Smooth compact supports are dense in Schwartz space

Apply [F2] successively in each coordinate and use [F3] to regroup derivatives; for complex functions apply the real rule to the four real products. This gives ${\mathrm{\partial }}^{\beta }(uv)={\sum }_{\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)({\mathrm{\partial }}^{\gamma }u)({\mathrm{\partial }}^{\beta -\gamma }v)$, where $\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)={\prod }_{j<n}\left(\genfrac{}{}{0px}{}{{\beta }_j}{{\gamma }_j}\right)$. Also, on $\mathrm{\mid }x\mathrm{\mid }\ge R$, $\mathrm{\mid }x\mathrm{\mid }\le {\sum }_{j<n}\mathrm{\mid }{x}_j\mathrm{\mid }$ gives $\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\delta }f(x)\mathrm{\mid }\le R^{-1}{\sum }_{j<n}{p}_{\alpha +e_j,\delta }(f)\mathrm{.}$ Indeed multiply the left side by $\mathrm{\mid }x\mathrm{\mid }$ and bound each $\mathrm{\mid }{x}_j{x}^{\alpha }{\mathrm{\partial }}^{\delta }f\mathrm{\mid }$ by its seminorm.

For $u={\chi }_R-1$, [F1] makes the undifferentiated product-rule term vanish on $\mathrm{\mid }x\mathrm{\mid }\le R$ and have coefficient at most one elsewhere. For $\gamma \mathrm{\ne }0$, ${\mathrm{\partial }}^{\gamma }{\chi }_R$ is supported on $R\le \mathrm{\mid }x\mathrm{\mid }\le 2R$, with supremum $R^{-\mathrm{\mid }\gamma \mathrm{\mid }}{C}_{\gamma }$, $C_{\gamma }=\mathrm{\parallel }{\mathrm{\partial }}^{\gamma }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$. Boundedness follows from continuity on its compact support. Hence for $R\ge 1$ the formula in step 1.1 gives $p_{\alpha \beta }(({\chi }_R-1)f)\le R^{-1}{\sum }_{j<n}{p}_{\alpha +e_j,\beta }(f)+{\sum }_{0\mathrm{\ne }\gamma \le \beta }\left(\genfrac{}{}{0px}{}{\beta }{\gamma }\right)C_{\gamma }{R}^{-1-\mathrm{\mid }\gamma \mathrm{\mid }}{\sum }_{j<n}{p}_{\alpha +e_j,\beta -\gamma }(f)\mathrm{.}$ This tends to zero. The product is smooth with compact support inside the dilated support of $\chi$; each weighted derivative is bounded on that compact set, so ${\chi }_{R}f\in C_c^{\mathrm{\infty }}\subseteq \mathcal{S}$. Taking the explicit integers $R=1,2,\dots$ proves density.