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\title[Existence of continuous eigenfunction]{Existence of continuous
  eigenfunctions for the Dyson model in the critical phase}

\author{Anders Johansson, Anders \"Oberg, and Mark Pollicott}
\date{}
\begin{document}

\maketitle
\begin{abstract}
  We prove that there exists a continuous eigenfunction of the transfer operator
  defined by potentials for the so-called Dyson model for all inverse critical
  temperatures that are strictly less than the critical inverse temperature.
  This includes all cases when the potential does not have summable variations,
  the classical condition that ensures the existence of a continuous
  eigenfunction of the transfer operator. As a consequence the inverse critical
  temperatures for the one-sided and the two-sided models are the same. We use
  the random cluster model for our method of proof, so the results also hold in
  this wider context. In particular, we base our conclusions of this paper by
  recent results of Duminil-Copin, Garban, and Tassion \cite{duminil}.
\end{abstract}
\def\h{h}


\section{Introduction}\noindent

It is well-known \cite{walters1} that there exists a continuous and strictly
positive eigenfunction $h$ for any transfer operator defined on a symbolic shift
space with a finite number of symbols such that the potential has summable
variations. Here we prove the existence of a continuous eigenfunction for the
important special class of Dyson potentials up to the critical phase, when the
potential does not satisfy the condition of summable variations. We stress that
it is the continuity that is the main difficulty. To see that there is a
measurable eigenfunction follows from a simple application of the martingale
convergence theorem.

More precisely, let $\T$ be the left shift on the space $\X=S^{{\mathbb Z}_+}$,
where $S$ is a finite set. Let $\phi:\X\to \RR$ be a continuous function defined up
to an additive constant. We refer to $\phi$ as the \emph{one-point potential}. A
\emph{Gibbs measure} $\nu\in\CM(\X)$ for $\phi$ is one that minimises the free energy
$P(\nu;\phi)=\nu(\phi)-H(\nu)$, where $H(\nu)$ denotes the entropy
$\lim_{n\to\infty} H(\nu\vert_{\CF_n})/n$ per time unit. This measure $\nu$ can also be
obtained as the Gibbs measure on $\X$ obtained from the full potential
$$ \Phi(x)=\sum_{k=0}^\infty \phi(\T^k x) $$
on $\X$. A Gibbs measure $\nu$ is also obtained as an eigenmeasure of the dual of
the \emph{transfer operator} $\mc L=\mc L_\phi$ defined on continuous functions by
\begin{equation}\label{trans}
  \mc{L} f(x)= \sum_{y\in T^{-1}x} e^{\phi(y)}\, f(y).
\end{equation}
Such an eigenmeasure $\nu$ satisfies $\mc{L}^* \mu=\lambda \nu$, for some maximal positive
eigenvalue $\lambda>0$.


The \emph{equilibrium measure} $\mu$ is a minimiser of $P(\mu;\phi)$ among all
\emph{translation invariant} measures $\mu\in{\CM}_\T(\X)$. Taking the natural
extension of $\mu$, we can also represent $\mu$ as a translation invariant measure
on the \emph{two-sided} space $\ol\X = S^\ZZ$ and we can alternatively construct
the measure $\mu\in\CM_\T(\ol\X)$ as the Gibbs measure to the full two-sided potential
\[
  \ol \Phi(x)=\sum_{k=-\infty}^\infty \phi(\T^k x)
\]
defined on $x\in\ol\X$.
If there exists a continuous eigenfunction $h(x)$ to the transfer operator, the
measure $\mu$ can be recovered as the Doeblin measure \cite{berger2} (a.k.a.\
$g$-measure in the terminology of Keane \cite{keane}) corresponding to the
\emph{Doeblin function} $g(x)$ where
\begin{equation}\label{g}
  g(x)= \frac{h(x) e^{\phi(y)}}{\lambda h(\T x)}.
\end{equation}

