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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – reuns
    Oct 1, 2017 at 19:51
  • $\begingroup$ 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. $\endgroup$
    – Pedro
    Oct 13, 2017 at 4:36

1 Answer 1


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 $1/x^{\alpha}.$

So in particular,

  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.

  • $\begingroup$ Thank you very much, this answered all of my questions. $\endgroup$
    – WaldoRozir
    Oct 1, 2017 at 20:09

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