I want to show the following statement:

if $u\in W^{1,p}(\Omega)$, then $u_-$,$u_+$ and $|u|\in W^{1,p}(\Omega)$ with $$D(u_+) = Du\cdot I_{u>0}\qquad\text{and}\qquad D(|u|)=Du\cdot \text{sign}(u)\qquad \text{a.e.}$$ Furthermore, if $u,\, v \in W^{1,p}(\Omega)$, then $\max\{u,v\} \in W^{1,p}(\Omega)$.

My hunch: the main difficulty is to show $u_+ \in W^{1,p}(\Omega).$ The rest can be shown by vector space property: $$u_- = -u+u_+,\\ |u| = u_-+u_+,\\ \max\{u,v\} = \frac{u+v}{2}+\frac{|u-v|}{2}.$$

To show $u_+ \in W^{1,p}(\Omega)$, I construct an estimation function: $$u_\epsilon = (\sqrt{u^2+\epsilon^2}-\epsilon)\cdot I_{u>0}.$$

My first question:

How can I show such an estimation function is in $W^{1,p}(\Omega)$ for any $\epsilon>0$?

My second question:

How to use the estimation function to show $u_+ \in W^{1,p}(\Omega)$? By sending $\epsilon \downarrow 0$?

My third question:

Could it be true for general $W^{1,p}(\Omega)$?


Sorry fot the mess in notation $\partial_i=(\cdot)_{x_i}$

1) Let $F:\mathbb{R}\to\mathbb{R}$, with $F\in C^1$, $|F'|\leq C$ and $F(0)=0$, then for every $u\in W^{1,p}(\Omega)$ we have $F(u)\in W^{1,p}(\Omega)$ and $(F(u))_{x_i}=F'(u)u_{x_i}$.

To prove this let $u_k \in C^{\infty}(\Omega)\cap W^{1,p}(\Omega)$ with $\| u_k-u\|_{1,p}\to 0$, then since $F(0)=0$ we have $|F(u)|\leq C|u|$ and so $F(u)\in L^p$, moreover $|F(u_k)-F(u)|\leq C|u_k-u|$ so $F(u_k) \to F(u)$ in $L^p$. Now since the $u_k$ are smooth the chain rule applies and gives $|(F(u_k))_{x_i}|=|F'(u_k)\partial_iu_k| \leq C|\partial_iu_k|$, (notice that this last sequence has an $L^p$ limit, namely $|\partial_iu|$) and since $F'$ is continous and (a subsequence) $u_k(x) \to u(x)$, that $F(u_k(x))\to F(u(x))$, $\partial_iu_k(x)\to \partial_iu$ a.e. and by a generalization of DCT that $F'(u_k)\partial_iu_k \to F'(u)\partial_iu$ in $L^p$. This gives the desired conclusion.

2) In your problem let $F_{\varepsilon}(t)=\sqrt{t^2+\varepsilon^2}-\varepsilon$ and apply the same arguments, noting that $|F_{\varepsilon}'|\leq 1$ and $F_{\varepsilon}(t) \to tI_{(0,\infty)}$ and $F'_{\varepsilon} \to I_{(0,\infty)}$. More expicitly $|F_\varepsilon(u)|\leq |u|$ and $|F_\varepsilon' (u)\partial_iu|\leq |\partial_iu|$.

  • $\begingroup$ thanks for the answer. How did you obtain $|F(u)|\leq|u|$? I guess you mean $|F(u)|\leq C\cdot |u|$ $\endgroup$ – newbie May 8 '12 at 2:16
  • $\begingroup$ @newbie Yes, I've edited. Hopefully that's the only mistake... $\endgroup$ – Jose27 May 8 '12 at 2:21
  • $\begingroup$ I think the subsequent inequality $|F(u_k)-F(u)|\leq \cdots$, but it isn't substantial. I have updated the question a little bit, please help me out if you don't mind. Cheers. $\endgroup$ – newbie May 8 '12 at 2:30
  • $\begingroup$ @newbie The answer to the third question is yes, and it'll become clear once you do the second question. For this use the hints I gave in 2), the goal is to prove that if $v_\varepsilon=F_\varepsilon(u)$ then $v_\varepsilon \to u_+$ in $L^p$ and $v_\varepsilon' \to DuI_{u>0}$ in $L^p$, and the ingredients are all there, just mimic the arguments given in 1) to prove said convergence. $\endgroup$ – Jose27 May 8 '12 at 2:56

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