The existence of a continuous eigenfunction $h$ to $\mc{L}$, such that
$\mc{L}h=\lambda h$, is however not automatic in same way the existence of the
eigenmeasure $\nu$ is. If one assumes summable variations of $\phi$,
\begin{equation}\label{sum}
  \sum_{n=1}^\infty \var_n (\phi)<\infty,
\end{equation}
where $\var_n(\phi)=\sup_{x\sim_n y}|\phi(x)-\phi(y)|$ ($x\sim_n y$ means that $x$ and $y$
coincide in the first $n$ entries), then the existence of a continuous
eigenfunction $h(x)$ follows from the typical ``cone-argument'' used in {\em
  e.g.}, Walters \cite{walters1}, which to date is the only known method for
providing the existence of a continuous eigenfunction. (??) In this paper, we
prove in a lemma that if there is a translation invariant measure $\mu$ which is
absolutely continuous with respect to the Gibbs measure $\mu$ the Radon--Nikodym
derivative $h(x)=\dfrac{\partial\mu}{\partial\nu}$ is an eigenfunction of $\mscr L$.

Both the question of uniqueness and the question about the a continuous
eigenfunction seems related to the smoothness of the potential. One possibility
for the existence of a continuous eigenfunction in general could be the
\emph{Berbee condition}. Berbee proved (\cite{berbee89}) that the Gibbs measure
$\nu$ and the equilibrium measure $\mu$ are unique if we have
\begin{equation}\label{berbee}
  \sum_{n=1}^\infty e^{-r_1-r_2-\cdots-r_n}=\infty,
\end{equation}
where $r_n=\var_n \phi$.


In \cite{johob} it was proved that uniqueness of Doeblin measures follows
whenever
$$ \sum_{n=1}^\infty (\var_n g)^2<\infty, $$
and more recently Berger et al.\ \cite{berger2} proved that for such Doeblin
measures uniqueness follows if $\var_n \log g <2/\sqrt{n}$ (see also
\cite{johob3} for a slightly stronger condition).
This connects to Dyson's counterexample in \cite{dyson} where it is shown that
there are examples of multiple equilibrium measures for $\phi$ whenever
$$\sum_{n=1}^\infty (\var_n \phi)^{1+\epsilon}<\infty,$$
where $\epsilon>0$.

Dyson's example is for a two-sided model, but we showed in \cite{johob4} that
the inverse critical temperature $\beta_c^+$ satisfies $\beta_c^+\leq 8\beta_c$, where $\beta_c$ is
the inverse critical temperature for the two-sided model, and this show that
Dyson's example of multiple equilibrium measures can be formulated for a
one-sided model as above.

Since we have the ``translation'' via \eqref{g} between the case for general
potentials and for Doeblin measures, we may guess that the existence of a
continuous eigenfunction cannot be moved very far away from the summability of
variations condition for a potential.

Here, however, we study only the Dyson potential: Fix $\alpha>1$ and $\beta>0$. Let the
one-point long-range Ising potential $\phi=\phi_{\alpha,\beta}$ be given by
$$\phi(x_0, x_1,\ldots)=x_0\cdot \beta \sum_{j=1}^\infty \frac{x_j}{j^\alpha},$$
and define the one-sided and two-sided Dyson potentials, $\Phi:\X\to \RR$ and
$\bar\Phi:\ol\X\to\RR$ as above. Let $\mu$ and $\nu$ be the Gibbs measures on $\CM(\X)$
and $\CM(\ol\X)$, corresponding to $\Phi$ and $\ol\Phi$, respectively.

We then have for $\alpha>2$
$$ \mu= h\nu, $$
where $h>0$ is a H\"older continuous eigenfunction. We are interested in the
boundary cases where $1<\alpha\le2$, when there exists a unique equilibrium measure for
$\ol\Phi$ for $\beta<\beta_c$ \cite{ACCN} and multiple equilibrium measures when $\beta>\beta_c$,
where the critical parameter $\beta_c=\beta_c(\alpha)>0$ for $1\le\alpha\le2$. We show here that the
same uniqueness properties holds for the ``one-sided'' Gibbs measure $\nu$.

In this case the summable variations condition is not satisfied for neither
$\ol\Phi$ nor $\Phi$; hence we may have multiple eigenmeasures for $\mc{L}^*$. In
this context we have $\var_n(\phi)=O(\frac{1}{n})$, but as we noted earlier, in the
general setting above, is not clear that there exists a continuous eigenfunction
even in the cases we have a unique equilibrium measure

We prove that if the inverse critical temperature $\beta$ is strictly smaller than
the critical inverse temperature $\beta_c$, which will be seen to be the same for
the two-sided and one-sided models, then we have a continuous eigenfunction of
$\mc{L}$.

