Why this function is in this Sobolev space?

Question

Let $\Omega \subset R^n (n \geq 2)$ a bounded domain with smooth boundary. Let $\phi \in W^{1,p}(\Omega)\cap L^{\infty}(\Omega) \ (2 \leq p < \infty)$ with $\phi^{+}$ different from the null function. Let $j_0 \in N$ the smallest natural number such that $j_0 \geq \| \phi\|_{L^{\infty}(\Omega)}.$ Let $ u \in W^{1,p}(\Omega)$ with $u - \varphi \in W^{1,p}_{0}(\Omega).$

I am reading a paper and the paper says that $ (|u| - j)^{+} \in W^{1,p}_{0}(\Omega)$ for $ j \in N$ with $j > j_0$. I have no idea of how prove this. Someone could help me ?

Thanks you for your attention.

You may check Stampacchia Theorem. From it follows that $|u-\varphi|,(u-\varphi)^+, (u-\varphi)^-\in W_0^{1,p}$. Also if $a\leq 0\leq b$ then $(u-\varphi-b)^+\in W_0^{1,p}$ and $(u-\varphi -a)^-\in W_0^{1,p}$ — Svetoslav, Oct 01 '15 at 20:53
I can not see it right now, can you write some short proof ? — Svetoslav, Oct 01 '15 at 21:05
for a moment I thought that I had the proof, but when I started to write, I found a problem in my reasoning — math student, Oct 01 '15 at 21:06
Well, this happens :) Still, if you find a proof, please tell it here. — Svetoslav, Oct 01 '15 at 21:12
After I wrote my commentary "I found a proof =)" , I started to write the proof =) — math student, Oct 01 '15 at 21:33

score 1 · Answer 1 · answered Oct 01 '15 at 21:30

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Maybe this can help someone one day. Since I saw this argument in a paper, maybe it is important.

Since $u - \varphi \in W^{1,p}_{0}(\Omega)$, we have

$|u - \varphi| \in W^{1,p}_{0}(\Omega).$

Note that $||u(x)| - |\varphi (x)|| \leq |u(x) - \varphi (x)|,$ for all $x$ then $||u| - |\varphi|| \in W^{1,p}_{0}(\Omega) $. For $j>j_0$ we have for all x that $|\varphi(x)| < j$, which implies $- |\varphi (x)| > -j$.

For all $x$ we have $\max(|u(x)| - |\varphi (x)| , 0) \geq \max(|u| - j, 0) \geq 0 \ (*)$.

Since $ |u| - |\varphi| \in W^{1,p}_{0}(\Omega) \rightarrow (|u| - |\varphi|)^{+} \in W^{1,p}_{0}(\Omega) \ (**) $. From $(*)$ and (**) we have $(|u| - j)^{+} \in W^{1,p}_{0}(\Omega) .$

answered Oct 01 '15 at 21:30

math student

4,566

Could you specify paper mentioned in the answer, as well as the one that prompted the question? – epimorphic Oct 01 '15 at 22:21
ok =) . The paper is this http://arxiv.org/pdf/1202.5264v2.pdf and the question is related with the beginning of page 13 – math student Oct 01 '15 at 22:29
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The step from $u-\varphi$ to $|u-\varphi|$ is nontrivial, btw. – Oct 02 '15 at 02:38
@NormalHuman The step from $u-\varphi\in W_0^{1,p}(\Omega)$ to $|u-\varphi|\in W_0^{1,p}(\Omega)$ follows from Stampacchia Theorem: The function $F(x)=|x|$ is Lipschitz continuous and $F(0)=0$ and therefore $F,o,(u-\varphi)\in W_0^{1,p}(\Omega)$. For me the step $||u(x)|-|\varphi(x)||\leq |u(x)-\varphi(x)|$ for all $x\Rightarrow ||u|-|\varphi||\in W_0^{1,p}(\Omega)$ is not clear, as well as the last line. – Svetoslav Jan 03 '16 at 14:03

Why this function is in this Sobolev space?

1 Answers1