\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{$1<c<\sqrt2$}
\define\qq#1{\bar q_{#1}(x)}
\define\ct{$\sqrt2<c<2$}
\define\cc{{$c=\sqrt2$}}
\define\fh#1{f^h_{{#1},N}(x)}
\define\Qb#1{\bar Q_{#1}(x)}
\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\rn{(R^2)^{2^n}}
\define\rk{(R^2)^{2^k}}
\define\im#1{\int_{\Om#1}}
\define\fformg{\vf\in\Gamma_n(r,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{\left(\frac{c^n}T+\frac{A_{n+1}}2\right)}
\define\cga{\left(\frac{c^n}{2T}+\frac{g_{n+1}}{4M}-\frac{A_{n+1}}4\right)}
\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\bb{\beta}
\define\a{\alpha}
\define\qm{{\bold Q}_{n,M}}
\define\qbb{ \bar{\bold Q}_{n,M}  }
\define\tm{{\bold T}_M}
\define\tmb{\bold{\bar T}_M}
\define\ttm{\bold{\tilde T}_M}
\define\ttmb{\bold{\tilde{\bar T}}_M}
\define\norm{\phi}
\define\reg#1{\varphi_{#1}(f_{#1})}
\define\regg#1{\varphi_{#1}(f)}
\define\bbb#1{\bar\beta_{#1}}
\define\der#1{\frac{d^{#1}}{dt^{#1}} }
\define\four#1{\tilde \varphi_{#1}(f_{#1})(t)}
\define\tmn{\bold{\tilde T}_{M_n} }
\define\ph#1{\psi_{#1}(t)}
\define\roott{\left(\frac n{\bbb{n}}\right)^{1/3}}
\define\bbh#1{\hat\beta_{#1}}
\define\expo{\exp\left\{-\frac1{\bb}x^2\right\} }
\define\expt{\exp\left\{-\frac1{2\bb}x^2\right\} }
\define\fx{f(x)}
\define\px{P(x)}
\define\bs{ {\sqrt\bb} }
\define\pxu{P(x,u)}
\define\becs{\exp\left\{-\frac1{2\bb}{r^2}
-\a2^{(n-1)/2}\left[\left(M- {2^{-n/2}}r\right)^2-
\left(M+ 2^{-n/2}x\right)^2\right]\right\} }
\define\becsl{\exp\left\{-\frac1{2\bb}{\bar r^2}
-\bar{\bar\a}2^{n/2}\left[\left(M-
2^{-(n+1)/2}\bar r\right)^2-
\left(M+ 2^{-(n+1)/2}x\right)^2\right]\right\}  }
\define\abb{\bar{\bar\a}}
\define\becsv{\exp\left\{-\frac1{2\bb}{\bar r^2}
-{\bar\a}2^{n/2}\left[\left(M-
2^{-(n+1)/2}\bar r\right)^2-
\left(M+ {2^{-(n+1)/2}}x\right)^2\right]\right\}  }
\define\rb{\bar r}
\define\mb{{\bar M}}
\topmatter
 
\title The large-scale limit of Dyson's hierarchical
vector-valued
model at low temperatures. \\
\rm The marginal case \cc.
\endtitle
\endtopmatter
\vskip-0.7truecm
\centerline{P.M. Bleher${}^1$ and P. Major${}^2$}
\vskip-0.6cm
$$
\align
&\text{{\kis
${\mats{}^1}$ Keldysh Institute for  Applied Mathematics of the
Academy of Sciences of the U.S.S.R.,}}\\ \vspace{-1.5mm}&\text{{\kis
Moscow A--47 Miusskaya Square 4, U.S.S.R.}}\\
&\text{{\kis
${\mats{}^2}$ Mathematical Institute of the Hungarian Academy
of Sciences,}}\\ \vspace{-1.5mm}&\text{{\kis Budapest H--1364 P.O.B.
127, Hungary}}
\endalign
$$
\vskip-0.2cm
 
\centerline
{\it Dedicated to Professor R.L. Dobrushin for his sixtieth birthday}
 
\bigskip
 
\noindent
{\bf Summary:} In this paper we construct the equilibrium states of
Dyson's vector-valued hierarchical model with parameter \cc{} at low
temperatures and
describe their large-scale limit. The analogous problems for \ct{} and
\co{} were solved in our papers [1] and [2]. In the present case the
large-scale limit is similar to the case \ct, i.e. it is a Gaussian
self-similar field with long-range dependence in the direction
orthogonal to and a field consisting of independent Gaussian random
variables in the direction parallel with the magnetization. The main
difference
between the two cases is that now the normalizing factor in the
direction of the magnetization contains, beside the square-root of the
volume, a logarithmic term too.
 
