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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 .

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

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