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Ising Model, Hierarichical Model, Triviality, Renormalization Group, Gaussian fixed point, Critical trajectory, Newman's inequality
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\title{
Triviality of hierarchical Ising model in four dimensions
}
\author{
Takashi Hara
\\ {\small Graduate School of Mathematics, Nagoya University,
} \\ {\small Chikusa-ku, Nagoya 464-8602, Japan
} \\ {\small e-mail address: {\tt hara@math.nagoya-u.ac.jp}
} \\ \and
Tetsuya Hattori
\\ {\small Graduate School of Mathematics, Nagoya University,
} \\ {\small Chikusa-ku, Nagoya 464-8602, Japan
} \\ {\small e-mail address: {\tt hattori@math.nagoya-u.ac.jp}
} \\ {\small URL: {\tt http://www.math.nagoya-u.ac.jp/\~{ }hattori/}
} \\ \and
Hiroshi Watanabe
\\ {\small Department of Mathematics, Nippon Medical School,
} \\ {\small 2--297--2, Kosugi, Nakahara, Kawasaki 211--0063, Japan
} \\ {\small e-mail address: {\tt watmath@nms.ac.jp}
}
}
\date{}
\begin{document}
\maketitle
\begin{center}{\bf Abstract} \end{center}
\small
Existence of critical renormalization group trajectory
for a hierarchical Ising model in 4 dimensions is shown.
After 70 iterations of renormalization group transformations,
the critical Ising model is mapped into
a vicinity of the Gaussian fixed point.
Convergence of the subsequent trajectory
to the Gaussian fixed point
is shown by power decay of the effective coupling constant.
The analysis in the strong coupling regime
is computer-aided
and Newman's inequalities on truncated correlations
are used to give mathematical rigor to the numerical bounds.
In order to obtain a criterion for convergence
to the Gaussian fixed point,
characteristic functions and Newman's inequalities are
systematically used.
\normalsize
%\tableofcontents
\section{Introduction and main result.}
\secl{hier}
% NOTATION (HIERARCHICAL MODEL)
Dyson's Hierarchical spin system
is an equilibrium statistical mechanical system defined as follows
\cite{dyson,sinai,ce,gk1,kw4}.
Let $\Lambda$ be a positive integer,
and denote the $2^\Lambda$ variables (spin variables) $\phi_\theta$,
Hamiltonian $H_\Lambda$, and
the expectation values $\langle\cdot\rangle$, respectively, by
\eqsb
\phi_\theta&=& \phi_{\theta_\Lambda,...,\theta_1}\,,
\ \ \theta=(\theta_\Lambda,...,\theta_1)\in\{0,1\}^\Lambda,
\\
H_\Lambda(\phi)&=&-\dfrac12\sum_{n=1}^\Lambda\left(\dfrac c4\right)^n
\sum_{\theta_\Lambda,...,\theta_{n+1}}
\left(\sum_{\theta_n,...,\theta_1}
\phi_{\theta_\Lambda,...,\theta_1}\right)^2,
\\
\langle F\rangle_{\Lambda,\den{}} &=&
\dfrac1{Z_{\Lambda,\den{}}}\int d\phi F(\phi)\exp(-\beta H_\Lambda(\phi))
\prod_{\theta}\den{}(\phi_\theta),
\\
Z_{\Lambda,\den{}} &=& \int d\phi \exp(-\beta H_\Lambda(\phi))
\prod_{\theta}\den{}(\phi_\theta),
\eqse
where $\den{}$ is a single spin measure density normalized as
\eqsb
\int_\reals \den{}(x)dx=1.
\eqse
In the following,
we shall fix the so far arbitrary normalization of the spin variables by
\eqnb\eqna{beta} \beta=\dfrac1c-\dfrac12\,. \eqne
Hierarchical models are so designed that
the block-spin renormalization group transformation ${\cal R}$
has a simple form. In fact,
${\cal R}$ is a non-linear transformation of
functions on $\reals$, defined as follows.
Define the block spins $\phi'$ by
\[
\phi'_\tau =
\dfrac{\sqrt c}{2}\sum_{\theta_1=0,1}\phi_{\tau\theta_1}
\,,\ \tau=(\tau_{\Lambda-1},...,\tau_1).
\]
If a function $F(\phi)$ depends on $\phi$ through $\phi'$ only,
namely, if there is a function $F'(\phi')$ on the block spins such that
\[
F(\phi)=F'(\phi'),
\]
then it holds that
\[
\langle F\rangle_{\Lambda,\den{}}
= \langle F'\rangle_{\Lambda-1,{\cal R}\den{}}\,,
\]
where
\eqnb\eqna{rg0}
{\cal R}\den{}(x) =\const\exp(\dfrac\beta2 x^2)\int_\reals
\den{}(\dfrac{x}{\sqrt c}+y)\den{}(\dfrac{x}{\sqrt c}-y)\,dy,\
x\in\reals.
\eqne
%MOTIVATION
Note that
\eqnb
\eqna{gaussfp}
\den{G}(x) = \const \exp(-\dfrac{1}{4}x^2)
\eqne
is a fixed point of ${\cal R}$, which we shall refer to as
the density function of the massless Gaussian measure.
By looking into the asymptotics of e.g., susceptibility for the
hierarchical massless Gaussian model defined by \eqnu{gaussfp},
and comparing it with that of standard nearest neighbor
massless Gaussian models on $d$-dimensional regular lattice,
we see that the dimensionality $d$ of the system may be identified
(at least for the Gaussian fixed point) as
\eqnb
\eqna{hierdim}
c = 2^{1-2/d}
\ \ \ \ (\beta=\dfrac12(2^{2/d}-1)).
\eqne
We shall extend the correspondences to hierarchical models with
arbitrary measures, and use the terminology $d$-{\it dimensional
hierarchical models} whenever \eqnu{hierdim} holds.
Asymptotic properties of the renormalization group trajectories
\eqnb\eqna{RG}
\den{N} = {\cal R}^N\den{0}\,,\ N=0,1,2,\cdots,
\eqne
are extensively investigated in a `weak coupling regime' i.e.,
in a ^^ neighborhood' of $\den{G}$ \cite{sinai,ce,gk1,gk2,gk3}.
In particular,
it is known
that, if $d\ge4$, then there are no non-Gaussian fixed points
in a ^^ neighborhood' of $\den{G}$\,, and that
a `continuum limit' constructed from a critical trajectory
with an initial function in a `neighborhood' of $\den{G}$
is trivial (Gaussian).
However, in order to study asymptotic properties
of strongly coupled models,
we have to analyze trajectories \eqnu{RG}
with initial functions in a `strong coupling regime'
far away from the Gaussian fixed point.
As a typical example, we consider in this paper
the hierarchical Ising model,
which is defined by the Ising spin measure density
parameterized by $s\ge 0$:
\eqnb
\eqna{Ising}
% \eqna{isinginitialdist}
\den{{\rm I},s}(x)=\dfrac12(\delta(x-s)+\delta(x+s)),
\eqne
which may be regarded as a strong coupling limit of
the $\phi^4$ measures:
\[
\den{\mu,\lambda}(x)=\const \exp(-\mu x^2 -\lambda x^4),
\ \ \ \mu=-2\lambda s^2, \ \lambda\to\infty.
\]
Here and in the following, we
use the standard notation $\delta(x-s)\,dx$ denoting
a probability measure with unit mass on a single point $x=s$.
Hierarchical Ising model has an infinite volume limit $\Lambda\to\infty$,
if $00$), and has a phase transition, if $12$)
\cite{dyson}.
% RESULT
It has been widely believed without proof that
the hierarchical Ising model in $d\ge4$ dimensions has
a critical trajectory converging to the Gaussian fixed point
and that
the `continuum limit' of the hierarchical Ising model
in $d\ge4$ dimensions will be trivial.
In this paper, we prove this fact.
In the present analysis, it is crucial that
the critical Ising model is mapped into a weak coupling regime
after a {\it small} number of renormalization group transformations
(in fact, 70 iterations for $d=4$).
Moreover,
using a framework essentially different from that of \cite{sinai,gk2},
we see in the weak coupling regime that
the `effective coupling constant' of a critical model decays
as $c_1/(N+c_2)$ after $N$ iterations in $d=4$ dimensions
(exponentially for $d>4$).
Our framework in the weak coupling regime is designed especially
for a critical trajectory starting at the strong coupling regime
so that the criterion of convergence to the Gaussian fixed point
can be checked numerically with mathematical rigor.
Corresponding results, triviality of $\phi^4_4$ spin model on
regular lattice (`full model'),
are much far harder, and a proof of triviality of
Ising model on $4$ dimensional regular lattice is, though widly believed,
still open.
We should here note the excellent and hard works of \cite{gkfullLH,gkfullCMP}
where the existence of critical trajectory in the weak coupling regime
(near Gaussian fixed point; `weak triviality') is solved
by rigorous block spin renormalization group transformation.
Our main theorem is the following:
\thmb
\thma{main}
If $d\ge 4$ (i.e. $c \geq \sqrt{2}$),
there exists a ^^ critical trajectory'
converging to the Gaussian fixed point
starting from the hierarchical Ising models. Namely,
there exists a positive real number $s_c$
such that if $\den{N}$, $N=0,1,2,\cdots$, are defined by \eqnu{RG} with
$\dsp \den{0} = \den{{\rm I},s_c}$\,, then
the sequence of measures $\den{N}(x)\,dx$, $N=0,1,2,\cdots$,
converges weakly to the massless Gaussian measure $\den{G}(x)\,dx$.
\thme
\remb
Our proof is partially computer-aided and shows for $d=4$ that
\[ s_c \in [1.7925671170092624,1.7925671170092625].\]
% s is denoted by K in [Hara] of th3.2.0, 1999.11.27.
\reme
\bigskip
In the following sections, we give a proof of \thmu{main}.
We will concentrate on the case $d=4$, since the cases $d>4$
can be proved along similar lines (with weaker bounds).
%-------------------------
\section{Strategy.}
%-------------------------
\seca{strategy}
% CONTENTS
The proof of \thmu{main} is decomposed into two parts:
\thmu{basin}(analysis in the weak coupling regime) and
\thmu{strongcoupling} (analysis in the strong coupling regime).
They are stated in \secu{proofmain},
and their proofs are given
in \secu{blehersinai} and \secu{comp}, respectively.
\thmu{main} is proved at the end of this section assuming them.
\itmb
\item
In \thmu{basin},
we control the \rg\ flow in a weak coupling regime
by means of a {\it finite} number of
truncated correlations (Taylor coefficients
of logarithm of characteristic functions),
and, in terms of the truncated correlations,
we give a criterion, a set of sufficient conditions,
for the measure to be in a domain of attraction
of the Gaussian fixed point.
\item
In \thmu{strongcoupling},
we prove,
by rigorous computer-aided calculations, that
there is a trajectory whose initial point is an Ising measure
and for which the criterion in \thmu{basin} is satisfied after
a small number of iterations.
\itme
The first part (\thmu{basin}) is essentially the Bleher-Sinai argument
\cite{BS1,BS2,sinai}.
However, the criteria introduced in the references \cite{sinai,gk2}
seem to be difficult to handle when `strong coupling constants'
are present in the model, as in the Ising models.
In order to overcome this difficulty,
we use characteristic functions of single spin distributions
and Newman's inequalities for truncated correlations.
The second part (\thmu{strongcoupling}) is
basically simple numerical calculations
of truncated correlations up to 8 points to ensure the criterion.
The results are double checked by Mathematica and \verb|C++| programs,
and furthermore they are made mathematically rigorous
by means of Newman's inequalities.
It should be noted that rigorous computer-aided proofs
are employed in \cite{kw4} to Dyson's hierarchical model in $d=3$ dimensions,
to prove, with \cite{kw3}, an existence of a non-Gaussian fixed point.
(The `physics' are of course different between $d=3$ and $d=4$.)
We also focus on a complete mathematical proof,
by combining rigorous computer-aided bounds with
mathematical methods such as Newman's inequalities and
the Bleher--Sinai arguments.
