Notes: Consider $f,f',f''\in {L_1}_({\mathbb{R}})$
The Parseval equality verifies:
$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{-it\xi}dt$
$\hat{f}(\xi)=O(|\xi|^{-2}),\:\:\:|\xi|\to\infty$
Then integrating by parts:
$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f(t)e^{-it\xi}dt=\frac{e^{-it\xi}}{-i\xi}f(t)|_{-\infty}^{\infty}+\frac{1}{i\xi\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f'(t)e^{-it\xi}dt=\frac{1}{i\xi\sqrt{2\pi}}\int_\limits{-\infty}^{\infty}f'(t)e^{-it\xi}dt$, since $\lim_{t\to\infty}=0\:\:,f\in {L_1}_{(\mathbb{R})}$
Question:
1) How does the author knows $\lim_{t\to\infty}=0$? Is it a property of ${L_1}_{(\mathbb{R})}$ space?
2) $\hat{f}(\xi)=O(|\xi|^{-2}),\:\:\:|\xi|\to\infty$ means that $\hat{f}(\xi)$ has the same asymptotic behaviour as $\frac{1}{{(\xi)}^2}$ as $|\xi|\to\infty$? Right?
Thanks in advance!
$f(x)=f(0)+\int_0^{x} f'(t)dt$. Since $f' \in L^{1}$ we see that $\lim_{x \to \infty} f(x)$ exists. If this limit $c$ is not zero then it is very easy to see that $f$ is not integrable. Hence $f(x) \to 0$ as $x \to \infty$. Similarly $f(x) \to 0$ as $x \to -\infty$. $\hat {f} (\xi)=O(|\xi|)^{-2}$ means that $|\xi|^{2}\hat {f} (\xi)$ is bounded. It is possible that this bounded functions tends to 0. So $\hat {f}$ may tend to 0 much faster than $\frac 1 {|\xi|^{2}}$.