(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 $p > 1$.

  • $\begingroup$ 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 $s$. $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.