Summability of double partial derivatives of $\frac{1}{|x|}$ in dimension $3$

Question

I know the fact that its laplacian is equal to the Dirac delta function. However, is it true that its partial derivatives of order 2 belong to $L^{1}(\mathbb{R}^3)$? And how should I show that, in case this is true? I had thought of calculating derivatives in standard coordinates and then pass to spherical coordinates to see if something got simpler, but I am not sure this would work.

Answer 1

The derivatives of order 2 of $u(x)=\frac{1}{|x|}$does not belong to $L^1$. The sum $\left(\partial^2_1+ \partial^2_2+\partial^2_3 \right) u= \delta_0 \notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).

As you mentioned an other option is to compute these derivatives: $$\partial_i u =\frac{x_i}{|x|^3}$$ so: $$\partial_{ij}^2 u = \begin{cases}\frac{1}{|x|^3}-\frac{3x_i^2} {|x^5|} \text{ if } i=j \\ -3\frac{x_i x_j}{|x|^5} \text{ if } i \neq j \end{cases}$$ but none of these functions are in $L^1$. For example using spherical coordinates you obtain $$\iiint_{\mathbb{R}^3} \frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =\int_0^\infty \int_0^{2 \pi} \int_0^\pi \frac{r^2 \sin(\theta)^2 |\cos(\phi) \sin(\phi)|}{r^5} r^2 \sin(\theta) d\theta d\phi dr=\int_0^\infty \frac{1}{r} \left(\int_0^{2\pi}\int_0^\pi \sin(\theta)^3 \frac{1}{2} |\sin(2\phi)| d\theta d\phi \right) dr=+\infty$$.

The same behavior emerges in all cases: a singularity like $\frac{1}{r}$ which is not integrable.