Why is $f_\epsilon(u) \in W_0^{1,2}(\Omega)$?

Question

For $\epsilon>0$ let $f_\epsilon(u)=\sqrt{\epsilon^2+u^2}-\epsilon$

One calculates that $\nabla f_\epsilon(u)=\frac{u}{\sqrt{\epsilon^2+u^2}}\nabla u $ , for $\epsilon$ to 0 this term goes to $\nabla |u|$

and one finds$ f_\epsilon(u) \in W_0^{1,2}(\Omega)$.

Answer 1

Ok so it is clear that $f_{\epsilon, u}$ is compactly supported from the fact that $u$ is.

Now we do $\int_{\Omega}|f_{\epsilon, u}|^2(x)dx+\int_{\Omega}|\nabla f_{\epsilon, u}|^2(x)dx$. It is our goal to show these integrals converge.

$\int_{\Omega}|f_{\epsilon, u}|^2(x)dx = \int_{\Omega}u^2 - 2\epsilon \sqrt{\epsilon^2+u^2} \leq \int_{\Omega}u^2 < \infty$ by defn of $u \in W_0^{1,2}$

$\int_{\Omega}|\nabla f_{\epsilon, u}|^2(x)dx = \int_{\Omega}\frac{u^2}{{u^2+\epsilon^2}}||\nabla u||_2^2 dx \leq \int_{\Omega}||\nabla u||_2^2dx <\infty$ again by $u \in W_0^{1,2}$