Proof of Tensor Powers Scale the Wasserstein Distance by the Square Root of N, and the One-Particle Marginal is Lipschitz with Constant $N^{-1/2}$

Each result cited is universally quantified over the data in its own statement. For $m\in\{q,qN\}$, couplings $\Pi(\cdot,\cdot)$, the quadratic cost $I$ and $W_{2}$ are those of Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §coupling, Couplings of Two Probability Measures on Euclidean Space and Their Quadratic Cost §cost and The Quadratic Wasserstein Distance on Euclidean Space §distance read with $m$ in place of $d$ (Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation §dimensions), $\mathrm{pr}_{1},\mathrm{pr}_{2}:\mathbb{R}^{m+m}\to\mathbb{R}^{m}$ are the coordinate projections of Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections, and $c_{m}(z)=\lVert\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z)\rVert^{2}$ on $\mathbb{R}^{m+m}$ is Borel and nonnegative by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §functions, so that $I(\pi)=\int c_{m}\,d\pi$. The block maps of $\mathbb{R}^{qN}$ are written $\mathfrak{p}_{k}:\mathbb{R}^{qN}\to\mathbb{R}^{q}$, and those of $\mathbb{R}^{(q+q)N}$, namely Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §blocks read with $q+q$ in place of $q$, are written $\hat{\mathfrak{p}}_{k}:\mathbb{R}^{(q+q)N}\to\mathbb{R}^{q+q}$; the sibling results Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points, Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts, Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals, The One-Particle Marginal of a Probability Measure on the Configuration Space and Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts are applied with $q+q$ in place of $q$ where these maps occur. Composites of Borel maps are Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps, and push-forwards and their change of variables are those of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward.

Step 1 (Claim 1). Let $P,P'\in\mathcal{P}_{2}(\mathbb{R}^{qN})$; then $P^{[1]},P'^{[1]}\in\mathcal{P}_{2}(\mathbb{R}^{q})$ by Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §moments. By Existence of an Optimal Coupling of Two Probability Measures with Finite Second Moment (with $m=qN$) there is $\gamma\in\Pi(P,P')$ with $I(\gamma)=W_{2}(P,P')^{2}$. For $k\in[N]$ let $S_{k}=(\mathfrak{p}_{k}\circ\mathrm{pr}_{1},\mathfrak{p}_{k}\circ\mathrm{pr}_{2}):\mathbb{R}^{qN+qN}\to\mathbb{R}^{q+q}$, Borel by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §pairing; by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections, $\mathrm{pr}_{i}\circ S_{k}=\mathfrak{p}_{k}\circ\mathrm{pr}_{i}$ for $i=1,2$. Let $J:\mathbb{R}^{qN+qN}\to\mathbb{R}^{(q+q)N}$ send $z$ to the configuration $[S_{1}(z),\dots,S_{N}(z)]$ of Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §configuration, so that $\hat{\mathfrak{p}}_{k}\circ J=S_{k}$. Each component of $J$ is a component of some $S_{k}$ (Block Indices: Enumerating an Initial Segment of Length qN by Blocks and Positions §bijection), hence Borel, so $J$ is Borel by claims 2 and 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. Put $\Gamma=J_{\#}\gamma\in\mathcal{P}(\mathbb{R}^{(q+q)N})$ and $\bar\gamma=\Gamma^{[1]}\in\mathcal{P}(\mathbb{R}^{q+q})$.

For $B\in\mathcal{B}(\mathbb{R}^{q})$, The One-Particle Marginal of a Probability Measure on the Configuration Space §marginal, $\hat{\mathfrak{p}}_{k}\circ J=S_{k}$, $\mathrm{pr}_{1}\circ S_{k}=\mathfrak{p}_{k}\circ\mathrm{pr}_{1}$ and $\gamma\in\Pi(P,P')$ give

and likewise $\bar\gamma(\mathrm{pr}_{2}^{-1}(B))=P'^{[1]}(B)$. So $\bar\gamma\in\Pi(P^{[1]},P'^{[1]})$.

