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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.

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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

1 Answer 1

up vote 1 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.

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