A Modified Poincaré Inequality in Discrete Settings

Created on October 20, 2025

2025   ·   formatting   math

This blog introduces the modified Poincaré inequality. Notably, analogous to the fact that log-Sobolev inequality implies Poincaré inequality, modified log-Sobolev inequality implies modified Poincaré inequality.

We first introduce some basic concepts. Let $\rho$ and $\mu$ be two distributions on \([n]=\{ 1,\cdots,n \}\) denoted by their probability mass functions. Assume that $\rho$ and $\mu$ have full support on $[n]$.

Definition 1 ($\mu$-variance). Let $f:[n]\to \mathbb{R}$ be a function and $\mu$ be a probability distribution on $[n]$. The variance of $f$ with respect to $\mu$ is defined as

\[\operatorname{Var}_\mu[f] = \mathbb{E}_\mu[(f-\mathbb{E}_\mu[f])^2] = \mathbb{E}_\mu[f^2]-(\mathbb{E}_\mu[f])^2,\]

where \(\mathbb{E}_\mu[f]=\sum_{x=1}^n \mu(x) f(x).\)

Definition 2 ($\mu$-covariance). Let $\mu$ be a probability distribution on $[n]$. The covariance of two functions $f,g:[n]\to\mathbb{R}$ is defined by

\[\operatorname{Cov}_\mu[f,g]=\mathbb{E}_\mu[(f-\mathbb{E}_\mu[f])(g-\mathbb{E}_\mu[g])]=\mathbb{E}_\mu[fg]-\mathbb{E}_\mu[f]\mathbb{E}_\mu[g].\]

By Cauchy-Schwarz inequality, we have

\[\operatorname{Cov}^2_\mu [f,g]\leq \operatorname{Var}_\mu[f] \operatorname{Var}_\mu[g].\]

Definition 3 ($\mu$-Entropy). Let $f:[n]\to \mathbb{R}$ be a function and $\mu$ be a probability distribution on $[n]$. Define

\[\operatorname{Ent}_\mu[f]=\mathbb{E}_\mu[f\log f]-\mathbb{E}_\mu[f]\log(\mathbb{E}_\mu[f]).\]

It can be easily verified that \(\operatorname{Ent}_\mu[\frac{\rho}{\mu}]=D_\mathrm{KL}(\rho\Vert \mu)\) for a probability distribution $\rho$ on $[n]$.

Definition 4 (Dirichlet Form, see Bobkov & Tetali (2006)). Let $([n],\mu)$ be a finite probability space, and let $P:[n]\times [n]\to [0,+\infty)$ be a non-negative function, called a kernel in the sequel. For all functions $f,g$ on $[n]$, one may define the associated Dirichlet form by

\[\mathcal{E}_\mu(f,g)=\frac{1}{2} \int \sum_{y=1}^n (f(x)-f(y))(g(x)-g(y))P(x,y) \, \mathrm{d}\mu(x)=\frac{1}{2} \sum_{x,y=1}^n (f(x)-f(y))(g(x)-g(y))P(x,y)\mu(x).\]

We further define corresponding gradient operator \(\nabla f(x)=\{\frac{1}{\sqrt{2}}(f(x)-f(y))\sqrt{P(x,y)}\}_{y\in [n]}\), so we can write out the gradient formula as follows:

\[\mathcal{E}_\mu(f,g)=\int \langle \nabla f(x),\nabla g(x)\rangle\, \mathrm{d}\mu(x).\]

Definition 5 (Log-Sobolev Inequality). Let $\mu$ be a probability distribution on $[n]$. We say $\mu$ satisfies the log-Sobolev inequality (LSI) with a constant $\alpha$ if for all function $f:[n]\to \mathbb{R}$ with $\operatorname{Ent}_\mu[f^2]<\infty$,

\begin{equation} \label{def:LSI} \frac{\alpha}{2}\operatorname{Ent}_\mu[f^2]\leq \mathcal{E}_\mu(f,f). \end{equation}

Definition 6 (Modified Log-Sobolev Inequality, see Bobkov & Tetali (2006)). Let $\mu$ be a probability distribution on $[n]$. We say $\mu$ satisfies the modified log-Sobolev inequality (MLSI) with a constant $\alpha$ if for all function \(f:[n]\to \mathbb{R}_{+}\) with \(\operatorname{Ent}_\mu[f]<\infty\),

\begin{equation} \label{def:MLSI1} 2\alpha \operatorname{Ent}_\mu[f]\leq \mathcal{E}_\mu(f,\log f). \end{equation}

In general the LSI is strictly stronger than MLSI, i.e., LSI implies inequality MLSI (see, e.g., Chewi (2023)). As he noted, in many cases, LSI is too strong in that it does not hold with a good constant $\alpha$; hence, MLSI is often the more appropriate inequality for the discrete setting.

Definition 7 (Poincaré Inequality) Let $\mu$ be a probability distribution on $[n]$. We say $\mu$ satisfies Poincaré inequality (PI) with a constant $\alpha$ if for all functions $f: [n] \to \mathbb{R}$, it holds that

\begin{equation} \label{def:PI} \alpha \cdot \operatorname{Var}_\mu [f] \leq \mathcal{E}_\mu(f,f). \end{equation}

We know that LSI implies PI with the same constant (see, e.g., Ledoux (2001)) and MLSI implies PI (see, e.g., Bobkov & Tetali (2006)).