\beginsection{1. Introduction}
 
First we briefly describe the model we are investigating. Dyson's
hierarchical model is a one-dimensional classical spin model on
the lattice $\zb=\{1,2,\dots\}$. Its Hamiltonian function depends on a
parameter $a$, \ $1<a<2$, and is defined as
$$
\ch(\s)=-\sum_{i\in\zb}\sum\Sb j\in \zb\\j>i\endSb
d(i,j)^{-a}\s(i)\s(j)\,,   \tag 1.1
$$
where  $d(i,j)=2^{n(i,j)-1}$, and
$$
n(i,j)=\min\left\{ n,\text{ there exists some $k$ such that
}(k-1)2^n<i,j\le k2^n\right\}.
$$
We are dealing with vector-valued models, where
$\s(j)\in R^p$ with some $p\ge2$. If $x\in R^p$ and $y\in R^p$
then $xy$ denotes scalar product. We consider models with the free
measure $\nu$,
$$
\frac {d\nu}{dx}(x)=p_0(x)=p_0(x,t)=C(t)\exp\left\{-
\frac{x^2}2-\frac t4|x|^4\right\},\qquad x\in R^p,  \tag 1.2
$$
where  $t>0$ 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 $0<T<T_0$ and $0<t<t_0$ then the limit
$M=\lim_{n\to\infty}
M_n>0$ 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 $0<t<t_0$ and $0<T<T_0$ the functions $q_n(x)$ defined by
formulas (2.2) and (2.2$'$) satisfy the following relations:
 
There are some  $M=M(T,t)>0$  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 $0<x<M$ then for all $\e>0$ there are  some thresholds
$t_0=t_0(\e)$, \ $T_0=T_0(\e)$ and a real number
$r_n$, \
$C_1n2^{-n/2}<M-r_n<C_2n2^{-n/2}$ with $C_2>C_1>0$
such that if $0<t<t_0$ and $<T< T_0$ then
$$
2^{-n/2}\sqrt n\bar q_n(x,T)\le
K\exp\left\{-\beta
\frac{2^n}n( x-M)^2\right\}\quad\text
{for  } r_n<x<M\tag2.7
$$
and
$$
\align
2^{-n/2}&\sqrt n\bar q_n(x,T)\le
K\exp\left\{-(1-\e)2^{n/2}\left(r_n^2-x^2\right)-\beta
\frac{2^n}n( r_n-M)^2\right\}\\&\qquad\text
{for  } 0<x<r_n\,.\tag2.7$'$
\endalign
$$
\endproclaim
(With some extra-work it can be shown that the number $r_n$ can be
chosen in the form $r_n=M-Cn2^{-n/2}$  with some $C>0$. 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
$$
\align
g_N&=g_N(N,h_N)=\frac{2^Nh_N}T\,\tag2.12\\
g_n&=g_n(N,h_N)=\frac{g_{n+1}}2  + \frac{2^{n/2}}T\bar M\qquad \text
{for } n<N \tag2.12${}^{\prime}$\\
A_N&=A_N(N,h_N)=0\tag2.13\\
A_n&=A_n(N,h_N)=\frac{A_{n+1}}4+\dfrac{\left(\dfrac{2^{n/2}}T
+\dfrac{A_{n+1}}2\right)^2}
{\dfrac{2^{(n+2)/2}}T+\dfrac{g_{n+1}}{\bar M}-A_{n+1}} \qquad \text {for
} n<N\,,\tag2.13${}^{\prime}$ \endalign
$$
where  $\bar M$ is defined in (1.9), and $M$  and  $T$  are the same
as in Theorem A.
 
