For simplicity, consider in $\mathbb{R}^3$, and the Fourier transform of the following function $$f=(x_1+x_2+x_3)(1+|x|^2+x_1^2(x_2^2+x_3^2)+x_2^2x_3^2)^{-t+is},~~ \frac12<t<1,~~s\in \mathbb{R}.$$

Note that the function is not in $L^1$, in fact $f$ behaves like $|x|^{-2t+1}$ at $\infty$. So the Fourier transform is taken in the sense of tempered distribution.

Note also that the function is not radial symmetric, however $f$ is symmetric in the sense that if any of the variables are interchanged, one obtains the same function. I'm interested in the quantitative behavior of $\hat{f}$ near the origin. Especially, for what values of $\sigma_i\ge 0$,$i=1,2,3,$ such that $$ \hat{f}(\xi)\leq C|\xi_1|^{-\sigma_1}|\xi_2|^{-\sigma_2}|\xi_3|^{-\sigma_3},~~|\xi|<1. $$ The following is an "unsuccessful" approach. First let's estimate the Fourier transform of $f(x)=1/{(1+|x|^2+x_1^2(x_2^2+x_3^2)+x_2^2x_3^2)^{t+is}}$. Then one can write $$ f(x)=\frac{1}{(B^2+A^2x_1^2)^{t+is}}, $$ where$ A^2(x_2, x_3)=1+x_2^2+x_3^2$, $B^2(x_2, x_3)=A^2+x_2^2 x_3^2$. Recall that \begin{align}\label{equ4.3} \mathcal{F}^{-1}(\frac{1}{(1+|x|^2)^z})(\xi)=\pi^{-\frac{n}{2}}2^{-(\frac{n}{2}+z-1)}\frac{1}{\Gamma(z)}|\xi|^{z-\frac{n}{2}}K_{\frac{n}{2}-z}(|\xi|), \end{align} where $K_v$ denotes the modified Bessel function of the second kind. $K_v$ satisfies \begin{align}\label{equ4.4} \frac{d}{dz}(z^{-v}K_v(z))=z^{-v}K_{v+1}(z), ~~~ \forall v\in \mathbb{C}. \end{align}

Combine the above two formula and by scaling, one has \begin{align}\label{equ4.6} I(\xi)&\triangleq \int_{\mathbb{R}^3}{e^{ix\cdot\xi}f(x)dx}\nonumber\\ &=\pi^{-\frac{1}{2}}2^{\frac12-z}\frac{1}{\Gamma(z)} \int_{\mathbb{R}^{2}}{e^{ix'\cdot\xi'}B^{1-2z}A^{-1}(\frac{B}{A}|\xi_1|)^{z-\frac12}K_{\frac12-z}(\frac{B}{A}|\xi_1|)dx'}\nonumber\\ &=\pi^{-\frac{1}{2}}2^{\frac12-z}\frac{|\xi_1|^{z-\frac12}}{\Gamma(z)} \int_{\mathbb{R}^{2}}{e^{ix'\cdot\xi'}A^{-2z}(x')(\frac{B}{A})^{\frac12-z}K_{\frac12-z}(\frac{B}{A}|\xi_1|)}dx', \end{align} In order to take out the factor $\frac{B}{A}$ from $K_{\frac12-z}(\frac{B}{A}|\xi_1|)$, we can use the following multiplication theorem of Bessel function \begin{align}\label{equ4.7} \lambda^{\nu}K_{\nu}(\lambda z)=\sum_{l=0}^{\infty}\frac{(-1)^l}{l!}(\frac{(\lambda^2-1)z}{2})^lK_{\nu-l}(z),~~~\lambda>0. \end{align} However, the estimate is fine with respect to $\xi_1$, but it get trouble with $\xi'$ since now it becomes \begin{align} I(\xi)&=c_z\sum_{l=0}^{\infty}\frac{(-1)^l}{2^ll!}(\int_{\mathbb{R}^{n-1}}{e^{ix'\cdot\xi'}\frac{x_2^{2l}x_3^{2l}}{A^{2(z+l)}}} ~~~dx')|\xi_1|^{z-\frac12+l}K_{\frac12-z-l}(|\xi_1|) \end{align} It seems that the integral involved in the series above is too singular that doesn't imply the convergence. I'm not sure how to overcome this issue.

Any reference or comment about this topic is welcome!

  • $\begingroup$ Do you require $P(x) \geq 0$ for all $x$? In general the function $f$ is not defined on some $x$. Think about $f(x) = (1 - x^2)^{-1}$. $\endgroup$ – WhatsUp May 30 '16 at 13:18
  • $\begingroup$ P=0, f=1, $\hat f=\delta$ $\endgroup$ – Piero D'Ancona May 30 '16 at 15:41
  • $\begingroup$ @WhatsUp, I did require $P\ge 0$. Fixed it. Thanks. $\endgroup$ – Tomas May 31 '16 at 6:28
  • $\begingroup$ @PieroD'Ancona, Thanks, I'm more interested in the case that $P\ne 0$. $\endgroup$ – Tomas May 31 '16 at 6:30
  • $\begingroup$ What I'm saying is that you always have singularities. This is just the simplest case $\endgroup$ – Piero D'Ancona May 31 '16 at 10:39

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