%% This document created by Scientific Word (R) Version 2.0 %% Starting shell: mathart1 \documentclass[12pt,thmsa]{article}% \usepackage{amssymb} \usepackage{sw20lart} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{graphicx}% \setcounter{MaxMatrixCols}{30} %TCIDATA{OutputFilter=latex2.dll} %TCIDATA{Version=4.00.0.2312} %TCIDATA{TCIstyle=Article/art4.lat,lart,article} %TCIDATA{CSTFile=article.cst} %TCIDATA{LastRevised=Tuesday, June 29, 2004 17:34:21} %TCIDATA{} \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{On the Polyharmonic Operator with a Periodic Potential}} \date{} \maketitle \begin{abstract} In this paper we obtain asymptotic formulas for eigenvalues and Bloch functions of the polyharmonic operator $L(r,q(x))=-\Delta^{r}+q(x),$ of arbitrary dimension $d$ with periodic, with respect to \ arbitrary lattice, potential $q(x),$ where $r\geq1.$ Then we prove that the number of gaps in the spectrum of the operator $L(r,q(x))$ is finite which is the generalisation of the Bethe -Sommerfeld conjecture for this operator. In particular, taking $r=1$ we get the proof of the Bethe -Sommerfeld conjecture for arbitrary dimension and arbitrary lattice. \end{abstract} \bigskip \ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ INTRODUCTION$ In this paper we consider the operator% \begin{equation} L(r,q(x))=-\Delta^{r}+q(x),\ x\in R^{d},\ d\geq2 \end{equation} with a periodic (relative to a lattice $\Omega$) potential $q(x)\in W_{2}% ^{s}(F),$ where $s\geq s_{0}=\frac{3d-1}{2}(3^{d}+d+2)+\frac{1}{4}d3^{d}+d+6,$ $F\equiv R^{d}/\Omega$ is a fundamental domain of $\Omega,$ and, without loss of generality, we assume that the measure $\mu(F)$ of $F$ is $1$. Let $L_{t}(r,q(x))$ be the operator generated in $F$ by (1) and the quasiperiodic conditions:% \begin{equation} u(x+\omega)=e^{i(t,\omega)}u(x),\ \forall\omega\in\Omega, \end{equation} where $t\in F^{\ast}\equiv R^{d}/\Gamma$ and $\Gamma$ is the lattice dual to $\Omega$, that is, $\Gamma$ is the set of all vectors $\gamma\in R^{d}$ satisfying $(\gamma,\omega)\in2\pi Z$ for all $\omega\in\Omega.$ It is well-known that the operator $L_{t}(r,q(x))$ has a discrete spectrum denoted by $\Lambda_{1}(t)\leq\Lambda_{2}(t)\leq....$These functions are called band functions of the operator $L$ and they form the spectrum $Spec(L)$ of the operator $r$, that is, $Spec(L)=\left\{ \Lambda_{n}(t):t\in F^{\ast },n=1,2,...\right\} .$ The corresponding eigenfunctions $\Psi_{1,t}% (x),\ \Psi_{2,t}(x),...,$ of $r_{t}$ are known as Block functions. In the case $q(x)=0$ these eigenvalues and eigenfunctions are $\mid\gamma+t\mid^{2r}$ and $e^{i(\gamma+t,x)}$ respectively, for $\gamma\in\Gamma$. The paper consists of three sections. In section 1 the eigenvalues $\left\vert \gamma+t\right\vert ^{2r}$, for big $\ \gamma\in\Gamma,$ are divided into two groups: non-resonance ones and resonance ones and for the perturbations of each group various asymptotic formulae are obtained. To describe the non-resonance and resonance eigenvalues $\left\vert \gamma+t\right\vert ^{2r}$ of the order of $\rho^{2r}$ ( written as $\left\vert \gamma+t\right\vert ^{2r}\sim\rho^{2r}$ which means that $c_{1}\rho<\left\vert \gamma+t\right\vert 1,$ is a linear conbination of $e^{i(\gamma+t+\gamma^{^{\prime}},x)}$ for $\gamma^{^{\prime}}\in \Gamma((k-1)\rho^{\alpha})\cup\{0\}$ with explicitly expressed coefficients. In section 3 we prove that the set $B$ has asymptotically ful measure in $R^{d}$ and contains the intervals $\{a+sb:s\in\lbrack-1,1]\}$ such that $\Lambda(a-b)<\rho^{2r},$ $\Lambda(a+b)>\rho^{2r}$\ and $\Lambda(\gamma+t)$ \ is continuous on this intervals. Hence there exists $\gamma+t$ for which $\Lambda(\gamma+t)=\rho^{2r}.