Let me consider an $L^1(\mathbb R^N)$ function $f$ such that $$ \int_{\mathbb R^N} f(x) dx =0. $$ Then I claim that the $N$-form $f(x) dx_1\wedge\dots\wedge dx_N$ is closed, i.e. there exists a vector field $X$ such that $$ \text{div} X=f. $$ I would like to pick-up an $X$ in $W^{1,1}(\mathbb R^N)$. Is it possible? Is there a somewhat constructive proof?
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1$\begingroup$ No, there is no reason why $X$ would be integrable, too. For example $f(x)=1/x^2$ for $x\ge 1$, $f(x)=-1/x^2$ for $x\le -1$, and interpolate linearly, say. So $X=\pm 1/x$ for $|x|\ge 1$. $\endgroup$– Christian RemlingJan 9, 2020 at 18:01
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$\begingroup$ @Christian Remling Thanks, you are absolutely right. I could ask more to the function $f$, say $f\in L^p$ so that $X=\Delta^{-1}\nabla f\in W^{1,p}$. Then if $p>N$, it should be true that the continuous function $X$ goes to zero at infinity and thus I could apply Green's Theorem to get an equivalence between null integral for $f$ and the existence of a vector field $X$ going to zero at infinity such that $f=\text{div} X$. $\endgroup$– BazinJan 10, 2020 at 9:56
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