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(The question was originally posted on MSE.)

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 on MSE using the Fourier convolution theorems:

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

Question: Is there a simple argument to conclude that $u^p \in \mathrm{H}^s(\mathbb{R})$ (and $\mathrm{H}^s(\mathbb{T})$) 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: $|\cdot|^p$ and $\cdot |\cdot|^{p-1} \in \mathrm{C}^{\lfloor p \rfloor}$ for $p > 1$.

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  • $\begingroup$ The claim you wish to conclude is false. Take a compactly supported smooth $u$ 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 high orders) although $u\in H^s(\mathbb R^n)$ for all $s$. $\endgroup$ – Joonas Ilmavirta Jul 13 '15 at 15:47
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Let $\frac12<s<1$, let $u\in H^s(\mathbb R)$, then also $u\in C_0$, and let $f$ be Lipschitz on $[-||u||_\infty \ ,+||u||_\infty]$ with $|f(x)-f(y)|\leq M|x-y|$. Then $$\int\int \frac{|f(u(t))-f(u(t'))|^2}{|t-t'|^{1+2s}}dt\ dt'\leq M^2\int\int\frac{|u(t)-u(t')|^2}{|t-t'|^{1+2s}}$$With $f(0)=0$ you also have $|f(u)|\leq M|u|$ so that $f(u)\in L^2$.

More generally, for $s=m+\sigma$ with $m$ (an integer) $\geq1$ and $0<\sigma<1$, use the equivalent definitions of $H^s$ as $\{u\in H^m: D^mu\in H^\sigma\}$ and of $H^\sigma$ as $\{v\in L^2: \int\int\frac{|v(t)-v(t')|^2}{|t-t'|^{1+2\sigma}}<\infty\}$, together with a development of $D^m (f(u))$. ($f\in W^{m+1,\infty}_{loc}$ with $f(0)=0$ is enough in this case).

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Hint at a partial answer to the revised question, extended to $p\in (0,\infty)$: for $\frac12<s<1$, $p\geq 1$ is necessary and sufficient.

The "if" part is straightforward using the double integral definition of $H^s$. The "only if" part will better be dealt with using Fourier series $i\ \sum \pm|n|^{-\alpha} e^{int}$ ($\pm :=$ sign of $n$) or $\sum |n|^{-\alpha} (e^{int}-1)$ that vanish at $0$ with a singularity.

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