$ It implies the validity of Bethe-Sommerfeld conjecture for $L(r,q(x)).$ Construction of the set $B$ consists of two steps. Step 1. We prove that all eigenvalues $\Lambda(\gamma+t)$, for $\mid\gamma \mid\sim\rho,$ of the operator $r_{t}(r,q(x)),$ lie in the $\varepsilon _{1}=\rho^{-d-2\alpha},$ neighborhood of a numbers $F(\gamma+t)=\mid \gamma+t\mid^{2r}+F_{k_{1}-1}(\gamma+t)$, $\lambda_{j}(\gamma+t)$ ( see (5), (7)), where $k_{1}=[\frac{d}{3\alpha}]+2.$ We call these number as the known part of the eigenvalues. Step 2. By eliminating the set of quasimomenta $\gamma+t$, for which the known parts $F(\gamma+t)$ of $\Lambda(\gamma+t)$ 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 constructed the set $B$ such that if $\gamma+t\in B$ then the following conditions (called simplicity conditions for eigenvalue $\Lambda(\gamma+t)\equiv\Lambda_{N(\gamma+t)}(t)$ ) hold \begin{equation} \mid F(\gamma+t)-F(\gamma^{^{\prime}}+t)\mid\geq2\varepsilon_{1},\text{ }% \end{equation} for $\gamma^{^{\prime}}\in K\backslash\{\gamma\},$ $\gamma^{^{\prime}}+t\in U^{1}(\rho^{\alpha_{1}},p)$ and% \begin{equation} \mid F(\gamma+t)-\lambda_{j}(\gamma^{^{\prime}}+t)\mid\geq2\varepsilon_{1}, \end{equation} for $\gamma^{^{\prime}}\in K,\gamma^{^{\prime}}+t\in E_{k}^{1}\backslash E_{k+1}^{1},j=1,2,...,$ where $K$ is the set of $\gamma^{^{\prime}}\in\Gamma$ satisfying $\mid F(\gamma+t)-\mid\gamma^{^{\prime}}+t\mid^{2r}\mid<\frac{1}% {3}\rho^{\alpha_{1}}$. Thus $B$ is the set of quasimomenta $\gamma+t\in U^{1}(\rho^{\alpha_{1}},p)$ satisfying the simplicity conditions (9), (10). By these conditions the eigenvalue $\Lambda(\gamma+t)$ can not coincide with other eigenvalues. The listed all results ( formulas (5), (7), construction and investigations of the simply set $B,$ the proof of asymptotic formulas (8) for Bloch function and implication the proof of the Bethe-Sommerfeld conjecture for arbitraty dimension and arbitrary lattices from this formulas )\ for the first time were proved in papers [12-15] for the Schrodinger operator $L(1,q(x)).$ The main difficults and the crucial point of papers [12-14] were the construction and investigations of the simple set $B$ of quasimomenta in neighborhood of the surface $\{\gamma+t\in U^{1}(\rho^{\alpha_{1}},p):F(\gamma+t)=\rho^{2}\}.$ 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 $q_{\gamma_{1}}(x)=\sum_{n\in Z}q_{n\gamma_{1}}e^{i(n\gamma_{1},x)}$ ( see ). So for construction of the simple set $B$ of quasimomenta we eliminate the vicinities of the diffraction planes ant the the sets connected with directional potential ( see (9), (10)). The simple sets $B$ of quasimomenta for the first time is constructed and investigated in  for $d=3$ and in  for $d\geq2$ ( if $d=2$ then for nonsmooth potential $q(x)\in L_{2}(F),$ if $d>2$ for smooth potential). Then Yu.E. Karpeshina proved ( see ,,) the convergense of pertubation series for a set, that is similar to $B$, of quasimomenta in the cases 1. $2r>d;$ 2. $4r>d+1,$ $(2r\leq d)$; 3. $d=3,r=1,$ and using it she proved the validity of \ the Bethe-Sommerfeld conjecture in these cases. In papers [2,3] asymptotic formulas for eigenvalues and Bloch function of the two and three dimensional operator $L_{t}(1,q(x))$ were obtained. In  asymptotic formulae for nonresonance eigenvalues of $L_{0}(1,q(x))$ were obtained. For the first time M.M. Skriganov [10,11] 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 and for the operator $L(r,q(x)),$ for $2r>d.