# Definition of Schwartz functions

I have two questions regarding the definition of Schwartz functions. So if we have the Schwartz space $$\mathcal{S}(\mathbb{R}^n)=\left\{ \phi \in C^\infty(\mathbb{R}^n) \,\Big|\, \forall \alpha, \beta \in \mathbb{N}_0^n: \; \sup_{x\in\mathbb{R}^n} |x^\alpha D^\beta \phi(x) | <\infty\; \right\}$$

1. Is it correct that the point why we have an $x^\alpha$ in front of the differential operator is to have a derivative that goes faster to $0$ than any polynomial grows?

2. Also does this mean that for an $\alpha$ there exist a $\beta$ such that for $\beta\leq$ derivatives the supremum is finite or must for any $\alpha$ the derivative always be finite, no matter what $\beta$ is?

• For every $\alpha,\beta \in\mathbb{N}$, $\|x^\alpha D^\beta \phi\|_{\infty} < \infty$. The Schwartz space is the natural space to ensure $f$ and its Fourier transform $\hat{f}$ and all their derivatives are smooth and integrable. Oct 1, 2017 at 19:51
• With respect to (i): Yes, because it implies $\displaystyle\lim_{|x|\to\infty}\|x\|^kD^\beta f(x)=0$ for all $k\in \mathbb N$ and all $\beta\in\mathbb{N}^n$. Here there is a proof. Oct 13, 2017 at 4:36

The idea is that $\phi$ lies in $\mathcal{S}(\mathbb R^n)$ provided $\phi$ and all of its partial derivatives (each $D^{\beta}\phi$) are 'rapidly decreasing,' in the sense that they decrease faster than every $1/x^{\alpha}.$
1. Yes, we want all of the partial derivatives to vanish sufficiently quickly. As an example, this prevents things like $f(x) = e^{-x}\sin(e^x),$ whose first derivative does not vanish as $x \rightarrow \infty.$
2. It means this holds for any choice of $\alpha$ and $\beta,$ so they do not depend on each other.