\magnification=\magstep1 \input amstex \documentstyle{amsppt} \hsize=16truecm \font \kis=cmr8 \font\small=cmmi8 \font\bfs=cmbx8 \font\its=cmti8 \font\mats=cmsy8 \leftheadtext\nofrills{{\rightline{\kis P. M. Bleher and P. Major}}} \rightheadtext\nofrills{\leftline{\kis Dyson's hierarchical model}} \parskip=3 pt \TagsOnRight \define\zb{\bold Z} \define\s{\sigma} \define\e{\varepsilon} \define\ch{\Cal H} \define\qb{\bar q} \define\bt{\bold T} \define\bbt{\bar \bt} \define\p#1{p_{#1}(x)} \define\q#1{q_{#1}(x)} \define\co{$1i\endSb d(i,j)^{-a}\s(i)\s(j)\,, \tag 1.1 $$ where $d(i,j)=2^{n(i,j)-1}$, and $$ n(i,j)=\min\left\{ n,\text{ there exists some $k$ such that }(k-1)2^n0$ is a sufficiently small number, and $C(t) $ is an appropriate norming constant which turns $p_0(x)$ into a density function. For the sake of convenience we shall work in the sequel with the number $c=2^{2-a}$ instead of the parameter $a$. We investigate the following problem: First we construct an equilibrium state $\mu=\mu(T) $ at low temperatures with magnetization in the direction of the first coordinate $x^{(1)}$ and then we want to describe its large-scale limit. In more detail, let $$ \sigma\,=\left\{\,\sigma(j)=\Bigl( \sigma^{(1)}(j),\dots,\sigma^{(p)}(j) \Bigr) \,\in R^p ,\, j\in \bold Z \right\} $$ be a random field with the distribution of the equilibrium state ${\mu=\mu(T)}$, and define for all $n=1$, 2, \dots the random field $$ \align \Cal R_n\sigma&=\left\{\Bigl(\Cal R_n\sigma^{(1)}(j),\dots ,\Cal R_n \sigma^{(p)}(j)\Bigr) \in R^p, \, j\in \bold Z\,\right\}, \tag1.3\\ \Cal R_n\sigma^{(1)}(j)&=\frac1{ A_n} \sum_{k=(j-1)2^n+1}^{j2^n} \left[\sigma^{(1)}(k)-E\sigma^{(1)}(k)\right],\quad j\in\bold Z\,, \tag1.4 \\ \Cal R_n\sigma^{(s)}(j)&=\frac1{B_n} \sum_{k=(j-1)2^n+1}^{j2^n} \sigma^{(s)}(k)\,,\quad j\in\bold Z , \quad s=2,\dots,p, \tag1.5 \endalign $$ where $A_n$ and $B_n$ are appropriate norming constants. We want to choose them in such a way that the finite dimensional distributions of the fields $\Cal R_n\s$ converge as $n\to\infty$, and also want to describe the limit field. Here $A_n$ is the norming constant in the direction of the magnetization and $B_n$ in the direction orthogonal to it. We have solved this problem for \ct{} in our paper [1] and for \co{} in [2]. In both cases we have to choose a ``critical'' normalization $B_n=2^nc^{-n/2}$ in the direction orthogonal to the magnetization, and the limit is a self-similar Gaussian field with long-range correlation. On the other hand, in the direction of the magnetization we have a different situation in the two cases. For \ct{} we have to choose $A_n=2^{n/2}$ and get a field of independent Gaussian variables for the limit. For \co{} the right choice in (1.4) is $A_n=2^nc^{-n}$, and the limit is a non-Gaussian field which we have described explicitly in [2]. Our aim in this paper is to solve this problem for \cc. The answer is very similar to the case \ct. Namely, we have to choose $B_n=2^nc^{-n/2}=2^{3n/4}$ and get a dependent Gaussian field in the direction orthogonal to the direction of the magnetization. In the direction of the magnetization we have to choose $A_n=2^{n/2}\sqrt n$, and the limit is a field consisting of independent Gaussian random variables. The main difference between the cases \ct{} and \cc{} is the appearance of a multiplying term $\sqrt n$ in the normalizing factor $A_n$ in the latter case. It is expected that translation invariant models with short-range interaction in the cases $d<4$, \ $d=4$ and $d>4$ show a behaviour similar to Dyson's model in the cases \co{}, \ \cc{} and \ct. Thus Dyson's model with \cc{} corresponds to four-dimensional translation invariant models. Let us formulate our results in more detail. In Theorem 1 formulated below we construct the equilibrium state whose large-scale limit will be investigated. Given some $h\in R^1$, \ $h\ge 0$, and a positive integer $n$ let us define the Gibbs measure $\mu^h_n=\mu^h_n(T,t)$ on $(R^p)^{2^n}$ with the density function $$ p^h_n(x_1,\dots,x_{2^n})= p^h_n(x_1,\dots,x_{2^n},t,T)\,,\quad x_j=\Bigl(x_j^{(1)},\dots,x_j ^{(p)}\Bigr)\in R^p\,,\quad j=1,\dots,2^n\,, $$ given by the formula $$ \align &p^h_n(x_1,\dots,x_{2^n}) \\&\qquad=Z^{-1}_n(T,t,h)\exp\biggl\{-\frac1T\biggl( -\sum^{2^n-1}_{i=1}\sum_{j=i+1}^{2^n}d(i,j)^{-3/2}x_ix_j-h\sum_{j=1}^ {2^n}x_j^{(1)}\biggr)\biggr\}\prod^{2^n}_{j=1}p_0(x_j,t)\,, \\ & \tag1.6 \endalign $$ where $$ Z_n=\int\exp\biggl\{-\frac1T\biggl(-\sum^{2^n-1}_{i=1} \sum_{j=i+1}^{2^n}d(i,j)^{-3/2}x_i x_j-h\sum_{j=1}^{2^n}x_j^{(1)}\biggr)\biggr\} \prod_{j=1}^{2^n}p_0(x_j,t)\,dx_j $$ is the grand partition function, and $p_0(x,t)$ is defined in (1.2). Let $p^h_n(x)=p^h_n(x,T)$ denote the density function of the average $2^{-n}\sum^{2^n}_{j=1}\sigma (j)$ of the $\mu^h_n$ distributed random vector $(\sigma (1),\dots, \sigma (2^n))$. Put $\mu_n=\mu^h_n$, \ $ p_n(x_1,\dots, x_{2^n})=p^h_n(x_1,\dots,x_{2^n})$ and $p_n(x)=p^h_n(x)$ in the case $h=0$. Let us introduce the functions $$ q_n(x) =q_n(x,T)=K_n \exp\biggl\{\frac{a_0}{2a_1}2^{n/2}x^2\biggr\}p_n\biggl( \sqrt{\frac T{a_1}}x\biggr) \tag1.7 $$ with $a_0=\frac2{2-\sqrt2}$, \ $a_1=a_0+1$ and the above defined functions $p_n(x)$, where the norming constant $K_n$ will be appropriately chosen. The function $q_n(x,T)$ is rotation invariant, i.e. the function $\qb_n(z,T)$, \ $z\in R^1$, defined by the formula $\qb_n(z,T)=q_n((z,0),T)$, \ $z\in R^1$, \ $0=(0,\dots,0)\in R^{p-1}$ satisfies the relation $q_n(x,T)=\qb_n(|x|, T)$. Choose the constant $K_n$ in (1.7) in such a way that $$ \int_0^\infty \qb_n(x,T)\,dx=1\,, $$ and define the numbers $$ M_n=\int_0^\infty x\qb_n(x,T)\,dx\,.