Let $u\in W^{1,p}(\Omega )$ where $\Omega \subset \mathbb R^d$ is open and connected. If $\nabla u=0$ do we have that $u$ is constant every where or a.e. only ?


Let $\Omega $ an open subset of $\mathbb R^d$ s.t $\Omega \subset \{0\leq x_d\leq 1\}$. Prove that there is $C>0$ s.t. for all $u\in \mathcal C_0^1(\Omega )$

$$\int_\Omega |\nabla u|^p\geq \int_\Omega |u|^p.$$


Suppose by contradiction, it doesn't hold. We can assume WLOG that $\Omega $ is connected

Q1) why such an hypothesis ? Does all open is a finite union of connected set ?

Then, there is $u_n$ s.t. $$\|\nabla u_n\|_{L^p(\Omega )}\to 0,\quad \|u_n\|_{L^p(\Omega )}=1.$$ By embedding theorem, there is a subsequence (still denoted $(u_n)$) that converge strongly to $u\in W_0^{1,p}(\Omega )$. Therefore, $$\nabla u=0,$$ and thus $u=C$ a.e. in $\Omega $. Since $u\in W_0^{1,p}(\Omega )$, we have that $u=0$ what contradict $\|u\|=1$.

Q2) Since $u\in W_0^{1,p}(\Omega )$ we can say that $u=0$ on the boundary, but it's measure is nulle. So how can they conclude that $u=0$ a.e. in $\Omega $ ?

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    $\begingroup$ The elements of Sobolev spaces are not defined pointwise, so we may only have almost-everywhere something. $\endgroup$ – Jack D'Aurizio Dec 25 '17 at 17:34
  • $\begingroup$ @JackD'Aurizio: Thanks a lot. I added more context, could you tell me why they conclude that $u=0$ a.e. since $u=0$ on the boundary that is of measure nulle ? $\endgroup$ – user386627 Dec 25 '17 at 17:49
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    $\begingroup$ Q1) Not every bounded, open set is a finite union of connected sets, but by adding to $\Omega$ at most countable segments you can make it path-connected, and such segments are neglected by $\int$ since they have measure zero. So you may just assume that $\Omega$ is connected at once. $\endgroup$ – Jack D'Aurizio Dec 25 '17 at 17:53
  • $\begingroup$ @JackD'Aurizio: I'm sorry, I don't really understand. $\Omega $ is a union at most countable of connected set ? $\endgroup$ – user386627 Dec 25 '17 at 18:04
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    $\begingroup$ That's straightfoward. $\Omega$ is the union of at most countable balls, the balls are connected sets, the connected components of $\Omega$ cannot be more than countable. $\endgroup$ – Jack D'Aurizio Dec 25 '17 at 18:12

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