Let $u\in BV(\mathbb{R}^N; \mathbb{R}^M)$. How does one prove that $$\operatorname{curl} Du = 0$$ holds in the sense of distributions?
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5$\begingroup$ Not completely sure what you mean by curl and $D$ here, but leaving that aside, the identity (for smooth functions) just follows from the fact that mixed partial derivatives can be taken in any order, which of course holds for distributions as well. $\endgroup$– Christian RemlingCommented Apr 13, 2019 at 20:50
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$\begingroup$ @ChristianRemling You're completely right. Thank you. $\endgroup$– RikuCommented Apr 13, 2019 at 21:03
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$\begingroup$ @Riku : Note that \operatorname{} has context-dependent spacing and \mathrm{} does not. Hence my edit. $\endgroup$– Michael HardyCommented Apr 14, 2019 at 5:14
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$\begingroup$ @ChristianRemling If you write down that in an answer. I'd gladly mark it as accepted. $\endgroup$– RikuCommented Apr 15, 2019 at 12:16
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$\begingroup$ @Riku: Thanks, but this remark doesn't seem to have enough substance for an answer, I'll leave it up as a comment. $\endgroup$– Christian RemlingCommented Apr 15, 2019 at 15:51
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