[proofplan]
We prove $\mathcal{F}: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ is a topological automorphism in four stages. First, two exchange identities (differentiation and multiplication) are established via integration by parts and the Leibniz integral rule. Second, iteration yields a formula expressing $\xi^\alpha \partial_\xi^\beta(\mathcal{F}f)$ as the Fourier transform of an $L^1$ function. Third, this formula combined with $\|\mathcal{F}h\|_{L^\infty} \le \|h\|_{L^1}$ bounds every Schwartz seminorm of $\mathcal{F}f$ by finitely many seminorms of $f$, proving the mapping property and continuity. Fourth, the [Fourier Inversion Theorem](/theorems/246) shows that $\mathcal{F}^{-1}$ (defined by the inverse integral) is a two-sided inverse, and the same analysis proves its continuity.
[/proofplan]
[step:Establish the differentiation exchange identity $\mathcal{F}(\partial_{x_j} f) = i\xi_j \mathcal{F}f$]
[claim:Differentiation exchange identity]
For every $f \in \mathcal{S}(\mathbb{R}^n)$, every $j \in \{1, \ldots, n\}$, and every $\xi \in \mathbb{R}^n$,
\begin{align*}
\mathcal{F}(\partial_{x_j} f)(\xi) &= i\xi_j \, \mathcal{F}f(\xi).
\end{align*}
[/claim]
[proof]
By Fubini's theorem (justified since $\partial_{x_j} f \in \mathcal{S} \subset L^1$), separate the $x_j$-integration:
\begin{align*}
\mathcal{F}(\partial_{x_j} f)(\xi) &= \int_{\mathbb{R}^{n-1}} e^{-ix' \cdot \xi'} \left(\int_{\mathbb{R}} \partial_{x_j} f(x', x_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j)\right) d\mathcal{L}^{n-1}(x'),
\end{align*}
where $x' = (x_1, \ldots, x_{j-1}, x_{j+1}, \ldots, x_n)$. Integrate the inner integral by parts in $x_j$. The boundary term $[f(x', x_j) e^{-ix_j \xi_j}]_{x_j = -\infty}^{+\infty}$ vanishes: since $f \in \mathcal{S}$, for any $N \geq 1$ we have $|f(x', x_j)| \leq C_N |x_j|^{-N}$ for $|x_j| \geq 1$, and $|e^{-ix_j \xi_j}| = 1$. The remaining term is
\begin{align*}
-\int_{\mathbb{R}} f(x', x_j)(-i\xi_j) e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j) &= i\xi_j \int_{\mathbb{R}} f(x', x_j) \, e^{-ix_j \xi_j} \, d\mathcal{L}^1(x_j).
\end{align*}
Reassembling the Fubini decomposition gives $\mathcal{F}(\partial_{x_j} f)(\xi) = i\xi_j \, \mathcal{F}f(\xi)$.
[/proof]
[/step]
[step:Establish the multiplication exchange identity $\partial_{\xi_j}(\mathcal{F}f) = -i\mathcal{F}(x_j f)$]
[claim:Multiplication exchange identity]
For every $f \in \mathcal{S}(\mathbb{R}^n)$, every $j \in \{1, \ldots, n\}$, and every $\xi \in \mathbb{R}^n$,
\begin{align*}
\partial_{\xi_j}(\mathcal{F}f)(\xi) &= -i \, \mathcal{F}(x_j f)(\xi).
\end{align*}
[/claim]
[proof]
We differentiate $\mathcal{F}f(\xi) = \int_{\mathbb{R}^n} f(x) e^{-ix \cdot \xi} \, d\mathcal{L}^n(x)$ under the integral sign via the Leibniz integral rule. We verify: (i) the integrand is differentiable in $\xi_j$ for each fixed $x$, with derivative $-ix_j f(x) e^{-ix \cdot \xi}$; and (ii) there exists an $L^1$ dominating function independent of $\xi$.
