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), 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 $c_{-1} < + \infty$?

## 1 Answer

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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.