[proofplan] We prove a Gaussian [Laplace transform](/page/Laplace%20Transform) bound with variance proxy $C$ and then convert it to the displayed tail estimate by optimizing Markov's inequality. Under the $T_1$ hypothesis, the Laplace bound is exactly the Bobkov-Götze dual formulation of the tensorized transport-[entropy inequality](/theorems/6729). Under the logarithmic Sobolev hypothesis, the Laplace bound follows from Herbst's argument applied to $e^{\theta F/2}$, using the pointwise estimate $\Gamma_n(F)\le 1$. Finally, the approximation hypothesis transfers the smooth bounded logarithmic Sobolev estimate to arbitrary integrable $1$-Lipschitz functions. [/proofplan] [step:Fix the product space and center the Lipschitz observable] Fix $n \in \mathbb{N}$. Let $\mu:=\nu_n$ denote the product probability measure on $E^n$, and let $\rho:=d_{2,n}$ denote the product metric. Let $F:E^n \to \mathbb{R}$ be a measurable $1$-Lipschitz function with \begin{align*} \int_{E^n}|F|\,d\mu < \infty. \end{align*} Define the centered function $G:E^n \to \mathbb{R}$ by \begin{align*} G(x)=F(x)-\int_{E^n}F\,d\mu \end{align*} for $x \in E^n$. Since subtracting a constant does not change Lipschitz constants, $G$ is also $1$-Lipschitz with respect to $\rho$, and \begin{align*} \int_{E^n}G\,d\mu = 0. \end{align*} [/step] [step:Derive the Laplace estimate from tensorized $T_1$] Assume the transport-entropy hypothesis. By tensorization, $\mu=\nu_n$ satisfies $T_1(C)$ with respect to $\rho=d_{2,n}$. By the Bobkov-Götze dual formulation of the $T_1$ transport-entropy inequality (citing a result not yet in the wiki: Bobkov-Götze dual formulation of $T_1$), applied to the integrable $1$-Lipschitz function $G:E^n\to\mathbb{R}$, for every $\theta \in \mathbb{R}$, \begin{align*} \int_{E^n}\exp(\theta G)\,d\mu \le \exp\left(\frac{C\theta^2}{2}\right). \end{align*} Thus the desired Laplace estimate holds under the $T_1$ hypothesis. [/step] [step:Use Herbst's argument for bounded smooth functions under the logarithmic Sobolev hypothesis] Assume the logarithmic Sobolev hypothesis, and first suppose that $F:E^n\to\mathbb{R}$ is bounded and smooth. Since $F$ is $1$-Lipschitz with respect to $\rho$, the hypothesis gives $\Gamma_n(F)\le 1$ pointwise. For $\theta>0$, define the positive bounded smooth function $H_\theta:E^n\to(0,\infty)$ by \begin{align*} H_\theta(x)=\exp(\theta G(x)). \end{align*} Define also the Laplace transform $Z:(0,\infty)\to(0,\infty)$ and its logarithm $\psi:(0,\infty)\to\mathbb{R}$ by \begin{align*} Z(\theta)=\int_{E^n}H_\theta\,d\mu, \end{align*} and \begin{align*} \psi(\theta)=\log Z(\theta). \end{align*} For every nonnegative integrable function $u:E^n\to[0,\infty)$ such that $u\log u$ is integrable, define the entropy functional $\operatorname{Ent}_\mu$ by \begin{align*} \operatorname{Ent}_\mu(u)=\int_{E^n}u\log u\,d\mu-u_\mu\log u_\mu, \end{align*} where $u_\mu:=\int_{E^n}u\,d\mu$. The tensorized logarithmic Sobolev inequality applied to $e^{\theta G/2}$ gives \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) \le 2C\int_{E^n}\Gamma_n(e^{\theta G/2})\,d\mu. \end{align*} Using the carré-du-champ chain rule for the map $t\mapsto e^{\theta t/2}$, \begin{align*} \Gamma_n(e^{\theta G/2})=\frac{\theta^2}{4}e^{\theta G}\Gamma_n(G). \end{align*} Since $G$ differs from $F$ by a constant, $\Gamma_n(G)=\Gamma_n(F)\le 1$. Therefore \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) \le \frac{C\theta^2}{2}\int_{E^n}e^{\theta G}\,d\mu = \frac{C\theta^2}{2}Z(\theta). \end{align*} On the other hand, boundedness of $G$ permits differentiation under the integral sign, so \begin{align*} Z'(\theta)=\int_{E^n}G e^{\theta G}\,d\mu. \end{align*} Since $\int_{E^n}G\,d\mu=0$, the entropy identity gives \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) = \theta Z'(\theta)-Z(\theta)\log Z(\theta). \end{align*} Dividing by $Z(\theta)>0$ yields \begin{align*} \theta\psi'(\theta)-\psi(\theta)\le \frac{C\theta^2}{2}. \end{align*} Equivalently, \begin{align*} \left(\frac{\psi(\theta)}{\theta}\right)' \le \frac{C}{2}. \end{align*} Because $G$ is bounded and centered, $\psi(\theta)/\theta \to 0$ as $\theta\downarrow 0$. Integrating from $0$ to $\theta$ gives \begin{align*} \frac{\psi(\theta)}{\theta}\le \frac{C\theta}{2}. \end{align*} Hence \begin{align*} \int_{E^n}\exp(\theta G)\,d\mu \le \exp\left(\frac{C\theta^2}{2}\right) \end{align*} for every $\theta>0$. [guided] We prove the Laplace transform estimate by converting the logarithmic Sobolev inequality into a differential inequality. The object to control is \begin{align*} Z(\theta)=\int_{E^n}\exp(\theta G)\,d\mu, \end{align*} where $G=F-\int_{E^n}F\,d\mu$ is centered. The logarithmic Sobolev inequality is designed for positive functions, so we apply it to the positive bounded smooth function $e^{\theta G/2}:E^n\to(0,\infty)$. The tensorized logarithmic Sobolev inequality states that \begin{align*} \operatorname{Ent}_\mu(h^2) \le 2C\int_{E^n}\Gamma_n(h)\,d\mu \end{align*} for bounded smooth positive functions $h:E^n\to(0,\infty)$. We take $h=e^{\theta G/2}$. Then $h^2=e^{\theta G}$, so \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) \le 2C\int_{E^n}\Gamma_n(e^{\theta G/2})\,d\mu. \end{align*} Now the carré-du-champ chain rule gives \begin{align*} \Gamma_n(e^{\theta G/2})=\frac{\theta^2}{4}e^{\theta G}\Gamma_n(G). \end{align*} The function $G$ differs from $F$ by a constant, so the carré-du-champ is unchanged: \begin{align*} \Gamma_n(G)=\Gamma_n(F). \end{align*} The point of using the product metric $d_{2,n}$ is now visible: the hypothesis says that every bounded smooth $1$-Lipschitz function for $d_{2,n}$ satisfies the dimension-free gradient bound $\Gamma_n(F)\le 1$. Therefore \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) \le \frac{C\theta^2}{2}\int_{E^n}e^{\theta G}\,d\mu = \frac{C\theta^2}{2}Z(\theta). \end{align*} We next rewrite the entropy in terms of $Z$. Since $G$ is bounded, differentiation under the integral sign is justified and \begin{align*} Z'(\theta)=\int_{E^n}G e^{\theta G}\,d\mu. \end{align*} By definition of entropy, \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) = \int_{E^n}e^{\theta G}\log(e^{\theta G})\,d\mu - Z(\theta)\log Z(\theta). \end{align*} Since $\log(e^{\theta G})=\theta G$, this becomes \begin{align*} \operatorname{Ent}_\mu(e^{\theta G}) = \theta Z'(\theta)-Z(\theta)\log Z(\theta). \end{align*} Writing $\psi(\theta)=\log Z(\theta)$, we have $\psi'(\theta)=Z'(\theta)/Z(\theta)$. Dividing the previous entropy bound by $Z(\theta)>0$ gives \begin{align*} \theta\psi'(\theta)-\psi(\theta)\le \frac{C\theta^2}{2}. \end{align*} The left side is exactly $\theta^2(\psi(\theta)/\theta)'$, so \begin{align*} \left(\frac{\psi(\theta)}{\theta}\right)'\le \frac{C}{2}. \end{align*} Finally, because $G$ is centered, the