Gaussian Concentration and Uniqueness of Equilibrium States in Lattice Systems

Abstract

We consider equilibrium states (that is, shift-invariant Gibbs measures) on the configuration space $S^{{\mathbb {Z}}^d}$ where $d\ge 1$ and S is a finite set. We prove that if an equilibrium state for a shift-invariant uniformly summable potential satisfies a Gaussian concentration bound, then it is unique. Equivalently, if there exist several equilibrium states for a potential, none of them can satisfy such a bound.

Appendix A

1.1 A.1 An Estimate

We first recall Young’s inequality for convolutions [3, p. 316]. Let and . Formally define their convolution by

If and , where $p,q\ge 1$, then where $r\ge 1$ is such that $1+r^{-1}=p^{-1}+q^{-1}$, then we have

Lemma A.1

Let $f:\Omega \rightarrow {\mathbb {R}}$ such that . Then for any $\Lambda \Subset {\mathbb {Z}}^d$ we have

Proof

Since $\delta _z(S_\Lambda f)\le \sum _{x\in \Lambda }\delta _{z-x}(f)$, we apply Young’s inequality with $r=2, p=2,q=1$, $u_x=\mathbb {1}_{\Lambda }(x)$, and $v_x=\delta _x(f)$ to get the desired estimate. $\square$

1.2 A.2 Proof Lemma 4.1

The version of this lemma in dimension $d=1$ is stated without proof in [19]. Since it is not completely obvious, we give it here for any $d\ge 1$.

We fix $\varepsilon >0$ and $k\ge 0$. The frequency of a pattern $p_k\in S^{\Lambda _k}$ in $\omega$ (see (14)) can rewritten as

$$\begin{aligned} {\mathfrak {f}}_{n,k}(\omega ;p_k)=\frac{1}{(2(n-k)+1)^d} \sum _{x\in \Lambda _{n-k}} \mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}}. \end{aligned}$$

By definition we have

$$\begin{aligned}&\Vert {\mathfrak {f}}_{n,k}(\omega ;\cdot )- {\mathfrak {f}}_{n,k}(\eta ;\cdot )\Vert _{{\scriptscriptstyle \mathrm {TV}}}=\nonumber \\&\frac{1}{2(2(n-k)+1)^d}\sum _{p_k \in S^{ \Lambda _k}} \left| \sum _{x\in \Lambda _{n-k}} \big (\mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}} -\mathbb {1}_{\{(\theta _x \eta )_{ \Lambda _k}=p_k\}}\big )\right| . \end{aligned}$$

(22)

Letting

$$\begin{aligned} {\mathcal {I}}_{\omega ,\eta ,n}=\big \{ x\in \Lambda _{n-k}: (\theta _x \omega )_{ \Lambda _k}=(\theta _x \eta )_{ \Lambda _k}\big \} \end{aligned}$$

we get

$$\begin{aligned}&\sum _{p_k \in S^{ \Lambda _k}}\left| \sum _{x\in \Lambda _{n-k}} \big (\mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}} -\mathbb {1}_{\{(\theta _x \eta )_{ \Lambda _k}=p_k\}}\big )\right| \\&\quad = \sum _{p_k \in S^{ \Lambda _k}} \left| \sum _{x\in {\mathcal {I}}_{\omega ,\eta ,n}^c} \big (\mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}} -\mathbb {1}_{\{(\theta _x \eta )_{ \Lambda _k}=p_k\}}\big )\right| \\&\quad \le \sum _{p_k \in S^{ \Lambda _k}} \sum _{x\in {\mathcal {I}}_{\omega ,\eta ,n}^c} \big |\mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}} -\mathbb {1}_{\{(\theta _x \eta )_{ \Lambda _k}=p_k\}}\big |\\&\quad \le \sum _{p_k \in S^{ \Lambda _k}} \sum _{x\in {\mathcal {I}}_{\omega ,\eta ,n}^c} \big (\mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}} +\mathbb {1}_{\{(\theta _x \eta )_{ \Lambda _k}=p_k\}}\big )\\&\quad = \sum _{x\in {\mathcal {I}}_{\omega ,\eta ,n}^c} \sum _{p_k \in S^{ \Lambda _k}}\big (\mathbb {1}_{\{(\theta _x \omega )_{ \Lambda _k}=p_k\}} +\mathbb {1}_{\{(\theta _x \eta )_{ \Lambda _k}=p_k\}}\big )\\&\quad = 2 \,\big | {\mathcal {I}}_{\omega ,\eta ,n}^c\big |. \end{aligned}$$

Hence we obtain from (22)

$$\begin{aligned} \Vert {\mathfrak {f}}_{n,k}(\omega ;\cdot )- {\mathfrak {f}}_{n,k}(\eta ;\cdot )\Vert _{{\scriptscriptstyle \mathrm {TV}}} \le \frac{\big | {\mathcal {I}}_{\omega ,\eta ,n}^c\big |}{(2(n-k)+1)^d}. \end{aligned}$$

(23)