\tag1.8 $$ Now we formulate the following \proclaim{Theorem 1} \it There are some thresholds $T_0>0$ and $t_0>0$ such that if $00$ exists, and $M^2=\frac{a_1(a_0-T)}{tT^2}+O(1)$ with $a_0=\frac2{2-\sqrt2}$ and $a_1=a_0+1$. Moreover, the following relation holds: Put $$ \bar M=\sqrt{\frac T{a_1}}M, \tag1.9 $$ and consider an arbitrary sequence of real numbers $h_n $, \ $ n= 0$,~1,~ 2, ~\dots such that $$ \frac{2\bar M}{2-\sqrt2} \left(\frac1{\sqrt2} \right)^n \le {h_n} \le D \left(\frac1{\sqrt2} \right)^n \tag 1.10 $$ with some $ \infty > D > \frac{2\bar M}{2-\sqrt2}$. Then the measures $\mu_n^{h_n}$ tend to a probability measure $\bar \mu =\bar \mu(t,T)$ on $(R^p)^{\bold Z}$. More precisely, for all $k \ge 0$ the measures $\mu^{h_n}_{k,n}$, the projections of the measures $\mu_n^h$ to $(R^p)^{2^k}$, converge to the projection of $\bar \mu$ to the first $2^k$ coordinates in variational metric as $n \to \infty$. The measure $\bar\mu$ does not depend on the choice of sequences ~$h_n$. \endproclaim The main result of this paper is the following \proclaim {Theorem 2}\it Let $\s=\left\{\sigma(n) =\left(\sigma^{(1)}(n),\dots,\sigma^{(p)}(n)\right) \in R^p,\; n\in\bold Z\right\}$ 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 (1.3), (1.4), (1.5) tend, with the choice $A_n=2^{n/2}\sqrt n$ and $B_n=2^{3n/4}$, to those of a Gaussian 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 numbers $k\ge0$ 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 $$ \aligned &h_k\left(x_1,\dots,x_{2^k}\right)\\ &\qquad =C(k)\exp\biggl\{-\frac1T \biggl[ \sum_{s=2}^p \biggl(\frac {2+\sqrt2}2\sum_{j=1}^{2^k} x_j^{(s)2}-\left(\sqrt2-1\right) \left(\frac{\sqrt2}4 \right)^k\biggl(\sum_{j=1}^{2^k} x_j^{(s)}\biggr)^2\\ &\qquad\qquad-\sum^{2^k-1}_{i=1}\sum_{j=i+1}^{2^k} d(i,j)^{-3/2} x_i^{(s)}x_j^{(s)} \biggr) +(6+2\sqrt2)M^2 \sum_{j=1}^{2^k} x_j^{(1)2}\biggr]\biggr\}. \endaligned \tag1.11 $$ \endproclaim It follows from the result in Appendix ~E of [2] that the measure constructed in Theorem ~1 is an equilibrium state. We restricted ourselves to the construction of equilibrium states for low temperatures where we are interested in their large-scale limit. The proofs of Theorems ~1 and~2 are based, similarly to the papers ~[1] and ~[2], on two analytic problems, where the action of an integral operator must be investigated. We formulate these problems in the next Section. \beginsection 2. The basic steps of the proof In this Section we discuss two analytical problems which play a central role in the proof of Theorems ~1 and ~2. The first one is connected with the asymptotical behaviour of the density function $p_n(x)$ of the average of a $\mu_n$ distributed vector defined after formula (1.6). It is proved (see e.g. Appendix ~A in [2]) that $p_n(x)$ satisfies the recursive relation $$ p_{n+1}(x)=C_n(T)\int \exp\left\{\frac{2^{n/2}}T \left(x^2-u^2 \right)\right\}p_n(x-u)p_n(x+u)\,du \tag2.1 $$ with the starting function $p_0(x)$ defined in formula (1.2). For us it is more convenient to work with the functions $q_n(x)$ defined in (1.7) instead of the functions $p_n(x)$. Simple calculation shows that relations (2.1) and (1.7) imply the recursive relations $$ q_{n+1}(x)=K_n\int \exp\{-2^{n/2}u^2\} q_n(x-u) q_n(x+u)\,du\tag2.2 $$ with the starting function $$ q_0(x)=q_0(x,T,t)=K_0\exp\left\{\frac{a_0-T}{2a_1}x^2 -\frac{tT^2}{4a_1 ^2}|x|^4\right\}, \tag2.2$'$ $$ where $a_0=\frac2{2-\sqrt2}$, \ $a_1=a_0+1$, and $K_n$ is an appropriate norming constant. (The numbers $a_0$ and $a_1$ will denote these numbers in the whole paper.) In Theorem A formulated below we describe the asymptotic behaviour of the function $q_n(x)$. We recall that we have introduced the functions $\bar q_n(z)=\bar q_n(z,T)$, $z\in R^1$ in Section ~1, and they satisfy the relation $q_n (x,T)=\bar q_n(|x|,T)$. \proclaim{ Theorem A}\it There are some thresholds $t_0$ and $T_0$ such that for $00$ and $n_0=n_0(T,t)>0$ such that for $n>n_0$ $$ \align 2^{-n/2}\sqrt nq_n(x,T)&=2^{-n/2}\sqrt n\bar q_n(|x|,T) \\&=\frac{\sqrt2M} {\sqrt{\pi}} \exp\left\{-\frac{2^{n+1}M^2}n (|x|-M)^2\right\}+r_n(x) \tag 2.3 \endalign $$ with $$ |r_n(x)|\le\frac K{\sqrt n}\tag2.3$'$ $$ and $$ \left|M^2-\frac{a_1(a_0-T)}{tT^2}\right|\le K\tag2.4 $$ with some $K>0$. Also the estimate $$ 2^{-n/2}\sqrt n\bar q_n(x,T)\le K\exp\left\{-\frac{ 2^{n/2}\mu}{\sqrt n} |x-M|\right\}\quad\text {for all } x>0\tag2.5 $$ holds with some $K>0$ and $\mu>0$ depending on $ T$ and $t$. \endproclaim For $x-M\gg 0$ or $x-M\ll0$ we need a better bound on $\bar q_n (x,T)$ than that given in ~(2.5). This is given in the following \proclaim{Proposition A} \it Under the conditions of Theorem A $$ 2^{-n/2}\sqrt n\bar q_n(x,T)\le K\exp\left\{-\beta \frac{2^n}n( x-M)^2\right\}\quad\text {for } x>M\tag2.6 $$ with some $\beta>0$ and $K>0$ depending on $t$ and $T$. If $00$ there are some thresholds $t_0=t_0(\e)$, \ $T_0=T_0(\e)$ and a real number $r_n$, \ $C_1n2^{-n/2}C_1>0$ such that if $00$. We do not prove it, because the slightly weaker statement formulated above is sufficient for our purposes. Also the dependence of the thresholds $t_0$ and $T_0$ on $\e$ can be dropped with the help of some additional investigation. We do not do it, because we do not need Proposition A with very small $\e$. What we need is that it holds with some $\e>0$ such that $1-\e>\frac {a_0}{2a_1}$.) Now we formulate the second problem we are interested in. Given some integers $0\le n\le N$ and positive real number $h>0$ consider the measure $\mu^h_N$ with density function $p^h_N(x_1,\dots,x_{2^N})$ defined in (1.6) (we replace the number $n$ by $N$ in it), and define its projection $\mu^h_{n,N}$ to the first $2^n$ coordinates $x_1,\dots,x_{2^n}$. We want to give a good asymptotic formula for the Radon--Nikodym derivative $\frac{d\mu^h_{n,N}}{d\mu_n}(x_1,\dots,x_{2^n})$, where $\mu_n$ is $\mu_n^h$ with $h=0$. It can be expressed explicitly by the following recursive integral formula: (See e.g. Appendix C in [2].) \proclaim{Formula for the Radon--Nikodym derivative} \it $$ \frac{d\mu^{h}_{n,N}}{d\mu_n}\left(x_1,\dots,x_{2^n}\right)= f^{h}_{n,N}\biggl(2^{-n}\sum^{2^n}_{j=1}x_j\biggr),\quad n \le N \tag2.8 $$ $$ \align f^{h}_{N,N}(x)&=K(N,h)\exp \left(\frac{2^Nhx^{(1)}}T\right)\tag2.9\\ \vspace{1\jot} f^{h}_{n,N}(x)&=K(n,N,h)\bold S_nf^{h}_{n+1,N}(x)\tag2.10 \endalign $$ with $$ \bold S_nf(x)=\int_{R^p}\exp\biggl(\frac{2^{n/2}}Txy\biggr)f\biggl(\frac{x+y}2 \biggr)p_n(y)\,dy \tag2.10${}^{\prime}$ $$ where $K(n,N,h)$ are appropriate norming factors and $p_n(x)$ is the density function of a $\mu_n$ distributed random vector. \endproclaim In Theorem B formulated below we give an asymptotic formula for the function $f_{n,N}^h(x)$ if $h=h_N$ satisfies the relation $$ \frac{2\bar M}{2-\sqrt2} \left(\frac1{\sqrt2} \right)^N \le {h_N} \le D \left(\frac1{\sqrt2} \right)^N \tag 2.11 $$ with some $ \infty > D > \frac{2\bar M}{2-\sqrt2}$. To formulate it we introduce the sequences $g_n$, $A_n$, \ $n=1$, ~2, ~\dots,~N defined by the recursive relations $$ \align g_N&=g_N(N,h_N)=\frac{2^Nh_N}T\,\tag2.12\\ g_n&=g_n(N,h_N)=\frac{g_{n+1}}2 + \frac{2^{n/2}}T\bar M\qquad \text {for } n0\}\tag2.14\\ \Omega^2_n& =\left\{x\in R^2,\,\bigl||x|-\bar M\bigr|<2^{-0.2n}\right\} \setminus \Omega^1_n \tag2.14${}^{\prime}$\\ \Omega^3_n&=\left\{x\in R^2,\,\bigl||x|-\bar M\bigr|\ge 2^{-0.2n}\right\}\,.