[proofplan] We first prove the stronger positive-function form for smooth functions with polynomial growth and bounded away from zero. The proof uses the Ornstein-Uhlenbeck semigroup: entropy is dissipated along the flow, and the dissipation is a Gaussian Fisher-information term for the evolved function. The gradient commutation formula and Cauchy-Schwarz contract this Fisher information by an exponential factor, and invariance of $\gamma_n$ turns the time integral of that factor into $1/2$. Finally, applying the positive-function estimate to a strictly positive regularization of $f^2$ and passing to the limit gives the stated logarithmic Sobolev inequality on the smooth polynomial-growth core. [/proofplan] [step:Define the Ornstein-Uhlenbeck semigroup and its generator] Let $C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ denote the [vector space](/page/Vector%20Space) of smooth real-valued functions on $\mathbb{R}^n$ whose partial derivatives of every order have at most polynomial growth. We use the convention that a probability measure $\mu$ on $\mathbb{R}^n$ satisfies $\operatorname{LSI}(1)$ if every admissible smooth real-valued function $f$ satisfies \begin{align*} \operatorname{Ent}_{\mu}(f^2)\leq 2\int_{\mathbb{R}^n}|\nabla f(x)|^2\,d\mu(x). \end{align*} For $t \in [0,\infty)$, define the Ornstein-Uhlenbeck operator $P_t: C^\infty_{\mathrm{pol}}(\mathbb{R}^n) \to C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ by \begin{align*} P_t h(x) := \int_{\mathbb{R}^n} h(e^{-t}x+\sqrt{1-e^{-2t}}\,y)\, d\gamma_n(y). \end{align*} For a vector field $v=(v_1,\dots,v_n) \in (C^\infty_{\mathrm{pol}}(\mathbb{R}^n))^n$, define $P_t v := (P_t v_1,\dots,P_t v_n)$. The generator on this smooth core is the linear differential operator $L: C^\infty_{\mathrm{pol}}(\mathbb{R}^n) \to C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ defined by \begin{align*} Lh := \Delta h - x\cdot \nabla h. \end{align*} For $t>0$, set $a_t:=e^{-t}$ and $b_t:=\sqrt{1-e^{-2t}}$. Differentiating the defining [Gaussian integral](/theorems/1140) under the integral sign gives \begin{align*} \partial_t P_t h(x)=\int_{\mathbb{R}^n}\nabla h(a_t x+b_t y)\cdot\left(-a_t x+\frac{a_t^2}{b_t}y\right)\,d\gamma_n(y). \end{align*} The differentiation is justified by polynomial-growth bounds on the first derivatives of $h$ and Gaussian integrability of polynomials. Applying the one-dimensional Gaussian [integration by parts](/theorems/210) identity in each coordinate to $y_i\partial_{x_i}h(a_t x+b_t y)$ gives \begin{align*} \int_{\mathbb{R}^n}y_i\partial_{x_i}h(a_t x+b_t y)\,d\gamma_n(y)=b_t\int_{\mathbb{R}^n}\partial_{x_i x_i}h(a_t x+b_t y)\,d\gamma_n(y). \end{align*} Summing over $i$ yields \begin{align*} \partial_t P_t h(x)=a_t^2P_t(\Delta h)(x)-a_t x\cdot P_t(\nabla h)(x)=\Delta P_t h(x)-x\cdot\nabla P_t h(x)=L P_t h(x). \end{align*} The identity at $t=0$ follows by taking the right derivative, using the same Gaussian [integration by parts](/theorems/2098) computation after passing to the limit $t\downarrow0$. For every $h \in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ and every $t \geq 0$, the Ornstein-Uhlenbeck gradient commutation identity, obtained by differentiating under the Gaussian integral, gives \begin{align*} \nabla