$ The Skriganov's method is based on the detail investigation of the arifmetical and geometrical properties of the lattice. B.E.J.Dahlberg \ and E.Trubowits  using an asymptotic of Bessel function, gave the simpler proof of this conjecture for the two dimensional Scrodinger operator. B. Helffer and A. Mohamed  by investigations the integrated density of states proved the validity of this conjecture for the Scrodinger operator for $d\leq4,$ for arbitrary lattice. Recently Parnovski and Sobelev  proved this conjecture for the operator $L(r,q(x)),$ for $8r>d+3.$ The method of this paper and papers [12-15] 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. For the operator $L(r,q),$ in order to avoid eclipsing the essence by technical details,\ we assume that $r\geq1.$ It can be replaced by $r>n_{s,d},$ where $n_{s,d}<1$ and depends on the smoothness $s$ of potential $q(x)\in W_{2}^{s}(R^{d}/\Omega)$ and the dimension $d$ of $R^{d}$ . In this paper for the different types of the measures of the subset $A$ of $R^{d}$ we use the same notation $\mu(A),$ because it will be clear which measure we mean. 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 $c_{\Gamma}\rho^{d-1},$ where $c_{\Gamma}$ is a constant depending on $\Gamma.$ Therefore the number of eigenvalues of $L_{t}(r,q)$ lying in $(a^{2r}-\rho^{2r-1},a^{2r}+\rho^{2r-1})$ is less than $c_{\Gamma}\rho^{d-1}.$ Besides we use the inequalities: 1. $\alpha_{1}+d\alpha<1-\alpha\,;$ 2. $d\alpha<\frac{1}{2}\alpha_{d};$ 3. $k_{1}\leq\frac{1}{3}(p-\frac{1}{2}(q(d-1));$ \ \ 4. $p_{1}\alpha_{1}\geq p\alpha$ 5. $3k_{1}\alpha>d+2\alpha;$\ \ \ \ 6. $\alpha_{k}+(k-1)\alpha<1;$ \ 7. $\alpha_{k+1}>2(\alpha_{k}+(k-1))\alpha,$ for $k=1,2,...,d,$ that can be easily verified by using the definitions $p=s-d,$ $\alpha_{k}=3^{k}\alpha,$ $\alpha=\frac{1}{q},$ $q=3^{d}+d+2,$ $k_{1}=[\frac{d}{3\alpha}]+2,$ $p_{1}\equiv\lbrack\frac{p}{3}]+1$ of the numbers $p,q,\alpha_{k},\alpha,k_{1},p_{1}.$ \section{Asymptotic Formulae for Eigenvalues} In this section we obtain the asymptotic formulae for nonresonance eigenvalues by iteration of the well-known formula% \begin{equation} (\Lambda_{N}-\mid\gamma+t\mid^{2r})b(N,\gamma)=(\Psi_{N,t}(x)q(x),e^{i(\gamm a +t,x)}). \end{equation} Putting the decomposition (3) of $q(x)$ into (11), we get \begin{equation} (\Lambda_{N}-\mid\gamma+t\mid^{2r})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 (11), (12) 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^{2r}}=% %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^{2r}}+O(\rho^{-p\alpha}) \end{equation} for every vector $\gamma^{^{\prime}}\in\Gamma$ satisfying the inequality \begin{equation} \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2r}\mid>\frac{1}{2}\rho ^{\alpha_{1}}, \end{equation} which is called the iterability condition. If \begin{equation} \mid\Lambda_{N}-\mid\gamma+t\mid^{2r}\mid<\frac{1}{2}\rho^{\alpha_{1}}, \end{equation} and $\mid\gamma+t\mid^{2r}$ is a non-resonance eigenvalue, i.e., $\gamma+t\in U^{r}(\rho^{\alpha_{1}},p)$ then \begin{equation} \mid\mid\gamma+t\mid^{2r}-\mid\gamma-\gamma_{1}+t\mid^{2r}\mid>\rho ^{\alpha_{1}},\mid\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2r}\mid>\frac{1}% {2}\rho^{\alpha_{1}},\text{ }% \end{equation} for all $\gamma_{1}\in\Gamma(p\rho^{\alpha}).