\tag2.14${}^{\prime\prime}$ \endalign $$ Clearly $\Omega^1_n\cup\Omega^2_n\cup\Omega^3_n =R^2$. Now we formulate the following \proclaim {Theorem B} \it For all $q$, \ $2^{-0.1}0$. Define the sequences $g_n$ and $A_n$, \ $0\le n\le N$, by formulas (2.12)--(2.13${}^{\prime}$) and put $\bar g_n=2^{-n/2}g_n$, \ $\bar A_n=2^{-n/2}A_n$. If $h_N$ satisfies relation (2.11) 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 $\bar g=\frac2{2-\sqrt2}\frac{\bar M}T$, and $\bar A=\frac{\sqrt2-1}{T}$. If the relations $N >N_0$ and $N>n^B$ also hold with some appropriate $N_0=N_0(\bar M,T,D)$ and $B=B(\bar M,T,D)$ then $|\bar g_n-\bar g|<4^{-n}$ and $|\bar A_n-\bar A|<4^{-n}$. \endproclaim The above results enable us to carry out a limiting procedure analogous to that in Sections ~6 and ~7 in Part ~II of ~[2], which leads to the proof of Theorems ~1 and ~2. The main step of this limiting procedure is to give a good estimate for the expression $$ p_n\biggl(2^{-n}\sum^{2^n}_{j=1}x_j \biggr)f^{h_N}_{n,N}\biggl( 2^{-n}\sum^{2^n}_{j=1}x_j\biggr) \,. \tag2.18 $$ Since we can express the function $p_n(x)$ through $q_n(x)$ Theorems ~A and ~B together with Proposition ~B enable us to give a good asymptotic formula for this expression in a typical domain around the point $(\bar M,0)\in R^2$. Then Theorem ~B together with Proposition ~A guarantee that the region outside this typical domain has a negligible effect. \beginsection 3. On Theorem A. The method of the proof The proof both of Theorem A and B is based on the ideas of [2]. Most proofs can be carried out in almost the same way, only the number $c$ must be replaced by $\sqrt2$. The proofs of such parts will be omitted, we only refer to the corresponding result in [2]. From now on the letters $C$, $C_1$, $K$ etc. will denote absolute constants. The same letter in different formulas may denote different constants if it is not stated otherwise. Let us introduce, similarly to Part I in [2], the numbers $M_n$ defined in (1.8) and the functions $$ f_n(x)=f_n(x,T)=2^{-n/2}\bar q_n\left(M_n+2^{-n/2}x,T\right), \tag3.1 $$ where the function $\bar q_n(x)$ was defined after formula (1.7). We shall deduce Theorem A from the following \proclaim{Theorem A$'$} \it Under the conditions of Theorem A the limit \ $\lim_{n\to \infty}M_n=M>0$ exists, and $$ M^2=\frac{a_1(a_0-T)}{tT^2}+R(T,t) \tag3.2 $$ with some $|R(T,t)|n_0$ $$ M_n=M+\frac{2+\sqrt2}{4M}2^{-n/2}+\delta (n)2^{-n/2},\qquad |\delta (n)|0$. The function $f_n$ satisfies the relations $$ \left|f_n(x,T)-\frac{ \sqrt2M}{\sqrt {n\pi}}\exp\left\{-\frac{2M^2}n x^2\right\}\right|<\frac Kn \qquad\text{for }x>-2^{n/2}M_n \tag3.4 $$ and $$ f_n(x,T)\le \frac{KM}{\sqrt n} \exp\{-\mu|x|\}\qquad \text{for } x>-2^{n/2}M_n \tag3.5 $$ for $n>n_0$ with some $\mu>0$ and $K>0$. \endproclaim To prove Theorem A$'$ let us introduce, similarly to [2], the operator $\qbb$; $$ \aligned \qbb f(x)&=\int\exp\left\{-2^{-n/2}{u^2}-v^2\right\} \\&\qquad f\left(2^{n/2}\left(\sqrt{ (M+2^{-(n+1)/2}x+2^{-n/2}u)^2+2^{-n/2}v^2}-M\right)\right) \\ &\qquad f\left(2^{n/2}\left(\sqrt{ (M+2^{-(n+1)/2}x-2^{-n/2}u)^2+2^{-n/2}v^2}-M\right)\right) du\,dv\,, \endaligned \tag3.6 $$ its standardization defined by the formula $$ \qm f(x)=\frac{\qbb f(x+m_n)} {\int_{-2^{(n+1)/2}M}^{\infty}\qbb f(x)\,dx} \tag3.7 $$ with $$ m_n=\frac{\int_{-2^{(n+1)/2}M}^{\infty} x\qbb f(x)\,dx} {\int_{-2^{(n+1)/2}M}^{\infty}\qbb f(x)\,dx} \tag3.7$'$ $$ together with their approximations $\tm$ and $\tmb$ given by the formulas $$ \tmb f(x)=\int e^ {-v^2} f\left(\frac x{\sqrt2}+u+\frac{v^2}{2M}\right) f\left(\frac x{\sqrt2}-u+\frac{v^2}{2M}\right) \,du\,dv \tag3.8 $$ and $$ \tm f(x)=\sqrt{\frac 2\pi} \tmb f\left(x-\frac{\sqrt2}{4M}\right). \tag3.8$'$ $$ The same calculation as that in (2.20) of [2] yields that the Fourier transforms of the operators $\tm$ and $\tmb$ defined by the formulas $\ttmb \tilde f=(\tmb f)\tilde{} $ and $\ttm \tilde f=(\tm f)\tilde{} $ satisfy the relation $$ \ttmb \tilde f(\xi)=\sqrt{\frac{\pi}2}\frac{ \tilde f\left(\frac{\xi}{\sqrt2}\right)^2 } {\sqrt{1+\frac{i{\xi}}{\sqrt2M}} } \tag3.9 $$ and $$ \ttm \tilde f(\xi)=\frac{\exp\left(\frac{i\sqrt2\xi}{4M}\right)} {\sqrt{1+\frac{i\xi}{\sqrt2M} }} \tilde f\left(\frac\xi{\sqrt2}\right)^2 . \tag3.9$'$ $$ The relation $$ \left(f_{n+1}(x),\,M_{n+1}\right)=\left(\bold Q_{n,M_n} f_n(x),\,M_n+2^{-(n+1)/2}m_n\right) \tag3.10 $$ holds with the starting pair $(f_0(x),\,M_0)$ defined by the relations $$ M_0=\int_0^\infty x\bar q_0(x)\,dx \qquad f_0(x)=\bar q_0(x-M_0)\,, \tag3.10$'$ $$ where the function $\bar q_0(x)$ was defined after formula (1.7) (with $n=0$). We have $$ \qm f(x)=\tm f(x)+\e_n(x)\,,\tag 3.11 $$ where $\e_n(x)$ is a small error term. We get a heuristic explanation of Theorem A$'$ by investigating the expression $\bold T^n_Mf(x)$ for large $n$ with a function $f(x)$ satisfying the relations $\int f(x)\,dx=1$ and $\int xf(x)\,dx=0$. Put $$ \varphi_k(\xi)=\log \tilde{\bold T}^k_M\tilde f(\xi)=\sum_{j=2}^{\infty} d_{j,k}\xi^k\,. $$ It follows from (3.9) that $$ d_{j,k+1}=2^{(2-j)/2}d_{j,k}+\frac{(-i)^j}{2j(\sqrt2M)^j}\,,\qquad j\ge2\,. $$ Hence $$ \lim_{n\to \infty}d_{j,n}=\frac{(-i)^j}{2j(\sqrt2M)^j(1-2^{(2-j)/2})} \qquad \text{for }j\ge3\,, $$ and $$ d_{2,n}=-\frac n{8M^2}+d_{2,0}\,. $$ The above relations imply that $$ \lim_{n\to\infty}\varphi_n\left(\frac\xi {\sqrt n}\right)=-\frac1{8M^2}\xi^2\,. $$ Since $f_n(x)$ behaves similarly to $\bold T^n_M f_0(x)$, the above calculation suggests that the expression $\sqrt nf_n(\sqrt nx)$ is asymptotically Gaussian with variance $\frac1{8M^2}$. We justify this heuristic argument similarly to the method of~[2]. First we show that if $t$ and $T$ are sufficiently small then for all not too large $n$ \ $f_n(x)$ is asymptotically normal with variance $\s=\frac{a_1}{2(a_0-T)}$. More precisely, we prove the following \proclaim{Proposition 1} \it For all positive integers $N\ge1$ there are some thresholds $t_0$ and $T_0$ such that if $0-2^{-n/2}M_n,\;j=0,1,2, \tag3.13 $$ and $$ |M_n-\hat M_0|\le