%-----------------------------
\subsection{Characteristic function.}
%-----------------------------
\seca{chrfcn}
Denote the characteristic function of the
single spin distribution $\den{N}$ as
\eqnb
\cha{N}(\xi)={\cal F}\den{N}(\xi)=
\int_\reals e^{\II\xi x} \den{N}(x)\,dx\,.
\eqne
The \rg\ transformation for $\cha{N}$ is
\eqnb\eqna{hierrecursion}
\cha{N+1}={\cal F}{\cal R}{\cal F}^{-1}\cha{N}\,,
\eqne
which has a decomposition
\eqnb
\eqna{RTS}
{\cal F}{\cal R}{\cal F}^{-1}={\cal T}{\cal S},
\eqne
where
\eqab
\eqna{Strans}
{\cal S}g(\xi)&=&g(\dfrac{\sqrt{c}}{2}\xi)^2,
\\
\eqna{Ttrans}
{\cal T}g(\xi)&=&\const \,\exp(-\dfrac{\beta}{2}\triangle)g(\xi),
\eqae
and the constant is so defined that
\[ {\cal T}g\,(0)=1\,. \]
The transformation \eqnu{hierrecursion} has same form as the $N=2$ case of
the Gallavotti hierarchical model \cite{gallavotti,kw,kw2}.
Note that only for $N=2$ the Gallavotti model is equivalent (by
Fourier transform) to the Dyson's hierarchical model.
We introduce a `potential' $V_N$ for the characteristic function $\cha{N}$
and its Taylor coefficients $\tru{n}{N}$ by
\eqab
\eqna{dualpot}
\cha{N}(\xi) &=& e^{-V_N(\xi)},
\\
\eqna{trunptfcn}
V_N(\xi) &=& \sum_{n=1}^\infty \tru{n}{N} \xi^{n}.
\eqae
(Note that $\cha{N}(0)=1$.)
The coefficient $\tru{n}{N}$ is called a truncated $n$ point correlation.
They are functions of Ising parameter $s$ in $\den{0}=\den{I,s}$\,,
but to simplify expressions,
we will always suppress the dependences on $s$ in the following.
In particular, for the initial condition $\den{0}=\den{I,s}$,
we have
\eqsb
&& \cha{0}(\xi)=\cha{I,s}(\xi)={\cal F} \den{I,s}(\xi) = \cos(s\xi),
\\
&& \tru{2}{0}= \frac12 s^2,
\ \tru{4}{0}=\frac1{12} s^4,
\ \tru{6}{0}=\frac1{45} s^6,
\ \tru{8}{0}=\frac{17}{2520} s^8,
\ \mbox{etc.},
\eqse
and
\eqsb
&& \den{1}(x)= {\cal R} \den{I,s}(x)=\const\left(e^{\beta c s^2/2}
\bigl \{ \delta(x-s\sqrt{c})+\delta(x+s\sqrt{c})\bigr \}
+2 \delta(x)\right),
\\
&& \cha{1}(\xi)=\frac1{1+k}(1+k \cos(\sqrt{c} s\xi)),
\ \mbox{ with }\ k=e^{\beta c s^2/2},
\\
&& \tru{2}{1}= k \ell,
\ \tru{4}{1}=\frac{k}{6} (2k-1) \ell^2,
\ \tru{6}{1}=\frac{k}{90} (16k^2-13k+1) \ell^3,
\\
&& \tru{8}{1}=\frac{k}{2520} (272k^3-297k^2+60k-1) \ell^4,
\ \mbox{etc.},
\ \mbox{ with }\ \ell=\frac{c s^2}{2(k+1)}\,.
\eqse
%-----------------------------
\subsection{Newman's inequalities.}
%-----------------------------
\seca{newman}
The function $V_N$ has a remarkable positivity property
and its Taylor coefficients obey Newman's inequalities
(for a brief review of relevant part, see \appu{Newman}):
\eqnb\eqna{Taylorbound}
0\le \tru{2n}{N}\le\dfrac{1}{n}(2\tru{4}{N})^{n/2},
\ n=3,4,5,\cdots.
\eqne
These inequalities follow from \cite[Theorem~3, 6]{newman},
since we have chosen the Ising spin distribution $\den{0}=\den{I,s}$
and the function of $\eta$ defined by
\eqnb
\int e^{ \eta x} h_N(x) dx =
\biggl \langle \exp \biggl(
\eta \Bigl ( \frac{\sqrt{c}}{2} \Bigr )^N
\sum_{\theta} \phi_{\theta} \biggr) \biggr \rangle_{N, h_{I,s}}
\eqne
has only pure imaginary zeros as is shown in \cite[Theorem~1]{newman}.
Note also that \eqnu{rg0} and \eqnu{Ising} imply
\eqnb\eqna{evenmeas} \tru{2n+1}{N}=0,\ n=0,1,2,\cdots. \eqne
The bounds \eqnu{Taylorbound} are extensively used in this paper.
We here note the following facts:
\itmb
\item
The right hand side of \eqnu{trunptfcn}
has a nonzero radius of convergence.
\item
It suffices to prove $\limf{N}\tru{4}{N}=0$
in order to ensure that $\tru{2n}{N}$, $n\ge 3$, converges to zero,
hence the trajectory converges to the Gaussian fixed point.
\itme
%---------------------------------
\subsection{Proof of \thmu{main}.}
%---------------------------------
\seca{proofmain}
Let $\den{0}=\den{I,s}$ and $d=4$.
Note the following simple observations on the `mass term' $\tru{2}{N}$,
which is the variance of $\den{N}(x)\,dx$.
\itmb
\item
$\tru{2}{N}$ is continuous in the Ising parameter $s$,
because $\den{N}(x)\,dx$
is a result of a finite number of \rg\ transformation \eqnu{rg0}.
\item
$\tru{2}{N}$ is increasing in $s$,
vanishes at $s=0$,
and diverges as $s\to\infty$.
\itme
We then put, for $N=0,1,2,\cdots$,
\eqab
\eqna{s0}
\underline{s}_{\,N}&=&\inf\{s>0\mid \tru{2}{N}\ge 1\},
\\
\eqna{s1}
\overline{s}_{N}&=&\inf\{s>0 \mid \tru{2}{N}\ge
\min\{1+\frac3{\sqrt2}\tru{4}{N},\;2+\sqrt2\} \}.
\eqae
%[We put $\overline{s}_{N}=\infty$
%if there are no $s$ satisfying the condition
%in the right hand side of \eqnu{s1}.]
Obviously, we have
\[
0< \underline{s}_{\,N}\le \overline{s}_{N}<\infty.
\]
Note also that
\eqnb
\eqna{criticalmass}
1\le \tru{2}{N}\le 1+\frac3{\sqrt2}\tru{4}{N}\,
\eqne
holds for $s\in [\underline{s}_{\,N},\overline{s}_{N}]$.
As is seen in \secu{blehersinai}, \eqnu{criticalmass} is necessary
for the model to be critical.
We call this \emph{a critical mass condition}.
The following theorem states our result in the weak coupling regime
and is proved in \secu{blehersinai}.
\thmb
\thma{basin}
Let $\den{0}=\den{I,s}$ and $d=4$.
Assume that there exist integers $N_0$ and $N_1$, satisfying $N_0\le N_1$,
such that, for $s\in [\underline{s}_{\,N_1},\overline{s}_{\,N_1}]$,
the bounds
\eqab
\eqna{smallnessofmu4}
0 \ \le& \tru{4}{N_0} &\le\ 0.0045,
\\
1.6 \trup{4}{N_0}{2} \ \le& \tru{6}{N_0} &\le\ 6.07 \trup{4}{N_0}{2},
\\
0 \ \le & \tru{8}{N_0} &\le\ 48.469 \trup{4}{N_0}{3},
\eqae
and
\eqnb
\eqna{aprioriforODE}
\tru{2}{N}< 2+\sqrt{2},\ \ N_0\le N< N_1\,,
\eqne
hold.
Then there exists an
$s_c\in [\underline{s}_{\,N_1},\overline{s}_{\,N_1}]$
such that if $s=s_c$ then
\eqsb
&&\lim_{N\to\infty}\tru{4}{N} = 0,
\\
&&\lim_{N\to\infty}\tru{2}{N} = 1\,.
\eqse
\thme
\remb
The original Bleher--Sinai argument takes $N_0=N_1$\,.
We include the $N_04$, the decay follows from \eqnu{mODE} and \eqnu{lODE}
with $d$-dependent coefficients,
namely, if we throw out the negative contributions
$-\tru{4}{}(t)$ and $-\tru{6}{}(t)$
to the right hand sides of \eqnu{mODE} and \eqnu{lODE}, respectively,
then we have upper bounds on $\tru{2}{}(t)$ and $\tru{4}{}(t)$.
This argument eventually yields exponential decay of $\tru{4}{N}$.
In case $d=4$, the situation is more subtle,
since the decay of $\tru{4}{N}$ is weak,
i.e., powerlike instead of exponential.
In order to derive the delicate bound on $\tru{4}{}(t)$,
a lower bound for $\tru{6}{}(t)$ must be incorporated,
which in turn needs an upper bound on $\tru{8}{}(t)$.
Thus, we have to deal with the equations \eqnu{mODE}--\eqnu{sODE}.
This is the principle of our estimation.
The result is the following:
\prpb\prpa{recineq}
Let $d=4$ and $N$ be a positive integer, and put
\eqab
\eqna{r}
r_N&=&
\dfrac{1}{1-(\sqrt2-1)(\tru{2}{N}-1)}=
\dfrac{1}{\sqrt2-(\sqrt2-1)\tru{2}{N}}\,,
\\
\eqna{zeta}
\zeta_N&=&\dfrac{\sqrt2r_N-1}{\sqrt2\tru{2}{N}}
=\dfrac{r_N}{\tru{2}{N}}-\dfrac{1}{\sqrt2\tru{2}{N}}\,.
\eqae
(i) If
\eqnb
\eqna{supple0}
\tru{2}{N} < 2+\sqrt{2},
\eqne
then
\eqab
\eqna{mubound}
\eqna{mB1}
\tru{2}{N+1} &\le& r_N\tru{2}{N}\,,
\\
\eqna{mb}
\tru{2}{N+1} &\ge& r_N\tru{2}{N}-3r_N^2\zeta_N\tru{4}{N}\,.
\eqae
(ii) If, furthermore,
\eqab
\eqna{supple1}
\dfrac{\tru{4}{N}}{4} &\ge&
\dfrac{15}{8\sqrt2}\zeta_N\tru{6}{N}+\dfrac{21}{4}\zeta_N^2\trup{4}{N}{2},
\\
\eqna{supple2}
\dfrac{\tru{6}{N}}{8\sqrt2}+\dfrac{1}{2}\zeta_N\trup{4}{N}{2}
&\ge& 24\zeta_N^3\trup{4}{N}{3}
+\dfrac{123}{8\sqrt2}\zeta_N^2\tru{4}{N}\tru{6}{N}
+\dfrac{7}{8}\zeta_N\tru{8}{N}\,,
\\
\eqna{supple3}
\dfrac{3}{2}\zeta_N\tru{4}{N} &\ge&
12\zeta_N^3\trup{4}{N}{2} + \dfrac{45}{8\sqrt2}\zeta_N^2\tru{6}{N}\,,
\eqae
then
\eqab
\eqna{mB2}
\tru{2}{N+1} &\le& r_N\tru{2}{N}-3r_N^2(\zeta_N\tru{4}{N}
-8\zeta_N^3\trup{4}{N}{2}-\dfrac{15}{4\sqrt2}\zeta_N^2 \tru{6}{N}),
\\
\eqna{lb}
\tru{4}{N+1} &\ge& r_N^4
(\tru{4}{N}-\dfrac{15}{2\sqrt2}\zeta_N\tru{6}{N}
-21\zeta_N^2 \trup{4}{N}{2}),
\\
\nonumber
\tru{4}{N+1} &\le& r_N^4
(\tru{4}{N}-\dfrac{15}{2\sqrt2}\zeta_N\tru{6}{N}-21\zeta_N^2 \trup{4}{N}{2}
\\
\eqna{lB}
&& +\dfrac{705}{2\sqrt2}\zeta_N^3\tru{4}{N}\tru{6}{N}
+447\zeta_N^4\trup{4}{N}{3}+\dfrac{105}{4}\zeta_N^2\tru{8}{N}),
\\
\eqna{nB}
\tru{6}{N+1} &\le&r_N^6(\dfrac{\tru{6}{N}}
{\sqrt2}+4\zeta_N\trup{4}{N}{2}),
\\
\eqna{nb}
\tru{6}{N+1} &\ge& r_N^6
(\dfrac{\tru{6}{N}}{\sqrt2}+4\zeta_N\trup{4}{N}{2}
-192\zeta_N^3\trup{4}{N}{3}
-\dfrac{123}{\sqrt2}\zeta_N^2\tru{4}{N}\tru{6}{N}-7\zeta_N\tru{8}{N}),
\\
\eqna{sB}
\eqna{sigmabound}
\tru{8}{N+1} &\le&
r_N^8(\dfrac{\tru{8}{N}}{2}+\dfrac{12}{\sqrt2}\zeta_N\tru{4}{N}\tru{6}{N}
+24\zeta_N^2\trup{4}{N}{3}).
