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Let $f$ be a complex function of class $C^{N}$ on $\mathbb R$, continuous and integrable on $ \mathbb R $. I would like to know why this function can check the following inequality: $$|\mathcal{F}(f)(x)|\leq C (1+|x|^{N})^{-1},$$ for a certain constant $ C>0 $, $ \forall x \in \mathbb R$, where $\mathcal{F}(f)$ is the Fourier transformation of $f$.

In fact, we have: $| (i\xi)^N \mathcal{F}(f)(\xi) | =|\mathcal{F}(f^{(N)})(\xi)| = |\int_{\mathbb R} f^{(N)}(y) e^{- i y \xi} \, dy| \leq \int_{\mathbb R} | f^{(N)}(y) | \, dy$, then $\,\, | \mathcal{F}(f)(\xi) | \leq \frac{C}{|\xi|^N}\,\,$ under the assumption of integrability of the derivatives of $ f $ (which we do not have in here). And so $ | \mathcal{F}(f)(\xi) | \leq \frac{C}{|\xi|^N} $ it's not $|\mathcal{F}(f)(\xi)|\leq \frac{C}{ 1+|\xi|^{N}}$ !!

Thank you in advance.

Z. Alfata
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    cIf you perform the same calculation with $(1+(i\xi)^N)\mathcal{F}(f),$ you get $\frac{C}{1+|\xi|^N}$ with $C=|f|{L^1}+|f^{(N)}|{L^1}$ as opposed to your constant, which is just $|f^{(N)}|_{L^1}$. You're correct that you probably need some integrability assumption on $f^{(N)}$. – WoolierThanThou Mar 21 '20 at 22:50

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Proof assuming that $f$ and $f^{(N)}$ are integrable: If $|\xi| \geq 1$ then $(1+|\xi|)^{N} \leq C|\xi|)^{N}$ for some $C_1$. If $|\xi| <1$ then $|\mathcal F f (\xi)| \leq \|f\|_1 \leq \frac {C_2} {(1+|\xi|)^{n} }$ for some $C_2$. Hence the inequality holds with a suitable $C$ in both cases.

Kavi Rama Murthy
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