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.