\magnification=\magstep1 \hsize=16.5truecm \input amstex \documentstyle{amsppt} \define\Om#1{\Omega_n^#1} \define\fn{f_{n,N}^{h_N}(x)} \define\mn{\mu_{n,N}^{h_N}} \define\mk{\mu_{k,N}^{h_N}} \define\gb{\bar g } \define\lhn{L_n(N,h_N)} \define\ab{\bar A} \define\e{\varepsilon} \define\rn{(R^2)^{2^n}} \define\rk{(R^2)^{2^k}} \define\co#1{c^{#1n}} \define\im#1{\int_{\Om#1}} \define\fformg{\vf\in\Gamma_n(r,x)} \define\s#1{S_n^{#1}f(x)} \define\obb{\hat\Omega_n^2} \define\ebb{\hat\varepsilon} \define\Omm#1{\Omega_{n+1}^#1} \define\eb{\bar\e } \define\ebx{\bar\e_n(x)} \define\vf{\varphi} \define\formg{\{\vf\in\Gamma_n(r,x)\}} \define\x#1{x^{(#1)}} \define\y#1{y^{(#1)}} \define\xo#1{x^{(2)#1}} \define\xy{\frac{x+y}2} \define\ca{(\frac{c^n}T+\frac{A_{n+1}}2)} \define\cga{(\frac{c^n}{2T}+\frac{g_{n+1}}{4M}-\frac{A_{n+1}}4)} \define\je{J_{n,\eb}} \define\jeb{\bar J_{n,\eb}} \define\gnx{g_n(\x1-M)+A_n\xo2} \define\ga{\frac{g_n}{2M}-A_n} \define\te{T_n^{\e}f(x)} \define\gan{\frac{g_{n+1}}{2M}-A_{n+1}} \define\ssb{\bar\sigma} \define\zz{\bold Z} \define\mb{\bar \mu} \define\zzd{\bold Z^d} \define\yn{Y_k(N)} \define\de{\delta} \leftheadtext\nofrills{{\rightline{\rm P. M. Bleher and P. Major}}} \rightheadtext\nofrills{\leftline{\rm The large-scale limit of Dyson's hierarchical model}} \topmatter \title The large-scale limit of Dyson's hierarchical\\ vector-valued model at low temperatures. \\ The non-Gaussian case. \\ \rm Part II. Description of the large-scale limit. \endtitle \author $^1$Bleher, P. M. and $^2$Major, P. \endauthor \affil $^1$Keldysh Institute of Applied Mathematics, Moscow A--47 Miusskaya Square 4, U.S.S.R.\\ $^2$Mathematical Institute of the Hungarian Academy of Sciences, Budapest H--1364 P.O.B. 127, Hungary \endaffil \endtopmatter \subheading{1. Introduction} In this paper we investigate the large-scale limit of the equilibrium state of Dyson's hierarchical vector valued $p$ dimensional, $p\ge2$, model with parameter $c$, $10 \right\}$. In this work we consider this function as the solution of equation (1.5). It is a density function which is positive for all~ $x$. Since the function $p_n(x,t)$ depends on $x$ only through $|x|$ we can define a function $\bar p_n(z)= \bar p_n(z,t,T)$, \ $z\in R^1$, such that $p_n(x)=\bar p_n (|x|)$ for all $x\in R^p$. Now we formulate the following \proclaim{Theorem A}\it If $10$ such that for all $00$ and $n_0=n_0(c,T,t)>0$ such that for $n>n_0$ $$ \align &c^{-n}p_n(x,T)=c^{-n}\bar p_n(|x|,T) \tag1.6\\&\quad=B\exp\left\{-\frac{a_0c^n}T M(|x|-M)\right\}g\left(\frac{a_1c^n}T M(|x|-M)\right)(1+r_n(x)) \endalign $$ for $-\eta nc^{-n}<|x|-M<\eta n^{1/\alpha}c^{-n}$ with some $B=B(c,T,t)>0$, \ $ \eta=\eta(c,T,t)>0$, and the error term $r_n(x)$ satisfies the inequality $$ |r_n(x)|\le Kq^n\tag1.7 $$ with some $K>0$ and $0M+\eta n^{1/\alpha}c^{-n}\tag1.8 $$ with some $\delta >0$,\ $ K>0$ and $L>0$ which depend on $c$, $ T$ and $t$. The solution of the equation (1.6) satisfies the inequality $$ 00\tag1.9 $$ with some $C>0$, \ $ A>0$. We have $$ c^{-n}\bar p_n(x,T)\le C_1\exp \{-C_2c^n|x-M|\} \quad\text {for} \quad 00$, \ $ C_2>0$ depending on $c$, \ $ T$ and $t$. We also have $$ M^2=\frac{a_0-T}{tT}+R(t,T) $$ with some $|R(t,T)|\le \text {const}.$, and such that $R(t,T)\to 0$ and $T\to 0$.\endproclaim Given some integers $N\ge k\ge 0$ we define the probability measure $\mu^h_{k,N}$ as the projection of the measure $\mu^h_N$ to the first $2^k$ coordinates, i.e. $\mu^ h_{k,N}$ is a probability measure on $(R^p)^{2^k}$, and for all measurable $A\subset (R^p)^{2^k}$ $ \mu^h_{k,N}(A)= \mu^h_N(A)=\mu^h_N\left( A\times (R^p)^{2^N-2^k}\right)$. Our first result is the following \proclaim{Theorem 1} \it Let the conditions of Theorem A be satisfied. Consider an arbitrary sequence of real numbers $h_N $, $ N= 0,1,2,\dots $ such that $$ \frac2{2-c} \frac MT \left(\frac c2 \right)^N \le \frac{h_N}T \le D \left(\frac c2 \right)^N \tag 1.11 $$ with some $ \infty > D > \frac2{2-c} \frac MT$, where $M$ and $T$ are the same as in Theorem A. Then the measures $\mu_N^{h_N}$ tend to a probability measure $\bar \mu =\bar \mu(t,T,c)$ on $(R^p)^{\bold Z}$. More precisely, for all $k \ge 0$ the measures $\mu^{h_N}_{k,N}$ converge to the projection of $\bar \mu$ to the first $2^k$ coordinates in variational metric as $N \to \infty$. The measure $\mu$ does not depend on the choice of sequences ~$h_N$. \endproclaim Then we prove the following \proclaim {Theorem 2}\it Let $\sigma =\left\{\sigma (n)=\left(\sigma^{(1)}(n), \dots,\sigma^{(p)}(n)\right) \in R^p,\; n\in\bold Z\right\}$ \ be a $\bar \mu$ distributed random field with the distribution $\bar \mu$ defined in Theorem 1. Then the finite dimensional distributions of the random fields $R_n\sigma$ defined in (1.1), (1.2), (1.3) tend to those of a random field $Y=\left(Y(n)=\left(Y^{(1)}(n),\dots, Y^{(p)}(n)\right)\in R^p, \; n \in \bold Z\right)$. For all $k \ge 0$ the density function $h_k\left(x_1,\dots,x_{2^k}\right)$, \ $ x_j=\left(x_j^{(1)},\dots,x_j^{(p)}\right)\in R^p$, of the random vector $(Y(1),\dots,Y(2^k))$ is given by the formula $$ \aligned &h_k\left(x_1,\dots,x_{2^k}\right)=C(k)\exp\biggl\{-\frac1T \sum_{s=2}^p\biggl(\frac1{2-c}\sum_{j=1}^{2^k} x_j^{(s)2}-\frac {2-c}c \biggl(\frac c4 \biggr)^k\biggl(\sum_{j=1}^{2^k} x_j^{(s)}\biggr)^2\\ &\qquad+\sum^{2^k-1}_{i=1}\sum_{j=i+1}^{2^k} U(i,j)x_i^{(s)}x_j^{(s)} \biggr)\biggr\}\prod _{j=1}^{2^k} g\biggl(\frac{4-c}{(2-c)T}\biggl(Mx_j ^{(1)} +\frac12 \sum_{s=2}^p x_j^{(s)2}\biggr)\biggr)\,, \endaligned \tag1.12 $$ where the function $g$ is defined in (1.5), the constant $M$ is the same as in Theorem A, and $C(k)$ is an appropriate norming constant.\endproclaim \par In Appendix E we prove the following \proclaim{Theorem B}\it The measure \ $\bar{\mu}=\bar{\mu}(T,t,c)$ \ constructed in Theorem 1 is a Gibbs state with Hamiltonian \ $\Cal H$ \ and free measure \ $\nu$ \ defined in formulas (1.1)--(1.3) of Part I at temperature \ $T$. \endproclaim Theorem B is very plausible. Its proof depends on a rather standard limiting procedure in statistical physics literature. Nevertheless, we have found no result which could have been directly applied in our case. We present the proof of Theorem B in Appendix E. \par Let us discuss the role of condition (1.11) in Theorem 1. The lower bound $$ h_N>\frac2{2-c}M\left(\frac2c\right)^N \tag 1.11${}^{\prime}$ $$ is essential in Theorem 1, it is needed to get a pure state with magnetization in the direction $e_1=(1,0,\dots,0)$ for the limit measure $\bar{\mu}$. If it were violated we would get a Gibbs state with Hamiltonian $\Cal H$ again for the limit, but this Gibbs state would be a mixture of Gibbs states with different directions of magnetization, and it is not natural to renormalize such a mixture. On the other hand the upper bound for $h_N$ in (1.11) seems not to be essential. We believe that the same limit measure $\bar{\mu}$ would be obtained for any sequence $h_N$, \ $h_N>0$ satisfying (1.11${}^{\prime}$) or with the help of the double limiting procedure $\mu^h=\lim_{N\to\infty}\mu_N^h$, \ $h>0$, \ $\bar{\mu}=\lim_{h\to 0} \mu^h $. This second way was chosen to construct the equilibrium state in the case $\sqrt20\}\tag2.6\\ \Omega^2_n& =\left\{x\in R^2,\,\bigl||x|-M\bigr|0$. Define the sequences $g_n$ and $A_n$, \ $0\le n\le N$, by formulas (2.4)--(2.5${}^{\prime}$) and put $\bar g_n=c^{-n}g_n$, \ $\bar A_n=c^{-n}A_n$. If $h_N$ satisfies relation (1.11) then $\bar g_N\ge \bar g_{N-1}\ge \dots \ge \bar g_0 \ge \bar g$ and $0=\bar A_N\le \bar A_{N-1} \le \dots \le \bar A_0 \le \bar A$ with $\bar g=\frac2{2-c}\frac MT$, and $\bar A=\frac{2-c}{cT}$. If the relations $N >N_0$ and $N>n^B$ also hold with some appropriate $N_0=N_0(c,M,T,D)$ and $B=B(c,M,T,D)$ then $|\bar g_n-\bar g|<4^{-n}$, $|\bar A_n-\bar A|<4^{-n}$. \endproclaim Proposition 1 together with the characterization of the asymptotic behaviour of the sequences $g_n$ and $A_n$ made in Lemma 1 supplies a good asymptotic formula for the Radon--Nikodym derivative $f_n$. Here $\Omega^1_n$ is the typical region, where we have a good asymptotic formula, in $\Omega^2_n$ and $\Omega^3_n$ we have only given an upper bound. Actually we are interested in the density function $$ p_n\biggl(2^{-n}\sum^{2^n}_{j=1}x_j \biggr)f^{h_N}_{n,N}\biggl( 2^{-n}\sum^{2^n}_{j=1}x_j\biggr) $$ of the measure $\mu^{h_N}_{n,N}$. The tail behaviour of the functions $p_n(x)$ and $f_n(x)$ together show that $2^{-n}\sum^{2^n}_{j=1}x_j$ is contained in $\Omega^3_n$ with a negligible small $\mu^{h_N}_{n,N}$ probability. It is contained in $\Omega^2_n$ also with a small probability, since in this domain $f_n(x)$ is small. To see it, let us observe that by Lemma 1 $$ \frac{g_n}{2M}-A_n\geq c^n\left(\frac{\bar g}{2M}-\bar A\right)= \frac{c^n}T\left(\frac1{2-c}-\frac{2-c}c\right)=\frac{c^n}T \frac{(4-c)(c-1)}{2-c}>0 \,, $$ hence the term $-\left(\frac{g_n}{2M}-A_n\right)c^{-0.8n}$ in the exponent of (2.8), makes this upper bound (2.8) sufficiently small for our purposes. \par In Section 7 of [6] we have given a heuristic argument for formula (2.7). The following remark explains the content of the estimate (2.8). \demo{Remark} If $x\in\Omega_n^1$ then $$ \align |x|&=\left(x^{(1)2}+x^{(2)2}\right)^{1/2}=x^{(1)}\left(1+ \frac{x^{(2)2}}{x^{(1)2}}\right)^{1/2}=x^{(1)}+\frac{x^{(2)2}} {2x^{(1)}}+O(x^{(2)4})\\&=x^{(1)}+\frac{x^{(2)2}}{2M}+O\left(x^{(2)4}+ x^{(2)2}|x^{(1)}-M|\right)=x^{(1)}+\frac{x^{(2)2}}{2M}+O(c^{-1.2n})\,, \endalign $$ hence $$ \align &\exp\left\{g_n(|x|-M)-\left(\frac{g_n}{2M}-A_n\right)x^{(2)2} \right\}\\ &\qquad=\exp\left\{g_n(x^{(1)}-M)+A_nx^{(2)2}+O(c^{-0.2n})\right\}\,. \endalign $$ \enddemo The above calculation shows that on the boundary of the domains $\Omega^1_n$ and $\Omega^2_n$ the right-hand side of formulas (2.7) and (2.8) have the same magnitude. The estimate (2.8) expresses the fact that this is the worst region, where the weakest upper bound can be given for $f_n(x)$\ in $\Omega^2_n$. \par With the help of Proposition 1, Lemma 1 and Theorem A we are able to carry out a limiting procedure which supplies Theorem 1. Moreover, it yields the following Proposition 2. Let $\bar{\mu}_n$ denote the projection of the measure $\bar{\mu}$ constructed in Theorem 1 to the first $2^n$ coordinates, i.e. let $\bar{\mu} _n$ be the measure on $(R^2)^{2^n}$ defined by the relation $\bar{\mu}_n(A)=\bar{\mu}\left(A\times(R^2)^\infty\right)$ for all measurable sets $A\subset (R^2)^{2^n}$. The following result holds true: \proclaim {Proposition 2}\it For all $q$, \ $c^{-0.2}M+c^{-0.4n}\,,\tag2.13${}^{\prime}$ \endalign $$ with $\bar g=\frac2{2-c}\frac MT$, \ $\bar A=\frac {2-c}{cT}$ and an appropriate norming constant $L_n$. This norming constant satisfies the relation $$ C_1\bar g$: \flushpar a) For $x\in\Omega^1_{n+1}$ $$ \align f(x)=&\exp\left\{g_{n+1}(x^{(1)}-M)+A_{n+1}x^{(2)2}+\varepsilon _{n+1}(x)\right\}\tag2.15 \\ &\sup_{x\in\Omega^1_{n+1}}|\varepsilon_{n+1}(x)|\leq q^{n+1}\,. \tag2.15${}^{\prime}$\endalign $$ b) For $x\in\Omega^2_{n+1}$ $$ 0\leq f(x)\leq\exp\left\{g_{n+1}(|x|-M)+\left(\frac{g_{n+1}}{2M}- A_{n+1}\right)c^{-0.8(n+1)}+q^{n+1}\right\}\,;\tag2.16 $$ c) For $x\in\Omega^3_{n+1}$ $$ 0\leq f(x)\leq \exp\left\{\frac{g_{n+1}}{2M}(|x|^2-M^2)\right\} \,\; \tag2.17 $$ then the function \ $S_nf(x)$ \ defined by (2.3${}^{\prime}$) satisfies, with the constants \ $g_n$ \ and \ $A_n$ \ defined by (2.4${}^{\prime}$) and (2.5${}^{\prime}$) with the above \ $g_{n+1}$ \ and \ $A_{n+1}$, the following relations with some appropriate norming constant \ $L_n$: \flushpar a) In the domain \ $\Omega^1_n$ $$ S_nf(x)=L_n\exp\left\{g_n(x^{(1)}-M)+A_nx^{(2)2}+\varepsilon_n(x) \right\}\tag2.18 $$ with $$ \sup_{x\in\Omega^1_n}|\varepsilon_n(x)|\leq q^n\,.