I have the function $$f(x)=\frac{e^{izxy}}{4\pixy}$$ with $y\in\mathbb{R}^3$ and $\Im z>0$. Let $s>\frac{1}{2}$. Clearly it is not in $H^{2,s}(\mathbb{R}^3)$ for the singularity of order $xy^{1}$. Now I consider the function $$F(x)=\frac{e^{izxy}1}{4\pixy}$$ so that $$\lim_{xy\to 0} F(x)=iz$$. Does $F(x)\in H^{2,s}(\mathbb{R}^3)$? According to me yes because $$\Delta_xF(x)=zf(z)\delta_y+\delta_y=zf(x)$$ where $\delta_y$ is the Dirac distribution in $y$. (I've used the relation ($\Deltaz)f(x)=\delta_y$)
