[proofplan] We first prove the inequality for bounded smooth functions by centering $f$ and applying the logarithmic Sobolev inequality to the perturbation $1+\varepsilon f$. A second-order expansion of the entropy at $\varepsilon = 0$ gives the variance term, while the Dirichlet side gives the gradient term. Dividing by $\varepsilon^2$ and letting $\varepsilon \to 0$ yields Poincare for bounded smooth functions. Finally, smooth truncations reduce the general $L^2$ case to the bounded case. [/proofplan] [step:Define entropy and restate the logarithmic Sobolev hypothesis] For every non-negative measurable function $H: \mathbb{R}^n \to [0,\infty)$ for which the following quantities are finite, define the entropy of $H$ with respect to $\nu$ by \begin{align*} \operatorname{Ent}_\nu(H) := \int_{\mathbb{R}^n} H\log H \, d\nu(x) - \left(\int_{\mathbb{R}^n} H \, d\nu(x)\right)\log\left(\int_{\mathbb{R}^n} H \, d\nu(x)\right). \end{align*} The assumption $\operatorname{LSI}(C)$ means that every smooth function $h: \mathbb{R}^n \to \mathbb{R}$ for which the quantities are finite satisfies \begin{align*} \operatorname{Ent}_\nu(h^2) \leq 2C \int_{\mathbb{R}^n} |\nabla h|^2 \, d\nu(x). \end{align*} [/step] [step:Reduce the bounded case to mean-zero functions] Let $f: \mathbb{R}^n \to \mathbb{R}$ be a bounded smooth function such that $|\nabla f| \in L^2(\nu)$. Define its $\nu$-mean $m \in \mathbb{R}$ by \begin{align*} m := \int_{\mathbb{R}^n} f \, d\nu(x), \end{align*} and define the centered smooth function $u: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} u(x) := f(x) - m. \end{align*} Then \begin{align*} \int_{\mathbb{R}^n} u \, d\nu(x) = 0, \end{align*} and, because subtracting a constant does not change the gradient, \begin{align*} \nabla u = \nabla f. \end{align*} Moreover, \begin{align*} \operatorname{Var}_\nu(f) = \int_{\mathbb{R}^n} u^2 \, d\nu(x). \end{align*} Thus it is enough in the bounded case to prove \begin{align*} \int_{\mathbb{R}^n} u^2 \, d\nu(x) \leq C \int_{\mathbb{R}^n} |\nabla u|^2 \, d\nu(x) \end{align*} for every bounded smooth function $u: \mathbb{R}^n \to \mathbb{R}$ with $\int_{\mathbb{R}^n} u \, d\nu(x)=0$. [/step] [step:Apply the logarithmic Sobolev inequality to $1+\varepsilon u$] Fix a bounded smooth function $u: \mathbb{R}^n \to \mathbb{R}$ satisfying \begin{align*} \int_{\mathbb{R}^n} u \, d\nu(x)=0. \end{align*} If $|\nabla u| \notin L^2(\nu)$, then the desired inequality for $u$ has infinite right-hand side and is immediate. Thus, for the perturbative argument, assume $|\nabla u| \in L^2(\nu)$. For $\varepsilon \in \mathbb{R}$, define the smooth function $g_\varepsilon: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} g_\varepsilon(x) := 1 + \varepsilon u(x). \end{align*} Since $u$ is bounded, $g_\varepsilon^2 \log(g_\varepsilon^2)$ is bounded for each fixed sufficiently small $\varepsilon$, so the entropy term is finite. Since $|\nabla u| \in L^2(\nu)$, the Dirichlet term is finite. Hence the assumed $\operatorname{LSI}(C)$ applies to $g_\varepsilon$. Also, \begin{align*} \nabla