\tag2.18${}^{\prime}$ $$ b) In the domain $\Omega^2_n$ $$ 0\leq S_nf(x)\leq L_n\exp\left\{g_n(|x|-M)-\left(\frac{g_n}{2M} -A_n\right)c^{-0.8n}+q^n\right\}\,.\tag2.19 $$ c) In the domain $\Omega^3_n$ $$ 0\leq S_nf(x)\leq L_n\exp\left\{\frac{g_n}{2M}(|x|^2-M^2)\right\}\,. \tag2.20 $$ \endproclaim \subheading{3. The proof of Lemma 1} The proof is a modification of that given for Lemma 1 in [5]. Simple calculation shows that $\bar g_n-\bar g =(\frac c2)^{N-n} (\bar g_N-\bar g)$. The statements of Lemma 1 about the sequence $g_n$ follow from this identity. To investigate $\bar A_n$ let us introduce the function $$ T(x,g)=\frac c4 x+\frac{\left(\frac1T+\frac c2 x\right)^2} {\frac2T+\frac{cg}M-cx}\,,\quad x\in R^1\,,\quad g\in R^1 \,. $$ Clearly, $\bar A_n=T(\bar A_{n+1},\bar g_{n+1})$. On the other hand $T(\bar A,\bar g)=\bar A$, and some calculation shows that $T$ has the following monotonicity properties: $T(x,g)g^{\prime}>\bar g$; and $T(x^ {\prime},g)>T(x,g)$ if $0\bar g$. (These properties follow e.g. from the relations $$ \frac{\partial}{\partial g}T(x,g)= -\frac cM \frac{\left(\frac1T+\frac c2 x\right)^2} {\left(\frac2T+\frac cMg-cx\right)^2}<0\,, $$ and $$ \frac{\partial}{\partial x}T(x,g)= \frac{c\left(\frac2T+\frac{cg}{2M}\right)^2} {\left(\frac2T+\frac cMg-cx\right)^2}>0\,, $$ and the fact that $T(x,g)$ has no singularity in the domain $\{(x,g),\,0\bar g\}$.) We have $0<\bar A_{N-1}<\bar A$, since $\bar A_{N-1}= T(0,\bar g_N)>0$ and $\bar A_{N-1}=T(0,\bar g_N)T(\bar A_{n+1}, \bar g_{n+1})=\bar A_n$, as we have claimed. \par The conditions $N>N_0$ and $N>n^B$ with sufficiently large $N_0$ and $B$ imply that $|\bar g _{\ell}-\bar g|<10^{-n}\ $ for all $0<\ell \leq \sqrt N$, and \ $\bar A-5^{-n}\bar A$. Hence the solutions of the equation $T(x,g^*)=x$ are very close to $A_1$ and $ A_2$. We claim that the monotonicity properties of the sequences $\bar g_n$ and $\bar A_n$ and the function $T(x,g)$ imply that $\bar A>\bar A_n\geq T^{\sqrt N-n}_{g^*} (0)$, where $T^k_{g^*}$ denotes the $k$-th iteration of the function $T(x,g^*)$ with fixed $g^*$ in the variable $x$. Indeed, $\bar A>\bar A_n$, $ 0 <\bar A_{\sqrt N} <\bar A$, and we get by induction that for all $\ell \geq 0 $ $ \bar A_{\sqrt N-\ell}\geq T^\ell_{g^*}(0)$, which implies the required statement with $\ell = \sqrt N-n$. \par To complete the proof of Lemma 1 it is enough to show that $T^n_{g^*}(0)$ tends exponentially fast in $n$ to the smaller solution $A^*$ of the equation $T_{g^*}(x)=x$. Since $T(x,g^*)$ is a convex increasing function (in the variable $x$) it is enough to show that $\frac {\partial T(x,g^*)}{\partial x}\leq \alpha <1$ for $x=A^*$ if $(A^*,g^*)$ is in a small neighbourhood of the point $(\bar A,\bar g)$. But this follows from the continuity of the function $\frac {\partial T(x,g)}{\partial x}$, and the fact that its value in the point $(\bar A,\bar g)$ equals $c^{-1}<1$. Lemma 1 is proved. \subheading{4. Some preparatory remarks to the proof of Proposition 1${}^{\prime}$} We shall prove the following estimates under the conditions of Proposition 1${}^{\prime}$. \par Put $$ S^i_nf(x)=\int_{\{y,\,\frac{x+y}2\in\Omega^i_{n+1}\}} \exp\left(\frac{c^n}Txy\right)f\left(\frac{x+y}2\right)p_n(y)\,dy\,, \quad i=1,2,3\,. $$ Then we have with some appropriate $\varepsilon^{\prime}= \varepsilon^{\prime}(c)$, \ $ \varepsilon^{\prime}>0$, and the same $q$ as in Proposition ~1${}^{\prime}$: \medpagebreak \flushpar In the domain $x\in\Omega^1_n$ $$ S^1_nf(x)=L_n\exp\left\{g_n(x^{(1)}-M)+A_nx^{(2)2}+\bar \varepsilon _n(x)+\ebb_n(x)\right\} \tag4.1 $$ with $$ \sup_{x\in\Omega^1_n}|\bar \varepsilon_n(x)|\leq q^{n+1}\,, \quad \sup_{x\in\Omega^1_n}| \ebb_n(x)| \leq Kc^{-0.2n}\,,\tag4.1${}^{\prime}$ $$ where $K$ does not depend on $n$, $$ \align &L_n \geq c^{-n}\exp\left(\frac{c^n}T M^2\right)\,, \tag4.1${}^{\prime\prime}$\\ &S^2_nf(x)\leq L_n\exp\left\{g_n(x^{(1)}-M)+A_nx^{(2)2}- \varepsilon ^{\prime}c^{0.2n}\right\}\,,\tag4.2\\ &S^3_nf(x)\leq L_n\exp\left\{g_n(x^{(1)}-M)+A_nx^{(2)2}-\frac16 c^{n/2} \right\}\,.\tag4.3 \endalign $$ \medpagebreak \flushpar In the domain $x\in\Omega^2_n$ $$ \align S^1_nf(x)&\leq L_n\exp\left\{g_n(|x|-M)-\left(\frac{g_n}{2M}-A_n \right)c^{-0.8n}+q^{n+1}+Kc^{-0.2n}\right\}\,,\tag4.4\\ S^2_nf(x)&\leq L_n\exp\left\{g_n(|x|-M)-\left(\frac{g_n}{2M}-A_n \right)c^{-0.8n}-\varepsilon^{\prime}c^{0.2n}\right\}\,,\tag4.5\\ S^3_nf(x)&\leq L_n\exp\left\{g_n(|x|-M)-\left(\frac{g_n}{2M}-A_n \right)c^{-0.8n}-\frac16 c^{n/2}\right\}\,.\tag4.6\\ \endalign $$ \medpagebreak \flushpar In the domain $x\in\Omega^3_n$ $$ S_nf(x)\leq L_n\exp\left\{\frac{g_n}{2M}(|x|^2-M^2)\right\}\,. \tag4.7 $$ We show that these estimates imply Proposition 1${}^{\prime}$. Indeed, for $x\in\Omega^1_n$ $$ S_nf(x)=S^{(1)}_nf(x)+S_n^{(2)}f(x)+S_n^{(3)}f(x)= L_n\exp\{g_n(x^{(1)}-M)+A_nx^{(2)2}+\varepsilon_n(x)\} $$ with $$ \sup_{x\in\Omega^1_n}|\varepsilon_n(x)|\leq \sup_{x\in\Omega^1_n} |\bar \varepsilon_n(x)|+\sup_{x\in\Omega^1_n}|\ebb _n(x)|+2\exp(-\varepsilon^{\prime}c^{0.2n})+ 2\exp\left(-\frac16 c^{n/2}\right)\,. $$ (Here we have exploited that $e^t<1+2|t|$ for small ~$t$.) Hence $$ \sup_{x\in\Omega^1_n}|\varepsilon_n(x)|\leq q^{n+1}+Kc^{-0.2n}+ 2\exp(-\varepsilon^{\prime}c^{0.2n})+ 2\exp\left(-\frac16 c^{n/2}\right)\leq q^n $$ if $c^{-0.2}0$. \endproclaim \demo{Proof of Lemma 2} \flushpar Part a). We have $$ 0\leq S_nf(x)\leq \int\exp\left\{\frac{c^n}T xy + \frac{g_{n+1}} {2M}\left(\left|\frac{x+y}2\right|^2-M^2\right)\right\}p_n(y)\,dy\,. $$ Clearly, $\max_{|y|=r} xy$ is taken in the point $y= \frac x{|x|} r$, and it equals $|x|r$. Similarly $\max_{|y|=r}(x+y)^2=(|x|+r)^2$. Hence $$ 0\leq S_nf(x)\leq2\pi\int_0^{\infty}\exp\left\{\frac{c^n}T|x|r+ \frac{g_{n+1}} {2M}\left[\left(\frac{|x|+r}2\right)^2-M^2\right]\right\}r\bar p_n (r)\,dr\,.\tag4.10 $$ Let us split up the integral into two parts, $\int^M_0$ and $\int^{\infty}_M$. Put $$ f_n(t)=c^{-n}\bar p_n(M+c^{-n}t)\,.\tag4.11 $$ It follows from (1.10) that $$ f_n(t)\leq C_1\exp(-C_2|t|)\qquad \text{for } -c^nM0 \tag4.11${}^{\prime\prime}$ $$ with some appropriate $C_1>0$, $C_2>0$ and $C_3>0$. (Relations (1.7) and (1.9) are needed in the domain $0<|x|<\eta n^{1/\alpha}c^{-n}$.) First we estimate the integral $\int ^M_0$. For $0\leq r\leq M$ we have $$ \frac {c^n}T|x|r +\frac{g_{n+1}}{2M}\left(\left(\frac{|x|+r}2 \right)^2-M^2\right)\leq \frac{c^n}T|x|M+\frac{g_{n+1}}{2M} \left(\left(\frac{|x|+M}2\right)^2-M^2\right)\,. $$ Hence (4.11) and (4.11${}^{\prime}$) imply that $$ \align \int^M_0\dots\, dr&\leq C_1 \exp\left\{\frac{c^n}T|x|M+\frac{g_{n+1}}2 \left[\left(\frac {|x|+M}2\right)^2-M^2\right]\right\}\\ &\qquad\int^0_{-c^nM}(M+c^{-n}t)e^{-C_2|t|}\,dt\\ &\leq C_4M\exp\left\{ \frac{c^n}T|x|M+\frac{g_{n+1}}{2M}\left[\left(\frac{|x|+M}2 \right)^2-M^2\right]\right\}\,. \endalign $$ Simple calculation yields the identity $$ \align &\frac{c^n}T|x|M+\frac{g_{n+1}}{2M}\left[\left(\frac{|x|+M}2 \right)^2-M^2\right]\\&\qquad=\frac{c^n}T M^2+\frac{g_n}{2M}(|x|^2-M^2)- \left(\frac{c^n}{2T}+\frac {g_{n+1}}{8M}\right)(|x|-M)^2\,. \endalign $$ Then, since $C_4 M0$. Hence $$ \int^{\infty}_{M+\e c^{-0.4n}}\dots \,dr \leq c^{-n}\exp\biggl\{ \frac{c^n}T M^2+\frac{g_n}{2M}(|x|^2-M^2)-2c^{n/2}\biggr\}\,. $$ The last inequality together with (4.16) imply that $$ R^{\e}_nf(x)\leq 4\pi c^{-n/2}\exp\biggl\{-2c^{n/2}+\frac{c^n}TM^2+ \frac {g_n}{2M}(|x|^2-M^2)\biggr\}\,.\tag4.17 $$ Now we estimate $Q^{\e}_nf(x)$. We have $$ 0\leq Q^{\e}_nf(x)\leq \int \Sb \{y,||y|-M|\leq c^{-0.4n},\\ {} \;\;\;xy\leq|x||y|-\e c^{-0.4n}\;\,\}\endSb \exp\biggl\{\frac{c^n}T xy+ \frac{g_{n+1}}{2M}\biggl[\left(\frac{x+y}2\right)^2-M^2\biggr]\biggr\} p_n(y)\,dy\,. $$ Since for $|y|=r$ \ $xy\leq r|x|-\e c^{-0.4n}$ and $\left(\frac{x+y}2\right)^2\leq\bigl(\frac{|x|+r}2\bigr)^2$ in the last integral, hence $$ \align Q^{\e}_nf(x)&\leq 2\pi\int^{\infty}_0\exp\biggl\{\frac{c^n}T \left(|x|r-\e c^{-0.4n}\right)+\frac{g_{n+1}}{2M}\biggl[\left(\frac {|x|+r}2\right)^2-M^2\biggr]\biggr\}r\bar p_n(r)\,dr\\ &\leq2\pi\exp\biggl(-\frac{\e c^{0.6n}}T\biggr)\int^{\infty}_0 \exp\biggl\{\frac{c^n}T|x|r+\frac{g_{n+1}}{2M} \biggl[\biggl(\frac{|x|+r}2\biggr)^2-M^2\biggr]\biggr\}\\ &\hskip 10truecm r\bar p_n(r)\,dr \,. \endalign $$ The last integral has already appeared in (4.10). We have estimated it in Part a), and bounded it by the right hand side of (4.15). Hence $$ \align Q^{\e}_nf(x)&\leq 2\pi c^n\exp\biggl\{-\frac{\e}Tc^{0.6n}+ \frac{c^n}TM^2+\frac{g_n}{2M}\left(|x|^2-M^2\right)\biggr\}\\ &\leq \frac12\exp\biggl\{-2c^{n/2}+\frac{c^n}TM^2+\frac{g_n}{2M} \left(|x|^2-M^2\right)\biggr\} \,. \endalign $$ This inequality together with (4.17) imply that $$ S^{\e}_nf(x)\leq\exp\biggl\{-2c^{n/2}+\frac{c^n}T M^2+ \frac{g_n}{2M}\left(|x|^2-M^2\right)\biggr\}\,. $$ Since $\frac1{2M}(|x|^2-M^2)=(|x|-M)+\frac1{2M}(|x|-M)^2$ and $\frac{g_n}{2M}(|x|-M)^2\leq Dc^{0.2n}$ if $||x|-M|\leq c^{-0.4n}$, hence the last inequality implies that under the conditions of Part ~b) $$ \align S^{\e}_nf(x)&\leq\exp\biggl\{-2c^{n/2}+\frac{c^n}TM^2+\frac{g_n}{2M} (|x|-M)+Kc^{0.2n}\biggr\}\\ &\leq\exp\biggl\{\frac {c^n}T M^2+g_n(|x|-M)-c^{n/2}\biggr\}\,, \endalign $$ as we have claimed. \enddemo \demo {Proof of Lemma 3} $$ \align 0\leq S_nf(x)&\leq \int_{\{y,\;||\frac{x+y}2|-M|\leq c^{-0.4(n+1)}\}} \exp\left\{\frac{c^n}Txy+g_{n+1}\left(\left|\frac{x+y}2 \right|-M\right)\right\}\\ &\qquad\qquad\qquad p_n(y)\,dy\,. \endalign $$ Since $$ \max_{|y|=r}\left\{\frac{c^n}Txy+g_{n+1}\left(\left|\frac {x+y}2\right|-M\right)\right\}\leq\frac{c^n}T|x|r+g_{n+1} \left[\left(\frac{|x|+r}2\right)-M\right]\,,$$ hence $$ S_nf(x)\leq 2\pi\int^{\infty}_0\exp\left\{\frac{c^n}T |x|r+g_{n+1}\left(\frac{|x|+r}2-M\right)\right\}r\bar p_n(r)\,dr\,. $$ Writing $\frac{c^n}T|x|r=\frac{c^n}TM^2+\frac{c^n}TM(|x|-M+r-M)+ \frac{c^n}T(|x|-M)(r-M)$ we get that $$ \align S_nf(x)\leq 2\pi\exp&\left\{\frac{c^n}TM^2+g_n(|x|-M)\right\}\\ &\int^{\infty}_0\exp\left\{g_n(r-M)+\frac{c^n}T(|x|-M)(r-M) \right\}r\bar p_n(r)\,dr\,. \endalign $$ The change of variables $r=M+c^{-n}t$ and the introduction of $\bar g_n=c^{-n}g_n$ yields that $$ \align S_nf(x)&\leq 2\pi\exp\left\{\frac {c^n}TM^2+g_n(|x|-M)\right\}\\ &\qquad\int^{\infty}_{-c^nM}\exp\left\{\bar g_nt + \frac{(|x|-M)}T t\right\}(M+c^{-n}t)f_n(t)\,dt\,. \endalign $$ Since $0 <\bar g_n 0$ independent of $n$. Hence $$ \align S_nf(x)&\leq 2\pi\bar K\exp\left\{\frac{c^n}T M^2+g_n(|x|-M)\right\}\\ &\leq K\exp\left\{\frac{c^n}T M^2+g_n(|x|-M)\right\}\,, \endalign $$ as we have claimed. \enddemo \subheading{5. The proof of Proposition 1${}^{\prime}$} In this Section we prove the estimates (4.1)--(4.7) which imply Proposition 1${}^{\prime}$. \medpagebreak a) The estimation of $\s1$ for $x\in\Om1$. It follows from (2.15) and (2.15${}^{\prime}$) that $$ \align \s1&=\int_{\{y,\,\frac{x+y}2\in\Omm1\}}\exp \Biggl\{\frac{c^n}Txy+ g_{n+1}\biggl(\frac{x^{(1)}+y^{(1)}}2-M\biggr)+A_{n+1}\\ &\qquad\biggl(\frac {x^{(2)}+y^{(2)}}2\biggr)^2 +\e_{n+1}\biggl(\frac{x+y}2\biggr)\Biggr\} p_n(y)\,dy\,. \endalign $$ Hence $$ \align \s1&=\exp(\ebx)\int_{\{y,\,\frac{x+y}2\in\Omm1\}}\exp \Biggl\{\frac{c^n}Txy+ g_{n+1}\left(\frac{x^{(1)}+y^{(1)}}2-M\right)\\ &\qquad+A_{n+1}\biggl(\frac {x^{(2)}+y^{(2)}}2\biggr)^2\Biggr\} p_n(y)\,dy\,. \endalign $$ with some $$ \sup_{x\in\Om1}|\ebx|\le q^{n+1}\,. \tag 5.1 $$ Let us rewrite the last expression in polar coordinate system. We get that $$ \s1=\exp(\ebx)\int_0^{\infty}I_n(r)\bar p_n(r)\,dr \tag 5.2 $$ with $$ \align I_n(r)=\int_{\formg}r\exp \Biggl\{\frac{c^n}T\left(x^{(1)}y^{(1)} + x^{(2)}y^{(2)}\right)&+ g_{n+1}\left(\frac{x^{(1)}+y^{(1)}}2-M\right)\\&+A_{n+1} \left(\frac {x^{(2)}+y^{(2)}}2\right)^2\Biggr\}\,d\vf \endalign $$ where $y^{(1)}=r\cos\vf$, \ $y^{(2)}=r\sin\vf$, \ $-\pi<\vf<\pi$, $y=(y^{(1)},y^{(2)})$ and $\Gamma_n(r,x)=\allowmathbreak \{\vf,\,\frac{x+y}2\in\Omm1\}$. We shall express $I_n(r)$ as an asymptotically Gaussian integral with respect to $\vf$. For this aim we give some bounds on $x^{(1)}$, \ $x^{(2)}$, \ $y^{(1)}$ and $y^{(2)}$ if $x\in\Om1$ and $\frac{x+y}2\in\Omm1$. We have $$ \aligned &|x^{(2)}|<\co{-0.4}\\ &|x^{(1)}-M|<\frac32\co{-0.4} \endaligned \aligned \qquad\qquad \text{if}\quad x\in\Om1\,. \endaligned \tag 5.3 $$ The second relation in (5.3) holds, since $$ \align &|x^{(1)}-M|=\bigl|(|x|^2-x^{(2)2})^{1/2}-M\bigr|= \left|\,|x|-\frac{x^{(2)2}}{2|x|}+O(x^{(2)4})-M\right|\\ &\qquad\le \bigl||x|-M\bigr|+\left| \frac{x^{(2)2}}{2|x|}+O(x^{(2)4})\right|\le \co{-0.4}+K\co{-0.8}\le\frac32\co{-0.4}\,. \endalign $$ Similarly, if $\frac{x+y}2\in \Omm1$ then $|\frac{x^{(2)} +y^{(2)}}2|<\co{-0.4}$, and $|\frac{x^{(1)}+y^{(1)}}2-M|<2\co{-0.4}$. Hence $$ |\y2|\le |-\x2|+|\x2+\y2|\le 3\co{-0.4}, \tag 5.4 $$ $$ |\y1-M|=|M-\x1|+|\x1+\y1-2M| \le6\co{-0.4}\, \tag 5.4${}^{\prime}$ $$ and $$ \align |r-M|=||y|-M|&=\left|(y^{(1)2}+ y^{(2)2})^{1/2}-M\right|\tag 5.4${}^{\prime\prime}$\\ &=\left|y^{(1)}+\frac{y^{(2)2}}{2\y1}+O(y^{(2)4})-M\right| \le 10\co{-0.4} \endalign $$ if $x\in \Om1$ and $\xy\in\Omm1$. In particular, (5.4${}^{\prime\prime}$) implies that $$I_n(r)=0 \qquad\text{if } |r-M|>10\co{-0.4},\text { and } x\in\Om1\,. \tag 5.5 $$ Furthermore, $|\vf|\le2|\sin\vf| =\frac2r|\y2|=O(\co{-0.4})$ and $\y1=r(1-\cos\vf )\allowmathbreak=r(1-\vf^2/2)+O(\co{-1.6})$, \ $\y2=r\sin\vf\allowmathbreak=r\vf+O(\co{-1.2})$ and $r=M+O(\co{-0.4})$ if $\fformg$. Hence $$ \align I_n(r)&=\left(1+O(\co{-0.2})\right) \int_{\formg}M\exp\Biggl\{\frac{c^n}T \left(\x1r\biggl(1-\frac{\vf^2}2\biggr)+\x2r\vf\right)\\ &\qquad +g_{n+1}\left(\frac{\x1+r(1-\frac{\vf^2}2)}2-M\right) +A_{n+1}\left(\frac{\x2+r\vf}2\right)^2\Biggr\} \,d\vf\\ &=\left(1+O(\co{-0.2})\right)M\exp \left\{\frac{c^n}Tr\x1 +\frac{g_{n+1}}2\biggl(\frac{\x1+r}2-M\biggr) +\frac{A_{n+1}}4 \xo2\right\}\\ &\qquad \int_{\formg}\exp\biggl\{-\biggl(\frac{c^n}{2T}\x1 +\frac{g_{n+1}}4-r \frac{A_{n+1}}4\biggr)r\vf^2\\ &\hskip6truecm + \biggl(\frac{c^n}T+\frac{A_{n+1}}2 \biggr)\x2 r\vf\biggr\}\,d\varphi\,. \endalign $$ Moreover, since $c^n\x1 r\vf^2=c^nM^2\vf^2\allowmathbreak+O(\co{-0.2})$, $g_{n+1}r\vf^2\allowmathbreak=g_{n+1}M\vf^2\allowmathbreak +O(\co{-0.2})$, $A_{n+1}r^2\vf^2\allowmathbreak =A_{n+1}M^2\vf^2\allowmathbreak+O(\co{-0.2})$ and $\ca\x2 r\vf \allowmathbreak=\ca \x2M\vf \allowmathbreak+O(\co{-0.2})$ in our case (observe that $g_{n+1}\le Dc^{n+1}$ and $A_{n+1}\le\ab c^{n+1}$), hence we make an error of order $O(\co{-0.2})$ by substituting $\x1$ and $r$ by $M$ in the last integral, i.e. $$ \align &I_n(r)=\left(1+O(\co{-0.2})\right) M \exp\left\{\frac{c^n}Tr\x1+\frac {g_{n+1}}2\biggl(\frac{\x1+r}2-M\biggr) +\frac{A_{n+1}}4\xo2\right\}\\ &\qquad\int_{\formg}\exp\left\{\cga M^2\vf^2+\ca M\x2\vf\right\}\,d\vf \,, \endalign $$ or equivalently $$ \align I_n(r)&=\left(1+O(\co{-0.2})\right) \exp\biggl\{\frac{c^n}Tr\x1+\frac {g_{n+1}}2\biggl(\frac{\x1+r}2-M\biggr)\\ &\qquad+\frac{A_{n+1}}4\xo2+ \frac{\ca^2\xo2}{4\cga}\biggr\}\\ &\qquad\int_{\formg}M\exp\left\{-M^2\cga (\vf-\gamma_n\x2)^2\right\}\,d\vf \,, \endalign $$ with $$ \gamma_n=\frac{\ca}{\cga}\,. $$ By (2.5${}^{\prime}$) we get from this relation that $$ \align &\tag 5.6\\ I_n(r)&=\left(1+O(\co{-0.2})\right) \exp\left\{\frac{c^n}Tr\x1+\frac {g_{n+1}}2\biggl(\frac{\x1+r}2-M\biggr)+A_n\xo2 \right\}\\&\qquad\int_{\formg}M\exp\left\{-M^2\cga \left(\vf-\gamma_n\x2\right)^2\right\}\,d\vf\,. \endalign $$ Since $$ \cga\ge c^n\left(\frac1{2T}+c\frac{\gb}{4M}-c\frac{\ab}4\right)=Kc^n \tag5.7 $$ with $K= \frac1{2T}+c\frac{\gb}{4M}-c\frac{\ab}4= \frac{c(4-c)}{4(2-c)T}>0$ relation (5.6) implies that $$ \align I_n(r)&\le\sqrt\pi\left(1+O(\co{-0.2})\right) \cga^{-1/2} \tag5.8\\ &\qquad\exp\left\{\frac{c^n}Tr\x1+\frac {g_{n+1}}2\left(\frac{\x1+r}2-M\right)+A_n\xo2 \right\} \endalign $$ Moreover, we claim that $$ \aligned I_n(r)&=\left(1+O(\co{-0.2})\right)\sqrt \pi\ \cga^{-1/2}\\ &\quad\exp\left\{\frac{c^n}Tr\x1+\frac {g_{n+1}}2\left(\frac{\x1+r}2-M\right)+A_n\xo2 \right\}\quad\text{if}\;|r-M|<\eb\co{-0.4} \endaligned \tag 5.9 $$ with some approppriate $\eb>0$. Because of (5.6) and (5.7) to prove (5.9) it is enough to show that there is some $\e=\e_0(\eb)>0$ such that $$ \bigl\{\vf\:|\vf-\gamma_n \x2|<\e_0\co{-0.4}\bigr\}\subset\Gamma_n(x,r) \qquad \text{if }\;|r-M|<\eb\co{-0.4}\,. \tag 5.10 $$ Since $A_{n+1}\le \ab c^{n+1}$ and $g_{n+1}\ge \gb c^{n+1}$ $$ \gamma_n=\frac{\ca}{2\cga}\le\frac{1 +\frac{2-c}2}{(1+\frac c{2-c}-\frac{2-c}2)M}=\frac{2-c}{cM}\,, $$ and we prove (5.10) by showing that $$ \left\{\frac{\x1+r\cos\vf}2,\, \frac{\x2+r\sin\vf}2\right\}\in\Omm1\tag 5.10${}^{\prime}$ $$ if $|\vf-\gamma_n\x2|<\e_0\co{-0.4}$, $|r-M|\le \eb\co{-0.4}$ and $x\in\Om1$. But in this case $$ \align &\left|\frac{\x2+r\sin\vf}2\right| \le \frac12\left(|\x2|+r|\vf|\right)\le\frac12 \left(|\x2|+r(\gamma_n|\x2|+|\vf-\gamma_n\x2|)\right)\\ &\qquad\le \frac12\biggl[\co{-0.4}+(M+\bar\e\co{-0.4}) \biggl(\frac{2-c}{cM}\co{-0.4} +\e_0\co{-0.4}\biggr)\biggr]\\ &\qquad\le \frac 1{\sqrt c}\co{-0.4}\le c^{-0.4(n+1)} \endalign $$ if $\e_0>0$ and $\eb>0$ are sufficiently small. We also get with the help of (5.3${}^{\prime}$) that $$ \align &\left|\frac{\x1+r\cos \vf}2-M\right|\le \left|\frac{\x1-M}2\right|+\frac r2\left|\cos\vf-1\right|+\left|\frac{r-M}2\right|\\ &\qquad\le\left|\frac{\x1-M}2\right| +\frac{r\vf^2}2+\left|\frac{r-M}2\right|\\ &\qquad\le\left|\frac{\x1-M}2\right|+\frac r2\left(\gamma_n|\x2| +\e_0\co{-0.4}\right)^2+ \left|\frac{r-M}2\right| \\ &\qquad\le\left(\frac34+\eb\right)\co{-0.4}+K\co{-0.8}\le c^{-0.1}c^{-0.4(n+1)} \endalign $$ if $\e_0>0$ and $\eb>0$ are sufficiently small. The above estimates imply (5.10${}^{\prime}$) hence also (5.10). Now we can estimate the term $\int_0^{\infty}I_n(r)\bar p_n(r)\,dr$. Relation (5.9) yields that $$ \align \int_{|r-M|<\eb \co{-0.4}}I_n(r)\bar p_n(r)\,dr &=\left(1+O(\co{-0.2})\right)\sqrt{\pi}\cga^{-1/2}\je\\ &\qquad\exp\left\{\frac{c^n}T\x1 M +\frac{g_{n+1}}4\left(\x1-M\right) +A_n\xo2\right\} \endalign $$ with $$ \align \je=\je(\x1)&=\int_{|r-M|<\eb\co{-0.4}} \exp\left\{\frac{c^n}T\x1 (r-M)+\frac{g_{n+1}}4(r-M) \right\}\bar p_n(r)\,dr\\ &=\int_{|t|<\eb\co{0.6}}\exp\left\{ \frac{t\x1}T+c\frac{\gb_{n+1}}4 t\right\}f_n(t)\,dt \endalign $$ with the function $f$ defined in (4.11). On the other hand, by (5.8) and (5.5) $$ \align &\int_{|r-M|>\eb \co{-0.4}}I_n(r)\bar p_n(r)\,dr\\ &\qquad\le\left(1+O(\co{-0.2})\right)\sqrt{\pi}\cga^{-1/2}\\ &\qquad\qquad\exp \left\{\frac{c^n}T\x1 M+\frac{g_{n+1}}4(\x1-M) +A_n\xo2\right\}\jeb \endalign $$ with $$ \jeb=\int_{10\co{0.6}>|t|>\eb\co{0.6}}\exp\left\{\frac{t\x1}T+ c\frac{\gb_{n+1}}4t\right\}f_n(t)\,dt\,. $$ Let us remark that $\je=\je(\x1)$ depends on $\x1$. We show that this dependence is very weak. Namely, since $$ \left|\frac {d\;\;{}}{d\x1}\je(\x1)\right|= \int_{|t|<\eb\co{0.6}}\exp\left\{\frac{t\x1}T+ c\frac{\gb_{n+1}}4t\right\}\frac{|t|}Tf_n(t)\,dt\le C<\infty $$ by (4.11${}^{\prime}$) and (4.11${}^{\prime\prime}$), and for $\x1=M$ the expression $$ \je(M)= \int_{|t|<\eb\co{0.6}}\exp\left\{\frac{tM}T+ c\frac{\gb_{n+1}}4t\right\}f_n(t)\,dt $$ satisfies the relation $$ 00$ for $|t|<1$ that follows from Theorem A. Hence $$ \je(\x1)=\left(1+O(\co{-0.4})\right)\je(M)\qquad\text{if }\;x\in\Om1\,. $$ Similarly, $$ 0\le\jeb=O\left(\exp(-K\co{0.6})\right)\,. $$ The above relations imply that $$ \align\int I_n(r)\bar p_n(r)\,dr &=\left(1+O(\co{-0.2})\right)\sqrt{\pi}\cga^{-1/2}\je(M)\\ &\qquad\exp\left\{\frac{c^n}T M^2+g_n(\x1-M) +A_n\xo2\right\} \,. \endalign $$ The last formula together with (5.1) and (5.2) imply (4.1) with $$ L_n=\sqrt{\pi}\cga^{-1/2}\je(M) \exp\left(\frac{c^n}TM^2\right)\,. $$ Since $\cga0\right\},\;\e>0, \\ &\bar\Om2= \Omega^1_{n,\e} -\Om1\,, \endalign $$ and $$ \obb = \left\{x,\bigl||x|-M\bigr|\le\co{-0.4}\right\}- \Omega^1_{n,\e}\,. $$ Clearly, $\Om2=\bar\Om2\cup\obb$. The domain $\Omega^1_{n,\e}$ is a slight enlargement of $\Om1$. It is not difficult to see by analizing the proof of relation (4.1) that for sufficiently small $\e>0$ $$ \s1=L_n\exp\left\{g_n(\x1-M) +A_n\xo2+\eb(x)+\ebb(x)\right\} $$ if $x\in\Omega^1_{n,\e}$ with some $|\ebx|0$ is appropriately chosen. The function $\s{\e}$ will be bounded with the help of Part b) of Lemma 2, and $T^{\e}_nf(x)$ similarly to $\s1$ in the case $x\in\Om1$. To apply Lemma 2 first we show that under the conditions of Proposition~1${}^{\prime}$ $$ 0\le f(x)\le2\exp\left\{\frac{g_{n+1}}{2M}\left(|x|^2-M^2\right) \right\}\qquad\text {for all }\;x\in R^2\,.