Local finite order characterization of distributions

Suppose $u$ is continuous, and fix $K$. By F2 its restriction is continuous at zero. Thus there are $m\ge 0$ and $\epsilon >0$ such that $p_m(\psi )<\epsilon$ implies $\mathrm{\mid }u(\psi )\mathrm{\mid }<1$: take the largest order and the smallest positive radius in a finite basic neighborhood contained in the inverse image of the open unit disk. If that intersection has no constraints, the whole space maps into the disk; linearity then makes $u$ zero on this stage, and $m=0,\epsilon =1$ works.

For $p_m(\phi )>0$, apply step 1.1 to $\epsilon \phi \mathrm{/}(2p_m(\phi ))$ to get $\mathrm{\mid }u(\phi )\mathrm{\mid }\le (2\mathrm{/}\epsilon )p_m(\phi )$. If $p_m(\phi )=0$, every positive real multiple of $\phi$ satisfies the same strict neighborhood inequality, so $t\mathrm{\mid }u(\phi )\mathrm{\mid }<1$ for every $t>0$, forcing $u(\phi )=0$. This proves the estimate for the fixed $K$, and the argument applies to every $K$ without selecting a family of pairs.

Conversely suppose the stated estimates hold. Fix $K$ and one witnessing pair. If $C_K=0$, the restriction is zero. Otherwise, for any $\delta >0$, the neighborhood $p_{m_K}<\delta \mathrm{/}{C}_K$ maps into $\mathrm{\mid }z\mathrm{\mid }<\delta$. Hence every restriction is continuous; F2 implies $u$ is continuous on $\mathcal{D}(\mathrm{\Omega })$, and F1 makes it a distribution. For empty $K$ the space is zero and $C_K=m_K=0$ suffice. $\mathrm{\square }$