If $u \in H^2(\mathbb{R}^3)$, does $r^{-1} u \in H^{\alpha}(\mathbb{R}^3)$ for some $\alpha > 0$?

Question

Let $u$ belong to the Sobolev space $H^1(\mathbb{R}^3)$. We have the classical Hardy inequality \begin{equation*} \int_{\mathbb{R}^3} \frac{|u|^2}{|x|^2} dx \le 4\int_{\mathbb{R}^3} |\nabla u(x)|^2 dx, \end{equation*} which is presented, for instance, on page 296 of Evan's PDE book (second edition). The proof relies on integration by parts only, i.e., it does not utilize the Fourier transform. My question is, if $u$ has additional Sobolev regularity, do we have that $|x|^{-1} u$ gains some Sobolev regularity? In particular I am interested in knowing whether $u \in H^2(\mathbb{R}^n)$, $n \ge 3$, implies $|x|^{-1}u \in H^{\alpha}(\mathbb{R}^n)$ for some $\alpha > 0$. Does the validity of this implication depend on the dimension?

I don't feel confident about tackling this question in a particular way. Perhaps the answer is negative because, if we think about $\alpha$ being a positive integer, additional regularity of $u$ does not necessarily help $D^{\beta}(|x|^{-1} u)$ be in $L^2$ near the origin, where $\beta$ is a multiindex with $|\beta| = \alpha$. So if the implication is true, it is likely only the case for certain $0 <\alpha < 1$. If the answer is positive one could try to show it with

$$\int |\xi|^{2\alpha} |\widehat{|x|^{-1}u}|^2 d\xi = \int |\xi|^{2\alpha} |\widehat{|x|^{-1}} \ast \widehat{u}|^2 d\xi< \infty, $$ where the hat denotes Fourier transform and the $\ast$ is convolution. Presumably the Fourier transform of $|x|^{-1}$ can be computed somewhat explicitly, depending on the dimension.

Answer 1

As noted by Christian Remling, the Fourier transform of $|x|^{-1}$ is $|\xi|^{-2}$, so the relevant integral you wish to bound is $$\iiint |\eta - \xi|^{-2} \bar{\hat{u}}(\xi) |\eta|^{2\alpha} \hat{u}(\zeta) |\eta - \zeta|^{-2} ~d\eta ~d\xi ~d\zeta $$ Do the $\eta$ integral first. This converges as long as $2\alpha < 1$. (At issue is the decay at infinity, if $2\alpha \geq 1$ the integrand is bounded below by $|\eta|^{-3}$ for large $|\eta|$; this issue with decay at infinity is directly tied to the lack of regularity of $|x|^{-1}$ at the origin.)

Writing $\mu = \frac12( \xi + \zeta)$ and $\rho \epsilon = \frac12 ( \xi - \zeta)$, where $\epsilon$ is a unit vector and $\rho$ a scalar, we see, with the change of variables $\eta = \mu + \rho \psi$ that we are down to estimating $$ \int \frac{|\eta|^{2\alpha}}{|\eta - \xi|^2 |\eta - \zeta|^2} ~d\eta = \int \frac{|\mu + \rho \psi|^{2\alpha}}{\rho |\psi -\epsilon|^2|\psi + \epsilon|^2} ~d\psi \leq C_1 \frac{|\mu|^{2\alpha}}{\rho} + C_2 \rho^{2\alpha - 1}. $$
Inserting it into the original integral (and taking moduli of the $u$ terms) we need to bound $$ \iint |\hat{u}(\xi)| |\hat{u}(\zeta)| \Big(\frac{|\xi + \zeta|^{2\alpha}}{|\xi - \zeta|} + |\xi - \zeta|^{2\alpha - 1} \Big) ~d\xi ~d\zeta \lesssim \\ \|u\|_{H^2}^2 \left(\iint \frac{\big(|\xi + \zeta|^{4\alpha} + |\xi - \zeta|^{4\alpha}\big) |\xi - \zeta|^{- 2}}{(1 + |\xi|^2)^2(1 + |\zeta|^2)^2} ~ d\xi~d\zeta \right)^{\frac12} $$ where the estimate is by Cauchy-Schwarz (once in $\xi$ and once in $\zeta$), and the initial $H^2$ bound on $u$ is assumed.

The integrability of the term inside the parentheses is not too hard to check. Split the integral into dyadic rectangles with $|\xi| \approx 2^k$ and $|\zeta|\approx 2^\ell$. When both $k,\ell \leq 0$ we are on a compact region, and the set $|\zeta - \xi| = 0$ has codimension 3 so the integral converges. When $k > \ell$ we have that $|\xi - \zeta| \approx |\xi + \zeta| \approx |\xi|$ and so the integral is bounded by $$ 2^{(4\alpha - 3)k} / (2^{-3\ell} + 2^{\ell}) $$ . When $k = \ell > 0$ we can replace the denominator by $|\xi|^{4}|\zeta|^4$. On the compact region the integral converges again due to $|\xi-\zeta| = 0$ having codimension 3, and the integrand scales like $2^{(4\alpha - 2 - 8)k}$. So the integral is bounded by $2^{(4\alpha - 4)k}$. The double summation in $k$ and $\ell$ therefore converges obviously.

The above sketch shows that $r^{-1} u$ is in $H^\alpha$ for $\alpha \in [0,\frac12)$. My comment shows that this range is sharp.