|q\beta-p|. \] \noindent Then for every $\beta$, rational or irrational, there is a sequence $\{q_n\}$ of positive integers with $q_0=1$ and a sequence $\{p_n\}$ of integers such that ${p_n\over q_n}$ are the best approximations to $\beta$, these sequences being finite for a rational $\beta$ and infinite if it is irrational. With some abuse of terminology we will most of the time apply the term "convergents" solely to the members of the sequence $\{q_n\}$, whereas the members of the sequence $\{{p_n\over q_n}\}$ will be referred to as "best approximations". In the standard notation \[ \|q\beta\|=\min_p|q\beta-p|. \] \noindent For any real $\beta$ the Continued fraction expansion above uniquely encodes the sequences $\{p_n\}$ and $\{q_n\}$ by the recursion rule: \begin{equation} \begin{array}{ll} p_{-2}=0,& p_{-1}=1, \\ \hfill \\ q_{-2}=1,& q_{-1}=0, \end{array} \label{inipq} \end{equation} \noindent and for $n\geq 0$ \begin{equation} \begin{array}{llll} p_n&=&a_n p_{n-1}+p_{n-2}, \\ \hfill \\ q_n&=&a_n q_{n-1}+q_{n-2}. \end{array} \label{recurpq} \end{equation} \noindent By the Dirichlet theorem (see Schmidt [1980]), every best approximation ${p_n\over q_n}$ to $\beta$ satisfies \begin{equation} \|q_n\beta\|=|q_n\beta-p_n|\leq{1\over q_{n+1}}. \label{dirr} \end{equation} \noindent Besides, there is another well-known fact about the consecutive best approximations, that we'll state as a Proposition: \begin{proposition} For $n=0,1..$ the quantities $q_n\beta-p_n$ and $q_{n+1}\beta-p_{n+1}$ have the opposite signs and \[ p_n q_{n+1}-q_n p_{n+1} =\pm1. \] \label{sign} \end{proposition} \noindent {\bf Proof:} See Schmidt [1980], pp. 10-11. $\Box$ \\ \medskip \noindent A number $\beta$ is called {\it badly approximable} if there exists a constant $00$ one has \begin{equation} \|q\beta\|>{c(\beta)\over q}. \label{bapr} \end{equation} \begin{proposition} $\beta$ is badly approximable if and only if all the terms in its Continued fraction expansion $\beta=[a_0,a_1,..]$ are uniformly bounded: \[ a_i\leq \bar{a},\ \ i=0,1,2,... \] \label{ba} \end{proposition} \noindent {\bf Proof:} See Schmidt [1980]. In fact, it follows from the two following important relations, which are the consequences of (\ref{fraction}), (\ref{recurpq}) and can be found in Schmidt [1980] respectively on pp. 10, 23, stating that for $n\geq2$ \begin{equation} {q_n\over q_{n-1}}=[a_n,a_{n-1},..,a_1], \label{rat} \end{equation} \noindent and for $n\geq0$ \begin{equation} \left|\beta-{p_n\over q_n}\right|={1\over q_n^2\left([a_{n+1},a_{n+2},..]+{1\over[a_n,a_{n-1},..,a_1]}\right)}, \label{approx} \end{equation} \noindent which gives the exact value for the difference between the irrational itself and its n-th best approximation. Moreover, (\ref{fraction}) tells us how to evaluate the right-hand part of (\ref{approx}), so that we can always take in the definition (\ref{bapr}) of badly approximable numbers \begin{equation} c(\beta)={1\over\max_{i>0}a_i+2}\ \Box \label{c} \end{equation} \noindent In particular, this Proposition together with (\ref{fraction}) imply that there is a continuum of badly approximable numbers (the argument is essentially the same as in the well-known proof of Cantor theorem on uncountability of real numbers). In addition given \begin{equation} \bar{a}=\max_{i>0}a_i\geq 1, \label{adef} \end{equation} \noindent all the 2-vectors $\vec{\omega}=\omega_+(1,\beta)$ will be diophantine with $\varpi=1$ and $\gamma={\omega_+\over 4+2\bar{a}}$ in the standard notation of (\ref{dio}), which implies that for a fixed $\bar{a}$ they form a zero measure set. Although in our analysis we will be mostly dealing with badly approximable numbers, the ultraviolet cutoff that we are going to apply to the series (\ref{sum}), as we have done previously, will eventually enable us to extend our result to the set of frequencies of asymptotically (as $\varepsilon\rightarrow 0$) full measure by letting $\bar{a}$ in (\ref{adef}) go to infinity as $\varepsilon\rightarrow 0$ and using instead of (\ref{adef}) the following less stringent definition: \begin{equation} \bar{a}(\varepsilon)=\max_{0 0,\ |q\beta-p|<{1\over2}} \exp \left( -(p+q)\sigma_0 - {\pi\omega_+|q\beta-p| \over2\sqrt{\varepsilon}}\right) \vec{A}_{\vec{k}}\sin( \dpr) \\ \hfill \\ &+& \sum_{p,q\in Z,\ q\geq0,\ |q\beta-p|\geq{1\over2}} \exp \left( -(p+q)\sigma_0 - { \pi\omega_+ |q\beta-p| \over2\sqrt{\varepsilon}} \right) \vec{A}_{\vec{k}}\sin(\dpr), \end{array} \label{trash1} \end{equation} \noindent where $\vec{k}=(-p,q)$, then for any positive $\delta_0$ and $\varepsilon$ small enough, the absolute value of the second sum will be bounded as: \begin{equation} |\Sigma_2|\leq \exp\left(-{\pi\omega_+\over4\sqrt{\varepsilon}}\right) \sum_{\vec{k}\in Z^2 }\left|\vec{A}_{\vec{k}}\right| e^{-|\vec{k}|\sigma_0} \leq \exp\left(-{\pi\omega_+\over 8 \sqrt{\varepsilon}}\right), \label{trash1bound} \end{equation} \noindent where we've used the assumption on the coefficients $\vec{A}_{\vec{k}}$ and the well known equality $\sum_{\vec{k}\in Z^l} e^{-|\vec{k}|\sigma_0}=\coth^{l} \left({\sigma_0\over2}\right)$. This estimate will turn out to be exponentially small compared to the asymptotics of the first sum $\Sigma_1$, to which we'll further restrict our attention. The first sum $\Sigma_1$ in (\ref{trash1}) has in fact only one index, since for each integer $q>0$ there is only one $p$ such that $|q\beta-p|<{1\over2}$. To this sum we can apply the ultraviolet cutoff, when $q$ becomes as large as $q_\varepsilon=[\varepsilon^{-{1\over 2}+\delta_0}]$, where $[\cdot]$ means an integer part, and $0<\delta_0\ll{1\over2}$ to be specified. We write \begin{equation} \begin{array}{lll} \Sigma_1 &=& \Sigma_1^{<}+\Sigma_1^{>} \\ \hfill \\ & \equiv & \left( \sum_{p,q\in Z,\ 1\leq q\leq q_\varepsilon, \ |q\beta-p|<{1\over2}} + \sum_{p,q\in Z,\ q> q_\varepsilon, \ |q\beta-p|<{1\over2}} \right) \\ \hfill \\ & \ & \exp\left(-(p+q)\sigma_0-\omega_+{\pi|q\beta-p| \over2\sqrt{\varepsilon}}\right) \vec{A}_{\vec{k}}\sin(\dpr), \end{array} \label{trash2} \end{equation} \noindent where again and henceforward $\vec{k}=(-p,q) $. It's easy to see (likewise in the preceding sections) that the ultraviolet part of the sum can be bounded as: \begin{equation} |\Sigma_1^{>}|\leq \exp\left(-\varepsilon^{-{1\over2}+\delta_0+\bar{\delta}}\right), \label{trash2bound} \end{equation} \noindent for any small $\bar{\delta}>0$ if $\varepsilon$ is small enough. Moreover, one can write \begin{equation} \begin{array}{lll} \Sigma_1^{<} & = & \bar{\Sigma}_1^{<}+\tilde{\Sigma}_1^{<} \\ \hfill \\ & \equiv & \sum_{n\geq0,\ q_n\leq q_\varepsilon} e^{-(p_n+q_n)\sigma_0- {\pi\omega_+\|q_n\beta\|\over2\sqrt{\varepsilon}}} A_{\vec{k}^n}\sin(\vec{k}^n\cdot\vec{\alpha}) \\ \hfill \\ & + & \sum_{n\geq0,\ q_{n-1}0$ there is one only $p$ such that $|\beta q-p|<{1\over2}$). The ultraviolet cutoff that we have previously made, enables us to define a "cutoff index" $n_\varepsilon$ in (\ref{adef1}) as: \begin{equation} n_\varepsilon\equiv\max\{ j>0:\ q_j\leq q_\varepsilon\}. \label{ieps} \end{equation} \noindent Suppose $q_{n-1},q_n$ stand for the two consecutive convergents to $\beta$. Then in $\tilde{\Sigma}^<_1$ we can estimate the absolute value of the sum of all the terms, corresponding to $q_{n-1}0$ provided that $\varepsilon$ is small enough. Moreover, one can see that we can let $\bar{a}$ in (\ref{adef1}) "slightly" depend on $\varepsilon$ if this still guarantees that the factor in (\ref{trash4bound}) is exponentially small and the estimates (\ref{trash1bound}), (\ref{trash2bound}) are exponentially small in comparison with the asymptotics that we are going to find. In our further estimate we will always keep $\bar{a}$, since we finally expect to let it grow as $\varepsilon\rightarrow 0$. So, we are basically left with $\bar{\Sigma}_1^<$ only to evaluate. Our main purpose, fulfilled in the succeeding section will be to show that it can be represented by its one single term plus an exponentially small error for the vast majority of the values of $\varepsilon$ small enough. \subsubsection{The Transition lemma} The focus of the whole discussion in this section is the Transition lemma, our principal number-theoretical tool to determine the asymptotics of the series (\ref{sum}), which states the simple fact that almost always there is a single term that accounts for its leading-order behavior, except for those values of $\varepsilon$ that sit in the "critical intervals" near the "critical values" , when the minimum of the exponent is not sharp enough to insure this. After completing the proof we'll make the necessary generalizations (in particular, we'll analyze what happens when $\varepsilon$ passes through one of the aforementioned, but not yet defined, critical values; we will also let the parameter $\bar{a}$ in (\ref{adef1}) depend on $\varepsilon$). First, we notice that from our