P_t h = e^{-t} P_t(\nabla h). \end{align*} The differentiation under the integral is justified because every derivative of $h$ has polynomial growth and every polynomial is integrable with respect to $\gamma_n$. The standard Gaussian measure is invariant under $(P_t)_{t\geq0}$: \begin{align*} \int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x) = \int_{\mathbb{R}^n} h(x)\, d\gamma_n(x). \end{align*} This Ornstein-Uhlenbeck invariance follows directly from the Gaussian representation of $P_t$: if $X$ and $Y$ are independent standard Gaussian random vectors in $\mathbb{R}^n$, then $e^{-t}X+\sqrt{1-e^{-2t}}Y$ is again standard Gaussian, since it has mean $0$ and covariance matrix $e^{-2t}I_n+(1-e^{-2t})I_n=I_n$. [/step] [step:Compute the entropy dissipation along the semigroup] Let $h \in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ be strictly positive and suppose that there is a constant $a>0$ such that $h(x)\geq a$ for every $x\in\mathbb{R}^n$. Define \begin{align*} m := \int_{\mathbb{R}^n} h(x)\, d\gamma_n(x). \end{align*} By invariance, $\int_{\mathbb{R}^n} P_t h\,d\gamma_n=m$ for every $t\geq0$. Recall that for a nonnegative Borel function $g: \mathbb{R}^n \to [0,\infty)$ with $g\log g\in L^1(\gamma_n)$, the Gaussian entropy is \begin{align*} \operatorname{Ent}_{\gamma_n}(g):=\int_{\mathbb{R}^n} g(x)\log g(x)\,d\gamma_n(x)-\left(\int_{\mathbb{R}^n}g(x)\,d\gamma_n(x)\right)\log\left(\int_{\mathbb{R}^n}g(x)\,d\gamma_n(x)\right). \end{align*} Define the entropy curve \begin{align*} \Phi: [0,\infty) \to \mathbb{R}, \qquad \Phi(t) := \operatorname{Ent}_{\gamma_n}(P_t h). \end{align*} Since $h\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$, the functions $P_t h$, $LP_t h$, and their derivatives have polynomial growth on compact $t$-intervals; since $h\geq a>0$, also $P_t h\geq a$ and $\log(P_t h)$ has at most logarithmic polynomial growth. These bounds give a Gaussian-integrable dominator for the derivative of the entropy integrand on compact $t$-intervals, so differentiation under the integral is justified by the [Dominated Convergence Theorem](/theorems/4). Thus \begin{align*} \Phi'(t) = \int_{\mathbb{R}^n} (L P_t h)(x)\log(P_t h(x))\, d\gamma_n(x). \end{align*} The Gaussian integration by parts identity for the generator $L$ says that, for $u,v \in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$, \begin{align*} \int_{\mathbb{R}^n} (Lu)(x)v(x)\, d\gamma_n(x) = -\int_{\mathbb{R}^n} \nabla u(x)\cdot \nabla v(x)\, d\gamma_n(x). \end{align*} This is the coordinate integration by parts formula for the Gaussian density, with the drift term $-x\cdot\nabla u$ cancelling the derivative of the density. We use it with $u=P_t h$ and $v=\log(P_t h)$ as follows: since $P_t h\geq a$ and $P_t h$ has polynomial growth with all derivatives, $v$ is smooth and $\nabla v$ has polynomial growth divided by a bounded-below denominator. Choose a smooth cutoff sequence $\rho_R: \mathbb{R}^n\to[0,1]$ with $\rho_R=1$ on $B(0,R)$, $\rho_R=0$ outside $B(0,2R)$, and $|\nabla\rho_R|\leq 2/R$; applying the identity to $v_R:=\rho_R v\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ and passing $R\to\infty$ by dominated convergence gives the same identity for $v=\log(P_t h)$. Applying this extended identity yields \begin{align*} \Phi'(t) = -\int_{\mathbb{R}^n} \frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x). \end{align*} Moreover $P_t h \to m$ pointwise and in every Gaussian $L^q$ space with finite $q$. To justify this, choose constants $C>0$ and $k\in\mathbb{N}$ such that $|h(z)|\leq C(1+|z|^k)$ for all $z\in\mathbb{R}^n$. For every $t\geq0$ and $x\in\mathbb{R}^n$, \begin{align*} |P_t h(x)|\leq C\int_{\mathbb{R}^n}\left(1+|e^{-t}x+\sqrt{1-e^{-2t}}y|^k\right)\, d\gamma_n(y)\leq C_1(1+|x|^k), \end{align*} where $C_1>0$ depends only on $C,k,n$ and the Gaussian $k$-th moment. For fixed $x$, the integrand $h(e^{-t}x+\sqrt{1-e^{-2t}}y)$ converges pointwise in $y$ to $h(y)$ and is dominated, for all sufficiently large $t$, by $C_2(1+|x|^k+|y|^k)$ for a constant $C_2>0$ depending only on $C$ and $k$. Dominated convergence with respect to $\gamma_n(y)$ gives $P_t h(x)\to m$. The bound $|P_t h(x)|\leq C_1(1+|x|^k)$, together with $P_t h\geq a$, implies that $(P_t h)\log(P_t h)$ is dominated by a fixed Gaussian-integrable polynomial-growth function. A second application of dominated convergence with respect to $\gamma_n(x)$ gives \begin{align*} \lim_{t\to\infty}\Phi(t)=\operatorname{Ent}_{\gamma_n}(m)=0. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $[0,\infty)$. For $S>0$, integrating the identity for $\Phi'$ over $[0,S]$ with respect to $\mathcal{L}^1$ gives \begin{align*} \Phi(0)-\Phi(S)=\int_0^S \int_{\mathbb{R}^n} \frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x)\, d\mathcal{L}^1(t). \end{align*} The integrand is nonnegative. Since $\Phi(S)\to0$, passing $S\to\infty$ and using the [Monotone Convergence Theorem](/theorems/509) gives \begin{align*} \operatorname{Ent}_{\gamma_n}(h)=\int_0^\infty \int_{\mathbb{R}^n} \frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x)\, d\mathcal{L}^1(t). \end{align*} [guided] The goal is to express entropy as the total amount of entropy dissipated by the Ornstein-Uhlenbeck flow. We start with a positive function $h$ bounded below by $a>0$ so that $\log(P_t h)$ is smooth and no singularity occurs in the quotient that will appear later. Define \begin{align*} m := \int_{\mathbb{R}^n} h(x)\, d\gamma_n(x). \end{align*} The invariance of $\gamma_n$ under $P_t$ gives \begin{align*} \int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x)=m. \end{align*} Thus the second term in the entropy of $P_t h$ is independent of $t$: \begin{align*} \left(\int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x)\right)\log\left(\int_{\mathbb{R}^n} P_t h(x)\, d\gamma_n(x)\right)=m\log m. \end{align*} Define the entropy curve $\Phi: [0,\infty) \to \mathbb{R}$ by \begin{align*} \Phi(t):=\operatorname{Ent}_{\gamma_n}(P_t h). \end{align*} Since $h$ and all its derivatives have polynomial growth, $P_t h$, $LP_t h$, and their derivatives have polynomial growth on compact $t$-intervals. The lower bound $h\geq a>0$ gives $P_t h\geq a$, so $\log(P_t h)$ has at most logarithmic polynomial growth. These bounds provide a Gaussian-integrable dominator for the derivative of the entropy integrand on compact $t$-intervals, so differentiating under the integral is justified by the [Dominated Convergence Theorem](/theorems/4). We also need the evolution equation for the semigroup. For $t>0$, set $a_t:=e^{-t}$ and $b_t:=\sqrt{1-e^{-2t}}$. Differentiating the Gaussian formula for $P_t h$ under the integral sign gives \begin{align*} \partial_t P_t h(x)=\int_{\mathbb{R}^n}\nabla h(a_t x+b_t y)\cdot\left(-a_t x+\frac{a_t^2}{b_t}y\right)\,d\gamma_n(y). \end{align*} The term involving $y$ is converted into a second derivative by Gaussian integration by parts: for each coordinate $i$, \begin{align*} \int_{\mathbb{R}^n}y_i\partial_{x_i}h(a_t x+b_t y)\,d\gamma_n(y)=b_t\int_{\mathbb{R}^n}\partial_{x_i x_i}h(a_t x+b_t y)\,d\gamma_n(y). \end{align*} After summing over $i$, this gives \begin{align*} \partial_t P_t h(x)=a_t^2P_t(\Delta h)(x)-a_t x\cdot P_t(\nabla h)(x). \end{align*} On the other hand, differentiating $P_t h$ with respect to the spatial variable gives $\nabla P_t h=a_tP_t(\nabla h)$ and $\Delta P_t h=a_t^2P_t(\Delta h)$, so \begin{align*} \partial_tP_t h(x)=\Delta P_t h(x)-x\cdot\nabla P_t h(x)=L P_t h(x). \end{align*} Therefore the chain rule gives \begin{align*} \Phi'(t)=\int_{\mathbb{R}^n} (L P_t h)(x)\bigl(\log(P_t h(x))+1\bigr)\, d\gamma_n(x)-\frac{d}{dt}(m\log m). \end{align*} The last derivative is $0$. Also, \begin{align*} \int_{\mathbb{R}^n} (L P_t h)(x)\, d\gamma_n(x)=\frac{d}{dt}\int_{\mathbb{R}^n}P_t h(x)\, d\gamma_n(x)=0. \end{align*} Therefore the $+1$ term drops out, and we obtain \begin{align*} \Phi'(t)=\int_{\mathbb{R}^n} (L P_t h)(x)\log(P_t h(x))\, d\gamma_n(x). \end{align*} Now apply the Gaussian integration by parts identity for the generator $L$ to the functions $u=P_t h$ and $v=\log(P_t h)$. This identity is the ordinary coordinate integration by parts formula for the Gaussian density, with the term $-x\cdot\nabla u$ in $L$ accounting for the derivative of that density. The stated identity is first proved for polynomial-growth smooth test functions; here $v$ has logarithmic polynomial growth, so we justify the extension. Choose a smooth cutoff sequence $\rho_R: \mathbb{R}^n\to[0,1]$ with $\rho_R=1$ on $B(0,R)$, $\rho_R=0$ outside $B(0,2R)$, and $|\nabla\rho_R|\leq 2/R$, and set $v_R:=\rho_R\log(P_t h)$. Then $v_R\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$, so integration by parts applies to $u$ and $v_R$. The terms involving $\nabla\rho_R$ vanish as $R\to\infty$ because Gaussian tails dominate every polynomial-growth factor, and the remaining terms converge by dominated convergence. Thus the identity applies to $u=P_t h$ and $v=\log(P_t h)$: \begin{align*} \int_{\mathbb{R}^n} (Lu)(x)v(x)\, d\gamma_n(x)=-\int_{\mathbb{R}^n}\nabla u(x)\cdot\nabla v(x)\, d\gamma_n(x). \end{align*} Since \begin{align*} \nabla \log(P_t h)(x)=\frac{\nabla P_t h(x)}{P_t h(x)}, \end{align*} we get \begin{align*} \Phi'(t)=-\int_{\mathbb{R}^n}\frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x). \end{align*} Finally, $P_t h$ converges to the constant $m$ as $t\to\infty$, and we spell out the domination needed for the entropy. Since $h$ has polynomial growth, choose constants $C>0$ and $k\in\mathbb{N}$ such that $|h(z)|\leq C(1+|z|^k)$ for all $z\in\mathbb{R}^n$. For every $t\geq0$ and $x\in\mathbb{R}^n$, \begin{align*} |P_t h(x)|\leq C\int_{\mathbb{R}^n}\left(1+|e^{-t}x+\sqrt{1-e^{-2t}}y|^k\right)\, d\gamma_n(y)\leq C_1(1+|x|^k), \end{align*} where $C_1>0$ depends only on $C,k,n$ and