Gauss–Ostrogradsky formula for Distributions Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $u\in W^{1,p}(\Omega)$, $v\in L^{p'}(\Omega)^N$ with $p\in (1,\infty)$. Let $\operatorname{div}(v)$ be the divergence of $v$ in the sense of distribtuion and suppose that $\operatorname{div}(v)=0$. Assume that $B$ is any ball in $\Omega$. How can one prove that $$\int_B v\cdot\nabla udx=\int_{\partial B}(v\cdot\nu)ud\Gamma$$ 
where $\nu$ is the outward normal and $\Gamma$ de surface measure in $\partial B$? 
This is claim $(7)$ from the paper of Zhikov : On the compansated compacness principle. I think I can prove it for almsot ball in $\Omega$ by using Fubini's theorem and mollifiers, but he estates the result for all $B$. Maybe there is a straighforward approach to it, but I can't see it.
Thank you
 A: After a long search, I found the answer in Constantin's book. I will post a more general answer here. Let $\Omega\subset\mathbb{R}^N$ be a bounded domain with Lipschitz boundary. Let $\gamma_{0,p}:W^{1,p}(\Omega)\to W^{1-1/p,p}(\partial\Omega)$ be the trace operator and $\ell_p:W^{1-1/p,p}(\partial\Omega)\to W^{1,p}(\Omega)$ the lift operator.
The lift operator is continuous and determined by, $u=\ell_p v$ $$
 \left\{ \begin{array}{ccc}
 \Delta_pu=0 &\mbox{ in $\Omega$} \\
  \gamma_{0,p}u=v &\mbox{ in $\partial\Omega$}
       \end{array} \right.
$$
Let $E_p=\{u\in L^p(\Omega)^N:\ \operatorname{div}u\in L^p(\Omega)\}$.
Note that $E_p$ is a Banach space with the norm $$\|u\|_E=\|u\|_p+\|\operatorname{div}u\|_p$$
Define $\gamma:E_p\to W^{-1+1/p,p}(\partial\Omega)^N$ by $$\langle \gamma (u),v\rangle_{W^{-1+1/p,p},W^{1-1/p',p'}}=\int_\Omega \ell_p(v)\operatorname{div}u+\int_\Omega \nabla(\ell_p (v))\cdot u $$
It follow that (by using the continuity of $\ell_p$) $$|\langle\gamma(u),v\rangle|\leq c\|u\|_E\|v\|_{W^{1-1/p',p'}} $$
which implies that $\gamma$ is a bounded linear operator. Now, we note that if $u\in C^\infty(\overline{\Omega})^N$ and $v\in C^\infty(\overline{\Omega})$, then, by the divergence theorem we have that $$\tag{1}\int_{\partial\Omega} (u\cdot\nu)v_{|\partial\Omega}=\int_\Omega v\operatorname{div}u+\int_\Omega \nabla v\cdot u $$ 
On the other hand, $$\tag{2}\langle \gamma (u),v_{|\partial\Omega}\rangle_{W^{-1+1/p,p},W^{1-1/p',p'}}=\int_\Omega \ell_p(v_{|\partial\Omega})\operatorname{div}u+\int_\Omega \nabla(\ell_p (v_{|\partial\Omega}))\cdot u $$
From $(1)$ and $(2)$, we conclude that (note that $v-\ell_p(v_{|\partial\Omega})\in W_0^{1,p'}(\Omega))$ $$\langle\gamma(u),v_{|\partial\Omega}\rangle=\int_{\partial\Omega} (u\cdot\nu)v_{|\partial\Omega},\ \forall\ v\in C^\infty(\overline{\Omega})$$
From the last equality, we conclude that $\gamma(u)=u\cdot\nu$ for every $w\in C^\infty(\overline{\Omega})^N$.