convenience assumption $\beta>0$ follows the fact that in its Continued fraction $a_0=0$ and all the numerators $p_n$ in the sequence $\{{p_n\over q_n}\}$ of the best approximations to $\beta$ for $n\geq1$ will be strictly positive ($p_0=a_0=0$). So further on we will be dealing with the series \begin{equation} \Sigma\equiv \sum_{0\leq n\leq n_\varepsilon } \exp\left( -k_n\sigma_0-{\pi\omega_+\|q_n\beta\|\over2\sqrt{\varepsilon}} \right) A_{\vec{k}^n}\sin(-p_n\alpha_1+q_n\alpha_2), \label{newsum} \end{equation} \noindent where the summation is taken over all the convergents under the cutoff. Here $\vec{k}^n=(-p_n,q_n) ,\ k_n=p_n+q_n$. Clear enough, $k_{-2}=k_{-1}=1$, and for $n\geq0$ the $k_n$'s abide the same recursion relations as (\ref{recurpq}) for $p_n,q_n$, namely \begin{equation} k_n=a_n k_{n-1}+k_{n-2}. \label{recurk} \end{equation} \noindent Our next step will be to find tight upper and lower bounds for the exponents in the series (\ref{newsum}), using (\ref{approx}). With some abuse of notation we will still apply the term "badly approximable" to all the numbers that satisfy (\ref{adef1}). We will think of $\bar{a}$ as a constant and in the end will simply check to what extent all the key estimates can digest the possible dependence of $\bar{a}$ on $\varepsilon$. Once again, we assume that \[ \beta=[a_1,a_2,...], \] \noindent is given by an infinite Continued fraction, such that all the $a_n$'s are uniformly bounded by some integer constant $\bar{a}$, for the moment independent of $\varepsilon$: \begin{equation} a_n\leq \bar{a},\ \ n=1,2,..,n_\varepsilon. \label{Brjuno} \end{equation} \noindent The exponents will equal \begin{equation} E_n=k_n\sigma_0+{\pi\omega_+\|q_n\beta\|\over2\sqrt{\varepsilon}}. \label{en} \end{equation} \noindent The key quantity we will have to deal with is going to be \begin{equation} {\cal D}^1_n= E_n-E_{n-1},\ \ n\geq 1, \label{Delta1} \end{equation} \noindent but in certain cases it will not be enough, and we'll have to compare the exponents whose indices differ by $2$ and consider a quantity \begin{equation} {\cal D}^2_n=E_n-E_{n-2}={\cal D}^1_n+{\cal D}^1_{n-1},\ \ n\geq 2. \label{delta2} \end{equation} \noindent We shall try to express ${\cal D}^1_n$ and ${\cal D}^2_{n}$ in terms of as few parameters as possible, the range thereof depending on $\bar{a}$, and then prove that for $\varepsilon$ small enough typically (in the sense of avoiding the sequence of relatively small intervals near the critical values for $\varepsilon$ to be defined shortly) for any badly approximable $\beta$, as far as the shape of the sequence $\{ E_n\}$ is concerned, we encounter either of the two situations described below. \begin{itemize} \item {\it Case 1 (V-shape):} If after a certain finite index $N_0$, the same for all $\beta$: $a_n>1,\ n\geq N_0$, or if $a_n=1\ \forall n\geq N_0$, for $n>N_0$ (the latter possibility pertains to numbers {\it equivalent} in the sense of their Continued fraction expansion to the "golden mean"), the sequence $\{E_n\}$ has a unique absolute minimum (V-shape) (depending on $\beta, \varepsilon$). \item {\it Case 2 (V,W-shape):} If a badly approximable $\beta$ does not fall into the above category, then depending on $\varepsilon$ small enough and $\beta$, the sequence $\{E_n\}$ has either one absolute minimum (V-shape), as described above, or two local minima for some $n=n_*$, $n=n_*+2$, (W-shape), out of which we will be able to find the absolute minimum implementing the quantity ${\cal D}^2_{n_*+2}$. The necessary conditions for these occurrences will be considered further in details. \end{itemize} \noindent Thus, despite the presence of the small divisors in $E_n$, their sequence usually behaves quite nicely, first decreasing monotonically, and then monotonically increasing in case of V-shape, which sometimes can bifurcate into W-shape, characterized by the presence of two, but never more, neighboring local minima as $\varepsilon$ changes. Finally, we will see that we can let the parameter $\bar{a}$ from (\ref{adef}) depend on $\varepsilon$ and extend the domain of $\beta$ to the set of asymptotically full measure. \medskip \noindent The only two simple formulas that we shall invoke will be (\ref{rat}) and (\ref{approx}). By the latter we have \begin{equation} \|q_n\beta\|={1\over q_n\left([a_{n+1},a_{n+2},..]+{1\over[a_n,a_{n-1},..,a_1]}\right)} \equiv {1\over q_n}\cdot{1\over z_n+x_n}, \label{nth} \end{equation} \noindent where we've defined for $n=1,2,..$ \begin{equation} \begin{array}{lll} x_n &\equiv& {1\over [a_n,a_{n-1},..,a_1]}={q_{n-1}\over q_n}, \\ \hfill \\ z_n &\equiv& [a_{n+1},a_{n+2},..]. \end{array} \label{xzdef} \end{equation} \noindent By their definitions, the above parameters $x_n,z_n$ are the functions of a large (infinite in case of $z_n$) number of integers, but (\ref{fraction}) actually tells us that they really "strongly" depend only on $a_n$ and $a_{n+1}$ respectively, so one "almost" has $x_{n+1}\simeq{1\over z_n}\simeq{1\over a_{n+1}}$. Because of that, during the proof we'll always deal with the upper or lower bounds for $x_n,z_n$, which will be expressed in terms of no more than two integer parameters: $a_n,a_{n-1}$ for the former and $a_{n+1},a_{n+2}$ for the latter. Using (\ref{fraction}) we can easily derive the following recursion relations for the parameters $x_n,z_n$ which will subsequently be rather useful: \begin{equation} \begin{array}{lll} x_n &=& {1\over a_n+x_{n-1}}, \\ \hfill \\ z_{n}&=& a_{n+1}+{1\over z_{n+1}}. \end{array} \label{xzrec} \end{equation} \noindent By (\ref{Brjuno}) we can easily establish for all $n$ the general uniform bounds for $x_n,z_n$ for all $n$ in terms of the parameter $\bar{a}$ from (\ref{adef}): \begin{equation} {1\over \bar{a}+1}{\cal Q}^1(x_{n_*},z_{n_*}) \] \noindent or \[ q_{n_*-1}>{\cal Q}^1(x_{n_*-1},z_{n_*-1}),\ \ q_{n_*}<{\cal Q}^1(x_{n_*},z_{n_*}). \] \noindent We assign a sign index $\chi(n)$ to each term of the sequence $\{q_n\}$ by the following rule: \[ \chi(n)=\mbox{sign}(q_n-{\cal Q}^1(x_n,z_n)). \] \noindent Thus, the transition index corresponds to a sign change in the sequence $\{\chi(n)\}$. Taking the partial derivatives of the function ${\cal F}(x,z)$ from (\ref{F1}) in $x,z$ we find \[ \begin{array}{lll} \partial_x {\cal F}(x,z)&=&{z-1\over(x+z)^2 (1-x)^2}(z-1+2x)>0, \\ \hfill \\ \partial_z {\cal F}(x,z)&=&{1\over (1-x)(x+z)^2}(x+1)>0, \end{array} \] \noindent if $x$ and $z$ lie in the intervals, specified by (\ref{xbound}), (\ref{zbound}). This enables us to limit the range of ${\cal F}(x,z)$ simply by taking the lower bounds for $x,z$ for the lower bound and the upper bounds for $x,z$ for the upper bound on ${\cal F}(x,z)$, and consequently on ${\cal Q}^1(x,z)$. We get: \begin{equation} {1 \over \bar{a}+2 }<{\cal F}(x,z)<\bar{a}, \label{fbound} \end{equation} \noindent We recall that the bounds (\ref{xbound}), (\ref{zbound}) for $x$ and $z$ have been determined with the relative tolerance $O(\bar{a}^{-1})$. Therefore, it's easy to see that the above interval for the values of ${\cal F}$ is $(1+O(\bar{a}^{-1}))$ times wider than its true range for the admissible values of $x$ and $z$. Then with the same tolerance we can write: \begin{equation} {\cal K}(\varepsilon)\sqrt{{1\over \bar{a}+2 }}<{\cal Q}^1(x,z)<{\cal K}(\varepsilon)\sqrt{\bar{a}}. \label{qbound} \end{equation} \noindent We need to keep a close eye on the tolerance to make sure that the true zeroes of the quantity ${\cal D}^1_n$ (considered as a function of a continuous variable $q$ and parameters $x,z$, whose range is specified by (\ref{xbound}), (\ref{zbound})), also fall into the interval (\ref{qbound}). In general, this will be our common strategy of neutralizing the remainder in (\ref{EN}), (\ref{deltan}), incorporated in several more cases below. Our inability to compute the exact values for the parameters $x_n,z_n$, which necessitates resorting to the estimates instead, becomes an advantage, for we will always insure that the tolerance of "measuring" $x_n$ and $z_n$, and the functions of these parameters that we deal with, is much larger than the possible damage due to the error terms. We shall estimate the possible error in the location of the true zero of the quantity ${\cal D}^1_n$, with $q_n$ considered as a continuous variable, resulting from the impact of the error term $O(q_{n-1}^{-1})$ in (\ref{deltan}) by observing that \[ {d\over dq}{\cal D}^1({\cal Q}^1(x_n,z_n),x_n,z_n)=2(1+\beta)\sigma_0(1-x_n), \] \noindent whereas the error term is \[ O(q_{n-1}^{-1})=O\left({1\over q_n x_n}\right), \] \noindent with $q_n$ satisfying (\ref{qbound}). This implies that the true zero of the quantity ${\cal D}^1_n$ from (\ref{deltan}) will be separated from ${\cal Q}^1(x_n,z_n)$ by at most a distance, which is $O(\bar{a}^{3\over2}\varepsilon^{1\over4})$ (see (\ref{qbound}) and (\ref{xbound})). We require that this distance not exceed the tolerance in the width of (\ref{qbound}), which works to our advantage. Indeed, due to the $O(\bar{a}^{-1})$ relative tolerance, the possible zeroes of ${\cal D}^1$ will sit at least $O(\varepsilon^{-{1\over4}}\bar{a}^{-{3\over2}})$ away from the boundary inside the interval (\ref{qbound}). Thus, if we require that \begin{equation} \sqrt{\varepsilon}\bar{a}^3\ll 1, \label{frst} \end{equation} \noindent this will be sufficient to insure that all the true zeroes of the function ${\cal D}^1_n$ will also respect the bounds (\ref{qbound}). We see that given a small $\varepsilon$, there is some integer $s(\varepsilon)$ and a finite number $r+1$ of convergents $q_s,q_{s+1},..,q_{s+r-1},q_{s+r}$ falling into the interval, given by (\ref{qbound}); then $s,..,s+r+1$ become the only possible candidates for the transition indices, and for all the convergents $q_{s-i}$, preceding $q_s$, we will have ${\cal D}^1_{s-i}<0$, $i=1,2,..,s-1$, and so their $\chi$-index will equal $-1$, whereas for all the convergents, following $q_{s+r}$, we will have ${\cal D}^1_{s+r+i}>0$, $i=1,2,..