Composition operators on fractional-order (periodic) Sobolev spaces

(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 $$p > 1$$.

• 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 $s$. Jul 13 '15 at 15:47

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).

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.