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 ?
Added
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 $ ?