$, and the $\chi$-index equal $+1$. Therefore, there always exists at least one transition point with a positive index. Suppose now, that $\varepsilon$ is such that the equality ${\cal Q}^1(x_{n},z_{n})=q_{n}$ never occurs. Otherwise we say that $\varepsilon$ attains one of its {\it critical values}. In fact, we aim to insure that ${\cal D}^1_n(q_n,x_n,z_n)\neq 0$ (otherwise we say that $\varepsilon$ attains its {\it true critical value}); this can happen only if $q_n$ lies in the interval, specified by (\ref{qbound}). Suppose, for some $\delta_1>0$, the value of $\varepsilon$ is such that \begin{equation} |{\cal Q}^1(x_{n},z_{n})- q_{n}|\geq 2q_n \varepsilon^{\delta_1}. \label{critint} \end{equation} \noindent Otherwise we say that $\varepsilon$ falls into a {\it critical interval}. We cannot take $\delta_1$ arbitrarily large, since the true critical value can be separated from a (computable) critical value by $O(\bar{a}^{3\over2}\varepsilon^{1\over4})$, so we must require \[ \bar{a}^{3\over2}\varepsilon^{1\over4}\ll q_n \varepsilon^{\delta_1}, \] \noindent what can be expressed in terms of $\varepsilon,\bar{a}$ only regarding (\ref{qbound}): \begin{equation} \bar{a}^2\varepsilon^{{1\over2}-\delta_1}\ll 1. \label{delta1} \end{equation} \noindent Roughly speaking (when $\bar{a}$ is $O(1)$), from (\ref{qbound}) we can see that the aforesaid equality can occur only when $q_n\sim\varepsilon^{-{1\over4}}$, which will also be a measure of the distance between $q_n$ and $q_{n-1}$, so if we choose $0<\delta_1<{1\over 4}$, then the critical intervals will have a small measure. We'll give it a strict formulation in the following Proposition. Moreover, if we avoid the critical intervals for the values of $\varepsilon$, then the difference between the exponents $E_n$ and $E_{n-1}$ is going to be of the order $O(\varepsilon^{-{1\over4}+\delta_1})$. \begin{proposition} The sequence $\{\varepsilon_n\}$ of the (true) critical values for $\varepsilon$ is nowhere dense as $n\rightarrow\infty$ and limits at zero, the relative measure of the critical intervals being bounded as $O(\bar{a}^2\varepsilon^{\delta_1})$, provided that $\delta_1$ satisfies (\ref{delta1}). \label{critical} \end{proposition} \noindent {\bf Proof:} Let's fix some small number $\varepsilon_*$. If ${\varepsilon_*\over16}\leq\varepsilon\leq\varepsilon_*$, then by (\ref{qbound}), the range of ${\cal Q}^1(x,z)$ will be bounded between \[ {K^*\over\sqrt{\bar{a}+2 }}\varepsilon_*^{-{1\over4}}<{\cal Q}^1(x,z) < 2K^*\sqrt{\bar{a}}\varepsilon_*^{-{1\over4}}. \] \noindent Suppose, the convergents $q_s,..,q_{s+N_*-1}$ fall into this interval. >From (\ref{xbound}), the maximum number $N_*$ of such convergents, which are the only ones for which the equality in question can occur, can be found from the following condition: \[ \left({\bar{a}+2 \over \bar{a}+1}\right)^{N_*}\leq2\sqrt{\bar{a}(\bar{a}+2 )}\leq2(\bar{a}+1), \] \noindent for $\min{q_n\over q_{n-1}}={\bar{a}+2 \over \bar{a}+1}$. so $N_*\leq 2(\bar{a}+1)^2$. In fact, Lochak [1993] can provide a much better estimate for this number. Suppose, $\varepsilon_{s+n}$ for $n=0,..,N_*-1$ are the critical values for $\varepsilon$ when \[ {\cal Q}^1(x_{s+n},z_{s+n})=\varepsilon_{s+n}^{-{1\over4}}K^* \sqrt{{\cal F}(x_{s+n},z_{s+n})}=q_{s+n}, \] \noindent then clearly $\varepsilon_{s+n}\geq {\varepsilon_*\over16},\ \forall n$. By (\ref{critint}), around each $\varepsilon_{s+n}$ we have to cut out an interval $\Delta\varepsilon_{s+n}$ to satisfy \[ q_{s+n}(1+2\varepsilon^{\delta_1})\leq K^*(\varepsilon_{s+n}- \Delta\varepsilon_{s+n})^{-{1\over4}}\sqrt{{\cal F}(x_{s+n},z_{s+n})} \] and \[ q_{s+n}(1-2\varepsilon^{\delta_1})\leq K^*(\varepsilon_{s+n}+ \Delta\varepsilon_{s+n})^{-{1\over4}}\sqrt{{\cal F}(x_{s+n},z_{s+n})}, \] \noindent what can be done for $\varepsilon_*$ small enough if \begin{equation} {\Delta\varepsilon_{s+n}\over\varepsilon_{s+n}}=16\varepsilon^{\delta_1}, \label{choice} \end{equation} \noindent provided that (\ref{delta1}) is satisfied, meaning that $q_{s+n}$ and ${\cal Q}^1(x_n,z_n)$ are very close to each other. Since the length of the interval of the values of $\varepsilon$ in consideration is ${15\over 16}\varepsilon_*$ and for all $n=0,..,N^*-1$ we have $\varepsilon_{s+n}\geq{\varepsilon_*\over 16}$, then the relative measure of $N^*$ critical intervals will be $O(\bar{a}^2\varepsilon^{\delta_1})$. The statement that the sequence of the (true) critical values accumulates at zero as $n\rightarrow\infty$ is quite obvious, just as the one that it is nowhere dense. In particular, the latter follows from the fact that having $\varepsilon$ vary in a small fixed interval, there will be a finite only number of convergents, for which, the critical values can lie inside this interval. $\Box$ \medskip \noindent Given a critical value of $\varepsilon$, we call its critical multiplicity the number of equalities ${\cal Q}^1(x_n,z_n)=q_n$ that take place. It's very unlikely that the multiplicity of a critical value be greater than one (a maximum value for it being the number of convergents sitting inside the interval, defined by (\ref{qbound})); if $\varepsilon$ lies outside the union of all the critical intervals, its critical multiplicity is certainly zero. Before we proceed we want to point out one more time that if the condition (\ref{frst}) holds, then the sequence $\{E_n\}$ decreases monotonically if $n ~~s+r$. Now we shall fix $\varepsilon$ outside the union of all the critical intervals and rescale the $q_n$'s and the function ${\cal Q}^1(x,z)$ by the factor ${\cal K}(\varepsilon)$. We will define the range of the rescaled function ${\cal Q}^1(x,z)$ where $x,z$ satisfy (\ref{xbound}), (\ref{zbound}) as an interval ${\cal I}_{\bar{a}}$: \begin{equation} {\cal I}_{\bar{a}}\equiv \left({1\over\sqrt{\bar{a}+2 }},\sqrt{\bar{a}}\right) \label{interval} \end{equation} \noindent Before we transit to our main result concerning the number of transition points, we'll have to express the quantity ${\cal D}^2_n$ in (\ref{delta2}) in terms of some parameters (a convenient set will be $q_{n-1},x_n,z_n$) as we've done it with ${\cal D}^1_n$. Using the recursion relations (\ref{xzrec}) we derive: \[ \begin{array}{lll} x_n &=& {1\over a_n +x_{n-1}}, \\ \hfill \\ z_{n-1} &=& a_n+ {1\over z_n}, \\ \hfill \\ q_n &=& q_{n-1}(a_n+x_{n-1}). \end{array} \] \noindent Substituting these relations into (\ref{delta1}) we get: \begin{eqnarray} {\cal D}^1_n\ \ \ & = &(1+\beta)\sigma_0 (a_n-1+x_{n-1})q_{n-1} -{\pi\omega_+\over2\sqrt{\varepsilon}} {1\over q_{n-1}} {z_n-1\over z_n x_n^{-1} +1}+O(q_{n-1}^{-1}), \nonumber \\ {\cal D}^1_{n-1} &=& (1+\beta)\sigma_0 (1-x_{n-1})q_{n-1} -{\pi\omega_+\over2\sqrt{\varepsilon}} {1\over q_{n-1}} { z_n(a_n-1)+1\over z_n x_{n}^{-1}+1 }+O(q_{n-2}^{-1}), \nonumber \\ {\cal D}^2_{n}\ \ \ &=& (1+\beta)\sigma_0 a_n q_{n-1} -{\pi\omega_+\over2\sqrt{\varepsilon}} {1\over q_{n-1}} {a_n z_n\over z_n x_{n}^{-1}+1} + O(q_{n-2}^{-1}). \label{delta2n} \end{eqnarray} \noindent In the same manner as we've done with ${\cal D}^1_n$, we throw away the supposedly small term $O(q_{n-2}^{-2})$ (which is not difficult to write up explicitly), and consider a function \[ {\cal D}^2(q,x,z)=(1+\beta)\sigma_0 q -{\pi\omega_+\over2\sqrt{\varepsilon}} {1\over q} {z\over 1+z/x} \] \noindent of a continuous variable $q$ and parameters $x,z$ which vary within the range specified by (\ref{xbound}), (\ref{zbound}); this function is zero if: \begin{equation} q={\cal Q}^2(x,z)={\cal K}(\varepsilon)\sqrt{{\cal G}(x,z)}, \label{qutwo} \end{equation} \noindent where ${\cal K}(\varepsilon)$ is the same as in (\ref{B}) and \begin{equation} {\cal G}(x,z)={z\over 1+z/x}. \label{G} \end{equation} \noindent It's easy to see that within the range of $x,z$ the partial derivatives of ${\cal G}$ never change their sign, namely $\partial_x {\cal G}(x,z)>0$ and $\partial_z {\cal G}(x,z)>0$. The range of ${\cal G}$ , fixed $a_n$ turns out to be: \begin{equation} {1\over a_n+2}< {\cal G}(x,z)<{1\over a_n}, \label{Gbound} \end{equation} \noindent with the relative tolerance in these inequalities being $O(\bar{a}^{-1})$. In particular, for we will need it most, for $a_n=1$ we can upgrade (\ref{Gbound}) to \begin{equation} {\bar{a}+2\over 3\bar{a}+4}< {\cal G}(x,z)<{\bar{a}+1\over \bar{a}+3}, \label{Gbound1} \end{equation} \noindent Then from (\ref{delta2}) if $q_{n-1}>{\cal Q}^2(x_{n},z_n)$, then $E_n>E_{n-2}$, and if $q_{n-1}<{\cal Q}^2(x_{n},z_n)$, then $E_n~~n_1$, if \[ n_2=n_1+1,\ a_{n_2}\geq2, \] or \[ n_2\neq n_1+1,\ a_{n_2}\geq1, \] \noindent then the necessary condition for $\chi(n_2)=-1$ is \[ q_{n_1}q_{n_2}<{\bar{a}\over \bar{a}+1}. \] \label{left} \end{proposition} \noindent {\it Remark:} The only possibility which is not covered by this Proposition, as well as the next one, is when simultaneously $n_2=n_1+1$ AND $a_{n_2}=1$.