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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?

LL 3.14
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HelloEveryone
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    Warning, $f(x)=0$ at a given point $x$ does not imply $f'(x)=0$ at this point. Similarly here, $Df(x)=0$ does not imply $\mathrm{div}(|Df|^c ,Df)(x)=0$ – LL 3.14 Jan 02 '23 at 09:50
  • Ah of course thank you, I somehow completely ignored (misunderstood) the divergence! – HelloEveryone Jan 02 '23 at 10:35

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