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What property does imply that a function $f \in L^1_{loc}(\Omega)$ ($\Omega \subset \mathbb{R}^n$) is weakly differentiable, namely there exists $g \in L^1_{loc}(\Omega)$ such that $\int_{\Omega} f\partial_i \phi = - \int_\Omega g\phi$, for all $\phi \in C^\infty_c(\Omega)$ and all $i=1,\ldots,n$? Clearly, if $f \in C^1_c(\Omega)$, then it is weakly differentiable, but is there a weaker assumption that will still imply it? Thank you.

Frank Zermelo
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    There indeed exists a complete characterization of weakly differentiable functions. In one dimension, they are exactly the absolutely continuous ones. In more dimensions, things are more complicated as one should check absolute continuity along every direction. But somebody did it. You can try looking for "APL property" (or something like that - I don't remember exactly) in the book on Sobolev spaces by Giovanni Leoni. – Giuseppe Negro Feb 23 '15 at 10:24
  • I found it: http://math.stackexchange.com/a/817440/8157 – Giuseppe Negro Feb 23 '15 at 10:31
  • By the way - the correct name is "ACL property", not "APL property". – Giuseppe Negro Feb 23 '15 at 10:34
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    The characterization if proposed as an exercise in the book written by Gilbarg and Trudinger, Exercise 7.8. If $\Omega$ is a domain, a function $u$ is weakly differentiable if and only if it is equivalent to a function $\bar{u}$ that is absolutely continuous on almost every line segments in $\Omega$ parallel to the coordinate axes and whose partial derivatives are locally integrable in $\Omega$. – Siminore Feb 23 '15 at 10:37
  • Thank you. That is really interesting. I did not know about the ACL property. – Frank Zermelo Feb 23 '15 at 13:36
  • For future reference, the characterization "weakly differentiable = absolutely continuous" is very misleading, if not wrong. A counterexample is $\mathbb{1}_{\mathbb{Q}\cap [0,1]} $ which is weakly differentiable but clearly not AC. As @Siminore luckily pointed out (thanks!) it's probably more like "weakly differentiable = its precise representative is absolutely continuous" (correct me if I'm wrong please). – rod Sep 25 '23 at 13:58

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