Smooth functions are weakly dense in distributions

If $\mathrm{\Omega }=\mathrm{\varnothing }$ use $f_j=0$ for every $j\in \mathbb{N}$. Otherwise set $K_0=\mathrm{\varnothing }$ and, for $j\ge 1$, set $K_j=\{x\in {\mathbb{R}}^n:\mathrm{\mid }x\mathrm{\mid }\le j,\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })\ge 1\mathrm{/}j\},$ with distance to the empty set interpreted as infinity. The distance function is one-Lipschitz, so F1 makes each $K_j$ compact as a closed bounded set; the strict inequalities $j<j+1$ and $1\mathrm{/}j>1\mathrm{/}(j+1)$ give $K_j\subseteq \mathrm{int}{K}_{j+1}$. If $L\subseteq \mathrm{\Omega }$ is compact, finitely many balls $B(x_i,r_i\mathrm{/}2)$ cover $L$ with $B(x_i,r_i)\subseteq \mathrm{\Omega }$; hence $L$ is bounded and has distance at least ${\min }_i(r_i\mathrm{/}2)>0$ from the complement, so $L\subseteq K_j$ for all sufficiently large $j$. Put ${\chi }_0=0$ and use Countable Choice with F1 to choose, for every $j\ge 1$, a cutoff ${\chi }_j$ equal to one near $K_j$. The product ${\chi }_{j}u$ vanishes outside $\mathrm{supp}{\chi }_j$ directly from its definition, so its support is compact and ambient closed. Let $w_j$ be its zero extension by F2. Apply F1 with $K=\{0\}$ and $\mathrm{\Omega }={\mathbb{R}}^n$ to obtain a nonnegative bump $\eta$ equal to one near zero. Its Lebesgue integral $c$ is finite and positive, so $\rho =\eta \mathrm{/}c$ is a unit-mass bump; compactness of its support gives $R>0$ with $\mathrm{supp}\rho \subseteq B(0,R)$. For each $j\ge 1$ with nonempty cutoff support, take $d_j$ to be the smaller of one and half the distance from $\mathrm{supp}{\chi }_j$ to ${\mathbb{R}}^n\setminus \mathrm{\Omega }$; for empty support put $d_j=1$. Then $d_j>0$ and $\mathrm{supp}{\chi }_j+\bar{B}(0,d_j)\subseteq \mathrm{\Omega }$. Set ${\epsilon }_j=\min (1\mathrm{/}j,d_j\mathrm{/}(2R))$.

Put $f_0=0$, and for $j\ge 1$ define $f_j=w_j\ast {\rho }_{{\epsilon }_j}$. F3 makes the latter smooth on all ${\mathbb{R}}^n$. If $x$ is outside the closed sum $\mathrm{supp}{\chi }_j+{\epsilon }_j\mathrm{supp}\rho$, the test $y\mapsto {\rho }_{{\epsilon }_j}(x-y)$ has support disjoint from $\mathrm{supp}{w}_j$. F4 gives $f_j(x)=0$. The sum is compact and lies in $\mathrm{\Omega }$ by step 1.1. Thus every $f_j{\mathrm{\mid }}_{\mathrm{\Omega }}$ belongs to $\mathcal{D}(\mathrm{\Omega })$, and F5 makes its integral functional a distribution.

For fixed $\psi \in \mathcal{D}(\mathrm{\Omega })$ extend it by zero to ${\mathbb{R}}^n$. For $j\ge 1$, apply F3's parameter-pairing lemma to $\psi (x){\rho }_{{\epsilon }_j}(x-y)$ and then affine substitution $x=y+{\epsilon }_{j}z$. This gives $\int f_j\psi =w_j({\psi }_{{\epsilon }_j})=u({\chi }_j{\psi }_{{\epsilon }_j})$, with ${\psi }_{\epsilon }(y)=\int \rho (z)\psi (y+\epsilon z)dz$. All sufficiently small $\epsilon$ have these test supports in one compact $L\subseteq \mathrm{\Omega }$. By step 1.1, $L\subseteq K_j$ for all sufficiently large $j$. Then ${\chi }_j{\psi }_{{\epsilon }_j}={\psi }_{{\epsilon }_j}$. For each multi-index $\alpha$, differentiating the compact integral and using $\int \rho =1$ gives ${\sup }_y\mathrm{\mid }{\mathrm{\partial }}^{\alpha }{\psi }_{\epsilon }(y)-{\mathrm{\partial }}^{\alpha }\psi (y)\mathrm{\mid }\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{L^1}{\sup }_{y,\mathrm{\mid }h\mathrm{\mid }\le \epsilon R}\mathrm{\mid }{\mathrm{\partial }}^{\alpha }\psi (y+h)-{\mathrm{\partial }}^{\alpha }\psi (y)\mathrm{\mid }\to 0$. Uniform continuity of every derivative of the compactly supported smooth $\psi$ proves the limit. F3's finite-order estimate on $L$ now gives $u({\psi }_{{\epsilon }_j})\to u(\psi )$, proving weak convergence. A convergent sequence meets every neighborhood of its limit, which proves density. Zero $u$ gives the zero sequence, and Countable Choice has precisely the uses in F5. $\mathrm{\square }$