\\ \noindent {\bf Proof:} We have ${\cal F}(x_{n_2},z_{n_2})={z_{n_2}-1\over(z_{n_2}+x_{n_2})(1-x_{n_2})}$. The maximum of $z_{n_2}$, is always smaller than $\bar{a}+1$ by (\ref{zbound}). First, we'll show that under the constraints of this Proposition the maximum of $x_{n_2}$ will be always smaller than $1-{q_{n_1}\over q_{n_2}}$ (which would be invalid if simultaneously $n_2=n_1+1$ and $a_{n_2}=1$). Indeed, if $a_{n_2}\geq2$ then always ${q_{n_1}\over q_{n_2}}<{1\over2}$, so $x_{n_2}={q_{n_1}\over q_{n_2}}<{1\over2}<1-{q_{n_1}\over q_{n_2}}$. Otherwise, if $n_2>n_1+1$, then there is an integer $n'$: $n_1 n_1$, if \[ n_2=n_1+1,\ a_{n_2}\geq2, \] or \[ n_2\neq n_1+1,\ a_{n_2}\geq1, \] \noindent then the necessary condition for $\chi(n_1)=1$ is \[ q_{n_1}q_{n_2}>{\bar{a}+1\over \bar{a}}. \] \label{right} \end{proposition} \noindent {\bf Proof:} Exactly in the same fashion as the previous proposition. We have ${\cal F}(x_{n_1},z_{n_1})={z_{n_1}-1\over(z_{n_1}+x_{n_1})(1-x_{n_1})}$. Substituting the minimum of $x_{n_1}$, which is greater than ${1\over \bar{a}+1}$, and the minimum of $z_{n_1}$, which is greater than $1+{q_{n_1}\over q_{n_2}}$, (which would be invalid if simultaneously $n_2=n_1+1$ and $a_{n_2}=1$) we get: \[ {\cal F}(x_{n_1},z_{n_1}) > {\bar{a}+1\over\bar{a}} {q_{n_1}\over q_{n_2}} {1\over 1+{1\over \bar{a}+1}+{q_{n_1}\over q_{n_2}}} > {\bar{a}+1\over\bar{a}} {q_{n_1}\over q_{n_2}}, \] \noindent consequently \[ {\cal Q}^1(x_{n_1},z_{n_1})>\sqrt{{\bar{a}\over \bar{a}+1}\cdot{q_{n_2}\over q_{n_1}}}. \] \noindent If $\chi(n_1)=1$, then $q_{n_1}>{\cal Q}^1(x_{n_1},z_{n_1})$, so the necessity of the condition $q_{n_1}q_{n_2}>{\bar{a}+1\over\bar{a}}$ follows. $\Box$\\ \noindent These two Propositions make the proof of our main result, which we call the Transition lemma, almost trivial, although the fact, stated in this Lemma, essentially tells us that if $\varepsilon$ is away from the union of the critical intervals, we can always find a unique absolute minimum of the sequence $\{E_n\}$. \begin{lemma}[The Transition lemma] Assume, $\varepsilon$ lies outside the union of all the critical intervals, and the convergents $q_s,..,q_{s+r}$ lie inside the interval ${\cal I}_{\bar{a}}$. If for some $ n_* \in\{s,..,s+r-1\}$ one has \[ a_{ n_*+2}=1, \] \[ \chi(n)=-1, \ s\leq n\leq n_* , \] \begin{equation} {1\over 1-x_{ n_*+1 }}< q^2_{ n_*+1 } ( 1+z_{ n_*+2 } (1+x_{ n_*+1}) )< {z_{ n_* +2}-1\over x_{ n_*+1 } }, \label{W} \end{equation} \noindent then $q_{ n_* +1}, q_{ n_* +3}$ are the only two transition points with the positive labels, and $q_{ n_* +2}$ is the only transition point with the negative label. In particular, the necessary condition for this to occur is \begin{equation} z_{ n_* +2}>{1\over 1- x_{ n_* +1}}. \label{Wnes} \end{equation} \noindent Otherwise, there is one and only one transition point $q_{ n_*+1 }$ with $n_* \in\{s-1,..,s+r\}$, and $\chi( n_*+1 )=+1$. \end{lemma} \noindent {\bf Proof:} Once again, we notice that all the points to the left of the interval ${\cal I}_{\bar{a}}$ carry negative labels and all the points to the right of it carry positive labels. That's why there is no point with the positive label inside ${\cal I}_{\bar{a}}$, or the only such a point is the last one, then the lemma holds trivially yielding V-shape. Otherwise, suppose, $q_{n_1}$ is the first point inside the interval ${\cal I}_{\bar{a}}$, such that $\chi(n_1)=+1$, then by definition it will be a transition point carrying the positive label. Also suppose that for $n_2>n_1$ the point $q_{n_2}$ also lies inside this interval and $\chi(n_2)=-1$, for otherwise $q_{n_1}$ would be the only transition point, and again we are in the Case 1 (V-shape). Then applying Propositions \ref{left} and \ref{right}, unless simultaneously $n_2=n_1+1$ AND $a_{n_2}=1$, we must have \[ {\bar{a}\over \bar{a}+1}>q_{n_1}q_{n_2}>{\bar{a}+1\over\bar{a}}, \] which is a contradiction. So the {\it only} possibility to have two such points inside ${\cal I}_{\bar{a}}$ will be the simultaneous fulfillment of two conditions: \begin{equation} \begin{array}{lll} n_2&=&1+n_1, \\ a_{n_2}&=&1, \end{array} \label{burbs} \end{equation} \noindent namely $q_{n_1}$ and $q_{n_2}$ must be neighbors, and the spacing between them must be pretty small, because $a_{n_2}=1$ implies $q_{n_2}=q_{n_1}+q_{n_1-1}$. By definition, $q_{n_1}$ will be a transition point with a positive label, and $q_{n_2}$ will be a transition point with a negative label. Moreover, there will be no points with negative labels to the right of $q_{n_2}$, because if some $q_{n_3}$ such that $n_3>n_2$ had $\chi(n_3)=-1$, then a pair of points $n_3$ and $n_1$ would satisfy the conditions of Propositions \ref{left}, \ref{right}, whose simultaneous application would lead to an absurd statement \[ {\bar{a}\over \bar{a}+1}>q_{n_1}q_{n_3}>{\bar{a}+1\over\bar{a}}. \] \noindent Hence, the only possibility to have more than one transition point is when for some $ n_* \in\{s-1,..,s+r-2\}$ the point $q_{ n_*+1 }$ is the leftmost one with a positive index, and the point $q_{ n_* +2}$ has a negative index; then all the points to the right of $q_{ n_* +2}$ will have positive indices, as it has just been shown. To have simultaneously $\chi( n_*+1 )=1$ and $\chi( n_* +2)=-1$ we must have \[ {\cal F}(x_{ n_*+1 },z_{ n_*+1 }) {1\over 1- x_{ n_*+1}} \] \noindent is the necessary condition for the W-shape to take place. This enables us to improve the lower bound for the rescaled value ${\cal Q}^2(x_{ n_*+2 },z_{ n_* +2})$ instead of (\ref{Gbound1}) for $a_{n_*+2}=1$ up to \begin{equation} {1\over\sqrt{2}}<{\cal Q}^2(x_{ n_*+2 },z_{ n_* +2})<\sqrt{\bar{a}+2 \over\bar{a}+4}. \label{boundqutwo} \end{equation} \noindent The proof of the Transition Lemma will be complete now if we recall that if the conditions (\ref{frst}), (\ref{critint}), and (\ref{delta1}) are satisfied, then the neglected error terms in (\ref{deltan}) and (\ref{delta2n}) cannot be of any harm. $\Box$ \medskip \noindent Suppose, we are in the case of W-shape, so the sequence $\{E_n\}$ has two local minima for $n=n_*$ and $n=n_*+2$. One of the necessary conditions is $a_{n_*+2}=1$, and we can use the function ${\cal D}^2$ to find the absolute minimum. Namely, we will have to compare $q_{n_*+1}$ and ${\cal Q}^2(x_{n_*+2}, z_{n_*+2})$, the latter necessarily falling into the limits of (\ref{boundqutwo}). Being always away from the critical intervals, we don't have to worry about the errors. The following Corollary states that if one of the indices $n_*$ or $n_*+2$ corresponds to the absolute minimum, then the other yields the second smallest member of the sequence $\{E_n\}$. \begin{corollary} Suppose, $n_*+2$ is the transition index, labeled $-1$ and $\varepsilon$ is small enough and away from the union of all the critical intervals. Then $E_{n_*+1} E_{n_*}$ and $E_{n_*+1}>E_{n_*+2}$, therefore if we prove the fact stated in this Corollary, this would mean that once we encounter W-shape, $E_{n_*}$ and $E_{n_*+2}$ are always the two smallest terms in the sequence $\{E_n\}$. Suppose, $E_{n_*-1} {\cal G}(x_{n_*+1}, z_{n_*+1})$. We'll show that this contradicts (\ref{W}). We compute ${\cal G}(x_{n_*+1}, z_{n_*+1})$ in terms of $x_{n_*+1}$ and $z_{n_*+2}$, using the recursion relations (\ref{xzrec}) and the necessary condition $a_{n_*+2}=1$ to substitute $z_{n_*+1}=1+{1\over z_{n_*+2}}$. This yields \[ q_{n_*}^2>{x_{n_*+1}(z_{n_*+2}+1)\over 1+z_{n_*+2}(1+x_{n_*+1})}. \] \noindent or, since $q_{n_*}=x_{n_*+1}q_{n_*+1}$, \[ q_{n_*+1}^2>{1\over x_{n_*+1}}{ z_{n_*+2}+1\over 1+z_{n_*+2}(1+x_{n_*+1})}, \] \noindent whereas (\ref{W}) implies that \[ q_{n_*+1}^2<{1\over x_{n_*+1}}{ z_{n_*+2}-1\over 1+z_{n_*+2}(1+x_{n_*+1})}, \] \noindent which is in the obvious contradiction with the previous statement. Now, suppose $E_{n_*+3} 1$, which