first-order term of $Z(\theta)$ at $\theta=0$ vanishes: \begin{align*} \int_{E^n}G\,d\mu=0. \end{align*} Thus $\psi(\theta)/\theta\to 0$ as $\theta\downarrow 0$. Integrating the differential inequality from $0$ to $\theta$ gives \begin{align*} \frac{\psi(\theta)}{\theta}\le \frac{C\theta}{2}. \end{align*} Multiplying by $\theta$ and exponentiating yields \begin{align*} \int_{E^n}\exp(\theta G)\,d\mu = e^{\psi(\theta)} \le \exp\left(\frac{C\theta^2}{2}\right). \end{align*} This is the desired dimension-free Laplace estimate for bounded smooth $F$. [/guided] [/step] [step:Pass from bounded smooth approximants to general Lipschitz functions] Continue under the logarithmic Sobolev hypothesis. Let $F:E^n\to\mathbb{R}$ be any measurable $1$-Lipschitz function with \begin{align*} \int_{E^n}|F|\,d\mu<\infty. \end{align*} By the approximation hypothesis, there exists a sequence of bounded smooth $1$-Lipschitz functions $F_k:E^n\to\mathbb{R}$ such that \begin{align*} \int_{E^n}|F_k-F|\,d\mu \to 0. \end{align*} Define $G_k:E^n\to\mathbb{R}$ by \begin{align*} G_k(x)=F_k(x)-\int_{E^n}F_k\,d\mu. \end{align*} Then $G_k\to G$ in $L^1(\mu)$, because \begin{align*} \int_{E^n}|G_k-G|\,d\mu \le 2\int_{E^n}|F_k-F|\,d\mu. \end{align*} In particular, $G_k\to G$ in $\mu$-probability. From the bounded smooth case, for every $s\ge 0$, \begin{align*} \mu(\{x\in E^n:G_k(x)\ge s\}) \le \exp\left(-\frac{s^2}{2C}\right). \end{align*} Fix $r>0$ and $0<s<r$. Since \begin{align*} \{x\in E^n:G(x)\ge r\} \subset \{x\in E^n:G_k(x)\ge s\} \cup \{x\in E^n:|G_k(x)-G(x)|>r-s\}, \end{align*} we obtain \begin{align*} \mu(\{x\in E^n:G(x)\ge r\}) \le \exp\left(-\frac{s^2}{2C}\right) +\mu(\{x\in E^n:|G_k(x)-G(x)|>r-s\}). \end{align*} Letting $k\to\infty$ and using convergence in probability gives \begin{align*} \mu(\{x\in E^n:G(x)\ge r\}) \le \exp\left(-\frac{s^2}{2C}\right). \end{align*} Letting $s\uparrow r$ gives \begin{align*} \mu(\{x\in E^n:G(x)\ge r\}) \le \exp\left(-\frac{r^2}{2C}\right). \end{align*} For $r=0$, the same inequality is immediate because the right-hand side equals $1$. Thus the desired tail estimate holds for all integrable $1$-Lipschitz $F$ under the logarithmic Sobolev hypothesis. [/step] [step:Convert the Laplace estimate into the Gaussian tail bound] It remains only to record the common final step. In either the $T_1$ case or the logarithmic Sobolev case, we have shown that for every $\theta>0$, \begin{align*} \int_{E^n}\exp(\theta G)\,d\mu \le \exp\left(\frac{C\theta^2}{2}\right). \end{align*} For $r\ge 0$, Markov's inequality applied to the nonnegative [random variable](/page/Random%20Variable) $x\mapsto \exp(\theta G(x))$ gives \begin{align*} \mu(\{x\in E^n:G(x)\ge r\}) \le \exp(-\theta r)\int_{E^n}\exp(\theta G)\,d\mu. \end{align*} Therefore \begin{align*} \mu(\{x\in E^n:G(x)\ge r\}) \le \exp\left(\frac{C\theta^2}{2}-\theta r\right). \end{align*} For $r>0$, choose $\theta=r/C$, which is positive. This gives \begin{align*} \mu(\{x\in E^n:G(x)\ge r\}) \le \exp\left(-\frac{r^2}{2C}\right). \end{align*} For $r=0$, the inequality reads $\mu(\{G\ge0\})\le 1$, which is true. Substituting the definition of $G$ gives \begin{align*} \mu\left(\left\{x\in E^n:F(x)-\int_{E^n}F\,d\mu\ge r\right\}\right) \le \exp\left(-\frac{r^2}{2C}\right). \end{align*} Since $\mu=\nu^{\otimes n}$ and $n\in\mathbb{N}$ was arbitrary, the same constant $C$ works for every product dimension. [/step]