[proofplan] We first prove the entropy tensorization estimate for a non-negative [test function](/page/Test%20Function) on a finite product probability space. The only structural input is the convexity of entropy as a functional of the density, which lets us pass from a two-factor decomposition to the full $n$-factor inequality by induction. We then apply this tensorization to $g=f^2$ and use the one-coordinate logarithmic Sobolev inequality on each frozen-coordinate slice of $f$. Finally, we replace the individual constants by their maximum and explain the closure extension by approximation and lower semicontinuity of entropy. [/proofplan]

[step:Define the coordinate slices and conditional entropy terms] For $x=(x_1,\dots,x_n)\in E$, define $x_{-i}\in E_{-i}:=\prod_{j\neq i}E_j$ by deleting the $i$th coordinate, and define $\nu_{-i}:=\bigotimes_{j\neq i}\nu_j$. For $f\in C_c^\infty(E)$ and fixed $x_{-i}\in E_{-i}$, define the slice map by \begin{align*} f_{x_{-i}}:E_i\to\mathbb{R}. \end{align*} It is given by \begin{align*} x_i\mapsto f(x_1,\dots,x_n). \end{align*} Since $f$ is smooth and compactly supported on the finite product manifold or Euclidean product $E$, each slice $f_{x_{-i}}$ belongs to $C_c^\infty(E_i)$. For a non-negative integrable function $g:E\to[0,\infty)$ with $g\log g\in L^1(\nu)$, define the $i$th conditional entropy function by \begin{align*} H_i:E_{-i}\to[0,\infty]. \end{align*} It is given by \begin{align*} x_{-i}\mapsto \operatorname{Ent}_{\nu_i}\bigl(g_{x_{-i}}\bigr), \end{align*} where $g_{x_{-i}}:E_i\to[0,\infty)$ is the slice $x_i\mapsto g(x_1,\dots,x_n)$. For $g=f^2$, compact support and smoothness imply that all integrals below are finite. [/step]

[step:Prove entropy tensorization on a finite product probability space] [claim:Entropy tensorizes over finite products] Let $\nu=\nu_1\otimes\cdots\otimes\nu_n$ be a finite product of probability measures. For every non-negative integrable $g:E\to[0,\infty)$ with $g\log g\in L^1(\nu)$, one has \begin{align*} \operatorname{Ent}_\nu(g)\leq \sum_{i=1}^n \int_{E_{-i}} \operatorname{Ent}_{\nu_i}(g_{x_{-i}})\,d\nu_{-i}(x_{-i}). \end{align*} [/claim] [proof] It is enough to prove the two-factor inequality and then iterate. Let $(X,\mu)$ and $(Y,\rho)$ be probability spaces, let \begin{align*} h:X\times Y\to[0,\infty) \end{align*} be integrable, and assume $h\log h\in L^1(\mu\otimes\rho)$. Define \begin{align*} a:X\to[0,\infty) \end{align*} by \begin{align*} x\mapsto \int_Y h(x,y)\,d\rho(y). \end{align*} The integrability hypothesis $h\log h\in L^1(\mu\otimes\rho)$ and the lower bound for $s\mapsto s\log s$ ensure that the positive and negative parts occurring below are integrable. Hence the [Fubini theorem](/theorems/513) applies to the entropy decomposition. By the definition of entropy and the product measure, \begin{align*} \operatorname{Ent}_{\mu\otimes\rho}(h)=\int_X \operatorname{Ent}_\rho(h(x,\cdot))\,d\mu(x)+\operatorname{Ent}_\mu(a). \end{align*} It remains to bound the second term by the entropy in the $X$ variable. We use the convexity of the relative entropy integrand. For a probability measure $\mu$, define \begin{align*} \Psi:[0,\infty)\times(0,\infty)\to\mathbb{R} \end{align*} by \begin{align*} (s,r)\mapsto s\log(s/r)-s+r. \end{align*} For fixed $r>0$, take $0\log(0/r)=0$. The function $\Psi$ is jointly convex on $[0,\infty)\times(0,\infty)$. Therefore, for every non-negative $u:X\to[0,\infty)$ with $u\log u\in L^1(\mu)$, \begin{align*} \operatorname{Ent}_\mu(u)=\inf_{r>0}\int_X \Psi(u(x),r)\,d\mu(x). \end{align*} Indeed, the infimum in $r$ is attained at \begin{align*} r=\int_X u\,d\mu. \end{align*} Apply this formula to $a=\int_Y h(\cdot,y)\,d\rho(y)$. Convexity of $\Psi$ and the fact that $\rho$ is a probability measure give, for every integrable map $r:Y\to(0,\infty)$, \begin{align*} \int_X \Psi\left(\int_Y h(x,y)\,d\rho(y),\int_Y r(y)\,d\rho(y)\right)\,d\mu(x)\leq \int_Y\int_X \Psi(h(x,y),r(y))\,d\mu(x)\,d\rho(y). \end{align*} The iterated integral is justified by Tonelli for the non-negative function $\Psi(h(x,y),r(y))$. To pass from the right-hand side to pointwise entropies, choose for each $m\in\mathbb{N}$ a positive [simple function](/page/Simple%20Function) $r_m:Y\to(0,\infty)$ whose values approximate, within $1/m$ on each level set, the scalar minimizers in the variational formula for $\operatorname{Ent}_\mu(h(\cdot,y))$. Monotone approximation of measurable non-negative functions and the variational formula then give \begin{align*} \operatorname{Ent}_\mu(a)\leq \int_Y \operatorname{Ent}_\mu(h(\cdot,y))\,d\rho(y). \end{align*} Substituting this estimate into the entropy decomposition yields \begin{align*} \operatorname{Ent}_{\mu\otimes\rho}(h)\leq \int_X \operatorname{Ent}_\rho(h(x,\cdot))\,d\mu(x)+\int_Y \operatorname{Ent}_\mu(h(\cdot,y))\,d\rho(y). \end{align*} This is the two-factor tensorization inequality. For $n>2$, apply the two-factor inequality to the decomposition $E=(E_1\times\cdots\times E_{n-1})\times E_n$. The first term is then tensorized again over $E_1,\dots,E_{n-1}$. Induction over $n$ gives the asserted finite-product estimate. [/proof] [/step]