Let $\mu$ be the Gibbs equilibrium measure with respect
to the Dyson potential $\phi$ and let $\nu$ be the one-sided Gibbs measure.
Define
$$
h_n(x)=\frac{\mu[x_0,\ldots x_n]}{\nu[x_0,\ldots, x_n]},
$$
and consider the measurable function $h(x)=\limsup_{n\to \infty}h_n(x)$. 

\begin{thm}\label{main} 
  If $\beta<\beta_c$, i.e.\ in the subcritical phase, then $h(x)$ is a continuous
  function on $\X$ and it is also an eigenfunction of $\mc L = \mc{L}_\phi$.
\end{thm}

We conjecture that there exists a continuous eigenfunction for a potential
$\phi$ whenever that potential satisfies Berbee's condition \eqref{berbee}.

The next result is a corollary of Theorem 1:

\begin{thm}
The critical $\beta_c$ for the two-sided model 
is the same as for the one-sided model, i.e., 
$\beta_c^+=\beta_c$.
\end{thm}

This improves our result of Theorem 1 in \cite{johob4}
that $\beta_c^+\leq 8\beta_c$.

\section{The one-dimensional random cluster model and the Ising--Dyson model}

For a finite graph, let $\w(G)$ denote the number of connected components
(``clusters'') in the graph $G$. For simple graphs $G\subset \binom V2$ on an
countably infinite set $V$ of vertices, we consider the number of clusters
$\w(G)$ as a \emph{potential}. This means that the difference
$\Delta\w(G,F) = \w(G)-\w(F)$ is defined for any two graphs $F$ and $G$ that
coincide outside a \emph{finite} subset $\Lambda\subset \binom V2$.

A random graph $G\sim\alpha$ on a set of vertices $V$ is a probability
distribution $\alpha$ on the set $\{0,1\}^{\binom V2}$. The random cluster
models $\mscr R(V,p,q)$ (FK-model \cite{grimmet}), we consider are specified by a set 
of vertices $V$ and a probabilities $p(ij)\in[0,1]$, defined on the set of pairs
$ij\in \binom V2$. The model $\mscr R(V,p,q)$ is a Gibbs distribution on random
graphs, i.e.\ configurations i $\gamma\in\{0,1\}^{\binom V2}$, with 
non-continuous potential
$$
\phi(\gamma) = 
- \log q \cdot\w(\gamma) + 
\sum_{ij} \gamma(ij)\cdot \log p(ij) + (1-\gamma(ij))\cdot\log (1-p(ij)).
$$
We refer to the models $\mscr R(V,p,1)=\eta(\binom V2,p)$ as ``Bernoulli
percolation''.

We consider two versions of the one-dimensional Potts--Dyson random cluster
model: First $\mscr R(\ZZ_+,\beta,\alpha,q)$ , we let $V=\ZZ$ (or $V=\ZZ_{\ge0}$ and consider the
random cluster model $\mscr R(V,p_J,q)$ where
\begin{equation}\label{eq:Jdef}
  p_J(ij) = 1-e^{-J(ij)} \quad\text{where}\quad 
  J({ij}) = \frac \beta{|i-j|^\alpha}.
\end{equation}
When $q$

Percolation is the event that the random graph contains an infinite cluster. It
is well-known \cite{ACCN} that for all $\alpha\in(1,2]$ there exists a critical
$\beta_c=\beta_c(\alpha)$ such that, percolation does not occur with probability 1 for
$0<\beta<\beta_c$, while for $\beta\in(\beta_c,\infty)$ there is with probability 1 no infinite
cluster.

