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

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

• We need to know more about $\Omega$. Why is $f_{\epsilon}$ compactly supported? – George Dewhirst May 6 at 20:52
• $\Omega \subset \mathbb{R^n}$ is a domain . – AnabolicHorse May 6 at 20:55
• ok seems reasonable, but where is $f_{\epsilon}$ supported? – George Dewhirst May 6 at 20:56
• $f_\epsilon(u)$ should give an approximation of |u| with $u \in W_0^{1,2}(\Omega)$ – AnabolicHorse May 6 at 20:59
• 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 – Rhys Steele May 6 at 21:15

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 $$u \in W_0^{1,2}$$

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