Elimination of a singularity

Question

Let $z=a+ib$ with $b>0$; the function $$f(x)=\frac{e^{iz|x|}}{|x|}$$ is in $L^2(\mathbb{R}^3)$; in fact $$\int_{\mathbb{R}^3}|f(x)|^2dx=4\pi\int_{0}^{+\infty}\frac{e^{-2b\,r}}{r^2}r^2dr=\frac{2\pi}{b}<\infty$$ For the singularity in $x=0$ $f(x)\notin H^2(\mathbb{R}^3)$; now I take $$g(x)=\frac{e^{iz|x|}-1}{|x|}$$. So I've eliminated the singularity and now the function is in $H^{2,-s}$, with $s>\frac{1}{2}$, isn't it?

Answer 1

I'm not sure exactly what the question is. But you have changed the function into a different function that doesn't have the singularity. But the new function doesn't agree on the values with the old function on all the other points (it's an entirely new function) so it wasn't a case of a removable singularity if that's what you were asking.