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Consider the function $$ f(x_1,x_2)=|x_1x_2|^{-\alpha/2}\int_{\mathbb{R}} \frac{e^{it(x_1+u)}-1}{i(x_1+u)} \frac{e^{it(x_2-u)}-1}{i(x_2-u)} |u|^{-\beta}du. $$ It is known that $f(x_1,x_2)\in L^2(\mathbb{R^2})$ if $$ -1/2<\alpha<1, \quad-1/2<\beta<1,\quad \alpha+\beta>1/2. $$ The question is to find an expression for the $L^2$-defined Fourier transform $$ \hat{f}(w_1,w_2)=\int_{\mathbb{R^2}} f(x_1,x_2)e^{-ix_1w_1-ix_2w_2} dx_1dx_2. $$

When $\alpha$ and $\beta$ are nonnegative, using the facts (1) $\int_0^t e^{isx}ds=\frac{e^{itx}-1}{ix}$, (2) $\int_{\mathbb{R}}e^{ixw} |x|^{a} dx=c|w|^{-a-1}$ for $a\in(-1,0)$ and (3) $\int_{\mathbb{R}}e^{ixw} dx=2\pi\delta(w)$, I have guessed (probably can be justified by some regularization argument) that $\hat{f}(x_1,x_2)$ is up to some constant the following

  1. $\alpha>0$, $\beta>0$: $$ \int_0^t\int_0^t |u-v|^{\beta-1}|u-x_1|^{\alpha/2-1} |v-x_2|^{\alpha/2-1} du dv. $$

  2. $\alpha>1/2$, $\beta=0$: $$ \int_0^t |u-x_1|^{\alpha/2-1}|u-x_2|^{\alpha/2-1} du. $$

  3. $\alpha=0$, $\beta>1/2$: $$ |x_1-x_2|^{\beta-1} I\{0<x_1,x_2<t\}. $$

But what about the cases: $\alpha>1/2$, $\beta<0$ and $\alpha<0$, $\beta>1/2$? I don't even know how to heuristically guess the answer.

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  • $\begingroup$ try analytic continuation in your parameters alpha, beta, etc. If you have a formula in the region where all is well and integrals converge, then what you are looking for will be given by an explicit analytic continuation. Before you embark on this you should study the book by Gel'fand and Shilov, Generalized functions Vol 1. $\endgroup$ – Abdelmalek Abdesselam Apr 13 '16 at 13:52

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