We will mainly consider the subcritical case $1<\alpha\le 2$ and for invers $\beta$ in the
interval $0<\beta <\beta_c(\alpha,q=1)$, i.e. $\beta$ is in the subcritical region for Bernoulli
perocalition. We use the following lemma.
\begin{lem}\label{geometric-bound}
 Let $X=X(C_s)$ denote the size of the cluster $C_s$ 
 containing a specified vertex $s\in V$. Then there are constants 
 $K$ and $r$, $0<r<1$, such that $$\P(X\ge m)\le K r^{m}. $$ 
\end{lem}
\begin{proof}
  As surveyed by Panagiotis \cite{panagiotis} (Theorem 1.2.1), we know that it
  holds for Bernoulli percolation ($q=1$) when $0<\beta<\beta_c(q=1)$. Since
  $\mscr R(V,p,1)$ stochastically dominates $\mscr R(V,p,q)$, $q>1$, it also
  holds for the FK-model. Finally, Duminil-Copine (\cite{duminil}) proves that
  $\beta_c(\alpha,q)$ are the same for the case $q=1$ and $q>1$.
\end{proof}

\subsubsection{The extended random cluster model}

The extended \emph{random cluster model} with $\mscr R_e(V,p,q)$ can be obtained
as the joint distribution of the Potts-model spin-sequences $x\in A^{V}$, where
$A=\{1,\dots,q\}$, and a random graph $\gamma\in\{0,1\}^{\binom V2}$. One obtains the
extended joint distribution of $(x,\gamma)$ by first considering the pair $(x,\gamma)$
chosen independently: The spin sequence $x\in\{0,1,\dots,q\}^{V}$ according to the
uniform Bernoulli measure $x\sim\eta(V,\frac1q\cdot\ett)$ on the spin sequences and the
random graph $\gamma$ according to the Bernoulli measure
$\eta(\binom V2, p)=\mscr R(V,p,1)$. Then $\mu=\mscr R_e(V,p,q)$ is the distribution
of $(x,\gamma)$ \emph{conditioned on} $x$ and $\gamma$ being \emph{compatible}: That is,
the event $B(x,\gamma)$ that no spins $x_i=+1$ and $x_j=-1$ in $x$ are connected by a
path (edge) in $\gamma$. Equivalently, all clusters in $\mscr C(\g)$ are
monochromatic under the ``colouring'' $x:V \to A$. If $(x,\mu)\sim\mu=\mscr R_e(V,p,q)$
then the marginal distribution $\mu\circ x^{-1}$ of $x$ is the Potts model with
interactions given by $J(ij)=\log(1-p(ij))$ and the marginal distribution
$\mu\circ\gamma^{-1}$ of $\gamma$ is the random cluster model $\mscr R(V,p,q)$ described above.

Let $S\subset V$ be a finite subset. We need to establish a relation between the
random cluster distribution of the graph $\gamma\sim\mu$ with the probability $\mu([x]_S)$
of a spin cylinder $[x]_S = \{y \mid y\vert_S = x\vert_S\}$, when $(x,\gamma)$ has the random
cluster distribution $\mu = \mathscr R_e(V,p,q)$. It follows from teh definition
that the conditional distribution of $x$ given $\gamma$ is the uniform distribution
on $A^{\mscr C(\gamma)}$. That is, one obtaine $x$ from $\g$ by assigning spins to
the clusters in $\mscr C(\g)$ uniformly at random. Thus, provided $[x]_S$ and
$\gamma$ are compatible, it follows that the probability of $[x]_S$ equals
$q^{-\w_S(\gamma)}$ , where $\w_S(\gamma)$ is the number of clusters in $\mscr C(\gamma)$ that
intersect $S$. In other words,
\begin{equation}\label{eq:gcyl}
  \mu([x]_S) = \int q^{-\w_S(\gamma)} B([x]_S,\gamma) \d\mu(\gamma)
\end{equation}
where $B([x]_S,\gamma)$ indicates the event that $[x]_S$ is compatible with $\gamma$.