We now look for an upper bound for $\big | {\mathcal {I}}_{\omega ,\eta ,n}^c\big |$. If $(\theta _x \omega )_{ \Lambda _k}=p_k$ and $(\theta _x \eta )_{ \Lambda _k}\ne p_k$, then $\omega _y\ne \eta _y$ for at least one site $y\in \Lambda _k+x$. Such a y can produce as many as $(2k+1)^d$ sites such that $(\theta _x \omega )_{ \Lambda _k}=p_k$ and $(\theta _x \eta )_{ \Lambda _k}\ne p_k$. Hence

$$\begin{aligned} \big | {\mathcal {I}}_{\omega ,\eta ,n}^c\big |&\le (2k+1)^d \,\big |\{x\in \Lambda _{n-k}: \omega _x\ne \eta _x \}\big | \\&\le (2k+1)^d \, \big |\{x\in \Lambda _{n}: \omega _x\ne \eta _x \}\big | \\&\le (2k+1)^d \, \bar{\mathrm {d}}_{ \Lambda _n}(\omega ,\eta ). \end{aligned}$$

Hence (23) yields

$$\begin{aligned} \Vert {\mathfrak {f}}_{n,k}(\omega ;\cdot )- {\mathfrak {f}}_{n,k}(\eta ;\cdot )\Vert _{{\scriptscriptstyle \mathrm {TV}}} \le \frac{(2k+1)^d}{(2(n-k)+1)^d} \,\bar{\mathrm {d}}_{ \Lambda _n}(\omega ,\eta ). \end{aligned}$$

Obviously there exists $\breve{N}>k$ such that for all $n\ge \breve{N}$ we have

$$\begin{aligned} \left( \frac{2n+1}{2(n-k)+1}\right) ^d\le \frac{5}{4} \end{aligned}$$

therefore, if we take

$$\begin{aligned} \bar{\mathrm {d}}_{ \Lambda _n}(\omega ,\eta )\le \frac{2\varepsilon }{5(2k+1)^d}\, (2n+1)^d \end{aligned}$$

we finally obtain

$$\begin{aligned} \Vert {\mathfrak {f}}_{n,k}(\omega ;\cdot )- {\mathfrak {f}}_{n,k}(\eta ;\cdot )\Vert _{{\scriptscriptstyle \mathrm {TV}}} \le \frac{\varepsilon }{2} \end{aligned}$$

for all $n\ge \breve{N}$, which concludes the proof of the lemma.

1.3 A.3 A Bound on Relative Entropy

Recall that ‘$\log$’ stands for the natural logarithm. We were not able to find a reference for a proof of the following estimate, so we prove it for the reader’s convenience.

Lemma A.2

Let $\nu$ and $\mu$ be probability measures on a finite set A. Then

$$\begin{aligned} \sum _{a\,\in A} \nu (\{a\}) \left| \log \frac{\nu (\{a\})}{\mu (\{a\})}\right| \le H(\nu |\mu )+\frac{2}{{{\,\mathrm{\mathrm {e}}\,}}} \end{aligned}$$

(24)

where

$$\begin{aligned} H(\nu |\mu )=\sum _{a\,\in A} \nu (\{a\}) \log \frac{\nu (\{a\})}{\mu (\{a\})}. \end{aligned}$$

Proof

Define

$$\begin{aligned} A^{\scriptscriptstyle {-}}=\left\{ a\in A: \log \frac{\nu (\{a\})}{\mu (\{a\})}<0\right\} . \end{aligned}$$

Now

$$\begin{aligned} \sum _{a\,\in A} \nu (\{a\}) \left| \log \frac{\nu (\{a\})}{\mu (\{a\})}\right|&= \sum _{a\,\in A\backslash A^{\scriptscriptstyle {-}}} \nu (\{a\}) \log \frac{\nu (\{a\})}{\mu (\{a\})} + \sum _{a\,\in A^{\scriptscriptstyle {-}}} \nu (\{a\}) \log \frac{\mu (\{a\})}{\nu (\{a\})}\\&= H(\nu |\mu )+ 2\sum _{a\,\in A^{\scriptscriptstyle {-}}} \nu (\{a\}) \log \frac{\mu (\{a\})}{\nu (\{a\})}. \end{aligned}$$

By the concavity of the logarithm function and Jensen’s inequality we get

$$\begin{aligned} \sum _{a\,\in A^{\scriptscriptstyle {-}}} \nu (\{a\}) \log \frac{\mu (\{a\})}{\nu (\{a\})}&= \nu (A^{\scriptscriptstyle {-}}) \sum _{a\,\in A^{\scriptscriptstyle {-}}} \frac{\nu (\{a\})}{\nu (A^{\scriptscriptstyle {-}}) } \log \frac{\mu (\{a\})}{\nu (\{a\})}\\&\le \nu (A^{\scriptscriptstyle {-}}) \log \left( \frac{\mu (A^{\scriptscriptstyle {-}})}{\nu (A^{\scriptscriptstyle {-}})}\right) \\&\le \nu (A^{\scriptscriptstyle {-}}) \log \left( \frac{1}{\nu (A^{\scriptscriptstyle {-}})}\right) \le {{\,\mathrm{\mathrm {e}}\,}}^{-1} \end{aligned}$$

where we used the elementary inequality $-x\log x\le {{\,\mathrm{\mathrm {e}}\,}}^{-1}$,$x\ge 0$. Therefore we arrive at (24). $\square$