A sobolev function with zero differential is constant

Question

Let $U$ be an open subset of $\mathbb{R}^n$:

Suppose that $u \in W^{1,p}(U)$ satisfies $Du=0$ almost everywhere in $U$.

How to prove $u$ is constant a.e in $U$?

Answer 1

We have to work with mollifiers $(\rho_\epsilon)_{\epsilon > 0}$, where we can start with $\epsilon$ sufficiently small for each point in $U$ so that the function is well defined.

It is known that as $\epsilon \to 0$, we have in $W^{1,p}$, that $u * \rho_\epsilon \to u$. Since $D(u * \rho_\epsilon) = Du * \rho_\epsilon = 0$, we have that $u * \rho_\epsilon$ is constant, since it is a smooth function with derivative $0$. Let this constant be $c_\epsilon$.

Hence, $u$ is the limit of constant functions, hence is also constant, since convergence in $W^{1,p}$ implies convergence in $L^p$, which implies convergence a.e.

It's worth noting that this is true only if $\Omega$ is connected, otherwise the function could take different constants in different connected components.

Answer 2

Since $D u = 0$, you have $u \in W^{k,p}(\Omega)$ for all $k \ge 1$. By Sobolev embedding, $u$ is smooth. Now, it is easy to check that $u$ is constant (in each connected component of $\Omega$).