$ Hence the vector $\gamma -\gamma_{1}$ satisfies the iterability condition (14). Therefore, in (13) one can replace $\gamma^{^{\prime}}$ by $\gamma-\gamma_{1}$ and write $b(N,\gamma-\gamma_{1})=% %TCIMACRO{\dsum _{\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}{\Lambda_{N}-\mid \gamma-\gamma_{1}+t\mid^{2r}}+O(\rho^{-p\alpha}).$ Putting it into right side of (12) and isolating the terms with the coefficient $b(N,\gamma)$, i.e., isolating the case $\gamma_{1}+\gamma_{2}=0$ we get $(\Lambda_{N}-\mid\gamma+t\mid^{2r})b(N,\gamma)=% %TCIMACRO{\dsum _{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}% {\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2r}}+O(\rho^{-p\alpha})=$% $% %TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{\mid q_{\gamma_{1}}\mid^{2}b(N,\gamma)}{\Lambda_{N}-\mid\gamma -\gamma_{1}+t\mid^{2r}}+% %TCIMACRO{\dsum _{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}% {\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2r}}+O(\rho^{-p\alpha}).$ The last sum is taken under the additional 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 $b(N,\gamma^{^{\prime}})$ for $\gamma^{^{\prime}}=\gamma-\gamma_{1}-\gamma_{2},$ and then for $\gamma ^{^{\prime}}=\gamma-\gamma_{1}-\gamma_{2}-\gamma_{3},...,$ by its expression from (13) and isolating each time the terms with coefficient $b(N,\gamma)$ we obtain \begin{equation} (\Lambda_{N}-\mid\gamma+t\mid^{2r})b(N,\gamma)=A_{p_{1}}(\Lambda_{N}% ,\gamma+t)b(N,\gamma)+C_{p_{1}}+O(\rho^{-p\alpha}), \end{equation} where $A_{p_{1}}(\Lambda_{N},\gamma+t)=\sum_{k=1}^{p_{1}}S_{k}(\Lambda _{N},\gamma+t)$ , $S_{k}(\Lambda_{N},\gamma+t)=% %TCIMACRO{\dsum _{\gamma_{1},...,\gamma_{k}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},...,\gamma_{k}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{k}}q_{-\gamma_{1}-\gamma _{2}-...-\gamma_{k}}}{\prod_{j=1}^{k}(\Lambda_{N}-\mid\gamma+t-\sum_{i=1}% ^{j}\gamma_{i}\mid^{2r})},$% $C_{p_{1}}=\sum_{\gamma_{1},...,\gamma_{p_{1}+1}\in\Gamma(\rho^{\alpha})}% \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{p_{1}+1}}b(N,\gamma -\gamma_{1}-\gamma_{2}-...-\gamma_{p_{1}+1})}{\prod_{j=1}^{p_{1}}(\Lambda _{N}-\mid\gamma+t-\sum_{i=1}^{j}\gamma_{i}\mid^{2r})}.$ Here the sums in the expressions of $S_{k}$ and $C_{p_{1}}$ are taken under the additional conditions $\gamma_{1}+\gamma_{2}+...+\gamma_{s}\neq0$ for $s=1,2,...,k$ and for $s=1,2,...,p_{1}$ respectively. These conditions and the inclusion $\gamma_{i}\in\Gamma(\rho^{\alpha}),$ for $i=1,2,...,p_{1}$ imply the relation $\sum_{i=1}^{j}\gamma_{i}\in\Gamma(p\rho^{\alpha})$ for the sum in the denominators of the fractions in expressions of $S_{k}$ and $C_{p_{1}}%$. Therefore from the relation (16) it follows that the absolute values of the denominators of the fractions in expressions of $S_{k}$ and $C_{p_{1}}$ more 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 inequality 4. in the end of introduction) give \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} Taking into account that for $\Lambda_{N}$ we only used condition (15) we obtain \begin{equation} S_{k}(a,\gamma+t)=O(\rho^{-k\alpha_{1}}),\forall k=1,2,...,p_{1}% \end{equation} for all $a$ satisfying $\mid a-\mid\gamma+t\mid^{2r}\mid<\frac{1}{2}% \rho^{\alpha_{1}}.