Dobrushin's uniqueness Theorem

In Misure di Gibbs a volume infinito we have studied the question wheter given a potential $\Phi$ the space of Infinite-volume Gibbs measures is non-empty.

Following ideas from The Ising model, we defined a notion of first order phase transition based on non-uniqueness of Gibbs measures.

At the start of the proof of existence, which needs that the specification $\pi$ is quasilocal (continuous), we fixed a boundary condition $\omega \in {\mathbb{Z}}^d$. It tunrs out that uniqueness of Gibbs measure is the same as indipendence from boundary conditions: all the limiting measures are the same.

Lemma The following are equivalent:

1. Uniqueness holds: $\mathcal{G}(\pi )=\{\mu \}$
2. For any $\omega$, and all ${\Lambda }_n\uparrow {\mathbb{Z}}^d$ and all local function

$${\pi }_{{\Lambda }_n}(f\mid \omega )\to \mu (f)$$

Proof $(1⟹2)$ is trivial, if there exists only one Gibbs measure then regardless of the boundary condition every converging subsequence converges to $\mu$, so that the original sequence also converges to $\mu$. $(2⟹1)$ Take any $\mu ,\nu \in \mathbb{G}(\pi )$ and a local function $f$.

$$|\mu (f)-\nu (f)|=|\mu {\pi }_{\Lambda }(f)-\nu {\pi }_{\Lambda }(f)|=\left|\int {\pi }_{\Lambda }(f\mid \omega )\mu (d\omega )-\int {\pi }_{\Lambda }(f\mid \eta )\mu (d\eta )\right|$$ $$\le \int |{\pi }_{\Lambda }(f\mid \omega )-{\pi }_{\Lambda }(f\mid \eta )|\mu (d\omega )\nu (d\eta )$$

Since local functions are bounded, the integrand is dominanted by $2\Vert f{\Vert }_{\infty }$ and we can apply the Dominated convergence theorem to take out the limit

$$\le \underset{n\to \infty }{\lim }\int |{\pi }_{{\Lambda }_n}(f\mid \omega )-{\pi }_{{\Lambda }_n}(f\mid \eta )|\mu (d\omega )\nu (d\eta )=0$$

Since $\mu (f)=\nu (f)$ for all bounded function, $\mu =\nu$. $\square$

To state Dobrushin’s uniqueness theorem, we need to define some notions.

As usual, a distance between probability measures is the total variation. Here we compare spin distribution given two boundary conditions:

$$\Vert {\pi }_i(\cdot \mid \omega )-{\pi }_i(\cdot \mid {\omega }^{\prime }){\Vert }_{TV}:=\sum _{{\omega }_i=\pm 1}|{\pi }_i({\omega }_i\mid \omega )-{\pi }_i({\omega }_i\mid {\omega }^{\prime })|$$

We can introduce the sup of total variation between measures in $i$ that differ only on a single spin $j$

$$C_{ij}(\pi ):=\underset{\omega ,{\omega }^{\prime }{\omega }_k={\omega }_k^{\prime }k\ne j}{\sup }\Vert {\pi }_i(\cdot \mid \omega )-{\pi }_i(\cdot \mid {\omega }^{\prime })\Vert$$

and a global property

$$C(\pi ):=\underset{i\in {\mathbb{Z}}^d}{\sup }\sum _{j\in {\mathbb{Z}}^d}{C}_{ij}(\pi )$$

Def Let $f:\Omega \to \mathbb{R}$. We define the oscillation of $f$ at spin $i\in {\mathbb{Z}}^d$ as:

$${\delta }_i(f):=\underset{\omega ,\eta {\omega }_k={\eta }_{k}k\ne i}{\sup }|f(\omega )-f(\eta )|$$

and the total oscillation of $f$ as

$$\Delta (f):=\sum _{i\in {\mathbb{Z}}^d}{\delta }_i(f)$$

The intuition is that the total oscillation tells us how “far” $f$ is from beeing a constant (which have total oscillation zero).

Lemma Let $f$ be continuous and with finite total oscillation. Then $\Delta f\ge \sup f-\inf f$. Proof Let ${\omega }_s^n$ and ${\omega }_i^n$ be two sequences that converge to the sup and inf respectively, such that ${\omega }_s^n={\omega }_i^n$ outside a volume ${\Lambda }^c$. Then a $n$ large enough

$$\sup f-\inf f\le f({\omega }_s^n)-f({\omega }_i^n)+ϵ$$

since ${\omega }_s^n={\omega }_i^n$ are different only inside a finite volume $\Lambda$, we can bound the difference using local oscillation:

$$f({\omega }_s^n)-f({\omega }_i^n)\le \sum _{i\in \Lambda }{\delta }_i(f)$$

thus

$$\le \Delta (f)+ϵ\square$$

Theorem (Dobrushin’s uniqueness theorem) Let $\pi$ be a specification satisfying the condition:

$$C(\pi )<1$$

called the Dobrushin’s condition of weak dependence. Then $|\mathcal{G}(\pi )|=1$

Proof Let $\mu ,\nu \in \mathcal{G}(\pi )$ and let $f$ be a local function.