B(n)t^{1/2}T\,, \tag3.14 $$ where $\hat M_0^2=\frac{a_1(a_0-T)}{Tt^2}$, \ $\s^2=\frac {a_1}{2(a_0-T)}$, \ $\norm(x,\s)$ denotes the normal density function with expectation zero and variance $\s$, and $B(n)$ is some appropriate multiplying factor depending on $n$, but not on $t$ and $T$. \endproclaim If $t_0$ and $T_0$ are sufficiently small then $\hat M_0$ is very large, therefore (3.14) states that for fixed $n$ (depending on $t$ and $T$) $M_n$ is very close to $\hat M_0$. Then (3.12) gives a good Gaussian approximation of $f_n(x)$ and (3.13) a good bound on its tail behaviour. The proof of Proposition 1 is based on the observation that $M_0$ almost agrees with the positive maximum $\hat M_0$ of the function $\bar q_0(x)$, \ $f_0(x)$ is almost Gaussian, and we commit a small error by substituting the operator $\qbb$ for small $n$ by the operator $\hat T_n$, $$ \align \hat T_nf_n(x)&=C\int\exp\left\{-v^2-2^{-n/2}u^2\right\}f_n(x+u)f_n(x-u)\,du \,dv\\&= C\sqrt\pi\int\exp\left\{-2^{-n/2}u^2\right\}f_n(x+u)f_n(x-u)\,du\,. \endalign $$ Since the proof is almost the same as the proof of the corresponding result for \co{} given in Section 4 of Part I in [2] we omit it. By the same reason we omit the proof of its Corollary formulated below. To formulate this result first we have to introduce the following notion: \proclaim{Definition of the regularization of a function} \it Let us choose some fixed function $\varphi(x)\in C_0^{\infty}(R^1)$ such that $1\ge\varphi(x)\ge0$ for all $x\in R^1$, \ $\varphi(x)=1$ for $|x|<1$, and $\varphi(x)=0$ for $|x|\ge2$. Put $\varphi_n(x)=\varphi\left(\frac1{100}2^{-n/2}x\right)$. Given some function $f(x)$, \ $f(x)\ge0$, \ $\int f(x)\,dx<\infty$ we define its $n$-th regularization $\regg n$ as $\regg n(x)=\frac 1{A_n}\varphi_n(x+B_n)f(x+B_n)$ with $A_n=\int \varphi_n(x)f(x)\,dx$ and $B_n=\frac1{A_n}\int x\varphi_n(x)f(x)\,dx$, provided that the above formula is meaningful, i.e. $A_n>0$. \endproclaim Now we formulate the following \proclaim{Corollary of Proposition 1} \it Under the conditions of Proposition 1 we have for all $n\le N$ $$ |\tilde \varphi_n(f_n)(t+is)|\le\frac{\exp s^2}{1+\frac{t^2}{200}} \qquad \text{for }|s|<2,\;\,t\in R^1\,, $$ and $$ \left|\frac {d^j}{dx^j} f_n(x)\right|\le10^5\exp\left\{-\left|2x+ \frac{2^{-n/2}x^2}{M_n}\right|\right\}\qquad\text{for }x>-2^{n/2}M_n,\;j=0,1,2. $$ \endproclaim Let us fix some positive integer $N$, and define the sequences $\a_n$, $\bb_n$, $n=N$, ~N+1, ~\dots, as $$ \align \a_N&=\frac1{200}\,,\tag3.15 \\ \a_{n+1}&=\left(1-2^{-n/4}\right)\a_n+\frac{10^{-12}}{M_n^2} \qquad \text{for }n\ge N\tag3.15$'$ \endalign $$ and $$ \align \bb_N&=1\,,\tag3.16 \\ \bb_{n+1}&=\left(1+2^{-n/4}\right)\bb_n+\frac{10}{M_n^2} \qquad \text{for }n\ge N\,,\tag3.16$'$ \endalign $$ where $M_n$ is defined in (1.8). Now we define the following Properties $I(n)$ and $J(n)$. \proclaim{Property $I(n)$} \it Let $n\ge N$. The function $f(x)$ satisfies Property $I(n)$ (with the starting index $N$ and parameter $C$) if $$ \left|\frac {d^j}{dx^j} f(x)\right|\le \frac C{\beta_n^{(j+1)/2}}\exp\left\{-\frac 1{\sqrt{\beta_n}}\left|2x+ \frac{2^{-n/2}x^2}{M_n}\right|\right\}\qquad\text{for }x>-2^{n/2}M_n,\;j=0,1,2 $$ with the above defined sequence $\bb_n$ and the number $M_n$ defined in (1.8). \endproclaim \proclaim{Property $J(n)$} \it Let $n\ge N$. The function $f(x)$ satisfies Property $J(n)$ (with the starting index $N$) if $$ |\tilde \varphi_n(f)(t+is)|\le\frac{ \exp \{\bb_n s^2 \}}{1+\a_nt^2} \qquad \text{for }|s|<\frac2{\sqrt{\bb_n}},\;\,t\in R^1\,, $$ with the above defined sequences $\a_n$ and $\bb_n$. \endproclaim Now we formulate \proclaim{Proposition 2} \it The multiplying factor $C$ and the starting index $N$ can be chosen in Properties $I(n)$ and $J(n)$ in such a way that under the additional conditions $M_n>K$ with some universal constant $K$, \ $|M_n-M_{n-1}|<1$, \ $100n>\bb_n>\max(9M_n^{-2},\,4^{-n})$ Properties $I(n)$ and $J(n)$ for the function $f_n(x)$ imply Properties $I(n+1)$ and $J(n+1)$ for the function $f_{n+1}(x)$ (with the same parameters $N$ and $C$). Also the following relations hold true: $$ \align M_{n+1}=M_n+2^{-(n+1)/2}m_n,\qquad &m_n=-\frac{\sqrt2}{4M_n}+\gamma(n)\\ &\quad \text{with } \gamma(n)-2^{(n+1)/2}M_{n+1},\,\;j=0,1,2, \tag3.18 \endalign $$ and $$ \left|\frac{d^j}{dx^j}\bold T_{M_n}\reg n(x)\right|\le \frac{C_1 C^2}{\bb_{n+1}^{(j+1)/2}}\exp\left\{-\frac{2|x|}{\sqrt{\bb_{n+1}}}\right\}, \quad x\in R^1,\;j=0,1,2,3,4 \tag3.19 $$ with some absolute constant $C_1$. As a consequence, if $00$ and $T_0>0$ then Properties $I(n)$ and $J(n)$ hold for the functions $f_n(x)$ with some appropriate parameters $C$ and $N$, and these functions satisfy relations (3.17)--(3.19). Also the relation $\bb_n<100n$ holds. \endproclaim Proposition 2 is proved similarly to the analogous result for \co{} in Sections ~5 and ~6 in Part ~I of~ [2], only the number ~$c$ must be replaced by $\sqrt2$ everywhere. The expressions $\qm f(x)$, \ $\tm \regg n(x)$ and $\qm f(x)-\tm\regg n(x)$ can be bounded with the help of Property $I(n)$, as it is formulated in Proposition 3 and proved in Section 5 in Part I of [2]. This enables us to reduce the problem to the investigation of $\bold T_{M_n}\reg n(x)$, which can be done with the help of Property $J(n)$ and formula (3.9$'$). The only difference between the cases \co{} and \cc{} is that for \cc{} the condition $\bb<100$ must be replaced by the condition $\bb<100n$ when the operator $\qm$ is investigated. This is so, because we apply our estimates with $\bb=\bb_n$, and the sequence $\bb_n$ defined in (3.16), (3.16$'$) is of order $const.\,n$. (In the case \co{} it was bounded by a constant.) Nevertheless, this difference causes no problem. The condition $\bb<100$ was applied in [2] for such arguments as to show that the estimate (5.4) of that work implies Lemma 3 for large ~$x$. To make such a conclusion we need some upper bound on ~$\bb$, but the estimate $\bb<100n$ is sufficient for our purposes. Proposition 2 enables us to bound the error term $\e_n(x)$ in (3.11) when the operator $\bold Q_{n,M_n} $ is applied for $f_n(x)$. With the help of this estimate we can turn the heuristic argument after formula (3.11) into a rigorous proof. \beginsection 4. The proof of Theorem A We prove Theorem A by estimating the Fourier transforms $\four n$. Let us fix some constants $N$ and $C$ in such a way that Propositions ~1 and~2 hold with this choice of the parameters. Let us introduce the functions $\ph n=\log\four n$ and the numbers $\bbb n=-\left.