We next introduce the modified Poincaré inequality:

Definition 8 (Modified Poincaré Inequality) Let $\mu$ be a distribution on $[n]$. We say that $\mu$ satisfies modified Poincaré inequality (MPI) with constant $\alpha$ if for all function $f:[n]\to \mathbb{R}_{+}$, the following inequality holds:

\begin{equation} \label{MPI} 4\alpha \cdot \operatorname{Var}_\mu[\sqrt{f}]\leq \mathcal{E}_\mu(f,\log f). \end{equation}

We can prove that MLSI implies MPI with the same constant. Moreover, PI implies MPI because of the inequality $4(\sqrt{a}-\sqrt{b})^2\leq (\log a-\log b)(a-b)$ for tow positive $a$ and $b$. Thus as discussed above, we have the relations between these inequalities with the same constant:

\[\begin{array}{ccc} \mathrm{LSI} \ \eqref{def:LSI} & \Longrightarrow & \mathrm{PI} \ \eqref{def:PI} \\ \Downarrow & \nearrow & \Downarrow \\ \mathrm{MLSI} \ \eqref{def:MLSI1} & \Longrightarrow & \mathrm{MPI} \ \eqref{MPI}. \end{array}\]

Finally, we directly prove that MLSI implies MPI with the same constant.

Proof. If $\mu$ satisfies MLSI with constant $\alpha$, we have that for all function $f:[n]\to \mathbb{R}_{+}$, it holds that

\begin{equation}\label{lemmaeq:MLSI} 2\alpha\cdot \operatorname{Ent}_\mu[f]\leq \mathcal{E}_\mu(f,\log f). \end{equation}

Note that for a small $\epsilon>0$,

\[\begin{align} &\operatorname{Ent}_\mu [1+\epsilon \sqrt{f}]\notag\\ ={}&\mathbb{E}_\mu [(1+\epsilon \sqrt{f})\log(1+\epsilon\sqrt{f})]-\mathbb{E}_\mu[1+\epsilon \sqrt{f}]\log\mathbb{E}_\mu[1+\epsilon\sqrt{f}]\notag\\ ={}&\mathbb{E}_\mu [(1+\epsilon \sqrt{f})(\epsilon\sqrt{f}-\frac{\epsilon^2}{2}f+O(\epsilon^3)]-(1+\epsilon\mathbb{E}_\mu[\sqrt{f}])(\epsilon\mathbb{E}_\mu[\sqrt{f}]-\frac{\epsilon^2}{2}\mathbb{E}_\mu[\sqrt{f}]^2+O(\epsilon^3))\notag\\ ={}&\frac{\epsilon^2}{2}\operatorname{Var}_\mu [\sqrt{f}]+O(\epsilon^3)\label{ent}, \end{align}\]

and

\[\begin{align} &\mathcal{E}_\mu (1+\epsilon\sqrt{f},\log (1+\epsilon\sqrt{f}))\notag\\ ={}&\frac{1}{2}\sum_{x,y=1}^n \mu(x)P(x,y)(\epsilon(\sqrt{f(x)}-\sqrt{f(y)}))(\log(1+\epsilon\sqrt{f(x)})-\log(1+\epsilon\sqrt{f(y)}))\notag\\ ={}&\frac{1}{2}\sum_{x,y=1}^n \mu(x)P(x,y)\epsilon(\sqrt{f(x)}-\sqrt{f(y)})(\epsilon\xi_\epsilon(x,y)(\sqrt{f(x)}-\sqrt{f(y)}))\notag\\ ={}&\epsilon^2\cdot\frac{1}{2}\sum_{x,y=1}^n \mu(x)P(x,y)(\sqrt{f(x)}-\sqrt{f(y)})^2\xi_\epsilon(x,y)\notag\\ \leq{}&\frac{\epsilon^2}{4}\frac{1}{2}\sum_{x,y=1}^n \mu(x)P(x,y)(f(x)-f(y))(\log f(x)-\log f(y))\xi_\epsilon(x,y),\label{dirichlet} \end{align}\]

where $\xi_\epsilon(x,y)$ is between $1+\epsilon\sqrt{f(x)}$ and $1+\epsilon\sqrt{f(y)}$, and when $\epsilon\to 0$, $\xi_\epsilon(x,y)\to 1$. The inequality in \eqref{dirichlet} holds since $4(\sqrt{a}-\sqrt{b})^2=(\int_a^b t^{-1/2}\mathrm{d}t)^2\leq (\int_a^b 1^2 \mathrm{d}t)(\int_a^b t^{-1}\mathrm{d}t)=(a-b)(\log a-\log b)$ for all $a,b>0$, where the inequality is by Cauchy-Schwarz inequality. Thus, by applying \eqref{lemmaeq:MLSI} to $1+\epsilon \sqrt{f}$ where $\epsilon>0$ is small, combining with \eqref{ent} and \eqref{dirichlet} and letting $\epsilon$ tend to 0, we obtain that \(4\alpha \cdot \operatorname{Var}_\mu[\sqrt{f}]\leq \mathcal{E}_\mu(f,\log f).\) We conclude the proof.$\square$