\eqae
\prpe
%
\bigskip\par
%
The rest of this section is devoted to a proof of \prpu{recineq}.
%
%-----------------------------------
%\subsection{Proof of \prpu{recineq}.}
%-----------------------------------
\prfb
%-----------------------------
% Integral equations
%-----------------------------
Now, observe that $\bar{\tru{2}{}}(t)$ defined by
\eqnb
\eqna{mODE0}
\dfrac{d}{dt}\bar{\tru{2}{}}(t)=4\bar{\tru{2}{}}(t)^2,
\ \ \bar{\tru{2}{}}(0)=\dfrac{1}{\sqrt2}\tru{2}{N}\,,
\eqne
is an upper bound of $\tru{2}{}(t)$:
\eqnb\eqna{mubd0}
\tru{2}{}(t) \le \bar{\tru{2}{}}(t)
= \dfrac{\tru{2}{N}}{\sqrt2} \dfrac{1}{1-2\sqrt2\tru{2}{N} t}\,.
\eqne
This, at $\dsp t=\frac{\beta}2$ ($\dsp =\frac{\sqrt2-1}4$ for $d=4$)
implies \eqnu{mB1}.
Put
\eqsb
M(t)&=&\dfrac{1}{1-2\sqrt2\tru{2}{N} t}\,,
\\
m(t)&=&\bar{\tru{2}{}}(t)-\tru{2}{}(t).
\eqse
We have $ m(t) \ge 0$,
and \eqnu{supple0} implies that $M(t)$ is increasing in $t\in[0,\beta/2]$.
By a change of variable
$\dsp z=M(t)-1$ ($\dsp dz= 2\sqrt2\tru{2}{N} M(t)^2 dt$)
and by putting
\[
\hat m(z) = m(t)/M(t)^2,
\ \ \hat{\tru{4}{}}(z) = \tru{4}{}(t)/M(t)^4,
\ \ \hat{\tru{6}{}}(z) = \tru{6}{}(t)/M(t)^6,
\ \ \hat{\tru{8}{}}(z) = \tru{8}{}(t)/M(t)^8,
\]
we have, from \eqnu{mODE} -- \eqnu{sODE},
\eqab
\eqna{lInt}
\hat{\tru{4}{}}(z) &=& \dfrac{\tru{4}{N}}{4} +\dfrac{1}{\sqrt2\tru{2}{N}}
\int_0^z (-8\hat m(z)\hat{\tru{4}{}}(z)-15\hat{\tru{6}{}}(z))dz,
\\
\eqna{nInt}
\hat{\tru{6}{}}(z)&=& \dfrac{\tru{6}{N}}{8\sqrt2}
+\dfrac{1}{\sqrt2\tru{2}{N}}
\int_0^z (8\hat{\tru{4}{}}(z)^2-12\hat m(z)\hat{\tru{6}{}}(z)
-28\hat{\tru{8}{}}(z))dz,
\\
\eqna{sInt}
\hat{\tru{8}{}}(z)&=& \dfrac{\tru{8}{N}}{32} +\dfrac{1}{\sqrt2\tru{2}{N}}
\int_0^z (24\hat{\tru{4}{}}(z)\hat{\tru{6}{}}(z)
-16\hat m(z)\hat{\tru{8}{}}(z)
-45 \hat{\tru{10}{}}(z))dz,
\\
\eqna{mInt}
\hat m(z)&=&
\dfrac{1}{\sqrt2\tru{2}{N}} \int_0^z (6\hat{\tru{4}{}}(z)-2\hat m(z)^2)dz,
\eqae
%-----------------------------
% Bounds
%-----------------------------
The equations \eqnu{lInt}--\eqnu{mInt}
with positivity of $\tru{2n}{}(t)$ imply
\eqab
\eqna{upbd4}
\hat{\tru{4}{}}(z)&\le&\dfrac{\tru{4}{N}}{4},
\\
\eqna{upbd6}
\hat{\tru{6}{}}(z)&\le& \dfrac{\tru{6}{N}}{8\sqrt2}
+ \dfrac{1}{\sqrt2\tru{2}{N}}\int_0^z\ 8\hat{\tru{4}{}}(z)^2dz
\ \le
\ \dfrac{\tru{6}{N}}{8\sqrt2}+\dfrac{\trup{4}{N}{2}}{2\sqrt2\tru{2}{N}}z,
\\
\eqna{upbd8}
\hat{\tru{8}{}}(z)&\le& \dfrac{\tru{8}{N}}{32}+ \dfrac{1}{\sqrt2\tru{2}{N}}
\int_0^z\ 24\hat{\tru{4}{}}(z)\hat{\tru{6}{}}(z)dz
\ \le
\ \dfrac{\tru{8}{N}}{32}
+\dfrac{3}{8}\dfrac{\tru{4}{N}\tru{6}{N}}{\tru{2}{N}}z
+\dfrac{3}{4}\dfrac{\trup{4}{N}{3}}{\trup{2}{N}{2}}z^2,
\\
\eqna{upbdm}
\hat m(z)&\le& \dfrac{1}{\sqrt2\tru{2}{N}}\int_0^z\ 6\hat{\tru{4}{}}(z) dz
\ \le
\ \dfrac{3\tru{4}{N}}{2\sqrt2\tru{2}{N}}z.
\eqae
In particular, \eqnu{upbdm} at $\dsp t=\frac{\beta}2$
($\dsp z=M(\frac{\beta}2)-1=\sqrt2 r_n -1$ for $d=4$) implies
\eqnu{mb}.
Using \eqnu{upbd4}, \eqnu{upbd6}, \eqnu{upbdm} in \eqnu{lInt}, we have
\eqnb\eqna{lambdalbound}
\hat{\tru{4}{}}(z) \ge \dfrac{\tru{4}{N}}{4}
-\dfrac{15\tru{6}{N}}{16\tru{2}{N}}z
-\dfrac{21\trup{4}{N}{2}}{8\trup{2}{N}{2}}z^2.
\eqne
Using \eqnu{upbd6}, \eqnu{upbd8}, \eqnu{upbdm}, \eqnu{lambdalbound}
in \eqnu{nInt} and \eqnu{mInt} we further have
\eqab
\eqna{nulbound}
\hat{\tru{6}{}}(z)&\ge& \dfrac{\tru{6}{N}}{8\sqrt2}
+\dfrac{\trup{4}{N}{2}}{2\sqrt2\tru{2}{N}}z
-\dfrac{12\trup{4}{N}{3}}{\sqrt2\trup{2}{N}{3}}z^3
-\dfrac{123\tru{4}{N}\tru{6}{N}}{16\sqrt2\trup{2}{N}{2}}z^2
-\dfrac{7\tru{8}{N}}{8\sqrt2\tru{2}{N}}z,
\\
\eqna{mlbound}
\hat m(z)&\ge& \dfrac{3\tru{4}{N}}{2\sqrt2\tru{2}{N}}z
-\dfrac{6\trup{4}{N}{2}}{\sqrt2\trup{2}{N}{3}}z^3
-\dfrac{45\tru{6}{N}}{16\sqrt2\trup{2}{N}{2}}z^2.
\eqae
When $d=4$, $\dsp\beta=\frac{\sqrt2-1}{2}$ and
$\dsp z=M(\frac{\beta}2)-1=\sqrt2r_N-1$
($\dsp M(\frac{\beta}2)=\sqrt2r_N$).
Then the assumptions \eqnu{supple1} -- \eqnu{supple3}
of \prpu{recineq} imply
that the right hand sides of \eqnu{lambdalbound}, \eqnu{nulbound},
and \eqnu{mlbound} are non-negative at $\dsp t=\frac{\beta}2$\,.
On the other hand,
they are concave in $z$ for $z\ge0$\,.
Recall also that $z=M(t)-1$ is increasing in $t\in[0,\beta/2]$.
Therefore,
they are non-negative for all $\dsp t\in[0,\beta/2]$.
Using \eqnu{lambdalbound}, \eqnu{nulbound}, and \eqnu{mlbound}
in \eqnu{lInt}, we therefore have
\eqab
\hat{\tru{4}{}}(z) &\le& \dfrac{\tru{4}{N}}{4}
-\dfrac{1}{\sqrt2\tru{2}{N}}\times
\nonumber \\
&&\times \int_0^z\left(
8\left(
\dfrac{3\tru{4}{N}}{2\sqrt2\tru{2}{N}}z
-\dfrac{6\trup{4}{N}{2}}{\sqrt2\trup{2}{N}{3}}z^3
-\dfrac{45\tru{6}{N}}{16\sqrt2\trup{2}{N}{2}}z^2\right)
\left(
\dfrac{\tru{4}{N}}{4}
-\dfrac{15\tru{6}{N}}{16\tru{2}{N}}z
-\dfrac{21\trup{4}{N}{2}}{8\trup{2}{N}{2}}z^2\right)
\right.
\nonumber\\
&&\ \ \left. +15\left(
\dfrac{\tru{6}{N}}{8\sqrt2}
+\dfrac{\trup{4}{N}{2}}{2\sqrt2\tru{2}{N}}z
-\dfrac{12\trup{4}{N}{3}}{\sqrt2\trup{2}{N}{3}}z^3
-\dfrac{123\tru{4}{N}\tru{6}{N}}{16\sqrt2\trup{2}{N}{2}}z^2
-\dfrac{7\tru{8}{N}}{8\sqrt2\tru{2}{N}}z\right)\right)dz
\nonumber \\
&\le& \dfrac{\tru{4}{N}}{4}
-\dfrac{15\tru{6}{N}}{16\tru{2}{N}}z
-\dfrac{21\trup{4}{N}{2}}{8\trup{2}{N}{2}}z^2
+\dfrac{705\tru{4}{N}\tru{6}{N}}{32\trup{2}{N}{3}}z^3
+\dfrac{447\trup{4}{N}{3}}{16\trup{2}{N}{4}}z^4
+\dfrac{105\tru{8}{N}}{32\trup{2}{N}{2}}z^2.
\eqna{mu4detail}
\eqae
Recalling that at $t=\beta/2$\ \ ($z=M(\frac{\beta}2)-1=\sqrt2r_N-1$)
we have
\eqsb
\bar{\tru{2}{}}(\frac{\beta}{2})&=&r_N\tru{2}{N}\,,
\\
\tru{2}{N+1}&=& r_N\tru{2}{N}
-\hat m(\sqrt2r_N-1)M(\frac{\beta}{2})^2, \\
\tru{4}{N+1}&=&\hat{\tru{4}{}}(\sqrt2r_N-1)M(\frac{\beta}{2})^4,
\\
\tru{6}{N+1}&=&\hat{\tru{6}{}}(\sqrt2r_N-1)M(\frac{\beta}{2})^6,
\\
\tru{8}{N+1}&=&\hat{\tru{8}{}}(\sqrt2r_N-1)M(\frac{\beta}{2})^8,
\eqse
we see that
\eqnu{mlbound},
\eqnu{lambdalbound}, \eqnu{mu4detail},
\eqnu{upbd6}, \eqnu{nulbound},
\eqnu{upbd8} imply \eqnu{mB2} -- \eqnu{sigmabound},
respectively.