For (ii), $|-ix_j f(x) e^{-ix \cdot \xi}| = |x_j f(x)|$. Choose $N > n + 1$. Then
\begin{align*}
|x_j f(x)| &\leq \frac{\sup_{z}|z_j f(z)|(1+|z|)^N}{(1+|x|)^N} = \frac{C_{N,j}}{(1+|x|)^N},
\end{align*}
where $C_{N,j}$ is finite because $f \in \mathcal{S}$ decays faster than any polynomial. Since $N > n$, the function $(1+|x|)^{-N}$ is integrable, and the bound is independent of $\xi$. The Leibniz rule gives
\begin{align*}
\partial_{\xi_j}(\mathcal{F}f)(\xi) &= \int_{\mathbb{R}^n} (-ix_j) f(x) \, e^{-ix \cdot \xi} \, d\mathcal{L}^n(x) = -i \, \mathcal{F}(x_j f)(\xi).
\end{align*}
[/proof]
[/step]
[step:Derive the iterated exchange identities for all multi-indices]
[claim:Iterated exchange identities]
For every $f \in \mathcal{S}(\mathbb{R}^n)$ and all multi-indices $\alpha, \beta \in \mathbb{N}_0^n$,
\begin{align*}
\xi^\alpha \, \partial_\xi^\beta(\mathcal{F}f)(\xi) &= (-1)^{|\beta|} \, i^{-|\alpha|-|\beta|} \, \mathcal{F}\bigl(\partial^\alpha(x^\beta f)\bigr)(\xi).
\end{align*}
[/claim]
[proof]
We establish two intermediate identities and combine them.
*Iterated differentiation.* Induction on $|\alpha|$. The base case $|\alpha| = 0$ is the identity. For the inductive step, suppose $\mathcal{F}(\partial^{\alpha'} f) = (i\xi)^{\alpha'} \mathcal{F}f$ holds for all $|\alpha'| \leq k$. Let $|\alpha| = k + 1$ and write $\alpha = \alpha' + e_j$. Then $\partial^\alpha f = \partial_{x_j}(\partial^{\alpha'} f)$. Since $\partial^{\alpha'} f \in \mathcal{S}$ (the [Schwartz space](/page/Schwartz%20Space) is closed under differentiation), the differentiation exchange identity applies:
\begin{align*}
\mathcal{F}(\partial^\alpha f)(\xi) &= i\xi_j \, \mathcal{F}(\partial^{\alpha'} f)(\xi) = i\xi_j \cdot (i\xi)^{\alpha'} \, \mathcal{F}f(\xi) = (i\xi)^\alpha \, \mathcal{F}f(\xi).
\end{align*}
*Iterated multiplication.* Induction on $|\beta|$. The base case is the identity. For the inductive step, suppose $\partial_\xi^{\beta'} (\mathcal{F}f) = (-i)^{|\beta'|} \mathcal{F}(x^{\beta'} f)$ holds for all $|\beta'| \leq k$. Let $|\beta| = k + 1$ and write $\beta = \beta' + e_j$. Since $x^{\beta'} f \in \mathcal{S}$ (the Schwartz space is closed under polynomial multiplication), the multiplication exchange identity applies:
\begin{align*}
\partial_\xi^\beta(\mathcal{F}f)(\xi) &= \partial_{\xi_j}\bigl((-i)^{|\beta'|} \mathcal{F}(x^{\beta'} f)\bigr)(\xi) = (-i)^{|\beta'|}(-i) \, \mathcal{F}(x_j \cdot x^{\beta'} f)(\xi) = (-i)^{|\beta|} \, \mathcal{F}(x^\beta f)(\xi).
\end{align*}
*Combining.* From the iterated differentiation identity, $\xi^\alpha \mathcal{F}g = i^{-|\alpha|} \mathcal{F}(\partial^\alpha g)$. Apply this with $g = x^\beta f$ and use the iterated multiplication identity:
\begin{align*}
\xi^\alpha \, \partial_\xi^\beta(\mathcal{F}f)(\xi) &= (-i)^{|\beta|} \, \xi^\alpha \, \mathcal{F}(x^\beta f)(\xi) = (-i)^{|\beta|} \cdot i^{-|\alpha|} \, \mathcal{F}\bigl(\partial^\alpha(x^\beta f)\bigr)(\xi).