$ Thus finding the number $N$ such that $\Lambda_{N}$ is close to $\mid\gamma+t\mid^{2r}$ and $b(N,\gamma)$ is not very small then dividing both sides of (17) by $b(N,\gamma)$ we can get the asymptotic formulas for $\Lambda_{N}$ (see Theorem 1(a)). These formulas show that in the nonresonance case the eigenvalue of the perturbed operator $L_{t}(r,q(x))$ is close to the eigenvalue of the unperturbed operator $L_{t}(r,0).$ However in Theorem 1(c) we prove that if $\mid\gamma+t\mid^{2r}$ is resonance eigenvalue, say, if $\gamma+t\in(\cap_{i=1}^{k}V_{\gamma_{i}}^{r}(\rho^{\alpha_{k}}))$ for $k\geq1,$ where $\gamma_{1},\gamma_{2},...,\gamma_{k}$ are linearly independent vectors of $\Gamma(p\rho^{\alpha})$ and the length of $\gamma_{i}$ is not more than the length of other vectors in $\Gamma\cap\gamma_{i}R,$ then the corresponding eigenvalue of $L_{t}(r,q(x))$ is close to the eigenvalue of the matrix constructed as follows. Introduse the sets \begin{align*} B_{k} & =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}}\},\\ B_{k}(\gamma+t) & =\gamma+t+B_{k}\equiv\{\gamma+t+b:b\in B_{k},\},\\ B_{k}(\gamma+t,p_{1}) & =B_{k}(\gamma+t)+\Gamma(p_{1}\rho^{\alpha}). \end{align*} Denote by $h_{i}+t$ for $i=1,2,...,b_{k}$ the vectors of $B_{k}(\gamma +t,p_{1}),$ where $b_{k}\equiv b_{k}(\gamma_{1},\gamma_{2},...,\gamma_{k})$ is the number of the vectors of $B_{k}(\gamma+t,p_{1})$. Define the matrix $C(\gamma+t,\gamma _{1},\gamma_{2},...,\gamma_{k})=(c_{i,j})$ by the formula $c_{i,j}=q_{h_{i}-h_{j}},\forall i\neq j,c_{i,i}=\mid h_{i}+t\mid^{2r},$ $i,j=1,2,...,b_{k}.$ We use essentially the following obvious statement: if $x\in R^{d},$ $\mid x\mid\sim\rho,$ $\gamma_{1}\in\Gamma,$ $\mid x+\gamma_{1}\mid\sim\rho$ then \begin{equation} \mid x\mid^{2r}-\mid x+\gamma_{1}\mid^{2r}=a^{2(r-1)}(\mid x\mid^{2}-\mid x+\gamma_{1}\mid^{2}) \end{equation} where $a\sim\rho.$ Therefore $V_{\gamma_{1}}^{r}(\rho^{\alpha_{1}})\subset V_{\gamma_{1}}^{1}(\rho^{\alpha_{1}}),$% \begin{equation} (\cap_{i=1}^{k}V_{\gamma_{i}}^{r}(\rho^{\alpha_{k}}))\subset\cap_{i=1}% ^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}})), \end{equation}% \begin{equation} U^{1}(\rho^{\alpha_{1}},p)\subset U^{r}(\rho^{\alpha_{1}},p) \end{equation} \ for $r\geq1$ and $k=1,2,....$Taking into account these inclusions we consider the resonance eigenvalue \ $\mid\gamma+t\mid^{2r}$ for $\gamma +t\in(\cap_{i=1}^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}})).$ \begin{theorem} $(a)$ Suppose $\gamma+t\in U^{r}(\rho^{\alpha_{1}},p),$ $\mid\gamma\mid \sim\rho.$ If (15) and \begin{equation} \mid b(N,\gamma)\mid>c_{4}\rho^{-c\alpha}% \end{equation} 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 $F_{0}=0,$ $F_{s}=A_{s}(\mid\gamma+t\mid^{2r}+F_{s-1},\gamma+t),$ for $s=1,2,....$ $(b)$ For $\gamma+t\in U^{r}(\rho^{\alpha_{1}},p),$ $\mid\gamma\mid\sim\rho$ there exists an eigenvalue $\Lambda_{N}$ of $L_{t}(r,q(x))$ satisfying (5). $(c)$Suppose $\mid\gamma\mid\sim\rho,$ $\gamma+t\in(\cap_{i=1}^{k}% V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}}))\backslash E_{k+1}^{1},$ where $1\leq k\leq d-1.$ If (15) and (23) hold then there is an indeks $j$ 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 $C(\gamma+t,\gamma _{1},\gamma_{2},...,\gamma_{k}).$ $(d)$ Every eigenvalue $\Lambda(\gamma+t)\sim\rho^{2r}$ ( see (6) for the definition of $\Lambda(\gamma+t)$) of the operator $L_{t}(r,q(x))$ satisfies either (5) or (25) for $c=\frac{q(d-1)}{2},$ namely $\Lambda(\gamma+t)$ satisfies (5) for $\gamma+t\in U^{r}(\rho^{\alpha_{1}},p)$ and satisfies (25) for $\gamma+t\in E_{k}^{1}\backslash E_{k+1}^{1},$ where $k=1,2,...,d-1.