This completes a proof of \prpu{recineq}.
\qed
\prfe
%-----------------------------
\section{Bleher--Sinai argument.}
%-----------------------------
\seca{blehersinai}
In order to show \thmu{basin},
we confirm existence of a critical parameter $s=s_c$
by means of Bleher-Sinai argument,
and, at the same time,
we derive the expected decay of $\tru{4}{N}$.
In Bleher-Sinai argument,
monotonicity of $\underline{s}_N$ and $\overline{s}_N$
with respect to $N$ is essential.
%-----------------------------
% \subsection{Critical mass.}
%-----------------------------
\prpb
\prpa{criticalmass}
Let $d=4$\,.
Then the following hold.
\itmb
\item
If $\dsp\tru{2}{N}-1<0$ then $\tru{2}{N+1}<\tru{2}{N}$\,.
\item
If $\dsp \frac14>\tru{2}{N}-1 \ge\frac{3}{\sqrt{2}}\tru{4}{N}$ then
$\tru{2}{N+1}\ge\tru{2}{N}$\,.
\itme
\prpe
\prfb
Note that for both cases in the statement,
the assumption \eqnu{supple0} in \prpu{recineq} holds.
Hence, \eqnu{mB1}, with \eqnu{r} and monotonicity of $\tru{2}{N}$,
implies
\eqnb
\eqna{shrink}
\tru{2}{N}-1<0\ \Longrightarrow
\ r_N<1 \ \Longrightarrow\ \tru{2}{N+1}<\tru{2}{N}\,.
\eqne
Next we see that \eqnu{mb}, with \eqnu{r} and \eqnu{zeta}, implies
\eqnb
\eqna{expand0}
\dfrac{\tru{2}{N}-1}{\tru{4}{N}}
\ge\dfrac{3r_N(\sqrt2r_N-1)}{(2-\sqrt2)\trup{2}{N}{2}}
\ \Longrightarrow\ \tru{2}{N+1}\ge\tru{2}{N}\,.
\eqne
Put
\[ L_1(x)=\frac3{\sqrt{2} x (\sqrt{2}-(\sqrt{2}-1) x)^2}\,. \]
Then by straightforward calculation we see
\[ 1\le x\le \frac54\ \Longrightarrow
\ L_1(x)\le L_1(1)=\frac{3}{\sqrt{2}}\,, \]
and \eqnu{r} implies
\[
L_1(\tru{2}{N})=\dfrac{3r_N(\sqrt2r_N-1)}{(2-\sqrt2)\trup{2}{N}{2}}\,.
\]
Therefore \eqnu{expand0} implies that
\eqnb
\eqna{expand}
\frac14 > \tru{2}{N}-1 \ge \dfrac{3}{\sqrt2}\tru{4}{N}
\ \Longrightarrow\ \tru{2}{N+1}\ge\tru{2}{N}\,.
\eqne
\qed\prfe
\corb
\cora{criticalmass}
Let $d=4$\,.
Then, for the $\underline{s}_{\,N}$ defined in \eqnu{s0},
it holds that $\underline{s}_{\,N}\le \underline{s}_{\,N+1}$\,.
\core
\prfb
Since $\tru{2}{N}$ is increaisng in $s$,
if $s<\underline{s}_{\,N}$ then $\tru{2}{N}<1$, hence
\prpu{criticalmass} implies $\tru{2}{N+1}<\tru{2}{N}<1$\,,
further implying $s<\underline{s}_{\,N+1}$\,.
Hence the statement holds.
\qed\prfe
For later convenience, define
\eqab
\eqna{rstr}
r_N^* &=& \dfrac{1}{1-(\sqrt{2}-1)\dfrac{3}{\sqrt2}\tru{4}{N}}\,,
\\
\eqna{zetalstr}
\zeta_{*N} &=& 1-\dfrac{1}{\sqrt2}\,,
\\
\eqna{zetaustr}
\zeta_N^* &=&
\dfrac{\sqrt2r_N^*-1}{\sqrt2(1+\dfrac3{\sqrt2}\tru{4}{N})}\,,
\eqae
Then we see that if \eqnu{criticalmass}
holds, then we have, from \eqnu{r} and \eqnu{zeta},
\eqab
\eqna{rrstr}
1\ < & r_N & <\ r_N^*,
\\
\eqna{zetazetastr}
\zeta_{*N}\ < & \zeta_N & <\ \zeta_N^*.
\eqae
%-----------------------------
% \subsection{Control of higher order correlation functions.}
%-----------------------------
\prpb
\prpa{nearGauss}
Let $d=4$ and put
\[
\alpha_0=0.0045,\ \ \alpha_1=1.6,\ \ \alpha_2=6.07,\ \ \alpha_3=48.469\,.
\]
Assume that there exists an integer $N$ such that
\eqnu{supple0} and
\eqab
\eqna{nearGaussa4}
(0\ \le) &\tru{4}{N}&\le\ \alpha_0,
\\
\alpha_1\trup{4}{N}{2}\ \le &\tru{6}{N}& \le\ \alpha_2\trup{4}{N}{2},
\\
\eqna{nearGaussa8}
(0\ \le) &\tru{8}{N}& \le\ \alpha_3\trup{4}{N}{3},
\eqae
hold. Then \eqnu{supple1}--\eqnu{supple3} hold, and the following
also hold:
\eqab
\eqna{nearGauss4}
(0\ \le) &\tru{4}{N+1} &\le\ \tru{4}{N} (1-0.08\tru{4}{N})
\ \ \ \ (\le \alpha_0),
\\
\eqna{nearGauss6}
\alpha_1\trup{4}{N+1}{2}\ \le &\tru{6}{N+1}& \le\
\alpha_2\trup{4}{N+1}{2},
\\
\eqna{nearGauss8}
(0\ \le) &\tru{8}{N+1}& \le\ \alpha_3\trup{4}{N+1}{3}.
\eqae
\prpe
\prfb
For $x\ge0$ put
\eqab
\nonumber
\ell_r(x) &=& \dfrac{1}{1-(\sqrt{2}-1)\dfrac{3}{\sqrt2}\,x}\,,
\\
\nonumber
\ell_d(x) &=& 1-\dfrac{1}{\sqrt2}\,,
\\
\nonumber
\ell_u(x) &=&
\dfrac{\sqrt2\ell_r(x)-1}{\sqrt2(1+\dfrac3{\sqrt2}\,x)}\,,
\\
\eqna{L2}
L_2(x) &=& 1-(\dfrac{15}{2\sqrt2}\alpha_2\ell_u(x)+21\ell_u(x)^2)\,x.
\eqae
In particular, \eqnu{rstr}, \eqnu{zetalstr}, \eqnu{zetaustr} imply
\[ r_N^*=\ell_r(\tru{4}{N}),
\ \ \ \zeta_{*N}=\ell_d(\tru{4}{N}),
\ \ \ \zeta_N^*=\ell_u(\tru{4}{N}).
\]
By explicit calculation, we see that
\eqnb
L_2(x)>0,\ \ \ 0\le x\le \alpha_0\,.
\eqne
The right hand side of \eqnu{supple1} is then bounded from above
by
\[ \frac14 \tru{4}{N}(1-L_2(\tru{4}{N})) \le \frac14 \tru{4}{N}\,, \]
hence \eqnu{supple1} holds.
Similarly, \eqnu{supple3} is seen to hold
for $0\le \tru{4}{N}\le \alpha_0$,
if we note that
the right hand side of \eqnu{supple3} is bounded from above by
\[
\zeta_N \trup{4}{N}{2}(12\zeta_N^*+\frac{45}{8\sqrt{2}} \zeta_N^* \alpha_2)
\le \frac34(1-L_2(\tru{4}{N}))\tru{4}{N} \le \frac32 \tru{4}{N}\,.\]
The condition \eqnu{supple2} is seen to hold with similar argument,
if we note the right hand side is bounded from above by
\[
\zeta_N \trup{4}{N}{3} (24\zeta_N^2
+\dfrac{123}{8\sqrt2}\zeta_N\alpha_2
+\dfrac{7}{8}\alpha_3),
\]
while the left hand side is bounded from below by
\[
\trup{4}{N}{2}
(\dfrac{\alpha_1}{8\sqrt2}+\dfrac{1}{2}\zeta_N).
\]
Therefore, the conclusions of \prpu{recineq} hold,
in particular, \eqnu{lb}--\eqnu{sB} imply
\eqab
\eqna{lb'}
\tru{4}{N+1}&\ge&r_N^4 \tru{4}{N}
\left(1-(\dfrac{15}{2\sqrt2}\zeta_N\alpha_2+21\zeta_N^2)\tru{4}{N}\right),
\\
\eqna{lB'}
\dfrac{\tru{4}{N+1}}{\tru{4}{N}}&\le&r_N^4
\left(1-(\dfrac{15}{2\sqrt2}\zeta_N\alpha_1+21\zeta_N^2)\tru{4}{N}
+(\dfrac{705}{2\sqrt2}\zeta_N^3\alpha_2
+447\zeta_N^4+\dfrac{105}{4}\zeta_N^2\alpha_3)\trup{4}{N}{2}\right),
\\
\eqna{nB'}
\dfrac{\tru{6}{N+1}}{\trup{4}{N+1}{2}}&\le&
\left(\dfrac{\tru{4}{N}}{\tru{4}{N+1}}\right)^2
r_N^6\left(\dfrac{\alpha_2}{\sqrt2}+4\zeta_N\right),
\\
\eqna{nb'}
\dfrac{\tru{6}{N+1}}{\trup{4}{N+1}{2}}&\ge&
\left(\dfrac{\tru{4}{N}}{\tru{4}{N+1}}\right)^2
r_N^6\left(\dfrac{\alpha_1}{\sqrt2}+4\zeta_N-(192\zeta_N^3
+\dfrac{123}{\sqrt2}\zeta_N^2\alpha_2+7\zeta_N\alpha_3)\tru{4}{N}\right),
\\
\eqna{sB'}
\dfrac{\tru{8}{N+1}}{\trup{4}{N+1}{3}}&\le&
\left(\dfrac{\tru{4}{N}}{\tru{4}{N+1}}\right)^3
r_N^8\left(\dfrac{\alpha_3}{2}+\dfrac{12}{\sqrt2}\zeta_N\alpha_2
+24\zeta_N^2\right).
\eqae
%-----------------------------
% \subsubsection{Choice of $\alpha_2$}
%-----------------------------
Rewriting \eqnu{lb'}, using \eqnu{rrstr} and \eqnu{zetazetastr}, we have
\eqnb
\eqna{lambda/lambda}
\dfrac{\tru{4}{N}}{\tru{4}{N+1}} \le \dfrac{1}{r_N^4}
\dfrac{1}{1-(\dfrac{15}{2\sqrt2}\zeta_N\alpha_2+21\zeta_N^2)\tru{4}{N}}
\le \frac1{L_2(\tru{4}{N})}\,.
\eqne
This and \eqnu{nB'} imply
\[
\dfrac{\tru{6}{N+1}}{\trup{4}{N+1}{2}}
\le
\dfrac{\dfrac{1}{\sqrt2}\alpha_2+4\ell_u(\tru{4}{N})}
{L_2(\tru{4}{N})^{2}}\,.
\]
By explicit calculation, we see that
\[ 0\le x\le \alpha_0\ \Longrightarrow
\ \dfrac{\dfrac{1}{\sqrt2}\alpha_2+4\ell_u(x)}{L_2(x)^2}\le \alpha_2.\]
Therefore the upper bound in \eqnu{nearGauss6} holds.