implies that \[ {q_{n_*+2}^2\over x_{n_*+3}}>{1\over x_{n_*+2}}>1, \] \noindent which contradicts the previous statement. Thus, the Corollary is proved. $\Box$ \medskip \noindent We can prove another Corollary, which will illustrate that the situation with W-shape does happen indeed. \begin{corollary} The necessary condition for \[ \begin{array}{llr} a_{n_*+2}&=&1, \\ \chi(n_*+1)&=&-1, \\ \chi(n_*+2)&=&+1 \end{array} \] \noindent is \[ z_{n_*+2}<{1\over 1-x_{n_*+1}}. \] \label{Crr1} \end{corollary} \noindent {\em Remark:} Compare with the necessary condition (\ref{Wnes}) for W-shape to take place. If the latter is satisfied, for no values of $\varepsilon$ is V-shape possible with the absolute minimum at $n_*+1$. But when $\varepsilon$ is decreased smoothly, the absolute minimum shall move to the right towards larger values of $n$. Thus most likely one will encounter W-shape for a certain interval of the values of $\varepsilon$. \medskip \noindent {\bf Proof:} Using the recursion relations (\ref{xzrec}) and what is given, we can write \[ \begin{array}{lll} q_{n_*+1}^2 &<&{\cal F}(x_{n_*+1},z_{n_*+1})= {1\over (1+z_{n_*+2}(1+x_{n_*+1}))(1-x_{n_*+1})}, \\ \hfill \\ q_{n_*+2}^2 = (1+x_{n_*+1}))^2 q_{n_*+1}^2 &>&{\cal F}(x_{n_*+2},z_{n_*+2})= {(z_{n_*+2}-1) (1+x_{n_*+1}))^2 \over (1+z_{n_*+2}(1+x_{n_*+1}))}{1\over x_{n_*+1}}. \end{array} \] \noindent These two statements will lead to the contradiction unless the necessary condition, claimed in this Corollary, holds true. $\Box$ \medskip \noindent By the Transition lemma, the only two possible shapes for the sequence $\{E_n\}$ are such that it has either one or two local minima, and we can detect the absolute minimum if $\varepsilon$ is outside the union of all the critical intervals. Suppose, given $\varepsilon$, the absolute minimum occurs for $n=n_*$. If the convergents $q_s,..,q_{s+r}$ sit inside the interval ${\cal I}_{\bar{a}}$, then $n_*\in \{s-1,s,..,s+r\}$. Obviously, since this is a minimum $\chi(n_*)=-1$, $\chi(n_*+1)=1$; for all $s>n_*$, except for possibly $s=n_*+2$, in the case when $n_*$ corresponds to the left prong of W-shape, when $\chi(n_*+2)=-1$. Now we can do better than (\ref{qbound}) to get the estimates on the location of the convergent $q_{n_*}$, corresponding to the absolute minimum of the sequence $\{E_n\}$. A rough shot would be \[ {1\over (\bar{a}+1)\sqrt{\bar{a}+2 }}< q_{n_*}< \sqrt{\bar{a}}. \] \noindent but the following two simple complementary to each other Propositions show that this estimate can be significantly improved. \begin{proposition} Suppose, (\ref{frst}) is satisfied. If $q_{n_*}$ corresponds to the absolute minimum of the sequence $\{E_n\}$, then $q_{n_*}<2$. \label{prop1} \end{proposition} \noindent {\bf Proof:} We notice that if $q_{n_*}$ corresponds to the absolute minimum of the sequence $\{E_n\}$, then $\chi(n_*)=-1$. First, we assume that $a_{n_*}\geq 2$. In this case we have: \[ q_{n_*}^2<{\cal F}(x_{n_*},z_{n_*})<2, \] \noindent where we have used that if $a_{n_*}\geq 2$, then $x_{n_*}<{1\over2},$ and (\ref{F1}). Otherwise, suppose $a_{n_*}=1$. Then either for all $n {1\over2}$. So, suppose, all the points to the left of $q_{n_*}$ carry negative indices (also including $q_{n_*}$ itself) and $a_{n_*}=1$. Then we must have simultaneously $q_{n_*}^2<{\cal F}(x_{n_*},z_{n_*})$ and $q_{n_*-1}^2=(q_{n_*}x_{n_*})^2<{\cal F}(x_{n_*-1},z_{n_*-1})$. Under the assumption that $a_{n_*}=1$, we can write $z_{n_*-1}=1+{1\over z_{n_*}}$ and $x_{n_*-1}={1\over x_{n_*}}-1$ and then use (\ref{F1}) to obtain the following two inequalities that must be satisfied simultaneously: \[ q_{n_*}^2< {z_{n_*}-1\over z_{n_*}+x_{n_*}} {1\over 1-x_{n_*}}< {1\over 1-x_{n_*}}, \] and \[ q_{n_*}^2 < {1\over z_{n_*}+x_{n_*}}{1\over2x_{n_*}-1}<{1\over2x_{n_*}-1}. \] \noindent Together these inequalities result in $q_{n_*}^2<3$. At the end we notice than if (\ref{frst}) is satisfied, then the tolerance for all the inequalities exceeds any possible error that can come from the influence of the neglected terms in (\ref{deltan}), (\ref{delta2n}). $\Box$ \medskip \noindent The second Proposition is essentially the one that we have just proved turned upside down. \begin{proposition} Suppose, (\ref{frst}) is satisfied. If $q_{n_*}$ corresponds to the absolute minimum of the sequence $\{E_n\}$, then $q_{n_*+1}>{1\over2} $. \label{prop2} \end{proposition} \noindent {\bf Proof:} First, obviously $\chi(q_{n_*+1})=+1$. If $a_{n_*+2}\geq 2$, then $z_{n_*+1}>2$, and from (\ref{F1}) we see that since $\chi(n_*+1)=1$, we must have $q_{n_*+1}^2>{\cal F}(x_{n_*+1},z_{n_*+1})>{1\over3}$. Otherwise, suppose $a_{n_*+2}=1$. Suppose, we deal with W-shape and $q_{n_*}$ corresponds to the left local minimum. Then we must have $q_{n_*+1}^2>{\cal G}(x_{n_*+2}, z_{n_*+2})>{1\over2}$ by (\ref{boundqutwo}). Otherwise, all the points to the right of $q_{n_*+1}$ must have positive indices. Then we must have simultaneously $q_{n_*+1}^2>{\cal F}(x_{n_*+1},z_{n_*+1})$ and $q_{n_*+2}^2=\left({q_{n_*+1}\over x_{n_*+2}}\right)^2>{\cal F}(x_{n_*+2},z_{n_*+2})$. Under the assumption that $a_{n_*+2}=1$, we can write $z_{n_*+1}=1+{1\over z_{n_*+2}}$ and $x_{n_*+2}={1\over 1+x_{n_*+1}}$ and then use (\ref{F1}) to obtain the following two inequalities that must be satisfied simultaneously: \[ q_{n_*+1}^2 > {1\over z_{n_*+2}}, \] \noindent and \[ q_{n_*+1}^2 > {z_{n_*+2}-1\over z_{n_*+2}+1}, \] \noindent which entails $q_{n_*+1}^2>{1\over3}>{1\over4}$. At the end we notice than if (\ref{frst}) is satisfied, then the tolerance for all the inequalities exceeds any possible error that can come from the influence of the neglected terms in (\ref{deltan}), (\ref{delta2n}). $\Box$ \medskip \noindent The above two Propositions imply that \begin{equation} {\cal K}(\varepsilon){1\over 2(\bar{a}+1)}< q_{n_*}< 2{\cal K}(\varepsilon). \label{minloc} \end{equation} \noindent We suggest a couple of remarks illustrating some facts that can be derived from the preceding results. \medskip \noindent {\it Remark 1:} When none of $a_{s+1},..,a_{s+r}$ equals $1$, then we always have only one transition point, so the sequence $\{ E_n\}$ has V-shape, first decreasing monotonically until it reaches the absolute minimum for $n= n_*$ and afterwards increasing. Moreover, if $\beta$ is such a number that in its Continued fraction one never occurs, then instead of a bound (\ref{interval}) for the range of the rescaled function ${\cal Q}^1(x,z)$ we will have much better bounds, almost as tight as ${1\over\sqrt{2}}<{\cal Q}^1(x,z)<\sqrt{2}$, because in this case always $x<{1\over2}$, $z>2$. Which implies that the interval ${\cal I}_{\bar{a}}$ cannot contain more than one member of the sequence $\{q_n\}$, as for all $n$ one has $q_{n+1}>2 q_n$ if $a_{n+1}\geq2$. So in this case the sequence $\{E_n\}$ behaves very nicely. \medskip \noindent {\it Remark 2:} If $\beta={\sqrt{5}-1\over 2}=[1,1,1,...]$, namely the so-called "golden number", then for all $n$ the values of the parameters $z_n$ are the same, and of $x_n$ almost the same (different at $O(q_{n-1}^{-2})$), so the values of the function ${\cal Q}^1(x_n,z_n)$ for all $n$ fall into a very narrow interval compared to the spacing between the consecutive convergents $q_{n-1}$ and $q_n$. So for any (small) $\varepsilon$ we will be able to find some $n_*(\varepsilon)$ such that at least $q_{n_*-1}<{\cal Q}^1(x_n,z_n) {1\over\bar{a}+2}$. Thus, \begin{equation} {\cal K}(\varepsilon) (\bar{a}+2)^{-{3\over 2}}< q_{n_{**}}< {\cal K}(\varepsilon) (\bar{a}+2)^{1\over2}, \label{minloc2} \end{equation} \noindent and the same (in fact, tighter) bounds will hold for W-shape. \label{TL} \subsubsection{Conclusion of the proof of Theorem \ref{Main3}} Finally, using the results of the preceding discussion, we turn to the conclusion of the proof of Theorem \ref{Main3}. At first we will be concerned with the smallness conditions. Suppose that \[ \bar{a}=\bar{a}_0\varepsilon^{-\delta_2}, \] \noindent where $\delta_2>0$ is rather small, and $\bar{a}_0$ is a constant, independent of $\varepsilon$. To be sure that (\ref{frst}) and (\ref{delta1}) are satisfied if $\varepsilon$ is small enough, we do the following: for any $\delta_3>0$ and $\varepsilon$ small enough we require \begin{equation} \begin{array}{rrrrrrr} {1\over2}& &-3\delta_2 &-\delta_3 &\geq 0 , \\ \hfill \\ {1\over2}& -\delta_1 &-2\delta_2&-\delta_3 &\geq 0. \end{array} \label{scnd} \end{equation} \noindent Now we have to recall (\ref{trash1bound}), (\ref{trash2bound}), and (\ref{trash4bound}), and also the fact that the number of convergents under the cutoff is definitely smaller than ${1\over\sqrt{\varepsilon}}$. In the following relations the first inequality is to insure that the cutoff parameter $q_\varepsilon$, that we have introduced before getting (\ref{trash2}), lies to the right of the interval where $q_{n_*}$ and $q_{n_{**}}$ can be