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Gibbs measure, Brownian motion, pair potential
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\begin{document}

\title{Gibbs Measures Relative to Brownian Motion}
\author{Hirofumi Osada, Herbert Spohn}
\address{\newline
Graduate School of Mathematics, Nagoya University
\newline
Chikusa-ku, Nagoya, 464-8602, JAPAN }
\email{osada@@math.nagoya-u.ac.jp}
\address{\newline  Departments of Mathematics and Physics
\newline Technische Universit\"{a}t M\"{u}nchen
\newline D-80209 M\"{u}nchen, Germany}
\email{spohn@@mathematik.tu-muenchen.de}
\date{\today}
\keywords{Gibbs measure, Brownian motion, pair potential}
\subjclass{60K35,82B21}
\begin{abstract}
We consider Brownian motion perturbed by the exponential of an action.
The action is the sum of an external, one-body potential and a two-body
interaction potential which depends only on the increments. Under
suitable conditions on these potentials we establish existence and
uniqueness of the corresponding Gibbs measure. We also provide an
example where uniqueness fails because of a slow decay in the
interaction potential.
\end{abstract}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%
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\ND\xxxa{\frac{1}{n}\sum_{|i|\le  nT}\1(x_{\frac{i}{n}})}
\ND\xxxba{\frac{1}{2n^2}\sum_{|i|, |j|\le  nT}
\2(\frac{i-j}{n},x_{\frac{i}{n}}-x_{\frac{j}{n}})}
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\section{Introduction}
In its standard form the theory of Gibbs measures is formulated as a random
field over $\Z^d$ with general single site space and a product measure as
reference measure \cite{georgii}. Gibbs measures also
arise in the context of the
Euclidean version of quantum field theory \cite{glimm}.
In this case the set-up is somewhat modified. The Gibbs measure is defined
on ${\cal S}'(\R^d)$, the space of tempered distributions
over $\R^d$, and the reference measure is a suitable Gaussian measure
on ${\cal S}'(\R^d)$.
Considerable effort has been invested to construct such Gibbs measures,
one central problem being the control of the singular
behavior at short distances.
In our paper we will study a class of Gibbs measures which also originate
from quantum mechanics, but are in fact
stochastic processes (rather than fields).

The physical context is a single quantum particle subject to a
prescribed external potential and coupled to a free bosonic field.
For simplicity we consider the case of one-dimensional motion only.
Then the position of the particle at time $t$ is $X_t\in\R$ and the
external potential is given through a function $\map{\1}{\R}{\R}$.
The reference measure is the Brownian bridge, denoted by $\Wd$.
Since $X_t$ will be constructed as stationary process, we choose the
left end point at $-T$, $X_{-T}=\xi_{-T}$ and the right end point
at $T$, $X_T=\xi_T$. $\Wd$ is defined on $\CT$.
Let $\phi$ be the ${\cal S}'(\R^d)$ valued infinite-dimensional
stationary Ornstein-Uhlenbeck process,
defined as the Gaussian measure, $\mu_G$, with mean zero and covariance
\begin{align}\label{:01}&
\int\mu_G(d\phi) \phi(g,t)\phi(f,0)=
\int            \hat{g}(k)^* \omega(k)^{-1} e^{-\omega(k)|t|} \hat{f}(k)d^dk.
\end{align}
Here $\phi(g,t)=\int\phi(y,t)g(y)d^dy$, $\hat{}$ denotes Fourier
transform, and $\omega(k)\ge 0$, $\omega(k)=\omega(-k)$.
The path measure of physical interest is then given by
\begin{align}
&Z_T^{-1}\exp[
         -\int_{|t|\le T}\{
\1(X_t) + \int \lambda(y-X_t) \phi(y,t) d^dy \}dt ]
\mu_G(d\phi)  \Wd       (dX) ,
\label{:02}
\end{align}
where $Z_T$ is the normalization and $\lambda\in {\cal S}({\R}^d)$.
Clearly, the Gaussian integration can be carried out explicitly with the result
\begin{align}
&Z_T^{-1} \exp[-
        \int_{|t|\le T} \1(X_t)dt
       - \frac{1}{2} \int_{|t|,|s|\le T}    \2 (t-s, X_t-X_s)   dtds ]
        \Wd (dX)
\label{:03}
\\
\intertext{and}
&
\2 (t,x)= -\int |\hat{\lambda}(k)|^2 \frac{1}{\omega(k)}
e^{-\omega(k)|t|} \cos(k\cdot x)   d^d k.
\label{:04}
\end{align}

In our paper we abstract from the specific physical application and 
investigate Gibbs measure of the form \eqref{:03} with a single site 
potential $\1$ and a pair potential $\2$.  
This means we study probability measures on $\C$, whose conditional measures,
i.e. the distribution of $\{X_t;|t|\le T\}$ conditioned on the outside path
$\{\xi_t;|t|\ge T\}$, have the Gibbsian structure \eqref{:03}.

On a theoretical level the basic issues are the {\it existence} of a 
Gibbs measure with potentials as in \eqref{:03} and its {\it 
uniqueness}. For the existence, besides some regularity on $\2$,
it should suffice to require that the potential $\1$ is sufficiently 
confining. This can be ensured either by  $\1(x)\to\infty$ as $|x|\to\infty$
or by $\1$ being sufficiently attractive close to the origin.
To really prove the existence only two methods seem to be available: 
monotonicity, which is used in this paper, and compactness as 
technically encoded in the Ruelle superstability estimates
\cite{ruelle,lebowitz}. This latter method should work for a large 
class of potentials. It relies on the product
structure of the a priori measure, however,  and we have not succeeded 
in
generalizing it to the present context. 

Monotonicity uses the FKG and Brascamp-Lieb inequalities. They 
require a rather special pair potential of the form 
$\2(t,x) = \4(t) v(x)$ with $v$ convex. Clearly for the physical pair 
potential \eqref{:04} such a condition cannot be satisfied. 
Fortunately,
in many cases the so-called dipole approximation suffices, which in the 
functional integral corresponds to the quadratic approximation
\begin{align}\label{:05}&
	\2 (t,x) \simeq -\int |\hat{\lambda}(k)|^2 \frac{1}{\omega(k)}
e^{-\omega(k)|t|}(1 - \frac{1}{2}(k\cdot x)^{2})   d^d k
= -\sigma(t) + \4(|t|) x^2,
\end{align}
to which our results apply.

For the uniqueness of the Gibbs measure
the standard statistical mechanics intuition is that a bounded
interaction energy between right and left half line should provide a
sufficient condition no matter what $\1$, which 
under our assumptions means 
$\int_{0}^{\infty}\4(t)tdt<\infty$. This is precisely the criterion
under which we establish uniqueness, cf. Theorem 2.3 below. To 
convince ourselves that for a  
more slowly decaying interaction potential uniqueness may fail
we consider the case where $\1$ is a double well potential
and  $\2$ is of the form \eqref{:05} with $\rho(t) \cong t^{-\gamma}$,
$1 < \gamma \leq 1$,    for
large $t$ and show that there exist then at least two distinct Gibbs measures.

To give an outline: In the following section we list the precise
assumptions on $\1$, $\2$ and state our main results.
In Section 3 and 4 we establish existence and in Section 5 uniqueness
under suitable conditions. In Section 6 we present our example for
nonuniqueness.

\section{Main results}
We introduce a notion of right- and left-dominators.
%\marginpar{dfn:1.1}
\begin{dfn}\label{dfn:1.1}
Let $f=f(x)$ and $g=g(x)$ be functions with value on ${\R\cup\{\infty}\}$
defined on $\R$.
We say $g=g(x)$ is a right-dominator (resp. left-dominator) of $f$ if
\\ (D.1) $g$ is convex and finite on at least two distinct points.
\\ (D.2) $f-g$ is non-decreasing (resp.\ non-increasing) in $x$, where
we use the convention that $\infty-\infty=0$ in case of $f(x)=g(x)=\infty$.
\\ (D.3) There exists a constant $a>0$  such that $g''(x)\ge 2a$
a.e.\ $x\in\{g<\infty\}$.
\end{dfn}

If $f$ and $g$ are differentiable, then a right-dominator satisfies
$f'(x) \ge g'(x)$ and a left-dominator $f'(x) \le g'(x)$ with $g'$
strictly increasing. We say a dominator $g$ is symmetric around $m$
if $g(x- m)=g(|x-m|)$ for all $x$.
We remark, if $g$ is a right-dominator (resp. left-dominator)
of $f$ symmetric around $m$ (resp.\ $-m$), then there exist right-dominator
(resp. left-dominator) of $f$ symmetric around $n$ (resp.\ $-n$) for all
$n\ge m$. Indeed, we easily see
\begin{align*}
&R^+_{n}(g)(x)=
\begin{cases}
g(x)&\text{for }x\ge n\\ g(2n-x)&\text{for }x\le n
\end{cases},\\
&  R^-_{n}(g)(x)=
\begin{cases}
g(-2n-x)&\text{for }x\ge -n\\ g(x)&\text{for }x\le -n
\end{cases}
\end{align*}
are such dominators.

Let $\map{\1}{\R}{\R\cup\{\infty\}}$ be the external potential. Let
$I = (a_{-},a_{+})$ be an open interval, $-\infty \le a_{-}<a_{+} \le
\infty$. We assume that $\1(x) = \infty$ for $x \in \R \setminus I$,
which simply means that in \eqref{:03} the paths are restricted to
$\{ X_{t} \in I, |t| \le T \}$.
Let $\map{\2}{\R^2}{\R}$ be interaction potential. We also refer to
$(\1,\2)$ as potential.
\vspace{1ex}

\noindent
(A.1) {\sf Assumptions on the external potential $\1$.} \\
(P.1) $\1:I \to \R$ is locally integrable and bounded from below.\\
(P.2) $\1$ has a right-dominator symmetric
around $m$ and a left-dominator symmetric
around $- m$ for some $m\ge 0$.
\vspace{2ex}

\noindent
(A.2)  {\sf Assumption on interaction potential $\2$.}\\
\thetag{W.1} $\2(t,x)=\4(t)v(x)$ with $\4\ge 0$, $\4(t)=\4(|t|)$ and
$v(x)=v(|x|)$. \\
\thetag{W.2} $v(\cdot)$ is convex and piecewise smooth.
In addition, for some $p_0>1$
\begin{align}\label{:2c}&
u(x)= \text{ess.sup}_{y\in \R}  \frac{|v'(x)-v'(x-y)|}{1+|y|^{p_0}}
                                                        \end{align}
is finite and for each $\e>0$ there exists a $b=b(\e) \ge 0$ such that
\begin{align}\label{:2a}&
 u(x)   \le  \e \{v'(x)-v'(x-b)\}       \quad \text{ for all }x\in \R.
                                                        \end{align}
\thetag{W.3} $\4(|t|)\le\4_0(t)$, where $\map{\4_0}{\R^+}{\R^+}$ is an
integrable, convex and non-increasing function such
that $\4_0(t)>0$ for all $t$ and  $\4_0(0)<\infty$.

