# $\nabla u=0$ implies $u$ is constant everywhere or almost everywhere?

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 $u\in \mathcal C_0^1(\Omega )$

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

Proof

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 $\Omega$ ?

• The elements of Sobolev spaces are not defined pointwise, so we may only have almost-everywhere something. – Jack D'Aurizio Dec 25 '17 at 17:34
• @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 ? – user386627 Dec 25 '17 at 17:49
• 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. – Jack D'Aurizio Dec 25 '17 at 17:53
• @JackD'Aurizio: I'm sorry, I don't really understand. $\Omega$ is a union at most countable of connected set ? – user386627 Dec 25 '17 at 18:04
• 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. – Jack D'Aurizio Dec 25 '17 at 18:12