Is the $p-$Laplacian defined in case of a vanishing first derivative, that is $Df(x)=0$ in $x\in\mathbb{R}^n$ for some smooth $f$? In case of $p\geq 2$ it should, but for $p<2$ I am not sure as to how to this would be defined. If $p\geq2$ it should always be zero as $\Delta_pf = \mathrm{div}(|Df|^{p-2}Df)$ or do I misunderstand something? Or does it depend on the second derivative (the Hessian) as in case of $p=2$ for example it would be $\mathrm{Tr}(D^2f)$ and thus not even (not directly at least) depend on the first derivative?
2026-04-03 01:55:18.1775181318
Understanding the definition of the $p-$Laplacian
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