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For $(x,v),(x',v') \in \mathbb{R}^{2d}$, if we set $$r\left((x,v),(x',v')\right) = \alpha|x-x'| + |x-x' + \gamma^{-1}(v-v')|,$$ where $\alpha,\gamma \in (0,\infty)$ are fixed constants. It is mentioned (without any proof) in this paper [1], arXiv pdf link that $$|(x,v) - (x',v')|^2 \leq (1+\gamma)^2\max(1,\alpha^{-2})r\left((x,v),(x',v')\right)^2.$$ The advertised inequality can be found in page 26 (equation (5.15)), where the definition of $r$ appears in page 8 (equation (2.9)). Any help is greatly appreciated!

Edit: This inequality is ultimately used to control the new Wasserstein semi-metric $W_\rho(\mu,\nu)$ in terms of the standard $L^2$ Wasserstein distance $W^2(\mu,\nu)$. To be more precise, it is used to obtain (2.23) in Corollary 2.6 from (2.17) in Theorem 2.3.

[1]: Eberle, Andreas; Guillin, Arnaud; Zimmer, Raphael, Couplings and quantitative contraction rates for Langevin dynamics, ZBL07114709.

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    $\begingroup$ why the downvote...? I feel this is not a easy question. $\endgroup$
    – Fei Cao
    Jul 26, 2021 at 4:45
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    $\begingroup$ I have voted to reopen because the question has clear context. I mean, asking for help in a tricky step in a paper is an excellent question! The only way I could see that the question could be improved would be if the OP stated precisely where in the paper the problem step is - but noone commented to say this, so I don't know what people were after. $\endgroup$
    – user1729
    Jul 28, 2021 at 10:52
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    $\begingroup$ @user1729 I have seen some meta comments/posts claim a good source is reasonable context, but of course not everyone agrees. The paper is not that long, so on a $<1$ minute glance-through I found that the inequality in question is $(5.15)$, pg 26, where $r((x,v),(x'v'))$ is defined in $(2.9)$, pg 8 (on the arXiv ver.) $\endgroup$
    – Calvin Khor
    Jul 28, 2021 at 11:18
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    $\begingroup$ @FeiCao: the one sentence explaining that the inequality helps to control $W_\rho$ in terms of $W^2$ is the sort of thing that would add a lot of "context" to your post. I recommend adding it to the question body. $\endgroup$
    – Calvin Khor
    Jul 28, 2021 at 17:37
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    $\begingroup$ @CalvinKhor thanks, I have included what you suggested! $\endgroup$
    – Fei Cao
    Jul 28, 2021 at 17:45

1 Answer 1

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WLOG $x'=v'=0, g=\gamma, a =\alpha,$ , and $r(x,v)=a|x|+|x+g^{-1}v|$.

Now $$|(x,v)|^2=|x|^2+|v|^2,$$ and the first term is easy to bound: \begin{align} |x|^2 &= a^{-2} a^2|x|^2\\ &\le a^{-2} (a|x|+|x+g^{-1}v|)^2\\&= \max(1,a^{-2}) r(x,v)^2. \end{align}For the second:\begin{align}|v|^2 &= g^2 |g^{-1}v|^2 \\ &\le g^2(|x| +|x+g^{-1}v|)^2 \\ &\le g^2\Big(\fbox{$(\max(1,a^{-2}))^{1/2}a|x|$} +(\max(1,a^{-2}))^{1/2}|x+g^{-1}v|\Big)^2 \\ &= g^2\max(1,a^{-2})(a|x|+|x+g^{-1}v|)^2 \\ &= g^2\max(1,a^{-2}) r(x,v)^2. \end{align} Adding the two inequalities gives the result (actually, something stronger as $1+g^2\le (1+g)^2$): $$|(x,v)|^2 \le (1+g^2)\max(1,a^{-2}) r(x,v)^2.$$ In obtaining the boxed term, we used $$ a^{-2}\le \max(1,a^{-2}) \implies 1\le \max(1,a^{-2})a^2 \implies |x|^2 \le \max(1,a^{-2})a^2|x|^2. $$

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  • $\begingroup$ Thank you very much! But there is a bug, the line $|v|^2 = g^2 |g^{-1}v|^2 \le g^2|x|^2 + g^2|g^{-1}v+x|^2$ is wrong, you are missing a factor of $2$ I believe. $\endgroup$
    – Fei Cao
    Jul 22, 2021 at 4:46
  • $\begingroup$ @FeiCao Yes, that is an issue. I'll give it a think $\endgroup$
    – Calvin Khor
    Jul 22, 2021 at 4:47
  • $\begingroup$ @FeiCao I've fixed it, I believe $\endgroup$
    – Calvin Khor
    Jul 22, 2021 at 4:59
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    $\begingroup$ Thank you very much for editing my question and post such a beautiful solution! $\endgroup$
    – Fei Cao
    Jul 22, 2021 at 5:02

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