\subsubsection{The interface perturbation}

We now fix some notation before proceeding to prove that we have a continuous eigenfunction.
\begin{center}
\begin{tabular}{rp{0.8\textwidth}}
  $d\tilde\eta(\e)$ & The Bernoulli measure $2^{-|\e|}\ltimes d\eta(\e)$ \\
  $B_n$  & The indicator of the event $[x]_n$ compatible with $\gamma$. \\
  $B_n^+$  & The indicator of the event $[x]_n$ compatible with $\gamma_+$. \\
  $B'_n(\gamma)$ & The indicator of the event $[x]_n$ is
               compatible with $\gamma$ or not compatible with $\gamma_+$. \\
  $\hat B_n(\gamma_-,\e)$ & The indicator of the event that there are
                   no two edges from some cluster $C$ in $\gamma_-$
                   to a pair $i,j\in[0,n)$ having opposite spins,
                   i.e. such that $x_i x_j = -1$. \\
  $\hat X_n(\gamma_-,\e)$ & The indicator of the event that $Q_n=0$. \\
  $R_n(\gamma)$ & Correction term so that
              $$ d\mu(\gamma|B^+_n) = B'_n \cdot 2^{R_n(\gamma)} \ltimes d\nu(\gamma_-) \d\tilde\eta(\e) \d\nu(\gamma_+|B_n) $$ \\
  $Q_{>n}(\gamma_-,\e)$ & Number of edges in $\e$ from clusters of $\gamma_-$ to vertices in $(n,\infty)$
                      such that there is at least one more edge in $\e$ from the same cluster to $[0,\infty)$
                      preceding it in some order\\
  $Q(\gamma_-,\e)$ & Number of edges in $\e$ from clusters of $\gamma_-$ to vertices in $[0,\infty)$
                 such that there is at least one more edge in $\e$ from the same cluster to $[0,\infty)$
                 preceding it in some order\\
  $X(C)$ & For a cluster $C\subset \gamma_{-}$ it is the number of edges in $\e$ to $[0,\infty]$. \\
  $i(C)$ & For a cluster $C\subset \gamma_{-}$ it is the rightmost vertex, i.e.\ $i(C)=\max \{j\in C\}$. \\
  $\lambda(C)$ & The sum $\lambda(C) = \frac{\beta}{2} \sum_{j\in C} \frac 1j$. \\
\end{tabular}
\end{center}

We will consider two versions of the long range Potts model with the Dyson
potential $\phi_{\beta,\alpha}(x)$: as the marginal distribution of $x$ and the
corresponding random cluster model $\mscr R(V,p,2)$ as the marginal
distributions $\g$.

We now observe that a configuration $\g$ in the two-sided model
$\mscr R(\ZZ,\alpha,\beta)$ difference between the one-sided random cluster model
$\nu = \mscr R(V_+,J)$ and the usual two-sided model $\mu = \mscr R(V,J)$. We will
use that a configuration $\g$ can be factored as $\g = (\g_+, \e, \g_-)$, where
$\g_-$ is the induced graph $\g[V_-]$ on vertices $-j\in V_-=\ZZ_{<0}$ and
$\g_+=\g[V_+]$ is the graph induced on vertices $i\in V_+=\ZZ_{\ge0}$. The graph
$\e=\g\cap E(V_+,V-)$ consists of edges $ji$, $i\ge0$ and $j\ge 1$, connecting vertices
$-j\in V_-$ with vertices $i\in V_+$. Note that we often use positive indices $i,j$,
$i\ge0$ and $j>0$, as labels for edges in $\e$. Thus $J(ij)=\beta/(i+j)^\alpha$ with this
labelling.

Let
$$ \tilde \eta(\epsilon)= 2^{-|\epsilon|} \ltimes \eta (\epsilon). $$
Note that both $\eta(\cdot)$ and $\tilde\eta (\cdot)$ are Bernoulli measures. Let also
$$ d\tilde \nu_n(\gamma) = \frac{d\nu(\gamma_-)\otimes \tilde \eta(\epsilon)\otimes \nu(\gamma_+)}{\nu(B([x]_n,\gamma_+))}. $$

Let $R_n$ be the number of correcting edges, i.e.,
$$
R_n=\# \{ij\in \epsilon: \omega (\gamma_{<ij}+ij)=\omega (\gamma_{<ij})\},\quad j>n.
$$
Let $B_n(\g)=C([x]_n,\gamma)$ be the indicator of the event that cylinder $[x]_n$ is
compatible with graph $\g=(\g_-,\e,\g_+)$. Let also
$B'_n(\g) = B_n(\g) + (1-B_n(\g_+))$ indicate compatibility of $[x]_n$ with $\g$
or not compatible with $\g_+$. We then have
\[
  h_n(x) = \frac{\mu [x]_n}{\nu [x]_n}\propto \int B_n(\g) \cdot 2^{R_n(\g)} \; d\tilde \nu_n (\gamma)
\]

Note that
$$ R_n(\g) \le Q_n(\g_-,\e) $$
where $Q_n$ denotes the number of edges in $\e$ that connects a vertex $j\ge n$ to a
cluster in $\g_-$ with at least one more edge from the cluster to $[0,n-1]$. That is,
$$  Q_n=\# \{ij \in \epsilon \mid  \exists k\, \exists l\, kl\in\e,  i \sim_{\gamma-} k, k>i, j > n\}.$$
Notice that $Q_n$ only depends on $(\gamma_-,\epsilon)$.



