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 $0 0$ 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 $0 0$ 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 $|M-M_n| 0$ 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 $0 0$ the inequality $$ f_n(x)<\frac{10}{\sqrt{ \bbh n}}\exp\left\{-\frac1{2\bbh n}x^2\right\} \qquad\text{for }|x| 0$ and a large $L=L(\e)>0$. We are going to show that if $0 L\sqrt{\bbh n},\; n=0,1,2,\dots\,. \tag5.1$'$ $$ 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 $$ \align \fx&\le\frac{10}{\sqrt\bb}\expo \quad\text{for }|x| L{\sqrt \bb} \tag5.2$'$ \\ \fx&\le \frac C{\bs}\quad \text{for all }x\in R^1 \tag 5.2$''$ \endalign $$ 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 $$ \align \fx&\le\frac\e{\bs}\expt \qquad\text {for }-r 0$, the main contribution to the integral $\qbb\fx$ is given in a small neighbourhood of the point $(u,v)=(0,0)$. For $x<0$ this statement remains valid only for $x>-\bar r$. For $x<-\bar r$ the main contribution to this integral is given in a small neighbourhood of the points $(u,v)=(0,\pm v^{*})$ with $v^{*2}= 2^{n/2}\left\{(M-2^{-(n+1)/2}\bar r)^2-(M+2^{-(n+1)}x)^2\right\}$. \demo{Proof of Lemma 3} Define the function $$ 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 -\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 $$ \align \px&\le\frac{\e^{7/4}\sqrt\pi}{\bb}\expt \qquad \text{for } -\rb 0,x+\sqrt2u<-\rb\}$. This integral can be estimated in the following way: $$ \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 $$ We give an upper bound on the right-hand side of (5.11) by replacing $\abb$ with $\bar\a$ in it and multiplying the expression by $\exp\{-(\abb-\bar\a)|\rb+x|\}$. The integral in this expression can be estimated by the rather rough bound $|\rb+x|$. These estimates show that the right-hand side of (5.11) is much less than the expression at the right-hand side of (5.9$'$). To estimate the integral $\int P(x,u)\,du$ in the case $-2^{(n+1)/2}M -\rb\}$ observe that some calculation yields that $$ \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$'$ $$ On the other hand, we get by rewriting Proposition A for $g_n(x)$ that there are some numbers $ B>0$, \ $D>0$ and $R_n$, \ $-C_1n C_2>0$ such that $$ 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 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} \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|