\documentclass{article}% \usepackage{graphicx} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb}% \setcounter{MaxMatrixCols}{30} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=4.00.0.2312} %TCIDATA{CSTFile=LaTeX article (bright).cst} %TCIDATA{Created=Tue Apr 06 14:54:02 2004} %TCIDATA{LastRevised=Monday, February 28, 2005 11:14:10} %TCIDATA{} %TCIDATA{} \newtheorem{theorem}{Theorem} {} \newtheorem{acknowledgement}{Acknowledgement}[section] \newtheorem{algorithm}{Algorithm}[section] \newtheorem{axiom}{Axiom}[section] \newtheorem{case}{Durum}[section] \newtheorem{claim}{Iddia}[section] \newtheorem{conclusion}{Conclusion}[section] \newtheorem{condition}{Condition}[section] \newtheorem{conjecture}{Conjecture}[section] \newtheorem{corollary}{Sonu\c{c}}[section] \newtheorem{criterion}{Criterion}[section] \newtheorem{definition}{Tan\i m}[section] \newtheorem{example}{\"{O}rnek}[section] \newtheorem{exercise}{Al\i \c{s}t\i rma}[section] \newtheorem{lemma}{Lemma} {} \newtheorem{notation}{Notation}[section] \newtheorem{problem}{ } \newtheorem{proposition}{Proposition}[section] \newtheorem{remark}{Remark} \newtheorem{solution}{Solution}[section] \newtheorem{summary}{Summary}[section] \newenvironment{proof}[Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \renewcommand{\baselinestretch}{1.0} \oddsidemargin 1.0cm \evensidemargin 1.0cm \voffset -1cm \topmargin 0.1cm \headheight 0.5cm \headsep 1.5cm \begin{document} \author{O. A. Veliev\\{\small \ Dept. of Math, Fen-Ed. Fak, Dogus University.,}\\{\small Acibadem, Kadikoy, Istanbul, Turkey,}\\{\small \ e-mail: oveliev@dogus.edu.tr}} \title{\textbf{Asymptotic Analysis of the Periodic Schrodinger Operator}} \date{} \maketitle \begin{abstract} In this paper we obtain asymptotic formulas of arbitrary order for the Bloch eigenvalues and Bloch functions of the Schrodinger operator $-\Delta+q(x),$ of arbitrary dimension, with periodic, with respect to arbitrary lattice, potential $q(x)$. Moreover, we estimate the measure of the isoenergetic surfaces in the high energy region. \end{abstract} \section{Introduction} \bigskip In this paper we consider the operator% \begin{equation} L(q(x))=-\Delta+q(x),\ x\in\mathbb{R}^{d},\ d\geq2 \end{equation} with a periodic (relative to a lattice $\Omega$) potential $q(x)\in W_{2}% ^{s}(F),$ where $s\geq s_{0}=\frac{3d-1}{2}(3^{d}+d+2)+\frac{1}{4}d3^{d}+d+6,$ $F\equiv \mathbb{R}^{d}/\Omega$ is a fundamental domain of $\Omega.$ Without loss of generality it can be assumed that the measure $\mu(F)$ of $F$ is $1$ and $\int_{F}q(x)dx=0.$ Let $L_{t}(q(x))$ be the operator generated in $F$ by (1) and the conditions:% \begin{equation} u(x+\omega)=e^{i(t,\omega)}u(x),\ \forall\omega\in\Omega, \end{equation} where $t\in F^{\star}\equiv\mathbb{R}^{d}/\Gamma$ and $\Gamma$ is the lattice dual to $\Omega$, that is, $\Gamma$ is the set of all vectors $\gamma \in\mathbb{R}^{d}$ satisfying $(\gamma,\omega)\in2\pi Z$ for all $\omega \in\Omega.$ It is well-known that ( see ) the spectrum of the operator $L_{t}(q(x))$ consists of the eigenvalues $\Lambda_{1}(t)\leq\Lambda_{2}(t)\leq....$The function $\Lambda_{n}(t)$ is called $n$-th band function and its range $A_{n}=\left\{ \Lambda_{n}(t):t\in F^{\ast}\right\}$ is called the $n$-th band of the spectrum $Spec(L)$ of $L$ and $Spec(L)=\cup_{n=1}^{\infty}A_{n}$. The eigenfunction $\Psi_{n,t}(x)$ of $L_{t}(q(x))$ corresponding to the eigenvalue $\Lambda_{n}(t)$ is known as Bloch functions. In the case $q(x)=0$ these eigenvalues and eigenfunctions are $\mid\gamma+t\mid^{2}$ and $e^{i(\gamma+t,x)}$ for $\gamma\in\Gamma$. This paper consists of 4 section. First section is the introduction, where we describe briefly the scheme of this paper and discuss the related papers. In papers [13-17] for the first time the eigenvalues $\left\vert \gamma+t\right\vert ^{2}$, for big $\ \gamma\in\Gamma,$ were divided into two groups: non-resonance ones and resonance ones and for the perturbations of each group various asymptotic formulae were obtained. Let the potential $q(x)$ be a trigonometric polynomial% $\sum_{\gamma\in Q}q_{\gamma}e^{i(\gamma,x)},$ where $q_{\gamma}=(q(x),e^{i(\gamma,x)})=\int_{F}q(x)e^{-i(\gamma_{1},x)}dx,$ and $Q$ consists of a finite number of vectors $\gamma$ from $\Gamma.$ Then the eigenvalue $\left\vert \gamma+t\right\vert ^{2}$ \ is called a non-resonance eigenvalue if $\gamma+t$ does not belong to any of the sets $\{x\in\mathbb{R}^{d}:\mid\mid x\mid^{2}-\mid x+b\mid^{2}\mid<\mid x\mid^{\alpha_{1}}\},$ that is, if $\gamma+t$ lies far from the diffraction hyperplanes $\{x\in\mathbb{R}^{d}:\mid x\mid^{2}=\mid x+b\mid^{2}\},$ where $\alpha_{1}\in(0,1),$ $b\in\{b_{1}+b_{2}+...b_{m}:b_{1},b_{2},...b_{m}\in Q\},$ and $m$ is fixed integer (see [15-17]). If $q(x)\in W_{2}^{s}(F),$ then to describe the non-resonance and resonance eigenvalues $\left\vert \gamma+t\right\vert ^{2}$ of the order of $\rho^{2}$ ( written as $\left\vert \gamma+t\right\vert ^{2}\sim\rho^{2}$) for big paramater $\rho$ we write the potential $q(x)\in W_{2}^{s}(F)$ in the form \begin{equation} q(x)=\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}q_{\gamma_{1}}e^{i(\gamma _{1},x)}+O(\rho^{-p\alpha}), \end{equation} where $\Gamma(\rho^{\alpha})=\{\gamma\in\Gamma:0<$ $\mid\gamma\mid <\rho^{\alpha})\}$, $p=s-d,$ $\alpha=\frac{1}{q},$ $q=3^{d}+d+2,$ and the relation $\left\vert \gamma+t\right\vert ^{2}\sim\rho^{2}$ means that $c_{1}\rho<\left\vert \gamma+t\right\vert 1)$ is explicitly expresed by the potential $q(x)$ and eigenvalues of $L_{t}(0).$ Besides, we prove that if the conditions \begin{align} & \mid\Lambda_{N}(t)-\mid\gamma+t\mid^{2}\mid<\frac{1}{2}\rho^{\alpha_{1}},\\ & \mid b(N,\gamma)\mid>c_{4}\rho^{-c\alpha}% \end{align} hold, \ where $b(N,\gamma)=(\Psi_{N,t},e^{i(\gamma+t,x)}),$ $\Psi_{N,t}(x)$ is a normalized eigenfunction of $L_{t}(q(x))$ corresponding to $\Lambda_{N}(t),$ then $\$the following statements are valid: (a) if $\gamma+t$ is in the non-resonance domain, then $\Lambda_{N}(t)$ satisfies (5) for $k=1,2,...,[\frac{1}{3}(p-c)]$ ( see Theorem 1); (b) if $\gamma+t\in E_{s}\backslash E_{s+1},$ where $s=1,2,...,d-1,$ then% \begin{equation} \Lambda_{N}(t)=\lambda_{j}(\gamma+t)+O(\mid\gamma+t\mid^{-k\alpha}), \end{equation} where $\lambda_{j}$ is an eigenvalue of the matrix $C(\gamma+t)$ ( see (26) and Theorem 2). Moreover, we prove that every big eigenvalue of the operator $L_{t}(q(x))$ for all values of quasimomenta $t$ satisfies one of these formulae. For investigation of the Bloch function, in section 3, we find the values of quasimomenta $\gamma+t$ for which the corresponding eigenvalues are simple , namely we construct the subset $B$ of $U(\rho^{\alpha_{1}},p)$ with the following properties: Pr.1. If $\gamma+t\in B,$ then there exists a unique eigenvalue, denoted by $\Lambda(\gamma+t),$ of the operator $L_{t}(q(x))$ satisfying (5). This is a simple eigenvalue of $L_{t}(q(x))$. Therefore we call the set $B$ the simple set of quasimomenta. Pr.2. The eigenfunction $\Psi_{N(\gamma+t)}(x)\equiv\Psi_{\gamma+t}(x)$ corresponding to the eigenvalue $\Lambda(\gamma+t)$ is close to $e^{i(\gamma +t,x)}$, namely \begin{equation} \Psi_{N}(x)=e^{i(\gamma+t,x)}+O(\mid\gamma+t\mid^{-\alpha_{1}}), \end{equation}% \begin{equation} \Psi_{\gamma+t}(x)=e^{i(\gamma+t,x)}+\Phi_{k-1}(x)+O(\mid\gamma+t\mid ^{-k\alpha_{1}}),\text{ }k=1,2,...\text{ ,}% \end{equation} where $\Phi_{k-1}$ is explicitly expresed by $q(x)$ and the eigenvalues of $L_{t}(0).$ Pr.3. The set $B$ contains the intervals $\{a+sb:s\in\lbrack-1,1]\}$ such that $\Lambda(a-b)<\rho^{2},$ $\Lambda(a+b)>\rho^{2},$\ and $\Lambda(\gamma+t)$ \ is continuous on these intervals. Hence there exists $\gamma+t$ such that $\Lambda(\gamma+t)=\rho^{2}$ for $\rho\gg1.$ It implies that there exist only a finite number of gaps in the spectrum of $L,$ that is, it implies the validity of Bethe-Sommerfeld conjecture for arbitrary dimension and for arbitrary lattice. Construction of the set $B$ consists of two steps. Step 1. We prove that all eigenvalues $\Lambda_{N}(t)\sim\rho^{2}$ of the operator $L_{t}(q(x))$ lie in the $\varepsilon_{1}=\rho^{-d-2\alpha}$ neighborhood of the numbers $F(\gamma+t)=\mid\gamma+t\mid^{2}+F_{k_{1}-1}(\gamma+t)$, $\lambda_{j}% (\gamma+t)$ ( see (5), (8)), where $k_{1}=[\frac{d}{3\alpha}]+2.$ We call these numbers as the known parts of the eigenvalues. Moreover, for $\gamma+t\in U(\rho^{\alpha_{1}},p)$ there is $\Lambda_{N}(t)$ satisfying $\Lambda_{N}(t)=F(\gamma+t)+o(\varepsilon_{1})$ ( see (5)) Step 2. By eliminating the set of quasimomenta $\gamma+t$, for which the known parts $F(\gamma+t)$ of $\Lambda_{N}(t)$ are situated from the known parts $F(\gamma^{^{\prime}}+t),$ $\lambda_{j}(\gamma^{^{\prime}}+t)$ ($\gamma ^{^{\prime}}\neq\gamma)$ of other eigenvalues at a distance less than $2\varepsilon_{1},$ we construct the set $B$ with the following properties: if $\gamma+t\in B,$ then the following conditions (called simplicity conditions for $\Lambda_{N}(t))$ hold \begin{equation} \mid F(\gamma+t)-F(\gamma^{^{\prime}}+t)\mid\geq2\varepsilon_{1}\text{ }% \end{equation} for $\gamma^{^{\prime}}\in K\backslash\{\gamma\},$ $\gamma^{^{\prime}}+t\in U(\rho^{\alpha_{1}},p)$ and% \begin{equation} \mid F(\gamma+t)-\lambda_{j}(\gamma^{^{\prime}}+t)\mid\geq2\varepsilon_{1}% \end{equation} for $\gamma^{^{\prime}}\in K,\gamma^{^{\prime}}+t\in E_{k}\backslash E_{k+1},j=1,2,...,$ where $K$ is the set of $\gamma^{^{\prime}}\in\Gamma$ satisfying $\mid F(\gamma+t)-\mid\gamma^{^{\prime}}+t\mid^{2}\mid<\frac{1}% {3}\rho^{\alpha_{1}}$. Thus $B$ is the set of $\gamma+t\in U(\rho^{\alpha_{1}% },p)$ satisfying the simplicity conditions (11), (12). As a consequence of these conditions the eigenvalue $\Lambda_{N}(t)$ does not coincide with other eigenvalues. To prove this, namely to prove the Pr.1 and (9), we show that for any normalized eigenfunction $\Psi_{N}(x)$ corresponding to $\Lambda_{N}(t)$ the following equality holds: \begin{equation} \sum_{\gamma^{^{\prime}}\in\Gamma\backslash\gamma}\mid b(N,\gamma^{^{\prime}% })\mid^{2}=O(\rho^{-2\alpha_{1}}). \end{equation} For the first time in [15-17] we constructed the simple set $B$ with the Pr.1 and Pr.3., though in those papers we emphasized the Bethe-Zommerfeld conjecture. Note that for this conjecture and for Pr.1, Pr.3. it is enough to prove that the left-hand side of (13) is less than $\frac{1}{4}$ ( we proved this inequality in [15-17] and as noted in Theorem 3 of  and