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I often come across instances in texts where people calculate the weak derivative of $|u|^s$ for $s>1$ as $s|u|^{s-1} \operatorname{sign}(u) \partial_x u$ for some $u\in W^{1,s}(\Omega)$.

However, as far as I understand, to apply such a chain rule, one needs $|u|^s$ to be globally Lipschitz on $\mathbb{R}$, which is not the case.

I am relatively new to the field of Sobolev spaces and PDEs and am not yet familiar with common techniques.

Rather than spending too much time pondering this problem without guidance, I thought I would ask this question here.

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    it is weak derivative, why do you need global Lipschitz. – Yimin Dec 24 '23 at 17:52
  • @Yimin why you think that we can neglect global Lipschitz here. See: https://math.stackexchange.com/questions/1110231/chain-rule-in-the-sobolev-space-w1-p – MackeyTopology Dec 24 '23 at 19:58
  • The Lipschitz condition enforces the weak derivative in $L^s$, if that is what OP needs, then it is true. My understanding is, for some PDE related problems. the weak derivative is supposed to be just in some $L^{p}$, $p>1$ (dimension at least 2), then the Lipschitz condition might be unnecessary. @MackeyTopology – Yimin Dec 24 '23 at 21:09
  • @Yimin how do you verify that this is the weak derivative. Can you provide an answer. Regardless in which space this may exist. Just want to have a hint. – MackeyTopology Dec 29 '23 at 21:51
  • @MackeyTopology I assume you have an a.e. pointwise converging sequence to $u$ (And a.e derivative to $\nabla u$), also convergent in $W^{1,s}$ norm. Then by Rellich Kondrachov theorem, the "weak derivative" is in some L^p if the dimension is not 1. So the sequence's corresponding derivative is also in $L^p$ bounded from above, then use theorem 3 of https://www.jstor.org/stable/2319009?seq=3#page_scan_tab_contents – Yimin Dec 29 '23 at 23:12

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