%-----------------------------
% \subsubsection{Choice of $\alpha_3$}\secl{gamma}
%-----------------------------
In a similar way, we note that \eqnu{sB'} and \eqnu{lambda/lambda} imply
\[
\dfrac{\tru{8}{N+1}}{\trup{4}{N+1}{3}}
\le
\dfrac{\dfrac{1}{2}\alpha_3+\dfrac{12}{\sqrt2}\ell_u(\tru{4}{N})\alpha_2
+24\ell_u(\tru{4}{N})^{2}}{L_2(\tru{4}{N})^{3}}\,.
\]
By explicit calculation, we see that
\[ 0\le x\le \alpha_0\ \Longrightarrow
\ \dfrac{\dfrac{1}{2}\alpha_3+\dfrac{12}{\sqrt2}\ell_u(x)\alpha_2
+24\ell_u(x)^2}{L_2(x)^2}\le \alpha_3.\]
Therefore \eqnu{nearGauss8} holds.
%-----------------------------
% \subsubsection{Upper bound of $\alpha_1$}\secl{alphaupper}
%-----------------------------
Similarly, from \eqnu{nb'} and \eqnu{lB'}, we have
\eqsb
\lefteqn{\dfrac{\tru{6}{N+1}}{\trup{4}{N+1}{2}}}
\\ &\ge&\dfrac{1}{\ell_r(\tru{4}{N})^{2}}
\\ && \hspace*{-40pt} \times
\dfrac{\dfrac{\alpha_1}{\sqrt2}+4\ell_d(\tru{4}{N})
-(192\ell_u(\tru{4}{N})^{3}
+\dfrac{123}{\sqrt2}\ell_u(\tru{4}{N})^{2}\alpha_2
+7\ell_u(\tru{4}{N})\alpha_3)
\tru{4}{N}}{\left(1-(\dfrac{15}{2\sqrt2}
\ell_d(\tru{4}{N})\alpha_1+21\ell_d(\tru{4}{N})^{2})\tru{4}{N}
+(\dfrac{705}{2\sqrt2}\ell_u(\tru{4}{N})^3\alpha_2+447\ell_u(\tru{4}{N})^{4}
+\dfrac{105}{4}\ell_u(\tru{4}{N})^{2}\alpha_3)\trup{4}{N}{2}\right)^2}
\\
&\ge&\alpha_1\,,
\eqse
if $\dsp 0\le \tru{4}{N} \le \alpha_0$\,.
Therefore the lower bound in \eqnu{nearGauss6} holds.
%-----------------------------
% \subsubsection{Lower bound of $\alpha_1$}\secl{alphalower}
%-----------------------------
Finally, from \eqnu{lB'}, we have, again with similar argument,
\eqsb
\dfrac{\tru{4}{N+1}}{\tru{4}{N}}&\le&
\ell_r(\tru{4}{N})^4
\left(1-(\dfrac{15}{2\sqrt2}\ell_d(\tru{4}{N})\alpha_1
+21\ell_d(\tru{4}{N})^{2})
\tru{4}{N}
\right. \\ &&\ \ \ \left.
+(\dfrac{705}{2\sqrt2}\ell_u(\tru{4}{N})^{3}\alpha_2
+447\ell_u(\tru{4}{N})^{4}
+\dfrac{105}{4}\ell_u(\tru{4}{N})^{2}\alpha_3)\trup{4}{N}{2}\right)
\\
&\le&1-0.08\tru{4}{N}\,,
\eqse
if
$\dsp 0\le \tru{4}{N} \le \alpha_0$\,.
Therefore \eqnu{nearGauss8} holds.
\qed\prfe
\corb
\cora{nearGauss}
Let $d=4$, and assume that
%$\overline{s}_{N}<\infty$ and that
for some $N$
the assumptions \eqnu{nearGaussa4} -- \eqnu{nearGaussa8}
in \prpu{nearGauss} hold for
all $s$ satisfying $\underline{s}_{\,N}\le s\le \overline{s}_{N}$\,,
where $\underline{s}_{\,N}$ and $\overline{s}_{N}$ are
defined in \eqnu{s0} and \eqnu{s1}.
Then it holds that $\overline{s}_{\,N+1}\le \overline{s}_{N}$\,.
\core
\prfb
By \eqnu{nearGaussa4},
$\dsp 1+\frac3{\sqrt2}\tru{4}{N} <2+\sqrt2$\,,
if $\underline{s}_{\,N}\le s\le \overline{s}_{N}$\,.
Hence, by \eqnu{s1},
\[
\overline{s}_{N}=
\inf\{s>0 \mid \tru{2}{N}\ge 1+\frac3{\sqrt2}\tru{4}{N} \},
\]
and, from monotonicity of $\tru{2}{N}$ in $s$, \eqnu{supple0} holds
if $s\le \overline{s}_{N}$\,.
Continuity of $\tru{2}{N}$ and $\tru{4}{N}$ in $s$ imply
\[ \tru{2}{N} = 1+\frac3{\sqrt{2}}\tru{4}{N}\,,\
\mbox{ if }\ s=\overline{s}_{\,n}\,. \]
(In particular, we may assume that $\dsp \frac54> \tru{2}{N}$\,.)
Hence \prpu{criticalmass} implies
\eqnb
\eqna{cor1}
\tru{2}{N+1}\ge 1+\frac3{\sqrt{2}}\tru{4}{N}\,,
\ \ \mbox{ for } s=\overline{s}_{N}\,.
\eqne
By assumptions at $s=\overline{s}_{N}$,
we see, from \prpu{nearGauss},
that $\dsp \tru{4}{N+1}\le \tru{4}{N}$,
which, with \eqnu{cor1}, implies
\[ \tru{2}{N+1}\ge 1+\frac3{\sqrt{2}}\tru{4}{N+1}\,. \]
This proves $ \overline{s}_{\,N+1}\le \overline{s}_{N}$\,.
\qed\prfe
%-----------------------------
% \subsubsection{Catching the critical point. III. Bleher-Sinai argument}
% \label{subsub-BS1}
%-----------------------------
\prfofb{\protect{\thmu{basin}}}
Note first that \coru{criticalmass} implies
\eqnb
\eqna{contractionLower}
\underline{s}_{\,N}\le \underline{s}_{\,N+1}\,,\ \
N=N_1,N_1+1,N_1+2,\cdots.
\eqne
With assumptions of the theorem and by induction on $N$,
\prpu{nearGauss} implies
that for any $s$ satisfying
$\underline{s}_{\,N_1}\le s\le \overline{s}_{\,N_1}$\,,
the bounds
\eqnu{nearGaussa4} -- \eqnu{nearGaussa8} hold for $N=N_1$\,.
Hence \coru{nearGauss} implies
$\overline{s}_{\,N_1+1}\le \overline{s}_{\,N_1}$\,.
Also since $s\le \overline{s}_{\,N_1}$ implies \eqnu{supple0} for $N=N_1$,
\prpu{nearGauss} implies that
\eqnu{nearGaussa4} -- \eqnu{nearGaussa8} hold for $N=N_1+1$
and $\underline{s}_{\,N_1+1}\le s\le \overline{s}_{\,N_1+1}$\,.
We can proceed with induction on $N$ and repeat this argument
to conclude that
\eqnu{nearGauss4} -- \eqnu{nearGauss8} hold for
$\underline{s}_{\,N}\le s\le \overline{s}_{N}$\,,
$N=N_1,N_1+1,N_1+2,\cdots$,
and
\eqnb
\eqna{contractionUpper}
\overline{s}_{\,N+1}\le \overline{s}_{N}\,,\ \ N=N_1,N_1+1,N_1+2,\cdots.
\eqne
The bounds \eqnu{contractionLower} and \eqnu{contractionUpper}
imply that a sequence of closed intervals on $\reals$
\[ [\underline{s}_{\,N_1},\overline{s}_{\,N_1}]\supset
[\underline{s}_{\,N_1+1},\overline{s}_{\,N_1+1}]\supset
[\underline{s}_{\,N_1+2},\overline{s}_{\,N_1+2}]\supset \cdots,\]
is contracting, hence there exists an $s_c$,
satisfying $\underline{s}_{\,N_1}\le s_c\le \overline{s}_{\,N_1}$,
such that
\[
\underline{s}_{\,N}\le s_c\le \overline{s}_{N}\,,\ \
N=N_1,N_1+1,N_1+2,\cdots.
\]
Hence, in particular, \eqnu{nearGauss4} holds for
all integer $N\ge N_1$ at $s=s_c$.
This implies
\[ \lim_{N\to\infty}\tru{4}{N} = 0\,, \]
at $s=s_c$.
Also we see that if $s=s_c$
then \eqnu{criticalmass} holds for all $N\ge N_1$\,.
Therefore we have
\[ \lim_{N\to\infty}\tru{2}{N} = 1\,, \]
at $s=s_c$.
This completes a proof of \thmu{basin}.
\qed\prfofe
%-----------------------------
\section{Strong coupling problem.}
%-----------------------------
\seca{comp}
We shall prove \thmu{strongcoupling} by
(computer-aided)
brute force evaluation of the Taylor coefficients of
$\cha{N}(\xi)$ instead of $V_N(\xi)$.
%-----------------------------
%\subsection{Taylor expansion of characteristic functions.}
\subsection{Taylor expansion.}
%-----------------------------
\seca{sub-hndn}
Define the Taylor coefficients $a_{n,N}$,
$n\in\pintegers$, of $\cha{N}$ by
\eqnb
\eqna{a}
\cha{N}(\xi) = \sum_{n=0}^\infty (-1)^n \frac1{n!}a_{n,N} \xi^{2n}.
\eqne
In particular, $a_{0,N}=\cha{N}(0)=1$\,.
Note also that
\[a_{n,N}\ge0,\ n\in\pintegers\,. \]
%-----------------------------
% \subsection{Dual Potential} \seca{sub-dual}
%-----------------------------
$\tru{n}{N}$ and $a_{n,N}$ are related, e.g., as
\[\arrb{l} \dsp
\tru{2}{N} = a_{1,N}\,,
\ \ \ \tru{4}{N} = \frac{a_{1,N}^2 - a_{2,N}}{2}\,,
\ \ \ \tru{6}{N} = \frac{a_{1,N}^3}{3} - \frac{a_{1,N} \, a_{2,N}}{2}
+ \frac{a_{3,N}}{6}\,,
\\ \dsp
\ \ \ \tru{8}{N} = \frac{a_{1,N}^4}{4} - \frac{a_{1,N}^2 \, a_{2,N} }{2}
+ \frac{a_{2,N}^2}{8} + \frac{a_{1,N} \,
a_{3,N}}{6} - \frac{a_{4,N}}{24}\,.
\arre \]
%-----------------------------
% \subsection{Initial Condition}
%-----------------------------
For Ising measure $\den{0}=\den{I,s}$,
\eqnb
\eqna{a0}
a_{n,0}
= (-1)^n \frac{n!}{(2n)!} \frac{d^{2n}\cha{0}}{d \, \xi^{2n}}(0)
= \frac{n!}{(2n)!} \int x^{2n} \, \den{I,s}(x) dx
=\frac{n!}{(2n)!}s^{2n},
\ \ n\in\pintegers\,.
\eqne
Note that
one of the Newman inequalities (see \eqnu{newman3}),
or the Gaussian inequalities, imply that
\eqnb
\eqna{hnbd.1}
a_{n,N}\le a_{1,N}^n=\trup{2}{N}{n},
\ \ n\in\pintegers\,.
\eqne
Define $b_{n,N}$, $n\in\pintegers$, by
\[
({\cal S} \cha{N})(\xi) = \cha{N}(\frac{\sqrt{c}}{2} \xi )^2
= \sum_{n=0}^{\infty} (-1)^n \frac1{n!} b_{n,N} \xi^{2n},
\]
where ${\cal S}$ is in \eqnu{Strans}.