in  the proof of this inequality does not differ from the proof of (13)). From (9) we got \ (10) (see ) . But in those papers these results are written briefly. The enlarged variant is written in  which can not be used as reference. In this paper we write these results in improved and enlarged form. The main difficulty and the crucial point of papers [15-17] were the construction of the simple set $B$ with the Pr.1.,Pr.3. This difficulty of the perturbation theory of $L(q(x))$ is of a physical nature and it is connected with the complicated picture of the chystal diffraction. If $d=2,3,$ then $F(\gamma+t)=\mid\gamma+t\mid^{2}$ and the matrix $C(\gamma+t)$ corresponds to the Schrodinger operator with directional potential $q_{\delta}(x)=\sum_{n\in Z}q_{n\delta}e^{in(\delta,x)}$ ( see ). So for construction of the simple set $B$ of quasimomenta we eliminated the vicinities of the diffraction planes and the sets connected with directional potential ( see (11), (12)). Besides,\ for nonsmooth potentials $q(x)\in L_{2}(\mathbb{R}^{2}/\Omega),$ we eliminated a set, which is described in the terms of the number of states ( see [15,19]). The simple sets $B$ of quasimomenta for the first time is constructed and investigated ( hence the main difficulty and the crucial point of perturbation theory of $L(q)$ is investigated) in  for $d=3$ and in [15,17] for the cases: 1. $d=2,$ $q(x)\in L_{2}(F);$ \ \ \ 2. $d>2,$ $q(x)$ is a smooth potential. Then, Yu.E. Karpeshina proved ( see [7-9]) the convergence of the perturbation series of two and three dimensional Schrodinger operator $L(q)$ with a wide class of nonsmooth potential $q(x)$ for a set, that is similar to $B$, of quasimomenta. In papers [3,4] the asymptotic formulas for the eigenvalues and Bloch function of the two and three dimensional operator $L_{t}(q(x))$ were obtained. In  the asymptotic formulae for the eigenvalues of $L_{0}(q(x))$ were obtained. In section 4 we consider the geometrical aspects of the simple sets. We prove that the simple sets $B$ has asymptotically full measure on $\mathbb{R}^{d}$. Moreover we construct a part of isoenergetic surfaces corresponding to $\rho^{2},$ which is smooth surfaces and has the measure asymptotically close to the measure of the Fermi surfaces $\{x\in R:\mid x\mid=\rho\}$ of the operator $L(0).$ The nonemptyness of the Fermi surfaces \ for $\rho\gg1$ implies the the validity of the Bethe-Sommerfeld conjecture. For the first time M.M. Skriganov [11,12] proved the validity of the Bethe-Sommerfeld conjecture for the Scrodinger operator for dimension $d=2,3$ for arbitrary lattice, for dimension $d>3$ for rational lattice. The Skriganov's method is based on the detail investigation of the arithmetic and geometric properties of the lattice. B.E.J.Dahlberg \ and E.Trubowits  using an asymptotic of Bessel function, gave the simple proof of this conjecture for the two dimensional Scrodinger operator. Then in papers [15-17] we proved the validity of the Bethe-Sommerfeld conjecture for arbitrary lattice and for arbitrary dimension by using the asymptotic formulas and by construction of the simple set $B,$ that is, by\ the method of perturbation theory. Yu.E. Karpeshina ( see [7-9]) proved this conjecture for two and three dimensional Schrodinger operator $L(q)$ for a wide class of singular potentials $q(x),$ including Coulomp potential, by \ the method of perturbation theory. B. Helffer and A. Mohamed , by investigations the integrated density of states, proved the validity of the Bethe-Sommerfeld conjecture for the Scrodinger operator for $d\leq4$ for arbitrary lattice. Recently L. Parnovski and A. V. Sobelev  proved this conjecture for $d\leq4.$ The method of this paper and papers [15-17] is a first and uniqie, for the present, by which the validity of the Bethe-Sommerfeld conjecture for arbitrary lattice and for arbitrary dimension is proved. In this paper for the different types of the measures of the subset $A$ of $\mathbb{R}^{d}$ we use the same notation $\mu(A).$ By $\mid A\mid$ we denote the number of elements of the set $A\subset\Gamma$ and use the following obvious fact. If $a\sim\rho,$ then the number of elements of the set $\{\gamma+t:$ $\gamma\in\Gamma\}$ satisfying $\mid\mid\gamma+t\mid-a\mid<1$ is less than $c_{5}\rho^{d-1}.$ Therefore the number of eigenvalues of $L_{t}(q)$ lying in $(a^{2}-\rho,a^{2}+\rho)$ is less than $c_{5}\rho^{d-1}.$ Besides, we use the inequalities:% \begin{align} \alpha_{1}+d\alpha & <1-\alpha\,,\ \ \ \ \ d\alpha<\frac{1}{2}\alpha _{d},\ \ \ k_{1}\leq\frac{1}{3}(p-\frac{1}{2}(q(d-1)),\\ p_{1}\alpha_{1} & \geq p\alpha,\ \ \ \ \ 3k_{1}\alpha>d+2\alpha ,\ \ \ \ \ \ \alpha_{k}+(k-1)\alpha<1,\nonumber\\ \alpha_{k+1} & >2(\alpha_{k}+(k-1))\alpha\nonumber \end{align} for $k=1,2,...,d,$ which follow from the definitions $p=s-d,$ $\alpha _{k}=3^{k}\alpha,$ $\alpha=\frac{1}{q},$ $q=3^{d}+d+2,$ $k_{1}=[\frac {d}{3\alpha}]+2,$ $p_{1}=[\frac{p}{3}]+1$ of the numbers $p,q,\alpha _{k},\alpha,k_{1},p_{1}.$ \section{Asymptotic Formulae for Eigenvalues} First we obtain the asymptotic formulas for the non-resonance eigenvalues by iteration of the formula% \begin{equation} (\Lambda_{N}-\mid\gamma+t\mid^{2})b(N,\gamma)=(\Psi_{N,t}(x)q(x),e^{i(\gamma +t,x)}), \end{equation} which is obtained from equation $-\Delta\Psi_{N,t}(x)+q(x)\Psi_{N,t}% (x)=\Lambda_{N}\Psi_{N,t}(x)$ by multiplying by $e^{i(\gamma+t,x)}).$ Introducing into (15) the expansion (3) of $q(x)$, we get \begin{equation} (\Lambda_{N}-\mid\gamma+t\mid^{2})b(N,\gamma)=\sum_{\gamma_{1}\in\Gamma (\rho^{\alpha})}q_{\gamma_{1}}b(N,\gamma-\gamma_{1})+O(\rho^{-p\alpha}). \end{equation} From the relations (15), (16) it follows that \begin{equation} b(N,\gamma^{^{\prime}})=\dfrac{(\Psi_{N,t}q(x),e^{i(\gamma^{^{\prime}}+t,x)} % )}{\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2}}=% %TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}b(N,\gamma^{^{\prime}}-\gamma_{1})}{\Lambda_{N}% -\mid\gamma^{^{\prime}}+t\mid^{2}}+O(\rho^{-p\alpha}) \end{equation} for all vectors $\gamma^{^{\prime}}\in\Gamma$ satisfying the inequality \begin{equation} \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2}\mid>\frac{1}{2}\rho ^{\alpha_{1}}. \end{equation} This inequality is called the iterability condition. If (6) holds and $\mid\gamma+t\mid^{2}$ is a non-resonance eigenvalue, i.e., $\gamma+t\in U(\rho^{\alpha_{1}},p),$ then \begin{equation} \mid\mid\gamma+t\mid^{2}-\mid\gamma-\gamma_{1}+t\mid^{2}\mid>\rho^{\alpha_{1 }% },\text{ }\mid\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2}\mid>\frac{1}{2}% \rho^{\alpha_{1}}\text{ }% \end{equation} for all $\gamma_{1}\in\Gamma(p\rho^{\alpha}).$ Hence the vector $\gamma -\gamma_{1}$ for $\gamma+t\in U(\rho^{\alpha_{1}},p)$ and $\gamma_{1}\in \Gamma(p\rho^{\alpha})$ satisfies (18). Therefore, in (17) one can replace $\gamma^{^{\prime}}$ by $\gamma-\gamma_{1}$ and write $b(N,\gamma-\gamma_{1})=% %TCIMACRO{\dsum _{\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}{\Lambda_{N}-\mid \gamma-\gamma_{1}+t\mid^{2}}+O(\rho^{-p\alpha}).$ Substituting this for $b(N,\gamma-\gamma_{1})$ into right-hand side of (16) and isolating the terms containing the multiplicand $b(N,\gamma)$, we get $(\Lambda_{N}-\mid\gamma+t\mid^{2})b(N,\gamma)=% %TCIMACRO{\dsum _{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}% {\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2}}+O(\rho^{-p\alpha})=$% $% %TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{\mid q_{\gamma_{1}}\mid^{2}b(N,\gamma)}{\Lambda_{N}-\mid\gamma -\gamma_{1}+t\mid^{2}}+% %TCIMACRO{\dsum _{\substack{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha %}),\\\gamma_{1}+\gamma_{2}\neq0}}}% %BeginExpansion {\displaystyle\sum_{\substack{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha }),\\\gamma_{1}+\gamma_{2}\neq0}}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}% {\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2}}+O(\rho^{-p\alpha}),$ since $q_{\gamma_{1}}q_{\gamma_{2}}=\mid q_{\gamma_{1}}\mid^{2}$ for $\gamma_{1}+\gamma_{2}=0$ and the last sum is taken under the condition $\gamma_{1}+\gamma_{2}\neq0.$ Repeating this process $p_{1}\equiv\lbrack \frac{p}{3}]+1$ times, i.e., in the last formula replacing $b(N,\gamma -\gamma_{1}-\gamma_{2})$ by its expression from (17) ( in (17) replace $\gamma^{^{\prime}}$ by $\gamma-\gamma_{1}-\gamma_{2}$) and isolating the terms containing $b(N,\gamma)$ etc., we obtain \begin{equation} (\Lambda_{N}-\mid\gamma+t\mid^{2})b(N,\gamma)=A_{p_{1}}(\Lambda_{N}% ,\gamma+t)b(N,\gamma)+C_{p_{1}}+O(\rho^{-p\alpha}), \end{equation} where $A_{p_{1}}(\Lambda_{N},\gamma+t)=\sum_{k=1}^{p_{1}}S_{k}(\Lambda _{N},\gamma+t)$ , $S_{k}(\Lambda_{N},\gamma+t)=% %TCIMACRO{\dsum _{\gamma_{1},...,\gamma_{k}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},...,\gamma_{k}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{k}}q_{-\gamma_{1}-\gamma _{2}-...-\gamma_{k}}}{\prod_{j=1}^{k}(\Lambda_{N}-\mid\gamma+t-\sum_{i=1}% ^{j}\gamma_{i}\mid^{2})},$% $C_{p_{1}}=\sum_{\gamma_{1},...,\gamma_{p_{1}+1}\in\Gamma(\rho^{\alpha})}% \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{p_{1}+1}}b(N,\gamma -\gamma_{1}-\gamma_{2}-...-\gamma_{p_{1}+1})}{\prod_{j=1}^{p_{1}}(\Lambda _{N}-\mid\gamma+t-\sum_{i=1}^{j}\gamma_{i}\mid^{2})}.$ Here the sums for $S_{k}$ and $C_{p_{1}}$ are taken under the additional conditions $\gamma_{1}+\gamma_{2}+...+\gamma_{s}\neq0$ for $s=1,2,...,k$ and $s=1,2,...,p_{1}$ respectively. These conditions and the inclusion $\gamma _{i}\in\Gamma(\rho^{\alpha})$ for $i=1,2,...,p_{1}$ imply the relation $\sum_{i=1}^{j}\gamma_{i}\in\Gamma(p\rho^{\alpha})$. Therefore from the second inequality in (19) it follows that the absolute values of the denominators of the fractions in $S_{k}$ and $C_{p_{1}}$ are greater than $(\frac{1}{2}% \rho^{\alpha_{1}})^{k}$ and $(\frac{1}{2}\rho^{\alpha_{1}})^{p_{1}}$ respectively. Hence the first inequality in (4) and $p_{1}\alpha_{1}\geq p\alpha$ ( see the fourth inequality in (14)) yield \begin{equation} C_{p_{1}}=O(\rho^{-p_{1}\alpha_{1}})=O(\rho^{-p\alpha}),\text{ }S_{k}% (\Lambda_{N},\gamma+t)=O(\rho^{-k\alpha_{1}}),\forall k=1,2,...,p_{1}. \end{equation} Since we used only the condition (6) for $\Lambda_{N},$ it follows that \begin{equation} S_{k}(a,\gamma+t)=O(\rho^{-k\alpha_{1}}) \end{equation} for all $a\in\mathbb{R}$ satisfying $\mid a-\mid\gamma+t\mid^{2}\mid<\frac {1}{2}\rho^{\alpha_{1}}.