero gives the identity derivative. Only finitely many cutoffs occur for each fixed test. No associativity is asserted merely from pairwise existence without the stated two-compact-factor hypothesis. $\mathrm{\square }$