= {1\over Z} \TR (q^2 e^{-\b H}). \ee Let a countable variety (a chain) of copies of the oscillator described above be considered. The oscillators in the chain are assumed to interact in such a way that the potential energy of the interaction between any two of them depends only on their displacements and their position in the chain. This dependence can be characterized as hierarchical. To describe it, we introduce \be \label{10.1} \Lan = \{ s \in \N \vert \ 2^n (j-1) +1 \leq s \leq 2^n j \}, \ j \in \N , \ n \in \Z_{+} . \ee Then \be \label{10.2} \Lan = \bigcup_{s\in \Lambda_{l,j}} \Lal, \ l = 1,2, \dots , n. \ee The interaction between the oscillators located at $j\in \N$ and $j' \in \N$ depends on the number of the hierarchy level on which both these points are located in the same subset $\Lambda_{n,s}$, that is on $n(j,j') = \min \{n \ \vert \ j\in \Lambda_{n,s} \ {\rm and} \ j' \in \Lambda_{n,s} \}$. It appears as a part of the local Hamiltonian $H_{\Lan }$ which describes the dynamics of the particles located in $\Lan$. The family of all such operators $\{ H_{\Lan }, n\in \Z_{+}, j\in \N \}$ is defined recursively by means of (\ref{10.2}): \be \label{10.4} H_{\Lan} = - \half (2^{\delta} -1) 2^{-n(1+\delta)} (\sum_{s\in \Lan} q_s )^2 + \sum_{s\in \Lambda_{1,j}} H_{\La}, \ee where $\delta$ was introduced in (\ref{3.7}), and the starting element $H_{\Lambda_{0,j}}$ is given by (\ref{2.6}). Each $H_{\Lan}$ acts in the space $\HoL$ which is the tensor product of copies of ${\Hos}$ indexed by $s\in \Lan $. The Gibbs state of the particles located in $\Lan$ may be described by means of temperature Green functions which are defined as follows. For $0\leq \tau_1 \leq \tau_2 \leq \dots \leq \tau_l $ , $l \in \N$, we set \beq \label{10.5} & & \Gamma^{\b , \Lan}_{A_1 , \dots , A_l } (\tau_1 , \dots , \tau_l ) = \frac{1}{Z_{\Lan}} \\ & & \TR \left\{A_1 \exp(-(\tau_2 - \tau_1 )H_{\Lan}) \dots A_l \exp( -(\b -\tau_l + \tau_1) H_{\Lan}) \right\},\nonumber \eeq where $$ Z_{\Lan} = \TR \left\{\exp( -\b H_{\Lan}) \right\}, $$ all $A_i , \ i=1, \dots , l$ are densely defined in $\HoL$, and such that the trace in (\ref{10.5}) exists. Put \be \label{10.6} A_i = A_{\Lan} = 2^{- \half n(1+\delta)}\sum_{s\in \Lan }q_s , \ \forall i. \ee Note that $\Lan$ consists of $2^n $ points, which means that the normalization of such sums is abnormal (comparing with that used in central limit theorems for weakly dependent random elements) due to the presence of the factor $2^{- n \delta /2}$ with $\delta >0$. Therefore, if the dependence between the displacements $q_s$ is weak, then one may expect that the sequence $\{ A_{\Lan}, n\in \N \}$ is degenerated at zero. The latter means that the sequences of Green functions defined by (\ref{10.5}) with such $A_i$ converge to zero for almost all values of their variables and for all $l\in \N$. To simplify the notations, we denote \be \label{10.7} \Gamma^{(n)}_{2l} (\tau_1 , \dots , \tau_{2l}) = \Gamma^{\b , \Lan }_{A_1 , \dots , A_{2l}} (\tau_1 , \dots , \tau_{2l}), \ee where $A_i$ are given by (\ref{10.6}). The following assertion establishes the relationship between the operator and functional integral approaches in quantum statistical mechanics. It directly follows from \cit{AHK} and was used e.g. in \cit{AKK}. \begin{Pn} \label{0.0pn} Let the sequence of functions $\{F_n , n\in \Z_{+} \}$ be set by Definition \ref{3.1df}. Then for all $\vp \in L^2 (I_{\b})$ \be \label{0.0} F_n (\vp) = \sum_{l=1}^{\infty} \frac{1}{2l!