Weak derivatives equals zero

Question

Im just learning sobovel space, I was wondering if the weak derivate holds similar things of the original derivate.

Let $U\subset \mathbb{R^n}$ is a open set, and $u\in W^{1,p}$

if $$Du=0 \ \ a.e$$

implies $u=\mathrm{const}?$

Answer 1

If $\phi$ is a $C^\infty$ function with compact support then the convolution $u*\phi$ is also smooth, and $D(u*\phi)=(Du)*\phi=0$. So $u*\phi$ is locally constant by the usual calculus argument. Now let $\phi$ be a standard mollifier and let it converge to a delta, so that $u*\phi$ converges to $u$ a.e. (or in $L^1_{\text{loc}}$ if you prefer). So $u$ is locally constant.