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I consider, for $s>\frac{1}{2}$, the space $L^{2,-s}(\mathbb{R}^3)=\bigg\lbrace{f: \int_{\mathbb{R}^3}|f(x)|^2(1+|x|^2)^{-s}<\infty\bigg\rbrace}$ and I have to show that the function $$f(x)=\frac{e^{i\sqrt{\lambda+i\varepsilon}|x|}}{|x|}$$ converges in the norm of the considered space to$$g(x)\frac{e^{i\sqrt{\lambda}|x|}}{|x|}$$when $\varepsilon\to 0$ where $\lambda>0$ and from complex analysis $\sqrt{\lambda+i\varepsilon}=\sqrt{\lambda_\varepsilon}+i\frac{\varepsilon}{2\sqrt{\lambda_\varepsilon}}$ with $\lambda_\varepsilon=\frac{\sqrt{\lambda^2+\varepsilon^2}+\lambda}{2}$, so $\lambda_\varepsilon\to\lambda$ and $$e^{i\sqrt{\lambda+i\varepsilon}}=e^{i\sqrt{\lambda_\varepsilon}}e^{-\frac{\varepsilon}{2\sqrt{\lambda_\varepsilon}}}\to e^{i\sqrt{\lambda}}$$. Then I consider the integral $$\int_{\mathbb{R}^3}\frac{|e^{i\sqrt{\lambda+i\varepsilon}}-e^{i\sqrt{\lambda}}|^2}{|x|^2}\frac{1}{(1+|x|^2)^s}dx $$ The integrand tend to $0$ when $\varepsilon\to 0$ almost everywhere and it is majorized by $$\frac{1}{|x|^2(1+|x|^2)^s}$$ integrable. So by the theorem of dominated convergence I have the thesis. Is it right?

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Yes, it is correct, as $$\int_{\{a\leqslant |x|\leqslant b\}}\frac 1{|x|^2(1+|x|^2)^s}dx=C\int_a^b\frac{r^{3-1}}{r^2(1+r^2)^s}ds=C\int_a^b\frac{ds}{(1+r^2)^s},$$ where $C$ is an universal constant.

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