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Let $u\in W^{1,p}(U)$ such that $Du=0$ a.e. on $U$. I have to prove that $u$ is constant a.e. on $U$.

Take $(\rho_{\varepsilon})_{\varepsilon>0}$ mollifiers. I know that $D(u\ast\rho_{\varepsilon})=Du\ast\rho_{\varepsilon}$, so $u\ast\rho_{\varepsilon}(x)=c $ for every $x\in U$, since it is a smooth function.

How can I conclude?

avati91
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2 Answers2

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By (well-known) properties of mollification, $u*\rho_\epsilon\rightarrow u$ in $W^{1,p}$ as $\varepsilon \rightarrow 0$. Since $u*\rho_\epsilon = c_\varepsilon$ is a constant for each $\varepsilon$, and $u*\rho_\epsilon\rightarrow u$, $u$ is the limit of constant functions and must be constant as well (e.g. because convergence in $L^p$ implies convergence a.e.).

you should note that

i) this is only true for each connected component of $U$

ii) strictly speaking $u*\rho_\epsilon$ is only defined on $U_\varepsilon = \{x\in U: d(x,\mathbb{R}^n\backslash U)>\varepsilon\}$, so you first get the result for any such domain, but then it it true for $U$ if $\varepsilon$ tends to $0$.

Thomas
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  • Is the limit of constant functions always constant? (if it exists) – avati91 Oct 05 '14 at 11:42
  • @avati91 'Limit' always refers to a definition of convergence or a topology, e.g. pointwise convergence or convergence in some norm. The answer to your question depends on the notion of convergence which is used, but in most cases the answer would be yes (again, on each component of the domain of definition). – Thomas Oct 05 '14 at 15:20
  • @avati91, yes, the weak limit of constant functions here is constant. Take a decreasing sequence $\epsilon_n$ that tends to zero. We know that $u\ast \rho_{\epsilon_n}$ converges to $u$ in $L^p_{\text{loc}}(U)$. But unless the numbers $c_{\epsilon_n}$ form a convergent sequence, this $L^p_{\text{loc}}(U)$ convergence cannot happen. Therefore, the constants $c_{\epsilon_n}$ must converge as numbers. Then it is easy to see that the limit of $u\ast \rho_{\epsilon_n}$ in $L^p_{\text{loc}}(U)$ is $\lim_n c_{\epsilon_n}$. – Chuwei Zhang Jan 26 '16 at 03:18
  • Does "convergence in Lp implies convergence a.e" hold in general? In mollification case $u\ast \rho_\varepsilon \to u$ a.e. holds indeed (c.f. Theorem 9.13 and Theorem 7.15 in Zygmund's book, Measure and Integral for $L^1$ case and this(https://math.stackexchange.com/questions/1096491/almost-everywhere-convergence-of-convolution-with-mollifiers#comment2234778_1096491) for $L^p$ case ) – user74489 Mar 13 '18 at 15:32
  • @user74489 See this question: https://math.stackexchange.com/questions/138043/does-convergence-in-lp-implies-convergence-almost-everywhere – Thomas Mar 13 '18 at 17:48
  • Yes, I knew Davide Giraudo's answer in your link. But now I am in addition interested about whether $u\star \rho_\varepsilon \to u$ as $\varepsilon\searrow 0$ almost everywhere, and finding some resource. – user74489 Mar 15 '18 at 01:02
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    I just wanted to comment about a delicate point regarding connectedness in your argument. (I think it might help future readers). The point is that $u \star \rho_{\epsilon}$ is only constant on $U_{\epsilon}$ if we know $U_{\epsilon}$ is connected. This is not always the case, even for arbitrarily small $\epsilon$ (think of a sequence of "forks" with bases which become thinner and thinner). Thus a more local argument is needed, like the one in here. This shows $u$ is locally a.e constant..., – Asaf Shachar May 14 '18 at 06:00
  • and now a gluing argument is needed (see here) in order to finish the proof. – Asaf Shachar May 14 '18 at 06:00
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Suppose $u, \text{weak } u' \in L^p_{loc}(U)$ (this is the general definition of $W^{1,p}(U))$. By $u'=0$ a.e., we know after mollification, $u_{\epsilon}' = C_{\epsilon}$ a.e. in $\Omega_{\epsilon}$. Since $u_{\epsilon} \to u$ in $L^p_{loc}$, then for any compact subset $W$ of $\Omega$, we know the $L^p$ convergence on $W$. Hence $C_{\epsilon} \to u$ on any compact subset. This immediately shows that $C_{\epsilon}$ is a bounded sequence and by Bolzano-Weierstrass, $C_\epsilon$ admits a convergent subsequence. Fix this convergent subsequence, and WLOG assume $C_{\epsilon} \to C$ as a sequence of real number. On $W$, $\vert\vert u-C \vert\vert_p \leq \vert\vert u-C_\epsilon \vert\vert_p + \vert\vert C_\epsilon - C\vert\vert_p$, the right hand side tends to $0$ by $L^p_{loc}$ convergence of $u_{\epsilon}$ and $W$ being of finite measure. Then $u=C$ a.e. on $W$. If here we assume $U$ is a bounded open domain, then by $\bar{\Omega}_\epsilon$ compact in $U$ and increasing to $U$, we can conclude the result by taking $\epsilon \to 0$.

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