$ Thus finding $N$ such that $\Lambda_{N}$ is close to $\mid\gamma+t\mid^{2}$ and $b(N,\gamma)$ is not very small, then dividing both sides of (20) by $b(N,\gamma),$ we get the asymptotic formulas for $\Lambda_{N}$. \begin{theorem} $(a)$ Suppose $\gamma+t\in U(\rho^{\alpha_{1}},p),$ $\mid\gamma\mid\sim\rho.$ If (6) and (7) hold, then $\Lambda_{N}$ satisfies formulas (5) for $k=1,2,...,[\frac{1}{3}(p-c)],$ where \begin{equation} F_{s}=O(\rho^{-\alpha_{1}}),\forall s=0,1,..., \end{equation} and $F_{0}=0,$ $F_{s}=A_{s}(\mid\gamma+t\mid^{2}+F_{s-1},\gamma+t)$ for $s=1,2,....$ $(b)$ For $\gamma+t\in U(\rho^{\alpha_{1}},p),$ $\mid\gamma\mid\sim\rho$ there exists an eigenvalue $\Lambda_{N}$ of $L_{t}(q(x))$ satisfying (5). \end{theorem} \begin{proof} $(a)$ To prove (5) in case $k=1$ we divide both side of (20) by $b(N,\gamma)$ and use (7), (21). Then we obtain \begin{equation} \Lambda_{N}-\mid\gamma+t\mid^{2}=\mid\gamma+t\mid^{2}+O(\rho^{-\alpha_{1}}). \end{equation} This and $\alpha_{1}=3\alpha$ ( see the end of the introduction) imply that formula (5) for $k=1$ holds and $F_{0}=0.$ Hence (23) for $s=0$ is also proved. Moreover, from (22), we obtain $S_{k}(\mid\gamma+t\mid^{2}% +O(\rho^{-\alpha_{1}}),\gamma+t)=O(\rho^{-\alpha_{1}})$ for $k=1,2,....$ Therefore (23) for arbitrary $s$ follows from the definition of $F_{s}$ by induction. Now we prove (5) by induction on $k$. Suppose (5) holds for $k=j$, that is, $\Lambda_{N}=\mid\gamma+t\mid^{2}+F_{k-1}(\gamma+t)+O(\rho^{-3k\alpha}).$ Substituting this into $A_{p_{1}}(\Lambda_{N},\gamma+t)$ in (20) and dividing both sides of (20) by $b(N,\gamma),$ we get \begin{align*} \Lambda_{N} & =\mid\gamma+t\mid^{2}+A_{p_{1}}(\mid\gamma+t\mid^{2}% +F_{j-1}+O(\rho^{-j\alpha_{1}}),\gamma+t)+O(\rho^{-(p-c)\alpha})=\\ & \mid\gamma+t\mid^{2}+\{A_{p_{1}}(\mid\gamma+t\mid^{2}+F_{j-1}% +O(\rho^{-j\alpha_{1}}),\gamma+t)-\\ A_{p_{1}}( & \mid\gamma+t\mid^{2}+F_{j-1},\gamma+t)\}+A_{p_{1}}(\mid \gamma+t\mid^{2}+F_{j-1},\gamma+t)+O(\rho^{-(p-c)\alpha}). \end{align*} To prove $(a)$ for $k=j+1$ we need to show that the expression in curly brackets is equal to $O(\rho^{-(j+1)\alpha_{1}}).$ It can be checked by using (4), (19), (23) and the obvious relation \begin{align*} & \frac{1}{\prod_{j=1}^{s}(\mid\gamma+t\mid^{2}+F_{j-1}+O(\rho^{-j\alpha_{1}% })-\mid\gamma+t-\sum_{i=1}^{s}\gamma_{i}\mid^{2})}-\\ & \frac{1}{\prod_{j=1}^{s}(\mid\gamma+t\mid^{2}+F_{j-1}-\mid\gamma +t-\sum_{i=1}^{s}\gamma_{i}\mid^{2})}\\ & =\frac{1}{\prod_{j=1}^{s}(\mid\gamma+t\mid^{2}+F_{j-1}-\mid\gamma +t-\sum_{i=1}^{s}\gamma_{i}\mid^{2})}(\frac{1}{1-O(\rho^{-(j+1)\alpha_{1}}% )}-1) \end{align*} $=O(\rho^{-(j+1)\alpha_{1}})$ for $s=1,2,...,p_{1}.$ $(b)$ Let $A$ be the set of indices $N$ satisfying (6). Using (15) and Bessel inequality, we obtain% $\sum_{N\notin A}\mid b(N,\gamma)\mid^{2}=\sum_{N\notin A}\mid\dfrac{(\Psi _{N}(x),q(x)e^{i(\gamma+t,x)})}{\Lambda_{N}-\mid\gamma+t\mid^{2}}\mid ^{2}=O(\rho^{-2\alpha_{1}})$ Hence, by the Parseval equality, we have $\sum_{N\in A}\mid b(N,\gamma )\mid^{2}=1-O(\rho^{-2\alpha_{1}}).$ This and the inequality $\mid A\mid \frac {1}{2}(c_{5})^{-1}\rho^{-\frac{(d-1)q}{2}\alpha}$, that is, (7) holds for $c=\frac{(d-1)q}{2}$ . Thus $\Lambda_{N}$ satisfies (5) due to $(a)$ \end{proof} Theorem 1 shows that in the non-resonance case the eigenvalue of the perturbed operator $L_{t}(q(x))$ is close to the eigenvalue of the unperturbed operator $L_{t}(0).$ However, in theorem 2 we prove that if $\gamma+t\in\cap_{i=1}% ^{k}V_{\gamma_{i}}(\rho^{\alpha_{k}})\backslash E_{k+1}$ for $k\geq1,$ where $\gamma_{1},\gamma_{2},...,\gamma_{k}$ are linearly independent vectors of $\Gamma(p\rho^{\alpha}),$ then the corresponding eigenvalue of $L_{t}(q(x))$ is close to the eigenvalue of the matrix constructed as follows. Introduce the sets: $B_{k}\equiv B_{k}(\gamma_{1},\gamma_{2},...,\gamma_{k})=\{b:b=\sum_{i=1}% ^{k}n_{i}\gamma_{i},n_{i}\in Z,\mid b\mid<\frac{1}{2}\rho^{\frac{1}{2}% \alpha_{k+1}}\},$% \begin{equation} B_{k}(\gamma+t)=\gamma+t+B_{k}=\{\gamma+t+b:b\in B_{k}\}, \end{equation} $B_{k}(\gamma+t,p_{1})=\{\gamma+t+b+a:b\in B_{k},\mid a\mid\frac{1}% {5}\rho^{\alpha_{k+1}}, \end{equation} where$\gamma^{^{\prime}}\in\Gamma(\rho^{\alpha}),\gamma_{j}^{^{\prime}}% \in\Gamma(\rho^{\alpha}),j=1,2,...,s$and$s=0,1,...,p_{1}-1.$\end{lemma} \begin{proof} The inequality$p>2p_{1}$( see the end of the introduction) and the conditions of the lemma 1 imply that$h-\gamma^{^{\prime}}-\gamma_{1}^{^{\prime}}-\gamma_{2}^{^{\prime}}% -...-\gamma_{s}^{^{\prime}}+t\in B_{k}(\gamma+t,p)\backslash B_{k}(\gamma+t)$for all$s=0,1,...,p_{1}-1.$It follows from the definitions of$B_{k}% (\gamma+t,p),B_{k}$that ( see (25))$h-\gamma^{^{\prime}}-\gamma_{1}^{^{\prime}}-\gamma_{2}^{^{\prime}}% -...-\gamma_{s}^{^{\prime}}+t=\gamma+t+b+a,$where \begin{equation} \mid b\mid<\frac{1}{2}\rho^{\frac{1}{2}\alpha_{k+1}},\mid a\mid\frac{1}{5}\rho ^{\alpha_{k+1}}. \end{equation} To prove (29) we consider two cases: Case 1.$a\in P$, where$P=Span\{\gamma_{1,}\gamma_{2},...,\gamma_{k}\}.$Since$b\in B_{k}\subset P,$we have$a+b\in P.$This with the third relation in (28) imply that$a+b\in P\backslash B_{k}$,i.e.,$\mid a+b\mid\geq\frac {1}{2}\rho^{\frac{1}{2}\alpha_{k+1}}$. Consider the orthogonal decomposition$\gamma+t=y+v$of$\gamma+t,$where$v\in P$and$y\bot P.$First we prove that the projection$v$of any vector$x\in\cap_{i=1}% ^{k}V_{\gamma_{i}}(\rho^{\alpha_{k}})$on$P$satisfies \begin{equation} \mid v\mid=O(\rho^{(k-1)\alpha+\alpha_{k}}). \end{equation} For this we turn the coordinate axis so that$Span\{\gamma_{1,}\gamma _{2},...,\gamma_{k}\}$coincides with the span of the vectors$e_{1}% =(1,0,0,...,0)$,$e_{2}=(0,1,0,...,0),...,e_{k}$. Then$\gamma_{s}% =\sum_{i=1}^{k}\gamma_{s,i}e_{i}$for$s=1,2,...,k$. Therefore the relation$x\in\cap_{i=1}^{k}V_{\gamma_{i}}(\rho^{\alpha_{k}})$implies that $\sum_{i=1}^{k}\gamma_{s,i}x_{i}=O(\rho^{\alpha_{k}}),s=1,2,...,k;\text{ }% x_{n}=\frac{\det(b_{j,i}^{n})}{\det(\gamma_{j,i})}\text{, }n=1,2,...,k,$ where$x=(x_{1},x_{2},...,x_{d}),\gamma_{j}=(\gamma_{j,1},\gamma _{j,2},...,\gamma_{j,k},0,0,...,0),b_{j,i}^{n}=\gamma_{j,i}$for$n\neq j$and$b_{j,i}^{n}=O(\rho^{\alpha_{k}})$for$n=j.$Taking into account that the determinant$\det(\gamma_{j,i})$is a volume of the parallelepiped$\{\sum_{i=1}^{k}b_{i}\gamma_{i}:b_{i}\in\lbrack0,1],i=1,2,...,k\}$and using$\mid\gamma_{j,i}\mid2(\alpha_{k}+(k-1)\alpha)$( see the \ seventh inequality in (14)), and the obvious equalities$(y,v)=(y,a)=(y,b)=0,$% \begin{equation} \mid\gamma+t+a+b\mid^{2}-\mid\gamma+t\mid^{2}=\mid a+b+v\mid^{2}-\mid v\mid^{2}, \end{equation} we obtain the estimation (29). Case 2.$a\notin P.$First we show that \begin{equation} \mid\mid\gamma+t+a\mid^{2}-\mid\gamma+t\mid^{2}\mid\geq\rho^{\alpha_{k+1}}. \end{equation} Suppose, to the contrary, that it does not hold. Then$\gamma+t\in V_{a}% (\rho^{\alpha_{k+1}}).$On the other hand$\gamma+t\in\cap_{i=1}^{k}% V_{\gamma_{i}}(\rho^{\alpha_{k+1}})$( see the conditions of Lemma 1). Therefore we have$\gamma+t\in E_{k+1}$which contradicts the conditions of the lemma. \ So (33) is proved. Now, to prove (29) we write the difference$\mid\gamma+t+a+b\mid^{2}-\mid\gamma+t\mid^{2}$as the sum of$d_{1}\equiv \mid\gamma+t+a+b\mid^{2}-\mid\gamma+t+b\mid^{2}$and$d_{2}\equiv\mid \gamma+t+b\mid^{2}-\mid\gamma+t\mid^{2}.$Since$d_{1}=\mid\gamma+t+a\mid ^{2}-\mid\gamma+t\mid^{2}+2(a,b),$it follows from the inequalities (33), (28) that$\mid d_{1}\mid>\frac{2}{3}\rho^{\alpha_{k+1}}$. On the other hand, taking$a=0$in (32) we have$d_{2}=\mid b+v\mid^{2}-\mid v\mid^{2}.$Therefore (30), the first inequality in (28) and the \ seventh inequality in (14) imply that$\mid d_{2}\mid<\frac{1}{3}\rho^{\alpha_{k+1}},\mid d_{1}\mid-\mid d_{2}\mid>\frac{1}{3}\rho^{\alpha_{k+1}},$that is, (29) holds \end{proof} \begin{theorem}$(a)$Suppose$\mid\gamma\mid\sim\rho,\gamma+t\in(\cap_{i=1}^{k}% V_{\gamma_{i}}(\rho^{\alpha_{k}}))\backslash E_{k+1},$where$k=1,2,...,d-1.$If (6) and (7) hold, then there is an index$j$such that \begin{equation} \Lambda_{N}=\lambda_{j}(\gamma+t)+O(\rho^{-(p-c-\frac{1}{4}d3^{d})\alpha}), \end{equation} where$\lambda_{1}(\gamma+t)\leq\lambda_{2}(\gamma+t)\leq...\leq\lambda _{b_{k}}(\gamma+t)$are the eigenvalues of the matrix$C(\gamma+t,\gamma _{1},\gamma_{2},...,\gamma_{k})$defined in (26).$(b)$Every eigenvalue$\Lambda_{N}(t)\sim\rho^{2}$of the operator$L_{t}(q(x))$satisfies either (5) or (34) for$c=\frac{q(d-1)}{2}.$\end{theorem} \begin{proof}$(a)$Writing the equation (16) for all$h_{i}+t\in B_{k}(\gamma+t,p_{1}),$we obtain% \begin{equation} (\Lambda_{N}-\mid h_{i}+t\mid^{2})b(N,h_{i})=\sum_{\gamma^{^{\prime}}\in \Gamma(\rho^{\alpha})}q_{\gamma^{^{\prime}}}b(N,h_{i}-\gamma^{^{\prime}% })+O(\rho^{-p\alpha}) \end{equation} for$i=1,2,...,b_{k}$( see (25) for definition of$B_{k}(\gamma+t,p_{1})$). It follows from (6) and lemma 1 that if$(h_{i}-\gamma^{^{\prime}}+t)\notin B_{k}(\gamma+t,p_{1}),$then $\mid\Lambda_{N}-\mid h_{i}-\gamma^{^{\prime}}-\gamma_{1}-\gamma_{2}% -...-\gamma_{s}+t\mid^{2}\mid>\frac{1}{6}\rho^{\alpha_{k+1}},$ where$\gamma^{^{\prime}}\in\Gamma(\rho^{\alpha}),\gamma_{j}\in\Gamma (\rho^{\alpha}),j=1,2,...,s$and$s=0,1,...,p_{1}-1.$Therefore, applying the formula (17)$p_{1}$times, using (4) and$p_{1}\alpha_{k+1}>p_{1}% \alpha_{1}\geq p\alpha$( see the\ fourth inequality in (14)), we see that if$(h_{i}-\gamma^{^{\prime}}+t)\notin B_{k}(\gamma+t,p_{1}),$then \begin{equation} b(N,h_{i}-\gamma^{^{\prime}})=\nonumber \end{equation}% \begin{equation} \sum_{\gamma_{1},...,\gamma_{p_{1}-1}\in\Gamma(\rho^{\alpha})}\dfrac {q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{p_{1}}}b(N,h_{i}-\gamma^{^{\prime }% }-\sum_{i=1}^{p_{1}}\gamma_{i})}{\prod_{j=0}^{p_{1}-1}(\Lambda_{N}-\mid h_{i}-\gamma^{^{\prime}}+t-\sum_{i=1}^{j}\gamma_{i}\mid^{2})}+ \end{equation}% $+O(\rho^{-p\alpha})=O(\rho^{p_{1}\alpha_{k+1}})+O(\rho^{-p\alpha}% )=O(\rho^{-p\alpha}).