[step:Apply one-coordinate logarithmic Sobolev inequalities to the frozen slices]Apply the entropy tensorization claim to $g:E\to[0,\infty)$, $x\mapsto f(x)^2$. This gives \begin{align*} \operatorname{Ent}_\nu(f^2)\leq \sum_{i=1}^n \int_{E_{-i}}\operatorname{Ent}_{\nu_i}\bigl(f_{x_{-i}}^2\bigr)\,d\nu_{-i}(x_{-i}). \end{align*} For each fixed $i$ and $x_{-i}$, the slice $f_{x_{-i}}$ belongs to $C_c^\infty(E_i)$, so the assumed one-coordinate $\operatorname{LSI}(C_i)$ gives \begin{align*} \operatorname{Ent}_{\nu_i}\bigl(f_{x_{-i}}^2\bigr)\leq 2C_i\int_{E_i}|\nabla_i f(x_1,\dots,x_n)|^2\,d\nu_i(x_i). \end{align*} Integrating this estimate over $E_{-i}$ with respect to $\nu_{-i}$ and applying Tonelli to the non-negative gradient-square integrand yields \begin{align*} \int_{E_{-i}}\operatorname{Ent}_{\nu_i}\bigl(f_{x_{-i}}^2\bigr)\,d\nu_{-i}(x_{-i})\leq 2C_i\int_E|\nabla_i f(x)|^2\,d\nu(x). \end{align*} Summing over $i$ proves \begin{align*} \operatorname{Ent}_\nu(f^2)\leq 2\sum_{i=1}^n C_i\int_E|\nabla_i f(x)|^2\,d\nu(x). \end{align*}[/step]

[guided]The point of tensorization is that the entropy of $f^2$ under the product measure can be controlled by entropies in one coordinate at a time. We apply the tensorization estimate to the non-negative function $g:E\to[0,\infty)$ given by $x\mapsto f(x)^2$. Because $f\in C_c^\infty(E)$, the function $g$ is integrable and $g\log g$ is integrable with respect to $\nu$. The tensorization estimate gives \begin{align*} \operatorname{Ent}_\nu(f^2)\leq \sum_{i=1}^n \int_{E_{-i}}\operatorname{Ent}_{\nu_i}\bigl(f_{x_{-i}}^2\bigr)\,d\nu_{-i}(x_{-i}). \end{align*} Now fix an index $i\in\{1,\dots,n\}$ and fix the remaining coordinates $x_{-i}\in E_{-i}$. The function to which we apply the coordinate logarithmic Sobolev inequality is not $f$ on the whole product, but the one-variable slice \begin{align*} f_{x_{-i}}:E_i\to\mathbb{R}. \end{align*} It is given by \begin{align*} x_i\mapsto f(x_1,\dots,x_n). \end{align*} This slice is smooth in the $E_i$ variable. Its support is contained in the projection of $\operatorname{supp} f$ onto $E_i$, hence is compact. Therefore $f_{x_{-i}}\in C_c^\infty(E_i)$, exactly the test class allowed in the hypothesis. The assumed $\operatorname{LSI}(C_i)$ on $(E_i,\nu_i)$ applies to this slice and gives \begin{align*} \operatorname{Ent}_{\nu_i}\bigl(f_{x_{-i}}^2\bigr)\leq 2C_i\int_{E_i}|\nabla_i f_{x_{-i}}(x_i)|^2\,d\nu_i(x_i). \end{align*} By definition of the coordinate gradient, $\nabla_i f_{x_{-i}}(x_i)$ is exactly the gradient of $f$ in the $E_i$ variable at the point $x=(x_1,\dots,x_n)$. Thus \begin{align*} \int_{E_i}|\nabla_i f_{x_{-i}}(x_i)|^2\,d\nu_i(x_i)=\int_{E_i}|\nabla_i f(x_1,\dots,x_n)|^2\,d\nu_i(x_i). \end{align*} We now integrate this inequality over all frozen values $x_{-i}$ with respect to $\nu_{-i}$. Since $\nu=\nu_{-i}\otimes\nu_i$ after the coordinates are ordered with $E_i$ last, Tonelli's theorem applies to the non-negative gradient-square integrand and identifies the iterated integral with the full product integral: \begin{align*} \int_{E_{-i}}\operatorname{Ent}_{\nu_i}\bigl(f_{x_{-i}}^2\bigr)\,d\nu_{-i}(x_{-i})\leq 2C_i\int_E|\nabla_i f(x)|^2\,d\nu(x). \end{align*} Substituting this estimate into the tensorized entropy bound and summing over $i$ gives \begin{align*} \operatorname{Ent}_\nu(f^2)\leq 2\sum_{i=1}^n C_i\int_E|\nabla_i f(x)|^2\,d\nu(x). \end{align*} This is the refined tensorized logarithmic Sobolev inequality, with the individual coordinate constants still visible.[/guided]

