Smooth compactly supported functions of an open set are dense in $L^2$

If $\mathrm{\Omega }=\mathrm{\varnothing }$, use $\phi =0$. Otherwise, for $m=0,1,2,\dots$, when $\mathrm{\Omega }\mathrm{\ne }{\mathbb{R}}^n$, set $K_m=\{x:\mathrm{\mid }x\mathrm{\mid }\le m+1,\mathrm{dist}(x,{\mathbb{R}}^n\setminus \mathrm{\Omega })\ge 1\mathrm{/}(m+1)\}$; when $\mathrm{\Omega }={\mathbb{R}}^n$, set $K_m={\bar{B}}_{m+1}(0)$. By [F2] each $K_m$ is compact and contained in $\mathrm{\Omega }$, the sets increase, and ${\bigcup }_m{K}_m=\mathrm{\Omega }$: in the proper-open-set case every point has positive distance from the complement by openness.

Let $F:={\mathbf{1}}_{\mathrm{\Omega }}f\in L^2({\mathbb{R}}^n)$ be the zero extension of [F1] and let $F_m:={\mathbf{1}}_{K_m}F$. Since $K_m\uparrow \mathrm{\Omega }$, the sequence $\mathrm{\mid }F-F_m{\mathrm{\mid }}^2=\mathrm{\mid }F{\mathrm{\mid }}^2{\mathbf{1}}_{\mathrm{\Omega }\setminus K_m}$ decreases pointwise to $0$ and is bounded by the integrable function $\mathrm{\mid }F{\mathrm{\mid }}^2$; hence $\mathrm{\parallel }F-F_m{\mathrm{\parallel }}_{L^2}^2={\int }_{{\mathbb{R}}^n}\mathrm{\mid }F-F_m{\mathrm{\mid }}^2\to 0$ by dominated convergence, and $F_m\to F$ pointwise. Equivalently, the integrals ${\int }_{K_m}\mathrm{\mid }F{\mathrm{\mid }}^2$ increase to $\mathrm{\parallel }F{\mathrm{\parallel }}_{L^2}^2$ by monotone convergence, so the same limit follows. Choose $m$ with $\mathrm{\parallel }F-F_m{\mathrm{\parallel }}_{L^2}<\delta \mathrm{/}2$.

Put $G=F_m$ for the $m$ selected in step 2.1, with a representative zero outside $K_m$. Then $\mathrm{\parallel }F-G{\mathrm{\parallel }}_2<\delta \mathrm{/}2$. If $K_m=\mathrm{\varnothing }$, $G=0$ and $\phi =0$ already has error less than $\delta$; hence assume $K_m\mathrm{\ne }\mathrm{\varnothing }$. Consider every pair $(y,r)$ with $y\in K_m$, $0<r\le 1$ and $B_r(y)\subseteq \mathrm{\Omega }$. The balls $B_{r\mathrm{/}2}(y)$ cover $K_m$, so compactness gives a finite nonempty subcover $B_{r_j\mathrm{/}2}(y_j)$; put ${\epsilon }_0={\min }_j{r}_j\mathrm{/}4>0$. If $y\in K_m$ and $\mathrm{\mid }h\mathrm{\mid }\le {\epsilon }_0$, some covering ball gives $\mathrm{\mid }y+h-y_j\mathrm{\mid }<r_j\mathrm{/}2+{\epsilon }_0<r_j$, so $y+h\in \mathrm{\Omega }$. Thus $K_m+{\bar{B}}_{{\epsilon }_0}(0)\subseteq \mathrm{\Omega }$. This sumset is compact: it is bounded, and for a point $x$ outside it the continuous function $y\mapsto \mathrm{\mid }x-y\mathrm{\mid }$ attains on $K_m$ a minimum greater than ${\epsilon }_0$, so its complement is open.

Choose the unit-mass smooth bump $\rho$ constructed in [F3], and put $G_{\epsilon }=G\ast {\rho }_{\epsilon }$ for $0<\epsilon <{\epsilon }_0$. By [F5] this is smooth globally. If $x\mathrm{\notin }{K}_m+{\bar{B}}_{\epsilon }(0)$, every $y\in K_m$ has ${\rho }_{\epsilon }(x-y)=0$, while $G(y)=0$ off $K_m$; hence the defining integral is zero. Its support therefore lies in the compact set $K_m+{\bar{B}}_{\epsilon }(0)\subseteq \mathrm{\Omega }$, so $G_{\epsilon }{\mathrm{\mid }}_{\mathrm{\Omega }}\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$.

By [F6], $\mathrm{\parallel }{G}_{\epsilon }-G{\mathrm{\parallel }}_2\to 0$. Choose $0<\epsilon <{\epsilon }_0$ with this norm less than $\delta \mathrm{/}2$ and set $\phi =G_{\epsilon }{\mathrm{\mid }}_{\mathrm{\Omega }}$. Since $F,G,G_{\epsilon }$ vanish outside $\mathrm{\Omega }$, $\mathrm{\parallel }\phi -f{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}=\mathrm{\parallel }{G}_{\epsilon }-F{\mathrm{\parallel }}_2\le \mathrm{\parallel }{G}_{\epsilon }-G{\mathrm{\parallel }}_2+\mathrm{\parallel }G-F{\mathrm{\parallel }}_2<\delta$.

Density of $C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$ in $L^2(\mathrm{\Omega })$ follows because $f$ and $\delta$ were arbitrary in step 5.1. Consequently $H_0^1(\mathrm{\Omega })$ is dense in $L^2(\mathrm{\Omega })$: given $g\in L^2(\mathrm{\Omega })$ and $\eta >0$, step 5.1 supplies $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$ with $\mathrm{\parallel }\phi -g{\mathrm{\parallel }}_{L^2}<\eta$, and $\phi \in H_0^1(\mathrm{\Omega })$ by [F7]. Finally let $h\in L^2(\mathrm{\Omega })$ satisfy $(h,v{)}_{L^2}=0$ for every $v\in H_0^1(\mathrm{\Omega })$. By density choose classes $v_k\in H_0^1(\mathrm{\Omega })$ with $\mathrm{\parallel }{v}_k-h{\mathrm{\parallel }}_{L^2}\to 0$; then Cauchy--Schwarz [F8] gives $\mathrm{\mid }(h,h)\mathrm{\mid }=\mathrm{\mid }(h,h-v_k)\mathrm{\mid }\le \mathrm{\parallel }h{\mathrm{\parallel }}_{L^2}\mathrm{\parallel }h-v_k{\mathrm{\parallel }}_{L^2}\to 0$, so $\mathrm{\parallel }h{\mathrm{\parallel }}_{L^2}^2=(h,h)=0$ and $h=0$.