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Let $\Omega \subset \mathbb{R}^N$ is open , $u\in C^1(\Omega)$.

How can we prove that : $i\in \{1,...,N\}$

$$\int_{\Omega}\frac{\partial u}{\partial x_i}(x)v(x)dx=-\int_{\Omega}u(x)\frac{\partial v}{\partial x_i}(x)dx,\forall v \in C_c^1(\Omega).$$

Note : We already know that Green theorem (divergence theorem) can be use if $\Omega$ open , bound has piecewise smooth boundary and the $\Omega$ we have is no information about boundary.

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    It is not true. You have $\int u_{x_i} v \, dx = -\int uv_{x_i} \, dx$. The boundary regularity does not matter since $v$ is compactly supported (you can use the support of $v$ as the domain). – 2017-01-16
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    But if this is not true , then we cann't prove that strong derivative and weak derivative are the same in $C^1(\Omega)$. My idea to prove this is find some $U$ open set that $\partial U$ are piecewise smooth boundary and $supp v \subset U \subset \Omega$. – 2017-01-16
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    I think your question is not worded correctly. Do you mean to say that if $\int_\Omega u_{x_i}v \, dx = 0$ for all $v$, then $u_{x_i}=0$? This is the kind of thing you would need to prove weak and strong derivatives are the same. – 2017-01-16
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    Ok let me edit the question – 2017-01-17

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Since $v$ has compact support, you can replace $\Omega$ by a ball of sufficiently large radius $R$, i.e.

$$\int_\Omega u_{x_i}v \, dx = \int_{B(0,R)} u_{x_i}v \, dx$$

and

$$\int_\Omega uv_{x_i} \, dx = \int_{B(0,R)} uv_{x_i} \, dx$$

provided the support of $v$ is contained in $B(0,R)$. Both $u$ and $v$ are defined to be zero in $B(0,R)\setminus \Omega$. The ball has a smooth boundary so you can use the form of Green's Theorem that you are familiar with.

Basically, if one of the functions is compactly supported, then you can ignore any boundary regularity issues.

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    Thank you ! Nice solution . – 2017-01-17