Is this function in the Sobolev space $H^{2,-s}(\mathbb{R}^3)$?

Question

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 ($-\Delta-z)f(x)=\delta_y$)