Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f$ be in the Shwartz space $\mathcal S(\Bbb R)$.
Why does the $\mathcal S$-norm $$ \|f\|_{a,b}=\sup_{x \in \mathbb R} |x^af^{(b)}(x)|, \text{ for } a,b \in \Bbb Z_+, $$ implies that $f$ vanish at infinity?

The norm gives a bound on $$ \lim_{|x| \to \infty} f(x) $$ but that doesn't show the function vanish.

This post raised this question.

share|cite|improve this question
Well, by definition $xf(x)$ is bounded, so ... – Olivier Bégassat Nov 9 '12 at 15:05
@OlivierBégassat I see. Thank you. – Nicolas Essis-Breton Nov 9 '12 at 15:08
up vote 2 down vote accepted

Let $C_a:=\sup_{x\in\Bbb R}|x^af(x)|$. As $f$ is in the Schwarz space, $C_a\in\Bbb R$. So we have for all $x\in\Bbb R$ and $a\in\Bbb Z_+$, $$|f(x)|\leqslant \frac{C_0+C_a}{1+|x|^a}.$$ In particular, $|f(x)|=O(|x|^{-a})$ for all positive integer $a$, at $ \pm\infty$ (and a constant depending on $a$), so the convergence is faster than polynomial.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.