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Dec 9, 2023 at 0:07 history became hot network question
Dec 8, 2023 at 18:03 vote accept Nate River
Dec 8, 2023 at 17:53 answer added user479223 timeline score: 12
Dec 8, 2023 at 17:06 comment added user479223 Weak derivative is always a function by definition. I'm cooking something up...
Dec 8, 2023 at 17:06 comment added Nate River Well that means the weak derivative must not even be a function..
Dec 8, 2023 at 16:53 comment added Nate River Yes the weak derivative (if it even exists) must definitely be unbounded in $L^1$ on every subinterval. Definitely. I think.
Dec 8, 2023 at 16:45 comment added Nate River Hm this is the Lipschitz case - for which I think it is easy to construct a differnetiable a.e. example. Holder though… @user479223
Dec 8, 2023 at 16:20 comment added Nate River I managed to get it differentiable at a point. As usual it is not trivial, if even possible to extend the local construction to a.e.
Dec 8, 2023 at 16:06 history asked Nate River CC BY-SA 4.0