$ Hence (35) has the form% $(\Lambda_{N}-\mid h_{i}+t\mid^{2})b(N,h_{i})=\sum_{\gamma^{^{\prime}}% }q_{\gamma^{^{\prime}}}b(N,h_{i}-\gamma^{^{\prime}})+O(\rho^{-p\alpha }),i=1,2,...,b_{k},$ where the sum is taken under the conditions$\gamma^{^{\prime}}\in\Gamma (\rho^{\alpha})$and$h_{i}-\gamma^{^{\prime}}+t\in B_{k}(\gamma+t,p_{1})$. It can be written in matrix form $(C-\Lambda_{N}I)(b(N,h_{1}),b(N,h_{2}),...b(N,h_{b_{k}}))=O(\rho^{-p\alpha}) ,$ where the rigth-hand side of this system is a vector having the norm$\mid\mid O(\rho^{-p\alpha})\mid\mid=O(\sqrt{b_{k}}\rho^{-p\alpha})$. Now, taking into account that$\gamma+t\in\{h_{i}+t:i=1,2,...,b_{k}\}and (7) holds, we have \begin{align} c_{4}\rho^{-c\alpha} & <(\sum_{i=1}^{b_{k}}\mid b(N,h_{i})\mid^{2}% )^{\frac{1}{2}}\leq\parallel(C-\Lambda_{N}I)^{-1}\parallel\sqrt{b_{k}}% c_{6}\rho^{-p\alpha},\\ \max_{i=1,2,...,b_{k}} & \mid\Lambda_{N}-\lambda_{i}\mid^{-1}=\parallel (C-\Lambda_{N}I)^{-1}\parallel>c_{4}c_{6}^{-1}b_{k}^{-\frac{1}{2}}% \rho^{-c\alpha+p\alpha}. \end{align} Sinceb_{k}$is the number of the vectors of$B_{k}(\gamma+t,p_{1}),$it follows from the definition of$B_{k}(\gamma+t,p_{1})$( see (25)) and the obvious relations$\mid B_{k}\mid=O(\rho^{\frac{k}{2}\alpha_{k+1}}),\mid\Gamma(p_{1}\rho^{\alpha})\mid=O(\rho^{d\alpha})$and$d\alpha<\frac {1}{2}\alpha_{d}$( see the end of introduction), we get \begin{equation} b_{k}=O(\rho^{d\alpha+\frac{k}{2}\alpha_{k+1}})=O(\rho^{\frac{d}{2}\alpha_{d }% })=O(\rho^{\frac{d}{2}3^{d}\alpha}),\forall k=1,2,...,d-1 \end{equation} Thus formula (34) follows from (38) and (39).$(b)$Let$\Lambda_{N\text{ }}(t)$be any eigenvalue of order$\rho^{2}$of the operator$L_{t}(q(x)).$Denote by$D$the set of all vectors$\gamma \in\Gamma$satisfying (6). From (15), arguing as in the proof of Theorem 1($b$), we obtain$\sum_{\gamma\in D}\mid b(N,\gamma)\mid^{2}=1-O(\rho^{-2\alpha_{1}}).$Since$\mid D\mid=O(\rho^{d-1})$( see the end of the introduction), there exists$\gamma\in D$such that$\mid b(N,\gamma)\mid>c_{7}\rho^{-\frac{(d-1)}{2}}=c_{7}\rho^{-\frac {(d-1)q}{2}\alpha}$, that is, condition (7) for$c=\frac{(d-1)q}{2}$holds. Now the proof of$(b)$follows from Theorem 1$(a)$and Theorem 2$(a),$since either$\gamma+t\in U(\rho^{\alpha_{1}},p)$or$\gamma+t\inE_{k}% \backslash E_{k+1}$for$k=1,2,...,d-1$( see (42) in Remark 1) \end{proof} \begin{remark} Here we note that the non-resonance domain$U(c_{8}\rho^{\alpha_{1}},p)$has an asymptotically full measure on$\mathbb{R}^{d}$in the sence that$\frac{\mu(U\cap B(\rho))}{\mu(B(\rho))}$tends to$1$as$\rho$tends to infinity, where$B(\rho)=\{x\in\mathbb{R}^{d}:\mid x\mid=\rho\}.$Clearly,$B(\rho)\cap V_{b}(c_{8}\rho^{\alpha_{1}})$is the part of sphere$B(\rho),$which is contained between two parallel hyperplanes$\{x:\mid x\mid^{2}-\mid x+b\mid^{2}=-c_{8}\rho^{\alpha_{1}}\}$and$\{x:\mid x\mid^{2}-\mid x+b\mid^{2}=c_{8}\rho^{\alpha_{1}}\}.$The distance of these hyperplanes from origin is$O(\frac{\rho^{\alpha_{1}}}{\mid b\mid}).$Therefore, the relations$\mid\Gamma(p\rho^{\alpha})\mid=O(\rho^{d\alpha}),$and$\alpha_{1}+d\alpha<1-\alpha( see the first inequality in (14)) imply \begin{align} \mu(B(\rho)\cap V_{b}(c_{8}\rho^{\alpha_{1}})) & =O(\frac{\rho^{\alpha _{1}+d-2}}{\mid b\mid}),\text{ }\mu(E_{1}\cap B(\rho))=O(\rho^{d-1-\alpha})\\ \mu(U(c_{8}\rho^{\alpha_{1}},p)\cap B(\rho)) & =(1+O(\rho^{-\alpha}% ))\mu(B(\rho)). \end{align} Ifx\in\cap_{i=1}^{d}V_{\gamma_{i}}(\rho^{\alpha_{d}}),$then (31) holds for$k=d$and$n=1,2,...,d.$Hence we have$\mid x\mid=O(\rho^{\alpha _{d}+(d-1)\alpha}).$It is imposible, because of$\alpha_{d}+(d-1)\alpha<1$( see the \ sixth inequality in (14)) and$x\in B(\rho).$It means that$(\cap_{i=1}^{d}V_{\gamma_{i}}(\rho^{\alpha_{k}}))\cap B(\rho_{0})=\emptyset$for$\rho_{0}\gg1$. Thus, for$\rho_{0}\gg1,$we have \begin{equation} \mathbb{R}^{d}\cap\{\mid x\mid>\rho_{0}\}=(U\cup(\cup_{s=1}^{d-1}% (E_{s}\backslash E_{s+1})))\cap\{\mid x\mid>\rho_{0}\}. \end{equation} \end{remark} \begin{remark} Here we note some properties of the known parts$\mid\gamma+t\mid^{2}+F_{k}(\gamma+t)$(see Theorem 1)and$\lambda_{j}% (\gamma+t)$( see Theorem 2) of the non-resonance and resonance eigenvalues of$L_{t}(q(x))$. Denoting$\gamma+t$by$x$, where$\mid\gamma+t\mid\sim\rho,\gamma+t\in U(\rho^{\alpha_{1}},p),$we prove \begin{equation} \frac{\partial F_{k}(x)}{\partial x_{i}}=O(\rho^{-2\alpha_{1}+\alpha}),\forall i=1,2,...,d;\forall k=1,2,... \end{equation} by induction on$k.$For$k=1$the formula (43) follows from (4) and \begin{equation} \frac{\partial}{\partial x_{i}}(\dfrac{1}{\mid x\mid^{2}-\mid x-\gamma_{1}% \mid^{2}})=\dfrac{-2\gamma_{1}(i)}{(\mid x\mid^{2}-\mid x-\gamma_{1}\mid ^{2})^{2}}=O(\rho^{-2\alpha_{1}+\alpha}), \end{equation} where$\gamma_{1}(i)$is the$i$-th component of the vector$\gamma_{1}% \in\Gamma(p\rho^{\alpha})$hence is equal to$O(\rho^{\alpha}).$Now suppose that (43) holds for$k=s.$Using this and (23), replacing$\mid x\mid^{2}$by$\mid x\mid^{2}+F_{s}(x)$in (44) and evaluting as above we obtain $\frac{\partial}{\partial x_{i}}(\dfrac{1}{\mid x\mid^{2}+F_{s}-\mid x-\gamma_{1}\mid^{2}})=\dfrac{-2\gamma_{1}(i)+\frac{\partial F_{s}% (x)}{\partial x_{i}}}{(\mid x\mid^{2}+F_{s}-\mid x-\gamma_{1}\mid^{2})^{2}% }=O(\rho^{-2\alpha_{1}+\alpha}).$ This formula together with the definition of$F_{k}$give (43) for$k=s+1.$Now denoting$\lambda_{i}(\gamma+t)-\mid\gamma+t\mid^{2}$by$r_{i}(\gamma+t)$we prove that \begin{equation} \mid r_{i}(x)-r_{i}(x^{^{\prime}})\mid\leq2\rho^{\frac{1}{2}\alpha_{d}}\mid x-x^{^{\prime}}\mid,\forall i. \end{equation} Clearly$r_{1}(x)\leq r_{2}(x)\leq...\leqr_{b_{k}}(x)$are the eigenvalue of the matrix$C(x)-\mid x\mid^{2}I\equiv C^{^{\prime}}(x),$where$C(x)$is defined in (26). By definition, only the diagonal elements of the matrix$C^{^{\prime}% }(x)$depend on$x$and they are$\mid x\mid^{2}-\mid x-a_{i}\mid^{2}=2(x,a_{i})-\mid a_{i}\mid^{2},$where$a_{i}=h_{i}+t-\gamma-t$and$h_{i}+t\in B_{k}(\gamma+t,p_{1}).$It follows from the definitions of$B_{k}(\gamma+t,p_{1})$( for$kd+2\alpha$( see the fifth inequality in (14)), and notations$F(\gamma+t)=\mid\gamma+t\mid^{2}+F_{k_{1}-1}(\gamma+t)$,$\varepsilon_{1}=\rho^{-d-2\alpha}$( see Step 1 in introduction), we obtain \begin{equation} \Lambda_{N}(t)=F(\gamma+t)+o(\varepsilon_{1}). \end{equation} Let$\Psi_{N}$be any normalized eigenfunction corresponding to$\Lambda_{N}$. Since the normalized eigenfunction is defined up to constant of modulas$1,$without loss of generality it can assumed that$\arg b(N,\gamma)=0,$where$b(N,\gamma)=(\Psi_{N},e^{i(\gamma+t,x)}).$Therefore to prove (9) it suffices to show that (13) holds. To prove (13) first \ we estimate \$\sum _{\gamma^{^{\prime}}\notin K}\mid b(N,\gamma^{^{\prime}})\mid^{2}$and then$\sum_{\gamma^{^{\prime}}\in K\backslash\{\gamma\}}\mid b(N,\gamma^{^{\prime}% })\mid^{2},$where$K$is defined in (11), (12). Using (46), the definition of$K, and (15), we get \begin{align} & \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2}\mid>\frac{1}{4}% \rho^{\alpha_{1}},\text{ }\forall\gamma^{^{\prime}}\notin K,\\ \sum_{\gamma^{^{\prime}}\notin K} & \mid b(N,\gamma^{^{\prime}})\mid ^{2}=\parallel q(x)\Psi_{N}\parallel^{2}O(\rho^{-2\alpha_{1}})=O(\rho ^{-2\alpha_{1}}).\nonumber \end{align} If\gamma^{^{\prime}}\in K$, then by (46) and by definition of$K,$it follows that \begin{equation} \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2}\mid<\frac{1}{2}\rho ^{\alpha_{1}}% \end{equation} Now we prove that the simplicity conditions (11), (12) imply \begin{equation} \mid b(N,\gamma^{^{\prime}})\mid\leq c_{4}\rho^{-c\alpha},\text{ }% \forall\gamma^{^{\prime}}\in K\backslash\{\gamma\}, \end{equation} where$c=p-dq-\frac{1}{4}d3^{d}-3.$If for$\gamma^{^{\prime}}+t\in U(\rho^{\alpha_{1}},p)$and$\gamma^{^{\prime}}\in K\backslash\{\gamma\}$the inequality in (49) is not true, then by (48) and Theorem 1(a), we have \begin{equation} \Lambda_{N}=\mid\gamma^{^{\prime}}+t\mid^{2}+F_{k-1}(\gamma^{^{\prime}% }+t)+O(\rho^{-3k\alpha}) \end{equation} for$k=1,2,...,[\frac{1}{3}(p-c)]=[\frac{1}{3}(dq+\frac{1}{4}d3^{d}+3)].$Since$\alpha=\frac{1}{q}$and$k_{1}\equiv\lbrack\frac{d}{3\alpha}]+2<\frac{1}{3}(dq+\frac{1}{4}d3^{d}+3)$, \ the formula (50) holds for$k=k_{1}.$Therefore arguing as in the prove of (46), we get$\Lambda_{N}-F(\gamma^{^{\prime}}+t)=o(\varepsilon_{1})$. \ This with (46) contradicts (11). Similarly, if the inequality in (49) does not hold for$\gamma^{^{\prime}}+t\in(E_{k}\backslash E_{k+1})$and$\gamma^{^{\prime}% }\in K,$then by Theorem 2(a) \begin{equation} \Lambda_{N}=\lambda_{j}(\gamma^{^{\prime}}+t)+O(\rho^{-(p-c-\frac{1}{4}% d3^{d})\alpha}), \end{equation} where$(p-c-\frac{1}{4}d3^{d})\alpha=(dq+3)\alpha>d+2\alpha$. Hence we have$\Lambda_{N}-\lambda_{j}(\gamma^{^{\prime}}+t)=o(\varepsilon_{1}).$This with (46) contradicts (12). So the inequality in (49) holds. Therefore, using$\mid K\mid=O(\rho^{d-1}),q\alpha=1,$we get \begin{equation} \sum_{\gamma^{^{\prime}}\in K\backslash\{\gamma\}}\mid b(N,\gamma^{^{\prime}% })\mid^{2}=O(\rho^{-(2c-q(d-1))\alpha})=O(\rho^{-(2p-(3d-1)q-\frac{1}{2}% d3^{d}-6)\alpha}). \end{equation} If$s=s_{0},$that is,$p=s_{0}-d$then$2p-(3d-1)q-\frac{1}{2}d3^{d}-6=6.$Since$\alpha_{1}=3\alpha,$the equality (52) and the equality in (47) imply (13). Thus we proved that the equality (9) holds for any normalized eigenfunction$\Psi_{N}$corresponding to any eigenvalue$\Lambda_{N}$\ satisfying (5). If there exist two different eigenvalues or multiple eigenvalue satisfying (5), then there exist two orthogonal normalized eigenfunction satisfying (9), which is imposible. Therefore$\Lambda_{N}$\ is a simple eigenvalue. It follows from Theorem 1(a) that$\Lambda_{N}$satisfies (5) for$k=1,2,...,[\frac{p}{3}],$because the inequality (7) holds for$c=0$( see (9)). \end{proof} \begin{remark} Since for$\gamma+t\in B$\ there exists a unique eigenvalue satisfying (5), (46) we denote this eigenvalue by$\Lambda(\gamma+t).$Since this eigenvalue is simple, we denote the \ corresponding eigenfunction by$\Psi_{\gamma +t}(x).$By Theorem 3 this eigenfunction satisfies (9). Clearly, for$\gamma+t\in B$\ there exists a unique index$N\equiv N(\gamma+t)$such that$\Lambda(\gamma+t)=\Lambda_{N(\gamma+t)}$) and$\Psi_{\gamma+t}(x)=\Psi _{N(\gamma+t)}(x)).$\end{remark} Now we prove the asymptotic formulas of arbitrary order for$\Psi_{\gamma +t}(x).$\begin{theorem} If$\gamma+t\in B$and$\mid\gamma+t\mid\sim\rho,$then the eigenfunction$\Psi_{\gamma+t}(x)\equiv\Psi_{N(\gamma+t)}(x)$corresponding to the eigenvalue$\Lambda_{N}\equiv\Lambda(\gamma+t)$satisfies formulas (10), for$k=1,2,...,n$, where$n=[\frac{1}{6}(2p-(3d-1)q-\frac{1}{2}d3^{d}-6)],\Phi_{0}(x)=0,\Phi_{1}(x)=% %TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}e^{i(\gamma+t+\gamma_{1},x)}}{(\mid\gamma+t\mid^{2}% -\mid\gamma+\gamma_{1}+t\mid^{2})},$and$\Phi_{k-1}(x)$for$k>2$is a linear combination of$e^{i(\gamma +t+\gamma^{^{\prime}},x)}$for$\gamma^{^{\prime}}\in\Gamma((k-1)\rho^{\alpha})\cup\{0\}$with coefficients (58), (59). \end{theorem} \begin{proof} By Theorem 3, formula (10) for$k=1$is proved. To prove formula (10) for arbitrary$k\leq n$we prove the following equivalent relations \begin{equation} \sum_{\gamma^{^{\prime}}\in\Gamma^{c}(k-1)}\mid b(N,\gamma+\gamma^{^{\prime}% })\mid^{2}=O(\rho^{-2k\alpha_{1}}), \end{equation}% \begin{equation} \Psi_{N}=b(N,\gamma)e^{i(\gamma+t,x)}+\sum_{\gamma^{^{\prime}}\in \Gamma((k-1)\rho^{\alpha})}b(N,\gamma+\gamma^{^{\prime}})e^{i(\gamma +t+\gamma^{^{\prime}},x)}+H_{k}(x), \end{equation} where$\Gamma^{c}(m)\equiv\Gamma\backslash(\Gamma(m\rho^{\alpha})\cup\{0\})$and$\parallel H_{k}(x)\parallel=O(\rho^{-k\alpha_{1}}).