[proofplan] We first prove the result for bounded smooth $F$, because then the exponential test functions required in the logarithmic Sobolev inequality are admissible and all exponential moments are finite. Applying the logarithmic Sobolev inequality to $g=e^{\beta F/2}$ gives an [entropy inequality](/theorems/6729) for $e^{\beta F}$, which is converted into a differential inequality for the logarithmic moment generating function. Integrating that differential inequality is the Herbst argument and yields the sub-Gaussian [Laplace transform](/page/Laplace%20Transform) bound. Finally, an exponential Markov estimate gives the upper and lower tails, and the stated truncation hypothesis lets the bounded result pass to unbounded $F$. [/proofplan] [step:Apply the logarithmic Sobolev inequality to the exponential test function] Assume first that $F$ is bounded. Define the mean $m\in\mathbb{R}$ by \begin{align*} m:=\mathbb{E}_{\mu}[F]=\int_{\mathbb{R}^n}F(x)\,d\mu(x). \end{align*} For each $\beta\in\mathbb{R}$, define the smooth bounded function \begin{align*} g_{\beta}\colon \mathbb{R}^n\to(0,\infty),\qquad g_{\beta}(x):=e^{\beta F(x)/2}. \end{align*} Since $F$ is bounded and smooth, $g_{\beta}$ is smooth and bounded, and the logarithmic Sobolev inequality is applicable to $g_{\beta}$. Its gradient is \begin{align*} \nabla g_{\beta}(x)=\frac{\beta}{2}e^{\beta F(x)/2}\nabla F(x). \end{align*} Because $F$ is smooth and $L$-Lipschitz, its directional derivatives satisfy $|\nabla F(x)\cdot v|\leq L|v|$ for every $x\in\mathbb{R}^n$ and $v\in\mathbb{R}^n$; taking the supremum over $|v|=1$ gives $|\nabla F(x)|\leq L$. Therefore, \begin{align*} 2C\int_{\mathbb{R}^n}|\nabla g_{\beta}(x)|^2\,d\mu(x)\leq \frac{C\beta^2L^2}{2}\int_{\mathbb{R}^n}e^{\beta F(x)}\,d\mu(x). \end{align*} Since $g_{\beta}^2=e^{\beta F}$, the logarithmic Sobolev inequality gives \begin{align*} \operatorname{Ent}_{\mu}(e^{\beta F})\leq \frac{C\beta^2L^2}{2}\int_{\mathbb{R}^n}e^{\beta F(x)}\,d\mu(x). \end{align*} [/step] [step:Convert the entropy estimate into a Laplace transform bound] For $\beta\in\mathbb{R}$, define the moment generating function \begin{align*} Z\colon \mathbb{R}\to(0,\infty),\qquad Z(\beta):=\int_{\mathbb{R}^n}e^{\beta F(x)}\,d\mu(x). \end{align*} Because $F$ is bounded, the functions $x\mapsto F(x)e^{\beta F(x)}$ are bounded on compact $\beta$-intervals, so differentiation under the integral sign is justified by the [Dominated Convergence Theorem](/theorems/4), and \begin{align*} Z'(\beta)=\int_{\mathbb{R}^n}F(x)e^{\beta F(x)}\,d\mu(x). \end{align*} Define the centered logarithmic moment generating function \begin{align*} \varphi\colon \mathbb{R}\to\mathbb{R},\qquad \varphi(\beta):=\log Z(\beta)-\beta m. \end{align*} By the definition of entropy, \begin{align*} \operatorname{Ent}_{\mu}(e^{\beta F})=\beta Z'(\beta)-Z(\beta)\log Z(\beta). \end{align*} Dividing the entropy estimate by $Z(\beta)>0$ gives \begin{align*} \beta\frac{Z'(\beta)}{Z(\beta)}-\log Z(\beta)\leq \frac{C\beta^2L^2}{2}. \end{align*} Since $\varphi'(\beta)=Z'(\beta)/Z(\beta)-m$, the left-hand side is $\beta\varphi'(\beta)-\varphi(\beta)$. Thus, for every $\beta\in\mathbb{R}$, \begin{align*} \beta\varphi'(\beta)-\varphi(\beta)\leq \frac{C\beta^2L^2}{2}. \end{align*} For $\beta\neq0$, set \begin{align*} q(\beta):=\frac{\varphi(\beta)}{\beta}. \end{align*} Then \begin{align*} q'(\beta)=\frac{\beta\varphi'(\beta)-\varphi(\beta)}{\beta^2}\leq \frac{CL^2}{2}. \end{align*} Also $\varphi(0)=0$ and $\varphi'(0)=0$, so $q(\beta)\to0$ as $\beta\to0$. If $\beta>0$, integration from $0$ to $\beta$ gives \begin{align*} q(\beta)\leq \frac{CL^2}{2}\beta. \end{align*} Multiplying by $\beta>0$ yields \begin{align*} \varphi(\beta)\leq \frac{CL^2\beta^2}{2}. \end{align*} If $\beta<0$, integration from $\beta$ to $0$ gives \begin{align*} -q(\beta)\leq -\frac{CL^2}{2}\beta. \end{align*} Hence $q(\beta)\geq \frac{CL^2}{2}\beta$, and multiplying by the negative number $\beta$ reverses the inequality: \begin{align*} \varphi(\beta)\leq \frac{CL^2\beta^2}{2}. \end{align*} Therefore, for every $\beta\in\mathbb{R}$, \begin{align*} \log\int_{\mathbb{R}^n}e^{\beta(F(x)-m)}\,d\mu(x)\leq \frac{CL^2\beta^2}{2}. \end{align*} [guided] The entropy estimate becomes useful only after rewriting it as a differential inequality for exponential moments. Define \begin{align*} Z\colon \mathbb{R}\to(0,\infty),\qquad Z(\beta):=\int_{\mathbb{R}^n}e^{\beta F(x)}\,d\mu(x). \end{align*} The boundedness of $F$ is the point of the reduction: for each fixed compact interval of $\beta$ values, both $e^{\beta F}$ and $F e^{\beta F}$ are bounded by constants depending on that interval. The [Dominated Convergence Theorem](/theorems/4) therefore applies to the corresponding difference quotients and permits differentiation under the integral sign, so \begin{align*} Z'(\beta)=\int_{\mathbb{R}^n}F(x)e^{\beta F(x)}\,d\mu(x). \end{align*} Let \begin{align*} m:=\int_{\mathbb{R}^n}F(x)\,d\mu(x) \end{align*} and define \begin{align*} \varphi\colon \mathbb{R}\to\mathbb{R},\qquad \varphi(\beta):=\log Z(\beta)-\beta m. \end{align*} This is the logarithm of the centered exponential moment: \begin{align*} \varphi(\beta)=\log\int_{\mathbb{R}^n}e^{\beta(F(x)-m)}\,d\mu(x). \end{align*} Now compute the entropy of $e^{\beta F}$. By definition, \begin{align*} \operatorname{Ent}_{\mu}(e^{\beta F})=\int_{\mathbb{R}^n}\beta F(x)e^{\beta F(x)}\,d\mu(x)-Z(\beta)\log Z(\beta). \end{align*} Using the formula for $Z'(\beta)$, this becomes \begin{align*} \operatorname{Ent}_{\mu}(e^{\beta F})=\beta Z'(\beta)-Z(\beta)\log Z(\beta). \end{align*} The previous step proved \begin{align*} \operatorname{Ent}_{\mu}(e^{\beta F})\leq \frac{C\beta^2L^2}{2}Z(\beta). \end{align*} Because $Z(\beta)>0$, division by $Z(\beta)$ gives \begin{align*} \beta\frac{Z'(\beta)}{Z(\beta)}-\log Z(\beta)\leq \frac{C\beta^2L^2}{2}. \end{align*} Since \begin{align*} \varphi'(\beta)=\frac{Z'(\beta)}{Z(\beta)}-m, \end{align*} we have \begin{align*} \beta\varphi'(\beta)-\varphi(\beta)=\beta\frac{Z'(\beta)}{Z(\beta)}-\log Z(\beta). \end{align*} Thus \begin{align*} \beta\varphi'(\beta)-\varphi(\beta)\leq \frac{C\beta^2L^2}{2}. \end{align*} The key move in Herbst's argument is to divide by $\beta^2$ and recognize a derivative. For $\beta\neq0$, define \begin{align*} q(\beta):=\frac{\varphi(\beta)}{\beta}. \end{align*} Then \begin{align*} q'(\beta)=\frac{\beta\varphi'(\beta)-\varphi(\beta)}{\beta^2}\leq \frac{CL^2}{2}. \end{align*} Because $\varphi(0)=0$ and $\varphi'(0)=0$, the quotient $q(\beta)$ tends to $0$ as $\beta\to0$. For $\beta>0$, integrating the inequality for $q'$ from $0$ to $\beta$ gives \begin{align*} q(\beta)\leq \frac{CL^2}{2}\beta. \end{align*} Multiplying by $\beta>0$ gives \begin{align*} \varphi(\beta)\leq \frac{CL^2\beta^2}{2}. \end{align*} For $\beta<0$, integrating from $\beta$ to $0$ gives \begin{align*} -q(\beta)\leq -\frac{CL^2}{2}\beta, \end{align*} so \begin{align*} q(\beta)\geq \frac{CL^2}{2}\beta. \end{align*} Multiplication by the negative number $\beta$ reverses the inequality and again yields \begin{align*} \varphi(\beta)\leq \frac{CL^2\beta^2}{2}. \end{align*} Therefore, for every real $\beta$, \begin{align*} \log\int_{\mathbb{R}^n}e^{\beta(F(x)-m)}\,d\mu(x)\leq \frac{CL^2\beta^2}{2}. \end{align*} [/guided] [/step] [step:Optimize the exponential moment bound for the upper tail] Let $t\geq0$ and let $\beta>0$. We use the exponential moment estimate directly through the following pointwise indicator bound, so no separate tail theorem is being invoked. On the measurable set \begin{align*} A_t:=\{x\in\mathbb{R}^n:F(x)-m\geq t\}, \end{align*} we have $\mathbb{1}_{A_t}(x)\leq e^{-\beta t}e^{\beta(F(x)-m)}$. Integrating this pointwise inequality with respect to $\mu$ gives \begin{align*} \mu(A_t)\leq e^{-\beta t}\int_{\mathbb{R}^n}e^{\beta(F(x)-m)}\,d\mu(x). \end{align*} Using the Laplace transform estimate, \begin{align*} \mu(A_t)\leq \exp\left(-\beta t+\frac{CL^2\beta^2}{2}\right). \end{align*} The quadratic function \begin{align*} \beta\mapsto -\beta t+\frac{CL^2\beta^2}{2} \end{align*} is minimized over $\beta>0$ at $\beta=t/(CL^2)$ when $t>0$, and the same bound at $t=0$ follows by taking any $\beta>0$. Substitution gives \begin{align*} \mu(A_t)\leq \exp\left(-\frac{t^2}{2CL^2}\right). \end{align*} [/step] [step:Apply the same argument to the lower tail] The map \begin{align*} -F\colon \mathbb{R}^n\to\mathbb{R} \end{align*} is smooth and $L$-Lipschitz, and its mean is $-m$. Applying the upper-tail estimate just proved to $-F$ gives, for every $t\geq0$, \begin{align*} \mu\left(\{x\in\mathbb{R}^n:-F(x)+m\geq t\}\right)\leq \exp\left(-\frac{t^2}{2CL^2}\right). \end{align*} This is exactly \begin{align*} \mu\left(\{x\in\mathbb{R}^n:F(x)-m\leq -t\}\right)\leq \exp\left(-\frac{t^2}{2CL^2}\right). \end{align*} Taking the union of the upper-tail and lower-tail events and using subadditivity of the probability measure $\mu$ gives \begin{align*} \mu\left(\{x\in\mathbb{R}^n:|F(x)-m|\geq t\}\right)\leq 2\exp\left(-\frac{t^2}{2CL^2}\right). \end{align*} [/step] [step:Pass from bounded functions to admissible unbounded truncations] Now suppose $F$ is not necessarily bounded. Let $(F_k)_{k\in\mathbb{N}}$ be bounded smooth truncations of $F$ satisfying the truncation hypothesis in the statement. For each $k\in\mathbb{N}$, let \begin{align*} F_k\colon \mathbb{R}^n\to\mathbb{R} \end{align*} be the corresponding truncation and define its mean $m_k\in\mathbb{R}$ by \begin{align*} m_k:=\int_{\mathbb{R}^n}F_k(x)\,d\mu(x). \end{align*} Since $F_k$ is bounded, smooth, and satisfies $|\nabla F_k(x)|\leq L$ for $\mu$-a.e. $x\in\mathbb{R}^n$, the bounded case gives, for every $\beta\in\mathbb{R}$, \begin{align*} \int_{\mathbb{R}^n}e^{\beta(F_k(x)-m_k)}\,d\mu(x)\leq \exp\left(\frac{CL^2\beta^2}{2}\right). \end{align*} The theorem statement assumes exactly the truncation passage needed here: the logarithmic Sobolev inequality is applied to the bounded smooth truncations, and the limits of the relevant means and exponential moments are justified by monotone convergence or by the [Dominated Convergence Theorem](/theorems/4), according to the chosen truncation scheme. Hence, for each fixed $\beta\in\mathbb{R}$, \begin{align*} \int_{\mathbb{R}^n}e^{\beta(F_k(x)-m_k)}\,d\mu(x)=e^{-\beta m_k}\int_{\mathbb{R}^n}e^{\beta F_k(x)}\,d\mu(x)\to e^{-\beta m}\int_{\mathbb{R}^n}e^{\beta F(x)}\,d\mu(x). \end{align*} Hence the same Laplace transform bound holds for $F$: \begin{align*} \int_{\mathbb{R}^n}e^{\beta(F(x)-m)}\,d\mu(x)\leq \exp\left(\frac{CL^2\beta^2}{2}\right). \end{align*} Repeating the exponential Markov optimization for $F$ gives the upper-tail estimate. Applying the same truncation argument to $-F$, using the admissible bounded smooth truncations of $-F$ from the statement, gives the lower-tail estimate. Combining the two tails by subadditivity of the probability measure $\mu$ gives \begin{align*} \mu\left(\{x\in\mathbb{R}^n:|F(x)-\mathbb{E}_{\mu}[F]|\geq t\}\right)\leq 2\exp\left(-\frac{t^2}{2CL^2}\right). \end{align*} This completes the proof. [/step]