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Let $W^{1,p}(U)$ be the Sobolev space, where $U$ is a connected bounded domain in $\mathbb{R}^n$ and $u\in W^{1,p}(U)$ satisfying $Du=0$ a.e. in $U$. Then $u$ is constant a.e. in $U$.

I don't know how to prove this. Especially, I don't know how to use "connected".

Please guide me.

yamada taro
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  • Exactly the same as for usual derivatives but using weak derivatives. Simply rewrite all the proof for the elementary calculus version. It need to be connected because otherwise the values could jump, so it would be constant in each connected component. – user40276 Jul 04 '14 at 07:42

1 Answers1

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Let $\phi_\epsilon$ denote a mollifier and let $u_\epsilon = u*\phi_\epsilon$ is a smooth function in $\Omega_\epsilon := \{x \in \Omega \mid \mathop{\rm dist}(x, \partial\Omega) > \epsilon\}$. As $Du_\epsilon = Du*\phi_\epsilon = 0$ in $\Omega_\epsilon$, $u_\epsilon$ is locally constant in $\Omega_\epsilon$. Hence, as $u_\epsilon \to u$ allmost everywhere, $u$ is locally constant in $\Omega$. As $\Omega$ is connected, $u$ is constant allmost everywhere.

martini
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  • Pardon me if I comment on your old answer, but I'd like to ask you about a variation of the problem showed in this image https://i.imgur.com/yXma55w.png As you can see, in that problem there isn't the assumption $u\in W^{1,p}(U)$, actually it is not written at all which is the domain of u... is the domain of $u$ implicitly assumed somewhere in the problem? – sound wave Jun 12 '21 at 14:45