Then
\eqnb
\eqna{bn-rec}
b_{n,N} = \left( \frac{c}{4} \right)^n
\sum_{\ell =0}^n \binom{n}{\ell}\, a_{\ell,N} \, a_{n-\ell,N}\,,
\ \ n\in\pintegers\,.
\eqne
With \eqnu{hnbd.1} we have,
\eqnb
\eqna{anbd.1}
b_{n,N} \le \left( \frac{c \tru{2}{N}}{2} \right)^n,
\ \ n\in\pintegers\,.
\eqne
%-----------------------------
% \subsection{Numerical Calculation} \seca{sub-anbn}
%-----------------------------
Next define $\tilde{a}_{n,N}$, $n\in\pintegers$, by
\[
\sum_{m=0}^\infty \frac{1}{m!}
\left( - \frac{\beta}{2} \right)^m \frac{d^{2m}}{d \xi^{2m}}
{\cal S} \cha{N}\,(\xi)
= \sum_{n=0}^\infty (-1)^n \frac{1}{n!}
\tilde{a}_{n,N}\xi^{2n}.
\]
Then
\eqnb
\eqna{an-rec}
\tilde{a}_{n,N}
= \sum_{m=0}^{\infty} \left( \frac{\beta}{2} \right)^m \,
b_{m+n,N}
\frac{(2m +2n)! n!}{m! (m+n)! (2 n)!}\,,
\ \ n\in\pintegers\,,
\eqne
and \eqnu{Ttrans} implies
\[
\cha{N+1}(\xi)
= \frac1{\tilde{a}_{0,N}}
\sum_{n=0}^\infty (-1)^n\frac{1}{n!}
\tilde{a}_{n,N} \xi^{2n},
\]
where we fixed the constant in the definition of ${\cal T}$
by $\cha{N+1}(0)=1$\,.
Comparing this
with \eqnu{a} we obtain a recursion relation in $N$ for $a_{n,N}$:
\eqnb
\eqna{arec}
a_{n,N+1}=\frac{\tilde{a}_{n,N}}{\tilde{a}_{0,N}}\,,
\ \ n\in\pintegers,\ N\in\pintegers\,.
\eqne
%-----------------------------
\subsection{Truncation.}
%-----------------------------
\seca{sub-tr.sc}
We will evaluate a finite number, say $M$, of $a_{n,N}$'s
($n=1,2,\cdots,M$) explicitly with aid of computer calculations,
by evaluating $a_{n,N}$, $n>M$, ^^ theoretically'.
For this,
we need to give bounds of series in \eqnu{bn-rec} and \eqnu{an-rec}
in terms of sums of finite terms.
The following proposition serves for this purpose.
\prpb
\prpa{truncation}
Let $M$ be a positive integer, and define
\[\bll_{n,N}\,,\ \bu_{n,N}\,,
\ \ n=0,1,2,\cdots,2M, \]
and
\[\tilde{\al}_{n,N}\,,\ \tilde{\au}_{n,N}\,,\ \al_{n,N}\,,\ \au_{n,N}\,,
\ \ n=0,1,2,\cdots,M, \]
inductively in $N\in\pintegers$, by
\[ \al_{n,0}=\au_{n,0}=\frac{n!}{(2n)!} s^{2n},
\ \ n=0,1,2,\cdots,M, \]
and
\eqnb
\eqna{bn-rec2}
\bll_{n,N} = \left( \frac{c}{4} \right)^n \, \times
\left\{ \arrb{ll} \dsp
\sum_{\ell=0}^n \, \binom{n}{\ell} \al_{\ell,N} \, \al_{n-\ell,N}
\,, & 0\le n \le M,
\\ \AHFA \dsp
\sum_{\ell=n-M}^M \,\binom{n}{\ell} \al_{\ell,N} \, \al_{n-\ell,N}
\,, & M < n \le 2M,
\arre \right.
\eqne
%
\eqnb
\eqna{bu.4}
\bu_{n,N} =
\left\{ \arrb{ll} \dsp
\left( \frac{c}{4} \right)^n \,
\sum_{\ell =0}^n \, \binom{n}{\ell} \au_{\ell,N} \, \au_{n-\ell,N}
\,, & 0\le n \le M,
\\ \AHFA \dsp
\min \left\{
\left( \frac{c}{4} \right)^n \,
\sum_{n-M \le \ell \le M} \,\binom{n}{\ell} \au_{\ell,N} \, \au_{n-\ell,N}
+ \overline{\Delta \bu}_{n,N},
\ \left( \frac{c \au_{1,N}}{2} \right)^n\right\}
\,, & M < n \le 2M,
\arre \right.
\eqne
%
\eqnb
\eqna{an-rec3}
\tilde{\al}_{n,N}
= \sum_{m=0}^{2M-n} \left( \frac{\beta}{2} \right)^m \,
\bll_{m+n,N} \frac{(2m +2n)! n!}{m! (m+n)! (2 n)!}\,,
\ \ 0\le n \le M,
\eqne
%
\eqnb
\eqna{an-rec4}
\tilde{\au}_{n,N} =
\sum_{m=0}^{2M-n} \left( \frac{\beta}{2} \right)^m \,
\bu_{m+n,N} \frac{(2m +2n)! n!}{m! (m+n)! (2 n)!}
+ \overline{\Delta \au}_{n,N}\,,
\ \ 0\le n \le M,
\eqne
%
\eqnb
\eqna{atildea}
\al_{n,N+1} = \frac{\tilde{\al}_{n,N}}{\tilde{\au}_{0,N}}\,,
\ \ \ \au_{n,N+1} = \frac{\tilde{\au}_{n,N}}{\tilde{\al}_{0,N}}\,,
\ \ 1\le n \le M,
\eqne
and
\[ \al_{0,N+1}=\au_{0,N+1}=1, \]
where we put
\eqnb
\eqna{Delb-bd.8'}
\overline{\Delta \bu}_{n,N}=
2 \left( \frac{c \, \au_{1,N}}{4} \right)^n \,
\binom{n}{n-M-1} \times \frac{1}{1 - \frac{n-M}{M+1} e^{-1/(M+1)}}
\times \frac{\au_{M,N}}{\al_{1,N}^{M}}\,,
\eqne
and
\eqnb
\eqna{Dela-bd.8'}
\overline{\Delta \au}_{n,N}=
\left( \frac{1}{2 \beta} \right)^n \,
\frac{\left( \beta c \au_{1,N} \right)^{2M+1} }{1 - 2 \beta c \au_{1,N}}
\binom{N}{n} \times \frac{\au_{M,N}}{\al_{1,N}^M}\,.
\eqne
If for an integer $N_1$ it holds that
\eqnb
\eqna{apriorifortruncation}
\au_{1,N}<\frac1{2 \beta c}\,,\ \ 0\le N\le N_1\,,
\eqne
then $a_{n,N}$, $b_{n,N}$, $\tilde{a}_{n,N}$,
$n\in\pintegers$, $N\in\pintegers$,
defined inductively by
\eqnu{a0}, \eqnu{bn-rec}, \eqnu{an-rec}, \eqnu{arec},
satisfy, for all $N\le N_1$\,,
\eqab
&& \bll_{n,N}\le b_{n,N}\le \bu_{n,N}\,,\ \ n=0,1,2,\cdots,2M,
\nonumber \\
&& \tilde{\al}_{n,N}\le \tilde{a}_{n,N}\le \tilde{\au}_{n,N}\,,
\ \ n=0,1,2,\cdots,M,
\nonumber \\
\eqna{truncatedbound}
&& \al_{n,N}\le a_{n,N}\le \au_{n,N},
\ \ n=0,1,2,\cdots,M.
\eqae
\prpe
The rest of this subsection is devoted to a proof of this proposition.
\prfb
The claimed bounds on $a_{n,N}$ in \eqnu{truncatedbound} hold for $N=0$.
We proceed by induction on $N$, and assume that
they hold for $N$.
By comparing \eqnu{bn-rec} with
\eqnu{bn-rec2}, and noting that $a_{n,N}$ are non-negative,
we see that the lower bound for $b_{n,N}$ in \eqnu{truncatedbound}
holds.
Assume for a moment that the upper bound for $b_{n,N}$
in \eqnu{truncatedbound} also holds.
Then comparing \eqnu{an-rec} with
\eqnu{an-rec3}, we see
that the lower bound for $\tilde{a}_{n,N}$ in \eqnu{truncatedbound} holds.
If the upper bound for $\tilde{a}_{n,N}$ also holds,
then \eqnu{arec} and \eqnu{atildea} imply that
the bounds for $a_{n,N+1}$ in \eqnu{truncatedbound} also hold.
Hence we are left with proving
the upper bounds for $b_{n,N}$ and $\tilde{a}_{n,N}$ in
\eqnu{truncatedbound}.
%-----------------------------
\paragraph*{Upper bound on $b_{n,N}$\,.}
%-----------------------------
%\seca{sub-bn-bd}
Note first that if $n\le M$, then
\[ b_{n,N}=
\left( \frac{c}{4} \right)^n \,
\sum_{\ell =0}^n \, \binom{n}{\ell} a_{\ell,N} \, a_{n-\ell,N}
\le
\left( \frac{c}{4} \right)^n \,
\sum_{\ell =0}^n \, \binom{n}{\ell} \au_{\ell,N} \, \au_{n-\ell,N}
=\bu_{n,N}\,, \]
hence $b_{n,N}\le \bu_{n,N}$ holds.
Also, \eqnu{anbd.1} implies
\[
b_{n,N} \le \left( \frac{c \tru{2}{N}}{2} \right)^n
\le \left( \frac{c \au_{1,N}}{2} \right)^n, \]
hence it suffices to prove
\eqnb
\eqna{bu.8}
b_{n,N} \le
\left( \frac{c}{4} \right)^n \,
\sum_{n-M \le \ell \le M} \,\binom{n}{\ell} \au_{\ell,N} \, \au_{n-\ell,N}
+ \overline{\Delta \bu}_{n,N},
\ \ M < n \le 2M.
\eqne
To prove \eqnu{bu.8}, first note
\eqab
\Delta \bu_{n,N}&=&
b_{n,N}
-\left( \frac{c}{4} \right)^n \,
\sum_{n-M \le \ell \le M} \,
\binom{n}{\ell} \au_{\ell,N} \, \au_{n-\ell,N}
\nonumber \\
&\le&
\left( \frac{c}{4} \right)^n \,
\sum_{0\le\ell < n-M \; {\rm or} \; M < \ell\le n} \,
\binom{n}{\ell} a_{\ell,N} \, a_{n-\ell,N}\,.
\eqna{bu.6}
\eqna{delb-def}
\eqae
Using the Newman inequalities \eqnu{newman3}
we see that if $\ell >M$
\eqnb
\eqna{au-apri.imp1}
a_{\ell,N} \le a_{M,N} a_{\ell-M,N} \le a_{M,N} \, a_{1,N}^{\ell-M}.
\eqne
Hence
\eqab
\Delta \bu_{n,N}
&\le&
\left( \frac{c}{4} \right)^n \,
\left(
\sum_{0\le\ell < n-M}
\binom{n}{\ell} a_{\ell,N} \, a_{M,N} \, a_{1,N}^{n-\ell-M}
+
\sum_{M < \ell\le n}
\binom{n}{\ell} a_{M,N} \, a_{1,N}^{\ell-M} a_{n-\ell,N}
\right)
\nonumber \\
&\le&
2
\left( \frac{c\,a_{1,N}}{4} \right)^n \,
\frac{a_{M,N}}{a_{1,N}^{M}} \sum_{\ell=0}^{n-M-1} \binom{n}{\ell}\,,
\eqna{bn-err.1}
\eqae
where we also used \eqnu{anbd.1}.
Write the summation in the right hand side as
\eqab
\sum_{\ell =0}^{n-M-1} \binom{n}{\ell}
&=& \binom{n}{n-M-1} \, \left[
1 + \frac{n-M-1}{M+2} + \frac{n-M-1}{M+2} \frac{n-M-2}{M+3}
\right.