\vspace{2ex}

We remark $v''\ge 0$, $v'(x)=-v'(-x)$ is non-decreasing and $u(x)=u(-x)$ by
definition.
\eqref{:2c} implies that there exists $C_{v}$ such that
\begin{align*}
&v(x)\le C_{v}(1+|x|)^{p_0+1} \quad \text{ for all }x\in\R
%\label{W.3'}
.\end{align*}
We easily see \eqref{:2a} is satisfied if $v(x)=|x|^a$ for some $a\ge 1$ or
more generally,
$v(x)=\sum_{i=1}^{N}a_i|x|^{p_i}$ with $ a_i>0,\  p_i\ge 1$.
\vspace{2ex}

To define a Gibbs measure (DLR measure) we first introduce the
energy $H_{T}(X,\xi)$ of the path $X=\{X_t;|t|\le T\}$,
$X\in C([-T,T];\R])$, for given outside
path $\xi=\{\xi_t;|t|\ge T\}$, $\xi\in C(\{|t|\ge T\};\R)$.
This energy is given by
\begin{align}\begin{split}
H_T(X,\xi)=&\xxxfa  \\
           &+\xxxfb.
\end{split}\label{;12}\end{align}
Note that $H_T$ can take the value $\infty$ but is bounded from below.

Let $\Wd$ be the path measure of the Brownian bridge on $\CT$ with boundary
condition $\Wd(X_{\pm T}=\xi_{\pm T})=1$ and let $\EW$ be the corresponding
expectation.
We define
\begin{align}
&Z_{T,\xi}=\EW[\exp[-H_{T}(X,\xi)]]
\label{;..a}
\end{align}

Let $\map{\pi_{T}}{\C}{\CT}$ and $\map{\pi_{T}^*}{\C}{C(\{|t|\ge T\};\R)}$
be  projections. For a probability measure $\mu$ on $\C$ we denote by $\Md$
the probability on $\CT$ given by
$$\Md=\mu(\pi_{T}\in\, \cdot\, | \pi_{T}^*=\xi).$$
Here $\mu(\cdot|\pi_{T}^*=\xi)$ is the regular conditional
probability with respect to the $\sigma$-field $\sigma[\pi_{T}^*]$, evaluated
by the value $\pi_{T}^*(X)=\xi$.

%\marginpar{dfn:G}
\begin{dfn}\label{dfn:G}
A probability measure $\mu$ on $\C$ is called Gibbs measure
with external potential $\1$ and interaction potential $\2$ if its conditional
measures are given by
\begin{align}
& \Md =Z_{T,\xi}^{-1}\exp[-H_T(X,\xi)] d\Wd
\label{;G}              \end{align}
for $\mu(\pi_T^*\in\cdot)$-a.s. $\xi\in\C$.
\end{dfn}
To establish the existence of a Gibbs measure we have to go
through a finite volume construction.
For this purpose we define $\Mf$ by
\begin{align} &\label{;11a}
\Mf=(Z_{T,\xi})^{-1}\exp[-H_T(X,\xi)] d\Wd.
\end{align}
as a measure on $\CT$, where we assume that $\|\xi\|_{\infty}<\infty$
with $\|\cdot\|_{\infty}$ the sup-norm on $\C$.
Let $\Eg$ denote the expectation with respect to $\Mf$.

%\marginpar{thm:1}
\begin{thm}[Localization] \label{thm:1}
Assume \thetag{A.1}, \thetag{W.1} and \thetag{W.2}.
Then there exists $\a>0$ such that,
for all $\xi$ with $\|\xi\|_{\infty}<\infty$,
there exists a constant $C_{1}$,
depending only on $a$ in \thetag{D.3}, $m$ in \thetag{P.2}, $\2$  and
$\|\xi\|_{\infty}$, satisfying
\begin{align}\label{;13}
&\Eg[e^{\a |X_t|^2}]\le C_{1} \quad\text{ for all }|t|\le T<\infty.
\end{align}
In particular, the finite dimensional distributions of $\{\Mf\}_{T}$ are
tight. \end{thm}

%\marginpar{rem:fv}
\begin{rem}\label{rem:fv}
The finite volume construction can be carried through also for free boundary
condition. In this case one has the energy
\begin{align}
   &H_T(X)=\xxxfa, %\label{;1f}
\end{align}
and define a sequence of probability measures $\Nf$ by
\begin{align}
& \Nf=(Z_{T,\xi,\circ})^{-1}\exp[-H_T(X)] d\Wd,
\label{;11b}
\end{align}
with $Z_{T,\xi,\circ}$ the normalization. Theorem \ref{thm:1} also holds in
this case.
\end{rem}
%\vspace{2ex}
%\marginpar{ex:2}

\begin{ex}\label{ex:2}
(1) By Lemma \ref{l:d1} we see the following satisfy \thetag{A.1}:
\begin{itemize}
\item $\1(x)=|x|^q+p(x)+\ell(x)$, where $q\ge 2$, $p$ is a polynomial whose
degree is less than or equal to $q-1$, and $\ell$ is a Lipschitz continuous
function.
\item $\1(x)=e^{|x|}+q(x)$, $q$ is a function whose derivative $q'$ is at
most polynomial growth order.
\end{itemize}
(2) An example of $\2$ satisfying \thetag{A.2} is:
\begin{itemize}
\item $w(t,x)=(1+|t|)^{-\a}|x|^{\b}$ with $\a>1$ and $\b \ge1$.
\end{itemize}
\end{ex}

%\vspace{1pc}
Using the moment bounds from Theorem \ref{thm:1} we prove
%\marginpar{thm:2}
\begin{thm}[Existence of Gibbs measure]\label{thm:2}
Assume \thetag{A.1} and \thetag{A.2}.
Then there exists a Gibbs measure for $(\1,\2)$.
\end{thm}

%\marginpar{rem:1}
\begin{rem} \label{rem:1}
(1) $\1$ may depend on time $t$. \\
(2) Once Theorem \ref{thm:2} is established, it is easy to see
that there exists at least one translation invariant Gibbs measure
for $(\1,\2)$.
\end{rem}

As for statistical mechanics in one dimension the uniqueness can be
linked to a sufficiently fast decay of the interaction energy.
%\marginpar{thm:3}

\begin{thm}[Uniqueness of Gibbs measure]\label{thm:3}
Let \thetag{A.1} and  \thetag{A.2} hold. If
\begin{align}
&\int_{0}^{\infty} t \4(t)dt<\infty,
\label{;w6}
\end{align}
then there exists exactly one translation invariant Gibbs measure for
$(\1,\2)$ satisfying for some $p_2> p_0+1$
\begin{align}\label{:w7}&
\int|X_t|^{p_2}  d\mu <\infty.
                                                        \end{align}
Moreover, any limit points of $\{\Nf\}_{T}$ or $\{\Mf\}_{T}$
for $\xi$ with $\|\xi\|_{\infty}<\infty$ as $T\to\infty$ are unique and,
henceforth, translation invariant.
\end{thm}

%\marginpar{rem:2}
\begin{rem}\label{rem:2}
 Theorem \ref{thm:3} is a corollary of Theorem \ref{thm:u4}, which
states a stronger uniqueness result. There we will prove
that a {\em tame} Gibbs measure is unique, c.f.  Definition \ref{dfn:tame}
for the notion of tame.
% !!This is a remark which cannot be well understood at this stage.!!
%(3) To prove the uniqueness we need a certain control of the growth order
%of $v$. Indeed, if $v(x)=0$ for $|x|<1/3$ and $v(x)=\infty$
%for $|x|\ge 1/3$, and $\4(t)>0$ for all $t$, and $\1=|x|^2$, then
%Gibbs measures are not unique. We easily see this by constructing Gibbs
%measures $\mu^n$ as limits of finite volume Gibbs measures with
%boundary conditions $\xi^n$ such that $\xi^n_t\equiv n$ ($n\in \Z$).
%For under $\mu^n$ almost sure paths $X_t$ are trapped between $n-1/3$
%and $n+1/3$ for all $t$.
\end{rem}

%\vspace{2ex}
To discuss nonuniqueness let us consider the particular potential
$\1(x)=\beta(x^4-x^2)$, $\beta>0$, where $\beta$ is a parameter which
controls the depth of the two minima at $x=\pm 1/\sqrt{2}$,
and $\2(t,x)=\a(1+|t|)^{-\gamma}x^2$, where $\a >0$ controls the strength
and $\gamma$ the range of the interaction, $\gamma >1$.
For $\a=0$ a typical path will fluctuate around one minimum and then
rapidly cross to the other minimum. The average waiting time in one
minimum is approximately equal to the inverse spectral gap of the symmetric
operator $-\frac{1}{2}\frac{d^2}{dx^2}+\1(x)$ on $L^2(\R,dx)$. As $\a$
increases, this waiting time becomes longer. If $\gamma \le 2$
and if $\a$ is sufficiently strong, then even for $T\to\infty$
boundary conditions persist: if $\xi_{-T} = 1 = \xi_{T}$, then in the
limit measure, the path will spend more time close to the right minimum.

\begin{thm}\label{thm:4}
Let $\1(x)=\beta(x^4-x^2)$ and $\2(t,x)=\a(1+|t|)^{-\gamma}x^2$.
If $1<\gamma \le 2$, then we can choose $\a,\beta>0$ such that there
are at least two
translation invariant
 Gibbs measures for $(\1,\2)$.
\end{thm}

\section{Proof of Theorem \ref{thm:1}. }
Let $E^{\1,\2}_{T,\xi}$ and $E^{\1,\2}_{T,\xi,\circ}$ be expectations with
respect to $\Mf$ and $\mu^{\1,\2}_{T,\xi,\circ}$, respectively.

%\marginpar{lem:2.1}
\begin{lem}\label{lem:2.1}
Let $\3$ be a right-dominater of $\1$.
Let $F$ be a non-negative, continuous function on
$\CT$ such that $F$ is increasing in the sense that
$F(X)\le F(Y)$ whenever $X\le Y$ on $[-T,T]$.
Suppose $\xi\le\eta$. Then
\begin{align}
&\label{;21}
 E^{\1,\2}_{T,\xi}[F(X)]   \le
 E^{\3,\2}_{T,\eta}[F(X)].
\\&
 E^{\1,\2}_{T,\xi,\circ}[F(X)]\le
 E^{\3,\2}_{T,\eta,\circ}[F(X)].
\label{;21b}
\end{align} \end{lem}

\begin{pf} We introduce discretizations $\Me$ and $\Mg$ of $\Mf$ and
$\Ma_{T,\eta}^{\3,\2}$, respectively. Let $L(n,T)=[-T,T]\cap \{\Z /n\}$.
Let $\Me$ and $\Mg$ be the probability measures on $\CnT$ given by
\begin{align}\label{;25}&
   \Me=(Z_1)^{-1}e^{-m_0-m_1}
   \prod_{|i|< nT}dx_{\frac{i}{n}}\times
   \delta_{\xi_{-T}}(dx_{-T})\times\delta_{\xi_{T}}(dx_{T})\\
&\Mg=(Z_2)^{-1}e^{-m_0-m_2}
        \prod_{|i|< nT}dx_{\frac{i}{n}} \times
        \delta_{\eta_{-T}}(dx_{-T})\times\delta_{\eta_{T}}(dx_{T})\ ,
\notag
                                                        \end{align}
where $Z_1$ and $Z_2$ are normalizations, and $m_0$, $m_1$ and $m_2$ are
given by
\begin{align*}%\label{:z1}
m_0=&\xxxba
        +\frac{n}{2}\sum_{-nT\le  i <  nT}  |x_{\frac{i}{n}}-x_{\frac{(i+1)}{n}}|^2
                                                        ,\\
m_1=    & \xxxa +  \xxxbb                       ,       \\
m_2=    &  \frac{1}{n}\sum_{|i| \le  nT}\3(x_{\frac{i}{n}}) +
        {\frac{1}{n^2}\sum_{|i|\le nT <  |j|}
        \2(\frac{i-j}{n},x_{\frac{i}{n}} -      \eta_{\frac{j}{n}})}.
                                                                \end{align*}
We note the square terms in the definition of $m_0$ come from the Brownian
bridge.

