# Asymptotic behavior of Fourier transform

Consider the function $f:\mathbb{R}^3\rightarrow \mathbb{R}$, $f(x) = |x|^{-1}$. It is locally integrable, and its distributional Fourier transform is $F(f)(k) = g(k) = 4\pi/|k|^2$. Intuitively, the long-range $|k|^{-2}$ decay of $g(k)$ comes from the singularity at $x=0$ in $f$.

Is this true? I.e., if $\chi$ is the characteristic function of an open, bounded set containing $0$, and consider $h = F(\chi f)$. Is it true that $h = O(|k|^{-2})$ as $|k|\rightarrow +\infty$?

At least, if the open bounded set in question is a ball of radius $R$, a direct computation gives $$h(k) \propto \frac{sin^2(|k|R/2)}{|k|^2} = O(|k|^{-2})$$

In this case, the hypothesis is verified, since $R$ was arbitrary.

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Asking that $\chi$ be the characteristic function of an arbitrary open set creates difficulties, related to the fact that $F(\chi)$ decays slowly itself, that do not relate to the main phenomenon here. If instead we ask that $\chi=1$ in a neighborhood of zero, and is $C^\infty$ with compact support, then $h(k)$ is proportional to the convolution $$h(k) \propto \int_{\mathbb R^3} \frac{F(\chi)(z)}{|k-z|^2}\,d^3z.$$ Because $F(\chi)$ is smooth and decays rapidly now, and $1/|k-z|^2$ is in $L^1_{\rm loc}$ and is bounded by, e.g., $4/|k|^2$ for $|k-z|\ge |k|/2$, one can perform the estimate $$|h(k)|\le C \left( \frac4{|k|^2}\|F(\chi)\|_{L^1} + I(|k|) \sup_{|z-k|<|k|/2} |F(\chi)(z)|\right) = O(|k|^{-2}),$$ where $$I(k) = \int_{|z-k|<|k|/2} |z-k|^{-2}d^3z \propto |k|.$$ Note $$\sup_{|z-k|<|k|/2} |F(\chi)(z)| \le \sup_{|z|\ge |k|/2}|F(\chi)(z)|$$ and this decays rapidly with $|k|$.
Thanks! I am also very interested in the case where $\chi$ is the characteristic function of, say, a box. This is because I am working with Fourier series of $1/r$ in the box: $F(\chi f)$ is analytic (Paley-Wiener theorem), and the Fourier coefficients are in fact $\beta F(\chi f)(\alpha k)$ for some constants $\alpha,beta$, depending on box size and normalization. –  Nemis L. Oct 1 '13 at 12:54