Let $\xi \in \mathbb{R}$ and $f$ belong to the Schwartz space on $\mathbb{R}$. I know, that Schwartz functions are rapidly decreasing, but I am not very familiar with them. My question is, why is it true that $$\lim_{x \to \infty} f(x)(e^{-2\pi i x \xi} - e^{2\pi i x \xi}) = 0$$ I mean we know that for any $k \in \mathbb{N}$ $$\sup_\mathbb{x \in R} |x|^k |f(x)| < \infty$$
Proving expression involving Schwartz function tends to zero
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schwartz-space
2 Answers
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Use a quick majoration :
$$\left| f(x) (e^{-2\pi i x \xi} - e^{2\pi i x \xi} ) \right| = \left| f(x) \right| \left| (e^{-2\pi i x \xi} - e^{2\pi i x \xi} ) \right| $$
$$ \leq |f(x)| \left( \left| e^{-2\pi i x \xi} \right| + \left| e^{2\pi i x \xi} \right| \right) $$
$$ \leq |f(x)| \left( 1+1 \right) $$
And, by definition, $|f(x)|$ converge to 0
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$|f(x)(e^{-2i\pi x \xi} - e^{2i \pi x \xi})|\le 2|f(x)| \to 0$