# A special case of the Divergence theorem

I am interested in the following statement:

Let $F$ be a vector field in $\mathbb{R}^n$ that is $C^1$-smooth in a domain $U$, continuous up to the boundary $\partial U$, and vanishing on $\partial U$. Then $\int_U div(F)=0$.

Naively, if $U_n$ is a smooth domain approximating $U$ from inside and $|F|\leq 1/n$ on $\partial U_n$, then the regular divergence theorem gives $|\int_{\partial U_n} F\cdot N| \leq Area(\partial U_n)/n$ which might very well diverge. So it looks like one needs a finite perimeter condition. However, is there a known example of a construction in a case of non-finite perimeter where the integral of the divergence is not zero?

• do you allow unbounded sets? Nov 29, 2013 at 20:08
• Sure, but $U$ bounded would be better. Nov 29, 2013 at 23:11
• If you let $F(x,y)=(xy,0)$ and $U=\{(x,y):y>0\}$,then that would be an easy counterexample when $U$ is unbounded, but in that case I would want $F$ to "vanish at infinity" as well. Nov 30, 2013 at 5:06

I will show that if $U$ is open and bounded, $f \in C^1 (U) \cap C(\bar{U})$, $f=0$ on $\partial U$ and $v \in \mathbb{R}^n$, then $$\int_U \partial_v f = 0,$$ where $\partial_v f$ is the directional derivative of $f$ in the direction $v$.
The proof goes as follows. First, the statement can be proved in the one-dimensional case by writing the open set $U$ as a countable collection of intervals, applying the fundamental theorem on each interval.
If $U$ is unbounded, the statement still holds if the directional derivative $\partial_v f$ is absolutely integrable: $\int_U \lvert \partial_v f\rvert < \infty$.
In fact it can be proved under these conditions the extension of $f$ to the whole space $\mathbb{R}^n$ by $0$ is weakly differentiable.
• Yes. It would also work under the assumption that $f \in W^{1, 1}_0 (U)$, where $W^{1, 1}_0 (U)$ is the completion of $C^1$ functions with compact support with respect to the Sobolev norm $\int_U \vert \nabla u\vert$. Dec 11, 2013 at 11:16
• Actually, this might not work after all. Because Fubini's Theorem might not hold for this $f$. Apr 2, 2014 at 20:34