If $u\in H^1(\Omega )$ then $|u|\in H^1(\Omega )$.

Question

Let $\Omega $ a smooth domain and $u\in H^1(\Omega )$. Prove that $$|u|\in H^1(\Omega ).$$

The proof goes as following : We define $$v_\varepsilon=\sqrt{\varepsilon+u^2}.$$ Then by the chain rule $$\nabla v_\varepsilon=\frac{u\nabla u}{\sqrt{\varepsilon+u^2}},$$ and $$v_\varepsilon\to |u|\quad \text{and}\quad \nabla v_\varepsilon=\frac{u\nabla u}{\sqrt{\varepsilon+v^2}}\to \frac{u}{|u|}\nabla u$$ in $L^2(\Omega )$. The claim follow.

Questions

1) $\Omega $ is not supposed bounded. So why $\sqrt{\varepsilon+u^2}\in L^2(\Omega )$ ? Isn't it wrong ? To me $\int_\Omega v_\varepsilon^2=\int_\Omega (\varepsilon+u^2)$ and there is no reason for it to be finite.

2) If $v_\varepsilon$ is indeed in $L^2(\Omega )$ then why $$v_\varepsilon\to |u|\quad \text{and}\quad \nabla v_\varepsilon=\frac{u\nabla u}{\sqrt{\varepsilon+v^2}}\to \frac{u}{|u|}\nabla u$$ in $L^2(\Omega )$ give us $|u|\in H^1(\Omega )$ ?

Answer 1

Ad 2. More precisely, we have the following convergence in $L^2(\Omega)$: $$ v_\varepsilon \to |u|, $$ $$ \nabla v_\varepsilon \to V := \begin{cases} \operatorname{sgn} (u) \nabla u & \text{where } u \neq 0 \\ 0 & \text{where } u=0 \end{cases}. $$ We only need to check the definition of weak derivatives. For any test function $\varphi \in C_c^\infty(\Omega,\mathbb R^n)$ $$ \int_\Omega v_\varepsilon \operatorname{div} \varphi = - \int_\Omega \nabla v_\varepsilon \varphi, $$ in the limit $$ \int_\Omega |u| \operatorname{div} \varphi = - \int_\Omega V \varphi. $$ This proves that $V$ is the weak gradient of $|u|$. Since $V \in L^2(\Omega)$, this means $u \in H^1(\Omega)$.

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Ad 1. This indeed doesn't work for unbounded $\Omega$. One can repeat the reasoning for any bounded subdomain $\Omega' \subseteq \Omega$ instead. This way, we get $|u| \in H^1(\Omega')$ with $\nabla |u| = V$ in $\Omega'$. Since $V$ lies in $L^2(\Omega)$, the same conclusion holds also for $\Omega$.

Answer 2

You're right that the trick with the $\epsilon$ doesn't work if $\Omega$ has infinite Lebesgue measure.

First, observe that $|u| = u^{+} + u^{-}$, where $u^{+}(x) = \max\{u(x),0\}$ and $u^{-}(x) = \max\{-u(x),0\}$. Next, show that $u^{+} \in H^{1}(\Omega)$ and $D(u^{+}) = Du \chi_{\{u > 0\}}$ and similarly for $u^{-}$. Your result now follows.