I have spent my spare time picking up Fourier analysis on $\mathcal{S}(\mathbb{R})$. Instead of rewriting everything to accommodate notation and proofs on $\mathcal{S}(\mathbb{R}^n)$, I think it would be nice to have a result saying that any $f \in \mathcal{S}(\mathbb{R}^n)$ is a limit of elements in $\text{span}(\phi_1 \otimes \cdots \otimes \phi_n)$ where $\phi_i \in \mathcal{S}(\mathbb{R})$.
If this were true, then I think I could just make an extension argument for all the usual operators like the position, momentum, convolution and Fourier transform.
One idea I had is showing that
$$C_c^{\infty}(\mathbb{R}^n) \subset \text{cl } \text{span}(C_c^{\infty}(\mathbb{R})^{\otimes n})$$ where the closure is with respect to the locally convex topology on $\mathcal{S}(\mathbb{R}^n)$ and then show that the compactly supported smooth functions are dense in Schwartz space.
Or is there some other typical method?
Edit
I wonder if I can say that
$\mathcal{S}(\mathbb{R}^n) = \text{span}(\mathcal{A})$ iff for each $\alpha, \beta \in \mathbb{N}_0$, the closure of $X^{\beta} \partial^{\alpha} \mathcal{A}$ with respect to the uniform norm generates $C_0(\mathbb{R}^n)$ and then apply the complex Stone-Weierstrass theorem on locally compact spaces to the tensor product of Schwartz functions on $\mathbb{R}$.
Best Answer
We will need some facts from the convolution theory. Let function $\omega\in C_c^\infty(\mathbb{R}^n)$ such that $\omega\geq 0$ and $\int_{\mathbb{R}^n}\omega(x)\, dx=1$. Denote $\omega_\varepsilon(x):=\varepsilon^{-n}\omega(x/\varepsilon)$.
If $f\in L^1_{\operatorname{loc}}(\mathbb{R}^n)$ then $f*\omega\in C^\infty(\mathbb{R}^n)$ and $D^\alpha (f*\omega)=f*(D^\alpha \omega)$ for all $\alpha\in \mathbb{N}^n_0$.
If $f\in C_c^\infty (\mathbb{R}^n)$ then $\sup\limits_{\mathbb{R}^n}|D^\alpha f-D^\alpha f*\omega_\varepsilon|\to 0$ at $\varepsilon \to 0$ for all $\alpha$ and there are $\varepsilon_0>0$ and compact set $K$ such shat $\operatorname{supp}f*\omega_\varepsilon,\operatorname{supp}f\subset K$, $\varepsilon\leq\varepsilon_0$ (i.e. $f*\omega_\varepsilon \to f$ in $C_c^\infty(\mathbb{R}^n)$).
$\int_{\mathbb{R}^n}|D^\alpha\omega_\varepsilon(x)|\, dx\leq c_\alpha\varepsilon^{-|\alpha|}$.
Fix $f\in C_c^\infty(\mathbb{R}^n)$. We will show that there is $$\{h^\ell\}\subset X:=\operatorname{span}\big(g\, |\, g(x)=g_1(x_1)...g_n(x_n), g_1,...,g_n\in C_c^\infty (\mathbb{R})\big)$$ such that $\sup\limits_{\mathbb{R}^n}|D^\alpha f-D^\alpha h^\ell|\to 0$ for all $\alpha$ and $\operatorname{supp}h^\ell,\operatorname{supp}f\subset K$ for all $\ell$ and for some compact set $K$.
Consider binary cubes $$Q^k_{i_1...i_n}=I^k_{i_1}\times...\times I^k_{i_n},\ k\in\mathbb{N}, i_j\in\mathbb{Z}$$ where $I^k_i=[i/2^k;(i+1)/2^k)$. Let $$f^k(x)=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\chi_{Q^k_{i_1...i_n}}(x)$$ where the number of syllables is finite since $\operatorname{supp}f$ is compact. It is easy to show that $$\delta_k:=\sup\limits_{\mathbb{R}^n}|f^k-f|\to 0.$$
Consider $\omega^1\in C_c^\infty(\mathbb{R})$ such that $\omega^1\geq 0$ and $\int_{\mathbb{R}}\omega^1(x)\, dx=1$. Then nonnegative function $\omega^n\in C_c^\infty(\mathbb{R}^n)$ defined by $$\omega^n(x)=\omega^1(x_1)...\omega^1(x_n)$$ belongs to $C_c^\infty (\mathbb{R}^n)$. Also $\int_{\mathbb{R}^n}\omega^n(x)\, dx=1$ and $\omega^n_\varepsilon(x)=\omega^1_\varepsilon(x_1)...\omega^1_\varepsilon(x_n)$.