\tag 5.12 $$ For $x\in\Omm1$ $$ \x1-M=|x|-M-\frac{\xo2}{2|x|}+O \left(\xo4\right)=|x|-\frac{\xo2}{2M}-M +O\left(\co{-1.2}\right)\,, $$ hence $$ 0\le f(x)\le\exp\left\{g_{n+1}(|x|-M)- \biggl(\frac{g_{n+1}}{2M}-A_{n+1}\biggr) \xo2+\e_{n+1}(x)+O(\co{-0.2})\right\}\,, $$ and since $\frac{g_{n+1}}{2M}-A_{n+1}>0$ hence $$ 0\le f(x)\le\frac32\exp\{g_{n+1}(|x|-M)\}\,. $$ This inequality together with the relation $$ |x|-M=\frac1{2M} \left[(|x|^2-M^2)-(|x|-M)^2\right]\le\frac1{2M}\left( |x|^2-M^2\right) $$ imply (5.12) for $x\in\Omm1$. Similarly, for $x\in \Omm2$ the relation $$ \align 0\le f(x)&\le\exp\left\{g_{n+1}(|x|-M) -\biggl(\frac{g_{n+1}}{2M}-A_{n+1} \biggr) c^{-0.8(n+1)}+q^{n+1} \right\}\\ &\le\frac32\exp\{g_n(|x|-M)\} \endalign $$ implies (5.12), and this relation also holds for $x\in\Omm3$ by relation (2.17). By (5.12) part b) of Lemma 2 can be applied for $\frac12 f(x)$, and it yields that $$ \s{\e}\le 2\exp\left\{\frac{c^n}TM^2+g_n(|x|-M)-c^{n/2}\right\} \qquad \text{if }x\in\Om2\;\;(\text{or }x\in\Om1)\,. $$ Since $L_n\ge c^{-n}\exp\{\frac{c^n}TM^2\}$, and $(\ga)\co{-0.8}=O(\co{0.2})\allowmathbreak\le\frac13c^{n/2}$, the last inequality implies that $$ \align \s{\e}\le L_n\exp&\left\{g_n(|x|-M)- \biggl(\ga\biggr)\co{-0.8}-\frac12c^{n/2}\right\} \tag 5.13\\&\qquad\qquad\qquad \qquad\qquad\text{if }x\in\Om2\;\;(\text{or }x\in\Om1)\,. \endalign $$ To estimate $T_n^{\e}f(x)$ first we show that if $x\in\Om2$, \ $y\in V_n^{\e}(x)$ and $\xy\in\Omm1$ then $$ \align &|\x2|<2\sqrt{\e}\co{-0.2},\;\; |\y2|<2\sqrt{\e}\co{-0.2},\;\; |\x1-M|\le2 \sqrt{\e}\co{-0.2},\tag 5.14 \\ &|\y1-M| \le2\sqrt{\e}\co{-0.2},\;\;\;\bigl||y|-M\bigr|\le \e\co{-0.4}\,. \endalign $$ Indeed, in this case $(x,y)\ge |x| |y|-\e\co{-0.4}$, and $$ \align 0&\le |x-y|^2=|x|^2+|y|^2-2(x,y)\le |x|^2+|y|^2-2|x||y|+\e\co{-0.4} \\ &=(|x|-|y|)^2+\e\co{-0.4} \le2(|x|-M)^2+2(|y|-M)^2 +\e\co{-0.4} \\ &\le2(1+\e)\co{-0.8}+\e\co{-0.4} \le2{\e}\co{-0.4}\,. \endalign $$ Hence $|\frac{\x2-\y2}2|\le\sqrt{\e} \co{-0.2}$ and $\frac12 |(\x1-M)-(\y1-M)|\allowmathbreak\le \sqrt{\e}\co{-0.2}$. Since $\xy\in\Omm1$, \ $|\frac{\x2+\y2 }2|\allowmathbreak\le c^{-0.4 (n+1)}\allowmathbreak\le \sqrt{\e}\co{-0.2}$ and $\frac12|(\x1-M)+(\y1-M)|\le\allowmathbreak\frac32c^{-0.4(n+1)} \allowmathbreak\le\sqrt{\e}\co{-0.2}$ by (5.3). These relations together with the definition of $V_n^{\e}(x)$ imply (5.14) that enables us to estimate $\te$ similarly to $\s1$ for $x\in\Om1$. We get that $$ \te\le(1+q^{n+1})\int_0^{\infty} I_n(r)\bar p_n(r)\,dr \tag 5.15 $$ with $$ \aligned I_n(r)&=\int_{\formg}r\exp\Biggl\{\frac{c^n}T \left(\x1\y1+\x2\y2 \right)\\ &\qquad+g_{n+1}\biggl(\frac{\x1+\y1}2-M\biggr) A_{n+1}\biggl(\frac{\x2+\y2}2 \biggr)^2\Biggr\}\,d\vf \,, \endaligned \tag 5.16 $$ where $\y1=r\cos\vf$, \ $\y2=r\sin\vf$, \ $y=(\y1,\y2)$ and $$ \Gamma_n (r,x)=\{\vf\:\xy\in\Omm1, y\in V_n^{\e}(x)\} . $$ Observe that by (5.14) $|r-M|<\e\co{-0.4}$, \ $ |\vf|\le2|\sin \vf|\allowmathbreak =\frac2r|\y2|\allowmathbreak\le\frac2r\sqrt{\e}\co{-0.2}$. Let us make the change of variables $z=\sin\vf$ in the integral $I_n(r)$. We have $\y2=rz$, \ $\y1=r(1-\frac{y^{(2)2}}{r^2})^{1/2}\allowmathbreak= r(1-\frac{y^{(2)2}}{2r^2})+O(y^{(2)4})$, \ $z\le 2\frac{\sqrt{\e}}r\co{-0.2}$, hence $|\y1-r(1-\frac{z^2}2)| \allowmathbreak \le K\e^2\co{-0.8}$ with some $K>0$ independent of $\e$. These relations imply that if $\fformg$ then $$ \align &\frac{c^n}T\left(\x1\y1+\x2\y2 \right) +g_{n+1}\biggl(\frac{\x1+\y1}2-M\biggr)+ A_{n+1}\biggl(\frac{\x2+\y2}2\biggr)^2 \\ &\quad \le \frac{c^n}Tr\left(\x1\biggl( 1-\frac{z^2}2\biggr)+\x2 z\right)+g_{n+1} \left(\frac{\x1+r(1-\frac{z^2}2)}2-M\right)\\ &\qquad\qquad+A_{n+1}\biggl(\frac{\x2+rz}2\biggr)^2+K\e^2\co{0.2} \endalign $$ with some $K>0$. Since $\frac{dz}{d\vf}>\frac12$ and $|r-M|<\e\co{-0.4}$ if $\Gamma_n(r,x)$ is not empty, the above inequality together with (5.16) imply that $$ I_n(r)=0\qquad\text{if } |r-M|\ge\e\co{-0.4}\tag 5.17 $$ and $$ I_n(r)\le3\exp(K\e^2\co{0.2})\hat I_n(r)\qquad \text{for }\; |r-M|<\e\co{-0.4}\tag 5.17${}^{\prime}$ $$ with $$ \align \hat I_n(r)&=\int_{-\infty}^{\infty}M\exp\Biggl\{ \frac{c^n}Tr\left(\x1\biggl(1-\frac{z^2}2\biggr)+\x2 z\right)\\ &\qquad+g_{n+1}\left(\frac{\x1 +r(1-\frac{z^2}2)}2-M\right) +A_{n+1}\biggl(\frac{\x1+rz}2\biggr)^2\Biggr\}\,dz\,. \endalign $$ The expression $\hat I_n(r)$ can be calculated explicitly, and we get that for $|r-M|<\e\co{-0.4}$ $$ \align \hat I_n=\sqrt{\pi}M\biggl(\frac{c^n}{2T}r\x1+&\frac{g_{n+1}}4r- \frac{A_{n+1}}4r^2\biggr)^{-1/2} \\ &\exp\left\{\frac{c^n\x1r}T+g_{n+1}\left(\frac{\x1 +r}2-M\right)+ A_n(\x1,r)\xo2\right\} \endalign $$ with $$ A_n(\x1,r)=\frac{A_{n+1}}4 +\frac{\ca^2}{2(\frac{c^n\x1}{Tr}+ \frac{g_{n+1}}{2r}-\frac{A_{n+1}}2)}\,. $$ Hence $$ \align \hat I_n(r)&\le Kc^{-n/2} \exp\left\{\frac{c^n\x1 r}T +g_{n+1}\left(\frac{\x1+r}2-M\right)+ A_n(\x1,r)\xo2\right\} \\ &=Kc^{-n/2} \exp\biggl\{\frac{c^n}TM^2+g_n(\x1-M)+A_n(\x1,r)\xo2\\ &\qquad\qquad+\biggl(\frac{c^n\x1}T+\frac{g_{n+1}}2\biggr) (r-M)\biggr\}\,. \endalign $$ Observe that $A_n=A_n(M,M)$, and if $x\in\obb$, \ $y\in V_n^{\e}(x)$ and $\xy\in\Om1$ then $$ \align &|A_n(\x1,r)-A_n|\le K_0c^n(|\x1-M|+|r-M|)\\ &\left|\x1-M-\biggl(|x|-M-\frac{\xo2}{2M}\biggr)\right|\le K_1\xo2(\xo2+|\x1-M|)\,, \endalign $$ hence (5.17${}^{\prime}$) implies that $$ \aligned I_n(r)&\le \bar Kc^{-n/2} \exp\Biggl\{K\e^2\co{0.2}+\frac{c^n}TM^2+g_n(\x1-M)+A_n\xo2\\ &\qquad+ K_0c^n(|\x1-M|+|r-M|) \xo2 +\left(\frac{c^n\x1}T+\frac {g_{n+1}}2\right)(r-M)\Biggr\}\\ &\le\bar Kc^{-n/2} \exp\Biggl\{K\e^2\co{0.2}+\frac{c^n}TM^2+g_n(|x|-M)-\left( \frac{g_n}{2M}-A_n\right)\xo2\\ &\qquad+\left[K_0c^n(|\x1-M|+|r-M|)+K_1g_n(\xo2+|\x1-M|)\right]\xo2\\ &\qquad+\left(\frac{c^n\x1}T+\frac{g_{n+1}}2\right)(r-M)\Biggr\} \endaligned \tag5.18 $$ if $|r-M|\le\e\co{-0.4}$. Since $\frac {g_n}{2M}-A_n>\alpha c^n$ with some $\alpha>0$ and $\xo2>(1+\e)\co{-0.4}$ if $x\in\obb$ $$ \align K\e^2\co{0.2}&+ \left[K_0c^n(|\x1-M|+|r-M|)+K_1g_n(\xo2+|\x1-M|)\right]\xo2 \\&-\left(\frac{g_n}{2M}-A_n\right)\xo2\\ &=K\e^2\co{0.2} -\left(\frac{g_n}{2M}-A_n+O\left(\co{0.8}\right)\right) \xo2\\ &\le K\e^2\co{0.2} -\left(\frac{g_n}{2M}-A_n+O\left(\co{0.8}\right)\right ) (1+\e)^2\co{-0.8} \\ &\le-\left(\frac{g_n}{2M}-A_n\right)(1+\e)\co{-0.8} \endalign $$ if $\e>0$ is chosen sufficiently small. This relation together with (5.18) imply that $$ \align I_n(r)&\le Kc^{-n/2}\exp\biggl\{\frac{c^n}TM^2+g_n(|x|-M)-(1+\e) \biggl(\frac{g_n}{2M}-A_n\biggr)\co{-0.8}\\ &\qquad+\biggl(\frac{c^n\x1}T+\frac{g_{n+1}}2\biggr)(r-M)\biggr\} \quad\text{if }|r-M|<\e \co{-0.4}\,. \endalign $$ With the help of this inequality, the estimate we have on the function $\bar p_n(r)$ and (5.17) we can bound the integral in (5.15) from above. We get that $$ \te\le L_n\exp\left\{g_n(|x|-M)-\biggl(1+\frac{\e}2\biggr) \biggl(\frac{g_n}{2M}-A_n\biggr)\co{-0.8}\right\}\,. $$ Since $\s1 \le\s{\e}+\te$ the last inequality together with (5.13) imply (4.4) for $x\in\obb$. \medpagebreak c) The estimation of $\s2$ for $x\in\Om1$ and $x\in\Om2$. We have $$ \align 0\le\s2&= \int_{\{y,\xy\in\Omm2\}} \exp\biggl(\frac{c^n}Txy\biggr)f\biggl(\xy\biggr)p_n(y)\,dy \\&\le\exp \left\{-\biggl(\gan\biggr)c^{-0.8(n+1)}+q^{n+1}\right\}\\ &\qquad\quad\int_{\{y,\xy\in\Omm2\}} \exp\left\{\frac{c^n}Txy+g_{n+1}\biggl(\frac{|x+y|}2-M\biggr) \right\}p_n(y)\,dy\,. \endalign $$ It follows from Lemma 3 that for $x\in\Om1\cup\Om2$ $$ \align 0\le\s2\le K\exp&\biggl\{-\biggl(\gan\biggr) c^{-0.8(n+1)}\\ &\qquad +q^{n+1} +\frac{c^n}TM^2+g_n(|x|-M)\biggr\}\,. \endalign $$ By Lemma 1 $g_{n+1}c^{-(n+1)}\ge g_nc^n$, \ $A_nc^{-n}\ge A_{n+1}c^{-(n+1)}$ i.e. $g_{n+1}\ge cg_{n}$ and $cA_n\ge A_{n+1}$. Hence $$ \biggl(\gan\biggr) c^{-0.8(n+1)}\ge \biggl(\ga\biggr)\co{-0.8}c^{0.2}\ge (1+\e)\biggl(\ga\biggr)\co{-0.8} $$ if $0<\e\e^{\prime}\co{0.2}$ with some appropriate $\e^{\prime}(\e)>0$ we get that $$ \biggl(\gan\biggr)c^{-0.8(n+1)} \ge \biggl(\ga\biggr)\co{-0.8}+\e^{\prime}\co{0.2} $$ Therefore $$ \align & \tag 5.19\\ \s2&\le K\exp\left\{\frac{c^n}TM^2+g_n(|x|-M)-\biggl(\ga\biggr)\co{-0.8} +q^{n+1}-\e^{\prime}\co{0.2}\right\} \\ &\le L_n\exp\left\{g_n(|x|-M)-\biggl(\ga\biggr)\co{-0.8} -\frac{\e^{\prime}}2\co{0.2} \right\} \,. \endalign $$ This is estimate (4.5) (with $\frac{\e^{\prime}}2$ instead of $\e^{\prime}$). For $x\in\Om1$ $$ \gnx=g_n(|x|-M)-\biggl(\ga\biggr) \xo2+O(\co{-0.2})\,, $$ hence (5.19) implies that for $x\in\Om1$ $$ \s2\le L_n\exp\left\{\gnx-\frac{\e^{\prime}}4\co{0.2}\right\}\,, $$ and this is relation (4.2) (with $\frac{\e^{\prime}}4$). \medpagebreak d) The estimation of $\s3$ for $x\in\Om1$ and $x\in \Om2$. Clearly $$ \s3 =\s{\e}+ \int_{\{y,\;\xy\in\Omm3\}\cap V_n^{\e}(x)} \exp\biggl(\frac{c^n}Txy\biggr)f\biggl(\xy\biggr)p_n(y)\,dy \,, $$ where $\s{\e}$ and $V_n^{\e}$ are defined in (4.8) and (4.8${}^{\prime}$). The term $\s{\e}$ is bounded in (5.13). On the other hand, we claim that there is some $\e_0=\e_0(c)$ such that if $0<\e<\e_0$ and $x\in \Om1\cup\Om2$ then the set $ \{y,\xy\in\Omm3\} \cap V_n^{\e}(x)$ is empty, hence the last integral is zero. We have to show that if $\bigl||x|-M\bigr|<\co{-0.4}$, \ $y\in V_n^{\e}$, i.e. $\bigl||y|-M\bigr|<\e\co{-0.4}$ and $xy>|x||y|-\e\co{-0.4}$ then $\xy\notin\Omm3$, i.e. $-c^{-0.4(n+1)}\allowmathbreak\le|\frac{x+y}2|-M\allowmathbreak \le c^{-0.4(n+1)}$. Estimation from above: $$ \left|\frac{x+y}2\right|-M\le \frac{|x|-M}2+\frac{|y|-M}2\le\frac{1+\e}2\co{-0.4} \le c^{-0.4(n+1)} $$ if $\frac{1+\e}2\le c^{-0.4}$, what holds for sufficiently small $\e$. Estimation from below: $$ \align &\left|\frac{x+y}2\right|^2-M^2=\frac{|x|^2+|y|^2+2xy}4-M^2\\ &\qquad\ge\frac{|x|^2+|y|^2+2|x||y|-2\e\co{-0.4}}4-M^2= \biggl(\frac{|x|+|y|}2\biggr)^2-M^2-\frac{\e}2\co{-0.4}\\ &\qquad\ge\left(\frac{M-\co{-0.4}+M-\e\co{-0.4}}2\right)^2-M^2- \frac{\e}2\co{-0.4}\\ &\qquad\ge -(1+\e)M\co{-0.4}-\frac{\e}2\co{-0.4}= -M\left(1+\e+\frac{\e}{2M}\right)\co{-0.4}\,. \endalign $$ Hence $$ \align \biggl|\frac{x+y}2\biggr|&\ge \left(M^2-M\biggl(1+\e +\frac{\e}{2M}\biggr)\co{-0.4}\right)^{1/2}\\ &\ge M-\frac{(1+\e+\frac{\e}{2M})}2\co{-0.4}-K\co{-0.8}\\ &=M-\left[\biggl(\frac12+\frac{\e}2\biggl(1+\frac1{2M} \biggr)\biggr) +K\co{-0.4}\right]\co{-0.4} \ge M-c^{-0.4(n+1)} \endalign $$ if $\frac12+\frac{\e}2(1+\frac1{2M})+K\co{-0.4} \le c^{-0.4}$, what we had to show. Hence $\s3=\s{\e}$, and (5.13) implies (4.6). To prove (4.3) we still have to remark that for $x\in \Om1$ $$ \align &g_n(|x|-M)-\biggl(\ga\biggr) \co{-0.8}-\frac12c^{n/2} \\ &\le g_n(|x|-M)-\biggl(\ga\biggr) \xo{2}-\frac12c^{n/2} \le g_n(\x1-M)+A_n \xo2-\frac16c^{n/2}\,. \endalign $$ \medpagebreak e) The estimation of $S_nf(x)$ for $x\in\Om3$. We get from (5.12) and Part a) of Lemma 2 that $$ 0\le S_nf(x) \le 2c^n\exp\left\{\frac{c^n}TM^2+ \frac{g_n}{2M}(|x|^2-M^2) -\frac{c^n}{3T}(|x|^2-M^2)\right \}\,. $$ For $x\in\Om3$ \ $\frac{c^n}{3T}(|x|^2-M^2)\ge\frac{c^{0.2n}}{3T}$, hence (4.1${}^{\prime\prime}$) implies that $$ \align 0\le S_nf(x) &\le 2c^n\exp\left\{\frac{c^n}TM^2+ \frac{g_n}{2M}(|x|^2-M^2) -\frac{\co{0.2}}{3T}\right\}\\ &\le L_n\exp \left\{\frac{g_n}{2M}(|x|^2-M^2)\right\}\,, \endalign $$ i.e. relation (4.7) holds, as we have claimed. Proposition 1${}^{\prime}$ is proved. \subheading{6. The proof of Theorem 1 and Proposition 2. Existence of the thermodinamical limit} First we show with the help of Proposition 1, Lemma 1 and Theorem A that for all $q$, $c^{-0.2}M+\co{-0.4} \tag 6.4 $$ $$ \fn\le L_n\exp \left\{\frac{\gb c^n}{2M}(|x|^2-M^2)\right\}\qquad\text{if }\;00$ and $|x|^2-M^2<0$, since here we apply that $\gb c^n0\,.