located, with the upper bound from (\ref{minloc2}). The second inequality insures that the factor in the right-hand side in (\ref{trash4bound}) can be rewritten as $\exp(-\varepsilon^{-\delta_3})$. The third one will be the restatement of (\ref{diffr}), provided that (\ref{scnd}) holds: \begin{equation} \begin{array}{rrrrrrr} {1\over4}&-\delta_0&-{1\over2}\delta_2&-\delta_3&\geq0, \\ \hfill \\ &\delta_0&-2\delta_2& - 2\delta_3 &\geq0, \\ \hfill \\ {1\over4}&-\delta_1 &-{3\over2}\delta_2 &-2\delta_3 &\geq0, \end{array} \label{thrd} \end{equation} \noindent Then, following our strategy of making estimates, we can rewrite the sum (\ref{newsum}) as follows: \begin{equation} \begin{array}{lll} \Sigma &=& 2A_{\vec{k}^*} \exp\left( -k_{*}\sigma_0-{\omega_+\|q_* \beta\|\pi\over2\sqrt{\varepsilon}} \right)\sin(\vec{k}^{*}\cdot\vec{\alpha}) \\ \hfill \\ & + & 2A_{\vec{k}^{**}} \exp\left( -k_{**}\sigma_0-{\omega_+\|q_{**} \beta\|\pi\over2\sqrt{\varepsilon}} \right)\sin(\vec{k}^{**}\cdot\vec{\alpha}) \\ \hfill \\ & \times & \left[1+O\left(e^{-\varepsilon^{-\delta_3}}\right)\right], \end{array} \label{newsum1} \end{equation} \noindent where $q_*=q_{n_*}$, $ p_* $ is the unique and positive numerator in the $n_*$th best approximation to $\beta$, which can be computed from the Continued fraction for $\beta$ using the recursion rule (\ref{recurpq}), $\vec{k}^*=(- p_* ,q_*) $, and $k_{*}= p_* +q_*$ and similarly for the second smallest term marked by double asterisks. Using (\ref{EN}), we can also estimate the minimum and the maximum for the exponents $E_{n_*}$ and $E_{n_{**}}$ when $n_*$ and $n_{**}$ are such that $q_{n_*}$ lies inside the interval, specified by (\ref{minloc}) and $q_{n_{**}}$ assumes the bounds (\ref{minloc2}). In addition, we know that $E_{n_{**}}$ exceeds $E_{n_*}$. While evaluating these exponents, we will keep in mind that provided that (\ref{frst}) is satisfied, the error term in (\ref{EN}) can be ignored. By (\ref{EN}) without the error term, $E_n$ reaches its minimum as a continuous function of $q$ when \[ q={\cal K}(\varepsilon)(x_n+z_n)^{-{1\over2}}, \] \noindent which equals \[ \left({2\sigma_0(1+\beta)\pi\omega_+\over\sqrt{\varepsilon}(x_n+z_n)}\right)^{{1\over2}}. \] \noindent We use the upper bound $x_n+z_n=\bar{a}+2$ from (\ref{xbound}) and (\ref{zbound}) and notice that $0<\beta<1$, which implies \begin{equation} \begin{array}{lll} E_{n_*} & \geq & \sqrt{{2\pi\omega_+\sigma_0\over (\bar{a}_0+2)}}\ \varepsilon^{-{1\over4}+{1\over2}\delta_2}. \end{array} \label{LB} \end{equation} \noindent To compute the maximum of the exponents $E_{n_*}$ and $E_{n_{**}}$ we substitute the (rescaled) upper bounds $q_{n_*}=2$ from (\ref{minloc}) and $q_{n_{**}}=\sqrt{\bar{a}+2}$ from (\ref{minloc2}) into the first term of (\ref{EN}). As for the second term, we notice that ${1\over q_n}{1\over(z_n+x_n)}<{1\over q_n}{2\over z_n+1}<{2\over q_{n+1}}$, so we can use the lower bound ${1\over2}$ for $q_{n_*+1}$ from Proposition \ref{prop2} for $E_{n_*}$ and the lower bound for $q_{n_{**}+1}$ for $E_{n_{**}}$; the latter can be easily seen to be $(\bar{a}+2)^{-{1\over2}}$. Thus, we obtain: \begin{equation} \begin{array}{lll} E_{n_*}&\leq & 8\sqrt{\pi\omega_+\sigma_0}\varepsilon^{-{1\over4}}, \\ \hfill \\ E_{n_{**}} &\leq & 8\sqrt{\pi\omega_+\sigma_0(\bar{a}_0+2)} \varepsilon^{-{1\over4}-{1\over2}\delta_2}. \end{array} \label{UB} \end{equation} \noindent where we have finally substituted $1$ for $\beta$. That's why the estimates (\ref{trash1bound}) and (\ref{trash2bound}) will be exponentially small in comparison with $|\Sigma|$ and negligible in the error analysis if we let $\bar{\delta}=\delta_3$ and require \[ {1\over4}-\delta_0-{1\over 2}\delta_2-3\delta_3\geq0. \] It remains to say that if (\ref{delta1}), or (\ref{scnd}) is satisfied, then according to Proposition \ref{critical}, the relative measure of the union of all the critical intervals will be $O(\varepsilon^{\delta_1-2\delta_2})$. The next thing to show will be that after we let the parameter $\bar{a}$ depend on $\varepsilon$, the frequencies to which this Theorem can be applied, form a set of asymptotically full measure. It's easy to see that our argument will not work for numbers $\bar{\beta}$ which are too close to rationals ${p\over q}$ for $q\leq q_\varepsilon=[\varepsilon^{-{1\over2}-\delta_0}]$. So, for all these rationals on the unit interval we will have to cut out their small neighborhoods, containing numbers $\bar{\beta}$ such that $|\bar{\beta}-{p\over q}|\leq{1\over (\bar{a}+2 )q^2}$. For each $q\leq q_\varepsilon$ there are at most $q$ choices for the numerator $p$. That's why the overall length of all the intervals to be cut out is bounded from above by the quantity \[ {1\over\bar{a}}\sum^{q_\varepsilon}_{q=1} {1\over q}\leq\varepsilon^{3\delta_2\over4} \] \noindent with our choice of $\bar{a}\sim\varepsilon^{-\delta_2}$. On the other hand, in what remains there are points not satisfying the diophantine condition (\ref{dio}) with $b={1\over2}$ (see (\ref{gaga})) and $\varpi=2-2\delta>1$; the relative measure of the latter won't exceed $\varepsilon^{1\over4}$. Thus, the relative measure of the frequencies to which either the KAM theorem, or the argument of this chapter cannot be applied, will be certainly bounded from above by $\varepsilon^{\delta_2\over2}$. Eventually, suppose $*$ is the index for the minimum exponent term and $**$ is the index for the second minimum exponent term in the series (\ref{sum}). We know that $\vec{\alpha}=\vec{0}$ is its true zero (in fact, there are more true zeroes as is stated by Theorem \ref{Main1}). The representation (\ref{newsum1}) hints us that there are many other zeroes of (\ref{sum}), which will be close to the points where the two closed trajectories on a 2-torus, given by $\vec{k}^*\cdot\vec{\alpha}=0$ and $\vec{k}^{**}\cdot\vec{\alpha}=0$ intersect. We will be able to prove it, by showing that the maximum error in (\ref{newsum1}) turns out to be exponentially small in comparison with the estimate for the quantity $\Upsilon$ that we are going to obtain further. The number of these additional zeroes will be at least $p_{*}q_{*}$, where ${p_*\over q_*}$ is the $n_*$th best approximation to $\beta$. The latter product can be bounded from below by $\beta\varepsilon^{-{1\over2}+2\delta_2+\d! elta_3}$, where we have used (\ref{minloc}). The same argument will be true for the intersections of the curves $\vec{k}^*\cdot\vec{\alpha}=\pi$ and $\vec{k}^{**}\cdot\vec{\alpha}=0$, etc., so we can multiply this estimate for the number of zeroes by at least $4$. One can easily evaluate $\Upsilon$ from its definition (\ref{msr}) applied to (\ref{newsum1}) as follows: \[ \begin{array}{lll} \Upsilon &=&4{\cal E}_*{\cal E}_{**}|k_1^*k_2^{**}-k_1^{**}k_2^*| |A_1^*A_2^{**}-A_1^{**}A_2^*| \\ \hfill \\ & \times & \left(1+O\left(e^{-\varepsilon^{-\delta_3}}\right) \right), \end{array} \] \noindent where the multiplier $4$ comes from parity. The splitting distance is given by the Melnikov function and a relatively small remainder, see (\ref{splittingdistance}). Besides, the contribution of the second and higher order terms due to the Fourier modes with either $k_1$ or $k_2$ equal zero will be $\sim e^{-C\varepsilon^{-{1\over2}}}$, hence, negligible. Therefore, if the above expression does not turn into zero for the Melnikov function, the remainder can be relegated into the error term. Apparently, this error won't influence the foregoing argument for the number of the new zeroes. Since $\vec{k}=(-p,q)$, in case of V-shape when $n_{**}=n_*\pm1$ we use Proposition \ref{sign} to claim that $|k_1^*k_2^{**}-k_1^{**}k_2^*|=1$. In case of W-shape the two local minima are indexed by $n_*$ and $n_*+2$, and moreover, $a_{n_*+2}=1$. Using the recursion rules (\ref{recurpq}), and Proposition \ref{sign} we see that in this case also $|k_1^*k_2^{**}-k_1^{**}k_2^*|=1$. As for the Melnikov function, in the notation of (\ref{sum}) we have: \[ \vec{A}_{\vec{k}}=\vec{k} {\pi|\mz|\over\varepsilon} \sum_{j=1}^{\nu_{\vec{k}}} A_{j\vec{k}} \prod^{j-1}_{l=1}\left({(\mz)^2\over\varepsilon}+4l^2\right), \] \noindent therefore, using the same reasoning we conclude that \[ |A_1^*A_2^{**}-A_1^{**}A_2^*|= \left({\pi\over\varepsilon}\right)^2 \prod_{\vec{k}\in\{\vec{k}^*,\vec{k}^{**}\}} |\mz|\sum_{j=1}^{\nu_{\vec{k}}} A_{j\vec{k}} \prod^{j-1}_{l=1}\left({(\mz)^2\over\varepsilon}+4l^2\right). \] \noindent The latter quantity is obviously nonzero. At this point, we notice that our choice of a positive $\beta$ meant nothing, except for the fact that in its Continued fraction $a_0=0$, which was insignificant. The other thing would be to change the multiplier $(1+\beta)$ for $(1+|\beta|)$ in formulas for $E_n$, ${\cal D}^{1,2}_n$, ${\cal K}$, etc. This concludes the proof of Theorem \ref{Main3}. $\Box$ \subsubsection{Shape evolution} In this section we demonstrate how the shape of