Compactly supported distributions extend to smooth functions

Take $\chi$ from F3. For any smooth $f$, $\chi f$ is a test. If ${\chi }^{\mathrm{\prime }}$ is another such cutoff, $(\chi -{\chi }^{\mathrm{\prime }})f$ is a test vanishing near $S$, so F2 gives equal pairings. Thus the formula is well-defined, linear, and agrees with $u$ on tests because $(1-\chi )\phi$ vanishes near $S$. If $f$ vanishes near $S$, so does $\chi f$, proving the additional vanishing assertion.

Apply F1 to $\chi f$. Let $m,C,L$ be its fixed estimate and put $L^{\mathrm{\prime }}=L\cup \mathrm{supp}\chi$, compactly inside $\mathrm{\Omega }$. The finite product formula gives $q_{L,m}(\chi f)\le 2^{m}A{q}_{L^{\mathrm{\prime }},m}(f)$, where $A={\max }_{\mathrm{\mid }\beta \mathrm{\mid }\le m}{\sup }_{L^{\mathrm{\prime }}}\mathrm{\mid }{\mathrm{\partial }}^{\beta }\chi \mathrm{\mid }<\mathrm{\infty }$. Hence $\mathrm{\mid }\tilde{u}(f)\mathrm{\mid }\le C{2}^{m}A{q}_{L^{\mathrm{\prime }},m}(f)$, proving continuity in the stated smooth topology.

Tests are dense in that topology. Indeed a basic neighborhood of a smooth $f$ imposes finitely many derivative bounds on compacts $K_1,\dots ,K_s$. Use F3 on their finite union to take $\eta =1$ near that union. Then the test $\eta f$ has exactly the same derivatives as $f$ on all these compacts, so it lies in the neighborhood. If there are no constraints, the zero test suffices. If two continuous linear extensions agree on tests, their difference has a closed zero kernel (the scalar target is Hausdorff) containing this dense set, hence vanishes on all smooth functions. This proves uniqueness without selecting a sequence of cutoffs. For empty $S$ take the zero extension; empty $\mathrm{\Omega }$ has only the zero function. $\mathrm{\square }$