If $u\in H^s$, is $v=u^2$ gaining or losing "regularity"? Let's consider a function $u\in H^s(\mathbb{R})$ for some $s\geq0$, where $H^s$ denotes the standard $L^2$-based Sobolev space. Now consider the function $v(x)=u^2(x)$. I am wondering if $v(x)$ is gaining or losing regularity with respect to $u(x)$. Specifically, I am wondering if $v\in H^m(\mathbb{R})$ with $m\geq s$ or $m\leq s$?. Is there any example of $u(x)$ such that $v(x)\notin H^s(\mathbb{R})$?
Edit: Does the fact that $s\geq0$ plays any role in these kind of properties?
 A: Since you have $\nabla v = 2 u\nabla u$, you can see that if $u$ is bounded by above and below, then $u$ and $v$ will have the same regularity (at least for $s=1$).
Then of course for high values of $u$, $v$ will be more singular. Assume for example that $u$ is compactly supported and $u(x) = 1/|x|^a$ near $0$ and smooth except in $0$. Then $v(x) = 1/|x|^{2a}$ is not even a locally integrable if $a>1/2$. More precisely, here $u\in H^{s}$ for $s<d/2-a$ and $v\in H^s$ for $s<d/2-2a$. So you are loosing regularity.
The effect is reversed for small values. $|x|^a$ is more "singular" compared to $|x|^{2a}$ (the case $a=1$ you even have $|x|^2$ infinitely smooth.)
A: Generally speaking you're going to lose regularity. The simplest example is s = 0. Then by the definition of $L^2$ you've got that $u \in H^0 = L^2$ but $u^2 \in L^1$. There are plenty of examples. Pick any nonnegative $g \in L^1(\mathbb R)$ but not in $L^2(\mathbb R)$ (See this question for examples of such $g$) and take $u = \sqrt g$.
Now consider $s =1$, so say that $ u \in H^1(\mathbb R)$. Then $u$ is weakly differentiable and the weak derivative is in $L^2$. By the weak product rule the derivative of $u^2$ is $2u'u$. Now $u \in H^1$ and $u' \in L^2$. It's not immediately clear whether or not the product will be in $L^2$, but I doubt it. 
Similar issues occur if $s > 1$.
There is a book called Theory of Sobolev Multipliers by Maz'ya and Shaposhnikova which will have a lot more detail and precise theorems.
