Uniform finite order bounds for pointwise bounded distributions

Fix compact $K$. For integers $j\ge 1$ put $E_j=\{\phi \in {\mathcal{D}}_K:\mathrm{\mid }u(\phi )\mathrm{\mid }\le j\text{ for every }u\in \mathcal{U}\}$. It is closed as an intersection of inverse images of closed disks under continuous restrictions. Pointwise boundedness implies ${\mathcal{D}}_K={\bigcup }_{j\ge 1}{E}_j$. The space contains zero, so it is nonempty, and F2–F4 give an $E_j$ with nonempty interior.

Take ${\phi }_0$ and a neighborhood ${\phi }_0+\{h:p_m(h)<\epsilon \}\subseteq E_j$ for some $m\ge 0,\epsilon >0$. Then ${\phi }_0\in E_j$ and for $p_m(h)<\epsilon$, linearity gives $\mathrm{\mid }u(h)\mathrm{\mid }\le \mathrm{\mid }u({\phi }_0+h)\mathrm{\mid }+\mathrm{\mid }u({\phi }_0)\mathrm{\mid }\le 2j$ for all $u$. For $p_m(h)>0$, scale $h$ by $\epsilon \mathrm{/}(2p_m(h))$ to obtain $\mathrm{\mid }u(h)\mathrm{\mid }\le (4j\mathrm{/}\epsilon )p_m(h)$. If $p_m(h)=0$, every positive multiple is in the neighborhood, forcing $u(h)=0$ by the same uniform bound. This gives the claimed estimate for the fixed $K$.

Let a net from $\mathcal{U}$ converge pointwise to a scalar-valued map $v$ on tests. Passing to limits in addition and scalar multiplication shows $v$ is complex-linear. Passing to the limit in the bound from step 2.1 gives $\mathrm{\mid }v(\phi )\mathrm{\mid }\le C{p}_m(\phi )$ on each ${\mathcal{D}}_K$. F1 proves $v$ is a distribution. The witnesses are obtained for one compact at a time, with no additional choice principle.

For a pointwise convergent sequence, every scalar sequence of evaluations is bounded: its convergent tail is bounded and its remaining finite set has a finite maximum. Thus its range is a pointwise bounded family, and step 3.1 applies. A general convergent scalar net need not be bounded over all its indices, so that reasoning is used only for sequences. For empty $\mathcal{U}$ take $C=m=0$; empty $K$ has zero test space and the same choice works. Dependent Choice was used precisely in F3 for the Baire step. $\mathrm{\square }$