Compactly supported distributions have global finite order

If $S=\mathrm{\varnothing }$, F2 gives $u=0$ and take $L=\mathrm{\varnothing }$, $m=C=0$, with the empty supremum zero. Otherwise take $\chi$ from F3 equal to one near $S$ and put $L=\mathrm{supp}\chi$. It is compactly inside $\mathrm{\Omega }$ and contains a neighborhood of $S$. For every test $\phi$, the test $(1-\chi )\phi$ vanishes near $S$, so $u(\phi )=u(\chi \phi )$ by F2.

Apply F1 on the single compact $L$ to obtain $C_0,m$ with $\mathrm{\mid }u(\psi )\mathrm{\mid }\le C_0{p}_m(\psi )$ on ${\mathcal{D}}_L$. The finite product rule gives ${\mathrm{\partial }}^{\alpha }(\chi \phi )={\sum }_{\beta \le \alpha }\left(\genfrac{}{}{0px}{}{\alpha }{\beta }\right){\mathrm{\partial }}^{\beta }\chi {\mathrm{\partial }}^{\alpha -\beta }\phi$; this follows by iterating the coordinate product rule, with coefficients combined by Pascal's identity. If $A={\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le m}{\sup }_L\mathrm{\mid }{\mathrm{\partial }}^{\beta }\chi \mathrm{\mid }$, then $p_m(\chi \phi )\le 2^{m}A{\max }_{\mathrm{\mid }\gamma \mathrm{\mid }\le m}{\sup }_L\mathrm{\mid }{\mathrm{\partial }}^{\gamma }\phi \mathrm{\mid }$. The derivatives of $\chi$ are bounded on $L$, so $A$ is finite. Together with step 1.1 this gives the asserted estimate with $C=C_0{2}^{m}A$.

If $\phi \in {\mathcal{D}}_K$ for another compact $K$, all its derivatives vanish off $K$, so the maximum over $L$ is at most $p_{m,K}(\phi )$. Thus the same exponent gives the global finite-order property (indeed the same $C$ works here). The proof selected only one cutoff and one finite-order witness pair and used no choice axiom. $\mathrm{\square }$