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How to prove if $u\in W^{1,p}$, then $|u|\in W^{1,p}$?

Since $|u|\in L_p$, I only need to show weak derivative of $|u|$ exists and $D|u| \in L_p$.

Can anyone give me some hint? Thanks!

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  • $\begingroup$ did you lose your interest in this question? no further questions from your side? $\endgroup$ Commented Feb 9, 2015 at 7:34
  • $\begingroup$ @Quickbeam2k1 Sorry for my slow response. For convenience, can we just define $f_{\epsilon}(x)=\sqrt{(x^2+\epsilon ^2)}$, then do the convergence thing? $\endgroup$
    – Sherry
    Commented Feb 10, 2015 at 2:17
  • $\begingroup$ the problem with that choice is that you don't obtain a pointwise limit for $f_\varepsilon'$ in $x=0$. If you already know that $Du=0$ a.e. on the set where $u=0$ then you can choose your $f_\varepsilon$. However, the standard proof (I haven't seen another proof) to show that is the proof indicated below. $\endgroup$ Commented Feb 10, 2015 at 6:31

2 Answers 2

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This is done by the celebrated Stampacchia Theorem. It can be found in books from Brezis, Kinderlehrer-Stampacchia and many more. However, let me give the quick idea on how to show this.

  1. You need the following chainrule: Let $f\in C^1(\mathbb{R})$ with bounded derivative. For $u\in W^{1,p}(\Omega)$ there holds: $D (f\circ u)= f'(u) D u$. This can be shown by approximating $u$ with smooth functions.

  2. Choose a certain approximation $f_\varepsilon(\cdot) \to (\cdot)_+$. Particularly, the approximation is such that $f_\varepsilon'(x)=0$ for $x\leq 0$ and $f_\varepsilon'(x)\leq C$. More precisely, one choice is $$f_\varepsilon(x) =\begin{cases}\sqrt{x^2+\varepsilon^2}-\varepsilon, &x>0\\ 0,& x\leq 0\end{cases}.$$ Then you can prove the chain rule for $f(\cdot)=(\cdot)_+$ passing with $\varepsilon\to 0$. You will obtain $D(u)_+=\chi_{u} Du$, where $$\chi_{u}(x)=\begin{cases}1,& u(x)>0\\0,&\text{elsewise}\end{cases}$$ (After the next step, you will see that $Du=0$ a.e. on the set where $u(x)=0$.)

  3. Perform Step 2 for $f(\cdot)=(\cdot)_-$ and note that $|\cdot|=(\cdot)_+ + (\cdot)_-$

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Recall that $|u|=u_+ + u_-$. Since Sobolev spaces are closed under addition, it suffices to show that $u_+,u_-\in W^{1,p}$. Now to show $u_+\in W^{1,p}$, just observe that $\|Du_+\|_p\leq \|Du\|_p$ to conclude.

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  • $\begingroup$ I don't follow your reasoning: The derivative of the Cantor function is $0$ a.e. but it's certainly not in any Sobolev space. How are you using $\| Du_+\|_p\leq \| Du\|_p$ to conclude? $\endgroup$
    – Jose27
    Commented Feb 8, 2015 at 8:35
  • $\begingroup$ I'm not seeing the issue - we are assuming that $u$ is in a Sobolev space, then showing $u_+$ is as well. For the Cantor function, $u=u_+$. $\endgroup$
    – pre-kidney
    Commented Feb 8, 2015 at 8:36
  • $\begingroup$ I gave a bad example. Could you show that $Du_+$ exists and is what it should be? I think this is where most of the difficulties come from and you're hand-waving these issues. $\endgroup$
    – Jose27
    Commented Feb 8, 2015 at 8:46
  • $\begingroup$ how would you show your estimate on $D u_+$ and $Du$? Particularly, how would you show that $u$ is weakly differentiable? $\endgroup$ Commented Feb 8, 2015 at 19:09

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