Let $\map{\iota}{\R^{L(n,T)}}{C([-T,T];\R)}$ denote the map defined by the
linear interpolation of $x=(x_{i/n})\in\R^{L(n,T)}$.
Let $\widetilde{\mu}_n^i=\mu_n^i\circ\iota^{-1}$ ($i=1,2$).
Then $\widetilde{\mu}_n^1$ and $\widetilde{\mu}_n^2$ converge to $\Mf$ and
$\Ma_{T,\eta}^{\3,\2}$ weakly in $C([-T,T];\R)$, respectively.
Hence \eqref{;21} is reduced to
\begin{align}\label{:zb}&
\int F(X) d\widetilde{\mu}_n^1  \le \int F(X) d\widetilde{\mu}_n^2
\quad \text{ for all } n
                                                        \end{align}
because $F$ is continuous and nonnegative. By definition we see \eqref{:zb}
is equivalent to
\begin{align}\label{:zc}&
\int F(\iota(x)) d\mu_n^1  \le \int F(\iota(x)) d\mu_n^2
\quad \text{ for all }n                                 .
\end{align}
In order to prove \eqref{:zc} we use Lemma \ref{lem:A.2}  in Appendix, and
Preston's FKG inequality (see Theorem 3 in \cite{pre}). Let
\begin{align*}
&a_i(t)=
\begin{cases}
\frac{1}{n}\1(t)+\frac{1}{n^2}\sum_j \2(\frac{i-j}{n},t-\xi_{\frac{j}{n}})
& \!\!\!\!\!\!\!\!\! \text{ for }|i|\le nT-1
\\
\frac{1}{n}\1(t)+\frac{1}{n^2}\sum_j \2(\frac{i-j}{n},t-\xi_{\frac{j}{n}})
+       \frac{n}{2} |t-\xi_{\pm (T+ \frac{1}{n})  }|^2
& \text{ for }i=\pm nT
\end{cases}
\\
&b_i(t)=
\begin{cases}
\frac{1}{n}\3(t)+ \frac{1}{n^2} \sum_j \2(\frac{i-j}{n},t-\eta_{\frac{j}{n}})
& \!\!\!\!\!\!\!\!\! \text{ for }|i|\le nT-1
\\
\frac{1}{n}\3(t)+\frac{1}{n^2}\sum_j \2(\frac{i-j}{n},t-\eta_{\frac{j}{n}})
                        +       \frac{n}{2} | t -\eta_{\pm (T+\frac{1}{n})}   |^2
& \text{ for }i=\pm nT
\end{cases}     \\
&c_{ij}(t)=
\begin{cases}
\frac{1}{2n^2}\2(\frac{i-j}{n},t)       + \frac{n}{4} t^2       & \text{ for }|i-j|=1  \\
\frac{1}{2n^2}\2(\frac{i-j}{n},t)                                               & \text{ for }|i-j|> 1
\end{cases}  \end{align*}
Here the sum $\sum_j$ is taken over $|j| \ge nT+1$.
Then the assumptions of Lemma \ref{lem:A.2} are fulfilled.
So we can apply Preston's FKG inequality (Theorem 3 in \cite{pre}) to
obtain \eqref{:zc}, which yields \eqref{;21}. The proof of \eqref{;21b} is
similar, so it is omitted.
\end{pf}

%\marginpar{lem:2.2}
\begin{lem}\label{lem:2.2}
Let $\3$ be a dominator of some function and $\3_{s}=a(x-b)^2$, where $a>0$
is a constant in \thetag{D.3} and $b\in\R$ is any number.
Let $G(x)=|x|^p$ with $p\ge 1$ or $G(x)=e^{\a |x|}$ or $G(x)=e^{\a |x|^2}$
with $\a>0$. Then
{\em
\begin{align}
&\label{;22}
E^{\3,\2}_{T,\eta}[G(X_t-E^{\3,\2}_{T,\eta}[X_t ])]\le
E^{\Qa,0}_{T,\eta}[G(X_t-E^{\Qa,0}_{T,\eta}[X_t ])]
\\&\label{;22a}
\Ef{\3,\2}{\eta,\circ}
[G(X_t-\Ef{\3,\2}{\eta,\circ}[X_t ])]\le
\Ef{\Qa,0}{\eta,\circ}
[G(X_t-\Ef{\Qa,0}{\eta,\circ}[X_t ])]
\end{align}
}
\end{lem}

\begin{pf} We first note that the cases of $G(x)=e^{\a|x|}$ and
$G(x)=e^{\a|x|^2}$ follows from the case of $G(x)=|x|^p$, $p\in\N$, because
of
$e^{\a|x|}=\sum (\a|x|)^p/p!$ and $e^{\a|x|^2}=\sum (\a|x|^2)^{p}/p!$. Now
we consider discretization of measures similarly as \eqref{;25}. Since
$\3$, $\Qa$ and $\2$ are convex functions, these discretized measures have
log-concave densities. Hence these inequalities for these discretized
measures follow from Brascump-Lieb inequality (see  \cite{bl} Theorem 5.1).

Hence taking the limit $n\to\infty$ completes the proof.
\end{pf}

Combining these lemmas we see the proof of Theorem \ref{thm:1} is reduced
to the bound on expectations: Let
\begin{align}
&\notag
C_{2.1}=\sup\{\Ef{\Qa,0}{\eta}\left[\exp \a |X_t-\Ef{\Qa,0}{\eta}[X_t
]|\right];
|t|\le T <\infty\}
\\
\notag
&
C_{2.2}=\sup\{\Ef{\Qa,0}{\eta}\left[\exp 2\a |X_t-\Ef{\Qa,0}{\eta}[X_t
]|^2\right];
|t|\le T <\infty\}.
\end{align}
Then, since $\mu^{\Qa,0}_{T,\eta}$ are Ornstein-Uhlenbeck processes,
$C_{2.1}$ is finite for all $\a>0$ and $C_{2.2}$ is finite for small
$\a>0$.
These values depend only on $\a$, $\|\eta\|_{\infty}$ and $\Qa$.
%\marginpar{lem:2.A}
\begin{lem}\label{lem:2.A}
Let $\1$, $\2$, $\3$, $\xi$ and $\eta$ be as in Lemma \ref{lem:2.1}.
\thetag{1}  Assume
\begin{align}\label{;2a1}&
C_{2.3}:=\sup\{|\Eh[X_t]|;|t|\le T<\infty\} <\infty.
\end{align}
Then for each $|t|\le T<\infty$
\begin{align}
\label{;2a2}&
\Eg \left[\exp[\a \max\{0,X_t\}] \right]   \le e^{\alpha C_{2.3}}\cdot C_{2.1}
\\
\label{;2a4}&
 \Eg \left[\exp[\a \max\{0,X_t\}^2]\right] \le e^{2\a C_{2.3}^2}\cdot C_{2.2}.
\end{align}
\thetag{2} The same result also holds for free boundary condition.
\end{lem}
\begin{pf}
Let $F(X)=\exp[\a \max\{0,X_t\}]$ and $G(x)=e^{\a|x|}$. Then
\begin{alignat*}{2}
\Ef{\1,\2}{\xi}[F(X)]
&\le  \Ef{\3,\2}{\eta}[F(X)]
       \qquad &&\text{by Lemma \ref{lem:2.1} }
       \\
&\le  \Ef{\3,\2}{\eta}[G(X_t)]
       \qquad &&\text{by }F(X)\le G(X_t)
       \\
&\le  e^{\alpha C_{2.3}}\cdot
       \Ef{\3,\2}{\eta}[G(X_t-\Ef{\3,\2}{\eta}[X_t ])] &&{}
       \\
&\le  e^{\alpha C_{2.3}}\cdot
       \Ef{\Qa,0}{\eta}[G(X_t-\Ef{\Qa,0}{\eta}[X_t ])]
       \quad&&\text{by Lemma \ref{lem:2.2} }.
\end{alignat*}
This implies \eqref{;2a2}. The proof of \eqref{;2a4} and \thetag{2} are
similar,
so we omit it. \end{pf}

\noindent{\bf  Proof of Theorem \ref{thm:1}.}
Let us take $\a$ so small in such a way that $C_{2.2}<\infty$.
By \thetag{P.2} there exists a right-dominator symmetric around $m>0$.
Let us take $b\ge  0$ in such a way that $b=\max\{m, \|\xi\|_{\infty}\}$.
Then by the remark after Definition \ref{dfn:1.1} there exists a
right-dominator $\3$ of $\1$ symmetric around $b$. Let $\eta  \equiv b$.
By symmetry
$$\Ef{\3,\2}{\eta}[X_t]=b\quad\text{ for all }t$$
in both boundary conditions, which means \eqref{;2a1}. So by Lemma
\ref{lem:2.A} we have \eqref{;2a4}.

By using a left-dominator we obtain also \eqref{;2a4} for $-X_t$. So
combining these two inequalities yields Theorem \ref{thm:1}.
\hfill$\Box$

%\marginpar{rem:31}
\begin{rem}\label{rem:31}(1)
In the proof of Theorem \ref{thm:1} we used only the convexity of $v$ in
the assumptions of $v$ in \thetag{W.2}. Accordingly, Theorem \ref{thm:1}
still holds, if we replace  \thetag{W.2} by the convexity of $v$ which may
take the value $\infty$ and if $Z_{T,\xi}$ in \eqref{;11a} is finite (this
is necessary to define $\Mf$).
\\
(2) Suppose $\1$ is time inhomogeneous; $\1=\1(t,x)$.
Suppose for each $t\in I$ there exist right- and left-dominators
$\3^{\pm}(t,\cdot)$ of $\1(t,\cdot)$ symmetric around $\pm m$ and
$\{\3^{\pm}\}''(t,x)\ge 2a$.
If we can take $m$ and $a$ are independent of $t\in I$, then the same
conclusion of Theorem \ref{thm:1} holds.
\end{rem}

\section{Proof of Theorem \ref{thm:2}}
In this section we will prove Theorem \ref{thm:2} in a slightly general
frame work.
%\marginpar{thm:42}
\begin{thm}\label{thm:42}
Assume \thetag{A.1}, \thetag{W.1},    \thetag{W.3} and
$v$ is convex such that $v=\infty$ for $|x|\ge a$   $(0<a\le \infty)$    and
\begin{align}\label{:2b}&
\text{$|v(x)| \le C_{4.1} e^{p_1|x|}$ for all $|x|< a$ for some
$C_{4.1},p_1>0$. }
                                                        \end{align}
Then there exists a Gibbs measure for $(\1,\2)$.
\end{thm}
%\marginpar{rem:41}
\begin{rem}\label{rem:41}
(1) Since \eqref{:2a} implies \eqref{:2b}, Theorem \ref{thm:2} follows from
Theorem \ref{thm:42}.\\
(2) Let $\1(x)=x^2$, $v(x)=0$ for $|x|<1$ and $v(x)=\infty$ for $|x|\ge 1$,
and $\4(t)=1$ for $|t|\le 1$ and $\4(t)=0$ for $|t|\ge 1$. Then the
assumptions of Theorem \ref{thm:42} are satisfied. \end{rem}