$The case$k=1$is proved due to (13). Assume that (53) is true for$k=m$. Then using (54) for$\ \ \ k=m,$and (3), we have$\Psi_{N}(x)(q(x))=H(x)+O(\rho^{-m\alpha_{1}}),$where$H(x)$is a linear combination of$e^{i(\gamma+t+\gamma^{^{\prime}},x)}$for$\gamma^{^{\prime}}\in\Gamma(m\rho^{\alpha})\cup\{0\}.$Hence$(H(x),e^{i(\gamma+t+\gamma^{^{\prime}},x)})=0$for$\gamma^{^{\prime}}% \in\Gamma^{c}(m).$So using (15) and the inequality in (47), we get% \begin{equation} \sum_{\gamma^{^{\prime}}}\mid b(N,\gamma+\gamma^{^{\prime}})\mid^{2}% =\sum_{\gamma^{^{\prime}}}\mid\dfrac{(O(\rho^{-m\alpha_{1}}),e^{i(\gamma +t+\gamma^{^{\prime}},x)})}{\Lambda_{N}-\mid\gamma+\gamma^{^{\prime}}% +t\mid^{2}}\mid^{2}=O(\rho^{-2(m+1)\alpha_{1}}), \end{equation} where the sum is taken under conditions$\gamma^{^{\prime}}\in\Gamma^{c}(m)$,$\gamma+\gamma^{^{\prime}}\notin K.$On the other hand, using$\alpha _{1}=3\alpha,$(52), and the definition of$n$( see Theorem 4), we get $\sum_{\gamma^{^{\prime}}\in K\backslash\{\gamma\}}\mid b(N,\gamma^{^{\prime}% })\mid^{2}=O(\rho^{-2n\alpha_{1}}).$ This with (55) implies (53) for$k=m+1.$Thus (54) is also proved. Here$b(N,\gamma)$and$b(N,\gamma+\gamma^{^{\prime}})$for$\gamma^{^{\prime}}% \in\Gamma((n-1)\rho^{\alpha})$can be calculated as follows. First we express$b(N,\gamma+\gamma^{^{\prime}})$by$b(N,\gamma)$. For this we apply (17) for$b(N,\gamma+\gamma^{^{\prime}}),$where$\gamma^{^{\prime}}\in\Gamma ((n-1)\rho^{\alpha}),$that is, in (17) replace$\gamma^{^{\prime}}$by$\gamma+\gamma^{^{\prime}}$. Iterate it$n$times and every times isolate the terms with multiplicand$b(N,\gamma).$In other word apply (17) for$b(N,\gamma+\gamma^{^{\prime}})$and isolate the terms with multiplicand$b(N,\gamma).$Then apply (17) for$b(N,\gamma+\gamma^{^{\prime}}-\gamma_{1})$when$\gamma^{^{\prime}}-\gamma_{1}\neq0.$Then apply (17) for$b(N,\gamma+\gamma^{^{\prime}}-\sum_{i=1}^{2}\gamma_{i})$when$\gamma ^{^{\prime}}-\sum_{i=1}^{2}\gamma_{i}\neq0,$etc. Apply (17) for$b(N,\gamma+\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i})$when$\gamma ^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}\neq0,$where$\gamma_{i}\in\Gamma (\rho^{\alpha}),j=3,4,...,n-1.$Then using (4) and the relations$\mid\Lambda_{N}-\mid\gamma+t+\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}% \mid^{2}\mid>\frac{1}{2}\rho^{\alpha_{1}}$( see (19) and take into account that$\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}\in\Gamma(p\rho^{\alpha}),$since$p>2n$),$\Lambda_{N}=P(\gamma+t)+O(\rho^{-n\alpha_{1}}),$where$P(\gamma+t)=\mid\gamma+t\mid^{2}+F_{[\frac{p}{3}]}(\gamma+t)$( see Theorem 3), we obtain% \begin{equation} b(N,\gamma+\gamma^{^{\prime}})=\sum_{k=1}^{n-1}A_{k}(\gamma^{^{\prime}% })b(N,\gamma)+O(\rho^{-n\alpha_{1}}), \end{equation} where$A_{1}(\gamma^{^{\prime}})=\dfrac{q_{\gamma^{^{\prime}}}}{P(\gamma +t)-\mid\gamma+\gamma^{^{\prime}}+t\mid^{2}}=\dfrac{q_{\gamma^{^{\prime}}}% }{\mid\gamma+t\mid^{2}-\mid\gamma+\gamma^{^{\prime}}+t\mid^{2}}+O(\frac {1}{\rho^{3\alpha_{1}}}),$% $A_{k}(\gamma^{^{\prime}})=% %TCIMACRO{\dsum _{\gamma_{1},...,\gamma_{k-1}}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},...,\gamma_{k-1}}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{k-1}}q_{\gamma^{^{\prime}% }-\gamma_{1}-\gamma_{2}-...-\gamma_{k-1}}}{\prod_{j=0}^{k-1}(P(\gamma +t)-\mid\gamma+t+\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}\mid^{2})}% =O(\rho^{-k\alpha_{1}}),$% \begin{equation} \sum_{\gamma^{\ast}\in\Gamma((n-1)\rho^{\alpha})}\mid A_{1}(\gamma^{\ast}% )\mid^{2}=O(\rho^{-2\alpha_{1}}),\sum_{\gamma^{\ast}\in\Gamma((n-1)\rho ^{\alpha})}\mid A_{k}(\gamma^{\ast})\mid=O(\rho^{-k\alpha_{1}}) \end{equation} for$k>1.$Now from (54) for$k=nand (56), we obtain \begin{align*} \Psi_{N}(x) & =b(N,\gamma)e^{i(\gamma+t,x)}+\\ & \sum_{\gamma^{\ast}\in\Gamma((n-1)\rho^{\alpha})}\sum_{k=1}^{n-1}% (A_{k}(\gamma^{\ast})b(N,\gamma)+O(\rho^{-n\alpha_{1}}))e^{i(\gamma +t+\gamma^{\ast},x)})+H_{n}(x). \end{align*} Using the equalities\parallel\Psi_{N}\parallel=1,\arg b(N,\gamma)=0,\parallel H_{n}\parallel=O(\rho^{-n\alpha_{1}})$and taking into account that the functions$e^{i(\gamma+t,x)},H_{n}(x),e^{i(\gamma+t+\gamma^{\ast },x)},(\gamma^{\ast}\in\Gamma((n-1)\rho^{\alpha}))$are orthogonal, we get$1=\mid b(N,\gamma)\mid^{2}+\sum_{k=1}^{n-1}(\sum_{\gamma^{\ast}\in \Gamma((n-1)\rho^{\alpha})}\mid A_{k}(\gamma^{\ast})b(N,\gamma)\mid^{2}% +O(\rho^{-n\alpha_{1}})),$\begin{equation} b(N,\gamma)=(1+\sum_{k=1}^{n-1}(\sum_{\gamma^{\ast}\in\Gamma((n-1)\rho ^{\alpha})}\mid A_{k}(\gamma^{\ast})\mid^{2}))^{-\frac{1}{2}}+O(\rho ^{-n\alpha_{1}})) \end{equation} (see the second equality in (57)). Thus from (56), we obtain \begin{equation} b(N,\gamma+\gamma^{^{\prime}})=(\sum_{k=1}^{n-1}A_{k}(\gamma^{^{\prime}% }))(1+\sum_{k=1}^{n-1}\sum_{\gamma^{\ast}}\mid A_{k}(\gamma^{\ast})\mid ^{2})^{-\frac{1}{2}}+O(\rho^{-n\alpha_{1}}). \end{equation} Consider the case$n=2.$By (58), (57), (59) we have$b(N,\gamma )=1+O(\rho^{-2\alpha_{1}}),b(N,\gamma+\gamma^{^{\prime}})=A_{1}(\gamma^{^{\prime}})+O(\rho^{-2\alpha _{1}})=\dfrac{q_{\gamma^{^{\prime}}}}{\mid\gamma+t\mid^{2}-\mid\gamma +\gamma^{^{\prime}}+t\mid^{2}}+O(\rho^{-2\alpha_{1}})$for all$\gamma ^{^{\prime}}\in\Gamma(\rho^{\alpha}).$These and (54) for$k=2$imply the formula for$\Phi_{1}$\end{proof} \section{Simple Sets and Fermi Surfaces} In this section we consider the simple sets$B$and construct a big part of the isoenergetic surfaces corresponding to$\rho^{2}$for big$\rho.$The isoenergetic surfaces of$L(q)$corresponding to$\rho^{2}$is the set$I_{\rho}(q(x))=\{t\in F^{\ast}:\exists N,\Lambda_{N}(t)=\rho^{2}\}.$In the case$q(x)=0$\ the isoenergetic surface$I_{\rho}(0)=\{t\in F^{\ast }:\exists\gamma\in\Gamma,\mid\gamma+t\mid^{2}=\rho^{2}\}$is the translation of the sphere$B(\rho)=\{\gamma+t:t\in F^{\ast},\gamma\in\Gamma,\mid\gamma+t\mid^{2}% =\rho^{2}\}$by the vectors$\gamma\in\Gamma.$We call$B(\rho)$the translated isoenergetic surfaces of$L(0)$corresponding to$\rho^{2}.$Similarly, we call the sets$P_{\rho}^{^{\prime}}=\{\gamma+t:\Lambda (\gamma+t)=\rho^{2}\}$and$P_{\rho}^{^{\prime\prime}}=\{t\in F^{\ast}:\exists\gamma\in\Gamma ,\Lambda(\gamma+t)=\rho^{2}\},$where$\Lambda(\gamma+t)=\Lambda_{N(\gamma +t)}(t)$is defined in Remark 3, the parts of translated isoenergetic surfaces and isoenergetic surfaces of$L(q).$In this section we construct the subsets$I_{\rho}^{^{\prime}}$and$I_{\rho}^{^{\prime\prime}}$of$P_{\rho}% ^{^{\prime}}$and$P_{\rho}^{^{\prime\prime}}$respectively and prove that the measures of these subsets are asymptotically equal to the measure of the isoenergetic surfaces$I_{\rho}(0)$of$L(0)$. In other word we construct a big part (in some sense) of isoenergetic surfaces$I_{\rho}(q(x))$of$L(q)$. As we see below the set$I_{\rho}^{^{\prime\prime}}$is a translation of$I_{\rho}^{^{\prime}}$by vectors$\gamma\in\Gamma$to$F^{\ast}$and the set$I_{\rho}^{^{\prime}}$lies in$\varepsilon$nieghborhood of the surface$S_{\rho}=\{x\in U(2\rho^{\alpha_{1}},p):F(x)=\rho^{2}\},$where$F(x)$is defined in Step 1 of introduction. Due to (46) ( replace$\gamma+t$by$x$in$F(\gamma+t)$) it is natural to call$S_{\rho}$the approximated isoenergetic surfaces in the nonresonance domain. Here we construct a part of the simple set$B$in nieghborhood of$S_{\rho}$that containes$I_{\rho}^{^{\prime}}$. For this we consider the surface$S_{\rho}$. As we noted in introduction \ ( see Step 2 and (11)) the non-resonance eigenvalue$\Lambda(\gamma+t)$does not coincide with other non-resonance eigenvalue$\Lambda(\gamma+t+b)$if$\mid F(\gamma+t)-F(\gamma+t+b)\mid>2\varepsilon_{1}$for$\gamma+t+b\in U(\rho^{\alpha_{1}},p)$and$b\in\Gamma\backslash\{0\}$. Therefore we eliminate \begin{equation} P_{b}=\{x:x,x+b\in U(\rho^{\alpha_{1}},p),\mid F(x)-F(x+b)\mid<3\varepsilon _{1}\} \end{equation} for$b\in\Gamma\backslash\{0\}$from$S_{\rho}$. Denote the remaining part of$S_{\rho}$by$S_{\rho}^{^{\prime}}.$Then we consider the$\varepsilon$neighbourhood$U_{\varepsilon}(S_{\rho}^{^{\prime}})=\cup_{a\in S_{\rho }^{^{\prime}}}U_{\varepsilon}(a)\}$of$S_{\rho}^{^{\prime}}$, where$\varepsilon=\frac{\varepsilon_{1}}{7\rho},U_{\varepsilon}(a)=\{x\in R^{d}:\mid x-a\mid<\varepsilon\}.$In this set the first simplicity condition (11) holds (see Lemma 2(a)). Denote by$Tr(E)=\{\gamma+x\in U_{\varepsilon}(S_{\rho}^{^{\prime}}):\gamma\in \Gamma,x\in E\}$and$Tr_{F^{\star}}(E)\equiv\{\gamma+x\in F^{\star}:\gamma\in\Gamma,x\in E\}$the translations of$E\subset R^{d}$into$U_{\varepsilon}(S_{\rho}^{^{\prime}})$and$F^{\star}$respectively. In order that the second simplicity condition (12) holds, we discart from$U_{\varepsilon}(S_{\rho}^{^{\prime}})$the translation$Tr(A(\rho))$of \begin{equation} A(\rho)\equiv\cup_{k=1}^{d-1}(\cup_{\gamma_{1},\gamma_{2},...,\gamma_{k}% \in\Gamma(p\rho^{\alpha})}(\cup_{i=1}^{b_{k}}A_{k,i}(\gamma_{1},\gamma _{2},...,\gamma_{k}))), \end{equation} where$A_{k,i}(\gamma_{1},...,\gamma_{k})=\{x\in(\cap_{i=1}^{k}V_{\gamma_{i}}(\rho^{\alpha_{k}})\backslash E_{k+1})\cap K_{\rho}:\lambda_{i}(x)\in(\rho^{2}-3\varepsilon_{1},\rho^{2}+3\varepsilon _{1})\},\lambda_{i}(x),b_{k}$is defined in Theorem 2 and \begin{equation} K_{\rho}=\{x\in\mathbb{R}^{d}:\mid\mid x\mid^{2}-\rho^{2}\mid<\rho^{\alpha _{1}}\}. \end{equation} As a result we construct the part$U_{\varepsilon}(S_{\rho}^{^{\prime}% })\backslash Tr(A(\rho))$of the simple set$B$(see Theorem 5(a)) which contains the set$I_{\rho}^{^{\prime}}$(see Theorem 5(c)). For this we need the following lemma. \begin{lemma}$(a)$If$x\in U_{\varepsilon}(S_{\rho}^{^{\prime}})$and$x+b\in U(\rho^{\alpha_{1}},p),$where$b\in\Gamma,$then \begin{equation} \mid F(x)-F(x+b)\mid>2\varepsilon_{1}, \end{equation} where$\varepsilon=\frac{\varepsilon_{1}}{7\rho},\varepsilon_{1}% =\rho^{-d-2\alpha},F(x)=\mid x\mid^{2}+F_{k_{1}-1}(x),k_{1}=[\frac {d}{3\alpha}]+2,$hence for$\gamma+t\in U_{\varepsilon}(S_{\rho}^{^{\prime}% })$the simplisity condition (11) holds.