$ \end{theorem} % %TCIMACRO{\TeXButton{Proof}{\proof}}% %BeginExpansion \proof %EndExpansion $(a)$ To prove (5) in case $k=1$ we divide both side of (17) by $b(N,\gamma)$ and use the estimations (18). Then we obtain \begin{equation} \Lambda_{N}-\mid\gamma+t\mid^{2r}=\mid\gamma+t\mid^{2r}+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 (24) for $s=0$ is also proved. Moreover by (19) we have $S_{k}(\mid\gamma+t\mid^{2r}+O(\rho^{-\alpha_{1}}),\gamma+t)=O(\rho ^{-\alpha_{1}})$ for $k=1,2,....$ Therefore (24) for arbitrary $s$ follows from the definition of $F_{s}$ by induction. Now we prove (5) for all $k$ by induction. If (5) is true for $k=j$, then putting the expression from (5) (for $k=j)$, for $\Lambda_{N\text{ }}$ into $A_{p_{1}}(\Lambda_{N},\gamma+t)$ in (17) and dividing both sides of (17) by $b(N,\gamma)$ we get \begin{align*} \Lambda_{N} & =\mid\gamma+t\mid^{2r}+A_{p_{1}}(\mid\gamma+t\mid^{2r}% +F_{j-1}+O(\rho^{-j\alpha_{1}}),\gamma+t)+O(\rho^{-(p-c)\alpha})=\\ & \mid\gamma+t\mid^{2r}+\{A_{p_{1}}(\mid\gamma+t\mid^{2r}+F_{j-1}% +O(\rho^{-j\alpha_{1}}),\gamma+t)-\\ A_{p_{1}}( & \mid\gamma+t\mid^{2r}+F_{j-1},\gamma+t)\}+A_{p_{1}}(\mid \gamma+t\mid^{2r}+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 $O(\rho^{-2(j+1)\alpha_{1}}).$ It can be cheked by using (4), (16), (24) and the obvious relation \begin{align*} & \frac{1}{\prod_{j=1}^{s}(\mid\gamma+t\mid^{2r}+F_{j-1}+O(\rho^{-j\alpha _{1}})-\mid\gamma+t-\sum_{i=1}^{s}\gamma_{i}\mid^{2r})}-\\ & \frac{1}{\prod_{j=1}^{s}(\mid\gamma+t\mid^{2r}+F_{j-1}-\mid\gamma +t-\sum_{i=1}^{s}\gamma_{i}\mid^{2r})}\\ & =\frac{1}{\prod_{j=1}^{s}(\mid\gamma+t\mid^{2r}+F_{j-1}-\mid\gamma +t-\sum_{i=1}^{s}\gamma_{i}\mid^{2r})}(\frac{1}{1-O(\rho^{-(j+1)\alpha_{1}}% )}-1) \end{align*} $=O(\rho^{-(j+1)\alpha_{1}})$, $\forall s=1,2,...,p_{1}.$ $(b)$ Let $A$ be the set of indices $N$ satisfying (15). Using (11) and Bessel inequality, we obtain% $\sum_{N\notin A}\mid b(N,\gamma)\mid^{2}=\sum_{N\notin A}\mid\dfrac{(\Psi _{N}(x),(q(x)e^{i(\gamma+t,x)})}{\Lambda_{N}-\mid\gamma+t\mid^{2r}}\mid ^{2}=O(\rho^{-2\alpha_{1}})$ Hence by the Parseval equality we have $\sum_{N\in A}\mid b(N,\gamma)\mid ^{2}=1-O(\rho^{-2\alpha_{1}}).$ This together with $\mid A\mid\frac{1}{2}(c_{\Gamma})^{-1}\rho^{-\frac{(d-1)q}{2}\alpha}$ , that is, (23) holds, for $c=\frac{(d-1)q}{2}$ . Thus $\Lambda_{N}$ satisfies (5) according to $(a)$. $(c)$Writing the equation (12) for all $h_{i}+t\in B_{k}(\gamma+t,p_{1}),$ we obtain% \begin{equation} (\Lambda_{N}-\mid h_{i}+t\mid^{2r})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 $i=1,2,...,b_{k}.$ First we show 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^{2r})}+ \end{equation}% $+O(\rho^{-p\alpha})=O(\rho^{-p\alpha}).$ For this, we apply the formula (13) $p_{1}$ times and use the inequality \begin{equation} \mid\Lambda_{N}-\mid h_{i}-\gamma^{^{\prime}}-\gamma_{1}-\gamma_{2}% -...-\gamma_{s}+t\mid^{2r}\mid>\frac{1}{6}\rho^{\alpha_{k+1}}, \end{equation} for $h_{i}+t\in B_{k}(\gamma+t,p_{1}),$ $(h_{i}-\gamma^{^{\prime}}+t)\notin B_{k}(\gamma+t,p_{1}),$ and $\gamma_{j}\in\Gamma(\rho^{\alpha}),$ where $j=1,2,...,s$ and $s=0,1,...,p_{1}-1.$ So now we prove (29). The relations $p>2p_{1}$ ( see the end of the introduction), $h_{i}+t\in B_{k}% (\gamma+t,p_{1})$, $(h_{i}-\gamma^{^{\prime}}+t)\notin B_{k}(\gamma+t,p_{1})$ and $\mid\gamma^{^{\prime}}\mid,\mid\gamma_{1}\mid,...,\mid\gamma_{p_{1}-1}% \mid<\rho^{\alpha}$ imply that $a_{s}\equiv h_{i}-\gamma^{^{\prime}}-\gamma_{1}-\gamma_{2}-...