\tag 6.7 $$ Similarly, $$ \im2\le\lhn \exp\left\{-\frac12\biggl(\frac{\gb}{2M} -\ab\biggr)\co{0.2}\right\}\,.\tag 6.7${}^{\prime}$ $$ (Observe that $\frac{\gb}{2M}-\ab>0$.) Define the number $T_n$, $$ T_n=\im1\exp\left\{\gb c^n(x^{(1)}-M)+\ab c^n x^{(2)2}\right\}p_n(x)\,dx\,.\tag 6.8 $$ It follows from Theorem A that $$ C_1c^{-n/2}n_0$ and $N>N_0(n,q)$ then $$ \left|\tilde\mu_n(A(n))-\mn (A(n))\right|\le Kq^n\tag 6.11 $$ with some $K>0$ independent of the set $A$. Indeed, $$ \align &\left|\tilde\mu_n(A(n))-\mn (A(n))\right|\le \int_{\Om2\cup\Om3}\fn p_n(x)\,dx\\ &\qquad+\int_{\tilde A(n)} \mu_n(\,dx_1\dots, dx_{2^n}) \biggl|f^{h_N}_{n,N}\biggl(2^{-n} \sum_{j=1}^{2^n}x_j\biggr)\\ &\qquad\qquad-L_n\exp\left\{\gb c^n2^{-n}\sum_{j=1}^{2^n} (x_j^{(1)}-M)+\ab c^n 4^{-n} \biggl(\sum_{j=1}^{2^n}x_j^{(2)}\biggr)^2\right\}\biggr| =I_1+I_2\,. \endalign $$ It follows from (6.5), (6.7) and (6.7${}^{\prime}$) that $$ I_1\le \exp\left(-\co{0.1}\right)\,.\tag 6.12 $$ On the other hand we increase the term $I_2$ by enlarging the domain of integration to the set $\{(x_1,\dots,x_{2^n}),\; \,2^{-n}\sum_{j=1}^{2^n}x_j\in \Om1\}$. Hence $$ I_2\le \im1\left|\fn -L_n\exp\left\{\gb c^n (x^{(1)}-M)+\ab c^nx^{(2)2}\right\}\right|p_n(x)\,dx\,. $$ The last inequality together with (6.2) and (6.2${}^{\prime}$) imply that $$ I_2\le2 q^n\mn\left((R^2)^{2^n}\right)=2q^n\;. $$ The last inequality together with (6.12) imply (6.11). Since for all $k\ge0$ and measurable sets $A\in \rk$, \ $k\le n\le N$ we have $\mn(A(n))=\mk(A)$, relation (6.11) implies that for all $\e>0$ there is some $N_0(\e)$ such that for $N>N_0(\e)$ and $N^{\prime}> N_0(\e)$ the relation $|\mn(A)-\mu^{h_{N^{\prime}}}_{k,N^{\prime}}(A)|<\e$ holds true. Let us emphasize that the threshold $N(\e)$ does not depend on the set $A$. Hence the last relation means that the limit $\bar\mu_k(A)=\lim_{N\to\infty}\mk(A)$ exists, and the convergence is uniform in $A$. This implies that $\mk\to\bar\mu_k$ in variational metric. To complete the proof of Theorem 1 we have to show that the measure $\bar\mu_k$ does not depend on the sequence $h_N$. But it is not difficult to see with the help of (6.11) that this statement holds, since $\bar\mu_k(A)= \lim_{n\to\infty}\tilde\mu(A(n))$, and the right hand side of the last expression does not depend on $h_N$. \demo{Proof of Proposition 2} Let $n>n_0$ and $N>N_0(n,q)$. Relations (6.1)--(6.5) hold for such pairs $n$ and $N$. By Theorem 1 the measures $\mn$ converge in variational metric to the projection $\bar\mu_n$ of the measure $\bar\mu$ to $\rn$ as $N\to \infty$. Since all measures are absolute continuous with respect to the measure $\mu_n$ the above convergence is equivalent to the convergence of the Radon--Nikodym derivatives $\frac{d\mn}{d\mu_n}=f_{n,N}^{h_N}$ to $\frac{d\bar\mu_n}{d\mu_n}=\bar f$ in $L_1$ norm in the space $(\rn,\mu_n)$ as $N\to\infty$. Since all functions $\fn$ satisfy (6.1)--(6.5) for $N>N_0(n,q)$, their limit $\bar f$ also has this property. Hence Proposition 2 holds true.\enddemo \subheading{7. The proof of Theorem 2. Existence of the large-scale limit} First we need some results about the transformation $Q_n=Q_n(k)$ of probability measures on $(R^2)^{2^{n+k}}$ to probability measures on $(R^2)^{2^k}$ to be defined below. First we define a transformation $Q_n=Q_n(k)$, \ $Q_n\: (R^2)^{2^{n+k}}\to (R^2)^{2^k}$ in the following way: For all $(x_1,\dots,x_{2^{n+k}})$, \ $x_j\in R^2$, $j=1,\dots,2^{n+k}$ $$ Q_n (x_1,\dots,x_{2^{n+k}})=(y_1,\dots,y_{2^k}),\qquad y_j= 2^{-n}\!\!\sum_{l=(j-1)2^n+1}^{j2^n}\!\! x_l,\qquad j=1,\dots,2^k. $$ This transformation induces the transformation $Q_n$ of probability measures on $(R^2)^{2^{n+k}}$ to probability measures on $(R^2)^{2^k}$ in a natural way. Namely, if $\nu$ is a probability measure on $(R^2)^{2^{n+k}}$ and $(\eta(1), \dots,\eta(2^{n+k}))$ is a $\nu $ distributed vector then $Q_n\nu$ is the distribution of the random vector $Q_n (\eta(1),\dots,\eta(2^{n+k}))$. In Theorem 2 we have to study an appropriately rescaled version of the measure $Q_n\bar \mu _{n+k}$. It is not difficult to see that relation (2.10) implies that $$ \frac{dQ_n\bar \mu_{n+k}} {dQ_n \mu_{n+k}} (x_1,\dots,x_{2^{k}})= \bar f_{n+k}\biggl(2^{-k}\sum_{j=1} ^{2^k}x_j\biggr).\tag 7.1 $$ We formulate below Theorem C which follows from the relatively simple Theorem 1 in [4]. For the sake of completeness we present its proof in Appendix B. \proclaim{Theorem C}\it The above defined measure $Q_n\mu_{n+k}=Q_n\mu_{n+k} (T,t)$ has a density function $h_k(x_1,\dots,x_{2^k})$ of the form $$ \align h_k(x_1,\dots,x_{2^k})=L(T,t,n,k)\exp&\left\{-\frac1T\Cal H_k\left(c^{n/2}x_1,\dots,c^{n/2}x_{2^k}\right)\right\} \prod_{j=1}^{2^k}p_n(x_j)\,\\ &\qquad\qquad\qquad x_j\in R^2,\quad j=1,\dots,2^k. \endalign $$ Here $\Cal H_k$ is the Hamiltonian function defined in (1.2${}^{\prime}$) of Part I, $p_n(x)$ is the function appearing in Theorem A, and $L$ is an appropriate norming constant.\endproclaim Formula (7.1) and Theorem C enable us to express the density function of the random vector $\{\Cal R_n \sigma^{(1)}(j), \, \Cal R_n \sigma^{(2)}(j),\;1\le j\le 2^k\}$ with the help of the functions $p_n(x)$ and $f_n(x)$, where the sequence $\{\sigma(j),\;j\in \bold Z\}$ is $\bar\mu$ distributed, and $\Cal R_n\sigma^{(1)}$, $\Cal R_n\sigma^{(2)}$ are defined in (1.2) and (1.3) of Part II. This density equals to $$ \align & \tag 7.2 \\ &h_{n,k}(x_1,\dots,x_{2^k})\\&\qquad=L_{n,k}\bar f_{n+k}\biggl(2^{-k}\sum_{j=1}^{2^k}\tilde x_j\biggr)\exp\left\{-\frac1T\Cal H_k\biggl(c^{n/2}\tilde x_1,\dots,c^{n/2}\tilde x_{2^k}\biggr)\right\} \prod_{j=1}^{2^k}p_n(\tilde x_j)\, \endalign $$ with $$ \tilde x = \tilde x(x)=\left(M+c^{-n}x^{(1)},c^{-n/2}x^{(2)}\right)\qquad \text{for }\;x=\left(x^{(1)},x^{(2)}\right)\,. \tag 7.2${}^{\prime}$ $$ Let us define the sets $W_n\subset R^2$ and $ \bar W_n\subset R^2$ by the formulas $$ \align &\bar W_n=\left\{(x^{(1)},x^{(2)}),\;M- \frac {n\eta} {c^{n}}<|x|0\right\} \\ &W_n=\left\{(x^{(1)},x^{(2)}),\;\tilde x(x)\in \bar W\right\}\,,\endalign $$ where $\eta$ and $\alpha$ are the same constants as in Theorem A, and $\tilde x(x)$ is defined in (7.2${}^{\prime}$). We shall show that there is some $n_0>0$ and $00\right\}\,, $$ and write $$ I_1=I_{1,1}+I_{1,2}=\int_{\Omega_n^1-\tilde\Omega_n^1}+ \int_{\tilde\Omega_n^1-\bar W_n}\,. $$ We get, similarly to the estimation of $I_3$ and $I_2$ that $I_{1,1}\le\exp(-Kc^{0.1n})$, and we can write by (2.11) and (2.11${}^{\prime}$) that $$\align I_{1,2}&\le 2L_n\int_{\tilde\Omega^1_n-\bar W_n} \exp\left\{\bar gc^n(x^{(1)}-M)+\bar Ac^nx^{(2)2}\right\}p_n(x)\,dx \\&\le 3L_n\int_{\tilde\Omega^1_n-\bar W_n} \exp\left\{\bar gc^n(|x|-M)-\biggl(\frac{\bar g_n}{2M}-A\biggr) c^nx^{(2)2}\right\}p_n(x)\,dx \,. \endalign $$ Then integrating first by $x^{(2)}$ we get that $$ I_{1,2}\le KL_nc^{-n/2}\biggl[\int_{M-c^{-0.4n}}^{M-\eta nc^{-n}}+\int_{M+\eta n^{1/\alpha}c^{-n}}^{M+c^{-0.4n}}\biggr] \exp\left(\bar gc^n(r-M)\right)\bar p_n(r)\,dr\le q^n $$ with the help of relations (1.8) and (1.10) in Theorem A. Let us emphasize that it was the multiplying term $q^n$ in (1.8) that enabled us to give an exponentially small bound for the second term in the last integral. The above estimates imply (7.3). To estimate the expression (7.2) in the case $x_j\in W_n$, \ $j=1,2,\dots,2^k$, we make some preparatory remarks. Put $$\ell_n(x)=c^n(|\tilde x(x)|-M)=c^n\left\{\left[(M+c^{-n}x^{(1)})^2 +c^{-n}x^{(2)2}\right]^{1/2}-M\right\} $$ We have $$ \align \ell_n(x)=x^{(1)}+\frac{x^{(2)2}}{2M}&+O\left(c^{-n} \Bigl(x^{(2)4}+|x^{(1)}| x^{(2)2}\Bigr)\right)\tag 7.4\\ &=x^{(1)}+\frac{x^{(2)2}}{2M}+O\bigl(c^{-0.8n}\bigr)\qquad \text{if} \quad x\in W_n\,, \endalign $$ because, as it is not difficult to see, $c^{-n/2}|x^{(2)}|\allowmathbreak0$ is chosen sufficiently small in Theorem A. Indeed, by Theorem A $$ p_n(\tilde x)=\exp \left\{-\frac{a_0M}T\ell_n(x)\right\} g\left(\frac{a_1M}T\ell_n(x)\right)\left(1+O(q^n)\right)\,, $$ $$ \exp\left\{-\frac{a_0M}T\ell_n(x)\right\}= \exp \left\{-\frac{a_0}T\biggl(Mx^{(1)} +\frac{x^{(2)2}}2\biggr)+O\left(c^{-0.8n}\right)\right\}\,, $$ and $$ \left|g\biggl(\frac{a_1}TM\ell_n(x)\biggr) -g\biggl(\frac{a_1}T(Mx^{(1)} +\frac{x^{(2)2}}2)\biggr)\right|=O(c^{-0.8n}) \tag 7.6 $$ by (7.4) and the boundedness of the function $\frac d{dx}g(x)$. (See Lemma 13 in Part I.) On the other hand, by Lemma 17 of Part I the relation $-2\eta nc^{-0.3n} $$ if $x\in W_n$, and $\eta$ is chosen in Theorem A sufficiently small. Hence (7.6) can be rewritten as $$ g\left(\frac{a_1}TM\ell_n(x)\right) =g\biggl(\frac{a_1}T\biggl(Mx^{(1)}+\frac {x^{(2)2}}2)\biggr)\left(1+O(c^{-n/2}\biggr)\right)\,. $$ The above relations imply (7.5). We also claim that $$ \align &\Cal H_k\left(c^{n/2}\tilde x_1,\dots,c^{n/2}\tilde x_{2^k}\right) \tag 7.7\\&\qquad=-\sum_{i=1}^{2^k-1}\sum_{j=i+1}^{2^k} U(i,j)\left[M(x_i^{(1)}+x_j^{(1)})+x_i^{(2)}x_j^{(2)}+c^nM^2\right] +O(q^n) \endalign $$ and $$ \bar f_{n+k}\biggl(2^{-k}\sum_{j=1}^{2^k}\tilde x_j\biggr)=L_{n+k}\exp \left\{\frac{c^k}{2^k}\bar g\sum_{j=1}^{2^k}x_j^{(1)}+ \frac{c^k}{4^k}\bar A\biggl(\sum_{j=1}^{2^k}x_j^{(2)}\biggr)^2 +O(q^n)\right\}\tag 7.8 $$ if $x_j\in W_n$, $j=1,\dots,2^k$. Indeed, $$ \align &\Cal H_k\left(c^{n/2}\tilde x_1,\dots,c^{n/2}\tilde x_{2^k}\right) \\&\qquad=-\sum_{i=1}^{2^k-1}\sum_{j=i+1}^{2^k} U(i,j)\left[c^n(M+c^{-n}x_i^{(1)})(M+c^{-n}x_j^{(1)}) +x_i^{(2)}x_j^{(2)}\right]\,, \endalign $$ hence to prove (7.7) it is enough to remark that in the last expression the terms $c^{-n}x^{(1)}_ix^{(1)}_j$ are negligibly small, since $c^{-n}x^{(1)}_ix^{(1)}_j=O(c^{-0.8n})$ if $x_i\in W_n$ and $x_j\in W_n$. To prove (7.8) we have to show that $2^{-k}\sum_{j=1}^{2^k}\tilde x_j\in \Omega^1_{n+k}$ if $\tilde x_j\in \bar W_n$ for all $j=1,\dots,2^k$ and then apply Proposition 2. We can write with the notation $\tilde x_j=(\tilde x^{(1)}_j,\tilde x^{(2)}_j)$ that $|2^{-k}\sum_{j=1}^{2^k}\tilde x_j^{(2)} |i\endSb \varphi(d(i,j))\s(i)\sigma(j),\qquad \sigma=\{\sigma(i),\;i\in \zz\}\,, \tag 8.1 $$ where $d(\cdot,\cdot)$ denotes the hierarchical distance on $\zz$ given in formula (1.1) of Part I. In this work we have considered models in the special case $\varphi(x)=|x|^{-a}$ with $a= 2-\frac{\log c}{\log2}$. One question we are going to discuss here is that which are the functions $\varphi$ for which Dyson's model with the Hamiltonian (8.1) has a phase transition at low temperatures. In the boundary case some more delicate phenomena appear which we also want to discuss. The behaviour of vector and scalar-valued models is different. First we discuss the vector-valued case. The quantities $M_n=M_n(T)$ considered in Part I can be defined in a natural way in the general case. The arguments of Part I suggest that the relation $$ M_{n+1}=M_n-\frac1{4^nM_n\varphi (2^n)} \tag 8.2 $$ holds true. The existence or non-existence of phase transition depends on whether $M=\lim_{n\to\infty}M_n$ equals to zero or not if $T$ is small, i.e. if $M_0$ is large. Hence formula (8.2) suggests that a phase transition at low temperatures occurs if and only if $\sum\frac1{4^n\varphi(2^n)}$ is convergent. Dyson has formulated the same conjecture in [13] and proved its convergent part in the special case when $\sigma(i)\in R^3$. He has also solved the problem for scalar-valued models. He proved that there is a phase transition at low temperatures if $\varphi (n)\ge C\frac{\log\log n}{n^2}$ with some $C>0$, and there is none if $\varphi (n)\frac{\log\log n}{n^2}\to 0$. Moreover, in the boundary case $\varphi (n)=C\frac{\log\log n}{n^2}$ the following Thouless effect occurs: There is some critical parameter $T_{cr.