\nonumber \\
&&
\left.
+ \frac{n-M-1}{M+2} \frac{n-M-2}{M+3} \frac{n-M-3}{M+4}
+ \cdots
\right].
\eqna{factor1}
\eqae
Noting that
\eqnb
\frac{a-x}{1+x} \le a e^{-2x},\ \ a \in (0, 1],\ x \in [0,1],
\eqne
we find,
by putting
$\dsp a = \frac{n-M}{M+1}$ and $\dsp \epsilon = \frac{1}{M+1}$\,,
\eqnb
\frac{n-M-k}{M+k+1}
= \frac{a - k \epsilon}{1 + k \epsilon}
\le a \, e^{-2 k \epsilon}.
\eqne
Hence \eqnu{factor1} has a bound
\[
\sum_{\ell =0}^{n-M-1} \binom{n}{\ell}
\le
\binom{n}{n-M-1} \times\sum_{k=0}^\infty a^k \, e^{-k(k+1) \epsilon}
\le \binom{n}{n-M-1} \times \frac{1}{1 - a e^{-\epsilon}}\,,
\ \ a = \frac{n-M}{M+1}\,,\ \epsilon = \frac{1}{M+1}\,,
\]
which implies
\eqnb
\eqna{Delb-bd.8}
\Delta \bu_{n,N} \le \,
\overline{\Delta \bu}_{n,N}\,,
\eqne
where $\overline{\Delta \bu}_{n,N}$ is defined in \eqnu{Delb-bd.8'}.
This proves \eqnu{bu.8}.
%-----------------------------
\paragraph*{Upper bound on $\tilde{a}_{n,N}$\,.}
%-----------------------------
% \seca{sub-an-bd}
Put
\eqab
\Delta \au_{\ell,N} &=&
\tilde{a}_{\ell,N}-
\sum_{m=0}^{2M-\ell} \left( \frac{\beta}{2} \right)^m \,
\bu_{m+\ell,N} \frac{(2m +2\ell)! \ell!}{m! (m+\ell)! (2 \ell)!}
\nonumber \\ &\le&
\sum_{m=2M+1-\ell}^{\infty}
\left( \frac{\beta}{2} \right)^m \,
b_{m+\ell,N}
\frac{(2m +2\ell)! \ell !}{m! (m+\ell)! (2 \ell)!}
\nonumber \\ &=&
\sum_{m=2M+1-\ell}^{\infty}
\left( 2 \beta \right)^m \,b_{m+\ell,N}
\frac{(2m +2\ell-1)!!}{(2m)!! \, (2 \ell-1)!!}\,.
\eqna{Dela-def}
\eqna{Dela-def2}
\eqae
Using \eqnu{au-apri.imp1} and \eqnu{anbd.1}, we see that if $n >2M$
\eqnb
b_{n,N}
= \left( \frac{c}{4} \right)^n \,
\sum_{\ell =0}^n \, \binom{n}{\ell} a_{\ell,N} \, a_{n-\ell,N}
\le \left( \frac{c}{4} \right)^n \,
\sum_{\ell =0}^n \, \binom{n}{\ell} a_{1,N}^n \,
\times
\frac{a_{M,N}}{a_{1,N}^M}
= \left( \frac{c\,a_{1,N}}{2} \right)^n \,\frac{a_{M,N}}{a_{1,N}^M}
\eqne
Therefore
\eqab
\Delta \au_{\ell,N}
& \le& \frac{a_{M,N}}{a_{1,N}^M}
\left( \frac{c\, a_{1,N}}{2} \right)^\ell \,
\sum_{m=2M+1-\ell}^{\infty}
\left( \beta c\,a_{1,N} \right)^m \,
\frac{(2m +2\ell-1)!!}{(2m)!! \, (2 \ell-1)!!}
\nonumber \\
&\le&
\frac{a_{M,N}}{a_{1,N}^M}
\left( \frac{c\, a_{1,N}}{2} \right)^\ell \,
\sum_{m=2M+1-\ell}^{\infty}
\left( \beta c\, a_{1,N} \right)^m \,
\binom{m+\ell}{\ell}
\nonumber \\
& =&
\frac{a_{M,N}}{a_{1,N}^M}
\left( \frac{c\, a_{1,N}}{2} \right)^\ell \,
\left( \beta c\, a_{1,N} \right)^{2M+1 - \ell} \,
\sum_{k=0}^{\infty}
\left( \beta c\, a_{1,N} \right)^k \,
\binom{2M+1+k}{\ell}
\nonumber \\
&=&
\frac{a_{M,N}}{a_{1,N}^M}
\left( \frac{1}{2 \beta} \right)^\ell \,
\left( \beta c\, a_{1,N} \right)^{2M+1} \,
\sum_{k=0}^{\infty}
\left( \beta c\, a_{1,N} \right)^k \,
\binom{2M+1+k}{\ell}\,.
\eqna{Dela-bd.2}
\eqae
Here,
\[
T_{2M+1, \ell}(r) =
\sum_{k=0}^{\infty}
\left( \beta c\, a_{1,N} \right)^k \,
\binom{2M+1+k}{\ell}
=\sum_{k=0}^{\infty}
r^k \,
\binom{2M+1+k}{\ell}
=\frac{1}{1-r} \sum_{m=0}^\ell \binom{2M+1}{\ell - m} q^m ,
\]
where $r= \beta c\, a_{1,N}$\,,
and $q = \frac{r}{1-r}$\,. By assumption $r < \frac12$\,.
The binomial coefficient in the summand is largest when $m=0$,
because $2M+1>2M\ge 2\ell$.
Therefore,
\eqnb
T_{2M+1, \ell}(r)
\le \frac{1}{1-r} \binom{2M+1}{\ell} \sum_{m=0}^{\ell} q^m
\le \frac{1}{1-r} \frac{1}{1-q} \binom{2M+1}{\ell}
= \frac{1}{1 - 2 r} \binom{2M+1}{\ell}\,.
\eqne
This proves
\eqnb
\eqna{Dela-bd.8}
\Delta \au_{\ell,N} \le
\left( \frac{1}{2 \beta} \right)^\ell \,
\frac{\left( \beta c\,a_{1,N} \right)^{2M+1} }{1 - 2 \beta c\,a_{1,N}}
\binom{2M+1}{\ell} \times \frac{a_{M,N}}{a_{1,N}^M}
\le \overline{\Delta \au}_{\ell,N}\,,
\eqne
where $\overline{\Delta \au}_{\ell,N}$ is defined in \eqnu{Dela-bd.8'}.
This proves
$\tilde{a}_{n,N}\le \tilde{\au}_{n,N}$\,.
\qed\prfe
\remb
We can ^^ improve' \prpu{truncation} by
employing (correct) bounds, in a similar way as
the term proportional to $\dsp \left( \frac{c \au_{1,N}}{2} \right)^n$
in \eqnu{bu.4}.
In actual calculations, we improve $\au_{n,N+1}$,
$n=1,2,\cdots,M$, in \eqnu{atildea},
the upper bounds for $a_{n,N+1}$'s,
using \eqnu{newman3} (as well as its special case \eqnu{anbd.1}).
To be more specific, we compare $\au_{4,N+1}$ in \eqnu{atildea}
with $\au_{2,N+1}^2$ and replace the definition if the latter is smaller.
Then we go on to ^^ improve' $\au_{6,N+1}$ by comparing
with $\au_{2,N+1} \au_{4,N+1}$, and so on.
Conceptually there is nothing really new here, but this
procedure improves the actual value of the bounds in \prpu{truncation}.
\reme
%-----------------------------
%\subsection{Explicit bounds from computer results.}
\subsection{Computer results.}
%-----------------------------
\seca{sec-num}
In this subsection we prove \thmu{strongcoupling} on computers using
\prpu{truncation}.
We double checked by Mathematica and
\verb|C++| programs on interval arithmetic.
Here we will give results from \verb|C++| programs.
Our program employs interval arithmetic, which gives rigorous bounds
numerically. The idea is to express a number by a pair of
^^ vector', which consists of an array of length $M$ of ^^ digits',
taking values in $\{0,1,2,\cdots,9\}$,
and an integer corresponding to ^^ exponent'.
To give a simple example, let $M=2$.
One can view that $0.0523$ is expressed on the program,
for example, as
$I_1=[5.2\times 10^{-2},5.3\times 10^{-2}]$,
and $3$ is expressed as
$I_2=[3.0\times 10^{0},3.0\times 10^{0}]$.
When the division $I_1/I_2$ is performed, our program routines
are so designed that they give correct bounds as an output.
Namely, the computer output of $I_1/I_2$ will be
$[1.7\times 10^{-2},1.8\times 10^{-2}]$.
We may occasionally lose the best possible bounds, but
the program is so designed that we never
lose the correctness of the bounds.
Thus all the outputs are rigorous bounds of the corresponding
quantities.
In actual calculation we took $M=70$ digits, which
turned out to be sufficient.
We also note that interval arithmetic is employed in \cite{kw4}
for hierarchical model in $d=3$ dimensions.
We took independent approach in programming ---
we focused on ease in implementing the interval arithmetic to
main programs developed for standard floating point calculations ---
so that structure and details of the programs are quite different.
However, our numerical calculations are `not that heavy' to require
anything special.
As will be explained below, we only need to consider 2 values for
the initial Ising parameter $s$:
\[s_{-}=1.7925671170092624,\ \mbox{ and }\ s_{+}=1.7925671170092625. \]
We perform explicit recursion on computers for each $s=s_{\pm}$
using \prpu{truncation}.
We summarize what is left to be proved:
\itmb
\item
$\dsp \au_{1,N}<\frac1{2 \beta c}$\,,
$0\le s\le s_{N_1}$\,, $0\le N\le N_1$\,, where $N_1=100$\,.
This condition is from \eqnu{apriorifortruncation},
imposed because we are going to do evaluation using
\prpu{truncation}.
Note that this condition is stronger than \eqnu{aprioriforODE}
in the assumptions in \thmu{strongcoupling}, because
$\dsp\frac1{2 \beta c}=\frac12(2+\sqrt2)=1.707\cdots$ for $d=4$\,.
\item
$\dsp s_{-}\le \underline{s}_{\,N_1}$ and
$\overline{s}_{\,N_1}\le s_{+}$\,.
To prove this, it is sufficient (as seen from the definitions
\eqnu{s0} and \eqnu{s1}) to prove
\eqnb
\eqna{spm}
\tru{2}{N_1}< 1,\ \mbox{ when }\ s=s_{-}\,,
\ \ \ \mbox{ and }\ \
\tru{2}{N_1}>1+\frac3{\sqrt2}\tru{4}{N_1}\,,\ \mbox{ when }\ s=s_{+}\,.
\eqne
\item
For any $s$ satisfying $s_{-}\le s\le s_{+}$, the bounds
\eqab
\eqna{final1}
(0 \ \le)& \tru{4}{N_0} &\le\ 0.0045,
\\
\eqna{final2}
1.6 \trup{4}{N_0}{2} \ \le& \tru{6}{N_0} &\le\ 6.07 \trup{4}{N_0}{2},
\\
\eqna{final3}
(0 \ \le)& \tru{8}{N_0} &\le\ 48.469 \trup{4}{N_0}{3},
\eqae
hold for $N_0=70$\,.
This condition comes from the assumptions in \thmu{strongcoupling}
(sufficient, if $\dsp s_{-}\le \underline{s}_{\,N_1}$ and
$\overline{s}_{\,N_1}\le s_{+}$).
\itme
We now summarize our results from explicit calculations.
\itmb
\item
We have $\dsp \au_{1,N}\le \frac12s_+^2=1.6066\cdots$,
$0\le s\le s_{+}$, $0\le N\le N_1$\,.
The largest value for $\au_{1,N}$ in the range of parameters
is actually obtained at $s=s_+$ and $N=0$\,.
