Assume there is a distribution in two variables $\mathcal{W}\in\mathcal{S}'(\mathbb{R}^2)$ with Fourier transform $\hat{\mathcal{W}}(\alpha,\beta)\equiv \int_{-\infty}^\infty e^{i(\alpha x+\beta y)} \mathcal{W}(x,y) \, dx \, dy$. Now assume a set of variables in $\mathbb{R}^4$ given by $\lbrace \mu_1,\mu_2,\nu_1,\nu_2\rbrace$. Does the distribution $\mathcal{W}$ exist, given that its Fourier transform is solution of the following equation? $$\frac{\hat{\mathcal{W}}(\nu_1,\mu_1) \hat{\mathcal{W}}(\nu_2,\mu_2)}{\hat{\mathcal{W}}(\nu_1+\nu_2,\mu_1+\mu_2)}=\frac{\sin(\nu_1\mu_2-\mu_1\nu_2)}{\nu_1\mu_2-\mu_1\nu_2}.$$
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2 Answers
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It seems your equation implies $\hat{W}(0,\mu_1)\hat{W}(0,\mu_2)=\hat{W}(0,\mu_1+\mu_2)$, which would mean that $\hat{W}(0,\mu)= e^{c\mu}$. This is not the Fourier transform of a valid marginal distribution of $y$.
answered Apr 11 at 8:16
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Denote $f(t) = \frac{\sin t}{t}$. Then your identity is that for any $u,v\in\mathbb{R}^2$, $$ \frac{\hat W(u) \hat W(v)}{\hat W(u+v)} = f(u\times v) $$ Therefore, for any $u,v,w\in\mathbb{R}^2$, $$ \frac{\hat W(u) \hat W(v) \hat W(w)}{\hat W(u+v+w)} = \frac{\hat W(u) \hat W(v)}{\hat W(u+v)} \cdot \frac{\hat W(u+v) \hat W(w)}{\hat W(u+v+w)} = f(u\times v) f((u+v)\times w) $$ and $$ \frac{\hat W(u) \hat W(v) \hat W(w)}{\hat W(u+v+w)} = \frac{\hat W(u) \hat W(v+w)}{\hat W(u+v+w)} \cdot \frac{\hat W(v) \hat W(w)}{\hat W(v+w)} = f((u+ v)\times w) f(v\times w) $$ Therefore we get $$ f(u\times v)f((u+v)\times w) = f(v\times w)f((u+v)\times w) $$ Note that for any $\alpha,\beta,\gamma\in\mathbb{R}$ there are $u,v,w\in\mathbb{R}^2$ such that $u\times v=\alpha$, $v\times w = \beta$, $u\times w = \gamma$, therefore it follows that for any $\alpha,\beta,\gamma\in\mathbb R$ we have $$ f(\alpha)f(\beta+\gamma) = f(\beta)f(\alpha+\gamma) $$ and it is easy to see that $f(t)=\frac{\sin t}{t}$ does not satisfy the last identity