$(b)$If$x\in U_{\varepsilon}(S_{\rho}^{^{\prime}}),$then$x+b\notin U_{\varepsilon}(S_{\rho}^{^{\prime}})$for all$b\in\Gamma.(c)$If$E$is a bounded subset of$\mathbb{R}^{d}$, then$\mu(Tr(E))\leq \mu(E)$.$(d)$If$E\subset U_{\varepsilon}(S_{\rho}^{^{\prime}}),$then$\mu (Tr_{F^{\star}}(E))=\mu(E).$\end{lemma} \begin{proof}$(a)$If$x\in U_{\varepsilon}(S_{\rho}^{^{\prime}}),$then there exists a point$a$in$S_{\rho}^{^{\prime}}$such that$x\in U_{\varepsilon}(a)$. Since$S_{\rho}^{^{\prime}}\cap P_{b}=\emptyset$( see (60) and the definition of$S_{\rho}^{^{\prime}}$), we have \begin{equation} \mid F(a)-F(a+b)\mid\geq3\varepsilon_{1}% \end{equation} On the other hand, using (43) and the obvious relations$\mid x\mid<\rho+1,\mid x-a\mid<\varepsilon,\mid x+b-a-b\mid <\varepsilon,$we obtain \begin{equation} \mid F(x)-F(a)\mid<3\rho\varepsilon,\mid F(x+b)-F(a+b)\mid<3\rho\varepsilon \end{equation} These inequalities together with (64) give (63), since$6\rho\varepsilon <\varepsilon_{1}.(b)$If$x$and$x+b$lie in$U_{\varepsilon}(S_{\rho}^{^{\prime}}),$then there exist points$a$and$c$in$S_{\rho}^{^{\prime}}$such that$x\in U_{\varepsilon}(a)$and$x+b\in U_{\varepsilon}(c).$Repeating the proof of (65), we get$\mid F(c)-F(x+b)\mid<3\rho\varepsilon.$This, the first inequality in (65), and the relations$F(a)=\rho^{2},F(c)=\rho^{2}$(see the definition of$S_{\rho})$give$\mid F(x)-F(x+b)\mid<\varepsilon_{1},$which contradicts (63).$(c)$Clearly, for any bounded set$E$there exist only a finite number of vectors$\gamma_{1},\gamma_{2},...,\gamma_{s}$such that$E(k)\equiv (E+\gamma_{k})\cap U_{\varepsilon}(S_{\rho}^{^{\prime}})\neq\emptyset$for$k=1,2,...,s$and$Tr(E)$is the union of the sets$E(k)$. For$E(k)-\gamma _{k}$we have the relations$\mu(E(k)-\gamma_{k})=\mu(E(k)),E(k)-\gamma _{k}\subset E.$Moreover, by$(b)(E(k)-\gamma_{k})\cap(E(j)-\gamma_{j})=\emptyset$for$k\neq j.$Therefore$(c)$is true.$(d)$Now let$E\subset U_{\varepsilon}(S_{\rho}^{^{\prime}}).$Then by$(b)$the set$E$can be devided into a finite number of the pairwise disjoint sets$E_{1},E_{2},...,E_{n}$such that there exist the vectors$\gamma_{1}% ,\gamma_{2},...,\gamma_{n}$satisfying$(E_{k}+\gamma_{k})\subset F^{\star},(E_{k}+\gamma_{k})\cap(E_{j}+\gamma_{j})\neq\emptyset$for$k,j=1,2,...,n$and$k\neq j.$Using$\mu(E_{k}+\gamma_{k})=\mu(E_{k}),$we get the proof of$(d),$because$Tr_{F^{\star}}(E)$and$E$are union of the pairwise disjoint sets$E_{k}+\gamma_{k}$and$E_{k}$for$k=1,2,...,n$respectively \end{proof} \begin{theorem}$(a)$The set$U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))$is a subset of$B$. For every connected open subset$E$of$U_{\varepsilon }(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho)$there exists a unique index$N$such that$\Lambda_{N}(t)=\Lambda(\gamma+t)$for$\gamma+t\in E,$where$\Lambda(\gamma+t)$is defined in Remark 3. Moreover, \begin{equation} \frac{\partial}{\partial t_{j}}\Lambda(\gamma+t)=\frac{\partial}{\partial t_{j}}\mid\gamma+t\mid^{2}+O(\rho^{1-2\alpha_{1}}),\forall j=1,2,...,d. \end{equation}$(b)$For the part$V_{\rho}\equiv S_{\rho}^{^{\prime}}\backslash U_{\varepsilon}(Tr(A(\rho)))$of the approximated isoenergetic surface$S_{\rho}$the following holds \begin{equation} \mu(V_{\rho})>(1-c_{9}\rho^{-\alpha}))\mu(B(\rho)). \end{equation} Moreover,$U_{\varepsilon}(V_{\rho})$lies in the subset$U_{\varepsilon }(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))$of the simple set$B.(c)$The isoenergetic surface$I(\rho)$contains the set$I_{\rho}% ^{^{\prime\prime}},$which consists of the smooth surfaces and has the measure \begin{equation} \mu(I_{\rho}^{^{\prime\prime}})=\mu(I_{\rho}^{^{\prime}})>(1-c_{10}% \rho^{-\alpha})\mu(B(\rho)), \end{equation} where$I_{\rho}^{^{\prime}}$is a part of the translated isoenergetic surfaces of$L(q),$which is contained in the subset$U_{\varepsilon}(S_{\rho }^{^{\prime}})\backslash Tr(A(\rho))$of the simple set$B.$In particular the number$\rho^{2}$for$\rho\gg1$lies in the spectrum of$L(q),$that is, the number of the gaps in the spectrum of$L(q)$is finite, where$q(x)\in W_{2}^{s_{0}}(\mathbb{R}^{d}/\Omega),d\geq2,s_{0}=\frac{3d-1}{2}% (3^{d}+d+2)+\frac{1}{4}d3^{d}+d+6,$and$\Omega$is an arbitrary lattice. \end{theorem} \begin{proof}$(a)$To prove that$U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))\subset B$we need to show that for each point$\gamma+t$of$U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))$the simplicity conditions (11), (12) hold and$U_{\varepsilon}(S_{\rho}^{^{\prime}% })\backslash Tr(A(\rho))\subset U(\rho^{\alpha_{1}},p).$By lemma 2(a), the condition (11) holds. Now we prove that (12) holds too. Since$\gamma+t\in U_{\varepsilon}(S_{\rho}^{^{\prime}}),$there exists$a\in S_{\rho}^{^{\prime }}$such that$\gamma+t\in U_{\varepsilon}(a).$The inequality (65) and equality$F(a)=\rho^{2}$imply \begin{equation} F(\gamma+t)\in(\rho^{2}-\varepsilon_{1},\rho^{2}+\varepsilon_{1}) \end{equation} for$\gamma+t\in U_{\varepsilon}(S_{\rho}^{^{\prime}}).$On the other hand$\gamma+t\notin Tr(A(\rho)).$It means that for any$\gamma^{^{\prime}}% \in\Gamma,$we have$\gamma^{^{\prime}}+t\notin A(\rho).$If$\gamma ^{^{\prime}}\in K$and$\gamma^{^{\prime}}+t\in E_{k}\backslash E_{k+1},$then by definition of$K$( see introduction) the inequality$\mid F(\gamma +t)-\mid\gamma^{^{\prime}}+t\mid^{2}\mid<\frac{1}{3}\rho^{\alpha_{1}}$holds. This and (69) imply that$\gamma^{^{\prime}}+t\in(E_{k}\backslash E_{k+1})\cap K_{\rho}$( see (62) for the definition of$K_{\rho}$). Since$\gamma ^{^{\prime}}+t\notin A(\rho),$we have$\lambda_{i}(\gamma^{^{\prime}% }+t)\notin(\rho^{2}-3\varepsilon_{1},\rho^{2}+3\varepsilon_{1})$for$\gamma^{^{\prime}}\in K$and$\gamma^{^{\prime}}+t\in E_{k}\backslash E_{k+1}.$Therefore (12) follows from (69). Moreover, it is clear that the inclusion$S_{\rho}^{^{\prime}}\subset U(2\rho^{\alpha_{1}},p)$( see definition of$S_{\rho}$and$S_{\rho}^{^{\prime}}$) implies that$U_{\varepsilon}(S_{\rho}^{^{\prime}})\subset U(\rho^{\alpha_{1}},p).$Thus$U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))\subset B.$Now let$E$be a connected open subset of$U_{\varepsilon}(S_{\rho}^{^{\prime }})\backslash Tr(A(\rho)\subset B.$By Theorem 3 and Remark 3 for$a\in E\subset U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho)$there exists a unique index$N(a)$such that$\Lambda(a)=\Lambda_{N(a)}(a)$,$\Psi_{a}(x)=\Psi_{N(a),a}(x)$,$\mid(\Psi_{N(a),a}(x),e^{i(a,x)})\mid ^{2}>\frac{1}{2}$and$\Lambda(a)$is a simple eigenvalue. On the other hand, for fixed$N$the functions$\Lambda_{N}(t)$and$(\Psi_{N,t}(x),e^{i(t,x)})$are continuous in neighborhood of$a$if$\Lambda_{N}(a)$is a simple eigenvalue. Therefore for each$a\in E$there exists a neighborhood$U(a)\subset E$of$a$such that$\mid(\Psi_{N(a),y}(x),e^{i(y,x)})\mid ^{2}>\frac{1}{2}$, for$y\in U(a).$Since for$y\in E$there is a unique integer$N(y)$satisfying$\mid(\Psi_{N(y),y}(x),e^{i(y,x)})\mid^{2}>\frac {1}{2},$we have$N(y)=N(a)$for$y\in U(a).$Hence we proved that% \begin{equation} \forall a\in E,\exists U(a)\subset E:N(y)=N(a),\forall y\in U(a). \end{equation} Now let$a_{1}$and$a_{2}$be two points of$E$, and let$C\subset E$be the arc that joins these points. (Note that the open connected subset of$\mathbb{R}^{d}$is arcwise connected). Let$U(y_{1}),U(y_{2}),...,U(y_{k})$be a finite subcover of the open cover$\cup_{a\in C}U(a)$of the compact$C,$where$U(a)$is a neighborhood of$a$satisfying (70). By (70), we have$N(y)=N(y_{i})=N_{i}$for$y\in U(y_{i}).$Clearly, if$U(y_{i})\cap U(y_{j})\neq\emptyset,$then$N_{i}=N(z)=N_{j},$where$z\in U(y_{i})\cap U(y_{j})$. Thus$N_{1}=N_{2}=...=N_{k}$and$N(a_{1})=N(a_{2}).$To calculate the partial derivatives of the function$\Lambda(\gamma +t)=\Lambda_{N}(t)$we write the operator$L_{t}$in the form$-\triangle -(2it,\nabla)+(t,t).Then, it is clear that \begin{align} \frac{\partial}{\partial t_{j}}\Lambda_{N}(t) & =2t_{j}(\Phi_{N,t}% (x),\Phi_{N,t}(x))-2i(\frac{\partial}{\partial x_{j}}\Phi_{N,t}(x),\Phi _{N,t}(x)),\\ \Phi_{N,t}(x) & =\sum_{\gamma^{^{\prime}}\in\Gamma}b(N,\gamma^{^{\prime}% })e^{i(\gamma^{^{\prime}},x)}, \end{align} where\Phi_{N,t}(x)=e^{-i(t,x)}\Psi_{N,t}(x).$If$\mid\gamma^{^{\prime}}% \mid\geq2\rho,$then using$\Lambda_{N}\equiv\Lambda(\gamma+t)=\rho^{2}+O(\rho^{-\alpha}),$( see (46), (69)), and the obvious inequality$\mid\Lambda_{N}-\mid\gamma^{^{\prime}}-\gamma_{1}-\gamma_{2}-...-\gamma _{k}+t\mid^{2}\mid>c_{11}\mid\gamma^{^{\prime}}\mid^{2}$for$k=0,1,...,p,$where$\mid\gamma_{1}\mid<\frac{1}{4p}\mid\gamma^{^{\prime}}\mid,$and itarating (17)$p$times by using the decomposition$q(x)=\sum_{\mid\gamma _{1}\mid<\frac{1}{4p}\mid\gamma^{^{\prime}}\mid}q_{\gamma_{1}}e^{i(\gamma _{1},x)}+O(\mid\gamma^{^{\prime}}\mid^{-p}),we get \begin{align} b(N,\gamma^{^{\prime}}) & =\sum_{\gamma_{1},\gamma_{2},...}\dfrac {q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{p}}b(N,\gamma^{^{\prime}}% -\sum_{i=1}^{p}\gamma_{i})}{\prod_{j=0}^{p-1}(\Lambda_{N}-\mid\gamma ^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}+t\mid^{2})}+O(\mid\gamma^{^{\prime}}% \mid^{-p}),\\ b(N,\gamma^{^{\prime}}) & =O(\mid\gamma^{^{\prime}}\mid^{-p}),\text{ }\forall\mid\gamma^{^{\prime}}\mid\geq2\rho \end{align} By (74) the series in (72) can be differentiated term by term. Hence \begin{equation} -i(\frac{\partial}{\partial x_{j}}\Phi_{N,t},\Phi_{N,t})=\sum_{\gamma ^{^{\prime}}\in\Gamma}\gamma^{^{\prime}}(j)\mid b(N,\gamma^{^{\prime}}% )\mid^{2}=\gamma(j)\mid b(N,\gamma)\mid^{2}+\Sigma_{1}+\Sigma_{2}, \end{equation} where\Sigma_{1}=\sum_{\mid\gamma^{^{\prime}}\mid\geq2\rho}\gamma^{^{\prime}% }(j)\mid b(N,\gamma^{^{\prime}})\mid^{2},\Sigma_{2}=\sum_{\mid \gamma^{^{\prime}}\mid<2\rho,\gamma^{^{\prime}}\neq\gamma}\gamma^{^{\prime}% }(j)\mid b(N,\gamma^{^{\prime}})\mid^{2}.$By (13),$\sum_{2}=O(\rho^{-2\alpha_{1}+1}),\gamma(j)\mid b(N,\gamma )\mid^{2}=\gamma(j)(1+O(\rho^{-2\alpha_{1}}),$and by (74),$\sum_{1}% =O(\rho^{-2\alpha_{1}}).$Therefore (71) and (75) imply (66).$(b)$To ptove the inclusion$U_{\varepsilon}(V_{\rho})\subsetU_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))$we need to show that if$a\in V_{\rho},$then$U_{\varepsilon}(a)\subset U_{\varepsilon }(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho)).$This is clear, because the relation$a\in V_{\rho}\subset S_{\rho}^{^{\prime}}$implies that$U_{\varepsilon}(a)\subset U_{\varepsilon}(S_{\rho}^{^{\prime}})$and the relation$a\notin U_{\varepsilon}(Tr(A(\rho)))$implies that$U_{\varepsilon }(a)\cap Tr(A(\rho))=\emptyset.