Throughout this section we fix $\1$ and $\2$ and often suppress them from
the notation. Let $\Nz$ be the probability measure on $\CT$ given by
\eqref{;11b} obeying free boundary condition with $\xi\equiv 0$. Let
$\mu_T$ be its extention to $\C$ given by
\begin{align*}%\label{;3.1}
  &\mu_T=\Nz \quad\text{ on }\sigma[X_t;|t|\le T],
\\&\mu_T(X_u=0\text{ for all }|u|\ge T)=1.
\end{align*}

Let $\A=\{\aa=(a_j)_{j\in\N};a_j>0\, \}$. For $\aa,\bb\in\A$ we write
$\aa\le \bb$ if $a_j\le b_j$ for all $j\in\N$.
For $k\in\N$ and $\aa=(a_j)\in\A$ we set
$$
\Xi(k,\aa)=\{\xi\in\C;\, \int_{\R}  \4_0(u)  e^{p_1|\xi_u|}   du\le k,\, \
        |\xi_{\pm j}|\le a_j \text{ for all }j\in\N  \, \}.
$$
By Theorem \ref{thm:1} and \thetag{W.3} there exist increasing sequences
$k^n$ and $\aa^n \in \A$ such that
\begin{align}\label{;3.2}
&\liminf_{T\to\infty}\mu_T(\Xi(k^n,\aa^n))\ge 1-\frac{1}{n}
.\end{align}
We set $\Xi_n=\Xi(k^n,\aa^n)$. Note that by construction $\Xi_{n}\subset
\Xi_{n+1}$. Let
$$\mu_T^n(\cdot)=\mu_T(\cdot\cap\Xi_n).$$
Note that $\mu_T^n$ are measures on $\C$ whose total volumes are less than
or equal to $1$, and $\{\mu_T^n\}$ is increasing in $n$ and compatible in
the sense that
\begin{align}
&\mu_T^{n+1}(A)=\mu_T^{n}(A)\text{ for }A\subset\Xi_n.
\label{:x}
\end{align}

For $ S\le T\in\N$ we set
$$      \ww_{S,T}(t,x,\xi)=\int_{S\le |u|\le T} w(t-u,x-\xi_u) du       .$$
%\marginpar{lem:3.1}
\begin{lem}\label{lem:3.1}
There exists a constant $C_{4.2}$ such that
\begin{align}\label{;3.3}
&\ww_{S,T}(t,x,\xi) \le C_{4.2} e^{p_1|x|}  \end{align}
for all $\xi\in\Xi_n$ such that $\ww_{S,T}(t,x,\xi) <\infty$.
\end{lem}
\begin{pf}
Let $C_{4.1}$ be the constant given in \eqref{:2b}. Then
\begin{align*}
\ww_{S,T}(t,x,\xi)
   &\le \int_{S\le |u|\le T}\4(t-u)C_{4.1} e^{p_1|x-\xi_u|}du
\\ &\le C_{4.1} e^{p_1|x|} \int_{S\le |u|}\4(t-u)e^{p_1|\xi_u|}du
\\ &\le C_{4.1} e^{p_1|x|}  C_{4.3} k^n
    .\end{align*}
Here $C_{4.3}$ is a constant such that $\sup_{|t|\le S}\4(u-t)\le
C_{4.3}\4_0(u)$ for all $ |u| \ge S $. We remark  by \thetag{W.2}
$C_{4.3}<\infty$. Taking $C_{4.2}=C_{4.1} C_{4.3} k^n$ completes the proof.
 \end{pf}

Let ${\Wa}_{S,x}$ denote the Brownian bridge conditioned
$x= \\ (x_{-},
x_+)\in\R^2$ at $\pm S$, and $\ES$ its expectation.
Let $\map{Z}{\R^2}{\R}$ such that
$$
Z(x)=\ES\left[ e^{-H_S(X)-C_{4.2} \int_{|t|\le S}   \exp[p_1|X_t|] dt} \right].
$$
Let
\begin{align*}%\label{;}
& W_{S,T}(X,\xi)        = \int_{|t|\le S}\ww_{S,T}(t,X_t,\xi)dt \\
& Z_{S,T,\xi}           = \ET\!\left[e^{-H_S(X)-W_{S,T}}\right].
\end{align*}

%\marginpar{lem3.2}
\begin{lem}\label{lem:3.2}
Let $C_{4.4}=v(0)\int_{|t|\le S \le|u|}\4(t-u)dtdu$. Then
\begin{align}
\label{;3.2a}& W_{S,T} \ge C_{4.4} ,\\
&\label{;3.2b}
Z_{S,T,\xi}\ge Z(\xi_{-S},\xi_{S})\quad\text{ for all }\xi\in\Xi_n
.\end{align}\end{lem}

\begin{pf} Since $w(t,x)\ge \4(t)v(0)$, \eqref{;3.2a} is clear.
\eqref{;3.2b}  follows from \eqref{;3.3} and the definition of $Z(x)$
immediately.
\end{pf}

\begin{lem}\label{lem:3.3}
There exists a constant $C_{4.5}=C_{4.5}(n,S)$ such that
\begin{align}\label{;3.4}&
\sup_{S\le T} E_T \! \left[|X_s-X_t|^4;\Xi_n \right]  \le   C_{4.5}|s-t|^2
\end{align}
$\text{for all } |s|, |t|\le S$, where $E_T$ is the expectation with
respect to $\mu_T$.
\end{lem}

\begin{pf}
For $x=(x_-,x_+)\in\R^2$ let $f(x)=\ES\!\left[|X_s-X_t|^4e^{-H_S(X)}\right]$.
Since $H_S(X)\ge -C_{4.6}$ for some $C_{4.6}=C_{4.6}(S)\ge 0$,
there exists a constant $C_{4.7}=C_{4.7}(n,S)$ such that
\begin{align}
&f(x)\le C_{4.7} |t-s|^2 \quad\text{ for all }|s|,|t|\le S,\ |x_{\pm}|\le
a^n_{S}.
\label{;3q}
\end{align}
Here $\aa^n=(a^n_k)_{k\in\N}$.
Let $A(\xi)=\{X\in\C;\widetilde{X}\in\Xi_n\}$, where
$\widetilde{X}(t)=X(t)$ if $|t|\le S$ and $\widetilde{X}(t)=\xi(t)$ if
$|t|\ge S$.
Taking the conditional expectation with respect to $\sigma[X_t;|t|\ge S]$
we see
\begin{align*}%\label{;}
& E_T[|X_s-X_t|^4;\Xi_n]
\\&     =E_T\!\left[(Z_{S,T,\xi})^{-1}\ET\!\left[|X_s-X_t|^4
        e^{-H_S(X)-W_{S,T}(X,\xi)};{A(\xi)} \right] \right]
\\&     \le E_T\!\left[
        Z(\xi_{-S},\xi_{S})^{-1}
        \ET             \!\left[|X_s-X_t|^4
        e^{-H_S(X)};{A(\xi)}  \right]\right]    e^{-C_{4.4}},
\\&     \le \left\{ \int_{|x_{\pm}|\le a^n_S}
        Z(x)^{-1} f(x) m(x) dx \right\}      e^{-C_{4.4}}
,\\
\intertext{where $m(x)dx$
is the distribution of $\mu_T\circ( X_{-S},X_{S})^{-1}$,}
&       \le C_{4.7} |s-t|^{2}
         \left\{ \int_{|x_{\pm}|\le a^n_S} Z(x)^{-1}  m(x) dx \right\}  e^{-C_{4.4}}.
\end{align*}
Here we used Lemma \ref{lem:3.2} for the third line and \eqref{;3q} for the
fifth line. Hence $C_{4.5}:= C_{4.7} \left\{ \sup_{|x_{\pm}|\le a^n_S}
Z(x)^{-1} \right\}
        \left\{e^{-C_{4.4}}\right\}$
satisfies \eqref{;3.4}.
\end{pf}

\begin{lem}\label{lem:3.4}
There exists a probability measure $\mu$ on $\C$ and a
subsequence $\{\mu_{T'}\}$ of $\{ \mu_T\}_ {T\in\N}$ such that
$\limi{T'}\mu_{T'}=\mu$ weakly in $\C$.
\end{lem}

\begin{pf}
By Lemma \ref{lem:3.3} for each $n\in N$ we see $\mu_T^n$ is relatively
compact as measures on $\C$ with the vague topology.
So by diagonal method we can choose a subsequence $\mu_{T'}^{n}$ such that
$\mu_{T'}^{n}$  converge in $\C$ in the vague topology for all $ n\in\N$.
Recall that $\mu_T^n$ are compatible in $n$ and increasing as we see in \eqref{:x}.
Hence its increasing limit $\mu_{T'}=\limi{n}\mu_{T'}^{n}$ is also
convergent sequence of measures on $\C$ with the vague topology.
So let $\mu$ denote its limit.
Then $\mu$ is a measure on $\C$ whose total volume is less than or equal to 1.

Since $\Xi_{n}$ are closed sets in $\C$, we have
$$
\mu(\Xi_n)\ge  \limsup_{T'}\mu_{T'}(\Xi_n)\ge 1-\frac{1}{n}.
$$
We thus see that $\mu$ is a probability measure on $\C$ such that
\begin{align}\label{;3.7}&
\mu(\cup_n\Xi_n)=1.
\end{align}
Since $\mu$ is a probability measure, $\mu_{T'}$ converge to $\mu$ not only
in  the vague topology but also weakly in $\C$. This completes the proof.
\end{pf}

\noindent
{\bf Proof of Theorem \ref{thm:42}.}
Let $\Upsilon_n=\{\xi\in\C; \int_{|t|\ge T} e^{p_1|\xi_t|} \4_0(t)dt\le n \}$.
Then from \eqref{;3.7} and \thetag{W.2} we easily see
$\mu(\cup_n\Upsilon_n)=1$.
Let $\{\mu_U \}_{U }$ be a sequence of probability measures
converging to $\mu$ obtained by Lemma \ref{lem:3.4}.
(We change the notation here from $\mu_{T'}$ to $\mu_{U}$).
Let $E^{U}$ and $E^{\mu}$ denote the expectation with respect to $\mu_{U}$
and $\mu$, respectively.

