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May 21, 2020 at 15:01 vote accept Jingeon An-Lacroix
May 20, 2020 at 20:53 answer added Piotr Hajlasz timeline score: 14
May 20, 2020 at 19:48 comment added Bazin The Sobolev space $W^{1,1}$ of $f\in L^1$ such that the distribution derivative $f'$ belongs also to $L^1$ is a nice space (by the way included in $L^{\frac{d}{d-1}}$). The space $BV$ of $f\in L^1$ such that the distribution derivative $f'$ is a Radon measure is larger (and nice as well).
May 20, 2020 at 17:40 comment added Christian Remling In one dimension, $f'$ (distributional derivative) being locally integrable is equivalent to $f$ being absolutely continuous. The result is discussed in my distribution lecture notes here: math.ou.edu/~cremling/teaching/ln.html
May 20, 2020 at 17:09 history asked Jingeon An-Lacroix CC BY-SA 4.0