\der 2\ph n\right|_{t=0}$, provided that these quantities are well-defined, i.e. we can take logarithm in these expressions. We shall prove the following \proclaim{Lemma 1} \it If $00$ and $T_0>0$ then \newline a) $$ \align \bbb N&=\frac{a_1}{2(a_0-T)}+\delta(N) \qquad |\delta(N)|\le 4^{-N} \tag4.1 \\ \bbb {n+1}&=\bbb n+\frac1{4M_n^2}+\delta(n)\qquad |\delta(n)|\le2^{-n/4}\qquad \text {for }n\ge N \tag4.2 \endalign $$ b) For $|t|<\left(\frac n{\bb_n}\right)^{1/3}$ and $n\ge N$ \ $\ph n$ is well-defined, and $$ \left|\der 3\ph n\right|\le \frac2{M_n^3}+2^{-n/4}\qquad \text{for }|t|\le \roott \text{ and }n\ge N \tag4.3 $$ \endproclaim \demo{Proof of Lemma 1}Because of Proposition 1 \ $\four N$ is very close to the Fourier transform of the normal density function $\norm(x,\s)$ with $\s^2=\frac{a_1}{2(a_0-T)}$, and the analogous result also holds for its derivatives. This implies (4.2) and (4.3) for $n=N$, since if $\four N$ were exactly normal then we would have $\bbb N=\frac{a_1}{2(a_0-T)}$ and $\der 3\ph N=0$. We prove (4.2$'$) and (4.3) in the general case by induction from $n$ to $n+1$. Let us introduce the operator $\bold{\hat T}_n$ by the formula $\bold{\hat T}_n\psi(t)=\log\tmn\exp\psi(t)$. It follows from (3.9$'$) that $$ -\left.\der2\bold{\hat T}_n\ph n\right|_{t=0} =\frac1{4M_n^2}+\bbb n\,, \tag4.4 $$ and $$ \der 3\bold{\hat T}_n\ph n =\frac 1{\sqrt2}\psi_n\left(\frac t{\sqrt2}\right) +\frac{\sqrt2i}{16M_n^3\left(1+\dfrac{it}{\sqrt2M_n}\right)^3 }\,.\tag4.4$'$ $$ Since $M_n$ is very large, (4.4$'$) together with our inductive hypothesis imply that $$ \left|\der3\bold{\hat T}_n\ph n\right|\le \frac2{M_{n+1}^3}+\frac1{\sqrt2}2^{-n/4}\quad\text{if }|t|<\left(\frac{n+1}{\bbb{n+1}}\right)^{1/3}.\tag4.5 $$ Because of the identities $\bold{\hat T}_n\psi_n(0) =\left.\der {}\bold{\hat T}_n\ph N\right|_{t=0}=0$ it follows from (4.4) and (4.5) that $$ \align &\Re \bold{\hat T}_n \ph n\ge-\left(\bbb n+\frac1{4M_n^2}\right) \frac{t^2}2- \left(\frac2{M^3_{n+1}}+ \frac{2^{-n/4}}{\sqrt2}\right)\frac{|t|^3}6\ge-\frac n{10}\\ &\qquad\text{for }|t|\le \left(\frac{n+1}{\bbb {n+1}}\right)^{1/3}, \endalign $$ where $\Re$ denotes real part. (Observe that $1<\bbb n< n/10$.) This relation implies that $$ \left|\tmn\four n\right|\ge e^{-n/10} \qquad \text{for } |t|\le \left(\frac{n+1}{\bbb {n+1}}\right)^{1/3}. \tag4.6 $$ We get similarly, by expressing the derivatives of $\tmn\four n$ through $\psi_n(t) $ and its derivatives, that $$ \left|\der j\tmn\four n\right|\le e^{n/10}\qquad \text{for } |t|\le \left(\frac{n+1}{\bbb {n+1}}\right)^{1/3} \;j=1,2,3. \tag4.6$'$ $$ On the other hand, some calculation with the help of (3.18) yields that $$ \left|\der j \four {n+1}-\der j\tmn \four n\right|\le K2^{-n/2}\qquad\text{for }|t|\in R^1 \text { and }j=0,1,2,3\,.\tag4.7 $$ By expressing $\der 3\ph {n+1}$ and $\der 3\bold{\hat T}_n\ph n$ by the corresponding Fourier transforms we get that relations (4.6), (4.6$'$) and (4.7) imply that $$ \left|\der 3\ph{n+1}-\der 3\bold{\hat T}_n \ph n\right| \le \frac 1{100}2^{-n/4}. $$ The last relation together with (4.5) imply (4.3) for $n+1$. It can be proved similarly that $$ \left|\der 2\ph{n+1}\Bigl|_{t=0}-\der2\bold{\hat T}_n \ph n\Bigl|_{t=0}\right|\le 2^{-n/4} $$ which relation together with (4.4) imply (4.2$'$) for $n+1$. Lemma 1 is proved. \enddemo \demo {The proof of Theorem A$'$} It follows from Lemma A that $$ \four n=\exp\left\{-\frac{\bbb n}2 t^2+R_n(t)t^3\right\}\quad \text{with } |R_n(t)|<\frac2{M_n^3}+2^{-n/4} \quad \text{if } t<\roott. $$ Hence $$ \left|\int_{|t|<\roott} e^{-itx}\left[e^{-\frac{\bbb n t^2}2} -\four n\right]\,dt\right| \le 2\left(\frac 2{M_n^3}+2^{-n/4}\right)\frac1{\bbb n^2} \tag4.8 $$ On the other hand $$ \left|\int_{|t|>\roott} e^{-itx}e^{-\frac{\bbb n t^2}2}\,dt \right|\le \exp \left\{-\frac{n^{2/3}}2 \bbb n^{1/3}\right\}, \tag4.8$'$ $$ and by Property $J(n)$ and the relation $\a_n>10^{-14}\bbb n$ $$ \left|\int_{|t|>\roott} e^{-itx}\four n\,dt \right|\le 10^{14}n^{-1/3}\bbb n^{-2/3}. \tag4.8$''$ $$ Relations (4.8), (4.8$'$) and (4.8$''$) imply that $$ \align &\left|\varphi_n(f_n)(x)-\frac1{\sqrt{2\pi \bbb n}} \exp \left\{-\frac{x^2}{2\bbb n}\right\}\right|\\&\qquad\qquad= \left|\int e^{-itx}\left[\four n-\exp \{-\bbb n t^2\}\,dt\right]\right| \\ &\qquad\qquad \le\frac 2{\bbb n^2}\left(\frac2{M_n^3}+2^{-n/4}\right) +10^{14}n^{-1/3}\bbb n^{-2/3}+\exp \left\{-\frac{n^{2/3}}2\bbb n^{1/3}\right\} \\&\qquad\qquad\le \frac1{\bbb n^2}\left(\frac4{M_n^2} +2^{-n/4}\right)+2\cdot10^{14}n^{-1/3}\bbb n^{-2/3}. \tag4.9 \endalign $$ In relation (4.9) $\varphi_n(f)(x)$ can be replaced by $f_n(x)$, since for $|x|<2^{n/2}$ they are very close to each other by (3.8), and for $|x|>2^{n/2}$ both terms at the left-hand side of (4.9) are negligible small. (The norming constants $A_n$ and $B_n$ appearing in the regularization are almost 0 and 1.) Hence (4.9) implies that $$ \left|f_n(x)-\frac 1{\sqrt{2\pi\bbb n} }\exp\left\{-\frac{x^2}{2\bbb n} \right\}\right|\le \frac1{\bbb n}\left(\frac4{M_n^2}+2^{-n/5} +10^{15}\left(\frac{\bbb n}n\right)^{1/3} \right)\qquad\text {for }n\ge N\,. \tag4.10 $$ Since $\left|\bbb n-\frac n{4M^2}\right|<10$, hence $$ \left|\frac 1{\sqrt{2\pi\bbb n} }\exp\left\{-\frac1{2\bbb n} x^2\right\}-\frac{\sqrt2M}{\sqrt{\pi n}} \exp\left\{-\frac{2M^2}n x^2\right\}\right| \le \frac{const.