Weak derivative zero implies constant function

Question

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?

Answer 1

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 $0$.

Answer 2

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 $\epsilon \to 0$.