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Let $W^{1,p}(U)$ be the Sobolev space. Suppose that $U$ is connected bounded domain in $\mathbb{R}^n$ and $u \in W^{1,p}(U)$ satisfies $Du=0$ a.e. in $U$. How can I prove that $u$ is constant a.e. in $U$?

Pooya
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1 Answers1

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For a Dirac-family $(\varphi_\varepsilon)_\varepsilon$ with $\mathrm{supp}\, \varphi_\varepsilon \subseteq \overline{B_\varepsilon(0)}$ you have $D(u*\varphi_\varepsilon) = (Du)*\varphi_\varepsilon = 0$ on all open sets $O \subseteq U$ with $\mathrm{dist}(O,\partial U) > \delta$, with $\delta > \varepsilon$, so $u * \varphi_\varepsilon$ is constant on all connected ones. Also, $u*\varphi_\varepsilon \to u$ in $L^p$ for $p \geq 1$, in particular $u * \varphi_\varepsilon \to u$ almost everywhere for some subsequence. But this means $u$ is constant almost everywhere on all $O$ with $\mathrm{dist}(O,\partial U) > 0$, so it is constant almost everywhere on $U$.

Thomas
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  • Where do you use connectedness of $U$? –  Sep 27 '16 at 14:18
  • The $O$'s should be connected too, so $D(u\varphi)=0$ there implies $u\varphi$ constant. – Thomas Sep 28 '16 at 07:03
  • It is intuitively quite true. Do you have a brief topological argument that why each $O$ is connected also? –  Sep 28 '16 at 12:11
  • I clarified my answer above. The first equality holds for all $O$, but the function is only constant on connected ones. – Thomas Sep 28 '16 at 16:24
  • why $u\ast \varphi_{\varepsilon}\to u$ in $L^{\infty}$? – Mr.xue Mar 12 '21 at 10:57
  • However, we can avoid the question by choosing $\varepsilon $ as a countable series tends to $0$. – Mr.xue Mar 12 '21 at 11:03