} $ such that $M(T)=\lim_{n\to\infty}M_n(T)>0$ for $T\le T_{cr.}$ and $M(T)=0$ for $T>T_{cr.}$. The quantity $M(T)$ has a physical content, it is called the spontaneous magnetization. The interesting feature of the above result is that it states that the function $M(T)$ has a discontinuity at $T=T_{cr.}$. This particular behaviour of the spontaneous magnetization appears only in the boundary case $\varphi (n)=C\frac{\log\log n}{n^2}$. On the other hand, the Thouless effect occurs in some other models too, like in the one-dimensional Ising model with $\frac1 {|x-y|^2}$ interaction, in one-dimensional percolation models if the probability of the event that the points $i$ and $j$ are connected has the order $C(T)|i-j|^{-2}$, e.t.c.. In recent time several interesting papers appeared on this subject, (see e.g. [1], [2], [17]). On the other hand, there are some other interesting phenomena connected with the Thouless effect, like the irregular behaviour of the correlation function, whose investigation requires essentially new ideas. The appearance of phase transitions and the Thouless effect in scalar-valued models are connected with the behaviour of the sequence $M_n$. The quantity $M_{n+1}$ can be expressed asymptotically in a simple way with the help of $M_n$ and the function $\varphi$ in scalar-valued models too. But this formula is essentially different from his vector-valued counterpart, namely $$ M_{n+1}\sim M_n\left[1-\exp\biggl\{-\frac1T M_n^24^n\varphi(2^n)\biggr\}\right]\,. \tag 8.3 $$ In the particular case $\varphi(n)=\frac{\log\log n}{n^2}$ we have $$ _{n+1}\sim M_n\left[1-\exp \biggl\{-\frac1T M_n^2\log n\biggr\}\right]\,. \tag 8.3${}^{\prime}$ $$ Formula (8.3${}^{\prime}$) may help us to understand the cause of the Thouless effect, at least at a heuristic level. If $M_n(T)<\sqrt T$ for some $n$ then relation (8.3) implies that $M(T)=\allowmathbreak\lim_{n\to\infty} M_n(T)=0$, hence either $M(T)\ge \sqrt T$ or $M(T)=0$. Since $M(T)\ne 0$ for small $T$, this relation implies the discontinuity of the function $M(T)$. In vector-valued models relation (8.2) does not suggest such a behaviour. We expect however that some delicate effects appear in this case too, and we are going to study them in the future. Let us remark that the study of existence or non-existence of phase transitions at low temperatures seems to be an essentially simpler problem than the study of the Thouless effect and related questions. In the first problem it is enough to consider sufficiently low temperatures, and in the case of vector-valued models with Hamiltonian function of the form (8.1) for instance the method of the present paper works without any essential changes. In the second problem however, one has to study the behaviour of the model near the critical temperature, and this requires more work and new ideas. Another problem we are going to discuss here is the description of the large-scale limit of vector-valued equilibrium states with translation invariant Hamiltonian function. We have discussed its scalar-valued counterpart in Section 8 of our paper [6], and formulated our conjectures about it. Let us consider vector-valued models on the $d$-dimensional integer lattice with Hamiltonian function $$ \Cal H(\sigma)=-\sum\Sb |i-j|=1\\i,j\in \zzd\endSb \sigma(i)\sigma(j)-\sum_{i\in \zzd}p(\sigma(i)) $$ with $$ p(x)=-\frac t4|x|^4-\frac{|x|^2}2,\qquad t>0\,, $$ and the Lebesgue measure on $R^p$ with some $p\ge 2$ as the free measure of the model. The expression $\sigma(i)\sigma(j)$ in the above formulas denotes scalar product. Let $\{X(k)= (X_k^{(1)},\dots,X_k^{(p)}),k\in\zzd\}$ be a random field with the distribution of a (pure) equlibrium state with the above Hamiltonian function at a certain temperature $T$. If $d\ge3$ then there exists a spontaneous magnetization at sufficiently low temperatures. This is proved with the help of the infrared bounds (see e.g. [14]). In case of phase transition we consider that pure state for which the direction of the spontaneous magnetization is $e_1=(1,0,\dots,0)$, i.e. $E\,X_k^{(1)}=M>0$, and $E\,X_k^{(s)}=0$ for $s=2,\dots,p$. Define the ``renormalized'' random fields $\{Y_k(N)= \allowmathbreak(\,Y_k(N)^1,\dots,Y_k(N)^p\,)\}$, $N=1,2,\dots$ by the formulas $$ Y_k(N)^1=A(N)^{-1}\sum_{j\in D_k(N)} (X_j^{(1)}-EX_j^{(1)} ) \tag 8.4 $$ $$ Y_k(N)^s=B(N)^{-1}\sum_{j\in D_k(N)} X_j^{(s)},\qquad s=2,\dots,p, \tag 8.4${}^{\prime}$ $$ with $$ \align D_k(N)=\Bigl\{j=(j^{(1)},\dots,j^{(d)})\in\zzd,\quad &k^{(r)}N+1\le j^{(r)}\le (k^{(r)}+1)N,\\ &\qquad r=1\dots,d\Bigr\}\,. \endalign $$ We are interested in the question that for which choice of $A(N)$ and $B(N)$ the fields $Y_k(N)$ have a non-trivial limit, i.e. a limit which is not concentrated on a single configuration. We also want to describe the distribution of the limit field. Dyson's hierarchical model with parameter $c$, \ $14$, \ $d=4$ and $d<4$, and they correspond to the cases $c>\sqrt2$, $c=\sqrt2$ and $c<\sqrt2$ in Dyson's hierarchical model. In accordance with [5] we expect that for $d>4$ the large-scale limit exists at low temperatures with $A(N)=N^{d/2}$ and $B(N)=N^{(d+2)/2}$. The limit field $\{Y_k=(Y_k^{(1)},\dots, Y_k^{(p)}),\;k\in\zzd\}$ is such that the fields $\{Y_k^{(j)},\;k\in\zzd\}$, \ $j=1,2,\dots,p$ are independent, $\{Y_k^{(1)},\;k\in\zzd\}$ consists of independent identically distributed Gaussian random variables with zero mean, $\{Y_k^{(s)},\;k\in\zzd\}$, \ $s=2,\dots,p$, are massless free Gaussian fields, i.e. they have the same distribution as the field $$ Y_k=C\int\frac{\exp(ikx)}{|x|} \prod_{j=1}^d\biggl\{\frac{\exp(ix^{(j)})-1} {ix^{(j)}} \biggr\} W(\,dx),\qquad k\in \zzd, $$ with some $C>0$, where $W(\,dx)$ is a complex-valued white noise field on $R^d$ with the conjugation property $W(x)= \overline {W(-x)}$. (For the definition of $W(x)$ see e.g. [16].) The situation is similar in the case $d=4$, i.e. the investigation of Dyson's model suggests a similar behaviour with the only difference that a logarithmic factor appears in the normalizing term $A(n)$. More precisely, for $d=4$ the fields $Y_k^{(N)}$ defined in (8.4) and (8.4${}^{\prime}$ ) with $A(N)= N^2\sqrt{\log N}$ and $B(N)=N^{(d+2)/2}=N^3$ have a Gaussian limit as $N\to\infty$ which consists of independent components $s=1,\dots,p$, similarly to the case $d>4$. The limit of the fields $Y_k^{(s)}$, $s=2,\dots,p$, is a massless free field. The result of the present work motivates the following conjecture for $d=3$. \proclaim{Conjecture}\it For $d=3$ the large-scale limit exists at low temperatures with the normalizations $A(N)=N^2$ and $B(N)=N^{5/2}$. The large-scale limit has the same distribution as the random field $\{Y_k=(Y_k^{(1)},\dots,Y_k^{(p)}),\;k\in\zz^3\}$ defined by the formulas $$ Y_k^{(s)}=C\int\frac{\exp(ikx)}{|x|} \prod_{j=1}^3\frac{\exp(ix^{(j)})-1} {ix^{(j)}} W_s(\,dx),\quad k\in \zz^3\,, \; s=2,\dots,p\tag8.5 $$ and $$ \align & \tag8.6\\ Y_1^{(s)}=-\frac{C^2}{2M} \sum_{s=2}^p\iint\frac{\exp(ik(x+y))}{|x||y|} \prod_{j=1}^3\biggl\{\frac{\exp[i(x^{(j)}+y^{(j)})]-1} {i(x^{(j)}+y^{(j)})}\biggr\} &W_s(\,dx)W_s(\,dy),\\&k\in \zz^3\,, \endalign $$ where $C>0$ is an appropriate positive constant, and $W_s(\,dx)$, $s=1,\dots,p$ are independent complex valued white noise fields on $R^3$ with the conjugation property $W(x)=\overline{W(-x)}$. (For the definition of two-fold stochastic integrals with respect to a Gaussian field see e.g. [16]. Such an integral appears in the definition of $Y_1^{(s)}$.) \endproclaim The fields $Y_s^{(k)}$, \ $s=2,\dots,p$ defined in (8.5) are massless free fields, the field defined in (8.6) belongs to the class of self-similar fields constructed in Dobrushin's paper [11]. It is a quadratic functional of a Gaussian field, just as the corresponding field in Dyson's model for $12$ the large-scale limit has the same (trivial) behaviour as for $d=1$. When typing the final version of this work the authors learned about some recent results about the Kosterlitz--Thouless effect (see [22], [23]). The arguments of these works, also supported by computer simulation, suggest that the situation in two-dimensional translation invariant models is essentially different from what we had expected. In particular, the difference between the cases $p=2$ and $p>2$ in the models we have discussed at the end of this Section does nevertheless not occur. \heading Appendix\endheading \subheading {Appendix A. The proof of the basic recursive relations (2.1) and (2.1${}^{\prime}$) in Part I} Formula (2.1${}^{\prime}$) immediately follows from (1.4) in Part I with $n=0$. To prove (2.1) let us first observe that the recursive relation $$\align \Cal H_{n+1}\left(x_1,\dots, x_{2^{n+1}}\right)&= \Cal H_n\left(x_1,\dots, x_{2^n}\right)+ \Cal H_n\left(x_{2^n+1},\dots, x_{2^{n+1}}\right)\tag A1\\&\qquad\qquad -c^n \biggl(2^{-n}\sum_{i=1}^{2^n}x_i \biggr) \biggl(2^{-n}\sum_{j=2^n+1}^{2^{n+1}}x_j \biggr)\endalign $$ holds for $n\ge0$, where $$ \Cal H_n(x_1,\dots, x_{2^n})=-\sum_{i=1}^{2^n}\sum_{j=i+1}^{2^n} U(i,j)x_i x_j\;, $$ and $U(i,j)$ is defined by (1.1) and (1.2${}^{\prime}$) in Part I. By relation (1.4) in Part I $$ \align p_{n+1}(x,T)=\frac1{Z_{n+1}(T,t)} \int\exp \biggl\{-\frac1T&\Cal H_n(x_1,\dots,x_{2^{n+1}})\biggr\}\tag A2\\ &\delta\biggl( 2^{-(n+1)}\sum_{i=1}^{2^{n+1}}x_i-x\biggr) \prod_{i=1}^{2^{n+1}}p(x_i)\,dx_i\;, \endalign $$ where $ Z_{n+1}(T,t) $ is an appropriate norming constant, and $\delta\bigl( 2^{-(n+1)}\sum_{i=1}^{2^{n+1}}x_i-x\bigr)$ means that integration in (A2) is taken on the hyperplane $2^{-(n+1)}\sum_{i=1}^{2^{n+1}}x_i=x$ with respect to the Lebesgue measure. Let us fix some number $u$, and calculate the integral on the right-hand side of (A2) by integrating first on the hyperplane defined by the relations $2^{-n} \sum_{i=1}^{2^n}x_i=x+u$ and $2^{-n} \sum_{i=2^n+1}^{2^{n+1}}x_i=x-u$ and then by integrating by $u$. We get with the help of relations (A1) and (A2) that $$ \align p_n(x,T)&=\frac1{Z_{n+1}(T,t)}\int \exp\left\{\frac{c^n}T (x+u)(x-u)\right\}\\ &\left[\int\exp \left\{-\frac1T\Cal H_n(x_1,\dots,x_{2^n})\right\}\delta\left(2^{-n}\sum_{i=1}^{2^n} x_i-(x+u)\right)\prod_{i=1}^{2^n}p(x_i)\,dx_i\right] \\ &\biggl[\int\exp \left\{-\frac1T\Cal H_n(x_{2^n+1},\dots,x_{2^{n+1}})\right\} \delta\biggl(2^{-n}\sum_{i=2^n+1}^{2^{n+1}} x_i-(x-u)\biggr)\\ &\qquad\prod_{i=2^n+1}^{2^{n+1}}\!