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.
(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 $\mathbb R^3$.