Compactly supported distributions have global finite order

F1 immediately gives $\mathrm{\mid }({\mathrm{\partial }}^{\alpha }{\delta }_a)(\phi )\mathrm{\mid }\le {\sup }_{x\in \mathrm{\Omega },\mathrm{\mid }\beta \mathrm{\mid }\le m}\mathrm{\mid }{\mathrm{\partial }}^{\beta }\phi (x)\mathrm{\mid }$, an explicit global estimate of order $m$ and constant one. It vanishes on tests supported away from $a$. To see it is nonzero on every neighborhood of $a$, take by F3 a cutoff equal to one near $a$ and multiply it by $(x-a{)}^{\alpha }\mathrm{/}\alpha !$; its $\alpha$ derivative at $a$ is one. Thus the support is exactly $\{a\}$, consistent with F2's compact-support theorem.

Suppose $m\ge 1$ and fix compact $K$ containing a ball about $a$. Choose $h\in \mathcal{D}({\mathbb{R}}^n)$ supported in the unit ball and equal to $x^{\alpha }\mathrm{/}\alpha !$ near zero, by F3. For sufficiently small $\epsilon >0$, let ${\phi }_{\epsilon }(x)={\epsilon }^{m}h((x-a)\mathrm{/}\epsilon )$, whose support is in $K$. Then ${\mathrm{\partial }}^{\alpha }{\phi }_{\epsilon }(a)=1$, whereas for every integer $0\le q<m$, [step 1.1, F1, F3] $p_q({\phi }_{\epsilon })\le {\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le q}{\epsilon }^{m-\mathrm{\mid }\beta \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\beta }h{\mathrm{\parallel }}_{\mathrm{\infty }}⟶0.$ An order-$q$ bound would force a pairing of modulus one to tend to zero, a contradiction. Thus no smaller order works on $K$, or globally. For $m=0$, the upper bound and nonvanishing in step 1.1 prove exact order zero, since allowable orders are nonnegative integers. Compacts avoiding $a$ give the zero restriction, and compacts without a neighborhood of $a$ are excluded from the sharp local assertion. $\mathrm{\square }$