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?

  • 2
    $\begingroup$ Actually, $c=c_\varepsilon$. Now show that this has a convergent subsequence as $\varepsilon \rightarrow 0$ $\endgroup$
    – Thomas
    Commented Oct 5, 2014 at 8:33
  • $\begingroup$ How can I prove that $c=c_{\varepsilon}$? $\endgroup$
    – avati91
    Commented Oct 5, 2014 at 8:37
  • $\begingroup$ Your $c$ depends on $\varepsilon$ by definition. $\endgroup$
    – Thomas
    Commented Oct 5, 2014 at 9:50
  • $\begingroup$ I don't understand how to proceed $\endgroup$
    – avati91
    Commented Oct 5, 2014 at 10:18

2 Answers 2


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 $0$.

  • $\begingroup$ Is the limit of constant functions always constant? (if it exists) $\endgroup$
    – avati91
    Commented Oct 5, 2014 at 11:42
  • $\begingroup$ @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). $\endgroup$
    – Thomas
    Commented Oct 5, 2014 at 15:20
  • $\begingroup$ @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 $\lim_n c_{\epsilon_n}$. $\endgroup$ Commented Jan 26, 2016 at 3:18
  • $\begingroup$ 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(math.stackexchange.com/questions/1096491/…) for $L^p$ case ) $\endgroup$
    – user74489
    Commented Mar 13, 2018 at 15:32
  • 1
    $\begingroup$ 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..., $\endgroup$ Commented May 14, 2018 at 6:00

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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