\section{Proof of Theorem 1}\noindent

\begin{lem}
  If $\mu\in\CM_\T(\X)$ is a translation invariant measure which is absolutely
  continuous with the Gibbs measure $\nu$ for $\phi$ then the Radon--Nikodym
  derivative $h(x)=\dfrac{\partial\mu}{\partial\nu}(x)$ is an eigenfunction to the transfer
  operator $\mscr L =\mscr L_\phi$.
\end{lem}
\begin{proof}
  We can assume $\lambda=1$. By assumption, the measure $\mu= \h \nu$ is translation
  invariant, i.e., $\mu\circ T^{-1}=\mu$. Hence we have $(h\nu)\circ T^{-1}=h\nu$ and it
  suffices to show that $(h\nu) \circ T^{-1}=({\mathcal L}h)\nu$.

  Let $A$ be any Borel subset of $\X$. Then
  $$(h\nu)\circ T^{-1} (A)=\int_A \sum_{y: Ty=x} h(y)e^{\phi(y)}\; d\nu(x)=\int_A h(x)\; d\nu(x).$$
\end{proof}

We need to prove that the $h_n(x)$ converge to a continuous function $h(x)$ as
$n\to\infty$. It is enough to show that
$$ \sup_{x,m\ge n} \frac{h_m(x)}{h_n(x)} = 1 + o({1}) \quad\text{as $n\to\infty$}$$
since this shows that $\log h_n$ is a Cauchy-sequence.

Recall that
\begin{equation}\label{eq:3}
  h_m(x) =  \int B'_m 2^{R_m} \d \nu(\g_-)\d\tl\eta(\e) \d \nu(\g_+|B_m^+).
\end{equation}
Since $R_m$ is the number of edges in $\e$ that do not reduce (with respect to
some order) the number of components in $\g\triangleright [0,n]$, it is clear
that
\begin{equation}\label{eq:RleQ}
  R_m \le Q_{>m}
\end{equation}
where \(Q_{>m}\) is the number of edges $ij\in\e$ where $i>n$ and $-j$ belongs
to a cluster $C$ in $\g_-$ that sends at least one more edge to $[0,\infty)$.

For $n\le m$, we have
$$ \hat B_n \hat X_n \le B'_m \le \hat B_n $$
and, on account of \eqref{eq:RleQ}, it follows that
\begin{equation}\label{eq:4}
  \int \hat B_n \hat X_n \d \nu(\g_-) \d\tl\eta(\e)
  \le h_m(x) \le \int \hat B_n 2^{Q_{>n}} \d\nu(\g_-) \d\tl\eta(\e).
\end{equation}
We have used that $\hat B_n$, $\hat X_n$ and $Q_{>m}$ are independent of $\g_+$ and that
$$\int\d\nu(\g_+|B_m^+)=1. $$

Since both $\hat B_n$ and $\hat X_n$ are decreasing in $(\g_-,\e)$ it follows
from the FKG inequality that
\begin{equation}
  \label{eq:5}
  I_n \cdot \int \hat X_n \d\nu(\g_-) \d\tl\eta(\e)
  \le h_m(x)
  \le I_n \cdot \int 2^{Q_{>n}} \d\nu(\g_-) \d\tl\eta(\e).
\end{equation}
where
\begin{equation}
  \label{eq:6}
  I_n = \int \hat B_n \d\nu(\g_-) \d\tl\eta(\e).
\end{equation}

We prove the following lemma.
\begin{lemma}\label{lem:qn}
  The integral
  \[
    \int 2^{Q_{>n}} \d\nu(\g_-)\, \d\tilde\eta(\e) = 1+o(1).
  \]
  as $n\to\infty$.
\end{lemma}

From \eqref{eq:5} and this lemma we deduce that
\begin{equation}
  \label{eq:2}
  h_m(x) = (1+o(1)) I_n = h_n(x) \cdot (1+o(1))
\end{equation}
if $m\ge n$ as $n\to\infty$. It follows that $\log h_n(x)$ is a Cauchy sequence
and hence that the limit $h(x)$ is continuous.