the sequence $\{E_n\}$ changes as $\varepsilon$ goes to zero. Suppose, that $\varepsilon$ is away from the union of all the critical intervals, and $\vec{k}^*$ is the index of a mode, corresponding to the convergent $q_*=q_{n_*}$, giving the leading-order behavior of the Fourier series (\ref{sum}). Moreover, suppose we deal with more common V-shape and the second minimum exponent is indexed by $n_*-1$. Anent the critical values, it is very unlikely that they can have multiplicities higher than $1$, so for simplicity we assume that all the critical values are simple. When $\varepsilon$ is decreased and it first comes close to the critical value for the quantity ${\cal D}^2_{n_*+1}$ (because of (\ref{delta2}) it happens before the critical value for the quantity ${\cal D}^2_{n_*+1}$ is reached) , and when it passes this critical value, the second minimum exponent becomes the one indexed by $n_*+1$. Further, $\varepsilon$ passes the critical value for the quantity ${\cal D}^2_{n_*+1}$, and after it does this, the absolute minimum will be given by $E_{n_*+1}$ and the second minimum exponent will be $E_{n_*}$. Eventually, suppose we are in the situation stipulated by Proposition \ref{Crr1}, but the necessary condition for V-shape is not satisfied. This means that before $\varepsilon$ passes through the critical value for ${\cal D}^1_{n_*+1}$ it passes the one for ${\cal D}^1_{n_*+2}$, and another local minimum gets born, then the one for ${\cal D}^2_{n_*+2}$, so the right local minimum becomes the absolute minimum. Then, eventually, $\vareps! ilon$ passes through the critical value for ${\cal D}^1_{n_*+1}$, and the left local minimum gets annihilated, so W-shape transforms into V-shape. We see that although the class of frequencies that we consider is very large, the arithmetics is quite regular and predictable, in particular as far as the mutual location of the critical values is concerned. We believe, this accounts for some "jumping" phenomena, observed numerically in Sim\'{o} [1994]. %%%%%%%% \section{Appendix. Proof of the Extension Lemma} As we have already mentioned, we are not pursuing optimality as far as the value of $\mu_0$ is concerned, and most of our following estimates finally leading to the smallness condition of the Extension lemma can probably be improved. During the proof we'll be concerned with the unstable whisker only, that's why we will further drop the superscript $u$. As one can see from the unperturbed solution (\ref{Aunp}) the function $y_0(\tau)$ has simple poles at \begin{equation} \tau={\iota\over\sqrt{\varepsilon}}\left({\pi\over 2}+m\pi\right) \label{poles} \end{equation} \noindent for $m\in Z$. We denote \[ p_m={1\over\sqrt{\varepsilon}}\left({\pi\over 2}+m\pi\right). \] \noindent We recall that by Theorem \ref{KAM}, for $\Re\tau\leq -T^*$ and $|\Im\vec{\alpha}|\leq\sigma_1= \sigma_0-{1\over 2}\sqrt{\varepsilon}$, the phase trajectories on the perturbed unstable whisker are $O\left({\mu\over\mu_0}\right)$ close to the corresponding unperturbed ones. Recall the definition of the variations (\ref{var}), for which we will also use the following notation: \[ \delta\tilde{\Gamma}(\tau,\vec{\alpha},t_0)=(\xi,\eta,\vec{\varphi},\vec{\zeta}) =\tilde{\Gamma} (\tau,\vec{\alpha},t_0)-\tilde{\Gamma}_0(\tau,\vec{\alpha},t_0), \] \noindent The momenta components $(\eta,\vec{\zeta})$ are the time-derivatives of the coordinate components $(\xi,\vec{\varsigma})$, the evolution of the latter being described by the second order system of ODE's (\ref{2sys}). We will further insure by our choice of $\mu_0$ that all the variations remain small during their time-evolution. We follow Delshams and Seara [1992] in choosing the fundamental solution of the linearized system (\ref{2lin}). Namely we take $\xi_1(\tau)={1\over\cosh \sqrt{\varepsilon}\tau}$ for the first solution; as for the second one (which we will still denote $\xi_2(\tau)$ independently from Section \ref{expsm}), we do the following: for $ p_m-{\pi\over2\sqrt{\varepsilon}}\leq\Im\tau 0$. Just like (\ref{global}) (see the notation of (\ref{Aunp}), (\ref{ash}), and the convention of Section \ref{expsm}, according to which we omit the parameter dependencies), we study the equivalent system of integral equations (this time over a finite time interval): \begin{equation} \begin{array}{ll} \xi(\tau,\tau_0) = \xi^{in}(\tau,\tau_0)+\int^\tau_{\tau_0}G(\tau, s) [\varepsilon h(s,\xi)-\mu g(x_0(s)+\xi(s),\vec{\varphi}_0(s)+\vec{\varsigma}(s))ds, \\ \hfill \\ \vec{\varsigma}(\tau,\tau_0) = \vec{\varsigma}(\tau_0)+(\tau-\tau_0)\vec{\zeta}(\tau_0) - \mu\int^\tau_{\tau_0} (\tau-s)\vec{f}(x_0(s)+\xi(s),\vec{\varphi}_0 (s)+\vec{\varsigma}(s))ds, \end{array} \label{ieq} \end{equation} \noindent where \begin{equation} G(\tau,\tau_0) = \xi_2(\tau)\xi_1(\tau_0)-\xi_1(\tau)\xi_2(\tau_0), \label{grn} \end{equation} \noindent and \begin{equation} \xi^{in}(\tau,\tau_0) = -{\partial\over\partial\tau_0} G(\tau,\tau_0)\xi(\tau_0) + G(\tau,\tau_0)\eta(\tau_0). \nonumber \\ \label{ini} \end{equation} \noindent The integration in (\ref{ieq}) is carried out along horizontal lines in the complex plane, in other words, $\Im\tau=\Im\tau_0=-\Im t_0$. It's useful to notice that the "worst" possible values for the arguments of $G(\tau,\tau_0)$ when $\tau,\tau_0\in\Sigma'$ are $\tau_0=\pm T^*,\tau =\iota p_* =\iota({\pi\over 2\sqrt{\varepsilon}}-\sqrt{\varepsilon})$. Thus, for $\tau,\tau_0\in\Sigma'$, increasing $K_0$ if necessary, we have: \begin{equation} \max_{\tau,\tau_0\in\Sigma'}\left(|G(\tau,\tau_0)|,\left|{\partial\over\partial\tau} G(\tau,\tau_0)\right|\right)\leq K_0\varepsilon^{-{11\over4}}. \label{gres} \end{equation} \noindent We'll use the method of iteration to ascertain the existence and uniqueness of the solution of (\ref{ieq}) and, consequently, (\ref{flow}). Mathematically speaking, we will consider a Banach space ${\cal B}_1$ of functions of a real variable $t$, analytic in a complex parameter $t_0$ such that $\tau\in\Sigma'$ for $\tau=t-t_0$, also analytic in $\vec{\alpha}$ for $|\Im\vec{\alpha}|\leq \sigma_*$, equipped with the $sup$-norm. We'll show that for $|\mu|\leq\mu_0$, the latter to be defined, the nonlinear operator, defined by (\ref{ieq}), is uniformly bounded and is a contraction operator, proving the Existence and Uniqueness of the solution of (\ref{2sys}) in ${\cal B}_1$. We'll start the iterations by letting $\xi^0,\eta^0,\vec{\varsigma}^0,\vec{\zeta}^0$ be identically zero. Then we can write down the explicit expressions for the first iterates. \begin{equation} \begin{array}{l} \xi^1 = \xi^{in}-\mu\int^\tau_{\tau_0}G(\tau,s) g(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0))ds, \\ \hfill \\ \vec{\varsigma}^1 = \vec{\varsigma}(\tau_0)+\vec{\zeta}(t_0) (\tau-\tau_0) -\mu\int_{\tau_0}^\tau (\tau-s) \vec{f}(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0))ds. \end{array} \label{frstitr} \end{equation} \noindent Recall that $\vec{\varphi}_0=\alpha+\omega t$ >From the KAM theorem the values of $\xi,\eta,\vec{\zeta},\vec{\varsigma}$ at "time" $\tau_0=-T^*$ for the whole admissible range of parameters are $O\left(\mu\over\mu_0\right)$, the latter satisfying (\ref{choiceofp}). Then from (\ref{gres}) we'll have \begin{equation} \begin{array}{lll} |\xi^{in}|, \ |\dot{\xi}^{in}|&=&O\left({\mu\over\mu_0\varepsilon^{11\over4}}\right). \end{array} \label{estini} \end{equation} \noindent Moreover, since \begin{equation} \begin{array}{lll} \sin{x_0(\tau)}&=& {2\sinh\sqrt{\varepsilon}\tau\over\cosh^2\sqrt{\varepsilon}\tau}, \\ \hfill \\ \cos x_0(\tau)&=&1-{2\over\cosh^2\sqrt{\varepsilon}\tau}, \end{array} \label{sincos} \end{equation} \noindent then once again increasing $K_0$ if necessary we have for $\tau\in\Sigma'$ \begin{equation} \begin{array}{lll} |\sin x_0(\tau)|, |\cos x_0(\tau)|&\leq &K_0(1+|\xi_1^2(\tau)|) \\ \hfill \\ |g(x_0(\tau),\vec{\varphi}_0(\tau))|&\leq& K_0(1+|\xi_1^{2\nu_0}(\tau)|) \\ \hfill \\ |\vec{f}(x_0(\tau),\vec{\varphi}_0(\vec{\alpha},\vec{\omega},t))|&\leq& K_0{1\over\sqrt{\varepsilon}}(1+|\xi_1^{2\nu_0}(\tau)|) \end{array} \label{est24} \end{equation} \noindent for $\vec{\alpha}\in W_{\sigma_1}T^{n-1}$. The factor ${1\over\sqrt{\varepsilon}}$ is due to the fact that initially we have an estimate for the norm of the perturbation $F(x,\vec{\varphi})$ in (\ref{Aham}) only. Now, having the expressions (\ref{frstitr}) for the first iterates we need several estimates for the integrals involved. They will easily follow from the expressions for $\xi_1$ and $\xi_2$, and the following essentially obvious formula that we borrow from Delshams and Seara [1992]. Consider two complex variables $w_1,w_2$ such that $\Im w_1=\Im w_2$, $\Re w_2>\Re w_1$ and also a real number $p$. Denote for $l=1,2,...$ \[ \rho^{-l}_{[p,w_1,w_2]}\equiv\sup_{\Re w_1\leq\Re s\leq \Re w_2,\ \Im s=\Im w_1} {1\over\left|s-\iota p\right|^l}. \] \noindent For $l=..,-1,0$ we'll let $\rho^{-l}_{[p,w_1,w_2]}=1$. Then the following Proposition is easy to verify: \begin{proposition} For $l=1,2..