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Let $f\in\mathscr{S}(\mathbb{R}^3)$ (the Schwartz space), $k>0$ and consider the integral $$\varphi(k)=\int_{\mathbb{R}^3}\frac{e^{ik|x|}}{|x|}f(x)dx$$Is the estimation $|\varphi(k)|\leq\frac{M}{k^2}$ right?

I can write it as $$\int_{\mathbb{R}^3}\frac{e^{ik|x|}}{|x|}f(x)dx=\lim_{\varepsilon\to 0}\frac{1}{k^2}\int_{\mathbb{R}^3\setminus B_\varepsilon(0)}k^2\frac{e^{ik|x|}}{|x|}f(x)dx$$ (the limit equals the given integral because $f$ is in the Schwartz space using the dominated convergence) Now I can use the fact that if $x\neq 0$ $$-\Delta\bigg(\frac{e^{ik|x|}}{|x|}\bigg)=k^2\frac{e^{ik|x|}}{|x|}$$ so $$\lim_{\varepsilon\to 0}\frac{1}{k^2}\int_{\mathbb{R}^3\setminus B_\varepsilon(0)}k^2\frac{e^{ik|x|}}{|x|}f(x)dx=\lim_{\varepsilon\to 0}\int_{\mathbb{R}^3\setminus B_\varepsilon(0)}\bigg(-\Delta\frac{e^{ik|x|}}{|x|}\bigg)f(x)dx$$ At this point I can use the integration by parts to have $$=\frac{1}{k^2}\lim_{\varepsilon\to 0}\bigg\{-\int_{\mathbb{R}^3\setminus B_0(\varepsilon)}dx\frac{e^{ik\vert x\vert}}{\vert x\vert}(\Delta f)(x)-\int_{\partial B_0(\varepsilon)}d\sigma f\nabla\bigg(\frac{e^{ik\vert x\vert}}{\vert x\vert}\bigg)\cdot\nu+\int_{\partial B_0(\varepsilon)}d\sigma\frac{e^{ik\vert x\vert}}{\vert x\vert}\nabla f\cdot\nu\bigg\}$$ Using the regularity of $f$ (and of all its derivatives) the integral in the last equation are bounded, so I obtain that $$|\varphi(k)|\leq\frac{M}{k^2}$$.

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  • $\begingroup$ I'm just curoius, but is this true that $$ -\Delta \left( \frac{e^{ik|x|}}{|x|} \right) = k^2 \frac{e^{ik|x|}}{|x|} $$ holds for $x \neq 0$? $\endgroup$ – Sangchul Lee Apr 26 '13 at 9:43
  • $\begingroup$ If my calculation are correct, I think yes. $\endgroup$ – Mat Apr 26 '13 at 9:48
  • $\begingroup$ Oh, I forgot that we are working on $n = 3$. Indeed, my calculation also yields $$ -\Delta \left( \frac{e^{ikr}}{r} \right) = \left( \frac{n-3}{r^3} - \frac{ik(n-3)}{r^2} + \frac{k^2}{r} \right)e^{ikr}. $$ $\endgroup$ – Sangchul Lee Apr 26 '13 at 9:56
  • $\begingroup$ Yes your calculations are true. $\endgroup$ – Mat Apr 26 '13 at 10:08
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Your argument is correct. But here is a different approach which gives more precise information. First, observe that for any orthogonal matrix $M\in O(3)$ the function $ f(Mx)$ is also in the Schwartz class. Consider $ g(x)=\int_{O(3)}f(Mx)\,d\mu(M)$ where $\mu$ is the Haar measure on $O(3)$. The function $g$ is still in the Schwarz class, and is constant on every sphere with center at the origin. All this is a fancy way of saying that the average of $f$ over the sphere $|x|=r$ is a smooth function of $\mathbb r$ which extends (by even reflection) to a Schwartz function on $\mathbb R$. Let $F(r)$ be this average, and observe that $$\varphi(k)=4\pi\int_0^\infty rF(r)e^{ikr}\,dr \tag0$$ The problem has become one-dimensional.

Naturally, we want to extend the integral in (0) to all of $\mathbb R$. Recall that $F$ has a smooth even extension, $F(r)=F(-r)$. Thus, $$\operatorname{Re}\varphi(k)=4\pi\int_0^\infty rF(r)\cos kr \,dr = 2\pi \int_{-\infty}^\infty |r| F(r)\cos kr \,dr = 2\pi \widehat{(|r|F)}(k)\tag1$$ and $$\operatorname{Im}\varphi(k)=4\pi\int_0^\infty rF(r)\sin kr \,dr = 2\pi \int_{-\infty}^\infty r F(r)\sin kr \,dr = -2\pi i \widehat{( r F)}(k)\tag2$$ Since $rF $ is a Schwartz function, so is $\operatorname{Im}\varphi$, according to (2). Unfortunately, $\operatorname{Re}\varphi$ is not as nice since the absolute value in (1) spoils the regularity. Well, at least $|r|F$ is a continuous function with globally integrable, piecewise continuous derivative. It is a shown here that the Fourier transform of such a function is $O(1/k^2)$. In fact, a remark in the linked answer implies that $k^2 \varphi(k)\to 4\pi f(0)$ as $k\to\infty$, if I managed to keep the constants right.

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