Content-Type: multipart/mixed; boundary="-------------0010060258497" This is a multi-part message in MIME format. ---------------0010060258497 Content-Type: text/plain; name="00-397.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="00-397.keywords" Ising Model, Hierarichical Model, Triviality, Renormalization Group, Gaussian fixed point, Critical trajectory, Newman's inequality ---------------0010060258497 Content-Type: application/x-tex; name="paper-mparc.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="paper-mparc.tex" \documentclass{article} \newif\ifmparc\mparctrue %\mparcfalse \newif\ifwithoutams\withoutamsfalse \ifwithoutams \usepackage{latexsym} \else \usepackage{amsmath,amssymb,euscript} \fi \usepackage[dvips]{graphicx} \setlength{\textwidth}{6.4in} \setlength{\textheight}{9.5in} \setlength{\topmargin}{0in} \setlength{\headheight}{0in}\setlength{\headsep}{0in} 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of AMS-LaTeX. \def\eqab{\begin{eqnarray}}\def\eqae{\end{eqnarray}} \def\eqb{\begin{eqnarray}}\def\eqe{\end{eqnarray}} \newcommand{\eqarray}{\begin{eqnarray}}\newcommand{\enarray}{\end{eqnarray}} \newcommand{\nnb}{\nonumber \\} \def\eqsb{\begin{eqnarray*}}\def\eqse{\end{eqnarray*}} \newcommand{\eqarrstar}{\begin{eqnarray*}} \newcommand{\enarrstar}{\end{eqnarray*}} \def\arrb#1{\begin{array}{#1}}\def\arre{\end{array}} \def\tbla#1{\label{tb:#1}}\def\tblu#1{Table~\ref{tb:#1}} \def\prpl#1{\label{p:#1}}\def\prpr#1{Proposition~\ref{p:#1}} \def\prpa#1{\label{p:#1}}\def\prpu#1{Proposition~\ref{p:#1}} \def\leml#1{\label{l:#1}}\def\lemr#1{Lemma~\ref{l:#1}} \def\lema#1{\label{l:#1}}\def\lemu#1{Lemma~\ref{l:#1}} \def\thml#1{\label{t:#1}}\def\thmr#1{Theorem~\ref{t:#1}} \def\thma#1{\label{t:#1}}\def\thmu#1{Theorem~\ref{t:#1}} \def\corl#1{\label{c:#1}}\def\corr#1{Corollary~\ref{c:#1}} \def\cora#1{\label{c:#1}}\def\coru#1{Corollary~\ref{c:#1}} \def\secl#1{\label{s:#1}}\def\secr#1{Section~\ref{s:#1}} 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\newcommand{\bu}{\Bar{b}} \newcommand{\bll}{\underline{b}} % without amsmath \ifwithoutams \newcommand{\dfrac}[2]{\frac{\dsp #1}{\dsp #2}} \newcommand{\Bar}[1]{\overline{#1}} \def\binom#1#2{\left(\raisebox{-0.5\baselineskip}{% \shortstack{$\dsp #1$ \\ $\dsp #2$}}\right)} \fi \def\EE#1{{\rm E}\left[\;#1\;\right]} \newcommand{\II}{\sqrt{-1}} \def\kakenhi{Grant-in-Aid for Scientific Research (C)} \def\monbushou{the Ministry of Education, Science, Sports and Culture} \def\rg{renormalization group} \def\tru#1#2{\def\truone{#2}\ifx\truone\empty\mu_{#1}\else\mu_{#1,#2}\fi} % truncated #1-point function (at #2nd iteration). \def\trup#1#2#3{\def\truone{#2}\ifx\truone\empty\mu^{#3}_{#1} \else\mu^{#3}_{#1,#2}\fi} % truncated #1-point function (at #2nd iteration)(with power #3). \def\den#1{h_{#1}} % density function at #1st iteration. \def\cha#1{\hat{h}_{#1}} % characteristic function at #1st iteration. \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 % REFERENCES. \begin{thebibliography}{99} \def\CMP{Commun.\ Math.\ Phys.\ } \def\JSP{Journ.\ Stat.\ Phys.\ } \bibitem{BS1} P.~M.~Bleher, Ya.~G.~Sinai, {\it Investigation of the critical point in models of the type of Dyson's hierarchical model,} \CMP {\bf 33} (1973) 23--42. \bibitem{BS2} P.~M.~Bleher, Ya.~G.~Sinai, {\it Critical indices for Dyson's asymptotically hierarchical models,} \CMP {\bf 45} (1975) 247--278. \bibitem{ce} P.~Collet, J.-P.~Eckmann, {\it A renormalization group analysis of the hierarchical model in statistical physics,} Springer Lecture Note in Physics {\bf 74} (1978). \bibitem{dyson} F.~J.~Dyson, {\it Exisitence of a phase--transision in a one--dimensional Ising ferromagnet,} \CMP {\bf 12} (1969) 91--107. \bibitem{gallavotti} G.~Gallavotti, {\it Some aspects of the renormalization problems in statistical mechanics,} Memorie dell' Accademia dei Lincei {\bf 15} (1978) 23--59. \bibitem{gk1} K.~Gaw\c{e}dzki, A.~Kupiainen, {\it Triviality of $\phi^4_4$ and all that in a hierarchical model approximation,} \JSP {\bf 29} (1982) 683--699. \bibitem{gk2} K.~Gaw\c{e}dzki, A.~Kupiainen, {\it Non-Gaussian fixed points of the block spin transformation. 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Suppl.){\bf 73} (1999) 763--765. \bibitem{numerical3} J.~J.~Godina, Y.~Meurice, M.~B.~Oktay, {\it High-accuracy calculations of the critical exponents of Dyson's hierarchical model,} Physical Review D {\bf 59} (1999) 096002 (13pp.). \fi \bibitem{kw} H.~Koch, P.~Wittwer, {\it A non-Gaussian renormalization group fixed point for hierarchical scalar lattice field theories,} \CMP {\bf 106} (1986) 495--532. \bibitem{kw2} H.~Koch, P.~Wittwer, {\it On the renormalization group transformation for scalar hierarchical models,} \CMP {\bf 138} (1991) 537--568. \bibitem{kw3} H.~Koch, P.~Wittwer, {\it A nontrivial renormalization group fixed point for the Dyson--Baker hierarchical model,} \CMP {\bf 164} (1994) 627--647. \bibitem{kw4} H.~Koch, P.~Wittwer, {\it Bounds on the zeros of a renormalization group fixed point,} Mathematical Physics Electronic Journal {\bf 1} (1995) No.~6 (24pp.). \bibitem{newman} C.~M.~Newman {\it Inequalities for Ising models and field theories which obey the Lee--Yang theorem,} \CMP {\bf 41} (1975) 1--9. \bibitem{sinai} Ya.~G.~Sinai, {\it Theory of phase transition: rigorous results,} Pergamon Press, 1982. \end{thebibliography} \end{document} ---------------0010060258497 Content-Type: application/postscript; 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