Equivalence of the Poincar\'e inequality with a transport-chi-square inequality in dimension one

In this paper, we prove that, in dimension one, the Poincar\'e inequality is equivalent to a new transport-chi-square inequality linking the square of the quadratic Wasserstein distance with the chi-square pseudo-distance. We also check tensorization of this transport-chi-square inequality.

For q ≥ 1, the Wasserstein distance with index q between two probability measures µ and ν on R d is denoted by where the infimum is taken over all probability measures γ on R d × R d with respective marginals µ and ν. We also introduce the relative entropy and the chi-square pseudo distance H(ν|µ) = R d ln dν dµ (x) dν(x) if ν absolutely continuous w.r.t. µ +∞ otherwise Next, we precise the inequalities that will be discussed in the paper.
Definition 0.1 The probability measure µ on R d is said to satisfy the Poincaré inequality P(C) with constant C if ∀ϕ : R d → R C 1 with a bounded gradient, the transport-chi-square inequality T χ (C) with constant C if ∀ν probability measure on R d , W 2 (µ, ν) ≤ √ Cχ 2 (ν|µ).
Before proving Theorem 1.1, we state our second main result dedicated to the tensorization property of the transport-chi-square inequality. Its proof is postponed in Section 4.

Remark 1.3
According to Proposition 8.4.1 [1], if µ 1 and µ 2 respectively satisfy T H (C 1 ) and T H (C 2 ), then µ 1 ⊗ µ 2 satisfies T H (C 1 ∨ C 2 ). The constant that we obtain in the tensorization of the transport-chi-square inequality is larger than The proof of the one-dimensional implication P(C) ⇒ T χ (32C) in Theorem 1.1 relies on the two next propositions, the proof of which are respectively postponed in Sections 2 and 3. When d = 1, we denote by F (x) = µ((−∞, x]) and G(x) = ν((−∞, x]) the cumulative distribution functions of the probability measures µ and ν. The càg pseudo-inverses of G (resp. F ) is defined by When µ (resp. ν) admits a density w.r.t. the Lebesgue measure, this density is denoted by f (resp. g). Moreover, the optimal coupling in (0.1) is given by γ = du • (F −1 , G −1 ) −1 where du denotes the Lebesgue measure on (0, 1) so that W q q (µ, ν) = 1 0 (F −1 (u) − G −1 (u)) q du (see [9] p107-109). We take advantage of this optimal coupling to work with the cumulative distribution functions and check the following proposition. In higher dimensions, far less is known on the optimal coupling and this is the main reason why we have not been able to check whether the Poincaré inequality implies the transport-chi-square inequality. Proposition 1.4 If a probability measure µ on the real line admits a positive probability density f , then, for any probability measure ν on R, Notice that since, by (1.1) and Fubini's theorem, the stronger bound is a consequence of the Cauchy-Schwarz inequality.
• It is not possible to control R . Indeed for f (x) = 1 2 e −|x| and dν(x) = 1 2 e −|x−m| dx, one has W 2 2 (µ, ν) = m 2 , G(x) = e x−m 2 1 {x≤m} + (1 − e m−x 2 )1 {x>m} and for m > 0, Next, when the probability measure µ on the real line admits a positive probability density satisfying a tail assumption known to be equivalent to the Poincaré inequality (see Theorem 6.2.2 [1]), we are able to control the right-hand-side of (1.2) in terms of χ 2 2 (ν|µ). Then for any probability density g on the real line with cumulative distribution function G(x) = x −∞ g(y)dy, Remark 1.7 • The combination of these two propositions implies that any probability measure µ on the real line admitting a positive density f such that b < +∞ satifies T χ (16b).
• Proposition 1.6 is a generalization of the last assertion in Lemma 2.3 [6] where f is restricted to the class of probability densities f ∞ solving f ∞ (x) = −A(F ∞ (x)) on the real line with The constant b associated with any such density is finite by the proof of Lemma 2.1 [6]. Moreover, in order to investigate the long-time behaviour of the solution f t of the Fokker-Planck equation [6] first investigates the ex- belongs to (k, k + 1) and is such that e −x dx. One has, using ∀y ≥ 0, ln(1 + y) ≥ y 1+y by concavity of the logarithm and 1 + e−1 2 e − k+1 2 ≤ √ e for the inequality, On the other hand, since for k ≥ n and x ∈ [k, k Proof of Theorem 1.1 : The implication T χ (C) ⇒ P(C) is obtained by linearization of the transport-chi-square inequality T χ (C). For ν ε = (1 + εφ)µ with φ : R d → R a C 2 function compactly supported and such that R d φ(x)dµ(x) = 0, according to [8] p394, there is a finite constant K not depending on ε such that When T χ (C) holds, then W 2 (µ, ν ε ) ≤ ε C R d φ 2 (x)dµ(x) and taking the limit ε → 0, one deduces that . Let now ϕ, φ n : R d → R be C 2 functions compactly supported with φ n taking its values in [0, 1], equal to 1 on the ball centered at the origin with radius n and ∇φ n bounded by 1. Taking the limit n → ∞ in the inequality written with φ replaced by , one deduces that the Poincaré inequality P(C) holds for ϕ. The extension to C 1 functions ϕ with a bounded gradient is obtained by density.
To prove the converse implication, we now suppose that d = 1, µ satisfies the Poincaré inequality P(C) and that χ 2 (ν|µ) < +∞. We set µ n = ρ n ⋆ µ and ν n = ρ n ⋆ ν for n ≥ 1 where denotes the density of the centered Gaussian law with variance 1/n. For ϕ a C 1 function on R with a bounded derivative such that 0 where we used the Poincaré inequalities for the Gaussian density ρ n ([1] Théorème 1.5.1 p10) applied to ϕ and for µ applied to ρ n ⋆ ϕ for the second inequality then Jensen's inequality. The probability measure µ n admits a positive density w.r.t. the Lebesgue measure and satisfies P( 1+nC n ). According to Theorem 6.2.2 [1], this property is equivalent to the fact that the constant associated with µ n through (1.3) is b n ≤ 2 1+nC n . Combining Propositions 1.4 and 1.6, one deduces that W 2 2 (µ n , ν n ) ≤ 32 1 + nC n χ 2 2 (ν n |µ n ).
2 Proof of Proposition 1.4 To prove the proposition, one first needs to express the Wasserstein distance in terms of the cumulative distribution functions F and G instead of their pseudo-inverses : Proof of Lemma 2.1 : Let us first suppose that µ admits a positive continuous density f w.r.t. the Lebesgue measure. Using the change of variables (v, w) = (F (x), F (y)) for the third equality then the equivalence w < By symmetry, one deduces that (2.1) holds.
Proof of Proposition 1.4 : One has By Fubini's theorem and a similar argument, With (2.2) and (2.3), then using Cauchy-Schwarz inequality and the change of variables u = F (x), one deduces that when µ admits a positive density f w.r.t. the Lebesgue measure, then Recognizing that the second factor in the r.h.s. is equal to W 2 (µ, ν), one concludes that (1.4) holds as soon as W 2 (µ, ν) < +∞. To prove (1.4) without assuming finiteness of W 2 (µ, ν), one defines a sequence (G n ) n of cumulative distribution functions converging pointwise to G by setting As a consequence, for fixed x ∈ R, the sequence (|G n (x)−F (x)|) n∈N is non-decreasing and goes to |G(x)−F (x)| as n → ∞. By monotone convergence, one deduces that lim n→+∞ R As a consequence, denoting by ν n the probability measure with c.d.f. G n , W 2 2 (µ, ν n ) = 1 0 (F −1 (u)− G −1 n (u)) 2 du < +∞ and W 2 2 (µ, ν) ≤ lim inf n→∞ W 2 2 (µ, ν n ) by Fatou Lemma. One concludes by taking the limit n → +∞ in (1.4) written with (ν n , G n ) replacing (ν, G).

