Distributional second-order derivatives of $\frac{e^{-|x|}}{4\pi |x|}$ to show the solution of $u -\Delta u=f$ is in $H^2$

In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $$u - \Delta u=f$$ where $$f \in L^2 (\mathbb{\Omega})$$ belong to $$H^2 (\mathbb{\Omega}) \cap H^1 _0 (\Omega)$$, when $$\Omega$$ is a bounded domain of $$\mathbb{R}^n$$ with $$\partial \Omega$$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $$\Omega = \mathbb{R}^n$$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $$H^{1}_0 (\Omega)$$. Then, later on in the book, he proves that under the hypothesis of some regularity for $$\partial \Omega$$, the solution is in $$H^2(\Omega)\cap H^1 _0 (\Omega)$$, and he does that using the so called "method of Nirmberg's translations".

The fact is: I am dealing with the case $$\Omega= \mathbb{R}^3$$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $$\Omega_N = B(0,N)$$ where the proof of Brezis works.

The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $$H^{2}(\mathbb{R}^3)$$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:

$$u(x)= \left( \frac{e^{-|y|}}{4 \pi |y|} \ast f \right)(x)$$

But when I try to find a weak second order derivative of $$u$$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $$\epsilon$$), I get this very annoying term that doesn't allow me to conclude:

$$\int_{B(0, \epsilon)^c} \partial_{x_1}^2 \left(\frac{1}{4 \pi |x|} \right) e^{-|x|} \phi(x+y) dx=\int_{B(0, \epsilon)^c} \left(\frac{3x_1 ^2 - |x|^2}{4 \pi |x|^5} \right) e^{-|x|} \phi(x+y) dx$$

Where $$\phi \in C^{\infty}_0 (\mathbb{R}^3)$$. Now, I understand that something like this was forced to appear. Indeed, I know that $$\Delta (1/|x|)= \delta(x)$$, so this isn't a surprise. However, that expression over there does not converge when $$\epsilon \to 0$$ (or t least I think so, there should be a singularity like $$1/\rho$$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $$u$$ in which I put all the derivatives on the term $$e^{-|x|}/(4 \pi |x|)$$ of the convolution.

EDIT: Turns out I can just cut-off a ball of radius $$\epsilon$$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.

• Why wouldn't it converge? The $\exp(-\lvert x\rvert)$ is there to help for large $\lvert x\rvert$. – user10354138 Dec 1 '18 at 14:17
• I mean convergence for $\epsilon \to 0$. There I have a singularity of order $1/ \rho$, passing in spherical coordinates. I'll edit the question, though. – tommy1996q Dec 1 '18 at 14:26