g_\varepsilon(x) = \varepsilon \nabla u(x). \end{align*} Therefore \begin{align*} \operatorname{Ent}_\nu(g_\varepsilon^2) \leq 2C\varepsilon^2 \int_{\mathbb{R}^n} |\nabla u|^2 \, d\nu(x). \end{align*} [guided] The logarithmic Sobolev inequality controls the entropy of a square. To extract a variance inequality from it, we perturb the constant function $1$ in the direction $u$. Define $g_\varepsilon: \mathbb{R}^n \to \mathbb{R}$ by the rule \begin{align*} g_\varepsilon(x) := 1+\varepsilon u(x). \end{align*} This choice is useful because the constant function has zero entropy, and the first nonzero term in the entropy expansion around $1$ is quadratic in $\varepsilon$. We verify the hypotheses needed for the assumed $\operatorname{LSI}(C)$. The function $u$ is smooth and bounded, hence $g_\varepsilon$ is smooth and bounded for every $\varepsilon \in \mathbb{R}$. For sufficiently small $\varepsilon$, the function $g_\varepsilon^2 \log(g_\varepsilon^2)$ is also bounded because $u$ takes values in a bounded interval and the scalar function $s \mapsto s^2\log(s^2)$ extends continuously at $s=0$ by assigning value $0$. Thus the entropy term is finite. Since $|\nabla u| \in L^2(\nu)$, the Dirichlet term is finite. The gradient is \begin{align*} \nabla g_\varepsilon(x) = \varepsilon \nabla u(x), \end{align*} so \begin{align*} |\nabla g_\varepsilon(x)|^2 = \varepsilon^2 |\nabla u(x)|^2. \end{align*} Applying the assumed logarithmic Sobolev inequality to $g_\varepsilon$ gives \begin{align*} \operatorname{Ent}_\nu(g_\varepsilon^2) \leq 2C \int_{\mathbb{R}^n} |\nabla g_\varepsilon|^2 \, d\nu(x). \end{align*} Substituting the gradient identity yields \begin{align*} \operatorname{Ent}_\nu(g_\varepsilon^2) \leq 2C\varepsilon^2 \int_{\mathbb{R}^n} |\nabla u|^2 \, d\nu(x). \end{align*} [/guided] [/step] [step:Expand the entropy to second order] Define the scalar function $\Phi: \mathbb{R} \to \mathbb{R}$ by setting $\Phi(0) := 0$ and, for $s \neq 0$, \begin{align*} \Phi(s) := s^2\log(s^2). \end{align*} We use the following little-o notation: for a real-valued function $r$ defined near $0$, the expression $r(a)=o(a^2)$ as $a \to 0$ means $r(a)/a^2 \to 0$, and $r(a)=o(1)$ as $a \to 0$ means $r(a) \to 0$. For $s$ near $0$, Taylor expansion at $1$ gives \begin{align*} \Phi(1+s) = 2s + 3s^2 + s^2\rho(s) \end{align*} where $\rho: (-\delta,\delta) \to \mathbb{R}$ is a function satisfying $\rho(s) \to 0$ as $s \to 0$ for some $\delta>0$. Let $M := \sup_{x\in\mathbb{R}^n}|u(x)|$. For $|\varepsilon|M<\delta$, the remainder satisfies \begin{align*} \sup_{x\in\mathbb{R}^n}|\rho(\varepsilon u(x))| \to 0 \end{align*} as $\varepsilon \to 0$. Since $\nu$ is a probability measure and $u$ is bounded, this uniform Taylor remainder gives \begin{align*} \int_{\mathbb{R}^n} \varepsilon^2u(x)^2\rho(\varepsilon u(x)) \, d\nu(x)=o(\varepsilon^2). \end{align*} Therefore, integrating the expansion with $s=\varepsilon u(x)$ gives \begin{align*} \int_{\mathbb{R}^n} g_\varepsilon^2 \log(g_\varepsilon^2) \, d\nu(x) = 2\varepsilon \int_{\mathbb{R}^n} u \, d\nu(x) + 3\varepsilon^2 \int_{\mathbb{R}^n} u^2 \, d\nu(x) + o(\varepsilon^2). \end{align*} The mean-zero condition cancels the linear term, so \begin{align*} \int_{\mathbb{R}^n} g_\varepsilon^2 \log(g_\varepsilon^2) \, d\nu(x) = 3\varepsilon^2 \int_{\mathbb{R}^n} u^2 \, d\nu(x) + o(\varepsilon^2). \end{align*} Next, \begin{align*} \int_{\mathbb{R}^n} g_\varepsilon^2 \, d\nu(x) = \int_{\mathbb{R}^n} (1+2\varepsilon u+\varepsilon^2u^2) \, d\nu(x) = 1+\varepsilon^2\int_{\mathbb{R}^n} u^2 \, d\nu(x). \end{align*} Using $\log(1+t)=t+o(t)$ as $t \to 0$, we obtain \begin{align*} \left(\int_{\mathbb{R}^n} g_\varepsilon^2 \, d\nu(x)\right) \log\left(\int_{\mathbb{R}^n} g_\varepsilon^2 \, d\nu(x)\right) = \varepsilon^2 \int_{\mathbb{R}^n} u^2 \, d\nu(x) + o(\varepsilon^2). \end{align*} Subtracting the two expansions gives \begin{align*} \operatorname{Ent}_\nu(g_\varepsilon^2) = 2\varepsilon^2 \int_{\mathbb{R}^n} u^2 \, d\nu(x) + o(\varepsilon^2). \end{align*} [/step] [step:Let the perturbation parameter tend to zero] Combining the entropy expansion with the logarithmic Sobolev estimate gives \begin{align*} 2\varepsilon^2 \int_{\mathbb{R}^n} u^2 \, d\nu(x) + o(\varepsilon^2) \leq 2C\varepsilon^2 \int_{\mathbb{R}^n} |\nabla u|^2 \, d\nu(x). \end{align*} For $\varepsilon \neq 0$, divide by $2\varepsilon^2$: \begin{align*} \int_{\mathbb{R}^n} u^2 \, d\nu(x) + o(1) \leq C \int_{\mathbb{R}^n} |\nabla u|^2 \, d\nu(x). \end{align*} Letting $\varepsilon \to 0$ yields \begin{align*} \int_{\mathbb{R}^n} u^2 \, d\nu(x) \leq C \int_{\mathbb{R}^n} |\nabla u|^2 \, d\nu(x). \end{align*} By the centering reduction, the Poincare inequality holds for every bounded smooth function $f: \mathbb{R}^n \to \mathbb{R}$ with $|\nabla f| \in L^2(\nu)$. [/step] [step:Pass from bounded smooth functions to general smooth functions by truncation] Let $f: \mathbb{R}^n \to \mathbb{R}$ be smooth, assume $f \in L^2(\nu)$, and assume \begin{align*} \int_{\mathbb{R}^n} |\nabla f|^2 \, d\nu(x) < \infty. \end{align*} For each $R \geq 1$, choose a smooth even cutoff function $\eta_R: \mathbb{R} \to [0,1]$ such that $\eta_R(t)=1$ for $|t|\leq R$ and $\eta_R(t)=0$ for $|t|\geq 2R$. Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}$. Define the smooth truncation function $\theta_R: \mathbb{R} \to \mathbb{R}$ by \begin{align*} \theta_R(t) := \int_0^t \eta_R(s) \, d\mathcal{L}^1(s). \end{align*} Then $\theta_R$ is bounded because $\eta_R$ has compact support, $\theta_R(t)=t$ for $|t|\leq R$, $|\theta_R(t)|\leq |t|$ for all $t \in \mathbb{R}$ because $0\leq \eta_R\leq 1$, and $|\theta_R'(t)|=|\eta_R(t)|\leq 1$ for all $t \in \mathbb{R}$. Define the bounded smooth function $f_R: \mathbb{R}^n \to \mathbb{R}$ by \begin{align*} f_R(x) := \theta_R(f(x)). \end{align*} The bounded case gives \begin{align*} \operatorname{Var}_\nu(f_R) \leq C \int_{\mathbb{R}^n} |\nabla f_R|^2 \, d\nu(x). \end{align*} [guided] We now justify the truncation passage in detail. Let $f: \mathbb{R}^n \to \mathbb{R}$ be smooth, assume $f \in L^2(\nu)$, and assume \begin{align*} \int_{\mathbb{R}^n} |\nabla f|^2 \, d\nu(x) < \infty. \end{align*} For each $R \geq 1$, choose a smooth even cutoff function $\eta_R: \mathbb{R} \to [0,1]$ such that $\eta_R(t)=1$ for $|t|\leq R$ and $\eta_R(t)=0$ for $|t|\geq 2R$. Define $\theta_R: \mathbb{R} \to \mathbb{R}$ by \begin{align*} \theta_R(t) := \int_0^t \eta_R(s) \, d\mathcal{L}^1(s). \end{align*} Then $\theta_R$ is smooth and bounded, $\theta_R(t)=t$ for $|t|\leq R$, $|\theta_R(t)|\leq |t|$ for all $t \in \mathbb{R}$, and $|\theta_R'(t)|\leq 1$ for all $t \in \mathbb{R}$. Define $f_R: \mathbb{R}^n \to \mathbb{R}$ by $f_R(x):=\theta_R(f(x))$. The bounded Poincare inequality applies to $f_R$ and gives \begin{align*} \operatorname{Var}_\nu(f_R) \leq C \int_{\mathbb{R}^n} |\nabla f_R|^2 \, d\nu(x). \end{align*} The chain rule gives $\nabla f_R(x)=\theta_R'(f(x))\nabla f(x)$, hence $|\nabla f_R(x)|^2\leq |\nabla f(x)|^2$. Also $f_R(x)\to f(x)$ for every $x \in \mathbb{R}^n$ and $|f_R(x)|\leq |f(x)|$. Since $f \in L^2(\nu)$, the [Dominated Convergence Theorem](/theorems/8) applied with dominating function $4|f|^2$ gives $f_R\to f$ in $L^2(\nu)$. The variance map is continuous on $L^2(\nu)$, so $\operatorname{Var}_\nu(f_R)\to\operatorname{Var}_\nu(f)$. Applying the [Dominated Convergence Theorem](/theorems/8) again to $|\theta_R'(f)|^2|\nabla f|^2\leq |\nabla f|^2$ gives convergence of the Dirichlet terms. Passing to the limit in the bounded inequality yields \begin{align*} \operatorname{Var}_\nu(f) \leq C \int_{\mathbb{R}^n} |\nabla f|^2 \, d\nu(x). \end{align*} [/guided] By the chain rule, \begin{align*} \nabla f_R(x) = \theta_R'(f(x))\nabla f(x), \end{align*} and therefore \begin{align*} |\nabla f_R(x)|^2 \leq |\nabla f(x)|^2. \end{align*} Also $f_R(x) \to f(x)$ for every $x \in \mathbb{R}^n$, and $|f_R(x)| \leq |f(x)|$. Since $|f_R(x)| \leq |f(x)|$ and $f \in L^2(\nu)$, the [Dominated Convergence Theorem](/theorems/8) applied to $|f_R-f|^2 \leq 4|f|^2$ gives $f_R \to f$ in $L^2(\nu)$. The map $v \mapsto \operatorname{Var}_\nu(v)$ is continuous on $L^2(\nu)$ because \begin{align*} |\operatorname{Var}_\nu(v)-\operatorname{Var}_\nu(w)| \leq \|v-w\|_{L^2(\nu)}\bigl(\|v\|_{L^2(\nu)}+\|w\|_{L^2(\nu)}\bigr)+2\|v-w\|_{L^2(\nu)}\bigl(\|v\|_{L^2(\nu)}+\|w\|_{L^2(\nu)}\bigr). \end{align*} Hence \begin{align*} \operatorname{Var}_\nu(f_R) \to \operatorname{Var}_\nu(f). \end{align*} For each fixed $t \in \mathbb{R}$, the identity $\theta_R(t)=t$ holds for all $R>|t|$, hence $\theta_R'(t)=1$ for all $R>|t|$. Therefore $|\theta_R'(f(x))|^2|\nabla f(x)|^2 \to |\nabla f(x)|^2$ for every $x \in \mathbb{R}^n$. Applying the [Dominated Convergence Theorem](/theorems/8) to the pointwise convergent integrands $|\theta_R'(f)|^2|\nabla f|^2 \leq |\nabla f|^2$ gives \begin{align*} \int_{\mathbb{R}^n} |\nabla f_R|^2 \, d\nu(x) \to \int_{\mathbb{R}^n} |\nabla f|^2 \, d\nu(x). \end{align*} Passing to the limit in the bounded Poincare inequality for $f_R$ yields \begin{align*} \operatorname{Var}_\nu(f) \leq C \int_{\mathbb{R}^n} |\nabla f|^2 \, d\nu(x). \end{align*} This is the desired Poincare inequality with the same constant $C$. [/step]