$ and the constants $0 \Re\tau_0$, $\Im\tau_0=\Im\tau=-\Im t_0$. and $\vec{\alpha}$ such that $|\Im\vec{\alpha}|\leq\sigma_1$, there exist independent of $\varepsilon,\mu$ positive constants $K_1,K_2,K_3,K_4$, such that the following once differentiable in $\tau$ estimates take place: \[ \begin{array}{lll} \left| \int^\tau_{\tau_0} G(\tau,s) g(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0)ds \right| &\leq & K_1 {1\over\sqrt{\varepsilon}} \xi_1(\tau) \left(\rho^{-2\nu_0+3}_{[{\pi\over2},\sqrt{\varepsilon}\tau_0,\sqrt{\varepsilon}\tau]}+1\right) \\ \hfill \\ &+& K_2 {1\over\sqrt{\varepsilon}} \xi_2(\tau) \left(\rho^{-2\nu_0}_{[{\pi\over2},\sqrt{\varepsilon}\tau_0,\sqrt{\varepsilon}\tau]}+1\right), \\ \hfill \\ \left| \int_{t_0}^t (\tau-s) \vec{f}\left(x_0(s),\vec{\alpha}+\vec{\omega} (s+t_0)\right)ds \right| &\leq& K_3 \tau {1\over\varepsilon} \left(\rho^{-2\nu_0+1}_{[{\pi\over2},\sqrt{\varepsilon}\tau_0,\sqrt{\varepsilon}\tau]}+1\right) \\ \hfill \\ &+& K_4 {1\over\varepsilon\sqrt{\varepsilon}} \left(\rho^{-2\nu_0+2}_{[{\pi\over2},\sqrt{\varepsilon}\tau_0,\sqrt{\varepsilon}\tau]}+ 1\right), \end{array} \] \label{pr2} \end{proposition} \noindent {\bf Proof:} We note that $g(x,\vec{\varphi})$ is a trigonometric polynomial in $x$ of degree $\nu_0$. The proof follows if we simultaneously apply (\ref{grn}), (\ref{est24}) and Proposition \ref{pr1}, where we change $\sqrt{\varepsilon}\tau\rightarrow\tau$ under the integral signs.$\Box$ \medskip \noindent Again, let's redefine $K_0=\max(K_0,K_1,K_2,K_3,K_4)$. Proposition \ref{pr2} admits an easy corollary that yields the estimates in powers of $\varepsilon$. If $|\Re\tau|\leq \varepsilon^{1\over4}$ and ${\pi\over2\sqrt{\varepsilon}}-\varepsilon^{1\over4}\leq\Im\tau\leq p_* $, one can use the fact that $\xi_2$ is small near $\iota p$, since it has a second-order zero at this point, namely $\xi_2\leq K_0 \varepsilon$, again increasing $K_0$ if necessary. Otherwise, if either $|\Re\tau|\geq\varepsilon^{1\over4}$ or $\Im\tau\leq {\pi\over2\sqrt{\varepsilon}}-\varepsilon^{1\over4}$, we clearly have $\rho^{-l}_{[{\pi\over2},\sqrt{\varepsilon}\tau_0,\sqrt{\varepsilon}\tau]}\leq\varepsilon^{-{3l\over4}}$. \medskip \noindent This leads to the following simple fact: \begin{proposition} For $\tau_0,\tau,\vec{\alpha}$ as in Proposition \ref{pr2} the following estimates take place for $\nu_0\geq 1$: \[ \begin{array}{lll} \left| \int^\tau_{\tau_0} G(\tau,s) g\left(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0)\right)ds \right| &\leq &K_0\varepsilon^{-2\nu_0+{1\over2}}, \\ \hfill \\ \left| \int^\tau_{\tau_0} {\partial\over\partial \tau}G(\tau,s) g\left(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0)\right)d\sigma \right| &\leq &K_0 \varepsilon^{-2\nu_0+{1\over2}}, \\ \hfill \\ \left| \int_{\tau_0}^\tau (\tau-s) \vec{f}\left(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0)\right)ds \right| &\leq &K_0\varepsilon^{-2\nu_0-{1\over2}}, \\ \hfill \\ \left| \int_{t_0}^t \vec{f}\left(x_0(s),\vec{\alpha}+ \vec{\omega}(s+t_0)\right)d\sigma \right| &\leq & K_0\varepsilon^{-2\nu_0}. \end{array} \] \label{pr3} \end{proposition} \medskip \noindent The next Proposition gives us the bounds for the same integrals away from the pole when the reason for concern is the growth of $\xi_2$. \begin{proposition} Given $\tau_0, \tau$ from the same connected component of $\Sigma'\setminus\Sigma''$ such that $\Re\tau_0<\Re\tau,\ \Im\tau_0=\Im\tau$, and $\vec{\alpha}$ such that $|\Im\vec{\alpha}|\leq\sigma_1$, the following estimates take place for $\varepsilon$ small enough. \[ \begin{array}{lll} \left| \int^\tau_{\tau_0} G(\tau,s) g\left(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0) \right)ds \right| &\leq &K_0\varepsilon^{-1}, \\ \hfill \\ \left| \int^\tau_{\tau_0} {\partial\over\partial t}G(\tau,s) g(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0))ds \right| &\leq &K_0\varepsilon^{-1}, \\ \hfill \\ \left| \int_{t_0}^t (\tau-s)\vec{f}(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0))ds \right| &\leq& K_0\varepsilon^{-2}, \\ \hfill \\ \left| \int_{t_0}^t \vec{f}(x_0(s),\vec{\alpha}+\vec{\omega}(s+t_0))ds \right| &\leq &K_0\varepsilon^{-2}. \end{array} \] \label{pr4} \end{proposition} \noindent {\bf Proof:} Obvious from (\ref{est24}) and the definition of $T^*$, if one does not hesitate to increase $K_0$. $\Box$ \medskip \noindent Now we can work out the bounds for the first iterates. We fix $\tau_0=-T^*$. It's easy to see that in the worst possible case the first iterates (\ref{frstitr}) can be bounded as follows: \[ \begin{array}{lll} |\xi^1|,\ |\eta^1| &\leq & K_0|\mu| \left(\mu_0^{-1}\varepsilon^{-{11\over4}}+ \varepsilon^{-2\nu_0+{1\over2}}\right), \\ \hfill \\ |\vec{\varsigma}^1|,\ |\vec{\zeta}^1| &\leq & K_0|\mu|\left(\mu_0^{-1}\varepsilon^{-1 } +\varepsilon^{-2\nu_0-1}\right), \\ \end{array} \] \noindent where $\mu_0$ comes from (\ref{choiceofp}). Therefore, we can redefine \begin{equation} \mu_0=\varepsilon^{ \max({n\over2}+{13\over4}+ \varpi+2a+\delta, \ 2\nu_0+{1\over2}+\delta}) \label{munotone} \end{equation} \noindent to insure that the first iterates are smaller than $1$ for $\tau\in\Sigma',\ |\Im\vec{\alpha}|\leq\sigma_1$. We now define the iterative process as: \begin{equation} \delta\tilde{\Gamma}^{l+1}(\tau,\vec{\alpha},t_0)={\bar{\cal I}}\delta\tilde{\Gamma}^{l}, \ l=0,1,.., \label{itr} \end{equation} \noindent where $\bar{{\cal I}}$ is the non-linear operator, corresponding to the system of integral equations (\ref{ieq}). First of all for \[ h(\tau,\xi)=[\sin x_0(\tau)(\cos\xi-1)+\cos x_0(\tau)(\xi-\sin\xi)] \] \noindent we will clearly have: \begin{eqnarray} |h(\tau,\xi^{l})-h(\tau,\xi^{l-1})| & \leq & K_0|\xi_1^2(\tau)| (|\xi^{l-1}|+|\xi^{l}|) \cdot|\xi^{l}-\xi^{l-1}| \nonumber \\ |h(\tau,\xi)| & \leq & K_0|\xi_1^2(\tau)||\xi|^2. \label{ges} \end{eqnarray} \noindent Therefore, repeating the estimates that we have just performed, we can convince ourselves that there exists a ball in ${\cal B}_1$ near zero of the radius $\mu_0\sqrt{\varepsilon}$, the norm of $\bar{\cal I}$ is uniformly bounded by $1$, where $\mu_0$ comes from (\ref{munotone}). Besides if $\delta\tilde{\Gamma}^{l-1}, \delta\tilde{\Gamma}^l$ are the two subsequent iterates (lying in the ball above), then one can write up the following estimates for $|\Im\vec{\alpha}|\leq \sigma_2 $: \begin{equation} \begin{array}{lll} \left| g(x_0+\xi^{l}, \vec{\varphi}_0+\vec{\varsigma}^l) - g(x_0+\xi^{l-1}, \vec{\varphi}_0+\vec{\varsigma}^{l-1}) \right| & \leq & \varepsilon^{-{1\over2}}K_0(1+|\xi_1|)^{2\nu_0} \\ \hfill \\ & \times & \left(|\xi^{l}-\xi^{l-1}|+ |\vec{\varsigma}^{l}-\vec{\varsigma}^{l-1}| \right), \end{array} \label{dif1} \end{equation} \noindent and \begin{equation} \begin{array}{lll} \left| \vec{f}(x_0+\xi^{l},\vec{\varphi}_0+\vec{\varsigma}^l) - \vec{f}(x_0+\xi^{l-1},\vec{\varphi}_0+ \vec{\varsigma}^{l-1}) \right| & \leq & K_0 \varepsilon^{-1} (1+|\xi_1|)^{2\nu_0} \\ \hfill \\ & \times &\left(|\xi^{l}-\xi^{l-1}|+ |\vec{\varsigma}^{l}-\vec{\varsigma}^{l-1}| \right). \end{array} \label{dif2} \end{equation} \noindent Using (\ref{ieq}), (\ref{ges}), (\ref{dif1}), (\ref{dif2}), and Proposition \ref{pr3}, we can write, having possibly increased the constant $K_0$: \[ \begin{array}{lll} |\xi^{l+1}-\xi^l| &\leq & K_0 \varepsilon [(|\xi^l|+|\xi^{l-1}|)|\xi^l-\xi^{l-1}| \\ \hfill \\ &+&{|\mu|\over\mu_0\sqrt{\varepsilon}} (|\xi^l-\xi^{l-1}|+ |\vec{\varsigma}^l-\vec{\varsigma}^{l-1}|)] \end{array} \] \noindent and \[ \begin{array}{lll} |\vec{\varsigma}^{l+1}-\vec{\varsigma}^l| &\leq &K_0{|\mu|\over\mu_0\sqrt{\varepsilon}} [|\xi^l-\xi^{l-1}|+ |\vec{\varsigma}^l-\vec{\varsigma}^{l-1}|]. \end{array} \] \noindent The same estimates hold for the derivatives $\eta$ and $\vec{\zeta}$. This implies that for $\varepsilon$ small enough and $|\mu|\leq\mu_0\sqrt{\varepsilon}$, where $\mu_0$ is given by (\ref{munotone}) $\bar{\cal I}$ is a contraction operator ($\delta$ in (\ref{munotone}) takes care of all the constants, as usual), and therefore, the solution on the unstable manifold exists as is unique, as stated in the Lemma. Finally, we notice that the first iterates were $O\left({\mu\over\mu_0}\right)$, therefore if we multiply $\mu_0$ by $\varepsilon^{{1\over2}}$, it will insure that $\vec{\varsigma}$ will never become larger than ${1\over4}\sqrt{\varepsilon}$, so we never find ourselves outside the analyticity domain of $\vec{f}$ (owing to $\delta$ in the expression for $\mu_0$). This completes the proof of the Extension lemma.$\Box$ \section*{References} \begin{verse} Arnold, V.I. [1964] Instability of dynamical systems with several degrees of freedom. {\em Sov. Math Dokl.}, {\bf 5}, 581-585. Benettin, G., Galgani, L., Giorgilli, A. 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