Fourier transform on Schwartz space

Remarks

By the rules for differentiation and multiplication, $\xi^\alpha \partial^\beta \hat f$ is, up to constants, the transform of $\partial^\alpha (x^\beta f)$. That function is in $\mathcal{S}(\R^n) \subset L^1(\R^n)$, so its transform is bounded, which shows $\hat f \in \mathcal{S}(\R^n)$. The same estimates, with a factor $(1 + \abs{x})^{-n-1}$ to control the integral, give continuity. Bijectivity is the inversion formula $\mathcal{F}^{-1} \mathcal{F} f = f$, proved by pairing with a Gaussian of shrinking width. Also $\mathcal{F}^4 f = f$, since $\mathcal{F}^2 f(x) = f(-x)$.

Example: the Gaussian $e^{-\pi \abs{x}^2}$ is a Schwartz function and is fixed by $\mathcal{F}$.

On $\mathcal{S}(\R^n)$ the transform also preserves the $L^2$ norm, $\int \abs{\hat f}^2 = \int \abs{f}^2$. Since $\mathcal{S}(\R^n)$ is dense in $L^2(\R^n)$, this is how the transform is extended to a unitary map of $L^2(\R^n)$. By duality it also extends to tempered distributions. The space $L^1(\R^n)$ has no such property: the transform of an integrable function need not be integrable.

Going deeper

For a student

This is the cleanest setting for Fourier analysis on $\R^n$: all formulas (inversion, Plancherel, the convolution theorem, differentiation rules) hold on $\mathcal{S}(\R^n)$ without any conditions. Results on $L^p$ spaces are then obtained by density, and results on tempered distributions by duality.