$\||u|\|_{H^s(\mathbb R^n)} \le C \| u \|_{H^s(\mathbb R^n)}$ holds for even if $s$ is not an integer? Let $u: \mathbb R^n \ni x\mapsto u(x) \in \mathbb C.$
I would like to know that the inequality
$$
\||u|\|_{\dot H^s(\mathbb R^n)} \le C \| u \|_{\dot H^s(\mathbb R^n)}
$$
or
$$
\||u|\|_{H^s(\mathbb R^n)} \le C \| u \|_{H^s(\mathbb R^n)}
$$
for any $s\ge 0$ with some positive constant $C$ . Actually, I think I can prove the inequality holds for $s=0,1,2,\cdots$, However, I cannot sure the inequality holds even if $s$ is not an integer. Here $|\cdot |$ is the absolute value notation and $\dot H^s$, $H^s$ are the homogeneous Sobolev space, Sobolev space of order $s$ respectively. 
 A: For Sobolev spaces, I like to think about the case $n=1$ first, since there, the Sobolev functions can be easily characterized. Using this, we will show that the claim is false for $s \geq 2$.
To this end, take $f \in C_c^\infty(\Bbb{R})$ with $f(x) = x$ for $|x| < 1$. Then $|f(x)| = |x|$ for $|x| < 1$ and hence the weak derivative of $|f|$ is given by
$$
g(x) := \frac{d}{dx} |f(x)| = \begin{cases}1 & x > 0 \\ -1 & x < 0\end{cases}
$$
for $|x| < 1$.
Note that $g$ does not have a continuous representative. In particular, $g \notin H^1$, whence $|f| \notin H^s$ (otherwise, we would have $g \in H^{s-1} \subset H^1$), although $f \in C_c^\infty \subset H^s$ for every $s > 0$.
For $s = 1$, it is a classical result about Sobolev spaces that your claim is true (iirc) and for $s=0$ we have $H^s = L^2$, so that the claim also holds. A moment ago, I was tempted to claim that this implies your claim for $0 \leq s \leq 1$ by interpolation, but since the map $f \mapsto |f|$ is nonlinear, this will not work that easily...