For the sake of simpler notations let us restrict ourselves to the
case $R^p=R^2$. Let us define the domains
$$
\align
\Omega^1_n&=\{x\in R^2,\,\bigl||x|-\bar
M\bigr|<2^{-0.2n}\,,\;|x^{(2)}| <2^{-0.2n}\,,\;x^{(1)}>0\}\tag2.14\\
\Omega^2_n&     =\left\{x\in
R^2,\,\bigl||x|-\bar M\bigr|<2^{-0.2n}\right\} \setminus
\Omega^1_n \tag2.14${}^{\prime}$\\
\Omega^3_n&=\left\{x\in R^2,\,\bigl||x|-\bar M\bigr|\ge
2^{-0.2n}\right\}\,.\tag2.14${}^{\prime\prime}$
\endalign
$$
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  $q$, \ $2^{-0.1}<q<1$,  there is
some  $n_0=n_0(T,\bar M,D,q)$  such that if (2.11) holds, and
$N\ge n\ge n_0$  then the Radon--Nikodym derivative
$f_n(x)=f_{n,N}^{h_N}(x)$  defined by  formulas (2.9)--(2.10$'$)
satisfies the following relations:
\newline
 a) In the domain  $\Omega^1_n$
$$
f_n(x)=L_n\exp\left\{g_n\left(x^{(1)}-\bar M\right)+A_nx^{(2)2}+
\varepsilon_n(x)\right\}\tag2.15
$$
with
$$
\sup_{x\in\Omega^1_n}|\varepsilon_n(x)|\le q^n\,.
$$
b) In the domain  $\Omega_n^2$
$$
0\le f_n(x)\le L_n \exp\left\{g_n(|x|-\bar
M)-\biggl(\frac{g_n}{2\bar M}
- A_n\biggr)2^{-0.4n}+q^n\right\}\,.\tag2.16
$$
c) In the domain  $\Omega^3_n$
$$
0\le f_n(x)\le L_n \exp\left\{\frac{g_n}{2\bar
M}\left(|x|^2-\bar M^2\right) \right\}\,,\tag2.17
$$
where the numbers  $A_n$  and  $g_n$  are defined in
(2.12)--(2.13${}^{\prime}$),
and  $L_n=L_n(N,h_N)$  is an appropriate norming constant.
\endproclaim
We also need the following result which  describes the asymptotic
behaviour of the sequences $g_n$ and $A_n$ defined by
(2.12)--(2.13$'$).
\proclaim{Proposition B} \it Let us choose some integer  $N$  and
real number $h_N>0$. 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 $|R(T,t)|<const$. Moreover, there is some $n_0=n_0(t,T)$
such that for $n>n_0$
$$
M_n=M+\frac{2+\sqrt2}{4M}2^{-n/2}+\delta (n)2^{-n/2},\qquad
|\delta (n)|<K2^{-n/2}  \tag3.3
$$
with some $K>0$. 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 $0<T<T_0$ and
$0<t<t_0$ then for all $n\le N$
$$
\align
\left|\frac{d^j}{dx^j}[f_n(x)-\norm(x,\s)]\right|\le\frac{B(n)}
{\sqrt{M_n}}\exp&\left\{-2^{(n+2)/2}|x|\right\}  \\
&\quad\text{if }|x|<\log M_n,\;\,j=0,1,2 ,
\tag3.12
\endalign
$$
$$
\left|\frac{d^j}{dx^j}f_n(x)\right|\le B(n)\exp
\left\{-2^{n/2}\left|2x+\frac {2^{-n/2}x^2}{M_n}\right|\right\}\quad
\text{if }x>-2^{-n/2}M_n,\;j=0,1,2,
\tag3.13
$$
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:
$$
\align
M_{n+1}=M_n+2^{-(n+1)/2}m_n,\qquad
&m_n=-\frac{\sqrt2}{4M_n}+\gamma(n)\\
&\quad \text{with } \gamma(n)<C_12^{-n/2}\sqrt{\bb_n} \tag3.17
\endalign
$$
$$
\align
&\left|\frac {d^j}{dx^j} f_{n+1}(x) -\bold T_{M_n}\reg n(x)\right|\le
\frac {C_1C^4}{\bb_{n+1}^{(j+1)/2}} 2^{-n/2}
\\  \vspace{2\jot}&\qquad\left[
\exp\left\{-\frac1{\sqrt{\bb_{n+1}}}\left|2x+
\frac{2^{-(n+1)/2}x^2}{M_{n+1}}\right|\right\}+
\exp\left\{-\frac{2|x|}{\sqrt{\bb_{n+1}}}\right\} \right]\\
\vspace{2\jot}&\qquad\qquad \text{for
}x>-2^{(n+1)/2}M_{n+1},\,\;j=0,1,2, \tag3.18 \endalign
$$
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 $0<T<T_0$ and $0<t<t_0$ with some suficiently
small $t_0>0$ 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 $0<t<t_0$, \ $0<T<T_0$  with some sufficiently
small $t_0>0$ 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 $|M-M_n|<CM^{-1} 2^{-n/2}$. This estimate  together with (4.11)
imply
(2.3). The remaining statements of Theorem A also follow from Theorem
A$'$.
\enddemo
Theorem A gives a good Gaussian approximation only for large $n$. On
the other hand, the error term in (4.10) is  small for all $n\ge
N$. Beside this, Proposition ~1 yields a good Gaussian approximation
for all $n\le N$ if $\hat M_0$ is very large.
These observations imply the following
\proclaim{Corollary of Theorem A$'$} \it Define the sequence $\bbb n$ by
(4.1) and (4.2) for $n\ge N$ and $\bbb n=\bbb N$ for $n\le N$. For
all $\delta>0$ 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 $0<t<t_0$ and $0<T<T_0$. As a consequence, for arbitrary $L>0$ the
inequality
$$
f_n(x)<\frac{10}{\sqrt{ \bbh n}}\exp\left\{-\frac1{2\bbh n}x^2\right\}
\qquad\text{for }|x|<L\sqrt{ \bbh n},\;n=0,1,2,\dots  \tag4.13
$$
holds with the sequence
$$
\align
\bbh n&={10}+\frac n{M_n^{1/2}}
\qquad\text {for }0\le n\le N \tag4.14\\
\bbh {n+1}&=\bbh n\left(1+2^{-n/4}\right)+\frac1{8M_n^2}
\qquad\text {for }n\ge N \tag4.14$'$
\endalign
$$
if the conditions of (4.12) hold with a sufficiently small
$\delta=\delta(L)$.
\endproclaim
 
