[proofplan]
Part (1) is immediate from the definition of the Schwartz seminorms: differentiating $f$ shifts the derivative index, so each seminorm of $\partial^\gamma f$ equals a single seminorm of $f$. Part (2) requires the Leibniz rule to handle the interaction between the polynomial factor $x^\gamma$ and the derivatives in the seminorm; the resulting bound is a finite linear combination of seminorms of $f$, with indices shifted by the multi-index $\gamma$.
[/proofplan]
[step:Prove Part (1): differentiation maps $\mathcal{S}$ to $\mathcal{S}$ with a single-seminorm identity]
By the definition of the Schwartz seminorm,
\begin{align*}
\|\partial^\gamma f\|_{\alpha,\beta} &= \sup_{x \in \mathbb{R}^n} |x^\alpha \, \partial^\beta(\partial^\gamma f)(x)| = \sup_{x \in \mathbb{R}^n} |x^\alpha \, \partial^{\beta+\gamma} f(x)| = \|f\|_{\alpha, \beta+\gamma}.
\end{align*}
Since $f \in \mathcal{S}(\mathbb{R}^n)$, every seminorm $\|f\|_{\alpha, \beta+\gamma}$ is finite, so every seminorm of $\partial^\gamma f$ is finite, giving $\partial^\gamma f \in \mathcal{S}(\mathbb{R}^n)$.
The identity $\|\partial^\gamma f\|_{\alpha,\beta} = \|f\|_{\alpha, \beta+\gamma}$ shows that each output seminorm equals a single input seminorm, which gives continuity of $f \mapsto \partial^\gamma f$.
[/step]
[step:Prove Part (2): polynomial multiplication maps $\mathcal{S}$ to $\mathcal{S}$ with a finite seminorm bound]
Fix $\alpha, \beta \in \mathbb{N}_0^n$.
We must bound $\|x^\gamma f\|_{\alpha,\beta} = \sup_x |x^\alpha \, \partial^\beta(x^\gamma f(x))|$.
Apply the Leibniz rule:
\begin{align*}
\partial^\beta(x^\gamma f) &= \sum_{\delta \leq \beta} \binom{\beta}{\delta} (\partial^\delta x^\gamma)(\partial^{\beta-\delta} f).
\end{align*}
The term $\partial^\delta x^\gamma$ is nonzero only when $\delta \leq \gamma$ componentwise, in which case $\partial^\delta x^\gamma = \frac{\gamma!}{(\gamma-\delta)!} x^{\gamma-\delta}$, a monomial of degree $|\gamma - \delta|$.
For each such term:
\begin{align*}
\sup_{x} \left|x^\alpha \cdot \frac{\gamma!}{(\gamma-\delta)!} x^{\gamma-\delta} \cdot \partial^{\beta-\delta} f(x)\right| &= \frac{\gamma!}{(\gamma-\delta)!} \sup_x |x^{\alpha+\gamma-\delta} \, \partial^{\beta-\delta} f(x)| = \frac{\gamma!}{(\gamma-\delta)!} \|f\|_{\alpha+\gamma-\delta, \, \beta-\delta}.
\end{align*}
Summing over the finitely many $\delta \leq \beta$ with $\delta \leq \gamma$:
\begin{align*}
\|x^\gamma f\|_{\alpha,\beta} &\leq \sum_{\substack{\delta \leq \beta \\ \delta \leq \gamma}} \binom{\beta}{\delta} \frac{\gamma!}{(\gamma-\delta)!} \|f\|_{\alpha+\gamma-\delta, \, \beta-\delta}.
\end{align*}
Each seminorm on the right is finite since $f \in \mathcal{S}(\mathbb{R}^n)$.
The finite set $F = \{(\alpha+\gamma-\delta, \beta-\delta) : \delta \leq \beta, \, \delta \leq \gamma\}$ and the constant $C = \max_\delta \binom{\beta}{\delta} \frac{\gamma!}{(\gamma-\delta)!}$ give the claimed bound, establishing both $x^\gamma f \in \mathcal{S}(\mathbb{R}^n)$ and continuity.
[guided]
Why does polynomial multiplication preserve the Schwartz class?
The seminorm $\|x^\gamma f\|_{\alpha,\beta}$ asks for the supremum of $|x^\alpha \partial^\beta(x^\gamma f)|$.
The factor $x^\gamma$ interacts with $\partial^\beta$ through the Leibniz rule, producing terms of the form $(\partial^\delta x^\gamma)(\partial^{\beta-\delta}f)$.
Each derivative $\partial^\delta$ landing on $x^\gamma$ reduces the polynomial degree from $|\gamma|$ to $|\gamma - \delta|$ and introduces a factorial coefficient $\gamma!/(\gamma - \delta)!$.
Crucially, the resulting monomial $x^{\gamma - \delta}$ combines with the existing $x^\alpha$ factor to give $x^{\alpha + \gamma - \delta}$, which simply shifts the first index of the seminorm.
The second index shifts from $\beta$ to $\beta - \delta$ because the remaining derivatives land on $f$.
The sum is finite because $\delta$ ranges over multi-indices bounded by both $\beta$ and $\gamma$.
The constant $C$ absorbs the binomial and factorial coefficients, which depend only on $\alpha, \beta, \gamma, n$ and not on $f$.
This is why the map $f \mapsto x^\gamma f$ is continuous on $\mathcal{S}$: each output seminorm is bounded by a fixed finite combination of input seminorms.
[/guided]
[/step]