For $z\in\mathbb{R}^{qN+qN}$, Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear gives $c_{q}(S_{k}(z))=\lVert\mathfrak{p}_{k}(\mathrm{pr}_{1}(z)-\mathrm{pr}_{2}(z))\rVert^{2}$, so by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §inner-product, $c_{q}\circ S_{k}\le c_{qN}$ and $\sum_{k=1}^{N}c_{q}(S_{k}(z))=c_{qN}(z)$. By monotonicity (claim 1 of Linearity and Monotonicity of the Lebesgue Integral), $\int c_{q}\circ S_{k}\,d\gamma\le I(\gamma)<\infty$, so each nonnegative Borel $c_{q}\circ S_{k}$ is integrable with respect to $\gamma$. Hence, by the $[0,\infty]$ identity of Existence and Uniqueness of Tensor Powers, and the Average of the Block Marginals §average for $\Gamma$ and $f=c_{q}$, change of variables for $J$, and Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §linear,

By The Quadratic Wasserstein Distance on Euclidean Space §distance, $W_{2}(P^{[1]},P'^{[1]})^{2}\le I(\bar\gamma)$; multiplying by $N>0$ gives claim 1.

Step 2 (Lower bound in claim 2). Let $\rho,\rho'\in\mathcal{P}_{2}(\mathbb{R}^{q})$. Then $\rho^{\otimes N},\rho'^{\otimes N}\in\mathcal{P}_{2}(\mathbb{R}^{qN})$ by Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §moments, and $(\rho^{\otimes N})^{[1]}=\rho$, $(\rho'^{\otimes N})^{[1]}=\rho'$ by Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §marginal-of-tensor. Step 1 gives $N\,W_{2}(\rho,\rho')^{2}\le W_{2}(\rho^{\otimes N},\rho'^{\otimes N})^{2}$.

Step 3 (Upper bound in claim 2). By Existence of an Optimal Coupling of Two Probability Measures with Finite Second Moment (with $m=q$) there is $\pi\in\Pi(\rho,\rho')$ with $I(\pi)=W_{2}(\rho,\rho')^{2}$. Let $\pi^{\otimes N}\in\mathcal{P}(\mathbb{R}^{(q+q)N})$ be its tensor power (The Tensor Power of a Probability Measure on Euclidean Space §tensor with $q+q$ in place of $q$), and let $\mathrm{pr}_{1}^{\oplus},\mathrm{pr}_{2}^{\oplus}:\mathbb{R}^{(q+q)N}\to\mathbb{R}^{qN}$ be the product maps of Particle Blocks of the Configuration Space: Block Maps, Configurations, Product Maps and Diagonal Points §product-map of the projections $\mathrm{pr}_{i}:\mathbb{R}^{q+q}\to\mathbb{R}^{q}$ (with $q+q$ in place of $q$ and $q$ in place of $p$); they are Borel by Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §product-map. The pairing $T=(\mathrm{pr}_{1}^{\oplus},\mathrm{pr}_{2}^{\oplus}):\mathbb{R}^{(q+q)N}\to\mathbb{R}^{qN+qN}$ is Borel by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §pairing, with $\mathrm{pr}_{i}\circ T=\mathrm{pr}_{i}^{\oplus}$ by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections. Put $\Sigma=T_{\#}\pi^{\otimes N}\in\mathcal{P}(\mathbb{R}^{qN+qN})$. For $B\in\mathcal{B}(\mathbb{R}^{qN})$, Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §pushforward and $\pi\in\Pi(\rho,\rho')$ give

and likewise $\Sigma(\mathrm{pr}_{2}^{-1}(B))=\rho'^{\otimes N}(B)$; so $\Sigma\in\Pi(\rho^{\otimes N},\rho'^{\otimes N})$.

For $w\in\mathbb{R}^{(q+q)N}$, Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §inner-product, Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §linear and Particle Blocks: Linearity, Splitting of Inner Products, Product Maps and Diagonal Shifts §product-map ($\mathfrak{p}_{k}\circ\mathrm{pr}_{i}^{\oplus}=\mathrm{pr}_{i}\circ\hat{\mathfrak{p}}_{k}$) give

By Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §particle-laws and change of variables, $\int c_{q}\circ\hat{\mathfrak{p}}_{k}\,d\pi^{\otimes N}=\int c_{q}\,d\pi=I(\pi)<\infty$, so each $c_{q}\circ\hat{\mathfrak{p}}_{k}$ is integrable with respect to $\pi^{\otimes N}$. Hence, by change of variables for $T$, Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions §linear, the integration identity of The One-Particle Marginal of a Probability Measure on the Configuration Space §marginal and Tensor Powers and One-Particle Marginals: Particle Laws, Product Integrals, Push-Forwards, Moments, Product Maps and Diagonal Shifts §marginal-of-tensor,

By The Quadratic Wasserstein Distance on Euclidean Space §distance, $W_{2}(\rho^{\otimes N},\rho'^{\otimes N})^{2}\le I(\Sigma)=N\,W_{2}(\rho,\rho')^{2}$. Together with Step 2 and antisymmetry of the order of $\mathbb{R}$, this is claim 2.