Mollifier approximation in distributions

Choose $R>0$ with $\mathrm{supp}\rho \subseteq \bar{B}(0,R)$. For a nonempty compact test support $K\subseteq \mathrm{\Omega }$ choose $d>0$ such that $K+\bar{B}(0,2d)\subseteq \mathrm{\Omega }$. If $\epsilon R<d$, then $K\subseteq V_{\epsilon }$, even with a fixed compact neighborhood inside it. F2 gives smoothness there. Hence $f_{\epsilon }$ is Borel and bounded on every compact subset of its safe domain, so it is locally integrable; F6 types its integral functional as a regular distribution. No nonnegativity or symmetry of $\rho$ is required.

Apply F3 to $F(x,y)=\psi (x){\rho }_{\epsilon }(x-y)$, with parameter $x$ in $V_{\epsilon }$, integrating on a compact neighborhood of $K$ contained in that domain. Its slices have a common compact support in $\mathrm{\Omega }$ there. Outside $K$ the integrand is zero. Thus F3 and F6 give [step 1.1, F3, F6] $\int f_{\epsilon }(x)\psi (x)dx=u({\psi }_{\epsilon }),{\psi }_{\epsilon }(y)=\int {\rho }_{\epsilon }(x-y)\psi (x)dx\mathrm{.}$ By the affine substitution $x=y+\epsilon z$, justified for these compact smooth integrands by F4, ${\psi }_{\epsilon }(y)=\int \rho (z)\psi (y+\epsilon z)dz$. All these tests are supported in $K+\bar{B}(0,d)$.

Extend $\psi$ smoothly by zero to ${\mathbb{R}}^n$. Its every derivative is uniformly continuous. Differentiating the last compact integral (by uniform difference-quotient estimates, or F3's derivative clause) yields [step 2.1, F3, F4] $p_m({\psi }_{\epsilon }-\psi )\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{L^1}{\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\sup }_{y,\mathrm{\mid }h\mathrm{\mid }\le \epsilon R}\mathrm{\mid }{\mathrm{\partial }}^{\alpha }\psi (y+h)-{\mathrm{\partial }}^{\alpha }\psi (y)\mathrm{\mid }⟶0.$ Here unit mass subtracts $\psi (y)$ inside the integral; the displayed estimate is also the Riemann integral triangle estimate for continuous compact functions. F5 on the common compact support now gives $u({\psi }_{\epsilon })\to u(\psi )$. This proves the assertion with step 2.1. If the test or domain is empty, both sides are zero; $\epsilon =0$ is a limit endpoint, not a defined kernel. Countable Choice enters only through F3, F4 and the Lebesgue regular-distribution interpretation. $\mathrm{\square }$