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My questions come from the paper Logarithmic Sobolev inequalities for some nonlinear PDE’s written by F. Malrieu (May 2001) where author omitted a good amount of details to be filled. Suppose that $W$ is convex, even, with polynomial growth and $U$ is uniformly convex. Let $\mu_N$ to be the probability measure with density $$\mu_N = \frac{1}{Z_N}\,\exp\left(-\sum_{i=1}^N U(x_i) - \frac{1}{2N}\sum_{i,j=1}^N W(x_i-x_j)\right)$$ with $Z_N$ being a normalization constant rendering $\mu_N$ to be a probability density function, also let $\bar{u}$ be the unique minimizer (stationary measure) of the free energy functional defined by $$\eta(f) = \int f(x)\,\log f(x)\,\mathrm{d} x + \int U(x)\,f(x)\,\mathrm{d} x + \frac 12 \iint W(x-y)\,f(x)\,f(y)\,\mathrm{d}x\,\mathrm{d}y,$$ i.e., $\bar{u}$ is the unique solution of $$ \bar{u}(x) = \frac{1}{Z}\,\exp\left(U(x) - W*\bar{u}(x)\right) $$ with $Z = \int \exp\left(U(x) - W*\bar{u}(x)\right)\,\mathrm{d}x$. It is implicitly used in pp. 15 and pp.16 of the aforementioned paper that we expect to have $$\|\mu_{1,N} - \bar{u}\|_1 \leq \frac{K}{\sqrt{N}} \quad \textrm{and} \quad \mathrm{W}_2(\mu_{1,N},\bar{u}) \leq \frac{K}{\sqrt{N}}, \tag{a}$$ where $\mu_{1,N}$ represents the first marginal of $\mu_N$ and $K > 0$ is some fixed constant. Can anyone help me to figure out the claimed bounds on the $L^1$ distance and the $2$-Wasserstein distance ?


Remark: I personally do not think the proof of the first bound in (a) automatically yields the second inequality in (a)

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  • $\begingroup$ The proof of Prop 3.21 is a bit terse/technical. I agree that it does not rely on the first bound in (a). Instead, the triangle inequality is used twice and the term in question (the second term in (a)) is bounded by $\sup_{t \ge 0} \sqrt{E|X_t^{1,n}-\bar{X}_t^1|^2}$, which can itself be bounded by Theorem 3.3. $\endgroup$ Commented Jun 2, 2022 at 20:40
  • $\begingroup$ May I know whether you can clarify it into a formal answer? As I am not sure how the "2" appears, and also how can you bound $\mathrm{W}_2(\mu_{1,N},\bar{u})$ by $\mathrm{W}_2(\mu_t,\mu^{(1,N)}_t)$ ? Moreover, as I said I have no clue as to how the first inequality in (1) is obtained... $\endgroup$
    – Fei Cao
    Commented Jun 2, 2022 at 21:05

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Referring to the proof of Prop 3.21 in Malrieu 2001, after the triangle inequality is applied twice, two of the terms are bounded via the upper bound $$ W_2(u_t,u_t^{(1,N)}) \vee W_2(\mu_{1,N},\bar{u}) \le \sup_{s \ge 0} \sqrt{E|X_s^{1,N}-\bar{X}_s^1|^2} \tag{$\star$}$$ which accounts for the factor $2$. The reason this bound holds is because of the supremum over $s$, which indicates that it holds for all $s\ge0$ including $s=t$ and $s=\infty$ which gives (via the coupling characterization of the 2 Wasserstein distance) the upper bounds in ($\star$) on $W_2(u_t,u_t^{(1,N)})$ and $W_2(\mu_{1,N},\bar{u}) $, respectively. To finish, Theorem 3.3 is invoked.

For the related $L^1$ bound between the densities, one uses the analogous upper bound $$ \|u_t-u_t^{(1,N)}\|_1 \vee \|\mu_{1,N}-\bar{u}\|_1 \le \sup_{s \ge 0} \|u_s-u_s^{(1,N)}\|_1 \;. $$ To finish, Prop. 3.13 is invoked with $k=1$.

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  • $\begingroup$ Thank you very much! I think I now understand that part. May I know do you have a good way to prove the first bound of (a) in my original post? $\endgroup$
    – Fei Cao
    Commented Jun 2, 2022 at 22:40
  • $\begingroup$ Than you Professor! I guess the author missed a $2$ then... $\endgroup$
    – Fei Cao
    Commented Jun 2, 2022 at 23:01
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    $\begingroup$ The author seems to use $K$ as a rolling constant, so the 2 is probably built into $K$. $\endgroup$ Commented Jun 2, 2022 at 23:10
  • $\begingroup$ The issue is that the author included the $2$ in one place while not in the other place, such inconsistency can sometimes cause major confusion... By the way, in the statement of Proposition 3.17 the exponent of the exponential should be $-2\,\beta\,t$ instead of $-\beta\,t\,/\,2$. $\endgroup$
    – Fei Cao
    Commented Jun 3, 2022 at 0:13

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