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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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    @OlivierBégassat I see. Thank you.2012-11-09

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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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    Even $|f(x)| \in o(|x|^{-a})$ (small Oh) for $|x| \to \infty$.2016-08-17