# Bounding Euclidean norm by slanted 1-norm

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.

• why the downvote...? I feel this is not a easy question. Jul 26, 2021 at 4:45
• 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. Jul 28, 2021 at 10:52
• @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.) Jul 28, 2021 at 11:18
• @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. Jul 28, 2021 at 17:37
• @CalvinKhor thanks, I have included what you suggested! Jul 28, 2021 at 17:45

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 $$a^{-2}\le \max(1,a^{-2}) \implies 1\le \max(1,a^{-2})a^2 \implies |x|^2 \le \max(1,a^{-2})a^2|x|^2.$$
• 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. Jul 22, 2021 at 4:46