# $u \in W_0^{1,2}(\Omega) \Rightarrow |u| \in W_0^{1,2}(\Omega)$

For $$\epsilon>0$$ define $$g_\epsilon(u)=\sqrt{\epsilon^2+u^2}-\epsilon$$.

One finds $$\nabla g_\epsilon(u)=\frac{u}{\sqrt{\epsilon^2+u^2}}\nabla u$$ and $$g_\epsilon(u)\in W_0^{1,2}(\Omega)$$ .

Then it holds $$\int_{\Omega}|\nabla g_\epsilon(u)|^2dx \leq \int_{\Omega}|\nabla u|^2dx$$

For some sequence $${n_k} \subset \mathbb{N}$$ it follows

$$\int_{\Omega}|\nabla |u||^2dx=\int_{\Omega}\lim\limits_{k \rightarrow \infty}|\nabla g_\frac{1}{n_k}(u)|^2 dx=\lim\limits_{k \rightarrow \infty}\int_{\Omega}|\nabla g_\frac{1}{n_k}(u)|^2 dx \leq \int_{\Omega}|\nabla u|^2dx$$ (1)

I do not really understand what (1) says .

• At the end, $\int_\Omega |\nabla |u||^2\leq \int_\Omega |\nabla u|^2<\infty ,$ and thus $|u|\in W^{1,2}$. – Surb May 6 at 13:05
• but why is $|u| \in L^2$ ? – AnabolicHorse May 6 at 13:07
• and why does it follow that $|u| \in W_0^{1,2}$? – AnabolicHorse May 6 at 13:11
• 1) $u\in L^2\iff |u|\in L^2$. 2) $u|_{\partial \Omega }(x)=0\iff |u||_{\partial \Omega }(x)=0$ – Surb May 6 at 13:23
• I think that is the definition with the trace – AnabolicHorse May 6 at 13:26