[proofplan] We use the Bobkov-Götze dual characterization of $T_1$: a probability measure satisfies $T_1(C)$ exactly when every bounded $1$-Lipschitz [test function](/page/Test%20Function) has centered [Laplace transform](/page/Laplace%20Transform) bounded by $\exp(C\lambda^2/2)$. First we rescale each coordinate metric so that every coordinate measure satisfies $T_1(1)$. Then the tensorization of the Bobkov-Götze Laplace estimate gives the same variance proxy for the Euclidean product metric, and converting back by Bobkov-Götze yields the displayed $T_1(1)$ inequality. The $L^1$ product estimate is obtained by applying the same tensorized Laplace argument with coordinate Lipschitz constants bounded by $1$, which makes the variance proxies add as $\sum_i C_i$. [/proofplan] [step:Convert each coordinate inequality to the Bobkov-Götze Laplace form] For each $i \in \{1,\dots,n\}$, define the rescaled metric $\rho_i:X_i \times X_i \to [0,\infty)$ by \begin{align*} \rho_i(x_i,y_i) := \frac{d_i(x_i,y_i)}{\sqrt{C_i}}. \end{align*} Since $W_{1,\rho_i}(\nu_i,\mu_i)=C_i^{-1/2}W_{1,d_i}(\nu_i,\mu_i)$ for every Borel probability measure $\nu_i$ on $X_i$, the hypothesis $T_1(C_i)$ with respect to $d_i$ is equivalent to $T_1(1)$ with respect to $\rho_i$. By the Bobkov-Götze dual characterization of $T_1$ (citing a result not yet in the wiki: Bobkov-Götze dual characterization of Talagrand's $T_1$ inequality), this means that for every bounded Borel function $g_i:X_i \to \mathbb{R}$ satisfying \begin{align*} |g_i(x_i)-g_i(y_i)| \leq \rho_i(x_i,y_i) \end{align*} for all $x_i,y_i \in X_i$, and for every $\lambda \in \mathbb{R}$, \begin{align*} \int_{X_i} \exp\left(\lambda\left(g_i-\int_{X_i}g_i\,d\mu_i\right)\right)\,d\mu_i \leq \exp\left(\frac{\lambda^2}{2}\right). \end{align*} [/step] [step:Tensorize the Laplace estimate for the weighted Euclidean product metric] Define the weighted Euclidean product metric $\rho_2:X \times X \to [0,\infty)$ by \begin{align*} \rho_2(x,y) := \left(\sum_{i=1}^n \rho_i(x_i,y_i)^2\right)^{1/2}. \end{align*} By the definition of $\rho_i$, this metric is exactly the metric $d_2$ in the statement. Let $f:X \to \mathbb{R}$ be a bounded Borel function satisfying \begin{align*} |f(x)-f(y)| \leq \rho_2(x,y) \end{align*} for all $x,y \in X$. The tensorization of the Bobkov-Götze Laplace estimate over independent coordinates (citing a result not yet in the wiki: tensorization of the Bobkov-Götze Laplace estimate for Euclidean product metrics) gives, for every $\lambda \in \mathbb{R}$, \begin{align*} \int_X \exp\left(\lambda\left(f-\int_X f\,d\mu\right)\right)\,d\mu \leq \exp\left(\frac{\lambda^2}{2}\right). \end{align*} The hypotheses of this tensorization result are exactly the coordinate Laplace estimates from the previous step and the fact that $\mu$ is the product measure $\mu_1\otimes\cdots\otimes\mu_n$. Its Euclidean normalization is the reason the same variance proxy $1$ is preserved: the squared coordinate sensitivities of a $\rho_2$-$1$-Lipschitz function enter additively and are bounded by $1$. [guided] We now pass from one coordinate to the product. The product metric we use is not the $L^1$ sum of the coordinate metrics, but the Euclidean metric \begin{align*} \rho_2(x,y) := \left(\sum_{i=1}^n \rho_i(x_i,y_i)^2\right)^{1/2}. \end{align*} This choice is essential because subgaussian variance parameters combine by summing squared Lipschitz sensitivities. Let $f:X \to \mathbb{R}$ be bounded, Borel, and $1$-Lipschitz with respect to $\rho_2$. Thus for every pair $x,y \in X$, \begin{align*} |f(x)-f(y)| \leq \left(\sum_{i=1}^n \rho_i(x_i,y_i)^2\right)^{1/2}. \end{align*} Fixing all coordinates except the $i$-th one shows that each one-coordinate section is $1$-Lipschitz with respect to $\rho_i$. The tensorized Bobkov-Götze estimate is stronger than merely applying the one-coordinate estimate $n$ times: it records the squared coordinate sensitivities and uses the Euclidean bound to keep their total at most $1$. Because $\mu=\mu_1\otimes\cdots\otimes\mu_n$, the coordinate variables are independent under $\mu$. Applying the tensorization theorem for the Bobkov-Götze Laplace estimate to the coordinate estimates obtained above gives, for every $\lambda \in \mathbb{R}$, \begin{align*} \int_X \exp\left(\lambda\left(f-\int_X f\,d\mu\right)\right)\,d\mu \leq \exp\left(\frac{\lambda^2}{2}\right). \end{align*} This is precisely the centered subgaussian Laplace bound with variance proxy $1$ for every bounded $1$-Lipschitz function on the product space $(X,\rho_2)$. [/guided] [/step] [step:Convert the product Laplace estimate back to $T_1(1)$] The space $X$ is Polish because it is a finite product of Polish spaces, and $\rho_2$ is a Borel metric built from the Borel metrics $\rho_i$. The previous step proves the Bobkov-Götze Laplace estimate with constant $1$ for every bounded $\rho_2$-$1$-Lipschitz Borel function $f:X \to \mathbb{R}$. Applying the converse direction of the Bobkov-Götze dual characterization of $T_1$ gives \begin{align*} W_{1,\rho_2}(\nu,\mu)^2 \leq 2H(\nu \mid \mu) \end{align*} for every Borel probability measure $\nu$ on $X$. Since $\rho_2=d_2$, this is exactly $T_1(1)$ for the weighted Euclidean product metric in the statement. [/step] [step:Repeat the tensorized Laplace argument for the product $L^1$ metric] Define $d_{1,n}:X \times X \to [0,\infty)$ by \begin{align*} d_{1,n}(x,y) := \sum_{i=1}^n d_i(x_i,y_i). \end{align*} Let $f:X \to \mathbb{R}$ be bounded, Borel, and $1$-Lipschitz with respect to $d_{1,n}$. For each $i \in \{1,\dots,n\}$ and every fixed choice of the other coordinates, the one-coordinate section \begin{align*} g_i:X_i &\to \mathbb{R} \end{align*} obtained from $f$ by varying only $x_i$ satisfies \begin{align*} |g_i(x_i)-g_i(y_i)| \leq d_i(x_i,y_i). \end{align*} Thus the coordinate Bobkov-Götze estimate for $\mu_i$ with respect to $d_i$ contributes variance proxy $C_i$. Tensorizing the coordinate Laplace estimates for the product measure $\mu$ gives, for every $\lambda \in \mathbb{R}$, \begin{align*} \int_X \exp\left(\lambda\left(f-\int_X f\,d\mu\right)\right)\,d\mu \leq \exp\left(\frac{\lambda^2}{2}\sum_{i=1}^n C_i\right). \end{align*} Applying the Bobkov-Götze dual characterization once more yields \begin{align*} W_{1,d_{1,n}}(\nu,\mu)^2 \leq 2\left(\sum_{i=1}^n C_i\right)H(\nu \mid \mu) \end{align*} for every Borel probability measure $\nu$ on $X$. This is the asserted $T_1\left(\sum_{i=1}^n C_i\right)$ inequality for the product $L^1$ metric. [/step]