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Let $U \subset \Bbb R^n$ be open, $1 \le i \le n$. Then

$\int_U$ $d \phi(x) \over dx_i $ $d^n x = 0$ for every $\phi \in C_c^1(U)$.

Proof:

We can assume that $U = \Bbb R^n$. Furthermore, it is enough to prove the statement for $i = 1$. So let $\phi \in C_c^1(\Bbb R^n)$. We choose $R \in \Bbb R_+$ so that $Supp(\phi) \subset [R, -R]^n$ holds. Then, for every $(x_2, ..., x_n) \in \Bbb R^{n-1}$, it holds that

$\int_{\Bbb R}$ $d\phi \over dx_1$ $(x_1, ..., x_n) dx_1 = \phi(x_1, x_2, ..., x_n)_{x_1 = R}^{x_1 = -R} = 0$.

How does he conclude the last two steps here? Do integration and differentiation simply "eliminate" each other? If so, why do we still put in the values $R$ and $-R$? And why does this whole thing equal $0$?

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    Why don't you use by parts?? – 2017-01-17
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    This isn't my proof. – 2017-01-17
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    Well what happens to compactly supported functions at the boundary? – 2017-01-17
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    They equal $0$? – 2017-01-17
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    Exactly @Julian. – 2017-01-17
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    Ah, well, that makes sense. I'm still a little bit confused by the second to last equality though. How do the integral and differential operator interact here? It seems like they eliminate each other, but this can't be the whole truth since we still put in the values $R$ and $-R$. – 2017-01-17
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    What happens when you evaluate $ \int_{0}^{1} f'(x) $ ?? – 2017-01-17
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    Ah, you just get $f(x)_0^1.$ :-) – 2017-01-17

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