\beginsection 5. The proof of Proposition A
 
First we prove formula (2.6).
Choose an appropriately small $\e>0$ and a large $L=L(\e)>0$. We are
going to  show that if $0<t<t_0$ and $0<T<T_0$ with some
$t_0=t_0(\e,L)$, and $T_0=T_0(\e,L)$ then
$$
f_n(x)\le \frac{10}{\sqrt{\bbh n}}\exp\left\{-\frac1{2\bbh
n}x^2\right\}\quad\text{for }|x|<L\sqrt{\bbh n},\; n=0,1,2,\dots
\tag5.1
$$
and
$$
f_n(x)\le \frac{\e}{\sqrt{\bbh n}}\exp\left\{-\frac1{4\bbh
n}x^2\right\}\quad\text{for }|x|>L\sqrt{\bbh n},\; n=0,1,2,\dots\,.
\tag5.1$'$
$$
 
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
$$
\align
\fx&\le\frac{10}{\sqrt\bb}\expo \quad\text{for }|x|<L{\sqrt \bb}
\tag5.2                                             \\
\fx&\le\frac{\e}{\sqrt\bb}\expt \quad\text{for }|x|>L{\sqrt \bb}
\tag5.2$'$                                            \\
\fx&\le \frac C{\bs}\quad \text{for all }x\in R^1 \tag 5.2$''$
\endalign
$$
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 $|m_n|<CM_n^{-1}$ for $n\le N$, \ $|m_n|<\frac
1{2M}+C_12^{-n/2}$ for $n>N$ 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
$$
\align
\fx&\le\frac\e{\bs}\expt \qquad\text {for }-r<x<-L\bs \tag5.8\\
\fx&\le\frac\e{\bs}\becs\\&\qquad\text{for }-2^{-n/2}M<x<-r
\tag5.8$'$
\endalign
$$
Put $\bar \a=\min\bigl((1+\e)\a,1-\e^{1/8}\bigr)$, \
$\abb=(1+\e^{1/8})\bar\a$ and
$$
\bar r=\dfrac{\sqrt2\abb\bb M}{1+\abb \bb2^{-n/2}}.\tag5.8$''$
$$
If $\bar r<\sqrt 2r$ then
$$
\align
\qbb\fx&\le\frac{\e^{3/2}}{\bs}
\expt \qquad
\text{for } -\bar r <x<-2L\bs
\tag5.9\\
\qbb\fx&\le\frac{\e^{3/2}}{\bs}\becsv
\\
&\qquad \text{for
}-2^{(n+1)/2}M<x<-\bar r\,.    \tag5.9$'$
\endalign
$$
\endproclaim
The proof of Lemma   3  is similar to that of Lemma 2. The main
difference is that in Lemma 2, i.e. when $x>0$, 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
$$
K(x)=\cases \frac{10}{\sqrt\bb}\expo \qquad\text{for }|x|<-\sqrt{2\bb}L\\
\frac{\e}{\bs}\expt \qquad\text{for }x>L\bs\text{ or }-r<x<-\sqrt{2\bb}L\\
\frac{\e}{\bs}\becs  \\ \qquad \text {for } -2^{n/2}M<x<-r .
\endcases
$$
Some calculation shows that for fixed $x$ the function
$$
\bar K(x,v)=\exp\{-\abb v^2\}K^2(\ell^{\pm}_{n,M}(x,0,v))
$$
takes its
maximum in
the point $v=0$ for $x>-\frac{\rb}{\sqrt2}$ 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
$$
\align
