Let $f\in L^{2,s}(\mathbb{R}^3)=\bigg\lbrace u\bigg|\int_{\mathbb{R}^3}dx\,|u(x)|^2(1+|x|^2)^s<\infty\bigg\rbrace$ ($s>\frac{1}{2}$) I have to calculate this limit $$\lim_{|x-y|\to 0}\int_{\mathbb{R}^3}\frac{e^{i\lambda|x-x'|}}{|x-x'|}f(x')dx'$$ with $\lambda>0$ and $y\in\mathbb{R}^3$. I've thought to use the theorem of dominated convergence, so I note that the integrand converges to $$\frac{e^{i\lambda|y-x'|}}{|y-x'|}f(x')$$ but I don't succed in finding a majoration of it with an integrable function not depending on $|x-y|$. Some ideas?
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$\begingroup$ You're asking about a limit of integrals of functions which aren't integrable. $\endgroup$– Nik WeaverMar 12, 2013 at 21:38
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$\begingroup$ $$\int_{\mathbb{R}^3}\bigg|\frac{e^{i\lambda|x-x'|}}{|x-x'|}f(x')\bigg|\leq \int_{\mathbb{R}^3}\frac{1}{|x-x'|(1+|x'|^2)^{\frac{s}{2}}}f(x')(1+|x'|^2)^{\frac{s}{2}}$$ Using the Schwartz inequality I have the integrability. $\endgroup$– SueMar 12, 2013 at 21:45
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$\begingroup$ I forgot to underline that $s>\frac{1}{2}$ $\endgroup$– SueMar 12, 2013 at 21:51
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$\begingroup$ Oh so should that be $u^2(x)$ in your definition of $L^{2,s}$? $\endgroup$– Nik WeaverMar 12, 2013 at 22:47
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$\begingroup$ Yes sorry...I just correct!!! $\endgroup$– SueMar 12, 2013 at 22:49
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2 Answers
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I think the limit of the integrals does equal the integral of the limit. First of all, outside of a small ball about $y$ we have dominated convergence. So the issue is whether $$\lim_{|x-y| \to 0} \int_{ball(y,\epsilon)} \frac{e^{i\lambda|x-x'|}}{|x-x'|} f(x')dx'$$ equals the same integral with $y$ in place of $x$. Thus we can ignore the exponential factor --- when $x$ gets close to $y$ it's essentially constant on this ball --- and the weight $(1 + |x|^2)^s$ which governs behavior at infinity but is irrelevant near $y$.
Fix $\epsilon > 0$ and let $h_x(x') = \frac{1}{|x-x'|}\cdot \chi_{ball(y,\epsilon)}$. I claim that $h_x \to h_y$ weakly in $L^2$ as $x \to y$. This will imply that $\int h_xf \to \int h_yf$, which is all you need for the reasons I just explained. Well, the functions $h_x$ are uniformly bounded in $L^2$, so weak convergence will follow from weak convergence when integrated against a dense subset of $L^2$. But $\int h_xg \to \int h_yg$ for any $L^2$ function $g$ which is $0$ on a neighborhood of $y$, by dominated convergence, and such functions $g$ are dense, so we are done.
(In fact weak convergence plus the fact that $\|h_x\| \to \|h_y\|$ implies that $h_x \to h_y$ in norm, which presumably you could show by direct calculation.)
answered Mar 13, 2013 at 4:14
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I had an idea. I know that the function $$F(x)=\int_{\mathbb{R}^3}\frac{e^{i\lambda|x-x'|}}{|x-x'|}f(x')dx'\in H^{2,-s}(\mathbb{R}^3)$$ when $f\in L^{2,s}(\mathbb{R}^3) $for $s>\frac{1}{2}$ (I've proved this fact); moreover the space $H^{2,-s}(\mathbb{R}^3)$ is conteined in $C(\mathbb{R}^3)$ and so the considered limit exists.