Let ${\cal F}_T=\sigma[X_t;|t|\le T]$ and ${\cal F}_{T}^*=\sigma[X_t;|t|\ge
T]$.
Let $f$ be a bounded, continuous ${\cal F}_T$-measurable function.
We set
\begin{align*}
& F(\xi)=Z_{T,\xi}^{-1} \EW[f(X) e^{-H_{T}(X,\xi)}] \\
& F_{U}(\xi)=Z_{T,U,\xi}^{-1} \EW[f(X) e^{-H_{T,U}(X,\xi)}],
\end{align*}
where $H_{T,U}(X,\xi)=H_T(X)+W_{T,U}(X,\xi)$ and $Z_{T,U,\xi}$ is the
normalizing constant of Gibbs measure $\mu^{U}$. Note that
\begin{align}
&F_{U}(\xi) = E^{U}\!\left[f(X) | {\cal F}_{T}^*\right] (\xi)
\label{;z}
\end{align}
By \thetag{A.2} and the definition of  $\Upsilon_n$ we easily see that $F$
and $F_{U}$ are bounded and continuous in $\xi$ on $\Upsilon_n$. Moreover
\begin{align}
&\limi{U}\sup_{\xi\in\Upsilon_n} |F(\xi)-F_{U}(\xi)|=0
\label{;y}
\end{align}

Let $g$ be a bounded, continuous ${\cal F}_{T}^*$-measurable function
whose support is contained in $\Upsilon_n$.
Then $fg$ and $Fg$ are bounded continuous functions supported on $\Upsilon_n$.
Since $\mu_U$ converge to $\mu$ weakly in $\C$, we have
\begin{alignat*}{2}
 E^{\mu}[f(\xi) \cdot g(\xi)] &=\limi{U} E^{U}\!\left[f(\xi)\cdot g(\xi)
\right]&&
\\&
=\limi{U} E^{U}\!\left[E^{U}\!\left[f(\xi) | {\cal F}_{T}^*\right] g(\xi)
\right]&&
\\&
=\limi{U} E^{U}\!\left[ F_{U}(\xi)\cdot g(\xi) \right]&&\text{ by }\eqref{;z}
\\&
=\limi{U} E^{U}\!\left[ F(\xi)\cdot g(\xi) \right]&&\text{ by }\eqref{;y}
\\&
=E^{\mu}\!\left[ F(\xi) \cdot g(\xi)   \right]&&{}
\end{alignat*}
Here for the second line we used the assumption that $g$ is ${\cal
F}_{T}^*$-measurable. We thus  obtain
$$
 E^{\mu}\!\left[ f(\xi) \cdot g(\xi) \right] =
 E^{\mu}\!\left[ Z_{T,\xi}^{-1} \EW[f(X) e^{-H_{T}(X,\xi)}]g(\xi) \right]
$$
for all bounded, continuous  ${\cal F}_T$-measurable $f$, and  bounded,
continuous ${\cal F}_{T}^*$-measurable $g$ with support contained in
$\cup_n\Upsilon_n$.
Since $\mu(\cup_n\Upsilon_n)=1$, this implies  \eqref{;G}.
\hfill$\Box$

\section{Uniqueness of Gibbs measure}
In this section we prove the uniqueness of Gibbs measure.
We first recall Papangelou's uniqueness result in one time parameter
dimension.  This result is originally for discrete time parameter; we
modify it for continuous time case  suitable for our purpose.
In the later half of this section we will apply this to prove Theorem
\ref{thm:3}.

Let $\SS$ be a Polish space. We regard $\SS$ as the space of spins.
We set
$$\CC^-=C((-\infty,0];\SS), \quad \CC^+=C([0,\infty);\SS),
\quad \CC=\CC^-\ts\CC^+.
$$
We endow these spaces with the compact uniform topology.
These spaces are Polish spaces with this topology.

Let ${\cal I}=\{ [a,b];a  <  b \in\Z \}$.
For $I=[a,b],J=[c,d]\in\cal I$ with $I\subset J$ we set
$J\backslash I =[c,a]\cup [b,d]$.
For $\eta\in C(J\backslash I;\SS)$ and $\xi\in\CC$
we set $\eta\bullet\xi=(\xi^-,\xi^+)\in\CC$ by
\begin{align*}
&\eta\bullet \xi_t=
\begin{cases}
\xi^-_{t-(c-a)}         &\quad{ if }\quad t < c-a,\\
\eta_{t+a}      &\quad{ if }\quad c-a \le t\le 0,\\
\eta_{t+b}      &\quad{ if }\quad 0\le t\le d-b,\\
\xi^+_{t-(d-b)}         &\quad{ if }\quad d-b <  t.
\end{cases}
\end{align*}
Here and after, whenever we write $\eta\bullet \xi$,
both of $\eta_c=\xi^-_0$ and
$\eta_d=\xi^+_0$ are implicitly assumed.
For a subset $A\subset\R$ we denote by $\pi_A$ the projection
$\map{\pi_{A}}{C(\R;\SS)}{C(A;{\SS})}$ given by $\pi_{A}(X)=X|_A$.

%\marginpar{dfn:spe}
\begin{dfn}\label{dfn:spe}
Let $\CC_0 \subset \CC$ be a Borel subset.
Let $\dd$ be a metric on $\CC_0$.
We assume $(\CC_0,\dd)$ is a Polish space, and the topology induced by
$\dd$ is stronger than the relative topology as a subset of $\CC$ but the
$\sigma$-fields generated by the two topology in $\CC_0$ are the same.
Let $Q$=$\{Q_{I}(\cdot,\xi)\}_{I\in{\cal I},\xi\in\CC_0}$, where
$Q_{I}(\cdot,\xi)$ is a probability measures on $C(I;{\SS})$ for each
$\xi\in\CC_0$ and for each $A\in{\cal B}(C(I;\SS))$, $Q_{I}(A,\cdot)$ is
${\cal B}({\CC}_0)$-measurable.
The triplicate $(Q,\CC_0,\dd)$ is called a specification
if it satisfies the compatibility condition; for each $I\subset J\in\cal I$
and $A\in{\cal B}(C(I;\SS))$, $\xi\in\CC_0$
\begin{equation}
\int_{C(J\backslash I;\SS)}
Q_{I}(A,\eta\bullet\xi) Q_{J}(\pi_{J\backslash I}\in d\eta)
=Q_{J}(\pi_{I}\in A,\xi)
\label{:2}
\end{equation}
\end{dfn}
\vspace{1ex}

For $I\in\cal I$ we define $\map{\pi_{I}^*}{C(\R;\SS)}{\CC}$ by
\begin{align}&
\pi_{I}^*(X)=\begin{cases}
X_{t-b}\quad  &\text{if}\quad t\ge b \\
X_{t-a}\quad  &\text{if}\quad t\le a .
\end{cases}
\label{:1}
\end{align}
For $I\in\cal I$ we define $P_{I}^*(\cdot)=P(\pi_{I}^*\in\cdot)$.

%\marginpar{dfn:ad}
\begin{dfn}\label{dfn:ad}
We say a probability measure $P$ on $C(\R;\SS)$ is admitted by the
specification
$(Q,\CC_0,\dd)$ if
\begin{alignat}{2}
&P_{I}^*(\CC_0)=1&&\quad\text{for all }I\in\cal I,
\label{:3}\\
&P(\pi_{I}\in \cdot|\pi_{J}^*=\xi)=Q_{I,J}(\cdot,\xi)
&&\quad\text{for $P_{J}^*$-a.s. }\xi.
\label{:4}
\end{alignat}
Here we set $Q_{I,J}(\cdot,\xi)=Q_{J}(\pi_{I}\in \cdot,\xi)$ for $I\subset
J$. \end{dfn}

\vspace{1ex}
Let $\|\cdot\|_{\text{total}}$ denote the total variation of signed measure.
For $\xi=(\xi^-,\xi^+)\in\CC$ we set $\xi_0=(\xi^-_0,\xi^+_0)\in\SS^2$.
We assume $(Q,\CC_0,\dd)$ satisfies the following;

\vspace{1ex}
\noindent\thetag{U.1}
$Q$ is translation invariant;
$Q_{I}=Q_{I+a}$ for each $I\in\cal I$, $a\in\Z$.

\vspace{1ex}
\noindent\thetag{U.2}
$\lim_{n\to\infty} \|
Q_{I}(\cdot,\xi_n)-Q_{I}(\cdot,\xi)\|_{\text{total}}=0$  if
$\lim_{n\to \infty}\dd(\xi_n,\xi)=0$ and $\xi_{n,0}=\xi_0$.

\vspace{1ex}
\noindent\thetag{U.3}
For each $\xi,\eta\in\CC_0$ such that $\xi_0=\eta_0$,
$Q_{I}(\cdot,\xi)$ and
$Q_{I}(\cdot,\eta)$ are mutually absolutely continuous.

\vspace{1ex}
\noindent\thetag{U.4}
There exists a family of Borel subsets $\{M_{I}(m)\}_{I\in{\cal I},
m\in\N}$ in $C(I;{\SS})$ such that
$$M_{I}(m)\subset M_{I}(n)  \quad\text{ if $m\le n$, }$$
and satisfying the following;
for each $\e>0$,  $\dd$-compact set $\KK \subset\CC_0$, $m\in\N$, there
exists an $n\in\N$ such that
\begin{align*}&
Q_{I,J}(M_{I}(n),\eta\bullet\xi)>1-\e
%
\end{align*}
for all $I,J,K \in\cal I$ such that $I\subset J\subset K$,
$\eta\in M_{K\backslash J}(m)$ and $\xi\in \KK $. \\
Here $M_{K\backslash J}(m)=M_{[c,a]}(m)\times M_{[b,d]}(m)$ for $J=[a,b]$
and $K=[c,d]$.

\vspace{1ex}
\noindent\thetag{U.5}
For each $\e>0$, any $\dd$-compact set $\KK $, any $m,n\in\N$ there exist
$k\in\N$ such that
\begin{align}  \| Q_{I}^n(\cdot,\zeta\bullet\xi)-
Q_{I}^n(\cdot,\zeta\bullet\eta) \|_{\text{total}}<\e
\label{:7} \end{align}
for all $I\subset J\in\cal I$ with $|J\backslash I|_*\ge k$, $\zeta\in
M_{J\backslash I}(m)$ and $\xi,\eta\in \KK $ with $\xi_0=\eta_0$.
\\
Here $|J\backslash I|_*=\min\{a-c,d-b\}$ for $I=[a,b]$ and $J=[c,d]$, and
$Q_{I}^n$ is given by
\begin{align}
&Q_I^n(\cdot,\xi)=
\begin{cases}
\frac{Q_I(\cdot\cap M_I(n),\xi)}{Q_I(M_I(n),\xi)} &\quad\text{if }
Q_I(M_I(n),\xi)>0
\\
0 &\quad\text{otherwise.}
\end{cases}
\label{:8}
\end{align}
\vspace{1.5ex}

We remark, if $\xi_0\not=\eta_0$, then $Q_I(\cdot,\xi)$ and
$Q_I(\cdot,\eta)$ are always mutually singular because $Q$ is a
specification of a probability measure supported on continuous paths.

%\marginpar{dfn:tame}
\begin{dfn}\label{dfn:tame}
A probability measure $P$ on  $C(\R;\SS)$ admitted by the specification
$(Q,\CC_0,\dd)$ is said to be $\dd$-tame if for every $\e>0$ there exists a
$\dd$-compact set $\KK\subset\CC_0$ such that
\begin{align}
&P_{I}^*(\KK)\ge 1-\e \quad\text{for all }I\in {\cal I}.
\label{:9}
\end{align}
\end{dfn}

%\marginpar{rem:trans}
\begin{rem}\label{rem:trans}
Suppose that $P$ is $\Z$-translation invariant (i.e.\ $P$ is invariant
under the translation $X_t\to X_{t+z}$ for all $z\in \Z$) and that
$P_{I}^*(\CC_0)=1$ for each $I$.
Then $P$ is $\dd$-tame because $\CC_0$ is a Polish space.
\end{rem}

%\marginpar{thm:u1}
\begin{thm}\label{thm:u1}
Any specification $(Q,\CC_0,\dd)$ satisfying \thetag{U.1}--\thetag{U.5}
admits at most one tame probability measure on $(C(\R;\SS),{\cal
B}(C(\R;\SS)))$. Such a probability measure, if exists, is translation
invariant.
\end{thm}
This theorem was proved by Papangelou \cite{p1} when the time parameter is
discrete. We modify Hypotheses \thetag{H.1}--\thetag{H.5} in Papangelou to
\thetag{U.1}--\thetag{U.5} in order to fit our problem.
The proof of Theorem \ref{thm:u1} is similar to Theorem 5.1 in Papangelou.
So we omit it.

