Fractional Sobolev norm of the modulus of a function

Question

Let $H^s(\mathbf{R}^d)$ be the Sobolev space with $s \in (0,1]$. It is known that if $u \in H^1(\mathbf{R}^d)$ then also $|u| \in H^1(\mathbf{R}^d)$ with equality of the norms $\| u \|_{H^1} = \| |u| \|_{H^1}$. I would like to see if something similar holds in the fractional case $s \in (0,1)$. Using the Gagliardo seminorm $$ [u]_{H^s}^2 = \int_{\mathbf{R}^d} \int_{\mathbf{R}^d} \frac{|u(x) - u(y)|^2}{|x-y|^{d+2s}} \; dxdy $$ it is easy to show that if $u \in H^s$ then $|u| \in H^s$ with $\| |u| \|_{H^s} \leq \| u \|_{H^s}$, because $ ||u(x)| - |u(y)|| \leq |u(x) - u(y)|$. Is the converse inequality true, maybe up to a constant which does not depend on $u$? I was trying to show this using the characterization $$ \| u \|_{H^s} = \sup_{\| v \|_{H^{-s}} \leq 1} \int_{\mathbf{R}^d} uv \; dx $$ with an appropriate choice of $v$, but I'm stuck.

Answer 1

The converse inequality $$ \| u \|_{\dot{H}^s(\mathbb{R}^d)} \lesssim \| ~ |u| ~ \|_{\dot{H}^s(\mathbb{R}^d)} $$ is true for $s \in (1/2,1)$, but not below $1/2$.

This is proved in the paper Remark on the absolute value in Sobolev spaces by Pierre Gilles Lemarié-Rieusset.