The Lyapunov Function Method for Poincaré Inequality

An elegant functional inequality that unifies ODEs, PDEs, SDEs, functional analysis, and Riemannian geometry.

Poincaré (spectral gap) inequality (Bakry et al., 2008) is the first important family of functional inequalities that charaterizes the exponential convergence of a random variable towards the equilibrium.

Langevin diffusion

Suppose we are interested in the convergence of the stochastic differential equation

\begin{equation}\notag dx_t = -\nabla U(x_t)dt + \sqrt{2}dW_t, \end{equation}

where $\nabla U(\cdot)$ denotes the gradient of a energy function $U$ and $(W_t)_{t\geq 0}$ is a Brownian motion. Under mild growth conditions on $U(\cdot)$, $x$ converges to a stationary measure $\mu(x)\propto e^{-U(x)}$.

Define a family of operators $(P_t)_{t\geq 0}$ as follows

\begin{equation}\notag P_t(f(x)) = E[f(x_t)|$=x], \end{equation} where the expectation is taken over a particular set to denote the conditional density.

For a smooth test function $f(x)$, Itô’s formula implies that

\begin{equation}\notag d f(x_t) = \sqrt{2} \nabla f(x_t) dB_t + Lf(x_t)dt, \end{equation} where $L$ is the infinitesimal generator of the symmetric Markov Semigroup $P_t$

\begin{equation}\notag Lf=\lim_{t\rightarrow 0} \frac{P_t f -f }{t}=\Delta f - \langle\nabla U, \nabla f\rangle, \end{equation} where $\Delta$ denotes the Laplace operator.

Poincaré Inequality

We say the Gibbs measure $\mu$ satisfies a Poincaré equality with a constant $C$ if

\begin{equation}\notag Var_{\mu}(f)=\int f^2 d\mu -(\int f d\mu)^2 \leq C \xi(f), \end{equation} where $\xi$ is the Dirichlet form defined as

\begin{equation}\notag \xi(f)=\int \Gamma(f)d\mu. \end{equation}

$\Gamma$ is the Carré du Champ operator satisfying

\begin{equation}\notag \Gamma(f)=\frac{1}{2}(L(f^2)-2 f L(f)). \end{equation} Since $\mu$ is reversible for $P_t$, we have the invariance property $\int L(f)=0$ for all f in the Dirichlet domain. In other words, for symmetric $\mu$, we have

\begin{equation}\notag \xi(f)=\int \Gamma(f)d\mu=-\int f L(f) d\mu =\int (\nabla f)^2 d\mu. \end{equation} where the last inequality follows by integration by parts such that: $-\int f L(f) d\mu=-\int f\nabla (e^{-U(x)}\nabla f)dx=-\int f d(e^{-U(x)} \nabla f)=f e^{-U(x)} \nabla f|_{boundary} + \int (\nabla f)^2 d\mu.$

Variance Decay

Now we study the decay of variance

\begin{equation}\notag \Lambda(t)=Var_{\mu}(P_t f)= \int(P_t f)^2d\mu. \end{equation} Reacll $\xi(f)=-\int f L(f) d\mu$. Taking the derivative

\begin{equation}\notag \Lambda_t(t)=2\int P_t f L P_t f d\mu = -2 \xi(P_t f). \end{equation}

Combining the Poincaré equality, we have that

\begin{equation}\notag \Lambda(t)=Var_{\mu}(P_t f)\leq C \xi(P_t f)=-\frac{C}{2}\Lambda_t(t) \end{equation} This means that $\Lambda_t(t)\leq -\frac{2}{C} \Lambda(t)$. Including an integration factor $e^{\frac{2t}{C}}$, we have

\begin{equation}\notag \nabla (\Lambda(t) e^{\frac{2t}{C}})=\Lambda_t(t) e^{\frac{2t}{C}} + \Lambda(t) \frac{2}{C} e^{\frac{2t}{C}}\leq 0. \end{equation} Hence $\Lambda(t) e^{\frac{2t}{C}} \leq \Lambda(0)$. In other words,

\begin{equation}\notag Var_{\mu}(P_t f)\leq e^{-2t/C} Var_{\mu}(f). \end{equation}

How to identify the Poincaré constant

Despite the appealing formulation, identifying the best constant $C>0$ is in general not easy. In this blog, we will show a method for determining a crude estimate of such a constant.

