Let us look at the subspace of smooth complex functions of $L^2(\mathbb{R}^n,\mathbb{C})$, call $H^s$ the Sobolev spaces. By the diamagnetic inequality $\lvert \nabla \lvert\psi\rvert\rvert (x) \le \lvert\nabla \psi\rvert(x)$ (a proof is [here](https://physics.stackexchange.com/questions/100948/is-there-a-physical-intuition-for-diamagnetic-inequality)), we have
\begin{align\*}
\lVert\,\lvert\psi\rvert\,\rVert\_{H^s} \le c\_s \lVert \psi\rVert\_{H^s}
 \end{align\*}
for $s=1$ and $c\_1 = 1$, where $c\_1$ does not depend on $\psi$. It is also true for $s=0$ with $c\_0 = 1$. Do we have such a result for $s=-1$, with $c\_{-1} < + \infty$?



'Diamagnetic' inequality for negative Sobolev spaces
This is not true in the case of $H^{-1}(\Omega) = H\_0^1(\Omega)^\*$ (real spaces), where $\Omega \subset \mathbb R^d$ is bounded:


* In <https://math.stackexchange.com/questions/336834/decomposition-of-functionals-on-sobolev-spaces>, we see that $|\psi|$ cannot be defined for all $\psi \in H^{-1}(\Omega)$.
* Even if $|\psi|$ can be defined (as a measure), it might not belong to $H^{-1}(\Omega)$, see <https://math.stackexchange.com/questions/1402697/decomposition-of-measures-acting-on-sobolev-spaces>
* Even if $|\psi| \in H^{-1}(\Omega)$, we do not get a bound. On $\Omega = (0,1)$, you can consider $\psi\_n(x) = \sin(n x)$ as an element in $H^{-1}(\Omega)$. Then, $\|\psi\_n\|\_{H^{-1}} \to 0$, but $\||\psi\_n|\|\_{H^{-1}} \not\to 0$.


I would expect that the same arguments can be adopted to your situation.


