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This answer shows that a locally integrable function on a domain $\Omega$ that has continuous weak partial derivatives must be equal to a $C^{1}$ function a.e. My question is, does this hold for distributions in general? Specifically, if we have a distribution $u\in \mathcal{D}'(\Omega)$ such that each partial (distributional) derivative $\partial_{i}u$ is a continuous function, does this imply $u$ is a $C^{1}$ function?

This would seem to be true in one dimension, at least on an interval, since in this case the $C^{1}$ function $$v(x) = \int_{a}^{x}u'(x)\,dx,$$ has derivative $u'$ (classically and as a distribution), so $u$ is equal to a $C^{1}$ function plus some constant. But it is not obvious to me whether this extends to higher dimensions.

Justin Toyota
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  • We can define two function $v_1(x,y) = \int_{a_1}^x \partial_1 u(t,y) , dt$ and $v_2(x,y) = \int_{a_2}^y \partial_2 u(x,t) , dt$ but they don't have to be equal. – md2perpe Dec 01 '23 at 19:26

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It would seem the answer is yes. The answer posted in the question shows that the claim is true if the distribution in question is locally integrable, so we need only show that a distribution with continuous partial derivatives must be locally integrable. This is a corollary of this answer by Piotr Hajlasz, which shows that a distribution on $\Omega$ whose partial derivatives are all in $L_{\text{loc}}^{p}(\Omega)$ ($1\leq p\leq \infty$) must also lie in $L_{\text{loc}}^{p}(\Omega)$.

Justin Toyota
  • 328