} \int_{I^{2l}_{\b}} \Gamma_{2l}^{(n)} (\tau_1 , \dots , \tau_{2l} ) \vp (\t_1 ) \dots \vp (\t_{2l}) d\tau_1 \dots d\tau_{2l}. \ee In other words, the functions introduced in (\ref{49}) and those defined by (\ref{10.5}) -- (\ref{10.7}) are the same. \end{Pn} Now let us describe the possibilities for the models we consider to belong to the families introduced above. This will be done by a number of lemmas which are mainly proved in our paper \cit{AKK}. The proof is based on the lattice approximation method (we do not give here the technical details which can be found e.g. in \cit{Si}). \begin{Lm} \label{5.1lm} Let $\hU_n (0) $ be defined by (\ref{5.9}). Then $\hU_0 (0)$ as a function of $\b$ possesses the properties:\\ {\rm (i)} it is continuously differentiable on $(0, +\infty)$; \\ {\rm (ii)} $\lim_{\b \ra 0} \hU_0 (0) = 0$; \\ {\rm (iii)} for every positive $\b$, \be \label{5.22} \hU_0 (0) \geq \phi (\b) := \bf \left( \frac{\b}{4m}\right), \ee where the function $f$ is defined \cit{DLS} implicitly by the relation $$ f(y\tanh y) = {1\ooover y} \tanh y, \ \ y\in (0,+\infty ). $$ \end{Lm} \begin{Lm} \label{5.2lm} Let $a <0$, and $\phi$ be defined above, then $$ \lim_{\b \ra \infty} \phi (\b) \geq U_{\rm min} := \frac{4ma^2 }{9b^2 }. $$ \end{Lm} \begin{Co} \label{5.1co} Let the parameters $m$, $a$, $b$ meet the estimate \be \label{5.23} U_{\rm min} > \bu ; \ \ a<0, \ee then the condition {\rm (iii)} of Definition \ref{5.1df} is satisfied. Moreover, $\b_{0}^{-} < \b_{0}^{+} $. \end{Co} \begin{Lm} \label{5.3lm} Let \be \label{5.24} b < \frac{\sqrt2}{3} \bu (\bu -1), \ee then the condition (\ref{5.20}) is satisfied. \end{Lm} For $b\geq 0$, the operator $H$ (\ref{2.6}) is such that $H^{-1}$ is compact. Then there exists a complete set of its eigenfunctions $\{ \psi_{s} \}$, hence we may set $H\psi_{s} = \epsilon_{s} \psi_{s} $, $q_{s{s'}} = (\psi_{s},q \psi_{s'})_{\Ho} $, and \be \label{5.25} \Delta_ H = \min \{ \epsilon_s - \epsilon_{s-1} , \ s\in \N \}. \ee \begin{Lm} \label{5.4lm} Let \be \label{5.26} {1\over m\Delta_H^2} <1, \ee then the condition {\rm (iii)} of Definition \ref{5.2df} is satisfied. \end{Lm} Set \be \label{5.200}_{\rm class} = \frac{\int_{\R}q^2 \exp (-V(q)) dq}{\int_{\R} \exp (-V(q) dq}, \ee where $V$ is given by (\ref{2.6}) with $b>0$ and $a\in \R$. \begin{Lm} \label{5.20lm} Let \be \label{5.201} \b > \frac{\bu}{_{\rm class}}, \ee then there exists $m(\b)$ such that for $m\geq m(\b)$, the condition $\hU_0 (0) = \bu $ is satisfied. \end{Lm} \begin{Lm} \label{5.21lm} For given $b>0$ and $a\in \R$, there exists $m_{-}(a,b)$ such that for $m\ < m_{-}(a,b)$, the condition {\rm (iii)} of Definition \ref{5.2df} is satisfied. \end{Lm} Let us discuss the consequences of these lemmas having in mind the description of the families ${\MC}_{+}$, ${\MC}_{-}$. Thus in order to belong to ${\MC}_{+}$, the model must be "weakly" anharmonic, that is $b$ must be small enough (see (\ref{5.24})). For $\d$ close to $\half$, $\bu$ is close to one, then the anharmonicity parameter $b$ ought to be close to zero. Consider the case of negative $a$, and $b$ obeying the restriction (\ref{5.24}). Then the function $V$ has two minima (wells) located at the points $ \pm \hat{q}$, where $\hat{q} = \sqrt{-a/b}$. Their depth is $\hat{v} = - V(\hat{q}) = a^2 / 4b$. Combining (\ref{5.23}) and (\ref{5.24}), we obtain the following sufficient condition to belong to the family ${\MC}_{+}$: \be \label{5.202} m\hat{v} > \frac{3\sqrt{2}}{16} \bu^2 (\bu -1). \ee Therefore, even for small values of the oscillator's mass, one can place the corresponding model into ${\MC}_{+}$ by setting the wells of $V$ to be deep enough. For the case of arbitrary negative $a$, we can place the model into the family ${\MC}_{+}$ by letting the oscillator's mass to be large enough as it is prescribed by Lemma \ref{5.20lm}. On the other hand, such a model can be placed into the family ${\MC}_{-}$ by letting this mass to be small enough (Lemma \ref{5.21lm}). \section{\bf Proof of Lemmas}%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% First we describe some properties of the sequences of functions considered. \begin{Lm} \label{6.2lm} For arbitrary $n\in \N$, \be \label{6.8} \hU_{n}(0) \leq \k (\hU_{n-1}(0) ) \hU_{n-1}(0); \ee \beq \label{6.9} 0 & \leq & \hU_{n}(k) \leq \k (\hU_{n-1}(k)) \hU_{n-1}(k) \nonumber\\ & + & (1-\2m )(\k (\hU_{n-1}(k))^3 2^{2\d -1}\bb{X_{n-1}}, \ \ k \neq 0; \eeq \be \label{6.10} \hU_n (0) \geq \k(\hU_{n-1}(0)) \hU_{n-1}(0) + \half (1 - \2m ) (\k(\hU_{n-1}(0)))^3 2^{2\d -1} X_{n-1}; \ee \be \label{6.11} 2^{2\d -1} (\k (\hU_{n-1}(0)))^4 X_{n-1} \leq X_n < 0; \ee \be \label{6.120} 0\leq \hU_n (k) \leq \hU_n (0), \ee where $\k(u)$ and $X_n$ are defined by (\ref{5.12}) and (\ref{5.10}) respectively. \end{Lm} The proof of this Lemma was done in our paper \cit{AKK} by means of methods derived in \cit{Koz} within the lattice approximation approach (see e.g. \cit{Si}). \begin{Pn} \label{6.1pn} For every $n \in \Z_{+}$ and $k\in {\KC}$, $\hU_n (k)$ is a continuously differentiable function of $\b$. \end{Pn} The proof of this Proposition can be done recursively on the base of Lemma \ref{5.1lm}. The following inductive lemma is the main tool used below. \begin{Lm} \label{6.3lm} Let the model belong to the family ${\MC}_{+}$, and ${\IC}_n $, $n \in {\Z_{+}}$, be the triple of statements $(i_n^1 , i_n^2 , i_n^3 )$ \beq i_n^1 & = & \{ \exists \b_n^{+} \in \oo \ : \ \hU_n (0) =\bu \ {\rm for} \ \b = \b_n^{+} ; \ \hU_n (0) < \bu \ \forall \b < \b_n^{+} \}; \nonumber\\ i_n^2 & = & \{ \exists \b_n^{-} \in \oo \ : \ \hU_n (0) = 1 \ {\rm for} \ \b = \b_n^{-}; \ \hU_n (0) < 1\ \ \forall \b < \b_n^{-} \}; \nonumber\\ i_n^3 & = & \{ \forall \b \in (0, \b_n^{+} )\ : \ \bb{X_n} < \bw \}. \nonumber \eeq Then\\ (i) ${\IC}_0 $ is true;\\ (ii) ${\IC}_{n-1}$ implies ${\IC}_n$.