}n\,. \tag4.10$'$ $$ For large $n$ the term $\frac1{\bbb n}$ can be replaced by $\frac{5M^2}{n}$ in (4.10), hence (4.10) and (4.10$'$) imply (3.4). Relation (3.5) holds because of Property $I(n)$, and relations (3.2) and (3.3) can be deduced from Proposition 2 in the same way as the analogous result in [2] in Lemma 10 of Part I. Theorem A$'$ is proved. \enddemo \demo{The proof of Theorem A} By Theorem A$'$ and (3.1) $$ 2^{-n/2}\sqrt n\bar q_n(x,T)=\frac{\sqrt2M}{\sqrt \pi} \exp\left\{- \frac{2^{n+1}M^2}n(x-M_n)^2\right\} +r_n(x) \tag4.11 $$ with $$ |r_n(x)|<\frac K{\sqrt n} \,. $$ We have to check that an error of order $O\left(\frac1 {\sqrt n}\right)$ is committed if $M_n$ is replaced by $M$ in (4.11). We have $$ \align &\left| \exp\left\{-\frac{2^{n+1}M^2}n(x-M)^2\right\} - \exp\left\{-\frac{2^{n+1}M^2}n(x-M_n)^2\right\} \right| \\ &\qquad \le\exp\left\{- \frac{2^{n+1}M^2}n(x-M)^2\right\}\frac{2^{n+1}}n M^2\left|(x-M)^2-(x-M_n)^2\right|\\ &\qquad \le\frac{2^{n+1}M^2}{ n}|M-M_n|\left(2|x-M|+2|M-M_n|\right) \exp\left\{-\frac{2^{n+1}}nM^2(x-M)^2\right\}\\&\qquad\le \frac K{\sqrt n}\,, \endalign $$ since $|M-M_n|0$ some positive integer $N$ and thresholds $t_0>0$ and $T_0>0$ can be chosen in such a way that $$ \left|f_n(x)-\frac 1{2\sqrt{\pi\bbb n}}\exp\left\{-\frac1{2\bbb n}x^2 \right\}\right|<\delta \qquad \text{for all }n\ge 0 \text{ and } x\in R^1 \tag 4.12 $$ if $00$ the inequality $$ f_n(x)<\frac{10}{\sqrt{ \bbh n}}\exp\left\{-\frac1{2\bbh n}x^2\right\} \qquad\text{for }|x|0$ and a large $L=L(\e)>0$. We are going to show that if $0L\sqrt{\bbh n},\; n=0,1,2,\dots\,. \tag5.1$'$ $$ Since $\lim_{n\to\infty}\frac{\bbh n}n=\frac1{8M^2}$, relations (5.1) and (5.1$'$) imply (2.6). Because of the Corollary of Theorem A$'$ we may assume that relation (5.1) and relation (5.1$'$) for $L\sqrt {\bbh n}<|x|<3L\sqrt {\bbh n}$ hold. It is enough to apply this Corollary for $3L$, and to choose $L$ in such a way that $\exp\{-\frac{L^2}4\}<\frac\e{10}$. Moreover, it can be seen from the form of $f_0(x)$ that for $n=0$ (5.1$'$) holds for all $x>L\sqrt{\bbh 0}$. Hence it is enough to prove (5.1$'$) for $x>2L\sqrt{\bbh n}$ by induction from $n$ to $n+1$. We shall do it with the help of the following \proclaim{Lemma 2} \it If $\e>0$ and $L>L(\e)>0$ are appropriately chosen (in dependence of the number $C$ appearing in the conditions of this Lemma), $n$ is some non-negative integer, $M>K>0$ with an appropriate $K>0$ and $$ \align \fx&\le\frac{10}{\sqrt\bb}\expo \quad\text{for }|x|L{\sqrt \bb} \tag5.2$'$ \\ \fx&\le \frac C{\bs}\quad \text{for all }x\in R^1 \tag 5.2$''$ \endalign $$ then $$ \qbb\fx\le \frac{\e^{3/2}}{\bs}\expt \quad\text {for }x>2L\bs $$ \endproclaim \demo{Proof of Lemma 2} The proof applies the same ideas as that of Lemma ~19 in Part ~I of ~[2]. Let us introduce the functions $$ \ell^{\pm}_{n,M}(x,u,v)= 2^{n/2}\left(\sqrt{\left(M+2^{-(n+1)/2}x\pm 2^{-n/2}u\right)^2+2^{-n/2}v^2}-M\right), $$ $$ P(x,u)=\int \exp \left\{-v^2\right\}f\left(\ell^{+}_{n,M}(x,u,v)\right) f\left(\ell^{-}_{n,M}(x,u,v)\right) \,dv \tag5.3 $$ and $$ \px=P(x,0)\,. \tag5.3$'$ $$ Then $$ \qbb\fx=2\int_0^\infty \exp\left\{-2^{-n/2}u^2\right\}\pxu\,du \,,\tag5.4 $$ and by the Schwarz inequality $$ \pxu\le\left[P \bigl(x+\sqrt2u\bigr)P \bigl(x-\sqrt2u\bigr)\right]^{1/2}\,.\tag5.5 $$ Let us estimate $\px$. It follows from (5.2)--(5.2$''$) and the inequality $\ell_{n,M}^{\pm}(x,0,v) \ge\allowmathbreak \ell_{n,M}^{\pm}(x,0,0)$ that $$ \align \px&\le \frac{\e^2}{\bb}\sqrt\pi\expt\quad\text{for }x>\sqrt{2\bb}L \tag5.6\\ \px&\le \frac{100}{\bb}\sqrt\pi\expo \quad\text{for }|x|<\sqrt{2\bb}L \tag5.6$'$\\ \px&\le \frac{C^2\sqrt\pi}\bb\quad\text{for all }x\in R^1. \tag5.6$''$ \endalign $$ These estimates together with (5.4) and (5.5) imply that for $x\ge 2L\bs$ $$ \align \qbb \fx&\le2\int_{0}^{\frac x{\sqrt2}-L\bs} \frac {\e^2\sqrt\pi}{\bb}\exp \left\{-\frac{x^2}{2\bb}-\frac{u^2}\bb\right\}\,du \\ &\qquad +2\int_{\frac x{\sqrt2}-L\bs}^{\frac x{\sqrt2}+L\bs} \frac{10\e}{\bb}\sqrt\pi \exp\left\{-\frac{(x-\sqrt2u)^2}{\bb}-\frac{(x+\sqrt2u)^2}{2\bb} \right\}\,du \\ &\qquad +2\int_{\frac x{\sqrt2}+L\bs}^{\infty} \frac{C\e}{\bb}\sqrt\pi \exp\left\{-\frac{(x+\sqrt2u)^2}{2\bb} \right\}\,du\le \frac{\e^{3/2}}{\sqrt\bb}\expt \\ \endalign $$ if $L=L(\e)$ is sufficiently large. Lemma ~2 is proved. \enddemo Let us apply Lemma 2 with $\fx=f_n(x)$, \ $\bb=2{\bbh n}$ and $M=M_n$. Since $f_{n+1}(x)=\bold{ Q}_{n,M_n}f_n(x)\le C_1\bold{\bar Q}_{n,M_n}f_n(x+m_n)$ with some $C_1>0$ hence in order to carry out our inductive procedure it is enough to show that $$ C_1\sqrt\e \exp\left\{-\frac1{4\bbh n}(x+m_n)^2\right\}\le \exp\left\{-\frac1{4\bbh{n+1} }\right\}. $$ This can be deduced from the inequality $$ \bbh {n+1}(x+m_n)^2+K\bbh n \bbh {n+1}\ge \bbh n x^2\,\tag5.7 $$ with sufficiently large $K>0$ if $\e>0$ is chosen sufficiently small. Since $|m_n|N$ and $\bbh n>{10}$ one gets formula (5.7) with the help of simple calculation from (4.13) and (4.13$'$). The proof of formulas (2.7) and (2.7$'$) is based on the following \proclaim{Lemma 3} \it Let the function $\fx$ satisfy the conditions of Lemma 2. Let some numbers $r>0$, \ $\bb>0$ and $\a>0$ be given in such a way that $r>\bb>{10}$, \ $\bb<\frac9{10}Mn$ and $\frac1{100}<\a<1-\e^{1/8}$. Let us assume that the function $\fx$ satisfies, beside the conditions of Lemma ~2, the estimates $$ \align \fx&\le\frac\e{\bs}\expt \qquad\text {for }-r0$, the main contribution to the integral $\qbb\fx$ is given in a small neighbourhood of the point $(u,v)=(0,0)$. For $x<0$ this statement remains valid only for $x>-\bar r$. For $x<-\bar r$ the main contribution to this integral is given in a small neighbourhood of the points $(u,v)=(0,\pm v^{*})$ with $v^{*2}= 2^{n/2}\left\{(M-2^{-(n+1)/2}\bar r)^2-(M+2^{-(n+1)}x)^2\right\}$. \demo{Proof of Lemma 3} Define the function $$ K(x)=\cases \frac{10}{\sqrt\bb}\expo \qquad\text{for }|x|<-\sqrt{2\bb}L\\ \frac{\e}{\bs}\expt \qquad\text{for }x>L\bs\text{ or }-r-\frac{\rb}{\sqrt2}$ and in the points $\pm v^{*}$ satisfying the equation $\ell^{\pm}_{n,M}(x,0,v^*)=-\frac{\rb} {\sqrt2}$ for $x<-\frac{\rb}{\sqrt2}$. (At this point we need the condition $\bar r<\sqrt 2r$ which guarantees that the estimate (5.8) holds in the point $\frac{\rb}{\sqrt2}$.) The function $\px$ defined in formula (5.3) can be estimated in the following way: $$ \px\le \int \exp\left\{-\e^{1/8}v^2\right\}\bar K(x,v)\,dv \le \e^{-1/4}\sqrt\pi\sup_v \bar K(x,v). $$ Hence we obtain that $$ \align \px&\le\frac{\e^{7/4}\sqrt\pi}{\bb}\expt \qquad \text{for } -\rb0,x+\sqrt2u<-\rb\}$. This integral can be estimated in the following way: $$ \align &\int_ {\{u>0,x+\sqrt2u<-\rb\}}P(x,u)\,du\\ &\qquad\le\frac{\e^{7/4}\sqrt\pi}{\bb}\becsl\\ &\qquad\qquad\int_ {\{u>0,x+\sqrt2u<-\rb\}}\exp\left\{-2^{-n/2}(1-\abb )u^2\right\}\,du \,. \tag5.11 \endalign $$ We give an upper bound on the right-hand side of (5.11) by replacing $\abb$ with $\bar\a$ in it and multiplying the expression by $\exp\{-(\abb-\bar\a)|\rb+x|\}$. The integral in this expression can be estimated by the rather rough bound $|\rb+x|$. These estimates show that the right-hand side of (5.11) is much less than the