An $L^2$ bound for a function of the form $w(x)=|x|^{-1} - (g^2*|x|^{-1}*g^2)(x)$ It is stated in the paper [E. Lieb  and H.-T. Yau,  The Chandrasekhar Theory of Stellar Collapse as the Limit of Quantum Mechanics Commun. Math. Phys. 112, 147-174 (1987)] 
that this function is "clearly" in $L^2$:

$w(x)=|x|^{-1} - (g^2*|x|^{-1}*g^2)(x)$
with  $g(x)=\xi^{3/4} \exp(-\pi \xi x^2/2)$ and therefore $||g||_2 =1$.

It's bound is supposed to be $|| w||_2 \leq \xi^{-1/4} \sqrt{\frac{32}{3 \pi^{1/2}}}$.
By evaluating the spherical parts I got to the point that

$0 \leq w(x) \leq \int d^3a \int d^3 b\ \ g^2 (a) g^2 (b)\left[ \frac{1}{|x|} - \textrm{min}\left\{ \frac{1}{|x|}, \frac{1}{|a|}, \frac{1}{|b|}  \right\}\right]$

I can see that it's in $L^2$, but every estimate I make is too rough to yield the above bound. Any ideas on some good estimates or to evaluate it exactly?
 A: Without loss of generality, let $\xi = 1$.  Here $x$ and $y$ will denote elements in $\mathbb{R}^3$.
Now we go all the way back to Isaac Newton!  We have the following integral over the surface of the sphere (the "shell theorem" from classical mechanics),
$$\int_{\mathbb{S}^2} |r \omega - x | ^{-1} d\omega = 4\pi\min(r^{-1}, |x|^{-1}).$$
This gives a striking identity for the convolution of $|x|^{-1}$ with a radially symmetric function $\rho(x) = \rho(|x|)$.
$$\left(|x|^{-1} \ast \rho(x) \right)(x) = 4\pi \int_0^\infty \min(r^{-1},|x|^{-1}) \rho(r) r^2 \,dr.$$
Note that this particular identity works only in $\mathbb{R}^3$, but there is a more general version that you can find in Lieb and Loss "Analysis", page 249.
Now let
$$\rho(x) = (g^2 \ast g^2)(x) = \frac{1}{\sqrt{8}} \exp\left(-\frac{\pi |x|^2}{2}\right).$$The operation of convolution is commutative and associative, so we have
$$w(x) = |x|^{-1} - (|x|^{-1} \ast \rho)(x) = \int_\mathbb{R^3} |x|^{-1} \rho(y) \,d^3y - \int_\mathbb{R^3} |x-y|^{-1} \rho(y) \,d^3y \\= 4\pi \int_0^\infty |x|^{-1} \rho(r)r^2\,dr - 4\pi \int_0^\infty \min(r^{-1}, |x|^{-1}) \rho(r)r^2 \,dr = 4\pi \int_{|x|}^\infty (|x|^{-1}-r^{-1}) \rho(r)r^2 \,dr.$$
Now we can rewrite $w(x)$ in terms of complementary error function "erfc",
$$w(x) = \frac{\operatorname{erfc}(|x|\sqrt{\pi/2})}{|x|}.$$
Thus we have
$$\|w\|_2^2 = \int_{\mathbb{R}^3} |w(x)|^2 \,d^3x = \int_0^\infty 4\pi \operatorname{erfc}(r\sqrt{\pi/2})^2 \,dr = 8(\sqrt{2}-1).$$
So we conclude with an exact value for the $L^2$-norm of $w$,
$$\|w\|_2 = \sqrt{8(\sqrt{2}-1)}.$$