In the rest of this section we will prove Theorem \ref{thm:3} by using
Theorem \ref{thm:u1}. So we find out a suitable specification.
Let $\SS=\R$ and for $\xi=(\xi^+,\xi^-)\in\CC_0$ we set
$\dd(\xi)=\dd_0(\xi)+\dd_1(\xi)$, where
$\dd_{i}(\xi)=d_{i}(\xi^+)+d_{i}(\xi^-_{-\cdot})$ $(i=0,1)$ and
 for $\xi\in C([0,\infty);\R) $,  $d_{i}(\xi)$ $(i=0,1)$ are  defined by
\begin{align*}%\label{;14}
& d_0(\xi)=\left\{\sum_{i=0}^{\infty}  2^{-i}\min\{\sup_{i\le t<i+1}|\xi_t
|,1\}\right\} \\
& d_1(\xi)=  \left\{ \int_0^{\infty} \rho_0(t) |\xi_t|^{p_2}
dt\right\}^{1/p_2}.
                                                                                                                         \end{align*}
Here $p_2>p_0+1$.
We write the metric $\dd(\xi,\eta)=\dd(\xi-\eta)$ by the same symbol $\dd$.
Note that $\dd_0$ corresponds to the compact uniform topology; accordingly,
the topology induced by $\dd$ is stronger than the compact uniform
topology.
Let
$$\CC_0=\{\xi;\dd_1(\xi)<\infty\}.$$
For $I\in\cal I$ and $\xi\in\CC_0$ we set
\begin{align}\begin{split}
H_{I}(X,\xi)=
\int_{I}\varphi(X_t)dt & + \frac{1}{2}
\int_{I\ts I}w(t-u,X_t-X_u)dtdu \\
&+  \int_{I\ts (\R\backslash I)}w(t-u,X_t-\xi^I_u)dtdu
\end{split}
\label{:12}\end{align}
Here for $\xi=(\xi^-,\xi^+) \in\CC_0$ and $I=[a,b]\in\cal I$, we define
$\xi^I\in C((-\infty,a],[b,\infty);\R)$  by
\begin{align}
 \xi_t^I=&
\begin{cases}             \xi^-_{t-a}&\text{ for }t\le a\\
                                        \xi^+_{t-b}&\text{ for }t\ge b.
\end{cases}
\label{:13}
\end{align}
Similarly as \eqref{;11a}, for $I=[a,b]\in\cal I$ and $\xi\in\CC_0$ we
define $\Ma^{\1,\2}_{I,\xi}$ by
\begin{align}
& \Ma^{\1,\2}_{I,\xi}=(Z_{I,\xi})^{-1}
\exp[-H_I(X,\xi)] d\WI
\label{:10}
\end{align}
Here $Z_{I,\xi}$ is the normalization and $\WI$ is the Brownian bridge on
$C(I;\R)$ conditioned
$\xi_0=(\xi_0^-,\xi_0^+)$ on the boundary.
In case of free boundary condition we write $\Ma^{\1,\2}_{I,\xi,\circ}$.

Let $Q_{I}(\cdot,\xi)=\Ma^{\1,\2}_{I,\xi}$. Then it is clear that
$(Q,\CC_0,\dd)$ is a specification. We want to check $(Q,\CC_0,\dd)$
satisfies \thetag{U.1}--\thetag{U.5}. We take for $I=[a,b]\in\cal I$ and
$m\in\N$
\begin{align}   \label{:11}
&M_{I}(m)=\{X  \in  C(I;\R);\ |X_a|, |X_b|,  A(X) \le m \},
 \\ \intertext{where }
&A(X)=\sup_{0<t<b-a}\left \{
\frac{1}{t}\int_{[a,a+t]\cup[b-t,b]}   |X_t|^{p_2}   dt \right \} .
\notag
\end{align}
For $\xi\in\CC_0$ and $I\in\cal I$ we set ;
\begin{align}
\1_{I,\xi}(t,x)&=\1(x)+\int_{u\notin I}\2(t-u,x-\xi^I_u)du
\label{:17a}
\end{align}
Then $\1_{I,\xi}(t,x)$ is a time inhomogeneous free potential on $I$.
%\marginpar{lem:u3}
\begin{thm}\label{lem:u3}
Suppose \thetag{A.1}, \thetag{W.1}, \thetag{W.3} and \eqref{;w6}.
Suppose that $v$ is convex. Assume  for each $I=[a,b]$ and $\xi\in\CC_0$
\begin{align}
& \text{right- and left-dominators $\3^{\pm}$ of $\1_{I,\xi}(t,x)$ }
\label{;w5}     \\
&
\text{symmetric around $\pm m$}    \text{with $\{\3^{\pm}\}''\ge 2\hat{a}>0$ }
\notag  \\
&
\text{such that $m$ and $\hat{a}$ depends only on $\dd_1(\xi)$ and $|b-a|$
exist.}
\notag
\end{align}
Let $(Q,\CC_0,\dd)$ and $M_{I}(m)$ be as above.
Then \thetag{U.1}--\thetag{U.5} are satisfied. In particular, the
associated $\dd$-tame Gibbs measure is unique.
\end{thm}

\begin{pf}
\thetag{U.1} and \thetag{U.3} are clear.
We next prove \thetag{U.2}: Since $\xi_{n,0}=\xi_0$ for all $n$, we have
$\Wa_{I,\xi_n}=\Wa_{I,\xi}$ by definition.
So for \thetag{U.2} it is enough to prove
\begin{align}
&\limi{n}\int_{C(I;\R)}
|e^{-H_I(X,\xi)}-e^{-H_I(X,\xi_n)}|d\Wa_{I,\xi}=0
\label{:16}
\end{align}
By $\limi{n}\dd(\xi_n,\xi)=0$, \thetag{A.2} and $p_2>p_0+1$             we have
$$\text{$\limi{n}H_I(X,\xi_n)=H_I(X,\xi)$ for each $X$. }$$
By assumption $H_I(X,\xi_n)$ and $H_I(X,\xi)$ are bounded from below. So
applying Lebesgue convergence theorem we have \eqref{:16}.

We next prove \thetag{U.4}. Let $\KK$ be a $\dd$-compact set
and $C_{5.1}=\max \\ \{\dd_1(\xi);
\xi\in \KK\}$. Then $C_{5.1}<\infty$.

For $\xi\in \KK $ and $\eta\in M_{K\backslash J}(m)$ we regard
$\1_{J,\eta\bullet\xi}(t,x)$ as the time inhomogeneous free potential on
$J$ and consider the associated Gibbs measure
$$
\Ma^{\eta\bullet\xi}_{J}:=\Ma^{\1_{J,\eta\bullet\xi},\2}_{J,(\eta\bullet\xi)
^J,\circ}
$$
with interaction potential $\2$ on $C(J;\R)$.
Let $E^{\eta\bullet\xi}_{J}$ denote the expectation with respect to
$\Ma^{\eta\bullet\xi}_{J}$.

By Remark \ref{rem:31} and \eqref{;w5} there exists a constant $C_{5.2}$,
depending only on $C_{5.1}$ and $m$, such that
\begin{align}
&E^{\eta\bullet\xi}_{J}[e^{|X_t|}]\le C_{5.2}
\qquad\text{ for all }t\in J.
\label{:20} \end{align}
By \eqref{:20} for each $\e$ there exists an $n_{1}$ such that \begin{align}
&\mu^{\eta\bullet\xi}_{J}
(       \{X ;   \pi_{J\backslash I}(X)  \in
                \{\zeta ;               \dd_1(\zeta\bullet\eta\bullet\xi) \le n_{1} \}  ) \ge 1-\e/4.
\label{:21}
\end{align}
For $\xi\in \KK $, $\eta\in M_{K\backslash J}(m)$ and
$\zeta\in\{\zeta; \dd_1(\zeta\bullet\eta\bullet\xi) \le n_{1} \}$
we consider the time inhomogeneous free potential
$\1_{I,\zeta\bullet\eta\bullet\xi}(t,x)$ on $I$.
Then by \eqref{;w5} we see there exist right-dominators $\3$ depending only
on $n_{1}$ and $|b-a|$.
$(\xi,\eta,\zeta)$.
Let $X^+=X_t^+=\max\{0,X_t\}$ and set
$$F_I(X)=\max\{ X_a^+, X_b^+,  A(X^+)  \}.  $$
Then $F$ is increasing in the sense of Lemma \ref{lem:2.1}.
So by Lemma \ref{lem:2.1} we have
\begin{align}
&E^{\eta\bullet\xi}_{J}[F_I(X)]\le
E^{\3,\2}_{J,\eta\bullet\xi^{J},\circ}[F_I(X)].
\label{:22}
\end{align}
By \eqref{:21}, \eqref{:22} and the Chebyshev's inequality, there exists an
$n_{2}$ such that
$$Q_{I,J}(M^+(n_2),\eta\bullet\xi)>1-\e/2$$
Here $M^{+}(n)=\{F_I(X)\le n\}$.
By using left-dominator we see there exists an $n_{3}$ such that
$Q_{I,J}(M^-(n_3),\eta\bullet\xi)>1-\e/2$.
Here $M^{-}(n)=\{F_I(-X)\le n\}$.
Since $M_I(n)=M^+(n)\cap M^-(n)$, we conclude (U.4).

We finally prove \thetag{U.5}.
For this it is enough to show
\begin{align}
&\limi{k} \sup _{\II_k }
 \{
\int_{M_I(n)}   |
        e^{-H_I(X,\zeta\bullet\xi)}-
        e^{-H_I(X,\zeta\bullet\eta)}|
        d\Wa_{I,\xi}    \}=0.
                                             \label{:18}\end{align}
Here
$\II_k=\{(\xi,\eta,\zeta,I,J); \xi,\eta\in \KK \text{ with } \xi_0=\eta_0,
\zeta\in M_{J\backslash I}(m), \ I,J\in {\cal I}\  \text{ such that }
 |J\backslash I|_*\ge k \}$.
Since $H_{I}(X,\zeta\bullet\xi)$ is bounded from below,
\eqref{:18} is reduced to
\begin{align}\begin{split}
\limi{k} \sup _{ X\in M_{I}(n),   \II_k }
|H_{I}&(X,\zeta\bullet\xi)- H_{I}(X,\zeta\bullet\eta)|  =0.
\end{split}
\label{:14}
\end{align}
Here we set $H_{I}(X,\zeta\bullet\xi)- H_{I}(X,\zeta\bullet\eta)=0$ in case
of $H_{I}(X,\zeta\bullet\xi)= H_{I}(X,\zeta\bullet\eta)=\infty$.
Note that $H_{I}(X,\zeta\bullet\xi) =\infty \text{ (or, }
H_{I}(X,\zeta\bullet\eta)) =\infty$) if and only if $\int_{t\in
I}\1(X_t)dt=\infty$; so $H_{I}(X,\zeta\bullet\xi)=\infty$ if and only if
$H_{I}(X,\zeta\bullet\eta)=\infty$.
We see
\begin{align*}
&        \sup _{ X\in M_{I}(n),   \II_k }
         |H_{I}(X,\zeta\bullet\xi)-H_{I}(X,\zeta\bullet\eta)|
\\
& =\sup _{ X\in M_{I}(n),   \II_k }
 \left|\int_{t\in I, u\notin J}   \2(t-u,X_t-{\xi}^J_u) -
\2(t-u,X_t-{\eta}^J_u)dtdu\right|
 \\
& \le 2  \sup_{X\in M_I(n),             \II'_k }
        \left\{\int_{t\in I, u\notin J}  \left|
        \2(t-u,X_t-{\xi}^J_u)   \right |   dtdu   \right\},
\end{align*}
where $\II'_k=\{(\xi,I,J);\ \xi\in \KK,\ I,J\in{\cal I}
\text{ such that }|J\backslash I|_*\ge k\}$.
Here for the third line we used $(\zeta\bullet\xi)^I_u={\xi}^J_u$ for
$u\notin J$ and $(\zeta\bullet\xi)^I_u=\zeta_u$ for $u\in I$.
By \eqref{;w6} we have
$$
\limi{k} \sup _{ \ X\in M_I(n), \II'_k }
         \left\{        \int_{t\in I, u\notin J}        \left|
\2(t-u,X_t-{\xi}^J_u) \right | dtdu     \right\} =0.
$$
This implies \eqref{:14}, which completes the proof of \thetag{U.5}.