\\ \end{Lm} {\bf Proof}. ${\IC}_0$ is true in view of Definition \ref{5.1df} and Lemma \ref{5.1lm} statements (i) and (ii). Let us prove the implication. We set $\k (\hU_n (0) ) = \k_n$, then for $\b = \b_{n}^{+}$, $\k_n = \bk $ and $\k_n < \bk $ for $ \b < \b_n^{+}$. Remind that $\bk = \k (\bu )$ is defined by (\ref{5.12}) and obeys the estimate (\ref{5.13}). Let $ \b = \b_{n-1}^{+}$, then (\ref{5.14}), (\ref{6.10}), and the statement $i_{n-1}^3$ yield \beq \label{6.15} \hU_n (0) & \geq & \bk \bu + \half (1-\2m) \bk^3 2^{2\d -1} X_{n-1}\nonumber\\ & > & \bk \bu [ 1 - \sqrt{2} (1-\2m) \bk^2 2^{2\d -1} \bw ] > \bu . \eeq For $\b = \b_{n-1}^{-}$, the estimate (\ref{6.8}) gives \be \label{6.16} \hU_{n-1}(0) \leq 1. \ee Taking into account the continuity of $\hU_n (0) $ as a function of $\b$ (arising from Proposition \ref{6.1pn}), the estimates (\ref{6.15}) and (\ref{6.16}), one obtains that there exists at least one value $\tilde{\b}_n^{+} \in (\b_{n-1}^{-}, \b_{n-1}^{+})$ such that $\hU_n (0) = \bu $. The smallest such $\tilde{\b}_n^{+}$ is denoted by $\b_n^{+}$. That is, $\b_n^{+} = \inf \tilde{\b_n^{+}}$. The mentioned continuity of $\hU_n (0)$ yields $\hU_n (0) < \bu $ for $ \b < \b_n^{+}$. This implies that the statement $i_n^1 $ is true. Similarly, we can prove the existence of $ \b_n^{-} \in [\b_{n-1}^{-}, \b_{n-1}^{+})$ by putting $\b_n^{-} = \inf \{ \b \in [\b_{n-1}^{-}< \b_{n-1}^{+}) \ \vert \ \hU_n (0) = 1 \}$. Thus the statement $i_n^2 $ is also true. For $\b < \b_n^{+}$, we have $\k_n < \bk $ which yields \be \label{6.17} \bb{X_n } < 2^{2\d -1} \bk^4 \bb{X_{n-1}} \leq \bb{X_{n-1}} < \bw, \ee that is $i_n^3 $ is true. The proof is concluded with the remark that \be \label{6.18} [\b_n^{-}, \b_n^{+}] \subset [\b_{n-1}^{-}, \b_{n-1}^{+}] \subset \oo. \ee \kasten \begin{Lm} \label{6.4lm} Let $\Delta_H$ be defined by (\ref{5.25}), then for all $\b$ and $k\in {\KC}$: \be \label{6.40} \hU_0 (k) \leq {1\over m(\Delta_H^2 + k^2 )}. \ee \end{Lm} {\bf Proof}. For the operator $H$ (\ref{2.6}), we have Proposition \ref{0.0pn}. Then for $U_2 ^{(0)} (0, \t )$, one obtains from (\ref{0.0}) $$ U_2 ^{(0)} (0, \t ) = {1\over Z} \sum_{s,s'} (q_{s{s}'})^2 \exp ( -\t (\epsilon_s - \epsilon_{s'}) -\b \epsilon_{s'} ). $$ Inserting this expression into the definition of $\hU_0 (k) $ (\ref{5.9}), one gets \be \label{6.6} \hU_0 (k) = {1\over Z} \sum_{s,s'} (q_{s{s}'})^2 \frac{ (\epsilon_s - \epsilon_{s'}) (e^{-\b \epsilon_{s'}} - e^{-\b \epsilon_s })}{ (\epsilon_s - \epsilon_{s'})^2 + k^2 }. \ee Taking into account that $q_{ss}=0$ (due to $Z_2 $ - symmetry of $H$), one can omit the case $s=s'$ when summing in (\ref{6.6}). Then estimating $(\epsilon_s - \epsilon_{s'})^2 $ with $s\neq s'$ by $\Delta_H^{2}$ (\ref{5.25}), one arrives at \beq \hU_0 (k) & \leq & \frac{1}{\Delta_H^{2} + k^2 } {1\over Z} \sum_{s,s'} (q_{s{s}'})^2 (\epsilon_s - \epsilon_{s'}) (e^{-\b \epsilon_{s'}} - e^{-\b \epsilon_s }) \nonumber\\ & = & {1\over \Delta_H^{2} + k^2 } {1\over Z} \TR ( [q,[H,q]] e^{-\b H})\nonumber\\ & = & {1\over m( \Delta_H^{2} + k^2 )}. \eeq Here $[.,.]