$To prove (67) first we estimate the measure of$S_{\rho},S_{\rho}^{^{\prime}},U_{2\varepsilon}(A(\rho)), namely we prove \begin{align} \mu(S_{\rho}) & >(1-c_{12}\rho^{-\alpha})\mu(B(\rho)),\\ \mu(S_{\rho}^{^{\prime}}) & >(1-c_{13}\rho^{-\alpha})\mu(B(\rho)),\\ \mu(U_{2\varepsilon}(A(\rho))) & =O(\rho^{-\alpha})\mu(B(\rho))\varepsilon \end{align} ( see below, Estimations 1, 2, 3). The estimation (67) of the measure of the setV_{\rho}$is done in Estimation 4 by using Estimations 1, 2, 3.$(c)$In Estimation 5 we prove the formula (68). The Theorem is proved \end{proof} In Estimations 1-5 we use the notations:$G(+i,a)=\{x\in G,x_{i}>a\},G(-i,a)=\{x\in G,x_{i}<-a\},$where$x=(x_{1},x_{2},...,x_{d}),a>0.$It is not hard to verify that for any subset$G$of$U_{\varepsilon}(S_{\rho }^{^{\prime}})\cup U_{2\varepsilon}(A(\rho))$, that is, for all considered sets$G$in these estimations, and for any$x\in G$the followings hold \begin{equation} \rho-1<\mid x\mid<\rho+1,\text{ }G\subset(\cup_{i=1}^{d}(G(+i,\rho d^{-1})\cup G(-i,\rho d^{-1})) \end{equation} Indeed, if$x\in S_{\rho}^{^{\prime}},$then$F(x)=\rho^{2}$and by definition of$F(x)$( see Lemma 2) and (23) we have$\mid x\mid=\rho+O(\rho ^{-1-\alpha_{1}}).$Hence the inequalities in (79) hold for$x\in U_{\varepsilon}(S_{\rho}^{^{\prime}}).$If$x\inA(\rho),$then by definition of$A(\rho)$( see (61), (62)), we have$x\in K_{\rho},$and hence$\mid x\mid=\rho+O(\rho^{-1+\alpha_{1}})$. Thus the inequalities in (79) hold for$x\in U_{2\varepsilon}(A(\rho))$too. The inclusion in (79) follows from these inequalities. If$G\subset S_{\rho},$then by (43) we have$\frac{\partial F(x)}{\partial x_{k}}>0$for$x\in G(+k,\rho^{-\alpha})$. Therefore to calculate the measure of$G(+k,a)$for$a\geq\rho^{-\alpha}$we use the formula \begin{equation} \mu(G(+k,a))=\int\limits_{\Pr_{k}(G(+k,a))}(\frac{\partial F}{\partial x_{k}% })^{-1}\mid grad(F)\mid dx_{1}...dx_{k-1}dx_{k+1}...dx_{d}, \end{equation} where$\Pr_{k}(G)\equiv\{(x_{1},x_{2},...,x_{k-1},x_{k+1},x_{k+2}% ,...,x_{d}):x\in G\}$is the projection of$G$on the hyperplane$x_{k}=0.$Instead of$\Pr_{k}(G)$we write$\Pr(G)$if$k$is unambiguous. If$D$is$m-$dimensional subset of$\mathbb{R}^{m},$then to estimate$\mu(D),$we use the formula \begin{equation} \mu(D)=\int\limits_{\Pr_{k}(D)}\mu(D(x_{1},...x_{k-1},x_{k+1},...,x_{m}% ))dx_{1}...dx_{k-1}dx_{k+1}...dx_{m}, \end{equation} where$D(x_{1},...x_{k-1},x_{k+1},...,x_{m})=\{x_{k}:(x_{1},x_{2}% ,...,x_{m})\in D\}.$ESTIMATION 1. Here we prove (76) by using (80). During this estimation the set$S_{\rho}$is redenoted by$G.$If$x\in G,$then$x\notin V_{b}(\rho ^{\alpha_{1}})$for all$b\in\Gamma(p\rho^{\alpha}).$Since the rotation does not change the measure, we choose the coordinate axis so that the direction of a fixed$b\in\Gamma(p\rho^{\alpha})$coinsides with the direction of$(1,0,0,...,0),$that is,$b=(b_{1},0,0,...,0),b_{1}>0.$Then the relations$x\notin V_{b}(\rho^{\alpha_{1}}),\mid b\mida,$where$a=(\rho^{\alpha_{1}}-b_{1}^{2})(2b_{1})^{-1}>\rho^{\alpha}.$Therefore$G=G(+1,a)\cup G(-1,a).$Now we estimate$\mu(G(+1,a))$by using (80) for$k=1and \ the relations% \begin{align} \frac{\partial F}{\partial x_{1}} & >\rho^{\alpha},(\frac{\partial F}{\partial x_{1}})^{-1}\mid grad(F)\mid=\frac{\mid x\mid}{x_{1}}% +O(\rho^{-2\alpha}),\\ \Pr(G(+1,a)) & \supset\Pr(A(+1,2a)), \end{align} wherex\in G(+1,a),a>\rho^{\alpha},A=B(\rho)\cap U(3\rho^{\alpha_{1}},p).$Here (82) follows from (43). Now we prove (83). If$(x_{2},...,x_{d})\in \Pr_{1}(A(+1,2a)),$then by definition of$A(+1,2a)$there exists$x_{1}$such that \begin{equation} x_{1}>2a>2\rho^{\alpha},\text{ }x_{1}^{2}+x_{2}^{2}+...+x_{d}^{2}=\rho ^{2},\mid\sum_{i\geq1}(2x_{i}b_{i}-b_{i}^{2})\mid\geq3\rho^{\alpha_{1}}% \end{equation} for all$(b_{1},b_{2},...,b_{d})\in\Gamma(p\rho^{\alpha}).$Therefore for$h=\rho^{-\alpha}$we have$(x_{1}+h)^{2}+x_{2}^{2}+...+x_{q}^{2}>\rho^{2}+\rho^{-\alpha},(x_{1}% -h)^{2}+x_{2}^{2}+...+x_{q}^{2}<\rho^{2}-\rho^{-\alpha}.$This and (23) give$F(x_{1}+h,x_{2},...,x_{d})>\rho^{2}$,$F(x_{1}-h,x_{2},...,x_{d})<\rho^{2}$. Since$F$is a continuous function there is$y_{1}\in(x_{1}-h,x_{1}+h)$such that (see (84)) \begin{equation} y_{1}>a,F(y_{1},x_{2},...,x_{d})=\rho^{2},\mid2y_{1}b_{1}-b_{1}^{2}% +\sum_{i\geq2}(2x_{i}b_{i}-b_{i}^{2})\mid>\rho^{\alpha_{1}}, \end{equation} because the expression under the absolute value in (85) differ from the expression under the absolute value in (84) by$2(y_{1}-x_{1})b_{1},$where$\mid y_{1}-x_{1}\mid0$. It follows from the definitions of$S_{\rho},P_{b}$and$F(x)$( see the beginning of this section, (60), and Lemma 2(a)) that if$(x_{1},x_{2},...,x_{d})\in Gthen \begin{align} x_{1}^{2}+x_{2}^{2}+...+x_{d}^{2}+F_{k_{1}-1}(x) & =\rho^{2},\\ (x_{1}+b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2}+F_{k_{1}-1}(x+b) & =\rho^{2}+h, \end{align} whereh\in(-3\varepsilon_{1},3\varepsilon_{1}).$Subtracting (86) from (87) and using (23), we get \begin{equation} (2x_{1}+b_{1})b_{1}=O(\rho^{-\alpha_{1}}). \end{equation} This and the inequalities in (79) imply \begin{equation} \mid b_{1}\mid<2\rho+3,\text{ }x_{1}=\frac{b_{1}}{2}+O(\rho^{-\alpha_{1}}% b_{1}^{-1}),\mid x_{1}^{2}-(\frac{b_{1}}{2})^{2}\mid=O(\rho^{-\alpha_{1}}). \end{equation} Consider two cases. Case 1:$b\in\Gamma_{1},$where$\Gamma_{1}=\{b\in \Gamma:\mid\rho^{2}-\mid\frac{b}{2}\mid^{2}\mid<3d\rho^{-2\alpha}\}.$In this case using the last equality in (89), (86), (23), and taking into account that$b=(b_{1},0,0,...,0),\alpha_{1}=3\alpha,$we obtain \begin{equation} x_{1}^{2}=\rho^{2}+O(\rho^{-2\alpha}),\mid x_{1}\mid=\rho+O(\rho^{-2\alpha -1}),x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2}=O(\rho^{-2\alpha}). \end{equation} Therefore$G\subset G(+1,a)\cup G(-1,a),$where$a=\rho-\rho^{-1}$. Using (80), the obvious relation$\mu(\Pr_{1}(G(\pm1,a))=O(\rho^{-(d-1)\alpha})$(see (90)) and taking into account that the expression under the integral in (80) for$k=1$is equal to$1+O(\rho^{-\alpha})$(see (82) and (90)), we get$\mu(G(\pm1,a))=O(\rho^{-(d-1)\alpha}).$Thus$\mu(G)=O(\rho^{-(d-1)\alpha}).$Since$\mid\Gamma_{1}\mid=O(\rho^{d-1}),$we have \begin{equation} \text{ }\mu(\cup_{b\in\Gamma_{1}}(S_{\rho}\cap P_{b})=O(\rho^{-(d-1)\alpha +d-1})=O(\rho^{-\alpha})\mu(B(\rho)). \end{equation} Case 2:$\mid\rho^{2}-\mid\frac{b}{2}\mid^{2}\mid\geq3d\rho^{-2\alpha}.$Repeating the proof of (90), we get% \begin{equation} \mid x_{1}^{2}-\rho^{2}\mid>2d\rho^{-2\alpha},\text{ }\sum_{k=2}^{d}x_{k}% ^{2}>d\rho^{-2\alpha},\text{ }\max_{k\geq2}\mid x_{k}\mid>\rho^{-\alpha}. \end{equation} Therefore$G\subset\cup_{k\geq2}(G(+k,\rho^{-\alpha})\cup G(-k,\rho^{-\alpha })).$Now we estimate$\mu(G(+d,\rho^{-\alpha}))$by using (80). Redenote by$D$the set$\Pr_{d}G(+d,\rho^{-\alpha}).$If$x\in G(+d,\rho^{-\alpha}),$then according to (86) and (43) the under integral expression in (80) for$k=d$is$O(\rho^{1+\alpha}).$Therefore the first equality in \begin{equation} \mu(D)=O(\varepsilon_{1}\mid b\mid^{-1}\rho^{d-2}),\text{ }\mu(G(+d,\rho ^{-\alpha}))=O(\rho^{d-1+\alpha}\varepsilon_{1}\mid b\mid^{-1}) \end{equation} implies the second equality in (93). To prove the first equality in (93) we use (81) for$m=d-1$and$k=1$and prove the relations$\mu(\Pr_{1}% D)=O(\rho^{d-2}),$\begin{equation} \text{ }\mu(D(x_{2},x_{3},...,x_{d-1}))<6\varepsilon_{1}\mid b\mid^{-1}% \end{equation} for$(x_{2},x_{3},...,x_{d-1})\in\Pr_{1}D.$First relation follows from the inequalities in (79)). So we need to prove (94). If$x_{1}\in D(x_{2}% ,x_{3},...,x_{d-1})$then (86) and (87) holds. Subtructing (86) from (87), we get \begin{equation} 2x_{1}b_{1}+(b_{1})^{2}+F_{k_{1}-1}(x-b)-F_{k_{1}-1}(x)=h, \end{equation} where$x_{2},x_{3},...,x_{d-1}$are fixed . Hence we have two equations (86) and (95) with respect two unknown$x_{1}$and$x_{d}$. Using (43), the implicit function theorem, and the inequalities$\mid x_{d}\mid>\rho^{-\alpha },\alpha_{1}>2\alpha$from (86), we obtain \begin{equation} x_{d}=f(x_{1}),\text{ }\frac{df}{dx_{1}}=-\frac{2x_{1}+O(\rho^{-2\alpha _{1}+\alpha})}{2x_{d}+O(\rho^{-2\alpha_{1}+\alpha})}=-\frac{x_{1}}{x_{d}% }+O(\rho^{-\alpha_{1}}). \end{equation} Substituting this in (95), we get \begin{equation} 2x_{1}b_{1}+b_{1}^{2}+F_{k_{1}-1}(x_{1}+b_{1},x_{2},...,x_{d-1},f(x_{1}% ))-F_{k_{1}-1}(x_{1},...,x_{d-1},f)=h. \end{equation} Using (43), (96), the first equality in (89), and$x_{d}>\rho^{-\alpha}$we see that the absolute value of the derivative (w.r.t.$x_{1}$) of the left-hand side of (97) satisfies$\mid2b_{1}+O(\rho^{-2\alpha_{1}+\alpha})(1+\mid\frac{df}{dx_{1}}\mid)\mid=\mid2b_{1}+O(\rho^{-2\alpha_{1}+\alpha})(1+\mid\frac{x_{1}}{x_{d}}% \mid)+O(\rho^{-3\alpha_{1}+\alpha})\mid>b_{1}.$Therefore from (97) by implicit function theorem, we get$\mid\frac{dx_{1}}{dh}\mid<\frac{1}{\mid b\mid}.$This inequality and relation$h\in(-3\varepsilon_{1},3\varepsilon _{1})$imply (94). Thus (93) is proved. In the same way we get the same estimation for$G(+k,\rho^{-\alpha})$and$G(-k,\rho^{-\alpha})$for$k\geq2$. Hence$\mu(S_{\rho}\cap P_{b})=O(\rho^{d-1+\alpha}\varepsilon_{1}\mid b\mid^{-1}),$for$b\notin\Gamma_{1}.$Since$\mid b\mid<2\rho+3$( see (89)) and$\varepsilon_{1}=\rho^{-d-2\alpha}$, taking into account that the number of the vectors of$\Gamma$satisfying$\mid b\mid<2\rho+3$is$O(\rho^{d}),$we obtain$\mu(\cup_{b\notin\Gamma_{1}}(S_{\rho}\cap P_{b}))=O(\rho^{2d-1+\alpha }\varepsilon_{1})=O(\rho^{-\alpha})\mu(B(\rho)).$This, (91) and (76) give the proof of (77). ESTIMATION 3. Here we prove (78). Denote$U_{2\varepsilon}(A_{k,j}(\gamma _{1,}\gamma_{2},...,\gamma_{k}))$by$G,$where$\gamma_{1,}\gamma _{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha}),k\leq d-1,$and$A_{k,j}$is defined in (61). We turn the coordinate axis so that$Span\{\gamma_{1,}\gamma_{2},...,\gamma_{k}\}=\{x=(x_{1},x_{2},...,x_{k}% ,0,0,...,0):x_{1},x_{2},...,x_{k}\in\mathbb{R}\}$. Then by (31), we have$x_{n}=O(\rho^{\alpha_{k}+(k-1)\alpha})$for$n\leq k,x\in G$. This, (79), and$\alpha_{k}+(k-1)\alpha<1$( see the sixth inequality in (14)) give$G\subset(\cup_{i>k}(G(+i,\rho d^{-1})\cup G(-i,\rho d^{-1})),\mu(\Pr_{i}(G(+i,\rho d^{-1})))=O(\rho^{k(\alpha_{k}+(k-1)\alpha)+(d-1-k)})$for$i>k.$Now using this and (81) for$m=d,$we prove that \begin{equation} \mu(G(+i,\rho d^{-1}))=O(\varepsilon\rho^{k(\alpha_{k}+(k-1)\alpha )+(d-1-k)}),\forall i>k. \end{equation} For this we redenote by$D$the set$G(+i,\rho d^{-1})$and prove that \begin{equation} \mu((D(x_{1},x_{2},...x_{i-1},x_{i+1},...x_{d}))\leq(42d^{2}+4)\varepsilon \end{equation} for$(x_{1},x_{2},...x_{i-1},x_{i+1},...x_{d})\in\Pr_{i}(D)$and$i>k.