\px&\le\frac{\e^{7/4}\sqrt\pi}{\bb}\expt  \qquad \text{for }
-\rb<x<-\sqrt{2\bb}L \tag5.10 \\
\px&\le\frac {\e^{7/4}\pi}{\bb}\becsl
\\&\qquad\text{for }-2^{(n+1)/2}M<x<-\rb\,.   \tag5.10$'$
\endalign
$$
We estimate the integral in (5.4) with the help of (5.5), (5.6),
(5.6$'$), (5.10) and (5.10$'$). Let us first consider the case
$-2^{(n+1)/2}M<x<-\rb$ and integrate in the domain
$\{u>0,x+\sqrt2u<-\rb\}$.
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
$-2^{(n+1)/2}M<x<-\rb$ in the
domain $\{x+\sqrt2u>-\rb\}$ 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 $R_n$, \ $-C_1n<R_n<-C_2n$
with some $C_1>C_2>0$ 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
$$
\aligned
g_n(x)&\le \frac K{\sqrt
n}\exp\biggl\{-\frac{a_0}{2T}R_n\left(2\mb+2^{-n/2}R_n\right) -\frac
BnR_n^2\\
&\qquad-D(R_n-x)\left(2\mb+2^{-n/2}(R_n+x)\right)\biggr\}
\quad \text{for }-2^{-n/2}\mb<x<R_n.
\endaligned  \tag6.4$'$
$$
(We have to choose $B=\frac{\bb a_1}T$, \ $R_n=2^{n/2} \sqrt{\frac
T{a_1}}(r_n-M)$ and $D=\left(1-\e-\frac{a_0}{2a_1}\right)\frac{a_1}T$
in Proposition A. We may assume that $D>0$ 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}<q<1$
there are some thresholds $n_0$ and $N_0(n,q)$ such that if $n\ge
n_0$ and  $N\ge N_0(n,q)$ then
$$
\frac{d\mn}{d\mu_n}(x_1,\dots,x_{2^n})=
f_{n,N}^{h_N}\biggl(2^{-n}\sum_{j=1}^{2^n}x_j\biggl) \tag7.1
$$
with
$$
\fn= L_n\exp\left\{\gb 2^{n/2}(x^{(1)}-\bar M)+\ab 2^{n/2}
x^{(2)2}+
\e_n(x)\right\}\qquad \text{for } x\in \Om1\,, \tag 7.2
$$
where
$$
\sup_{x\in\Om1}|\e_n(x)|\le q^n\,,\tag 7.2${}^{\prime}$
$$
$$
\fn\le L_n\exp\left\{\gb 2^{n/2}(|x|-\bar M)-\biggl(\frac
{\gb}{2\bar M}-\ab\biggr)2^{0.1n}
+q^n\right\}\qquad\text{for  } x\in\Om2\tag 7.3
$$
$$
\fn\le L_n\exp\left \{\frac{\gb
2^{n/2}}{\bar M}(|x|^2-\bar M^2)\right\}\qquad\text{if
}\;x>\bar M+2^{-0.2n}         \tag 7.4
$$
$$
\fn\le L_n\exp \left\{\frac{\gb
2^{n/2}}{2\bar M}(|x|^2-\bar M^2)\right\}\qquad\text{if
}\;0<x<\bar M-2^{-0.2n}\tag 7.4${}^{\prime}$
$$
with the numbers $\bar g$ and $\bar A$ appearing in
Proposition B and some appropriate norming constant $L_n$ which
satisfies
the relation
$$
C_1<2^{-n/4}\bar L_n<C_2\qquad\text{with some
}\;0<C_1<C_2<\infty\,,\tag 7.5
$$
where $\bar L_n=L_n K_n\exp\left\{\frac{a_0}{2T}2^{n/2}\mb^2\right\}$
with the same norming constant $K_n$ as in (1.9).
 