\!p(x_i)\,dx_i\biggr]\,du \\ &=C_n\int\exp \left\{\frac{c^n}T(x^2-u^2)\right\} p_n(x+u)p_n(x-u)\,du\;, \endalign $$ as we have claimed. \subheading{Appendix B. The proof of Theorem C} Since the measure $\mu_{n+k}$ has the density function $$ \frac1Z_{n+k}\exp \left\{-\frac1T\Cal H_{n+k}(z_1,\dots,z_{2^{n+k}})\right\} \prod_{i=1}^{2^{n+k}}p(z_i) \, , $$ the density function of the measure $Q_n \mu_{n+k}$, the function $h_k(x_1,\dots,x_{2^k})$ equals to $$ \align h_k(x_1,\dots,x_{2^k})=\frac1Z_{n+k}\int \exp &\left\{-\frac1T\Cal H_{n+k}(z_1,\dots,z_{2^{n+k}})\right\}\tag B1\\ &\prod_{l=1}^{2^k}\delta\biggl(2^{-n}\! \! \sum_{j=(l-1)2^n+1}^{l2^n} \!\! z_j-x_l\biggr) \prod _{i=1}^{2^{n+k}} p(z_i)\,dz_i \,, \endalign $$ where $\prod_{l=1}^{2^k}\delta(2^{-n}\sum_{j=(l-1)2^n+1}^{l2^n} z_j-x_l) $ in the integral (B1) means that integration is taken on the hyperplane defined by the relations $2^{-n}\sum_{j=(l-1)2^n+1}^{l2^n}z_j=x_l$, \ $l=1,\dots,2^k$, with respect to the Lebesgue measure. The special structure of the hierarchical distance implies that $$ \Cal H_{n+k}(z_1,\dots,z_{2^{n+k}}) = \sum_{l=1}^{2^k}\Cal H_n(z_{(l-1)2^n+1},\dots,z_{l2^n})- \sum_{i=1}^{2^k}\sum_{j=i+1}^{2^k} c^nU(i,j)\bar z_i\bar z_j\,, $$ with $\bar z_i=2^{-n}\sum_{p=(i-1)2^n+1}^{i2^n} z_p$, \ $i=1\dots,2^k$. Hence relation (B1) can be rewritten as $$ \align h_k(x_1,\dots,x_{2^k})&=\frac1Z_{n+k}\exp \left\{ \frac1T \sum_{i=1}^{2^k}\sum_{j=i+1}^{2^k} c^nU(i,j)x_ix_j \right\}\\ &\qquad\cdot\prod_{l=1}^{2^k}\int \exp \left\{-\frac1T \Cal H_n(z_{(l-1)2^n+1},\dots,z_{l2^n})\right\}\\ &\qquad\qquad \delta\biggl(2^{-n}\sum_{j=(l-1)2^n+1}^{l2^n} z_j-x_l\biggr) \prod_{j=(l-1)2^n+1}^{l2^n}p(z_j)\,dz_j \\ &=C_{k,n}\exp \left\{-\frac1T\Cal H_k(c^{n/2}x_1,\dots, c^{n/2}x_{2^k})\right\}\prod_{l=1}^{2^k}p_n(x_l,T)\,, \endalign $$ as we have claimed. \subheading{Appendix C. The calculation of the Radon--Nikodym derivatives. The proof of formulas (2.1)---(2.3${}^{\prime}$) in Part II} For $n=N$ relations (2.1) and (2.2) of Part II immediately follow from formula (1.4) in Part II. Hence it is enough to prove our relations by induction from $n+1$ to $n$. Clearly, $$ \align P_{n+1}(x_1,\dots,x_{2^{n+1}})&=C_n P_n(x_1,\dots,x_{2^n}) P_n(x_{2^n+1},\dots,x_{2^{n+1}})\\ &\qquad \exp\left\{\frac{c^n}T \biggl(2^{-n}\sum_{j=1}^{2^n} x_j\biggr) \biggl(2^{-n}\!\sum_{j=2^n+1}^{2^{n+1}}\! x_j\biggr) \right\} \endalign $$ with some norming constant $C_n$. Given some measurable set $A\subset (R^p)^{2^n}$ define the cylindrical set $\tilde A \subset (R^p)^{2^{n+1}}$ as $\tilde A=A\times (R^p)^{2^n}$. By our inductive hypothesis for $n+1$ $$ \align \mu_{n,N}^{h_N}(A)&=\int_{\tilde A} f_{n+1,N}^{h_N} \biggl(2^{-(n+1)}\sum_{j=1}^{2^{n+1}} x_j \biggr) P_{n+1}(x_1,\dots,x_{2^{n+1}})\,dx_1\dots dx_{2^{n+1}}\\ &=C_n \int_{\tilde A} f_{n+1,N}^{h_N} \biggl(2^{-n}\biggl(\sum_{j=1}^{2^n}\frac{ x_j}2 + \sum_{j=2^n+1}^{2^{n+1}}\!\frac {x_j}2 \biggr) \biggr)\\ &\qquad\qquad P_n(x_1,\dots,x_{2^n}) P_n(x_{2^n+1},\dots,x_{2^{n+1}}) \\ &\qquad\qquad\exp\left\{\frac{c^n}T \biggl(2^{-n}\sum_{j=1}^{2^n} x_j \biggr) \biggl(2^{-n}\sum_{j=2^n+1}^{2^{n+1}} x_j\biggr)\right\} \,dx_1\dots dx_{2^{n+1}} \;. \endalign $$ Let us calculate the last integral by first integrating on the hyperplanes where $x_1,\dots,x_{2^n}$ and $y= \frac1{2^n}\sum_{j=2^n+1}^{2^{n+1}} x_j$ are fixed. Since $P_n(x_{2^n+1},\dots,x_{2^{n+1}})$ is the only term in the integrand which is not constant on such a hyperplane, and its integral equals $p_n(y)$ on it, we get that $$ \align \mu_{n,N}^{h_N}(A)&=C_n\int_{A\times R^p} f_{n+1,N}^{h_N} \biggl(2^{-n}\Bigl(\sum_{j=1}^{2^n} \frac{x_j}2 +\frac y2\Bigr) P_n(x_1,\dots,x_{2^n})\biggr)p_n(y) \\ &\qquad\exp\biggl\{\frac{c^n}T 2^{-n}(\sum_{j=1}^{2^n} x_j)y\biggr\}\,dx_1\dots\,dx_{2^n}dy\;. \endalign $$ Hence we get, by integrating first by the variable $y$ that $$ \mu_{n,N}^{h_N}(A) =C_n^{\prime}\int_AS_n f_{n+1,N}^{h_N} \biggl(2^{-n}\sum_{j=1}^{2^n} x_j \biggr) P_n(x_1,\dots,x_{2^n}) \,dx_1\dots\, dx_{2^n}\;. $$ Since this relation holds for all measurable sets $A\subset (R^p)^{2^n}$, it implies our inductive hypothesis for $n$. \subheading{Appendix D. On limit Gibbs states} Here we briefly describe the definition of limit Gibbs states (also called equilibrium states in the literature,) and discuss some important questions related to this definition. Limit Gibbs states are defined with the help of a Hamiltonian (often called energy) function, a free measure and a physical parameter, the temperature T. The Hamiltonian function is a formal series. Let us have a subset $\bold Z\subset \bold Z^d$ of the $d$-dimensional integer lattice and a closed set $K\subset R^p$ in the $p$-dimensional Euclidedan space. We consider a Hamiltonian function $\Cal H(\sigma)$ of the form $$ \Cal H(\sigma)=-\sum_{i,j\in\bold Z} U(i,j)\sigma(i)\sigma(j),\quad \sigma=\{\sigma(j),\;\sigma(j)\in K,\;j\in \bold Z\}, $$ where $U(\cdot,\cdot)$ is a given function, $ U:\;\bold Z\times\bold Z\to R^1$, and $\sigma(i)\sigma(j)$ denotes scalar product. (There is a more general definition of Hamiltonian functions, but this special class is sufficiently large for our purposes.) Given some finite set $V\subset \bold Z$, we define the energy function $\Cal H_V(\sigma)$ as $$ \Cal H_V(\sigma)=-\sum_{i,j\in V} U(i,j)\sigma(i)\sigma(j) \,,\qquad \sigma=\{\sigma(i),\;i\in V\}, \tag D1 $$ and the conditional energy in $V$ with respect to a configuration $\bar\sigma$ in $\bold Z-V$ as $$ \align \Cal H_V(\sigma|\bar \sigma)&=\Cal H_V(\sigma)- \sum_{i\in V}\sum_{j\in \bold Z-V} U(i,j)\sigma(i)\bar\sigma(j) ,\tag D2 \\ &\qquad\sigma=\{\sigma(i),\;i\in V\},\; \bar\sigma=\{\bar\sigma(i),\;i\in \bold Z-V\}, \endalign $$ provided that the last sum is convergent. Given some $h\in R^p$, we also introduce the energy of a configuration $\sigma$ in the volume $V$ with respect to the external field $h$ as $$ \Cal H_V^h(\sigma)=\Cal H_V(\sigma)-h\sum_{i\in V}\sigma(i).\tag D3 $$ Given some finite set $V\subset\zz$, a configuration $\ssb=\{\ssb(j),j\in\zz-V\}$ outside $V$, a Hamiltonian function $\Cal H(\sigma))$ and a free measure $P(\,dx)$ on $K$, we define the Gibbs measure in volume $V$ with respect to the external field $\ssb$ at temperature $T$ as the probability measure $\mu_{V,T}(\,\cdot\, |\ssb)$ on $K^V$ given by the formula $$ \mu_{V,T}(\sigma\in A|\ssb)=\frac1{Z(V,T,\ssb)}\int_A\exp \biggl\{- \frac1T\Cal H_V(\sigma|\ssb)\biggr\} \prod_{j\in V} P(\,d\sigma(j)), \tag D4 $$ where $A\in K^V$ is an arbitrary measurable set, $\sigma=\{\sigma(j),j\in V\}$ and $Z(V,T,\ssb)$ is an appropriate norming constant, provided that the above expression is meaningful. Now we formulate the following \proclaim{Definition of Gibbs states}\it A probability measure $\mb$ is a Gibbs state with Hamiltonian function $\Cal H$ and free measure $P$ at temperature $T$ if a $\mb$ distributed random field $\sigma (j),\;j\in\zz$ satisfies the following relation: For any finite set $V$ and measurable set $A\subset K^V$ the conditional probability of the event $\sigma\in A$, \ $\sigma=\{\sigma(j), j\in V\}$ with respect to the condition $ \{\sigma(j)=\ssb,\; j\in \zz-V\}$ with a configuration $\ssb=\{\ssb(j),\,j\in \zz-V\}$ equals to $$ \mb (\sigma\in A\,|\,\{\sigma(j),\,j\in \zz-V\}=\ssb)= \mu_{V,T}(A\,|\,\ssb) $$ with $\mb$ probability one, where $\mu_{V,T}$ is defined in (D4).\endproclaim The question arises whether Gibbs states on $K^{\zz}$ exist, and whether they are unique. A natural way to construct Gibbs states is to carry out the following procedure. Choose an increasing family of sets $V_n\subset\zz$, \ $\cup V_n=\zz$, fix a configuration $\ssb=\ssb^{(n)}= \{\ssb(j),\,j\in \zz-V_n\}$ for each $V_n$, and consider the measures $\mu_{V_n,T}(\,\cdot\,|\ssb)$ defined in (D4). Prove that under some mild restrictions there is a convergent subsequence of this sequence, and the limit of this subsequence is a Gibbs state. The problem about the uniqueness of Gibbs states is closely related to the question whether, in dependence of the choice of the external configuration $\ssb^{(n)}$, different limits can appear in the above construction. A slightly different, and often useful approach is to choose a sequence $h_n\in R^n$, $h_n\to 0$, and try to construct Gibbs states as the limit of a sequence of measures of the form $\mu_{V_n,T}^{h_n}$, where we define the probability measure $\mu_{V,T}^h$ as $$ \mu_{V,T}^h(A)= \frac1{Z(V,T,h)} \int_A\exp \biggl\{ -\frac1T\biggl(\sum_{i\in V}\sum_{j\in V}U(i,j)x_ix_j-h\sum_{i\in V}x_i\biggr)\biggr\} \prod_{i\in V}P(dx_i)\,.