-\gamma _{s}+t\in B_{k}(\gamma+t,p)\backslash B_{k}(\gamma+t)$ for all $s=0,1,...,p_{1}-1.$ Since $B_{k}(\gamma+t,p)=\gamma+t+B_{k}+\Gamma (p\rho^{\alpha})$ we have a decomposition $a_{s}=\gamma+t+b+a,$ where $b\in B_{k},a\in\Gamma(p\rho^{\alpha}),$ that is, \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} It follows from (20) that, to verify (31) it is enough to prove it for $r=1.$ To prove (31) for $r=1$ we consider two cases: Case 1. $a\in P\equiv Span\{\gamma_{1,}\gamma_{2},...,\gamma_{k}\}.$ Then the relations $a+b\in P,$ $\gamma+t+a+b\notin\gamma+t+B_{k}$ 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 orthogonal decomposition $\gamma+t=x+v$ of $\gamma+t,$ where $v\in P$ and $x\bot P.$ It is not hard to show that (see equality (40) in Remark 1) \begin{equation} \mid v\mid=O(\rho^{(k-1)\alpha+\alpha_{k}}). \end{equation} It together with above inequality for $\mid a+b\mid$ and the relations $\alpha_{k+1}>2(\alpha_{k}+(k-1)\alpha)$ ( see the inequality 7. in the end of the introduction), $\gamma+t=x+v,$ $(x,v)=(x,a)=(x,b)=0,$% \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} give us the estimation (31) for $r=1.$ 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 that this inequality does not hold. Then $\gamma+t\in V_{a}^{1}% (\rho^{\alpha_{k+1}}).$ On the other hand $\gamma+t\in\cap_{i=1}^{k}% V_{\gamma_{i}}^{1}(\rho^{\alpha_{k+1}}).$ Therefore we have $\gamma+t\in E_{k+1}^{1}$ which contradicts to the conditions of the Theorem 1(c). The difference $\mid\gamma+t+a+b\mid^{2}-\mid\gamma+t\mid^{2}$ can be written as the sum of the differences $d_{1}\equiv\mid\gamma+t+a+b\mid^{2}-\mid \gamma+t+b\mid^{2}$ and $d_{2}\equiv\mid\gamma+t+b\mid^{2}-\mid\gamma +t\mid^{2}.$ Since $d_{1}=\mid\gamma+t+a\mid^{2}-\mid\gamma+t\mid^{2}+2\langle a,b\rangle,$ it follows from the inequalities (34), (30) that $\mid d_{1}% \mid>\frac{2}{3}$ $\rho^{\alpha_{k+1}}$. Taking $a=0$ in (33) we have $d_{2}=\mid b+v\mid^{2}-\mid v\mid^{2}.$ Therefore (32), (30) and the inequality 7. in the end of the introduction 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, (31) holds. Thus in all cases (31) and hence (29) is proved. Therefore relation (28) follows from (4) and (29), since $p_{1}\alpha_{k+1}>p_{1}\alpha_{1}\geq p\alpha$ ( see the inequality 4. in the end of introduction). Hence (27) has 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 $h_{i}-\gamma^{^{\prime}}+t\in B_{k}(\gamma+t,p_{2})$. It can be written as $(C-\Lambda_{N}I)(b(N,h_{1}),b(N,h_{2}),...b(N,h_{b_{k}}))=O(\rho^{-p\alpha}) ,$ where the rigth 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 by (23) we have \begin{align} c_{4}\rho^{-c\alpha} & <(\sum_{h_{i}}\mid b(N,h_{i})\mid^{2})^{\frac{1}{2}% }\leq\parallel(C-\Lambda_{N}I)^{-1}\parallel\sqrt{b_{k}}c_{5}\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_{5}^{-1}b_{k}^{-\frac{1}{2}}% \rho^{-c\alpha+p\alpha}. \end{align} Using that $\mid B_{k}\mid=O(\rho^{\frac{k}{2}\alpha_{k+1}}),$ $\mid \Gamma(p_{1}\rho^{\alpha})\mid=O(\rho^{d\alpha})$ 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 the formula (25) follows from (36) and (37). $(d)$ Let $\Lambda_{N\text{ }}\equiv\Lambda(\gamma+t)$ be any eigenvalue of order $\rho^{2r}$ of the operator $L_{t}(r,q(x)).