[step:Replace the coordinate constants by the product constant] Define \begin{align*} C:=\max_{1\leq i\leq n}C_i. \end{align*} Since $C_i\leq C$ for every $i$, the refined estimate gives \begin{align*} \operatorname{Ent}_\nu(f^2)\leq 2C\sum_{i=1}^n\int_E|\nabla_i f(x)|^2\,d\nu(x). \end{align*} By definition of the product gradient on $E$, its squared norm is \begin{align*} |\nabla f(x)|^2=\sum_{i=1}^n|\nabla_i f(x)|^2. \end{align*} Therefore \begin{align*} \operatorname{Ent}_\nu(f^2)\leq 2C\int_E|\nabla f(x)|^2\,d\nu(x). \end{align*} This is precisely $\operatorname{LSI}(C)$ for the product measure on the test class $C_c^\infty(E)$. [/step]

[step:Pass the estimate to the Dirichlet closure by approximation] Assume the coordinate inequalities have been extended to the corresponding one-coordinate Dirichlet closures, and let $F$ belong to the closure of $C_c^\infty(E)$ under the norm $\|u\|_{L^2(\nu)}^2+\sum_{i=1}^n\int_E|\nabla_i u|^2\,d\nu(x)$. Choose a sequence $(f_k)_{k\in\mathbb{N}}$ in $C_c^\infty(E)$ such that $f_k\to F$ in this norm. Then $f_k\to F$ in $L^2(\nu)$, and for each $i$ the coordinate gradients $\nabla_i f_k$ converge in $L^2(\nu)$ to the corresponding weak Dirichlet gradient $\nabla_i F$. The already proved estimate gives, for every $k$, \begin{align*} \operatorname{Ent}_\nu(f_k^2)\leq 2\sum_{i=1}^n C_i\int_E|\nabla_i f_k(x)|^2\,d\nu(x). \end{align*} Since $f_k\to F$ in $L^2(\nu)$, after passing to a subsequence realizing the lower limit we have $f_k^2\to F^2$ in $L^1(\nu)$ and almost everywhere. The function $s\mapsto s\log s$ is bounded below on $[0,\infty)$, so Fatou's lemma applied to the non-negative functions $f_k^2\log(f_k^2)+\mathrm{e}^{-1}$ gives \begin{align*} \int_E F(x)^2\log(F(x)^2)\,d\nu(x)\leq \liminf_{k\to\infty}\int_E f_k(x)^2\log(f_k(x)^2)\,d\nu(x). \end{align*} Also, \begin{align*} \int_E f_k(x)^2\,d\nu(x)\to \int_E F(x)^2\,d\nu(x). \end{align*} Hence \begin{align*} \operatorname{Ent}_\nu(F^2)\leq \liminf_{k\to\infty}\operatorname{Ent}_\nu(f_k^2). \end{align*} Taking the lower limit in the smooth estimates and using the $L^2$ convergence of each gradient gives \begin{align*} \operatorname{Ent}_\nu(F^2)\leq 2\sum_{i=1}^n C_i\int_E|\nabla_i F(x)|^2\,d\nu(x). \end{align*} Thus the same tensorized logarithmic Sobolev estimate holds on the Dirichlet closure. This completes the proof. [/step]