\tag D5 $$ If $K$ is a compact subset of $R^p$ then standard results in probability theory imply the compactness of the measures $\mu_{V_n,T}(\,\cdot\,|\ssb)$ or of $\mu_{V_n,T}^{h_n}$ in weak topology, i.e. the existence of a convergent subsequence in this topology. (See e.g. [3].) Nevertheless, there are many interesting models, where the set $K$ is non-compact (e.g. $K=R^p$,) and in such cases a hard analysis is needed to prove the existence of such a convergent subsequence. (See e.g. [10] or [20] as an example.) In order to prove that the limit of the sequence of measures $\mu_{V_n,T}(\,\cdot\,|\ssb)$ (or $\mu_{V_n,T}^{h_n}$) is really a Gibbs state it is worth while to rewrite the definition of Gibbs states in an equivalent integral form. Let $f=f(x_{j_1},\dots,x_{j_k})$ and $g=g(x_{l_1},\dots,x_{l_k})$, $x_{j_i}\in R^p$, $x_{l_i}\in R^p$, be two bounded and continuous functions with finitely many arguments, $V=\{j_1,\dots,j_k\}$, $W=\{l_1,\dots,l_k\}$, $V\subset \zz$, $W\subset \zz$ such that $V\cap W=\emptyset$. The measure $\mb$ on $K^{\zz}$ is a Gibbs state if and only if $$ \int_{K^{\zz}}fg\,d\mb=\int_{K^{\zz-V}}\mu (f)g \,d\mb\tag D6 $$ for all functions $f$ and $g$ with the above properties, where $$ \align \mu(f)=\mu(f)(\ssb)=\int_{K^V} & f(\sigma(j_1),\dots,\sigma(j_k))\ \mu_{V,T}(\,d\sigma |\ssb) \tag D7\\ &\qquad \sigma=\{\sigma(j),j\in V\},\quad \ssb=\{\sigma(j),j\in \zz-V\} , \endalign $$ and $\mu_{V,T}$ is defined in (D4). Let us consider an arbitrary sequence of sets $V_n\subset \zz$, \ $\cup V_n=\zz$, and numbers $h_n\in R^p$, \ $h_n\to0$. Some calculation shows that for sufficiently large $n$ (if $V\subset V_n $, $W\subset V_n$) $$ \int_{K^{V_n}}fg\,d\mu_{V_n,T}^{h_n}=\int_{K^{V_n-V}} \mu^{h_n} (f)g \,d\mu_{V_n,T}^{h_n} \tag D8 $$ with $$ \align &\mu^{h_n}(f)=\mu_{V,V_n,T}^{h_n} (f;\sigma(j),j\in V_n-V) \\ &=\frac{ \int f(\sigma(j_1),\dots,\sigma(j_k)) \exp \bigl\{- \frac1T(\Cal H_{V,V_n}(\sigma) -h_n\sum\limits_{i\in V}\sigma(i)) \bigr\} \prod\limits_{i\in V} P(\,d\sigma(i))} {\int \exp\bigl\{- \frac1T(\Cal H_{V,V_n}(\sigma) -h_n\sum\limits_{i\in V}\sigma(i)) \bigr\} \prod\limits_{i\in V} P(\,d\sigma(i)) } \,, \endalign $$ where $$ \Cal H_{V,V_n}(\sigma)=\sum_{i\in V}\sum_{j\in V_n}\sigma(i)\sigma(j)\;. $$ If the sequence $\mu_{V_n,T}^{h_n}$ tends weakly to the measure $\mb$ then the left-hand side of (D8) converges to that of (D6). Hence to prove that the limit measure $\mb$ is a Gibbs state it suffices to establish the convergence of the right-hand side of (D8) to that of (D6). If the Hamiltonian function has a finite range interaction, i.e. there is some number $r>0$ such that $U(i,j)=0$ if $|i-j|\ge r$ then it is not difficult to see that $\mu^{h_n}(f)(\ssb)\to\mu(f)(\ssb)$, and the required convergence can be proved with the help of this relation. In case of infinite range interaction one must be more careful, especially if the state space $K$ is non-compact. Dyson's model which we are investigating is such a model. In Theorem 1 of Part II we have proved the weak convergence of the measures $\mu_N^{h_n}$ to $\mb$. (Actually, we have proved a stronger form of convergence.) In Appendix E we prove Theorem B, i.e. we show that the limit measure is a Gibbs state. In the proof we approximate Dyson's model with a model with finite range interaction, and this enables us to carry out the required limiting procedure. In Appendix E we restrict ourselves to Dyson's model, although the argument also works in more general cases. \subheading{Appendix E. The proof of Theorem B} We apply the argument of Appendix D. The proof of Theorem B can be completed by showing that also in the case of Dyson's model the right-hand side of (D8) tends to that of (D6). We formulate this statement in more detail. It suffices to consider the case when $V=\{1,2,\dots,2^k\}$, $W= \{2^k+1,2,\dots,2^m\}$ with some $0\le k< m$, i.e. $f=f(x_1,\dots,x_{2^k})$, $g=g(x_{2^k+1},\dots,x_{2^m})$ and $V_N=\{1,2,\dots,2^N\}$. (We apply the notation of Appendix D.) Introduce the functions $$ \align &p^h_{k,N}(x_1,\dots,x_{2^k}| x_{2^k+1},\dots,x_{2^N})\tag E1 \\ &\qquad=\frac { \exp\bigl\{- \frac1T(\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^N} U(i,j)x_ix_j- h\sum_{i=1}^{2^k}x_i) \bigr\} } { \int \exp \bigl\{- \frac1T(\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^N} U(i,j)x_ix_j- h\sum_{i=1}^{2^k}x_i) \bigr\} \prod_{i=1}^{2^k} p(x_i)\,dx_i }\,, \\ \vspace{1\jot} &\qquad\qquad\qquad x_j\in R^p,\;j=1,\dots,2^N \endalign $$ and $$ \align &p_{k}(x_1,\dots,x_{2^k}| x_{2^k+1},\dots)\tag E1${}^{\prime}$ \\ &\qquad=\frac { \exp\bigl\{- \frac1T\sum_{i=1}^{2^k}\sum_{j=i+1}^{\infty} U(i,j)x_ix_j\bigr\} } { \int \exp \bigl\{- \frac1T\sum_{i=1}^{2^k}\sum_{j=i+1}^{\infty} U(i,j)x_ix_j \bigr\} \prod_{i=1}^{2^k} p(x_i)\,dx_i } \,, \\ \vspace{1\jot} &\qquad\qquad\qquad x_j\in R^p,\;j=1,\dots, \endalign $$ where the function $U(\cdot,\cdot)$ is defined in (1.2) and $p(x)$ in (1.3) of Part I. Put $$ \align \mu^h_{k,N}(f)( x_{2^k+1},\dots,x_{2^N}) &=\int f(x_1,\dots,x_{2^k}) \tag E2 \\ &\qquad p^h_{k,N}(x_1,\dots,x_{2^k}| x_{2^k+1},\dots,x_{2^N}) \prod_{i=1}^{2^k} p(x_i)\,dx_i \endalign $$ and $$ \mu_k(f)( x_{2^k+1},\dots) = \int f(x_1,\dots,x_{2^k}) p_k(x_1,\dots,x_{2^k}| x_{2^k+1},\dots) \prod_{i=1}^{2^k} p(x_i)\,dx_i . \tag E2${}^{\prime}$ $$ The convergence of the right hand side of (D8) to that of (D6) is equivalent to the relation $$ \lim_{N\to\infty}\int g\mu_{k,n}^{h_N}(f)\,d\mu_N^{h_N}= \int g\mu_k(f)\,d\mb \tag E3 $$ in our case, where $ \mu_N^{h_N} $ and $\mb$ are the same probability measures on $(R^p)^{2^N}$ and $(R^p)^{\zz}$ as in Theorem 1 of Part I. To prove this relation let us introduce the sets $A(K,k,n,N)$ and $A(K,k,n)$,where $K\in R^1$, \ $K>0$, \ $k, n, N\in \zz$ and $k0$ some $n=n(\e)$ and $K=K(\e,n) $ can be chosen in such a way that $$ \mu^{h_N}_N \bigl ((x_{2^k+1},\dots,x_{2^N}) \notin A(K,k,n,N)\bigr)<\e \tag E4 $$ and $$ \mb \bigl (( x_{2^k+1},\dots,) \notin A(K,k,n,)\bigr)<\e\,. \tag E4${}^{\prime}$ $$ To prove relations (E4) and (E4${}^{\prime}$) let us first observe that there is a universal constant $C$, $C>0$, such that if $\s(j)$ is the $j$-th coordinate of a $\mu_N^{h_N}$ or $\mb$ distributed random variable then the inequality $E\sigma^2(j)0$ that $$ \mu^{h_N}_N \left\{(x_{2^k+1},\dots,x_{2^N}) \notin A(K,k,n,N)\right\}\le C\biggl(\frac{2^n-2^k}{K^2}+ \sum_{j=n}^N 2^{j(1-2\alpha)}\biggr)<\e $$ if first $n$ and then $K$ is chosen sufficiently large. The proof of (E4${}^{\prime}$) is the same. (We remark that relations (E4) and (E4${}^{\prime}$) hold with arbitrary $\alpha>0$ in the definition of the sets $A(\cdot,\cdot,\cdot)$. To prove it we have to apply the stronger statement $E\sigma^{2k}0$ there is some $N_0=N_0(\e,K,n)$ such that for $N>N_0$ $$ \align \biggl | p_{k,N}^{h_n}(x_1,\dots,x_{2^k}|x_{2^k+1}\dots,x_{2^N})&- p_{k,N_0}^{h_n}(x_1,\dots,x_{2^k}|x_{2^k+1}\dots,x_{2^{N_0}})\biggr| \tag E5\\ &\quad<\e\biggl(\sum_{i=1}^{2^k}|x_i|\biggr)\exp\biggl\{(2Kn+1) \sum_{i=1}^{2^k}|x_i|\biggr\} \endalign $$ and $$ \align \biggl | p_k(x_1,\dots,x_{2^k}|x_{2^k+1}\dots)&- p_{k,N_0}^0(x_1,\dots,x_{2^k}|x_{2^k+1}\dots,x_{2^{N_0}})\biggr | \tag E5${}^{\prime}$\\ &\quad<\e\biggl(\sum_{i=1}^{2^k}|x_i|\biggr)\exp\biggl\{(2Kn+1) \sum_{i=1}^{2^k}|x_i|\biggr\} \endalign $$ if $(x_{2^k+1},\dots,x_{2^N})\in A(K,k,n,N)$ and $(x_{2^k+1},\dots)\in A(K,k,n)$. (The constants $K$ and $n$ in formulas (E5) and (E5${}^{\prime}$) are the same as in the definition of the sets $A(K,k,n,N)$.) First we show that (E5) and (E5${}^{\prime}$) together with (E4) and (E4${}^{\prime}$) imply (E3), hence also Theorem B. Indeed, since $p_n(x)$ decreases at infinity faster than $\exp(-x^2/2)$ hence (E5) and (E5${}^{\prime}$) imply that $$ \left|\mu_{k,N}^{h_N}(f)(x_{2^k+1},\dots,x_{2^{N}})- \mu_{k,N_0}^0(f)(x_{2^k+1},\dots,x_{2^{N_0}}) \right| 0$ can be chosen arbitrary small, relation (E3) holds in its original form. We prove only (E5) the proof of (E5${}^{\prime}$) being the same. Let us first observe that for any $\eta>0$ there is some $N_0=N_0 (K,n,\eta)$ such that for $N\ge N_0$ and $(x_{2^k+1},\dots,x_{2^N})\in A(K,k,n,N)$ $$ \align & \tag E6 \\ &\left| \exp\biggl\{- \frac1T(\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^N} U(i,j)x_ix_j- h_N\sum_{i=1}^{2^k}x_i) \biggr\} - \exp\biggl\{- \frac1T\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^{N_0}} U(i,j)x_ix_j\biggr\}\right| \\ &=\exp\biggl\{- \frac1T\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^N} U(i,j)x_ix_j\biggr\}\\ &\qquad\left| \exp\left\{- \frac1T\biggl(\sum_{i=1}^{2^k}\sum_{j=2^{N_0}+1}^{2^N} U(i,j)x_ix_j- h_N\sum_{i=1}^{2^k}x_i\biggr) \right\} - 1\right|\\ &\le\eta \frac{\sum_{i=1}^{2^k}|x_i|}T\exp \left\{\frac{2Kn+\eta}T \sum_{i=1}^{2^k}|x_i|\right\} . \endalign $$ In the last relation we have applied the inequality $|e^x-1| \le |x|e^{|x|}$ together with the relations $h_N<\eta/2$, \ $ | \sum_{j=2^{N_0}+1}^{2^N}U(i,j)x_j|\le |\sum_{j=N_0}^{\infty} 2^{j(\alpha-1)}\bigl(\frac c4\bigr)^j|<\eta/2$ and $|\sum_{j=i+1}^{2^N}U(i,j)x_j|\le 2Kn$ if $N>N_0$, \ $j>2^k$, \ $(x_{2^k+1},\dots,x_{2^N})\in A(K,k,n,N)$ and $N_0$ is sufficently large. Integrating inequality (E6) with respect to the measure $\prod_{i=1}^{2^k}p(x_i)\,dx_i$ we get that $$ \align &\biggl|\int \exp\biggl\{- \frac1T\biggl(\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^N} U(i,j)x_ix_j- h_N\sum_{i=1}^{2^k}x_i\biggr) \biggr\}\prod_{i=1}^{2^k}p(x_i)\,dx_i \tag E7\\ &\qquad- \int \exp\biggl\{- \frac1T\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^{N_0}} U(i,j)x_ix_j\biggr\} \prod_{i=1}^{2^k}p(x_i)\,dx_i \biggr| 0$ can be taken arbitrary small if $N_0=N_0(K,n,\eta) $ is sufficiently large. Hence we prove (E5) by showing that $$ \int \exp\bigl\{- \frac1T\sum_{i=1}^{2^k}\sum_{j=i+1}^{2^{N_0}} U(i,j)x_ix_j\bigr\} \prod_{i=1}^{2^k}p(x_i)\,dx_i >D \tag E8 $$ with some $D >0$ on the set $A(K,k,n,N_0)$, i.e. the integral in (E8) is separated from zero. Here the constant $D$ may depend on $K$ and $n$ but not on $N_0$. Relation (E8) holds, since if $|x_j|<1$, $j=1,2,\dots,2^k$ and $(x_{2^k+1},\dots, x_{2^N})\in A(K,k,n,N_0)$ then the integrand in (E8) is separated from zero. \heading Table of Contents\endheading \vskip 0.5cm \subheading{Preface} \bigpagebreak \subheading{Part I} Limit theorems for the average spin. \item{1.}Introduction \item{2.}On the content of Theorem 1. Convergence to the solution of the fixed point equation. \item{3.}On the strategy of the proof. The inductive procedure. \item{4.}The proof of Proposition 1 and its Corollary. The first step of the inductive procedure. \item{5.}The proof of Proposition 2. The second step of the inductive procedure. \item{6.}The proof of Proposition 3 and some of its consequences. \item{7.}On the fixed point equation $T_M g=g$. \item{8.}The proof of Theorems 1 and 1$^{\prime}$. \item{9.}The proof of Theorem 2. The behaviour of the density of the average spin at infinity. \item{10.}The proof of Lemmas 17 and 18. \subheading{Part II}Description of the large-scale limit. \item{1.}Introduction. \item{2.}On the basic estimates needed during the proof. Reduction to integral equations. \item{3.}The proof of Lemma 1. \item{4.}Some preparatory remarks to the proof of Proposition 1$^{\prime}$. \item{5.}The proof of Proposition 1$^{\prime}$. \item{6.}The proof of Theorem 1 and Proposition 2. Existence of the thermodynamical limit. \item{7.}The proof of Theorem 2. Existence of the large-scale limit. \item{8.}Some open problems and conjectures. \subheading{Appendix} \item{A.}The proof of the basic recursive relations (2.1) and (2.1$^{\prime}$) in Part I. \item{B.}The proof of Theorem C. \item{C.}The calculation of the Radon--Nikodym derivatives. The proof of formulas (2.1)---(2.3${}^{\prime}$) in Part II. \item{D.}On limit Gibbs states. \item{E.}The proof of Theorem B. \bye