The second statement is clear from Theorem \ref{thm:u1}.
\end{pf}

%\marginpar{l:ua}
\begin{lem}\label{l:ua}
\thetag{A.1} and \thetag{A.2} imply \eqref{;w5}.
\end{lem}

\begin{pf}
Let $\2_{I,\xi}(t,x)=\int_{u\notin I}\2(t-u,x-\xi^I_u)du$.
For \eqref{;w5} it is enough to show $\2_{I,\xi}(t,x)$ satisfies \eqref{:a1};
for each $I=[a,b]$ ;
\begin{align}\label{:a1}
& \text{right- and left-dominators of $\2_{I,\xi}(t,x)$ symmetric around
$\pm m_1$}
\\
& \text{such that $m_1$ depends only on $b-a$ and $\dd_1(\xi^I)$ exist.}
 \notag
\end{align}
Indeed, if $d_1$ and $d_2$ are right-dominators of $\1$ and
$\2_{I,\xi}$
symmetric around $m$ and $m_1$, respectively,
then $R^+_{m_2}(d_1+d_2)$, where $m_2=\max \\ \{|m|,
|m_1|\}$,
is such a dominator.
Indeed, $R^+_{m_2}(d_1+d_2)$ is symmetric around $m_2$, and
$$R^+_{m_2}(d_1+d_2)'' \ge R^+_{m_2}(d_1)'' \ge d_1''.$$
Accordingly, $\hat{a}$ is taken to be that of $d_1$.
We can construct left-dominator similarly.

Let $\bar{\4}_t=\int_{u\notin I}\4(t-u)du$.
Let $\lambda_t$ be the probability measure on $\R$ given by
$$\lambda_t(A)=\bar{\4}_t^{-1} \int_{u\notin I}\4(t-u) 1_{A}(\xi_u)du.$$
Let
$ \bar{v}(t,x)=\int_{\R} (v(x-y)-v(x))\lambda_t(dy)$. Then
$\2_{I,\xi}(t,x)= \bar{\4}_t\cdot  v(x) +\bar{\4}_t \cdot \bar{v}(t,x)$.
We see
\begin{align}\label{:a5}
\2'_{I,\xi}(t,x)&=\bar{\4}_t \cdot (v'(x) + \bar{v}'(t,x))\\
&\ge     \bar{\4}_t \cdot (v'(x)  -  u(x) \{\sup_{t\in I}\int_{\R}\{1+ |y|
^{p_0}\}
\lambda_t(dy)\} ) ,\notag       \\
\intertext{by using \eqref{:2a} with $\e=1/\{\sup_{t\in I}\int_{\R}\{1+ |y|
^{p_0}\}    \lambda_t(dy)\}$, we have }
&\ge     \bar{\4}_t \cdot (v'(x)  - (v'(x)-v'(x-b))) \notag     \\
&=      \bar{\4}_t  \cdot v'(x-b)   .\notag
                                                        \end{align}
Then $\bar{\4}_t  \cdot v(x-b)$ is the right-dominator satisfying \eqref{;w5}.
We can construct the left dominator similarly.
\end{pf}

%\marginpar{thm:u4}
\begin{thm}\label{thm:u4}
Suppose \thetag{A.1} and \thetag{A.2} and \eqref{;w6}.
Then $\dd$-tame Gibbs measure is unique.
\end{thm}

\begin{pf}
The statement follows from Theorem \ref{lem:u3} and Lemma \ref{l:ua}.
\end{pf}

\noindent{\bf Proof of Theorem \ref{thm:3}.}
Since any translation invariant measures satisfying \eqref{:w7} are
$\dd$-tame, we have the first statement from Theorem~\ref{thm:u4}.
By Theorem~\ref{thm:1} any limit points $\mu$ of $\{\Nf\}_{T}$ or $\{\Mf\}$
for $\xi$ with $\|\xi\|_{\infty}<\infty$ are $\dd$-tame.
Hence by Theorem~\ref{thm:u4} again we obtain the second claim.
\qed

\section{An Example of Nonuniqueness}
In this section we establish by example that in our general set-up
there can be several Gibbs measures for a given potential. For the
sake of concreteness we choose
\begin{align}\label{:61}
&\1(x)=\beta(x^4-x^2), \quad \beta>0,\\
\intertext{and}
&\2(t,x)=\a \4(t)\frac{1}{2} x^2, \quad \a>0,
\quad \rho(t) = (1 +|t|)^{- \gamma}.
\label{:62}
\end{align}
By \eqref{;w6} we must require $\gamma>1$. Furthermore we need
\begin{align}
&1 < \gamma \le 2.
\label{:63}
\end{align}
Let $\za{\, \cdot\, }_b(T)$ be the expectation of the finite volume Gibbs
measure with outside path $\xi_t=b>0$ for $|t|\ge T$ and let $\za{\,
\cdot\, }_b$ be the limit measure.
%\marginpar{thm:61}
\begin{thm}\label{thm:61}
Let $\phi$ and $w$ be given by \eqref{:61}, \eqref{:62}.
Then there exist $\a, \beta, b>0$ such that
\begin{align}
&\za{X_0}_b \ge m^* >0
\label{:64}
\end{align}
\end{thm}
%\marginpar{rem:61}
\begin{rem}\label{rem:61}
Clearly $\za{X_0}_{-b}=- \za{X_0}_{b}$.
Thus \eqref{:64} establishes that there are at least two Gibbs
measures for the potential \eqref{:61}, \eqref{:62}.
\end{rem}
\begin{pf}
Working out the square
\begin{align}
&\frac{1}{2} \int_{|t|, |u| \le T} \4(t-u) (X_t-X_u)^2 dtdu=
\label{:65}\\
&\qquad
- \int_{|t|, |u| \le T} \4(t-u) X_t X_udtdu
+\int_{|t|\le T} X_t^2 (\int_{|u|\le T} \4(t-u) du) dt,
\notag \\
&\frac{1}{2} \int_{|t| \ge T} (\int_{|u|\le T} \4(t-u) (b -X_u)^2 du) dt=
\notag \\
& \qquad
 \int_{|u| \le T} \bar{\4}(u)   (  \frac{1}{2}  X_u^2 - b X_u + \frac{1}{2}
b^2 )  du,
\notag
\end{align}
where $\bar{\4}(u)= \int_{|t+u|\ge T} \4(t) dt $.
By Griffiths II \cite{simon} $\za{X_t}_{b}(T)$ is decreasing in
the strength of the quadratic part of the potential. Therefore
\begin{align}\label{:66}
 \za{X_{\tau}}_{b}(T)  &\ge
Z^{-1} \int\WT (dX) X_{\tau} \exp[-\intT(\1(X_t)+\a' X_t^2)dt \\
&\quad +\frac{\a}{2}\intT\intT \4(t-u) X_t X_u dtdu
   +b\a \intT \bar{\4} (u)X_u du  ]\notag \\
& =: \za{X_{\tau}}_b'(T)  \notag
\end{align}
with $\a'=4\a \int_0^{\infty}\4(t)dt$.
Here $\WT$ is the Brownian bridge on $\CT$ conditioned to $X_{\pm T}=b$.

Let $2T= (2N+1)\delta$ with integer $N$.
We introduce the block variables
\begin{align}
&M_n=\frac{1}{\delta} \int_{(n-\frac{1}{2})\delta}^{(n+\frac{1}{2})\delta}
dt X_t, \quad \text{ with }n=-N,\ldots,N.
\label{:67}
\end{align}
and use again Griffiths II to decrease the expectation
$\za{X_{\tau}}_{b}'(T)$.
Firstly, we replace $\WT$ by
$\prod_{n=-N}^{N}\Wa_{\delta[n-\frac{1}{2},n+\frac{1}{2}],0}$,
where $\Wa_{[\cdot,\cdot],0}$ is the Brownian bridge on the interval
$[\cdot,\cdot]$ with $0$ boundary condition.
Secondly we replace $\4(t-u)$ by $\4_{\delta}(t,u)$,
where $\4_{\delta}$ is the largest function on $[-T,T]^2$ such that
 $\4_{\delta}\le \4$, $\4_{\delta}$ is constant on each square
 $\delta([n-\frac{1}{2},n+\frac{1}{2})\times
             [m-\frac{1}{2},m+\frac{1}{2})) $,     $(m\not= n)$, and
$\4_{\delta}=0$ on
$\delta([m-\frac{1}{2},m+\frac{1}{2})\times
             [m-\frac{1}{2},m+\frac{1}{2})) $
$m=-N,\ldots,N$.
Thirdly, we replace $\bar{\4}$ by $\bar{\4}_{\delta}$,
where $\bar{\4}_{\delta}$ is the largest function such that
$\bar{\4}_{\delta}\le \bar{\4}$, and constant on each interval
$ [m-\frac{1}{2},m+\frac{1}{2})$.
We set $\4_{\delta}(n-m)=\4_{\delta}(t,u)$ for
$t,u\in  \delta([n-\frac{1}{2},n+\frac{1}{2})\times
                       [m-\frac{1}{2},m+\frac{1}{2}))    $
and  $\bar{\4}_{\delta}(m)=\bar{\4}_{\delta}(t)$ for
$t \in  \delta   [m-\frac{1}{2},m+\frac{1}{2})    $. Then
\begin{align}
&\za{X_0}_b'(T) \ge \za{M_0}_b'(T)  \ge  \za{M_0}_{b,N}.
\label{:68}
\end{align}
Here $\za{\, \cdot \,}_{b,N}$ is the expectation with respect to the
measure on $\R^{2N+1}$ defined by
\begin{align}\label{:69}
&Z^{-1} \prod_{n=-N}^N \mu(dM_n)\\
&\quad \quad
\exp\left[\frac{\a \delta^2}{2} \sum_{m,n=-N,\, m\not= n}^N
\4_{\delta}(n-m) M_nM_m +
\a b \delta \sum_{n=-N}^N  \bar{\4}_{\delta}(n) M_n  \right]
\notag .
\end{align}
We note that by construction $\4_{\delta}(n)>0$ and
$\4_{\delta}(n)\ZZ n^{-\gamma}$ for large $n$. Also
$\bar{\4}_{\delta}(n)=\sum_{m=N-n}^{\infty}\4_{\delta}(m)+\sum_{m=N+n}^{\inf
ty}\4_{\delta}(m)$.
The single site measure $\mu$ is the distribution of $M_0$ under the measure
\begin{align}
&Z^{-1} \Wa_{\delta[-\frac{1}{2},\frac{1}{2}],0}
\exp\left[-\int_{-\delta/2}^{\delta/2}
\left( \1(X_t) + \a'X_t^2\right)  dt \right]
\label{:610}
\end{align}

Let us choose $a>0$ such that
\begin{align}
&\mu([0,a])=2\mu([\sqrt{2}a,\infty)).
\label{:611}
\end{align}
Since $\mu$ is even and $\mu$([a,$\infty$)) strictly decreasing,
there is a unique $a = a(\alpha,\beta)$ satisfying \eqref{:611}.
We set $b=\delta a$.
By the inequality of Wells \cite{bricmont} we have
\begin{align}
&\za{M_0}_{b,N} \ge \za{\sigma_0}_{+,N},
\label{:612}
\end{align}
where $\za{\, \cdot \, }_{+,N}$ is the Ising measure with Hamiltonian
\begin{align}
& H_I =  -\frac{\a \delta^2 a^2}{2} \sum_{m\not =n}
\4_{\delta}(m-n)  \sigma_m\sigma_n
\label{:613}
\end{align}
on the interval $[-N,\ldots,N]$ with $+$ boundary conditions on
$\Z\backslash [-N, \\ \ldots,  N]$, $\sigma_m=\pm 1, m=-N,\ldots,N$.