expression at the right-hand side of (5.9$'$). To estimate the integral $\int P(x,u)\,du$ in the case $-2^{(n+1)/2}M-\rb\}$ observe that some calculation yields that $$ \align \expt&=\becsl\\ &\qquad \exp \left\{-\left(\frac1{2\bb}+2^{-(n+2)/2}\abb\right)(x+\rb)^2\right\} \,,\tag5.12 \endalign $$ because of the definition of $\rb$. Because of this identity the estimates (5.10) and (5.10$'$) enable us to estimate the integral $\int P(x,u)\,du$ in this case similarly to the estimation of (5.11), only in this case the last term in (5.12) helps us to bound the pre-exponential term. Similar calculations enable us to bound the integral (5.4) for $x>-\rb$ and to deduce the estimates (5.9) and (5.9$'$). Lemma 3 is proved. \enddemo Formulas (5.8) and (5.8$'$) hold for $\fx=f_0(x)$ with $\bb=2\bbh 0=20$, \ $\a=\frac1{100}$, \ $M=M_0$ and $r=\sqrt2\a\bb M$. If the conditions of Lemma 3 are satisfied for $f_n(x)$ with $M=M_n$, \ $\bb=2\bbh n$ and some $\a_n$ and $r_n$ then Lemma 3 gives an estimate on $\bold {\bar Q}_{n,M_n}f_n(x)$. An argument similar to that given after Lemma 2 gives an estimate when the operator $\bold {\bar Q}_{n,M_n}$ is replaced by $\bold { Q}_{n,M_n}$. In such a way we get by induction the estimates (5.8) and (5.8$'$) for $f_n(x)$ with $\bb= 2 \bbh n$, an increasing sequence $\a_n$ which tends to $1-\e^{1/8}$ and a number $r_n$ which is a small perturbation of the expression given in (5.8$''$). Since $\frac{\bbh n}n$ has a positive limit as $n\to \infty$, the number $r=r_n$ which appears in the estimates (5.8) and (5.8$'$) for $f_n(x)$ during this induction has the order $n$. By rewriting these estimates for $\bar q_n(x)$ with the help of (3.1) we obtain the estimates (2.7) and (2.7$'$) (with $\e^{1/8}$ instead of ~$\e$). \beginsection 6. The proof of Theorem B The proof of Proposition B is the same as that of Lemma 1 in Part II of [2], hence we omit it. The proof of Theorem B is also very similar to the method of Part II in [2], only the number $c$ must be replaced by $\sqrt2$ and $M$ by the constant $\mb$ defined in (1.9) everywhere. The main difference is that now we have a weaker control about the tail behaviour of the density function of the average spin $p_n(x)$. As a consequence, we can prove some estimates only in a weaker form. Nevertheless, they are sufficient for our purposes. Let us discuss this question in more detail. Introduce the functions $\bar p_n(x)$ and $g_n(x)$, ~$x\in R^1$ as $$ \align \bar p_n(x)&=K_n\exp\left\{\frac {a_0}{2a_1}2^{n/2}M^2\right\}p_n(\tilde x), \qquad \tilde x=(x,0)\in R^2, \tag6.1 \\ g_n(x)&=2^{-n/2}\bar p_n\left(\mb+2^{-n/2}x\right), \tag6.2 \\ \endalign $$ where $p_n(x)$ is defined after formula (1.6), the number $\mb$ in (1.9), and $K_n$ is the same norming constant as in (1.7). By formula (1.7) $$ g_n(x)=2^{-n/2}\exp\left\{-\frac{a_0}{2T}x \left(2\mb+2^{-n/2}x\right)\right\}\bar q_n \left(M+ 2^{-n/2}\sqrt{\frac{a_1}{T}}x\right), $$ hence Theorem A yields that $$ g_n(x)=\frac1{\sqrt n}\exp\left\{-\frac{a_0}{2T}x \left(\mb+2^{-n/2}x\right)\right\} \left[\frac{\sqrt2 M}{\sqrt\pi}\exp\left\{-\frac{2\mb^2}n x^2\right\}+R_n(x)\right] \tag 6.3 $$ with $$ \left|R_n(x)\right|\le \frac K{\sqrt n}. \tag6.3$'$ $$ On the other hand, we get by rewriting Proposition A for $g_n(x)$ that there are some numbers $ B>0$, \ $D>0$ and $R_n$, \ $-C_1nC_2>0$ such that $$ g_n(x)\le \frac K{\sqrt n}\exp\left\{-\frac{a_0}{2T}x\left(2\mb+2^{-n/2}x\right) -\frac Bnx^2\right\}\qquad \text{for }x>R_n \tag 6.4 $$ and $$ \aligned g_n(x)&\le \frac K{\sqrt n}\exp\biggl\{-\frac{a_0}{2T}R_n\left(2\mb+2^{-n/2}R_n\right) -\frac BnR_n^2\\ &\qquad-D(R_n-x)\left(2\mb+2^{-n/2}(R_n+x)\right)\biggr\} \quad \text{for }-2^{-n/2}\mb0$ by choosing $\e$ in (2.7) sufficiently small.) The estimates (6.4) and (6.4$'$) are the natural counterparts of the estimates (4.11$'$) and (4.11$''$) in Part II of [2]. The function $f_n(x)$ defined by formula (4.11) of that work is the analogue of our function $g_n(x)$. The bound given on $g_n(x)$ decreases at infinity slower than its counterpart in [2] because of the multiplying term $1/n$ in formula (6.4). Another, and even more important difference between the two cases is that in the points $x\sim- const.\,n$ relations (6.4) and (6.4$'$) give no better bound on the function $g_n(x)$ than $\exp \{Cn\}$ with some positive $C>0$. As a consequence, in several estimates we have to multiply the upper bound by an exponential term instead of a constant, as it is the case in [2]. But this estimates suffice for us, because in the final estimates we have a double exponential term which compensates this effect. Applying the same argument as in [2] we get that Theorem B follows from an analogue of Proposition 1$'$ in Part II of [2] which is obtained if $c$ is replaced by $\sqrt2$ and $M$ by $\mb$ in this result. For the sake of convinience we also make the following modification. From now on we shall work with the function $ K_n\exp\left\{\frac {a_0}{2a_1}2^{n/2}M^2\right\}p_n(x)$ instead of the original function $p_n(x)$ and we denote it in the same way. This modification influences only the norming constant $L_n$ in the Radon--Nikodym derivative. The proof of this modified version of Proposition 1$'$ of [2] is very similar to the original one. We have to estimate certain integral expressions in the domains $\Omega_n^i$, ~$i=1$ ~2,~3, defined in (2.14)--(2.14$''$). We rewrite these integrals in polar coordinate system and first estimate the integrals on a circle of fixed radius $r$. This can be done in the same way as in [2]. Then the integrals with respect to $r$ can be estimated with the help of formulas (6.4) and (6.4$'$) instead of formulas (4.11$'$) and (4.11$''$) in [2]. We get in such a way slightly weaker estimates than those in [2], but they suffice for our purposes. Lemmas~2 and ~3 of Part ~II of [2] remain valid after the replacement of $c$ and $M$ by $\sqrt2$ and $\mb$ in the following weaker form: In Lemma ~3 the multiplying term~$K$ and in Part~a) of Lemma~2 the multiplying term $c^n$ before the exponent must be replaced by $K^n$, where $K$ is some appropriate constant depending on $t$ and $T$. Also the estimates of Section~5 of Part~II of ~[2] remain valid. The only place where the argument of the proof must be slightly changed is Part~a) where $\bold S^1_n\fx$ is estimated for $x\in\Omega_n^1$. The argument of [2] works if we show that the expressions $\je(x_1)$ defined by the formula $$ \je(\x1) =\int_{|t|<\eb2^{0.3n}}\exp\left\{ \frac{t\x1}T+\sqrt2\frac{\gb_{n+1}}2 t\right\}g_n(t)\,dt\tag6.5 $$ with some sufficiently small $\eb>0$ satisfy the following relations: $$ \je(\x1)=\left(1+O(2^{-0.1n})\right)\je(M)\qquad\text{if }\;x\in\Om1\,, \tag6.6 $$ and $$ \je(\mb)>K_1>0. \tag6.7 $$ Relation (6.7) simply follows from (6.3) if we restrict the domain of integration in (6.5) to the domain $|t|<\frac1{3M}\sqrt{n\log n}$. (The corresponding estimate (5.11) in Part II of [2] also contained an upper bound on $\je(\mb)$, but we do not need this bound.) Then relation (6.5) follows from the following observations: The ratio of the integrands in the expressions $\je(\x1)$ and $\je(\mb)$ are closer to ~1 than $const.2^{-0.05n}$ if $|t|<2^{0.05n}$ and $\x1\in\Omega_n^1$ and therefore $|\x1-\mb|<2^{-0.2n}$, and the contribution of the domain $|t|>2^{0.05n}$ to these integrals is less than $\exp\{-const.2^{0.05n}\}$. The remaining part of the proof works with some natural modification of the proof given in~[2], hence we omit it. \beginsection 7. The proof of Theorems 1 and 2 To prove Theorem 1 first we show that for all $q$, \ $ 2^{-0.1}\bar M+2^{-0.2n} \tag 7.4 $$ $$ \fn\le L_n\exp \left\{\frac{\gb 2^{n/2}}{2\bar M}(|x|^2-\bar M^2)\right\}\qquad\text{if }\;0n_0$ and $N>N_0(n,q)$ then $$ \left|\tilde\mu_n(A(n))-\mn (A(n))\right|\le Kq^n $$ with some $K>0$ independent of the set $A$. Theorem 1 can be proved with the help of the above relation similarly to [2]. Moreover, this argument also yields the following \proclaim{Corollary of Theorem 1} \it Let $\bar \mu_n$ denote the projection of of the measure $\bar\mu$ constructed in Theorem 1 to $(R^2)^{2^n}$. There is some function $\bar f_n(x)$ such that $$ \frac{d\bar \mu_n}{d\mu_n}(x_1,\dots,x_{2^n})= \bar f_n\left(2^{-n}\sum_{j=1}^{2^n}x_j\right). $$ Let $n>n_0$ with some threshold $n_0>0$. Then relations (7.1)---(7.5) remain valid if $f_{n,N}^{h_N}$ is replaced by $\bar f_n(x)$ in them. \endproclaim Now we turn to the proof of Theorem 2. Let us introduce the Hamiltonian $\Cal H_k$ in the volume $(R^2)^{2^k}$ by the formula $$ \Cal H_k(x_1,\dots,x_{2^k})=-\sum_{i=1}^{2^k-1}\sum_{j=i+1}^{2^k} d(i,j)^{-3/2}x_ix_j\,. $$ Let $\s=\{\s(j)=(\s_1(j),\s_2(j)),\;j\in\bold Z\}$ be a $\bar \mu $ distributed vector and consider the random vector $\{\left(\Cal R_n \sigma^{(1)}(j), \, \Cal R_n \sigma^{(2)}(j)\right),\;1\le j\le 2^k\}$ defined by formulas (1.3)--(1.5) with $A_n=2^{n/2}\sqrt n$ and $B_n=2^{3n/4}$. The argument at the beginning of Section 7 in Part II of [2] also shows that the density function $h_{n,k}(x_1,\dots,x_{2^k})$ of this vector can be expressed in the following way: $$ \align &h_{n,k}(x_1,\dots,x_{2^k})\\&\qquad=L_{n,k}\bar f_{n+k}\biggl(2^{-k}\sum_{j=1}^{2^k}\tilde x_j\biggr)\exp\left\{-\frac1T\Cal H_k\biggl(2^{n/4}\tilde x_1,\dots,2^{n/4}\tilde x_{2^k}\biggr)\right\} \prod_{j=1}^{2^k}p_n(\tilde x_j)\, \\ & \tag7.10 \endalign $$ with $$ \tilde x=\tilde x(x)=\left( \bar M+2^{-n/2}\sqrt nx^{(1)},2^{-n/4}x^{(2)}\right)\qquad \text{for }\;x=\left(x^{(1)},x^{(2)}\right)\,. \tag 7.10${}^{\prime}$ $$ Let us define the sets $W_n\subset R^2$ and $ \bar W_n\subset R^2$ by the formulas $$ \align \bar W_n=\biggl\{(x^{(1)},x^{(2)}&),\;\bar M- \frac{\sqrt T} {8\bar M}2^{-n/2} \sqrt{n\log n}<|x|<\bar M+\frac{\sqrt T}{8\bar M} 2^{-n/2}\sqrt{n\log n}, \\&\qquad \qquad \qquad \qquad \qquad\qquad\qquad|x^{(2)}|<2^{-n/4}n^{1/5},\;x^{(1)}>0\biggr\} \\ W_n=\Bigl\{(x^{(1)},x^{(2)}&),\;\tilde x(x)\in \bar W\Bigr\}\,, \endalign $$ We claim that for all $j=1$,~2,~\dots,~$2^k$ $$ P\left(\,\bigl(\Cal R_n\sigma^{(1)}(j),\;\Cal R_n\sigma^{(2)}(j)\,\bigr)\notin W_n\right)\le n^{-1/100}\qquad \text{if }\,n\ge n_0.\tag 7.11 $$ and $$ \align h_{n,k}(x_1,\dots,x_{2^k}) &=h_{k}(x_1,\dots,x_{2^k})\left(1+O(n^{-1/9})\right)\\ &\qquad\text{if }x_j\in W_n\text{ for all }j=1,2,\dots,2^k,\tag7.12 \endalign $$ where $h_{k}(x_1,\dots,x_{2^k})$ is the function defined in (1.11) (with $s=2$). Relations (7.11) and (7.12) together imply Theorem 2. Relation (7.12) can be proved with the help of the following estimates: $$ \align p_n(x)&=K_n\left[\exp\left\{-\frac{2^{n+1}M^2a_1}{nT}\left(|x|-\bar M\right)^2 \right\}+O\left(\frac1{\sqrt n}\right) \right]\exp\left\{-\frac{a_0} {2T}2^{n/2}x^2\right\} \\ &= K_n\exp\left\{-\frac{2^{n+1}M^2a_1}{nT}\left(|x|-\bar M\right)^2 -\frac{a_0}{2T}2^{n/2}\left(x^{(1)2}+x^{(2)2}\right) +O\left(n^{-1/9}\right)\right\} \\ &=\bar K_n\exp\biggl\{-\frac{2^{n+1}M^2a_1}{nT}\left(x^{(1)}-\bar M\right)^2 -\frac{a_0}{2T}2^{n/2}\left(2\mb (x^{(1)}-\mb)+x^{(2)2}\right) \\&\qquad\qquad\qquad\qquad +O\left(n^{-1/9}\right) \biggr\} \qquad \text{if }x\in \bar W_n, \tag 7.13 \endalign $$ since in this case we can put the $O(\cdot)$ term into the exponent by appropriately decreasig the power of ~$n$ in it, $$ \frac{2^n}n(|x|-\bar M)^2=\frac{2^n}n\left(x^{(1)}-\mb\right)^2+O\left( n^{-3/10}\sqrt{\log n}\right)\qquad\text{for }x\in \bar W_n\,, \tag7.14 $$ and $$ 2^{n/2}x^{(1)2}=2^{n/2}\mb^2+2 \cdot2^{n/2}\mb(x^{(1)}-\mb) +O\left(2^{-n/2}n\log n\right) \qquad \text{for }x\in \bar W_n.\tag7.15 $$ We also have $$ \align \Cal H_k(2^{n/4}x_1,\dots,2^{n/4}x_{2^k})&= C_{k,n}-2^{n/2}\biggl[ \sum_{i=1}^{2^k-1}\sum_{j=i+1}^{2^k} d(i,j)^{-3/2}x_i^{(2)} x_j^{(2)}\\ &\qquad+ a_0\mb\sum_{i=1}^{2^k} \left(1-2^{-k/2}\right) (x_i^{(1)}-\mb) \biggr]+O\left(2^{-n/2}n\log n\right) \\ &\qquad\qquad \qquad \text{if }x_i\in \bar W_n,\;j=1,\dots,2^k,\tag7.16 \endalign $$ since $\sum_{j=1}^{2^k}d(i,j)^{-3/2}=a_0(1-2^{-k/2})$ for all $1\le i\le 2^k$, and $ 2^{n/2}(x^{(1)}_i-\mb)(x^{(1)}_j-\mb)= \allowmathbreak O(2^{-n/2}n\log n)$ in this case. Because of the Corollary of Theorem 1 and the relation $\bar g=\frac{a_0\mb}T$ $$ \align \bar f_{n+k}\left(2^{-k}\sum_{j=1}^{2^k}x_j\right)=C_{n,k}\exp&\biggl\{ 2^{n/2}\biggl(\frac{a_0\mb}T2^{-k/2}\sum_{j=1}^{2^k}(x^{(1)}-\mb) \\&\qquad+\bar A2^{-3k/2}\biggl(\sum_{j=1}^{2^k}x_j^{(2)}\biggr)^2 \biggr)+O(q^n)\biggr\} \\&\qquad \qquad\text{if }x_j\in \bar W_n,\;j=1,\dots,2^k. \tag7.17 \endalign $$ Relation (7.12) follows from (7.10), (7.13), (7.16) and (7.17). Relation (7.11) can be proved in the same way as it is done in Section 7 of Part II of [2], only the relations $\bigl||x|-M\bigr|