I'm confused about Schwartz Space. In particular I'm finding it difficult to show a function is / isn't in Schwartz space.

Definition:The Schwartz Class $\mathcal{S}(\mathbb{R}^{n})$ is the set of all functions $$f:\mathbb{R}^{n} \rightarrow \mathbb{C}$$

which satisfy $\rho_{\alpha,\beta}(f):= \sup_{x \in \mathbb{R}^{n}}|x^{\alpha}\partial^{\beta}f|< \infty$ for all multi-indices $ \alpha, \beta.$

I know that this means that the elements of the space are those continuously differentiable functions which, along with all their derivatives vanish at infinity faster than any inverse power of $x.$

Question: Why is it an inverse power of $x$ I don't see how this follows from the definition.

Question: Why is $|x|^{2} \notin \mathcal{S}(\mathbb{R}^{n})?$ I struggle to apply the definition of Schwartz space.

I know that I obviously need to show that it doesn't satisfy the definition but It's quite confusing to me.

Any help appreciated.


  • $\begingroup$ For a start, this inequality must hold for $\alpha=\beta=0$. Then is states that $\sup|f(x)|<\infty$, that is $f$ is bounded. Is $x\mapsto|x|^2$ bounded? $\endgroup$ Jul 9, 2018 at 18:24
  • $\begingroup$ @LordSharktheUnknown ah that makes sense! Thanks. $\endgroup$
    Jul 9, 2018 at 18:28
  • $\begingroup$ If $\sup ||x|^{N} f| = C < \infty$, then $|f(x)| \leq C |x|^{-N}$. $\endgroup$
    – fourierwho
    Jul 9, 2018 at 20:08
  • $\begingroup$ I don't see how this follows from the definition. In fact, this is equivalent to the definition. See this. $\endgroup$
    – Pedro
    Oct 29, 2018 at 7:26


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