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In wikipedia they say that : if $\Omega \subset \mathbb R^n$ is open and bounded, then there is $C>0$ s.t. for all $u\in W^{1,p}_0(\Omega )$, $$\|u\|_{L^p}\leq C\|\nabla u\|_{L^p}.$$

I agree that many of proofs use the fact that $u(x)=0$ for at least one $x\in \bar\Omega $. But using Reilleich-Kondrachov compacity theorem, it looks like that we can avoid this fact.

Indeed, suppose that there is $(u_n)\subset W^{1,p}(\Omega )$ s.t. $\|v\|_{L^p}=1$ and $\|\nabla v_n\|_{L^p}\leq \frac{1}{n}$. In particular, $(u_n)$ is bounded in $W^{1,p}(\Omega )$ and thus has a subsequence (still denoted $(u_n)$) that converges weakly. Let denote $u$ the limit. Since $\|\nabla u_n\|_{L^p}\to 0$, then $\|\nabla u\|_{L^p}=0$, and thus $u$ is constant a.e. The only constante function that is $L^p$ is $0$, therefore $u=0$. By Reilleich Kondrachov, $u_n\to 0$ strongly in $L^p$, this contradict that $\|u\|=1$. Is this wrong ?

user657324
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  • If you're in a bounded domain all constant functions are integrable. – Jose27 Apr 06 '19 at 19:10
  • On the other hand Rellich might fail in unbounded domains. – Jose27 Apr 06 '19 at 19:12
  • @Jose27: Oh, yes, you write. So, I need to have one $x\in \Omega $ s.t. $u(x)=0$... But since a function in $W^{1,p}(\Omega )$ is defined a.e., does it make really sens to say that there is $x$ s.t. $u(x)=0$ ? – user657324 Apr 06 '19 at 21:34

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