Composition operators on fractional-order Sobolev spaces

Question

Preliminaries: We know that the fractional-order Sobolev spaces $\mathrm{H}^s(\mathbb{R})$ and $\mathrm{H}^s(\mathbb{T})$ are closed under multiplication provided $s > 1/2$. This is proved for example in the following questions using the Fourier convolution theorems:

• Sobolev space is an algebra
• For which $s \in \mathbb{R}$, is $H^s(\mathbb{T})$ a Banach algebra?

Thus we infer that $u^p \in \mathrm{H}^s(\mathbb{R})$ for all $p \in \mathbb{Z}_+$ whenever $u \in \mathrm{H}^s(\mathbb{R})$. Similarly for $H^s(\mathbb{T})$.

Question: Is there a simple argument to conclude that $u^p \in \mathrm{H}^s(\mathbb{R})$ for all interpolating values $p \in [1, \infty)$, too?

More generally, from a MathOverflow question it is stated that the composition $f \circ u \in \mathrm{H}^s(\mathbb{R})$ for all real-valued $u \in \mathrm{H}^s(\mathbb{R})$ provided $f \in \mathrm{C}^{\lfloor s + 2\rfloor}(\mathbb{R})$ with $f(0) = 0$. I would like to see references to this result or an outline of a proof.

Revised question: As pointed out by Joonas Ilmavirta below, I had made an elementary mistake concerning the expression $u^p$ for $p \in (1, \infty) \setminus \mathbb{Z}_+$, which should have been $|u|^p$ instead. Then I ask:

For which $p \in (1, \infty) \setminus \mathbb{Z}_+$ and $s > \frac{1}{2}$ is $|u|^p$ or $u|u|^{p - 1}$ in $\mathrm{H}^s(\mathbb{T})$ whenever $u \in \mathrm{H}^s(\mathbb{T})$?

(Similarly for $\mathrm{H}^s(\mathbb{R})$, but the periodic case is most important.)

Note: $|\cdot|^p \in \mathrm{C}^{\lfloor p \rfloor}$ for $p > 1$.

Answer 1

The claim you wish to conclude is false. Pick $u\in C_0^\infty(\mathbb R^n)$ so that $u(0)=0$ but $\nabla u(0)\neq0$. If $p\geq1$ is not a natural number, then there is $s>0$ so that $u^p\notin H^s(\mathbb R^n)$ (because classical derivatives fail to exist to orders above $p$) although $u\in H^s(\mathbb R^n)$ for all $s$. This works in any dimension, including $n=1$.

There is, of course, also the problem of defining $u^p$ for many values of $p$. The conclusion above is valid, no matter how you define $u^p$ for $u<0$ (you can take any of $|u|^p$, $-|u|^p$ and $0$), since $x\mapsto x^p$ is not smooth.