Let $\psi\in H^1(\mathbb{R}^3)\cap W^{1/2,6}(\mathbb{R}^3)$. Is it true that $\psi\nabla\psi\in L^2(\mathbb{R}^3)$?
By Sobolev embedding, $W^{1/2,6}(\mathbb{R}^3)\hookrightarrow L^{p}(\mathbb{R}^3)$ for every $p<+\infty$. The embedding in $L^{\infty}(\mathbb{R}^3)$ fails in general, but in order to show the result it would be sufficient to prove that, if $\psi$ diverges at some point $x$, then $\nabla\psi$ is actually slighty better than $L^2(\mathbb{R}^3)$ in a neighborhood of $x$, which appears to be the case in some explicit example. Can this argument be made rigorous?