the Gaussian $k$-th moment. For fixed $x$, the expression $e^{-t}x+\sqrt{1-e^{-2t}}y$ converges to $y$. The integrand $h(e^{-t}x+\sqrt{1-e^{-2t}}y)$ is dominated, for all sufficiently large $t$, by $C_2(1+|x|^k+|y|^k)$ for a constant $C_2>0$, and this dominating function is integrable with respect to $\gamma_n(y)$. Therefore the [Dominated Convergence Theorem](/theorems/4) gives \begin{align*} P_t h(x)\to \int_{\mathbb{R}^n}h(y)\,d\gamma_n(y)=m. \end{align*} The lower bound $P_t h\geq a$ and the upper bound $|P_t h(x)|\leq C_1(1+|x|^k)$ imply that $(P_t h)\log(P_t h)$ is dominated by a fixed polynomial-growth function of $x$, hence by a $\gamma_n$-integrable function. The [Dominated Convergence Theorem](/theorems/4) with respect to $\gamma_n(x)$ gives \begin{align*} \lim_{t\to\infty}\Phi(t)=\operatorname{Ent}_{\gamma_n}(m)=0. \end{align*} For $S>0$, integrating the differential identity from $0$ to $S$ gives \begin{align*} \Phi(0)-\Phi(S)=\int_0^S \int_{\mathbb{R}^n}\frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x)\, d\mathcal{L}^1(t). \end{align*} The integrand is nonnegative, so the [Monotone Convergence Theorem](/theorems/509) applies as $S\to\infty$. Using $\Phi(S)\to0$, we obtain \begin{align*} \operatorname{Ent}_{\gamma_n}(h)=\int_0^\infty \int_{\mathbb{R}^n}\frac{|\nabla P_t h(x)|^2}{P_t h(x)}\, d\gamma_n(x)\, d\mathcal{L}^1(t). \end{align*} [/guided] [/step] [step:Bound the entropy dissipation using commutation and Cauchy-Schwarz] Fix $t\geq0$ and $x\in\mathbb{R}^n$. By the Ornstein-Uhlenbeck gradient commutation identity proved from the defining Gaussian integral, \begin{align*} |\nabla P_t h(x)|^2=e^{-2t}|P_t(\nabla h)(x)|^2. \end{align*} Write $Z_{t,x}: \mathbb{R}^n \to \mathbb{R}^n$ for the affine map defined by \begin{align*} Z_{t,x}(y):=e^{-t}x+\sqrt{1-e^{-2t}}\,y. \end{align*} Then \begin{align*} P_t(\nabla h)(x)=\int_{\mathbb{R}^n}\nabla h(Z_{t,x}(y))\, d\gamma_n(y), \end{align*} and \begin{align*} P_t h(x)=\int_{\mathbb{R}^n}h(Z_{t,x}(y))\, d\gamma_n(y). \end{align*} Let $L^2(\mathbb{R}^n,\gamma_n;\mathbb{R}^n)$ denote the [Hilbert space](/page/Hilbert%20Space) of equivalence classes of Borel maps $g: \mathbb{R}^n\to\mathbb{R}^n$ satisfying \begin{align*} \int_{\mathbb{R}^n}|g(y)|^2\,d\gamma_n(y)<\infty. \end{align*} Since $h>0$, the [Cauchy-Schwarz inequality](/theorems/432) in this Hilbert space, applied to the product \begin{align*} \frac{\nabla h(Z_{t,x}(y))}{\sqrt{h(Z_{t,x}(y))}}\sqrt{h(Z_{t,x}(y))} \end{align*} gives \begin{align*} |P_t(\nabla h)(x)|^2 \leq P_t\left(\frac{|\nabla h|^2}{h}\right)(x)P_t h(x). \end{align*} Therefore \begin{align*} \frac{|\nabla P_t h(x)|^2}{P_t h(x)}\leq e^{-2t}P_t\left(\frac{|\nabla h|^2}{h}\right)(x). \end{align*} [/step] [step:Integrate the dissipation estimate to prove the positive-function inequality] Using the entropy dissipation identity and the pointwise estimate from the previous step, the integrand is nonnegative because $h>0$. Therefore the nonnegative product-measure integration theorem, obtained from the [Monotone Convergence Theorem](/theorems/509) by approximation with increasing simple functions, justifies the space-time integration over $[0,\infty)\times\mathbb{R}^n$ and gives \begin{align*} \operatorname{Ent}_{\gamma_n}(h)\leq \int_0^\infty e^{-2t}\int_{\mathbb{R}^n}P_t\left(\frac{|\nabla h|^2}{h}\right)(x)\, d\gamma_n(x)\, d\mathcal{L}^1(t). \end{align*} By the Ornstein-Uhlenbeck invariance of $\gamma_n$ established from the Gaussian representation of $P_t$, \begin{align*} \int_{\mathbb{R}^n}P_t\left(\frac{|\nabla h|^2}{h}\right)(x)\, d\gamma_n(x)=\int_{\mathbb{R}^n}\frac{|\nabla h(x)|^2}{h(x)}\, d\gamma_n(x). \end{align*} Since \begin{align*} \int_0^\infty e^{-2t}\, d\mathcal{L}^1(t)=\frac{1}{2}, \end{align*} we obtain \begin{align*} \operatorname{Ent}_{\gamma_n}(h)\leq \frac{1}{2}\int_{\mathbb{R}^n}\frac{|\nabla h(x)|^2}{h(x)}\, d\gamma_n(x). \end{align*} Thus the positive-function logarithmic Sobolev inequality holds for every $h\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ bounded below by a positive constant. [/step] [step:Apply the positive-function inequality to $f^2+\varepsilon$] First suppose $f\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$. For $\varepsilon>0$, define \begin{align*} h_\varepsilon: \mathbb{R}^n \to (0,\infty), \qquad h_\varepsilon(x):=f(x)^2+\varepsilon. \end{align*} Then $h_\varepsilon\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ and $h_\varepsilon\geq\varepsilon$. Applying the positive-function inequality gives \begin{align*} \operatorname{Ent}_{\gamma_n}(f^2+\varepsilon)\leq \frac{1}{2}\int_{\mathbb{R}^n}\frac{|\nabla h_\varepsilon(x)|^2}{h_\varepsilon(x)}\, d\gamma_n(x). \end{align*} Since \begin{align*} \nabla h_\varepsilon(x)=2f(x)\nabla f(x), \end{align*} we have \begin{align*} \frac{|\nabla h_\varepsilon(x)|^2}{h_\varepsilon(x)}=\frac{4f(x)^2|\nabla f(x)|^2}{f(x)^2+\varepsilon}\leq 4|\nabla f(x)|^2. \end{align*} Therefore \begin{align*} \operatorname{Ent}_{\gamma_n}(f^2+\varepsilon)\leq 2\int_{\mathbb{R}^n}|\nabla f(x)|^2\, d\gamma_n(x). \end{align*} Let $\psi:[0,\infty)\to\mathbb{R}$ be defined by $\psi(r)=r\log r$, with the convention $\psi(0)=0$. Since $f\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$, the functions $\psi(f^2+\varepsilon)$ and $f^2+\varepsilon$ are dominated, for $0<\varepsilon\leq1$, by a fixed polynomial-growth function that is integrable with respect to $\gamma_n$. The [Dominated Convergence Theorem](/theorems/4) gives \begin{align*} \lim_{\varepsilon\downarrow0}\int_{\mathbb{R}^n}\psi(f(x)^2+\varepsilon)\, d\gamma_n(x)=\int_{\mathbb{R}^n}\psi(f(x)^2)\, d\gamma_n(x). \end{align*} Also $\int_{\mathbb{R}^n}(f(x)^2+\varepsilon)\, d\gamma_n(x)\to\int_{\mathbb{R}^n}f(x)^2\, d\gamma_n(x)$, and continuity of $r\mapsto r\log r$ on $[0,\infty)$ gives convergence of the normalization term in the entropy. Hence \begin{align*} \operatorname{Ent}_{\gamma_n}(f^2)\leq 2\int_{\mathbb{R}^n}|\nabla f(x)|^2\, d\gamma_n(x). \end{align*} [/step] [step:Identify the logarithmic Sobolev constant] The definition of $\operatorname{LSI}(1)$ for a probability measure $\mu$ is precisely the assertion that every admissible smooth function $f$ satisfies \begin{align*} \operatorname{Ent}_{\mu}(f^2)\leq 2\int_{\mathbb{R}^n} |\nabla f(x)|^2\, d\mu(x). \end{align*} Taking $\mu=\gamma_n$ and using the inequality just proved for every $f\in C^\infty_{\mathrm{pol}}(\mathbb{R}^n)$ shows that $\gamma_n$ satisfies $\operatorname{LSI}(1)$ on the smooth polynomial-growth core. [/step]