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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)$.

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  • $\begingroup$ We need to know more about $\Omega$. Why is $f_{\epsilon}$ compactly supported? $\endgroup$ – George Dewhirst May 6 at 20:52
  • $\begingroup$ $\Omega \subset \mathbb{R^n}$ is a domain . $\endgroup$ – AnabolicHorse May 6 at 20:55
  • $\begingroup$ ok seems reasonable, but where is $f_{\epsilon}$ supported? $\endgroup$ – George Dewhirst May 6 at 20:56
  • $\begingroup$ $f_\epsilon(u)$ should give an approximation of |u| with $u \in W_0^{1,2}(\Omega)$ $\endgroup$ – AnabolicHorse May 6 at 20:59
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    $\begingroup$ What is wrong with the answer to this given in your other question here? It would help answers if we knew what it is you dont understand about that $\endgroup$ – Rhys Steele May 6 at 21:15
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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 $u \in W_0^{1,2}$

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  • $\begingroup$ The calculation shows that $f\epsilon(u) \in W^{1,2}(\Omega)$ $\endgroup$ – AnabolicHorse May 6 at 21:32
  • $\begingroup$ Sure but $u \in W_0^{1,2}(\Omega)$, and wherever $u = 0$ so is $f_{\epsilon}u$ $\endgroup$ – George Dewhirst May 6 at 21:33
  • $\begingroup$ The fact that u is compactly supported plus the calculation show that $f\epsilon(u) \in W_0^{1,2}(\Omega)$? $\endgroup$ – AnabolicHorse May 6 at 21:34
  • $\begingroup$ yep that's right, the above space is just the compactly supported functions in $W^{1,2}(\Omega)$ $\endgroup$ – George Dewhirst May 6 at 21:35
  • $\begingroup$ thank you very much for your help . $\endgroup$ – AnabolicHorse May 6 at 21:36

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