Schwartz space is dense in L2

For $u\in C_c^{\mathrm{\infty }}$, every derivative vanishes off its compact support: outside the support, $u$ is zero on a neighbourhood. Thus $x^{\alpha }{\mathrm{\partial }}^{\beta }u$ is continuous with compact support, hence bounded, for all $\alpha ,\beta$. The empty-support case is the zero function. Therefore $C_c^{\mathrm{\infty }}\subseteq \mathcal{S}$.

If $u\in \mathcal{S}$, then [F2] gives $u\in L^1$, while $\mathrm{\mid }u(x)\mathrm{\mid }\le p_{00}(u)$ by the defining seminorm. Hence ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }u{\mathrm{\mid }}^2\le p_{00}(u){\int }_{{\mathbb{R}}^n}\mathrm{\mid }u\mathrm{\mid }<\mathrm{\infty }\mathrm{.}$ Thus every Schwartz function determines an $L^2$ class.

Given $f\in L^2$ and $\epsilon >0$, apply [F1] with $p=2$ to obtain $u\in C_c^{\mathrm{\infty }}$ with $\mathrm{\parallel }f-u{\mathrm{\parallel }}_2<\epsilon$. Step 1.1 puts this same $u$ in Schwartz space, and step 1.2 confirms that its class belongs to $L^2$. Every norm ball about $f$ therefore meets the Schwartz classes, proving density with precisely the countable-choice assumption of the smooth-density supplier.