I. Using the diagonal Cantor process, we find the sequence $\{\varepsilon(\ell)\}$ such that $\varepsilon(\ell)\to 0$ and $\delta^\alpha_\ell:=\sup\limits_{\mathbb{R}^n}|D^\alpha f-D^\alpha f*\omega^n_{\varepsilon(\ell)}|\to 0$ for all $\alpha$.
II. $f^k*\omega^n_\varepsilon\in X$. Indeed $$f^k*\omega^n_\varepsilon(x)=\int\limits_{\mathbb{R}^n}\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\chi_{Q^k_{i_1...i_n}}(y)\omega^n_\varepsilon(x-y)\, dy$$ $$=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\int\limits_{\mathbb{R}}...\int\limits_{\mathbb{R}}\chi_{I^k_{i_1}}(y_1)...\chi_{I^k_{i_n}}(y_n)\omega^1_\varepsilon(x_1-y_1)...\omega^1_\varepsilon(x_n-y_n)\, dy_1...dy_n$$ $$=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\prod\limits_{j=1}^n\int\limits_{\mathbb{R}}\chi_{I^k_{i_j}}(y)\omega^1_\varepsilon(x_j-y)\, dy=\sum\limits_{i_1...i_n}f\Big(\frac{i_1}{2^k},...,\frac{i_n}{2^k}\Big)\prod\limits_{j=1}^n\chi_{I^k_{i_j}}*\omega^1_\varepsilon(x_j)$$ where $\chi_{I^k_{i_j}}*\omega^1_\varepsilon\in C_c^\infty(\mathbb{R})$.
III. We have $$|D^\alpha f^k*\omega^n_\varepsilon(x)-D^\alpha f*\omega^n_\varepsilon(x)|\leq \int\limits_{\mathbb{R}^n}|f^k(y)-f(y)||D^\alpha\omega^n_\varepsilon(x-y)|\, dy\leq \delta_kc_\alpha\varepsilon^{-|\alpha|}.$$
IV. Using the diagonal Cantor process, we find the sequence $\{k(\ell)\}$ such that $\delta_{k(\ell)}\varepsilon(\ell)^{-|\alpha|}\to 0$ for all $\alpha$.
V. Finally let $h^\ell=f^{k(\ell)}*\omega^n_{\varepsilon(\ell)}$. We have $$|D^\alpha f^{k(\ell)}*\omega^n_{\varepsilon(\ell)}(x)-D^\alpha f(x)|\leq |D^\alpha f^{k(\ell)}*\omega^n_{\varepsilon(\ell)}(x)-D^\alpha f*\omega^n_{\varepsilon(\ell)}(x)|+|D^\alpha f*\omega^n_{\varepsilon(\ell)}(x)-D^\alpha f(x)|$$ $$\leq \delta_{k(\ell)}c_\alpha\varepsilon(\ell)^{-|\alpha|}+\delta^\alpha_\ell.$$ I.e. $\sup\limits_{\mathbb{R}^n}|D^\alpha h^\ell-D^\alpha f|\leq \delta_{k(\ell)}c_\alpha\varepsilon(\ell)^{-|\alpha|}+\delta^\alpha_\ell$ where $\delta_{k(\ell)}c_\alpha\varepsilon(\ell)^{-|\alpha|}+\delta^\alpha_\ell\to 0$ for all $\alpha$.
The validity of condition $$\operatorname{supp}h^\ell,\operatorname{supp}f\subset K$$ is easy to check.
Therefore $h^\ell\to f$ in $\mathcal{S}(\mathbb{R}^n)$ (this follows from the properties of the sequence $\{h^\ell\}$, and condition $\operatorname{supp}h^\ell\subset K$ plays an important role).