\item
Our calculations turned out to be accurate to obtain more than 40 digits
below decimal point correctly for $\tru{2}{100}$ and $\tru{4}{100}$
at $s=s_{\pm}$, which is more than enough to prove \eqnu{spm}.
In fact, we have
\[\arrb{l}
0.99609586499804791366176669341357334889503943
\le \al_{1,100}
\\ \le \tru{2}{100} \le \au_{1,100} \le
0.99609586499804791366176669341357334889503972,
\\ \ \mbox{ at }\ s=s_-\,,\arre \]
and
\[\arrb{l}
1.0131857903720691722396611098376636943838027
\le \al_{1,100}
\\ \le \tru{2}{100} \le \au_{1,100} \le
1.0131857903720691722396611098376636943838031,
\hbox{\vrule height0pt depth7pt width0pt}
\\
0.00281027097809098768088795100753480139767915
\le \frac12 (-\au_{2,100}+\al_{1,100}^2)
\\ \le \tru{4}{100} \le \frac12 (-\al_{2,100}+\au_{1,100}^2) \le
0.00281027097809098768088795100753480139767969,
\\ \ \mbox{ at }\ s=s_+\,.
\arre \]
\item
To prove \eqnu{final1} -- \eqnu{final3},
we note the following.
Let us write the $s$ dependences of $a_{n,N}$ and $\tru{n}{N}$
explicitly like $a_{n,N}(s)$ and $\tru{n}{N}(s)$.
For any integer $N$ and for any $s$ satisfying $s_{-}\le s\le s_{+}$,
the monotonicity of $a_{n,N}(s)$ with respect to $s$ implies
\eqnb
\eqna{pmbound}
\tru{4}{N}(s)= \frac12 (-a_{2,N}(s)+a_{1,N}(s)^2)
\le \frac12 (-a_{2,N}(s_-)+a_{1,N}(s_+)^2)=:\bar\mu_{4,N}\,.
\eqne
Hence if we can prove
\[ \bar\mu_{4,70}\le 0.0045, \]
then we have proved \eqnu{final1}.
In a similar way, sufficient conditions for \eqnu{final2} and \eqnu{final3}
are
\[
1.6\le \frac{\underline\mu_{6,70}}{\bar\mu_{4,70}^2}\,,
\ \ \ \frac{\bar\mu_{6,70}}{\underline\mu_{4,70}^2}\le 6.07\,,
\ \ \ \frac{\bar\mu_{8,70}}{\underline\mu_{4,70}^3}\le 48.469\,,
\]
with obvious definitions (as in \eqnu{pmbound} for
$\bar\mu_{4,N}$)
for $\underline\mu_{n,70}$ and $\bar\mu_{n,70}$\,.
The bounds we have for these quantities are
(we shall not waste space by writing too much digits):
\[
\bar\mu_{4,70} \le 0.004144,
\ \ 3.6459 \le \frac{\underline\mu_{6,70}}{\bar\mu_{4,70}^2}\,,
\ \ \frac{\bar\mu_{6,70}}{\underline\mu_{4,70}^2}\le 3.7542,
\ \ \frac{\bar\mu_{8,70}}{\underline\mu_{4,70}^3}\le 38.488.
\]
\itme
This completes a proof of \thmu{strongcoupling},
and therefore \thmu{main} is proved.
% ACKNOWLEDGEMENTS.
\section*{Acknowledgements}
The authors would like to thank Prof.~Y.~Takahashi for his
interest in the present work and for discussions.
Part of this work was done while T. Hara was at
Department of Mathematics, Tokyo Institute of Technology.
The researches of T.~Hara and T.~Hattori are partially supported
by \kakenhi\ of \monbushou.
% \pagebreak\par
%-----------------------------
\appendix
%-----------------------------
%-----------------------------
%\section{Newman's inequalities for moments of measures
%obeying Lee-Yang Theorem.}
\section{Newman's inequalities.}
%-----------------------------
\seca{Newman}
\label{sub-obs}
\label{sec-Taylor}
Let $X$ be a stochastic variable
which is in class ${\cal L}$ of \cite{newman}.
$X\in{\cal L}$ has Lee-Yang property, which states that the zeros of
the moment generating function $\EE{e^{H X}}$ are pure imaginary.
In fact, it is shown in \cite[Proposition~2]{newman} using Hadamard's
Theorem that $\EE{e^{H X}}$ has a following expression:
\eqnb\eqna{LeeYang}
\EE{e^{H X}} = e^{b H^2} \,
\prod_{j}\left(1 + \frac{H^2}{\alpha_j^2}\right),
\eqne
where $b$ is a non-negative constant and $\alpha_j$, $j=1,2,3,\cdots$,
is a positive nondecreasing sequence satisfying
$\dsp\sum_{j=1}^{\infty} \alpha_j^{-2} < \infty$.
Consequences of \eqnu{LeeYang}
in terms of inequalities among moments ($n$ point functions)
are given in \cite{newman}, among which we note the following.
\ittb
\item[1.~Positivity {\protect\cite[Theorem~3]{newman}.} ]
Put
\eqnb\eqna{mu}
\tru{2n}{}= - \frac1{(2n)!}
\left.\dfrac{d^{2n}}{d\xi^{2n}} \log \EE{e^{\II \xi X}}\right|_{\xi=0}\,.
\eqne
Then,
\eqnb\eqna{nonnegative}
\tru{2n}{} \ge 0,\ n=0,1,2,\cdots.
\eqne
(Note that \eqnu{LeeYang} implies $\tru{2n+1}{}=0$\,.)
\item[2.~Newman's bound {\protect\cite[Theorem~6]{newman}.}]
Put $v_{2n}= n \tru{2n}{}$. Then,
\eqnb\eqna{un-rel1}
v_{4n} \le v_{4}^{n},
\ \ v_{6} \le \sqrt{v_{4}v_{8}},
\ \ v_{4n+2} \le v_{6} \, v_{4}^{n-1} ,
\eqne
where the first and third inequalities follow from (2.10) of \cite{newman},
while the second one is (2.12) of \cite{newman}.
These imply $v_{2n} \le v_{4}^{n/2}$, $n \ge 2$, and therefore
\eqnb\eqna{un-rel2}
\tru{2n}{} \le \frac{(2 \tru{4}{})^{n/2}}{n}\,,\ n=2,3,4,\cdots.
\eqne
\itte
Furthermore, we will prove the following.
\prpb
Put $\dsp a_N=\dfrac{N!}{(2N)!} \EE{X^{2N}}$\,, $N\in \pintegers$. Then,
\eqnb
\eqna{newman3}
a_{M+N} \le a_M \, a_N \ \ N,M=0,1,2,\cdots.
\eqne
\prpe
\prfb
Put $y_j=\alpha_j^{-2} >0$\,. Then
\eqnb
\EE{e^{H X}} = e^{b H^2} \, \prod_{j} \left( 1 + H^2 \, y_j \right).
\eqne
Expand the infinite product to obtain
\eqnb
\prod_{j}
\left ( 1 + H^2 \, y_j \right)
= 1 + H^2 \sum_j y_j + \frac{H^4}{2!} \sum_{i, j}{}' y_i y_j
+ \frac{H^6}{3!} \sum_{i, j, k}{}' y_i y_j y_k
+ ...
= \sum_{n=0}^\infty \frac{H^{2n}}{n!} c_n ,
\eqne
with
\eqnb
\eqna{cn-def}
c_n = \sum_{i_1, i_2, ..., i_n}{}'
y_{i_1} y_{i_2}y_{i_3} ... y_{i_n}\,,
\eqne
where primed summations denote summations over non-coinciding indices.
Hence we have,
\eqnb
\EE{e^{H X}}
= \sum_{N=0}^\infty H^{2N} \sum_{m,n: m+n = N}
\frac{b^m}{m!} \frac{c_n}{n!}
= \sum_{N=0}^\infty H^{2N} \sum_{n= 0}^N
\frac{b^{N-n}}{(N-n)!} \frac{c_{n}}{n!}\,.
\eqne
Comparing with
$\dsp \EE{e^{H X}} = \sum_{N=0}^\infty \frac{a_N}{N!} H^{2N}$,
we obtain
\[ a_N = \sum_{n= 0}^N \binom{N}{n} b^{N-n} c_{n}\,. \]
Note that \eqnu{cn-def} implies
\eqnb
c_{n+m} \le c_m c_n\,,
\eqne
because the conditions of primed summations are weaker for
the left hand side. This with $b\ge 0$ implies
\eqsb
a_M \, a_N
&=&
\sum_{m= 0}^M \sum_{n= 0}^N \binom{M}{m} \binom{N}{n}
b^{M+N-m-n} \, c_{m} \, c_{n}
\\
&\ge&
\sum_{m= 0}^M \sum_{n= 0}^N \binom{M}{m} \binom{N}{n}
b^{M+N-m-n} \, c_{m+n}
\\
&=&
\sum_{\ell =0}^{M+N} b^{M+N-\ell} \, c_{\ell}
\sum_{\shortstack{
$m: 0 \le m \le M,$ \\
$\; \; 0 \le \ell - m \le N$}}^\ell
\binom{M}{m} \binom{N}{\ell-m}
\\
&=& \sum_{\ell =0}^{M+N} b^{M+N-\ell} \, c_{\ell}
\binom{M+N}{\ell}
= a_{M+N}\,,
\eqse
where, in the last line, we also used
\eqnb
\sum_{\shortstack{$m:\;0\le m\le M$,\\$\ 0\le\ell -m\le N$}}^\ell
\binom{M}{m} \binom{N}{\ell-m}
=\binom{M+N}{\ell}\,,
\eqne
which is seen to hold if we compare the coefficients of $x^{\ell}$
of an identity
$\dsp (1+x)^{M+N} = (1+x)^M (1+x)^N$.
\qed\prfe
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\end{document}
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1 -15 rmoveto (N) show 0 -7 rmoveto (1) show
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210 430 moveto (s=s) show 0 -7 rmoveto (c) show
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400 -23 moveto (m) show 0 -9 rmoveto (2) show
-35 395 moveto (m) show 0 -9 rmoveto (4) show
} def
/critical
{ 228 400 moveto
96 0 lineto
[4 4] 0 setdash
1.3 setlinewidth
stroke
} def
/calx {dup 132 mul 400 div 96 add exch 1500 exch div sub} def
/calX {dup 132 mul 400 div 96 add exch 1500 exch div add} def
/upper
{10
78 {5 add dup calX exch dup} repeat
pop
newpath
moveto 77 {lineto} repeat
[] 0 setdash
stroke
} def
/lower
{10
39 {10 add dup calx exch dup} repeat
pop
newpath
moveto 38 {lineto} repeat
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stroke
} def
/drawcircle
{dup 3 -1 roll dup 4 1 roll exch
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3 0 360 arc 1 setgray fill 0 setgray
} def
/drawcircles
{95 67 drawcircle
140 67 drawcircle
179 269 drawcircle
190 268 drawcircle
96 0 drawcircle
} def
/drawarrow
{
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newpath
228 425 moveto 0 -20 rlineto stroke
228 405 moveto -3 7 rlineto 6 0 rlineto closepath fill
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100 100 translate
makeframe
drawregion
printtext
critical
upper
lower
drawcircles
drawarrow
showpage
%%EOF
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#!/bin/csh -f
# Uuencoded gz-compressed .tar file created by csh script uufiles
# For more info (11/95), see e.g. http://xxx.lanl.gov/faq/uufaq.html
# If you are on a unix machine this file will unpack itself: strip
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# (uudecode ignores these header lines and starts at begin line below)
# Then say csh c-source.uu
# or explicitly execute the commands (generally more secure):
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# tar -xvf c-source.tar
# On some non-unix (e.g. VAX/VMS), first use editor to change filename
# in "begin" line below to c-source.tar-gz , then execute
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# gzip -d c-source.tar-gz
# tar -xvf c-source.tar
#
uudecode $0
chmod 644 c-source.tar.gz
gunzip -c c-source.tar.gz | tar -xvf -
rm $0 c-source.tar.gz
exit
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end
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