 
In the proof of this statement we argue just as in Section 6 of Part
II of [2]. Because of Theorem B and Proposition B the constants $g_n$
and $ A_n$ can be replaced by $\bar g2^{n/2}$ and $\bar A2^{n/2}$ in
(2.15)--(2.17) by slightly changing the error terms. To show that
$L_n=L_n(N,h_N)$ can be chosen
independently of $N$ and $h_N$ we observe that
a calculation analogous to that in Section 6 of [2] yields that
$$
\mu_{n,N}^{h_N}\left(\Omega_n^2\cup\Omega^3_n\right)\le
L_n(N,h_N)\exp\left\{-K2^{0.3n}\right\}
\tag 7.6
$$
and for
$$
T_n=\im1\exp\left\{\gb 2^{n/2}(x^{(1)}-\bar M)+\ab 2^{n/2}
x^{(2)2}\right\}p_n(x)\,dx\,.\tag 7.7
$$
the relation
$$
\mu_{n,N}^{h_n}(\Omega_1^n)=
L_n(N,h_N)T_n\left(1+O(q^n)\right),\qquad2^{-0.1}<q<1,\tag7.8
$$
holds. The estimate
$$
C_12^{-n/4}< K_n\exp \left\{
\frac{a_0}{2T}2^{n/2}\mb^2\right \}T_n< C_22^{-n/4}\quad \text{with
some }0<C_1<C_2<\infty\,.\tag 7.9
$$
also holds true.
 
The proof of (7.9) is similar to that of (6.9) in [2], only one has
to observe that $\bar g$ equals $\frac{a_0\mb}{T}$, i.e. $-1$ times
the coefficient of $x$    in
(6.3)--6.4$'$), and  this causes some cancellation if we express
$p_n(x)$
through $g_n(x)$ in the integral (7.7). Since
$\mu_{n.N}^{h_n}(R^2)=1$, relations
(7.6)  and (7.8) imply that $L_n$  can be chosen as $T_n^{-1}$, and
then (7.9) implies (7.5). Theorem 1 can be proved with the help of this
information in the following way:
 
Fix  some integer $k\ge 0$, and define for all $n\ge k$ and
measurable sets $A\subset \rk$ the
cylindrical set $A(n)=A\times (R^2)^{2^n-2^k}\subset \rn$.
Put
$$
\tilde \mu_n(A)=L_n\int_{\tilde A(n)}\exp\left\{\gb 2^{-n/2}
\sum_{j=1}^{2^n}(x^{(1)}_j-\bar M)+\ab
2^{-3n/2}\biggl(\sum_{j=1}^{2^n}
x_j^{(2)}\biggr)^2\right\}\prod_{j=1}^{2^n}p(x_j)\,dx_j
$$
with $\tilde
A(n)=A(n)\cap\{(x_1,\dots,x_{2^n}),\;2^{-n}\sum_{j=1}^{2^n}x_j\in
\Om1\}$.
We prove similarly to [2] that if $n>n_0$ 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
$\bigl||x|-M\bigr|<c^{-0.4n}$ and $|x^{(2)}|<c^{-0.45n}$
must be replaced by
$\bigl||x|-\bar M\bigr|<\frac{\sqrt T}{8\mb}2^{-n/2}\sqrt {n\log n}$
and $|x^{(2)}|<2^{-n/4}n^{1/5}$ in the definition of the set $\tilde
\Omega^1_n$. We get a weaker bound in (7.11) than the corresponding
estimate in [2], but it is sufficient for our purposes. Theorem 2 is
proved.
 
\bigpagebreak
 
\flushpar
{\bf References:}
 
\smallpagebreak
\advance \baselineskip by -2pt
\kis
\item   {[1]}Bleher, P.M., Major, P.: Renormalization of
Dyson's hierarchical vector-valued ${\mats\Phi^4}$-model at low
temperatures. \its Commun. Math. Phys. \bfs 95\kis, 487--532
(1984)
\item   {[2]}Bleher, P.M., Major, P.: The large-scale
limit of Dyson's hierarchical vector-valued model at low
temperatures. The non-Gaussian case. \its Annales de
l'Institut Henri Poincar\'e, Physique Th\'eorique\/ \bfs 49\kis,
Vol\. ~1 (1988)
 
\bye
 
