Schwartz space

Remarks

Example: the Gaussian $f(x) = e^{-\abs{x}^2}$ is a Schwartz function. Each derivative $\partial^\beta f$ is a polynomial times $e^{-\abs{x}^2}$, and a polynomial times $e^{-\abs{x}^2}$ is bounded on $\R^n$. Every test function on $\R^n$ is a Schwartz function. Non-example: $f(x) = 1/(1 + x^2)$ on $\R$ is smooth, but $p_{2, 0}(f) = \sup_x x^2/(1 + x^2) = 1$ while $p_{3, 0}(f) = \sup_x \abs{x}^3/(1 + x^2) = \infty$. The space is closed under differentiation, under multiplication by polynomials and under the Fourier transform, which is a bijection of $\mathcal{S}(\R^n)$ onto itself. Equivalently, $f$ is a Schwartz function if and only if $(1 + \abs{x})^N \abs{\partial^\beta f(x)}$ is bounded for every $N \in \N$ and every $\beta$.