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Hoi, consider $\displaystyle u= \frac{1}{|x|}e^{-|x|}$ for $x\in \mathbb{R}^3$, then one can see that $\Delta u = u$ for $|x|>0$ ( which one can see by transferring $u$ to spherical coordinates).

So can we then conclude: (*) $(1-\Delta)u$ is a distribution with support $= \left\{0\right\}$ and order $N=0$?

I want to either show that $((1-\Delta)u, \phi) = \phi(0)$ or show that (*) holds.

One of the 2 is enough to show that $(1-\Delta)u = \delta$. Thanks for any help.

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    $\begingroup$ How do you see $u$ as a distritubtion, I mean, how do you define $(u,\phi)$? I am asking it, because $u$ is not even locally integrable. $\endgroup$ – Tomás Apr 9 '13 at 18:13
  • $\begingroup$ I use it as $\int_{\mathbb{R}^3}u\phi$. Apparantly $cu$ with $c = 1/(4\pi)$ is the unique solution to $(1-\Delta)u= \delta$. $\endgroup$ – DinkyDoe Apr 9 '13 at 18:55
  • $\begingroup$ Your integral is not well defined. It can be $\infty$, so it not defines a distribution. You have to define in another way. $\endgroup$ – Tomás Apr 9 '13 at 19:12
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    $\begingroup$ @DinkyDoe: You are right for every $\phi$ with $ 0\notin\operatorname{supp }\phi$ and for such $\phi$ the integral is well-defined. But with the general definition of $(u,\phi)$, the Dirac delta enters the formula ... $\endgroup$ – Vobo Apr 9 '13 at 20:52
  • $\begingroup$ Thanks, however $(\delta, \phi) = \phi(0)$ and i would have to show the same: $((1-\Delta),\phi) = \phi(0)$ which is not so easy to see...So clearly $((1-\Delta),\phi) = \phi(0) =0$ holds true if $0\notin \text{supp} \ \phi$. So now what if $\phi(0)\neq 0$... why does then $((1-\Delta)u,\phi) = \phi(0)$ also hold true. $\endgroup$ – DinkyDoe Apr 10 '13 at 11:41
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(Side note for Tomás: $u$ is locally integrable, and therefore is a distribution).

Recall that $\Delta (|x|^{-1})=-4\pi\delta_0$ (if you don't know it already, see Laplacian of the potential function). Thus, if you can show that $\Delta (u -|x|^{-1}) = u$, the conclusion $(1-\Delta) u=-4\pi \delta_0$ will follow.

The function $u(x)-|x|^{-1} = (e^{-|x|}-1)|x|^{-1} = -1 + |x|/2+\dots$ is in $W^{2,1}$ locally, which implies that its distributional Laplacian is represented by its pointwise Laplacian. See Calculation of the Laplacian of a function in $\mathbb R^3$.

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  • $\begingroup$ You are right, i am doing wrong calculation. $\endgroup$ – Tomás Apr 10 '13 at 22:43
  • $\begingroup$ OK, i know now why $\Delta(|x|^{-1}) = -4\pi\delta_0$ holds... but why what you say implies the desired conclusion I dont follow $\endgroup$ – DinkyDoe Apr 11 '13 at 12:15
  • $\begingroup$ I can prove both the statements you make...but I dont see yet why it implies what we want. Thnx if you can explain that. $\endgroup$ – DinkyDoe Apr 11 '13 at 12:28

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