[proofplan]
We construct an explicit sequence of test functions converging to $f$ in the Schwartz topology by truncation: multiply $f$ by smooth cutoffs $\chi_k$ that equal $1$ on $B(0,k)$ and are supported in $B(0, 2k)$. The rapid decay of $f$ absorbs the polynomial factors arising from derivatives of $\chi_k$, ensuring every Schwartz seminorm of $f_k - f$ tends to zero.
[/proofplan]
[step:Define the truncation sequence]
Choose $\chi \in \mathcal{D}(\mathbb{R}^n)$ with $\chi(x) = 1$ for $|x| \leq 1$, $\mathrm{supp}(\chi) \subseteq B(0, 2)$, and $0 \leq \chi \leq 1$.
Define $\chi_k(x) := \chi(x/k)$ for $k \geq 1$.
Then $\chi_k \in \mathcal{D}(\mathbb{R}^n)$, $\chi_k = 1$ on $B(0, k)$, $\mathrm{supp}(\chi_k) \subseteq B(0, 2k)$, and $f_k := \chi_k f \in \mathcal{D}(\mathbb{R}^n)$.
[/step]
[step:Show $\|f_k - f\|_{\alpha,\beta} \to 0$ for every Schwartz seminorm]
Since $f_k - f = (\chi_k - 1)f$ and $\chi_k - 1$ vanishes on $B(0, k)$, only the region $|x| \geq k$ contributes to $\|f_k - f\|_{\alpha,\beta}$.
By the Leibniz rule,
\begin{align*}
\partial^\beta\bigl((1 - \chi_k) f\bigr) &= \sum_{\delta \leq \beta} \binom{\beta}{\delta} \bigl(\partial^\delta(1 - \chi_k)\bigr) \, (\partial^{\beta - \delta} f).
\end{align*}
**The $\delta = 0$ term.** We have $|x^\alpha (1 - \chi_k(x)) \, \partial^\beta f(x)| \leq |x^\alpha \, \partial^\beta f(x)| \cdot \mathbb{1}_{|x| \geq k}$.
Since $f \in \mathcal{S}$, the function $x \mapsto |x^\alpha \, \partial^\beta f(x)|$ is bounded and tends to $0$ as $|x| \to \infty$, so
\begin{align*}
\sup_{|x| \geq k} |x^\alpha \, \partial^\beta f(x)| &\to 0 \quad \text{as } k \to \infty.
\end{align*}
**The $|\delta| \geq 1$ terms.** The chain rule gives $\partial^\delta \chi_k(x) = k^{-|\delta|}(\partial^\delta\chi)(x/k)$, so $\partial^\delta(1 - \chi_k)$ is supported in $B(0, 2k) \setminus B(0, k)$ with
\begin{align*}
|\partial^\delta(1 - \chi_k)(x)| &\leq k^{-|\delta|} \|\partial^\delta\chi\|_{L^\infty}.
\end{align*}
Therefore
\begin{align*}
|x^\alpha \, (\partial^\delta(1 - \chi_k))(x) \, (\partial^{\beta-\delta} f)(x)| &\leq k^{-|\delta|} \|\partial^\delta\chi\|_{L^\infty} \cdot |x^\alpha \, \partial^{\beta-\delta} f(x)| \cdot \mathbb{1}_{k \leq |x| \leq 2k}.
\end{align*}
The factor $|x^\alpha \, \partial^{\beta-\delta} f(x)|$ is bounded by $\|f\|_{\alpha, \beta - \delta} < \infty$, and the factor $k^{-|\delta|} \to 0$ as $k \to \infty$.
Summing over $\delta$, each term in the Leibniz expansion tends to $0$, so $\|f_k - f\|_{\alpha, \beta} \to 0$.
[guided]
The key tension in this argument is between two competing effects:
differentiating $\chi_k$ produces factors of $k^{-|\delta|}$ that grow when $k$ is small but decay when $k$ is large, while the support of $(1 - \chi_k)$ recedes to infinity.
For the $\delta = 0$ term, no derivatives hit the cutoff, so we simply observe that $(1 - \chi_k)f$ is supported where $|x| \geq k$.
The rapid decay of $f \in \mathcal{S}$ ensures $\sup_{|x| \geq k} |x^\alpha \partial^\beta f(x)| \to 0$.
For the $|\delta| \geq 1$ terms, the chain rule gives $\partial^\delta \chi_k(x) = k^{-|\delta|}(\partial^\delta \chi)(x/k)$.
This derivative is supported in the annulus $k \leq |x| \leq 2k$ and has magnitude at most $k^{-|\delta|}\|\partial^\delta \chi\|_{L^\infty}$.
Multiplying by $|x^\alpha \partial^{\beta-\delta}f(x)|$, which is bounded by the finite Schwartz seminorm $\|f\|_{\alpha, \beta-\delta}$, gives a bound of $k^{-|\delta|}\|\partial^\delta \chi\|_{L^\infty}\|f\|_{\alpha,\beta-\delta}$.
Since $|\delta| \geq 1$, this vanishes as $k \to \infty$.
The Leibniz expansion is a finite sum (over $\delta \leq \beta$), so the total tends to zero.
[/guided]
[/step]
[step:Conclude density]
Since $\|f_k - f\|_{\alpha,\beta} \to 0$ for every $\alpha, \beta \in \mathbb{N}_0^n$, the sequence $\{f_k\} \subset \mathcal{D}(\mathbb{R}^n)$ converges to $f$ in the Schwartz topology.
[/step]