$ stands for the commutator. \kasten {\bf Proof of Lemma \ref{5.4lm}}. It follows directly from the estimate (\ref{6.40}) with $k=0$. {\bf Proof of Lemma \ref{5.5lm}}. Let some $u\in (1, \bu )$ be chosen and let $D_n $ denote the closure of the set $\{ \b \in [\b_n^{-}, \b_n^{+}] \ \vert \ 1 \leq \hU_n (0) \leq u \}$. It is nonempty: $ \ \b_n^{-} \in D_n $. For chosen $u$, we define $\ve >0$ by putting $\k (u) = 2^{\ve} $. Then this $\ve$ obeys the restriction \be \label{6.41} \ve < \bar{\ve } := \min \{ {1\over 6} \ , \ \frac{1-2\d }{4} \}, \ee which follows from (\ref{5.11}), (\ref{5.13}), and from the monotonicity of $\k (u) $ given by (\ref{5.12}). Then the set \be \label{6.42} D_{*} = \bigcap_{n\in \Z_{+}} D_n , \ee is nonempty and closed. Thus for $\b \in D_{*}$, we have for all $n\in \Z_{+}$; \beq \label{6.43} 1 \leq \hU_n (0) & \leq & u \ ; \\ \k (\hU_n (0)) & \leq & 2^{\ve} . \eeq Applying the estimate (\ref{6.11}), one gets \be \label{6.430} \bb{X_n } \leq 2^{(2\d -1 +4\ve )n} \bw, \ee with $ 2\d -1 +4\ve $ being negative (see (\ref{6.41})). This yields \be \label{6.44} X_n \ra 0, \ \ \forall \b \in D_{*}. \ee Now we prove that $ \hU_n (0) \ra 1 $ for $n \ra \infty $ and $\b \in D_{*}$. Rewrite the estimate (\ref{6.10}) as follows: \be \label{6.21} \hU_n (0) - 1 \geq \chi_{n-1} (\hU_{n-1}(0) -1) + (\p -1) \k_{n-1} T_{n-1}, \ee where $$ \chi_{n-1} = 1+ \frac{1-\2m}{1 - (1-\2m)\hU_{n-1}(0)}; $$ and \be \label{6.22} T_{n-1} = (\hU_{n-1}(0) -1)^2 + \half 2^{\d -1} \k_{n-1}^2 X_{n-1}. \ee Then $\chi_{n-1} > 2^{\d} $ for $\b \in D_{*}$, which is caused by the fact $\hU_n (0) \geq 1$. For such $\b $, let us prove that $T_n \leq 0$, $ \forall n\in \Z_{+}$. Assume that for some $p\in \N$, we had $T_{p-1} >0 $, then (\ref{6.21}) would yield $$ \hU_p (0) -1 > \chi_{p-1} (\hU_{p-1} (0) -1 ) \geq 2^{\d} ( \hU_{p-1}(0) -1). $$ In this case, one obtains from (\ref{6.22}) \beq T_p & = & (\hU_p (0) -1 )^2 + \half 2^{\d - 1} \k_p^2 X_p \nonumber\\ & > & 2^{2\d} (\hU_{p-1}(0) -1)^2 + \half 2^{\d -1} 2^{2\ve} 2^{2\d -1 +4\ve} X_{p-1}\nonumber\\ & = & 2^{2\d} \left[ (\hU_{p-1}(0)-1)^2 +\half 2^{\d -1} 2^{6\ve -1}X_{p-1} \right] \nonumber\\ & > & 2^{2\d } \left[(\hU_{p-1}(0)-1)^2 +\half 2^{\d -1}X_{p-1} \right] \nonumber\\ & \geq & 2^{2\d} \left[ (\hU_{p-1}(0) -1)^2 + \half 2^{\d -1} \k_{p-1}^2 X_{p-1} \right] \nonumber\\ & = & 2^{2\d} T_{p-1}. \nonumber \eeq To derive these estimates, we have used: (\ref{6.43}), (\ref{6.11}), (\ref{6.41}). Therefore, assuming the positivity of $T_{p-1}$, we have obtained the same property of $T_n $ for all $n\geq p $, as well as $ T_n \geq 2^{2n\d} T_{p-1}$ that being used in (\ref{6.21}) would yield $\hU_{n+1}(0) -1 > 2^{2n\d} T_{p-1} (\p -1)\k_p $. The latter runs counter to the boundedness of the sequence $\{\hU_n (0) \}$ (\ref{6.43}). Thus $T_n \leq 0$ for all $n\in \N$. Taking this into account as well as the estimate (\ref{6.430}), we obtain from (\ref{6.22}): \be \label{6.25} \hU_n (0) -1 = O( 2^{n(2\d -1 +4\ve)/2}). \ee Set $\b_{*} = \inf D_{*}$, then for $ \b = \b_{*}$ : $ \hU_n (0) \ra 1 \ , \ \ X_n \ra 0 $, which proves the first part of the lemma. For $\b < \b_{*}$, there exists $p\in \N$ such that (\ref{6.43}) does not occur. There are two possibilities of its failure: (i) $\hU_p (0) <1$; (ii) $ \hU_p (0) > u $. For chosen $\b$, (\ref{6.18}) implies $\b < \b_n^{+} \ \ \forall n\in \N$. Thus the second possibility is contradictory to the statement $i_p^1$ of Lemma \ref{6.3lm}. Hence it remains only $ \hU_p (0) <1$ which gives $\k_p <1$, and then the estimate (\ref{6.8}) yields $$ \hU_{p+1}(0) < \hU_p (0) < 1. $$ Repeating these arguments we conclude that the bounded sequence $\{ \hU_n (0) , \ n \geq p \}$ is monotone and hence convergent to some limit $\hU <1$. Note that $\k_n <1$, thus (\ref{6.8}) may be used to get $$ \hU_{n+1}(0) \leq \k (\hU_n (0) ) \hU_n (0) < \hU_n (0). $$ The left and the right hand parts of this double inequality converge to the same limit $\hU $, then the middle part ought to converge to this limit too. Therefore, $$ \hU = \k (\hU ) \hU, $$ that has only one solution: $ \hU =0 $, which is the limit of the sequence $\{ \hU_n (0) \}$ for $\b < \b_{*}$. As for the sequence $\{ X_n \}$, its convergence to zero for $\b < \b_{*}$ has been established above by the estimate (\ref{6.430}). For $Y_n$ given by (\ref{5.31}), Lemma \ref{5.7lm} implies $ \bb{Y_n } \leq \bb{X_n }$ which yields $ Y_n \ra 0$ for $\b \leq \b_{*}$. \kasten {\bf Proof of Lemma \ref{5.6lm}}. For every model from the family ${\MC}_{-}$, we have $\hU_n (0) <1$ for all positive $\b$. Therefore, the sequence $\{ \hU_n (0), n\in \Z_{+} \}$ is monotone and hence convergent to zero similarly to the case of $\b < \b_{*}$ just considered. The latter yields the same convergence of the sequence $\{ Y_n \}$, as it follows from Lemma \ref{5.10lm}. \kasten Few words ought to be said about the proof of Lemma \ref{5.7lm}. All inequalities (\ref{5.32}) -- (\ref{20.33}) are proven within the lattice approximation method mentioned earlier by proving similar estimates for classical systems of unbounded ($\vp^4 $ -- like) spins. The estimate corresponding to (\ref{5.32}) is proven in \cit{AKK} . The sign rule (\ref{5.33}) is proven by means of the Ising approximation of $\vp^4 $-spins (see e.g. \cit{Si}) on the base of Shlosman's results for the Ising model \cit{Shl}. The estimate (\ref{20.33}) corresponds to the Griffiths' inequalities valid for $\vp^4$-spin systems. {\bf Proof of Lemma \ref{5.8lm}}. For $\b < \b_{*}$ , we have proven $\hU_n (0) \ra 0 $ when $n \ra \infty $. Thus the estimate (\ref{6.120}) implies the asserted convergence to hold for such $\b$. The proof for the case $\b = \bs$ will be done as follows. First we prove that for $\b = \bs$, there exists a finite subset of $\KC$ , $\KC^{{\rm one}}$ such that for $k\in \KC^{{\rm one}}$, $\hU_n (k) \ra 1 $ when $n \ra \infty$, and $\hU_n (k) \ra 0$ for $k\in \KC^{{\rm zero}} := \KC \setminus \KC^{{\rm one}}$. Then we show that $\KC^{{\rm one}}$ consists of only one point: $ \KC^{{\rm one}} = \{ 0 \}$. Let $\a = -(2\d -1 +4\ve) >0$, and \be \label{6.45} v_n = 1 - 2^{-n\a}2^\d \bw , \ \ n \in \Z_{+}. \ee Then all $v_n$ are positive (see (\ref{5.14}) and $v_n \ra 1 $ when $n \ra \infty$. We set \beq \label{6.46} {\KC}_n & = & \{ k\in {\KC} \vert \ \hU_n (k) \leq v_n \}; \nonumber\\ \bar{\KC}_n = \KC \setminus {\KC}_n & = & \{ k\in \KC \vert \ \hU_n (k) > v_n \}. \eeq Let us prove that ${\KC}_{n-1} \subset {\KC}_n $ for all $n \in \N$. Applying the estimate (\ref{6.9}), we obtain \beq \label{6.47} & & \hU_n (k) - \hU_{n-1}(k) \leq - \frac{ 1- \2m }{ 1 - ( 1- \2m )\hU_{n-1}(k)} \nonumber\\ & & \left\{ \hU_{n-1}(k) (1 - \hU_{n-1}(k)) - \k (\hU_{n-1}(k))^2 2^{\d-1} 2^{-\a (n-1)} \bw \right\}. \eeq Let $k \in {\KC}_{n-1}$, then $ \k (\hU_{n-1}(k)) <1$ and \beq \label{6.61} & & \hU_{n-1}(k) (1- \hU_{n-1}(k)) - \k (\hU_{n-1}(k))^2 2^{\d-1}2^{- \a (n-1)} \bw \nonumber\\ & > & \hU_{n-1}(k) (1-\hU_{n-1}(k)) - 2^{- \a (n-1)} 2^{\d -1} \bw \nonumber\\ & := & p_{n-1}( \hU_{n-1}(k)). \eeq The polynomial $p_{n-1}(v)$ has two roots \be \label{6.60} v_{\pm }= \half \left\{ 1 \pm \sqrt{ 1- 2^{-\a (n-1)} 2^{\d+1} \bw}\right\}. \ee Due to (\ref{5.14}), both roots are real, and $ v_{+} > v_{n-1}$ (see (\ref{6.45}). Let $\hU_{n-1}(k) \leq v_{-}$, then for all $n\in \N$ , $\hU_{n-1}(k) < 2^\d \bw < 1$, which yields $\k (\hU_{n-1}(k)) <1$. Applying these estimates in (\ref{6.9}), we get $\forall n\in \N$ \beq \hU_n (k) & < & 2^\d \bw + 2^{2\d -1}(1-\2m ) \bw \nonumber\\ & = & 2^{\d -1}(1+\p )\bw \leq 1 - 2^\d \bw < v_n , \nonumber \eeq where we have used $ \bw \leq 2^{1-\d } (3 + \p )^{-1}$, which follows from (\ref{5.14}). Thus in this case $k \in {\KC}_n $. Now let us consider the case $ v_{-} < \hU_{n-1}(k) < v_{+}$. Then $p_{n-1}(\hU_{n-1}(k))$ in (\ref{6.61}) is positive, yielding (by (\ref{6.47})) $ \hU_n (k) < \hU_{n-1}(k) \leq v_{n-1} < v_n $. Hence $k\in {\KC}_n $ also in this case. Having in mind that $v_{+} > v_{n-1}$, we conclude that all possible values of $\hU_{n-1}(k)$, for $k\in {\KC}_{n-1}$, have been taken into account. Then $k \in{\KC}_{n-1}$ implies $k\in {\KC}_n $, i.e. ${\KC}_{n-1} \subset {\KC}_n $, for all $n\in \N$.\\ Set \be \label{6.48} {\KC}^{{\rm zero}} = \bigcup_{n\in \Z_{+}} {\KC}_n \ ; \hskip1cm {\KC}^{{\rm one}} = {\KC} \setminus {\KC}^{(0)} = \bigcap_{n \in\Z{+}} \bar{\KC}_n. \ee Then: \beq & & {\rm (i)}\ \ {\KC}^{{\rm one}} \ \ \ {\rm is} \ \ {\rm finite}; \nonumber\\ & & {\rm (ii)} \ \ \hU_n (k) \ra 1 \ , \hskip1cm \forall k\in {\KC}^{{\rm one}}; \nonumber\\ & & {\rm (iii)} \ \ \hU_n (k) \ra 0 \ , \hskip1cm \forall k\in {\KC}^{{\rm zero}}. \nonumber \eeq To prove (i), we note that $\bw < 2^{-1-\d }$, hence $v_0 \geq 1/2$. Then taking into account (\ref{6.40}), we get \beq \label{6.49} {\KC}_0 & \supset & \{ k \in {\KC} \vert \ k^2 \geq {2\over m} - \Delta_H^2 \}, \\ \bar{\KC}_0 & \subset & \{ k\in {\KC} \vert \ k^2 < {2\over m }- \Delta_H^2 \}. \nonumber \eeq The latter yields that $\bar{\KC}_0$ is finite, but ${\KC}_0 \subset {\KC}_1 \subset \dots \subset {\KC}_n \subset \dots \subset {\KC}^{{\rm zero}}$ implies $ {\KC}^{{\rm one}} \subset \bar{\KC}_0 $. This means ${\KC}^{{\rm one}}$ is finite as well. To prove (ii), we remark that accordingly to the definition of $\bar{\KC}_n $ and due to (\ref{6.120}), we have for $k \in \bar{\KC}_n$: $$ v_n < \hU_n (k) \leq \hU_n (0). $$ The latter implies that the value $k=0$ belongs to ${\KC}^{{\rm one}}$. For $\b = \bs$, we had $\hU_n (0) \ra 1$. But the sequence $\{v_n \}$ also converges to unity. This proves (ii). To prove (iii), we let $k\in {\KC}^{{\rm zero}}$. Therefore, there exists $n_0 \in \N$ such that $k\in {\KC}_{n-1}$ for all $ n>n_0 $. Then either $\hU_{n-1}(k) \leq v_{-} < 2^{-\a (n-1)} \p \bw$, or $ v_{-} < \hU_{n-1}(k) < v_{+}$. The latter yields $\hU_n (k) < \hU_{n-1}(k) $. The validity of the former assumption leads directly to the convergence we are proving. The validity of the latter one means that the sequence $\{ \hU_n (k) , \ n>n_0 \}$ is monotone and bounded from below by zero. This yields in turn that such a sequence converges to zero similarly to the case of $\b<\b_{*}$ where the convergence of the sequence $\{ \hU_n (0) \}$ to zero has been proved. The proof of the lemma is concluded by showing that ${\KC}^{(1)} = \{ 0 \}$. Up to now, we have proved that the sequences $\{ \hU_n (k) \vert k\in {\KC} \}$ converge to the characteristic function of the set ${\KC}^{{\rm one}}$. Thus the sequence of functions $\{ U_2^{(n)} (\t , \t' ) \vert \t , \t' \in I_{\b_{*}} \}$ converges in $L^2 (I^2_{\bs})$ to \be \label{6.50} U_2 (\t , \t' ) = {1\over \bs } \left\{ 1 + \sum_{k\in {\KC}^{{\rm one}} \setminus \{ 0 \}} \cos k(\t - \t') \right\}, \ee with finite ${\KC}^{{\rm one}}$. Therefore, there exists a subsequence $\{U_2^{(n_{j})} (\tau , \t' ) \}$ converging to this $U_2 (\t , \t' )$ almost everywhere on $ I^2_{\bs}$. Thus the sign rule (\ref{5.33}) yields in this case that $U_2 (\t , \t' )$ ought to be nonnegative also almost everywhere on $I^2_{\bs}$. It is not difficult to show that the only case where this nonnegativity is preserved is just ${\KC}^{{\rm one}} = \{ 0 \}$ (see (\ref{6.50}). \kasten {\bf Proof of Lemma \ref{5.9lm} }. Within the lattice approximation approach on the base of the Lieb - Sokal theorem \cit{LS}, we can show that for all $n\in \Z_{+}$, the function $f_n (z) := F_n (z\xi_1 ) $ is different from zero on the following set $ \{ z\in \C \ \vert {\rm Re}z \neq 0 \}$. This function is entire, even, and of the order of growth which is less then two. Then it should possess purely imaginary zeros, hence its Hadamard's representation ought to be exactly the same as (\ref{5.36}) \kasten \section{\bf Acknowledgments} The authors are extremely grateful to Professor Leonard Gross for his many stimulating remarks and constructive criticism on a previous version of this paper. 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