Let us suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is a smooth function. Furthermore, for every $\alpha$ multi-index, there exists $C_\alpha > 0$ such that $$ |D^\alpha f(\xi)| \leq \frac{C_\alpha}{(1+|\xi|)^{|\alpha|}}. $$ Does it follow that, for every $\alpha$, there exists $C'_\alpha > 0$ such that $$ |D^\alpha (\mathcal{F}^{-1}(f))(x)| \leq \frac{C'_\alpha}{|x|^{n+|\alpha|}} $$ where $\mathcal{F}^{-1}$ is the inverse Fourier transform (which exists since $f \in \mathcal{S}'$)? I tried to do it using the definition, but it is really messed up because $\mathcal{F}^{-1}$ is in general in $\mathcal{S}'$. For instance, if $f$ is a constant function, then its inverse transform is a dirac $\delta$, then I should give it a pointwise meaning, and I don't know when this is possible. Any help would be really appreciated.
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1A related problem. Are you considering Schwartz class? – Mhenni Benghorbal May 09 '13 at 13:48
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@Mhenni Benghorbal: $f$ is smooth and bounded, so $\mathcal{F}^{-1}(f)$ is in the space of tempered distributions $\mathcal{S}'$. I cannot say anything more though, since it might be something like a $\delta$. Random thoughts: if $f$ is compactly supported, $\mathcal{F}^{-1}$ is automatically smooth, but then bounding the derivatives of $f$ is no longer necessary. Also, I would like to understand whether if there is an approximation approach: i.e. you approximate something in $\mathcal{S}'$ by smooth (also $L^1$'d be enough) functions, and then take the limit in $\mathcal{S}'$-norm. – Second May 09 '13 at 13:59