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Dec 21, 2022 at 0:57 comment added Piotr Hajlasz @shuhalo You can find functions in $W^{1,n}$ that there are nowhere continuous and hence they cannot be differentiable at any point. However, existence of partial derivatives does not imply differentiability or even continuity. No contradiction here. The characterization os Sobolev spaces through absolute continuity on lines is the best geometric characterization you can get.
Dec 21, 2022 at 0:09 comment added shuhalo Thank you for this exposition. However, I am wondering how this can be compatible with the fact that there are nowhere differentiable functions in Sobolev spaces. For example: metaphor.ethz.ch/x/2021/fs/401-3462-00L/sc/extras/… Can you comment on that?
May 8, 2022 at 19:09 comment added Holden Lyu I’ve never learned this property before. As you said, it’s a totally different way to think about Sobolev functions and it’s so amazing. Thank you so much for your help!
May 8, 2022 at 13:12 comment added Piotr Hajlasz @user734979 It is correct provided if you are aware that you are using the Fubini theorem in your argument. I still like to think about Sobolev functions in terms of absolute continuity on lines since it gives you many results directly without necessity of approximation.
May 8, 2022 at 6:09 comment added Holden Lyu Thank you so much for your help! I’ve followed the way you told to prove it and it’s amazing. By the way, I’ve just updated my proof above, would it be a proof now?
May 8, 2022 at 6:06 vote accept Holden Lyu
May 7, 2022 at 23:33 history answered Piotr Hajlasz CC BY-SA 4.0