# Weak derivative zero implies constant function

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?

• Actually, $c=c_\varepsilon$. Now show that this has a convergent subsequence as $\varepsilon \rightarrow 0$ – Thomas Oct 5 '14 at 8:33
• How can I prove that $c=c_{\varepsilon}$? – avati91 Oct 5 '14 at 8:37
• Your $c$ depends on $\varepsilon$ by definition. – Thomas Oct 5 '14 at 9:50
• I don't understand how to proceed – avati91 Oct 5 '14 at 10:18

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 $0$.

• Is the limit of constant functions always constant? (if it exists) – avati91 Oct 5 '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 5 '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 3:18
• 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(math.stackexchange.com/questions/1096491/…) for $L^p$ case ) – user74489 Mar 13 '18 at 15:32
• @user74489 See this question: math.stackexchange.com/questions/138043/… – Thomas Mar 13 '18 at 17:48