Lemma 2.9.2A976F203FD66

Proof. Let $f\in \mathcal S$. We first prove the second rule in (2.9.3). By the mean value theorem, the difference quotients of $\xi \mapsto \mathrm e^{-2\pi \mathrm i x\cdot \xi }$ in the direction of $\xi _ j$ are bounded by $2\pi |x_ j|$, and $x_ jf$ is integrable. Hence dominated convergence allows us to differentiate under the integral sign: $\partial _{\xi _ j}\widehat f$ exists and is the transform of $-2\pi \mathrm i x_ jf$, which is continuous. Since $x_ jf\in \mathcal S$, induction shows that $\widehat f$ is smooth and that $\partial ^\beta \widehat f$ is the transform of $(-2\pi \mathrm i x)^\beta f$.

For the first rule, Fubini's theorem allows us to integrate in $x_ j$ before the other coordinates. For almost every value of the other coordinates, $f$ and $\partial _ jf$ are integrable on the corresponding line, and an integration by parts over $[-a,a]$ followed by $a\to \infty $ gives

because the boundary terms are bounded by $|f(x)|$ at $x_ j=\pm a$, and $f(x)\to 0$ as $|x|\to \infty $. Integrating over the other coordinates proves the first rule. Combining the two rules, we find that $\xi ^\alpha \partial ^\beta \widehat f$ is a constant multiple of the transform of $\partial ^\alpha (x^\beta f)$. This function belongs to $\mathcal S$ and is therefore integrable, so $\xi ^\alpha \partial ^\beta \widehat f$ is bounded by a multiple of its $L^1$ norm. Hence $\widehat f\in \mathcal S$.

For the Gaussian, consider first $d=1$, put $g(x)=\mathrm e^{-\pi x^2}$, which belongs to $\mathcal S(\mathbb R)$, and set $F=\widehat g$. Since $-2\pi \mathrm i xg=\mathrm i g'$, the two rules give $F'=\mathrm i\, \widehat{g'}=\mathrm i(2\pi \mathrm i\xi )F=-2\pi \xi F$, so $F(\xi )\mathrm e^{\pi \xi ^2}$ is constant. Moreover $F(0)=1$: squaring the positive integral $F(0)$ and using polar coordinates in the plane, where $s_1=2\pi $, gives $F(0)^2=2\pi \int _0^\infty r\mathrm e^{-\pi r^2}\, \mathrm d r=1$. Hence $F(\xi )=\mathrm e^{-\pi \xi ^2}$. In dimension $d$, Fubini's theorem writes the transform of $\mathrm e^{-\pi |x|^2}$ as a product of one-dimensional transforms, which proves the formula for $t=1$. For general $t$, the linear substitution $x=y/\sqrt t$ gives