\subsubsection*{Proof of Lemma~\ref{lem:qn}}

We condition on a fixed graph $\g_-$ with distribution $\nu_-$. Let $C$ be a
cluster of $\g_-$. Note that
$$ Q=(X(C_1) -1)_{+} +(X(C_2)-1)_{+} + \ldots $$
where $X(C)$, is a sum of independent Bernoulli variables
$$ X(C) = \sum_{-j\in C} \sum_{i=0}^\infty \e_{ji} $$
where
$$
\P(\e_{ij}=1) = \frac{1- \exp\{-\frac \beta{(i+j)^2}\}}{2}
$$
It follows that we can approximate $X(C)$ with a Poisson variable (***)
$\tl X(C) \sim \opn{Po}(\lambda(C))$ with
$$
\lambda(C) = \frac{\beta}{2} \sum_{j\in C} \frac 1j
\approx \frac{\beta}{2} \sum_{j\in C} \sum_{i=0}^\infty \P(\e_{ij}=1).
$$
Note that
\begin{equation}
  \label{eq:lambdabound}
    \lambda(C) \leq \log \left(1+\frac{|C|}{i(C)}\right)
\end{equation}
where $-i(C)=\max C$.

Order the clusters of $\g_-$ as $C_1,C_2,\dots$ etc. so that
$i(C_1)<i(C_2)<\dots$. For each cluster $C_i\in\mscr C(\gamma_-)$ we can from
stochastic dominance construct a random cluster $\tl C_i$ such that (i)
$C_i \subset \tl C_i$ and (ii) $i(\tl C_i)=i(C_i)$. We can further assume that the
$\tl C_i$s are \emph{independent} with the same distribution.

Let now
\[
  \tl Q = \sum_{C_i\inc\mscr C(\gamma_-)} (\tl X(\tl C_i) - 1)_+.
\]
where $\P(\tl X(\tl C)| \tl C) = \opn{Po}(\lambda(\tl C))$ which stochastically
dominates $X(C)$. For a poisson variable $X\sim\opn{Po}(\lambda)$ we have
\[
  \E(2^{(X-1)_+}) = \frac{\exp(\lambda(e^{\ln 2}-1)) + e^{-\lambda}}{2} = \cosh(\lambda)
\]

We then have
$$
\E(2^Q | \gamma_- ) \leq \prod \cosh (\lambda(C_i))\leq \prod \cosh (\lambda (\tilde C_i)).
$$
We obtain, since $i(C_k)\geq k$ and \qr{eq:lambdabound} and the independence of
$\tl C_k$, that
\begin{align}
  E(2^Q) &\leq \prod_{k=1}^\infty \E\left(\frac{1}{2}\left(1+\frac{Y}{k}+\frac{1}{1+\frac{Y}{k}}\right) \right) \\
         &\leq \prod_{k=1}^\infty \E\left(1+\frac{Y^2}{k^2} + \frac{Y^3}{k^3} + \dots \right).
\end{align}
where $Y$ has the common distribution of $|C_k|$. It is easy to see that this is
less than $\infty$ on account of Lemma~\ref{geometric-bound}, which states that the
distribution $Y$ has an exponentially decreasing bound for the upper tail.

Since $Q = \lim Q_n$ we obtain from this that
\[
  \E(2^{Q_{>n}}) = \E(2^{Q-Q_n}) = 1+o(1) \quad\text{as $n\to\infty$.} \qed
\]





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\noindent
Anders Johansson, Department of Mathematics, University of G\"avle,
801 76 G\"avle, Sweden. Email-address: ajj@hig.se\newline

\noindent
Anders \"Oberg, Department of Mathematics, Uppsala University, P.O.\
Box 480, 751 06 Uppsala, Sweden. E-mail-address:
anders@math.uu.se\newline

\noindent
Mark Pollicott, Mathematics Institute, University of Warwick,
Coventry, CV4 7AL, UK. Email-address: mpollic@maths.warwick.ac.uk

\end{document}