Thus altogether we have shown, for $b=\delta a$,
\begin{align}
&\za{X_0}_b(T) \ge \za{\sigma_0}_{+,N}.
\label{:614}
\end{align}
For the Ising model \eqref{:613} it is proved that
$\za{\sigma_0}_{+,N} \ge m^* >0$
uniformly in $N$ provided the coupling is sufficiently large
\cite{dyson,frohlich}.
To check that all conditions can be satified simultaneously,
we set $\delta=1$ and let $\a_c$ be the corresponding critical
coupling of the Ising model. We note that $a(\alpha,\beta)
\rightarrow 1$ as $\beta \rightarrow \infty$ for fixed $\alpha$.
Thus we choose $\alpha = 4 \alpha_c$ and $\beta$ so large that
$\alpha a(\alpha,\beta) \geq 2 \alpha_c$. Then
$\za{\sigma_0}_{+,N} \ge m^* >0$
uniformly in $N$. \end{pf}

With some extra effort one could discuss the phase diagram in the
$(\a,\beta)$ plane.
Since we just wanted to provide an example we do not enter into details.
We should mention however that one feature turns out to be qualitatively wrong,
which teaches us that we have not yet found the \lq\lq right'' proof.
Let us fix $\beta$ sufficiently large and the blocking interval $\delta$.
($\delta$ could be optimized too without changing the conclusions).
For small $\a$ we have $\za{X_0}_b=0$ for any choice of $b$.
According to Theorem 1 the bound
$$
\za{X_0}_b \ge m^* >0 \quad\text{ with } b=a\delta
$$
holds for sufficiently larger $\a$.
As we further increase $\a$, $\1(x)+\a'x^2$ obtains a single minimum.
Then $a$ will decrease and we can keep the effective coupling,
$\a \delta^2 / a^2$, no longer above criticality.
Thus our argument fails for sufficiently large coupling $\a$.
The reason is that the ferromagnetic inequalities refer only to the
off-diagonal
piece in \eqref{:65}.
No monotonicity seems to be available in $\4(t-u)$ for the full action
$\iint \4(t-u) (X_t-X_u)^2 dtdu$.

\section{Appendix}

%\marginpar{lem:A.1}
\begin{lem}\label{lem:A.1}
Let $\map{a}{\R}{\R\cup\{\infty\}}$ be a convex function such that
$a(t)=a(-t)$. \\
\thetag{1}      Then $\text{for all }s,t,u,v\ge 0  \text{ such that }s+t\le
u+v,\text{ and } s,t\le u$
\begin{align} &a(s)+a(t)\le a(u)+a(v) \quad
\label{;A.2}  \end{align}
\thetag{2} Let $x\wedge y=\min\{x,y\}$ and $x\vee y=\max\{x,y\}$.
Then for $x_1,x_2,y_1,y_2\in\R$
$$
a(x_1-x_2)+a(y_1-y_2)-a(x_1\wedge y_1-x_2\wedge y_2)-
a(x_1\vee y_1-x_2\vee y_2)\ge 0.
$$
\end{lem}
\begin{pf}
If $v \ge \max\{s,t \}$, then the claim is clear.
Hence without loss of generality, we can assume $v\le t$. Then
\begin{eqnarray}
a(u)-a(t) & =   \int_{0}^{u-t} a'(t+\theta)d\theta              \ge
                \int_{0}^{u-t} a'(t+\theta-(t-v))d\theta \notag\\
& = a(u-t+v)- a(v). \notag
\end{eqnarray}
Since $u-t+v\ge s$, we complete the proof of (1).

Let $I$ denote the left hand side of the inequality in \thetag{2}.
If $x_1\le y_1, x_2\le y_2$, or $x_1>y_1, x_2>y_2$, then $I=0$.
If $x_1\le y_1, y_2\le x_2$, then we divide the case into six cases;

(1) $y_1\le y_2$, \quad

(2) $x_1   \le   y_2  \le y_1  \le x_2$,\quad

(3) $x_1   \le   y_2  \le x_2  \le y_1  $,\quad

(4) $y_2   \le   x_1  \le y_1  \le x_2$,\quad

(5) $y_2   \le   x_1  \le x_2  \le y_1$,\quad

(6) $y_2   \le   x_2  \le x_1  \le y_1$,\quad  \newline
In Case (1), $I\ge 0$ follows from \eqref{;A.2}.
Indeed, we can take $s=y_2-x_1$, $t=x_2-y_1$, $u=x_2-x_1$ and $v=y_2-y_1$.
Case (2) also follows from \eqref{;A.2} by taking
$s=y_2-x_1$, $t=x_2-y_1$, $u=x_2-x_1$ and $v=y_1-y_2$.
The proof of (3)--(6) cases is similar, so we omit it.
The case of $x_2\le y_2, y_1\le x_1$ is reduced to the case of $x_1\le y_1,
y_2\le x_2$ by interchanging $x_1$ and $y_1$ by $x_2$ and $y_2$,
respectively.
\end{pf}

%\marginpar{lem:A.2}
\begin{lem}\label{lem:A.2}
Let $a_i,b_i,c_{ij}$ $(1\le i,j)$ be measurable functions on $\R$
such that $a_i-b_i$ are non decreasing for all $i$.
Assume $c_{ij}$ are convex and $c_{ij}(t)=c_{ji}(t)=c_{ij}(-t)$ for all
$i,j$, and $c_{ii}=0$. Let $\mu_i$ $(i=1,\ldots,n)$ be Radon measures on
$\R$. Set
$$
f (x)=\sum_{i=1}^na_i(x_i)+\sum_{i,j=1}^n c_{ij}(x_i-x_j),
\quad
g (x)=\sum_{i=1}^nb_i(x_i)+\sum_{i,j=1}^n c_{ij}(x_i-x_j),
$$
and
$$      A=\int_{\R^n}e^{-f }\prod_{i=1}^n\mu_i(dx_i),\quad
        B=\int_{\R^n}e^{-g }\prod_{i=1}^n\mu_i(dx_i).           $$
Assume $0<A,B<\infty$. Set
$F(x)=A^{-1}e^{-f (x)}$ and $ G(x)=B^{-1}e^{-g (x)}$.
Then
$$
F(x)\cdot G(y)\le F(x\wedge y)\cdot G(x\vee y).
$$
Here $x\wedge y=(\min\{x_i,y_i\})_{i=1,\ldots,n}$ and $x\vee
y=(\max\{x_i,y_i\})_{i=1,\ldots,n}$.
\end{lem}
\begin{pf}
Let $\Phi =f (x)+g (y)-f (x\wedge y)-g (x\vee y)$.
Then the claim is equivalent to $\Phi \ge 0$.
We write $\Phi =\sum_{i=1}^n I_i+\sum_{i,j=1}^{n}I_{ij}$, where
\begin{align*}
&I_i=a_i(x_i)+b_i(y_i)-a_i(x_i\wedge y_i)-b_i(x_i\vee y_i),\\
&I_{ij}=c_{ij}(x_i-x_j)+c_{ij}(y_i-y_j)\\
&\qquad\quad
-c_{ij}(x_i\wedge y_i-x_j\wedge y_j)-c_{ij}(x_i\vee y_i-x_j\vee y_j).
                                                        \end{align*}
Since $a_i-b_i$ is non decreasing, we have $I_i\ge 0$.
$I_{ij}\ge 0$ follows from Lemma \ref{lem:A.1} (2).
Collecting these completes the proof.
\end{pf}

%%
We next give a sufficient condition of the existence of symmetric dominators:
%\marginpar{l:d1}
\begin{lem}\label{l:d1}
Suppose $\1(x)<\infty$ for all $x\in\R$.
Suppose there exists a convex function $\Pb$ such that $\Pb''(x)\ge 2a$
a.e.\ $x$ for some $a>0$, symmetric around $m\in\R$ and $\1-\1_c$ is
absolutely continuous and there exist $n, b$ satisfying $0\le b \le n-|m|$
and
\begin{alignat}{2}
&  \1'(x)  \le \Pb'(x+b)       && \text{ for a.e.\  }x\le -n,  \label{:d1}\\
&  \1'(x) \le \Pb'(x+2n-b)\quad   &&\text{ for a.e.\  }x\ge -n, \label{:a2}\\
&  \1'(x) \ge \Pb'(x-b)         &&\text{ for a.e.\  }x\ge n,  \label{:a3}\\
&  \1'(x) \ge \Pb'(x-2n+b)  &&\text{ for a.e.\  }x\le n.   \label{:a4}
\end{alignat}
Then there exist right- and left-dominators symmetric around $\pm n$.
\end{lem}
\begin{pf}
For $0\le b\le n- |m| $ define
$$ R_{n,b}^{\pm}(\Pb)=R_n^{\pm}(\Pb(\cdot\mp b)) .$$
Then $R_{n,b}^{\pm}(\Pb)$ are convex and symmetric around $\pm n$.
By $\Pb''\ge 2a$  $\text{ a.e.\ } x$ we see
$$R_{n,b}^{\pm}(\Pb)''\ge 2a            \quad\text{ a.e.\ } x. $$
By \eqref{:d1}--\eqref{:a4} we see
\begin{align}
&R_{n,b}^{+}(\Pb)'(x) \le \1'(x) \le  R_{n,b}^{-}(\Pb)'(x) \quad\text{
a.e.\ } x.
\label{;p4} \end{align}
Hence $R_{n,b}^{\pm}(\Pb)$ are right- and left-dominators of $\1$ symmetric
around $\pm n$.
\end{pf}
%\marginpar{rem:d1}
\begin{rem}\label{rem:d1}
We can apply Lemma \ref{l:d1} to examples in Example \ref{ex:2} (1) by taking
$\Pb=|x|^q+|x|^2$ for the first one and $\Pb=e^{|x|}$ for the second.
\end{rem}

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