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A function $f\in W^{1,2}_{loc}$ is in the space $W^{1,2}_{-\tau}$ if $\int_{\mathbb{R}^n} f^2|x|^{2\tau-n}<\infty$ and $\int_{\mathbb{R}^n} |\partial_kf|^2|x|^{2\tau+2-n}<\infty$ for all $k=1,2,\dots n$.

Given a vector field $Z$ in a space of $W^{1,2}_{-\tau}$, $\tau>0$. Assume ($\mathbb{R}^n,\ g=g_{Euc}$), I would like to know if the following inequality holds. $$\int_{\mathbb{R}^n} |Z|^2||x|^{2\tau-n} \leq C \int_{\mathbb{R}^n} |L_Zg|^2|x|^{2\tau+2-n}.$$ Where $L_{Z}g$ refers to Lie derivative.

This inequality is actually found in a paper by J. Corvino and R. Schoen's paper on VEE (P.196-198). https://projecteuclid.org/download/pdf_1/euclid.jdg/1146169910 But for the vector field part, the proof is omitted.

Thank you very much!

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    $\begingroup$ That can't be right. There are such things as Killing vector fields, especially for $g_{Euc}$. Is there any restrictions on $\tau$ that would rule them out? Or maybe can you assume that $Z$ is orthogonal to the Killing fields? The estimate with $L_Zg$ replaced by $\nabla Z$ is essentially just some version of Hardy's inequality. $\endgroup$ Oct 23, 2017 at 20:02
  • $\begingroup$ Thank you for your reminder. I should add that $\tau>0$ is assumed. Hence, killing vector field is not in the weighted sobolev space. And hence, it cannot be a counter example to the inequality. $\endgroup$
    – Leonardo
    Oct 28, 2017 at 13:57
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    $\begingroup$ Am I wrong in believing that if you integrate the right side by parts twice, you'll get something like $$\int (2|\nabla X|^2 + |\nabla\cdot X|^2)|x|^{...} + \text{ lower order terms?}$$ $\endgroup$
    – Deane Yang
    Oct 28, 2017 at 19:24
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    $\begingroup$ With the given weights, approximate $X$ by compactly supported smooth vector fields in the $W^{1,2}_{-\tau}$ norm, and integrate by parts as @DeaneYang suggested. This shows that the right hand side is coercive on $|\nabla Z|^2$ up the lower order terms, all of which you can absorb into the left hand side by enlarging the constant, after applying the Hardy's inequality as in the scalar case. Finally take the limit. $\endgroup$ Oct 29, 2017 at 15:27
  • $\begingroup$ The right hand side is equal to $\int (\sum_{i,j} 2Z_{,ij}^2 +2Z_{i,j}Z_{j,i}) |x|^{a}$. For the square terms, I can use scalar version to control it. But I cannot get the idea for the cross term part. It seems that the cross term have no lower bound. $\endgroup$
    – Leonardo
    Oct 30, 2017 at 3:00

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