The Lyapunov Function Method for Poincaré Inequality

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.

Discussions

[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.