Proof of Proposition 1.6
Let us assume that b < +∞ and R (f −g) 2 f (x)dx < +∞. By integration by parts, for n ∈ N * , n −n For x larger than the median m of the density f , by definition of b, then by the equality x (g − f )(y)dy and Cauchy-Schwarz inequality, one has where the right-hand-side tends to 0 as x → +∞ by integrability of (f −g) 2 f on the real line.
m x dy f (y) = 0. Taking the limit n → ∞ in (3.1) and using again the definition of b, one deduces that is locally integrable on R since the first factor is integrable and the second one is locally bounded. Let a n < +∞ denote the integral of this function on [−n, n].
By Cauchy Schwarz inequality, , we obtain by integration by parts that for n ≥ |m|, Plugging this estimation in (3.3), one deduces that ∀n ≥ |m|, a n ≤ 1 {an>0} 2 + ε n a n Using that, according to the analysis of the boundary terms in the first integration by parts performed in the proof, lim n→+∞ ε n = 0 and that (a n ) n is non-decreasing, one may take the limit n → ∞ in this inequality to obtain One easily concludes with (3.2).
Proof : According to Theorem 1.1, µ satisfies P(C). By spatial translation, one may assume that R d ydµ(y) = 0. Applying the Poincaré inequality P(C) to the functions x = (x 1 , . . . , One easily concludes by summation of these inequalities.