I think there is an (possible) error in Villani's monograph titled "Hypocoercivity". To be specific, in page 48, he wrote "For the second term in (6.9), we use the identity $$\nabla\cdot(Df\nabla u) = f\nabla\cdot(D\nabla u) + \langle D\nabla u,\nabla f\rangle = f\nabla\cdot(D\nabla u) + f\langle D\nabla u,\nabla \log f\rangle.$$ So $$-\int f\left\langle \frac{C\nabla\cdot(Df\nabla u)}{f}, C'u\right\rangle = -\int f\left\langle C\nabla\cdot(Df\nabla u), C'u\right\rangle - \int f\left\langle C\left\langle D\nabla u,\nabla \log f\right\rangle_{\mathbb{R}^N},C'u\right\rangle_{\mathbb{R}^m}". $$ Here are some background: $C = (C_1,\ldots,C_m)$ and $C' = (C'_1,\ldots,C'_m)$ are $m$-turples of derivation operators on $\mathbb{R}^N$, $u = \log(f) + E$ with $E \in C^2(\mathbb{R}^N)$. I think the author made a serious mistake as components of $C$ are derivation operators on $\mathbb{R}^N$, thus (I think) we should have that \begin{align*} C\nabla\cdot(Df\nabla u) = C\left(f\nabla\cdot(D\nabla u) + f\langle D\nabla u,\nabla \log f\rangle\right) &= f\,C\nabla\cdot(Df\nabla u) + f\,C\left\langle D\nabla u,\nabla \log f\right\rangle \\ &\quad+ \mathrm{two~additional~terms} \end{align*} So why should we have $$-\int f\left\langle \frac{C\nabla\cdot(Df\nabla u)}{f}, C'u\right\rangle = -\int f\left\langle C\nabla\cdot(Df\nabla u), C'u\right\rangle - \int f\left\langle C\left\langle D\nabla u,\nabla \log f\right\rangle_{\mathbb{R}^N},C'u\right\rangle_{\mathbb{R}^m} $$ without two additional terms on the right hand side?
1 Answer
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On the PDF version on his website, he writes the left-hand side $$-\int f\left\langle C\left(\frac{\nabla\cdot(Df\nabla u)}{f}\right), C'u\right\rangle$$ which probably means it was a typo, the $C$ should be down the fraction. And there, you don't have this problem.
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1$\begingroup$ Thanks! I downloaded the version from google scholar, I didn't realize that he also have this material in his website $\endgroup$– Fei CaoCommented Mar 13, 2021 at 20:17
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1$\begingroup$ If it's available on Google Scholar, it usually means that it's open-access, which generally means the author is happy to share it, including the typo corrections. :) $\endgroup$ Commented Mar 13, 2021 at 20:22
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$\begingroup$ Thanks! I will keep that in mind $\endgroup$– Fei CaoCommented Mar 13, 2021 at 20:38
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$\begingroup$ If you don't mind, may I ask an additional question from the PDF version you just shared? In page 44 (toward the end of the proof of Lemma 32), I think Villani implicitly used the identity $\langle (CAu)\cdot(Au),C'u\rangle = \langle Au,\langle CAu, C'u\rangle \rangle$. As he mentioned earlier in page 42, $(CAu)\cdot(Au)$ means the the evaluation of the matrix $CAu$ on the vector $Au$, but if that's the case, I don't see why he can use the first identity I stated here (unless the matrix $CAu$ is symmetric...) $\endgroup$– Fei CaoCommented Mar 14, 2021 at 5:38