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Let $w\in W^{1,p}(\Omega)$ for some $p>1$ and $\Omega\subset\mathbb{R}^N$ be an open, bounded and connected Lipschitz domain. If $w>0$ a.e. on $\Omega$ is it true that $\nabla\left (\log w\right )\in L^1_{\text{loc}}(\Omega)$ and moreover:

$$\nabla\left (\log w\right )=\dfrac{\nabla w}{w}$$

The chain rule did not work since the natural logarithm is not a Lipschitz function.

This question is related to other questions of mine that are still unanswered (or have very unclear answer).

https://math.stackexchange.com/questions/4961238/how-to-prove-that-a-quotient-function-is-constant

Intriguing simple question about Sobolev space $W^{1,p}(\Omega)$

Here Iosif Pinelis gave an answer that is very unclear to me...and I feel that it is not true, because of this kind of counterexamples: https://math.stackexchange.com/questions/4960647/weak-derivative-of-a-quotient

The reason why I insist so much with it is that in many articles I see it used as a straightforward thing...but for months I didn't succeed in proving it. Here is one of the articles I talk about: J.Giacomoni - Quasilinear parabolic problems with variable exponents qualitative analysis and stabilization, Communications in Contemporary Mathematics, vol. 20, no.8(2018), page 25:

enter image description here

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  • $\begingroup$ Irrespective of integrability, it is not hard to prove that the identity holds pointwise a.e. e.g. using a cutoff argument. Depending on how your cited equation is used, this might already be good enough. $\endgroup$ Commented Feb 24 at 13:35
  • $\begingroup$ In the article it is said that from $\nabla (v_2/v_1)=0$ it follows that $v_2=cv_1$ a.e. on $\Omega$, which is not true as it seems... $\endgroup$ Commented Feb 24 at 13:38

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I believe the answer is no. Take $w(x) = |x|$ on $(-1,1)$. Then by the fact that the natural logarithm is continuously differentiable on $(0, \infty)$ and the classical chain rule, the classical derivatives of $w$ exist and equal $\frac{1}{x}$ on $(-1, 1)\setminus \{0\}$, which is not in $L^1_{\text{loc}}$.

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  • $\begingroup$ So the printscreen I put there is not true? $v_1$ and $v_2$ are like $w$ (as conditions) $\endgroup$ Commented Feb 24 at 13:09
  • $\begingroup$ If I take on $(-1,1)$ $v_1=|x|$ and $v_2(x)=\begin{cases} x, x>0\\ -2x, x<0\end{cases}$. Then $\dfrac{\nabla v_1}{v_1}=\dfrac{\nabla v_2}{v_2}=\dfrac{1}{x}$. But $\dfrac{v_1}{v_2}\in\{1,1/2\}$ Am I right? $\endgroup$ Commented Feb 24 at 13:14

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