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I have the function $$f(x)=\frac{e^{iz|x-y|}}{4\pi|x-y|}$$ 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 $|x-y|^{-1}$. Now I consider the function $$F(x)=\frac{e^{iz|x-y|}-1}{4\pi|x-y|}$$ so that $$\lim_{|x-y|\to 0} F(x)=iz$$. 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 ($-\Delta-z)f(x)=\delta_y$)

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Non-mathematical suggestion: you have several unregistered accounts, I recommend registering one of them and then getting a moderator to merge the rest into that one. –  Yemon Choi Mar 15 '13 at 9:12
Another non-mathematical suggestion: It seems that you have split a certain somewhat substantial mathematical problem into tiny pieces and keep on posting asking for solutions of these sub-problems. I doubt this is the most efficient way of solving your problem, and not necessarily the fairest way of dealing with other MO-users, either. Besides: some of your questions really are terribly technical and not really of large interest. –  Delio Mugnolo Mar 16 '13 at 4:43
Delio Mugnolo: We welcome technical questions at MathOverflow. However, I note that solutions tend to take time before appearing. –  S. Carnahan Mar 16 '13 at 5:45

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