$To prove (99) it is sufficient to show that if both$x=(x_{1},x_{2}% ,...,x_{i},...x_{d})$and$x^{^{\prime}}=(x_{1},x_{2},...,x_{i}^{^{\prime}% },...,x_{d})$are in$D,$then$\mid x_{i}-x_{i}^{^{\prime}}\mid\leq (42d^{2}+4)\varepsilon.$Assume the converse. Then$\mid x_{i}-x_{i}% ^{^{\prime}}\mid>(42d^{2}+4)\varepsilon$. Without loss of generality it can be assumed that$x_{i}^{^{\prime}}>x_{i}.$So$x_{i}^{^{\prime}}>x_{i}>\rho d^{-1}$( see definition of$D$). Since$x$and$x^{^{\prime}}$lie in the$2\varepsilon$neighborhood of$A_{k,j},$there exist points$a$and$a^{^{\prime}}$in$A_{k,j}$such that$\mid x-a\mid<2\varepsilon$and$\mid x^{^{\prime}}-a^{^{\prime}}\mid<2\varepsilon.$It follows from the definitions of the points$x,x^{^{\prime}},a,a^{^{\prime}}that the following inequalities hold:% \begin{align} \rho d^{-1}-2\varepsilon & 42d^{2}\varepsilon,\\ (a_{i}^{^{\prime}})^{2}-(a_{i})^{2} & >2(\rho d^{-1}-2\varepsilon )(a_{i}^{^{\prime}}-a_{i}),\nonumber\\ & \mid\mid a_{s}\mid-\mid a_{s}^{^{\prime}}\mid\mid<4\varepsilon,\forall s\neq i.\nonumber \end{align} On the other hand for points ofA_{k,j}$the inequalities in (79) hold, that is, we have$\mid a_{s}\mid<\rho+1,\mid a_{s}^{^{\prime}}\mid<\rho+1.$Therefore these inequalities and the inequalities in (100) imply$\mid\mid a_{s}\mid^{2}-\mid a_{s}^{^{\prime}}\mid^{2}\mid<12\rho\varepsilon$for$s\neq i$, and hence$\sum_{s\neq i}\mid\mid a_{s}\mid^{2}-\mid a_{s}^{^{\prime}}\mid^{2}% \mid<12d\rho\varepsilon<\frac{2}{7}\rho d^{-1}(a_{i}^{^{\prime}}-a_{i}),$% \begin{equation} \mid\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}\mid>\frac{3}{2}\rho d^{-1}\mid a_{i}^{^{\prime}}-a_{i}\mid. \end{equation} Now using the inequality (45), the obvious relation$\frac{1}{2}\alpha_{d}<1$( see the end of the introduction), the notations$r_{j}(x)=\lambda _{j}(x)-\mid x\mid^{2}$( see Remark 2),$\varepsilon_{1}=7\rho\varepsilon$( see Lemma 2(a)), and (101), (100), we get$\mid r_{j}(a)-r_{j}(a^{^{\prime}})\mid<\rho^{\frac{1}{2}\alpha_{d}}\mid a-a^{^{\prime}}\mid<\frac{1}{2}\rho d^{-1}\mid a_{i}^{^{\prime}}-a_{i}\mid,\mid\lambda_{j}(a)-\lambda_{j}(a^{^{\prime}})\mid\geq\mid\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}\mid-\mid r_{j}(a)-r_{j}(a^{^{\prime}})\mid>\rho d^{-1}\mid a_{i}^{^{\prime}}-a_{i}\mid>42d\rho\varepsilon>6\varepsilon _{1}.$The obtained inequality$\mid\lambda_{j}(a)-\lambda_{j}(a^{^{\prime}}% )\mid>6\varepsilon_{1}$controdicts with inclutions$a\in A_{k,j},a^{^{\prime}}\in A_{k,j},$since by definition of$A_{k,j}$( see (61)) both$\lambda_{j}(a)$and$\lambda_{j}(a^{^{\prime}})$lie in$(\rho^{2}% -3\varepsilon_{1},\rho^{2}+3\varepsilon_{1}).$Thus (99), hense (98) is proved. In the same way we get the same formula for$G(-i,\frac{\rho}{d}).$So$\mu(U_{2\varepsilon}(A_{k,j}(\gamma_{1,}\gamma_{2},...,\gamma_{k}% )))=O(\varepsilon\rho^{k(\alpha_{k}+(k-1)\alpha)+d-1-k}).$Now taking into account that$U_{2\varepsilon}(A(\rho))$is union of$U_{2\varepsilon}% (A_{k,j}(\gamma_{1,}\gamma_{2},...,\gamma_{k})$for$k=1,2,..,d-1;j=1,2,...,b_{k}(\gamma_{1},\gamma_{2},...,\gamma_{k}),$and$\gamma _{1},\gamma_{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha})$\ ( see (61)) and using that$b_{k}=O(\rho^{d\alpha+\frac{k}{2}\alpha_{k+1}})$( see (39)) and the number of the vectors$(\gamma_{1},\gamma_{2},...,\gamma_{k})$for$\gamma_{1},\gamma_{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha})$is$O(\rho^{dk\alpha}),$we obtain $\mu(U_{2\varepsilon}(A(\rho)))=O(\varepsilon\rho^{d\alpha+\frac{k}{2}% \alpha_{k+1}+dk\alpha+k(\alpha_{k}+(k-1)\alpha)+d-1-k}).$ Therefore to prove (78), it remains to show that$d\alpha+\frac{k}{2}\alpha_{k+1}+dk\alpha+k(\alpha_{k}+(k-1)\alpha)+d-1-k\le q d-1-\alpha$or% \begin{equation} (d+1)\alpha+\frac{k}{2}\alpha_{k+1}+dk\alpha+k(\alpha_{k}+(k-1)\alpha)\leq k \end{equation} for$1\leq k\leq d-1$. Dividing both side of (102) by$k\alpha$and using$\alpha_{k}=3^{k}\alpha,\alpha=\frac{1}{q},q=3^{d}+d+2$( see the end of the introduction) we see that (102) is equivalent to$\frac{d+1}{k}% +\frac{3^{k+1}}{2}+3^{k}+k-1\leq3^{d}+2.$The left-hand side of this inequality gets its maximum at$k=d-1.$Therefore we need to show that$\frac{d+1}{d-1}+\frac{5}{6}3^{d}+d\leq3^{d}+4,$which follows from the inequalities$\frac{d+1}{d-1}\leq3,d<\frac{1}{6}3^{d}+1$for$d\geq2.$ESTIMATION 4. Here we prove (67). During this estimation we denote by$G$the set$S_{\rho}^{^{\prime}}\cap U_{\varepsilon}(Tr(A(\rho))$. Since$V_{\rho }=S_{\rho}^{^{\prime}}\backslash G$and (77) holds, it is enough to prove that$\mu(G)=O(\rho^{-\alpha})\mu(B(\rho)).$For this we use (79) and prove$\mu(G(+i,\rho d^{-1}))=O(\rho^{-\alpha})\mu(B(\rho))$for$i=1,2,...,d$by using (80) ( the same estimation for$G(-i,\rho d^{-1})$can be proved in the same way). By (43), if$x\in G(+i,\rho d^{-1}),$then the under integral expression in (80) for$k=i$and$a=\rho d^{-1}$is less than$d+1.$Therefore it is sufficient to prove \begin{equation} \mu(\Pr(G(+i,\rho d^{-1}))=O(\rho^{-\alpha})\mu(B(\rho)) \end{equation} Clearly, if$(x_{1},x_{2},...x_{i-1},x_{i+1},...x_{d})\in\Pr_{i}(G(+i,\rho d^{-1})),$then$\mu(U_{\varepsilon}(G)(x_{1},x_{2},...x_{i-1},x_{i+1},...x_{d}))\geq 2\varepsilon$and by (81), it follows that \begin{equation} \mu(U_{\varepsilon}(G))\geq2\varepsilon\mu(\Pr(G(+i,\rho d^{-1})). \end{equation} Hence to prove (103) we need to estimate$\mu(U_{\varepsilon}(G)).$For this we prove that \begin{equation} U_{\varepsilon}(G)\subset U_{\varepsilon}(S_{\rho}^{^{\prime}}),U_{\varepsilon }(G)\subset U_{2\varepsilon}(Tr(A(\rho))),U_{\varepsilon}(G)\subset Tr(U_{2\varepsilon}(A(\rho))). \end{equation} The first and second inclusions follow from$G\subset S_{\rho}^{^{\prime}}$and$G\subset U_{\varepsilon}(Tr(A(\rho)))$respectively (see definition of$G$). Now we prove the third inclusion in (105). If$x\in U_{\varepsilon }(G),$then by the second inclusion of (105) there exists$b$such that$b\in Tr(A(\rho)),\mid x-b\mid<2\varepsilon.$Then by the definition of$Tr(A(\rho))$there exist$\gamma\in\Gamma$and$c\in A(\rho)$such that$b=\gamma+c$. Therefore$\mid x-\gamma-c\mid=\mid x-b\mid<2\varepsilon,x-\gamma\in U_{2\varepsilon}(c)\subset U_{2\varepsilon}(A(\rho)).$This together with$x\in U_{\varepsilon}(G)\subset U_{\varepsilon}(S_{\rho }^{^{\prime}})$(see the first inclusion of (105)) give$x\in Tr(U_{2\varepsilon}(A(\rho)))$( see the definition of$Tr(E)$in the beginning of this section), i.e., the third inclusion in (105) is proved. The third inclusion, Lemma 2(c), and (78) imply that$\mu(U_{\varepsilon}(G))=O(\rho^{-\alpha})\mu(B(\rho))\varepsilon.$This and (104) imply the proof of (103)$\diamondsuit$ESTIMATION 5 Here we prove (68). Divide the set$V_{\rho}\equiv V$into pairwise disjoint subsets$V^{^{\prime}}(\pm1,\rho d^{-1})\equiv V(\pm1,\rho d^{-1}),V^{^{\prime}}(\pm i,\rho d^{-1})\equiv V(\pm i,\rho d^{-1})\backslash (\cup_{j=1}^{i-1}(V(\pm j,\rho d^{-1}))),$for$i=2,3,...,d.$Take any point$a\in V^{^{\prime}}(+i,\rho d^{-1})\subset S_{\rho}$and consider the function$F(x)$( see Lemma 2(a)) on the interval$[a-\varepsilon e_{i},a+\varepsilon e_{i}],$where$e_{1}=(1,0,0,...,0)$,$e_{2}=(0,1,0,...,0),...$. By the definition of \$S_{\rho}$we have$F(a)=\rho^{2}$. It follows from (43) and the definition of$V^{^{\prime}}(+i,\rho d^{-1})$that$\frac{\partial F(x)}{\partial x_{i}}>\rho d^{-1}$for$x\in\lbrack a-\varepsilon e_{i},a+\varepsilon e_{i}].$Therefore \begin{equation} F(a-\varepsilon e_{i})<\rho^{2}-c_{15}\varepsilon_{1},\text{ }F(a+\varepsilon e_{i})>\rho^{2}+c_{15}\varepsilon_{1}. \end{equation} Since$[a-\varepsilon e_{i},a+\varepsilon e_{i}]\in U_{\varepsilon}(a)\subset U_{\varepsilon}(V_{\rho})\subset U_{\varepsilon}(S_{\rho}^{^{\prime}% })\backslash Tr(A(\rho))$( see Theorem 5$(b)$), it follows from Theorem 5$(a)$that there exists index$N$such that$\Lambda(y)=\Lambda_{N}(y)$for$y\in U_{\varepsilon}(a)$and$\Lambda(y)$satisfies (46)( see Remark 3). Hence (106) implies that \begin{equation} \Lambda(a-\varepsilon e_{i})<\rho^{2},\Lambda(a+\varepsilon e_{i})>\rho^{2}. \end{equation} Moreover it follows from (66) that the derivative of$\Lambda(y)$with respect to$i$-th coordinate is positive for$y\in\lbrack a-\varepsilon e_{i}% ,a+\varepsilon e_{i}].$So$\Lambda(y)$is a continuous and increasing function in$[a-\varepsilon e_{i},a+\varepsilon e_{i}].$Therefore (107) implies that there exists a unique point$y(a,i)\in\lbrack a-\varepsilon e_{i},a+\varepsilon e_{i}]$such that$\Lambda(y(a,i))=\rho^{2}.$Define$I_{\rho}^{^{\prime}}(+i)$by$I_{\rho}^{^{\prime}}(+i)=\{y(a,i):a\in V^{^{\prime}}(+i,\rho d^{-1})\}).$In the same way we define$I_{\rho }^{^{\prime}}(-i)=\{y(a,i):a\in V^{^{\prime}}(-i,\rho d^{-1})\}$and put$I_{\rho}^{^{\prime}}=\cup_{i=1}^{d}(I_{\rho}^{^{\prime}}(+i)\cup I_{\rho }^{^{\prime}}(-i)).$To estimate the measure of$I_{\rho}^{^{\prime}}$we compare the measure of$V^{^{\prime}}(\pm i,\rho d^{-1})$with the measure of$I_{\rho}^{^{\prime}}(\pm i)$by using the formula (80) and the obvious relations \begin{equation} \Pr(V^{^{\prime}}(\pm i,\rho d^{-1}))=\Pr(I_{\rho}^{^{\prime}}(\pm i)),\text{ }\mu(\Pr(I_{\rho}^{^{\prime}}(\pm i)))=O(\rho^{d-1}), \end{equation}% \begin{equation} (\frac{\partial F}{\partial x_{i}})^{-1}\mid grad(F)\mid-(\frac{\partial \Lambda}{\partial x_{i}})^{-1}\mid grad(\Lambda)\mid=O(\rho^{-2\alpha_{1}}). \end{equation} Here the first equality in (108) follows from the definition of$I_{\rho }^{^{\prime}}(\pm i)$. The second equality in (108) follows from the inequalities in (79), since$I_{\rho}^{^{\prime}}\subset U_{\varepsilon }(S_{\rho}^{^{\prime}})$. Formulas (43), (66) imply (109). Clearly, using (108), (109), and (80) we get$\mu(V^{^{\prime}}(\pm i,\rho d^{-1}% ))-\mu(I_{\rho}^{^{\prime}}(\pm i))=O(\rho^{d-1-2\alpha_{1}}).$On the other hand if$y\equiv(y_{1},y_{2},...,y_{d})\in I_{\rho}^{^{\prime}}(+i)\cap I_{\rho }^{^{\prime}}(+j)$for$i(1-c_{10}% \rho^{-\alpha}))\mu(B(\rho))\diamondsuit\$ \begin{thebibliography}{99} % \bibitem {b1}B.E.J. Dahlberg, E. Trubuwits, A Remark on two Dimensional Periodic Potential, \ \ Comment. Math. Helvetica 57, 130-134, (1982). \bibitem {b2}M.S.P. Eastham, \textquotedblright The Spectral Theory of Periodic Differential Equations,\textquotedblright\ Scotting Academic Press, Edinburgh, 1973. \bibitem {b3}J. Feldman, H. Knorrer, E. Trubowitz, The Perturbatively Stable Spectrum of the Periodic Schr\"{o}dinger Operator, Invent. 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