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Let $\Omega \subset \mathbb{R}$, bounded and regular. Prove that if $u \in H^1(\Omega)$, then $|u| \in H^1(\Omega)$?

$H^1(\Omega)=\{u \in L^2(\Omega) \mbox{ s.t } \partial_x u \in L^2(\Omega)\}$

Student
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1 Answers1

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Let $\delta>0$ and define $f_\delta:\mathbb{R}\to [0,\infty)$ by $$ f_\delta(x) = \left\{ \begin{array}{rl} \sqrt{x^2+\delta^2}-\delta &\mbox{ if $x\ge0$} \\ 0 &\mbox{ otherwise} \end{array} \right. $$

Note that $f_\delta\in C^1$ and $f_\delta(x)\to g(x)$ for all fixed $x$, where $g(x)=x$ if $x\ge 0$ and $g(x)=0$ if $x<0$. Consider the sequence $u_\delta=f_\delta (u)$. You can easily check that $f_\delta(u)\in H^1(\Omega)$ and $$\int_\Omega f_\delta(u)\frac{\partial\varphi}{\partial x_i}=\int_\Omega f'_\delta (u)\frac{\partial u}{\partial x_i}\varphi,\ \forall\ \varphi\in C_0^\infty(\Omega) \tag{1}$$

Now you can verify the hypothesis of Lebesgue theorem both for $f_\delta(u)$ and $f'_\delta(u)$ to conclude that $f_\delta (u)\to |u|$ in $L^2$ and $f'_\delta (u)\to \chi(u)$ where $\chi$ denotes the characteristic function of $u$ in the set where $u> 0$.

Remark: There is a more general result, which is called Stampacchia's theorem, and contains the previous result.

Remark 2: Note that $\Omega\subset\mathbb{R}$ is not necessary here.

Tomás
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  • It's correct thank you for sending me the name of the theorem, i read it and i thing it's enough to take $G(x)=|x|$ which is almost everywhere differentiable. – Student Feb 17 '14 at 15:26
  • Yes it is enough and this is consequence of Stampacchia's theorem, however, to prove it with this functions is more difficulty, That's why I regulirise it. – Tomás Feb 17 '14 at 15:43
  • This is a nice approach, because you also get the expression for the derivative (to show only that $G\circ u\in H^1(\Omega)$ for $G$-Lipschitz is easy). Do you know, whether there is such approach for the more general case of say $G$- Lipschitz with finite number of discontinuities (or infinite)? I mean, an approach, where you find $G_n(s)\to G(s)$ uniformly and also $G_n'(s)\to G'(s)$ poinwtise, so that you can use Lebesgue DCT on $G_n'(u(x))\to G'(u(x))$ – Svetoslav May 06 '16 at 19:54
  • Maybe, still care should be taken of the points $x\in\Omega$ where $u(x)=a_i$ for some $i$, where ${a_i}_{i=1}^N$ are the points of discontinuities for $G'$ ? – Svetoslav May 06 '16 at 19:58
  • Dear @Svetoslav , take a look on this post. You will find the answer for all your questions. Also, in Evans book of PDE, you can find a proof of this fact for $C^1$ functions. The proof for Lipschitz functions is more elaborate and makes use of the notion of absolutely continuous functions. – Tomás May 06 '16 at 22:22
  • Thanks! I read your old post and saw the reference in there. For $C^1$ functions we can use the Lebesgue DCT on the derivatives, because $G'$ is continuous and we then have pointwise a.e convergence ($G'(u_n(x))\frac{\partial u_n(x)}{\partial x_i}\to G'(u(x))\frac{\partial u(x)}{\partial x_i}$). It seems, that for just Lipschitz $G$ all proofs in the leterature use the ACL property. – Svetoslav May 07 '16 at 05:30