\end{align*}
The prefactor $(-i)^{|\beta|} \cdot i^{-|\alpha|} = (-1)^{|\beta|} i^{-|\alpha|-|\beta|}$ has modulus $1$.
[/proof]
[/step]
[step:Prove $\mathcal{F}$ maps $\mathcal{S}$ into $\mathcal{S}$ and is continuous]
We must show $\|\mathcal{F}f\|_{\alpha,\beta} := \sup_{\xi} |\xi^\alpha \partial_\xi^\beta(\mathcal{F}f)(\xi)| < \infty$ for every $\alpha, \beta$.
By the iterated exchange identities, $|\xi^\alpha \, \partial_\xi^\beta(\mathcal{F}f)(\xi)| = |\mathcal{F}(\partial^\alpha(x^\beta f))(\xi)|$. The bound $\|\mathcal{F}h\|_{L^\infty} \leq \|h\|_{L^1}$ gives
\begin{align*}
\|\mathcal{F}f\|_{\alpha,\beta} &\leq \|\partial^\alpha(x^\beta f)\|_{L^1}.
\end{align*}
By the Leibniz rule, $\partial^\alpha(x^\beta f)$ is a finite sum of terms of the form $c_\mu \, x^{\beta-\mu} \partial^{\alpha-\mu} f$, each a product of a polynomial and a Schwartz function, hence itself in $\mathcal{S} \subset L^1$. Bounding $\|x^{\beta-\mu} \partial^{\alpha-\mu} f\|_{L^1}$ using $(1+|x|)^{-N}$ decay (with $N = n+1$) gives
\begin{align*}
\|\mathcal{F}f\|_{\alpha,\beta} \leq C_{\alpha,\beta} \sum_{(\gamma,\delta) \in S_{\alpha,\beta}} \|f\|_{\gamma,\delta}
\end{align*}
for a finite set $S_{\alpha,\beta}$ and constant $C_{\alpha,\beta}$ independent of $f$. Since a [linear map](/page/Linear%20Map) between Frechet spaces is continuous if and only if every output seminorm is bounded by finitely many input seminorms, $\mathcal{F}$ is continuous.
[/step]
[step:Prove $\mathcal{F}$ is a topological automorphism via the Fourier Inversion Theorem]
Define
\begin{align*}
\mathcal{F}^{-1}g(x) &= \frac{1}{(2\pi)^n} \int_{\mathbb{R}^n} g(\xi) \, e^{ix \cdot \xi} \, d\mathcal{L}^n(\xi).
\end{align*}
This operator satisfies analogous exchange identities with $i$ replaced by $-i$: [integration by parts](/theorems/210) in $\xi_j$ gives $\mathcal{F}^{-1}(\partial_{\xi_j} g)(x) = -ix_j \, \mathcal{F}^{-1}g(x)$, and differentiation under the integral gives $\partial_{x_j}(\mathcal{F}^{-1}g)(x) = i \, \mathcal{F}^{-1}(\xi_j g)(x)$. The same reasoning as in the preceding steps shows $\mathcal{F}^{-1}$ maps $\mathcal{S}(\mathbb{R}^n)$ continuously into itself.
Every $f \in \mathcal{S}$ satisfies $f \in L^1$ and $\mathcal{F}f \in L^1$ (from the mapping property $\mathcal{F}: \mathcal{S} \to \mathcal{S}$ and $\mathcal{S} \subset L^1$). The [Fourier Inversion Theorem](/theorems/246) then gives $\mathcal{F}^{-1}(\mathcal{F}f) = f$ pointwise. By the same argument applied to $\mathcal{F}^{-1}$, $\mathcal{F}(\mathcal{F}^{-1}g) = g$ for every $g \in \mathcal{S}$. Therefore $\mathcal{F}$ is a bijection $\mathcal{S} \to \mathcal{S}$ with continuous inverse, i.e., a topological automorphism.
[/step]