We denote a Lyapunov function by $V$ if $V\geq 1$ and if there exist $\lambda>0, b\geq 0$ and some $R > 0$ such that for all $x$, the following drift condition holds

\begin{equation}\notag LV(x) ≤ -\lambda V(x) + b 1_{B(0, R)}(x). \end{equation}

By Theorem 1.4 [1], we show that if there exists a Lyapunov function $V(x)$ satisfying the drift condition, then $\mu $ satisfies a $L^2$ Poincaré inequality with constant $C_P=\frac{1}{\lambda}(1+b\kappa_R)$, where $\kappa_R$ is the L2 Poincaré constant of $\mu$ restricted to the ball B(0,R).

Given a smooth function $g$, we know that $Var_{{\mu}}(g)\leq \int (g-c)^2 d\mu$ for all $c$. In what follows, we set $f=g-c$, where $c$ is a constant to be selected later.

Next, we reformulate the drift condition and take an integral for $f^2$ with respect to $\mu$:

\begin{equation}\notag \int f^2 d\mu \leq \int \frac{-LV}{\lambda V} f^2 d\mu + \int f^2 \frac{b}{\lambda V}1_{B(0, R)}d \mu. \end{equation}

Control the first term $\int \frac{-LV}{\lambda V} f^2 d\mu$

Since $L$ is $\mu$-symmetric, by integration by parts, we get

$\int \frac{-LV}{V}f^2d \mu= \int \nabla\left(\frac{f^2}{V}\right) \nabla V d\mu$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\int \frac{f}{V} \nabla f \nabla V d\mu - \int \frac{f^2}{V^2} |\nabla V|^2 d\mu$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int |\nabla f|^2 d\mu - \int |\nabla f - \frac{f}{V} \nabla V|^2 d\mu$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq \int |\nabla f|^2 d\mu$

Control the second term $\int f^2 \frac{b}{\lambda V}1_{B(0, R)} d\mu$

Since $\mu$ satisfies a Poincaré inequality on $B(0, R)$ with a constant $\kappa_R$, we have

\begin{equation}\notag \int_{B(0, R)} f^2 d\mu\leq \kappa_R \int_{B(0, R)} |\nabla f|^2 d\mu + (1/\mu(B(0, R))) \left(\int_{B(0, R)} fd\mu\right)^2. \end{equation}

Fix $c=\int_{B(0, R)} gd\mu$. We have \begin{equation}\notag \int_{B(0, R)} \frac{f^2}{V}d\mu\leq \int_{B(0, R)} f^2 d\mu\leq \kappa_{R}\int_{B(0, R)} |\nabla f|^2d\mu. \end{equation} Eventually, we have

\begin{equation}\notag Var_{\mu}(f)=\int f^2 d\mu \leq \frac{1}{\lambda} (1+b\kappa_R) \int |\nabla f|^2 d\mu. \end{equation} In other words, the Poincaré inequality has a crude constant $C_p=\frac{1}{\lambda} (1+b \kappa_R)$.

Construction of the Lypunov function

Suppose we require one tail condition of the energy function $U(x)$, i.e. there exist $\alpha >0$ and $R\geq 0$ such that

Assumption $\langle x, \nabla U(x)\rangle \geq \alpha |x|$ for all $|x|\geq R$ (C1)

Now it is sufficient to build a Lyapunov function $V(x)=e^{\gamma |x|}$, where $|x|=\sqrt{\sum_{i=1}^n x_i^2}$.

Note that $\frac{\partial V(x)}{\partial x_i}= \gamma \frac{x_i}{|x|} V(x)$ and $\frac{\partial^2 V(x)}{\partial x_i^2}=\frac{\gamma}{|x|} V(x)+ \gamma^2 \frac{x_i^2}{|x|^2} V(x) - \gamma \frac{x_i^2}{|x|^3}V(x)$.

In the sequel, we have

$LV(x)=\gamma\left(\frac{n-1}{|x|}+\gamma-\frac{x}{|x|} \nabla U(x)\right) V(x)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq \gamma\left(\frac{n-1}{|x|} + \gamma -\alpha \right) V(x)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \leq -\gamma(\alpha-\gamma-\frac{n-1}{R}) V(x) + b 1_{B(0, R)}(x)$.

Hence $V(x)$ is a Lyapunov function provided

\begin{equation}\notag \lambda = \gamma(\alpha-\gamma-\frac{n-1}{R})>0, \end{equation}

which suffices to choose $\gamma<\alpha$, a large $R$ and assume the (C1) condition.


[1] The construction of Lyapunov function implies a tail decay for the distribution $\mu\propto e^{-U(x)}$ outside the ball $B(0, R)$.

[2] Obtaining a sharper estimate of Poincaré constant may require isoperimetric inequality.

  1. Bakry, D., F. Barthe, P. C., & Guillin, A. (2008). A Simple Proof of the Poincaré Inequality for A Large Class of Probability Measures. Electron. Comm. Probab., 13, 60–66.