$ Denote by $D$ the set of all vectors $\gamma\in\Gamma$ satisfying (15). From (11), arguing as in the proof of ($b$), 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), from definition of $\Lambda(\gamma+t)$ we get $\mid b(N,\gamma)\mid>c_{6}\rho^{-\frac{(d-1)}{2}}=c_{6}\rho^{-\frac {(d-1)q}{2}\alpha}$, that is, condition (23) for $c=\frac{(d-1)q}{2}$ holds. Now the proof follows from $(a)$ and $(c)$ since either $\gamma+t$ $\in U^{1}(\rho^{\alpha_{1}},p)$ or $\gamma+t\in$ $E_{k}^{1}\backslash E_{k+1}^{1}$ for $k=1,2,...,d-1$ ( see (41) in Remark 1). The Theorem is proved.$\diamondsuit$ \begin{remark} Here we note that the nonresonance domain $U^{r}(c_{7}\rho^{\alpha_{1}},p)$ has an asymptotically full measure on $R^{d}$ in the sence that $\frac {\mu(U\cap B(\rho))}{\mu(B(\rho))}$ tends to $1$ as $\rho$ tends to infinity, where $B(\rho)=\{x\in R^{d}:\mid x\mid=\rho\}.$ By (22) it is enough to prove this for $r=1.$ It is clear and well known that $B(\rho)\cap V_{b}^{1}% (c_{7}\rho^{\alpha_{1}})$ is the part of sphere $B(\rho)$ which is contained between two parallel hyperplanes $\{x:\mid x\mid^{2}-\mid x+b\mid^{2}=-c_{7}\rho^{\alpha_{1}}\}$and $\{x:\mid x\mid^{2}-\mid x+b\mid^{2}=c_{7}\rho^{\alpha_{1}}\}.$ The distance of these hyperplanes from origin are $\frac{c_{7}\rho^{\alpha_{1}}}{\mid b\mid}.$ This, the relation $\mid\Gamma(p\rho^{\alpha})\mid=O(\rho^{d\alpha}),$ and $\alpha_{1}+d\alpha<1-\alpha$ ( see the inequality 1. in the end of the introduction) imply \begin{align} \mu(B(\rho)\cap V_{b}^{1}(c_{7}\rho^{\alpha_{1}})) & \sim\frac{\rho ^{\alpha_{1}+d-2}}{\mid b\mid},\text{ }\mu(E_{1}^{1}\cap B(\rho))=O(\rho ^{d-1-\alpha})\\ \mu(U^{1}(c_{7}\rho^{\alpha_{1}},p)\cap B(\rho)) & =(1+O(\rho^{-\alpha}% ))\mu(B(\rho)). \end{align} Now we consider $E\equiv(\cap_{i=1}^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}% }))\cap B(\rho).$ 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 $x\in\cap_{i=1}^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}})$ 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 $\det(\gamma_{j,i})$ of the system is a volume of the parallelepiped $\{\sum_{i=1}^{k}b_{i}\gamma_{i}:b_{i}\in\lbrack0,1],i=1,2,...,k\}$ , hence greater than some constant $c$ and using that $\mid\gamma_{j,i}\mid \rho_{0}\}$ for $\rho_{0}\gg1$ is empty set. Therefore we have \begin{equation} R^{d}\cap\{\mid x\mid>\rho_{0}\}=(U^{1}\cup(\cup_{s=1}^{d-1}(E_{s}% ^{1}\backslash E_{s+1}^{1})))\cap\{\mid x\mid>\rho_{0}\}. \end{equation} \end{remark} \begin{remark} Here we note some properties of the known part $\mid\gamma+t\mid^{2r}+F_{k}(\gamma+t)$ ( for $k=1,2,...$ see Theorem 1) of the nonresonance eigenvalues of $L_{t}(r,q(x))$. Denoting $\gamma+t$ by $x$ , where $\mid\gamma+t\mid\sim\rho,$ $\gamma+t\in U^{1}(\rho^{\alpha_{1}},p),$ we prove \begin{equation} \frac{\partial F_{k}(x)}{\partial x_{i}}=O(\rho^{2-2r-2\alpha_{1}+\alpha }),\forall i=1,2,...,d;\forall k=1,2,... \end{equation} We prove (42) by induction with respect to $k.$ Using (20) one can easily verify that% \begin{align*} & \mid x\mid^{2r}-\mid x-\gamma_{1}\mid^{2r}\sim\rho^{2r-2}(\mid x\mid ^{2}-\mid x-\gamma_{1}\mid^{2}),\\ & \mid x\mid^{2r-2}-\mid x-\gamma_{1}\mid^{2r-2}\sim\rho^{2r-4}(\mid x\mid^{2}-\mid x-\gamma_{1}\mid^{2}), \end{align*} where $\mid\mid x\mid^{2}-\mid x-\gamma_{1}\mid^{2}\mid>\rho^{\alpha_{1}},$ since $x\notin V_{\gamma_{1}}^{1}(\rho^{\alpha_{1}}).$ Therefore% $\frac{\partial}{\partial x_{i}}(\dfrac{1}{\mid x\mid^{2r}-\mid x-\gamma _{1}\mid^{2r}})=$% \begin{equation} \dfrac{2rx_{1}(\mid x\mid^{2r-2}-\mid x-\gamma_{1}\mid^{2r-2})}{(\mid x\mid^{2r}-\mid x-\gamma_{1}\mid^{2r})^{2}})+ \end{equation}% $\dfrac{-2\gamma_{1}(i)\mid x-\gamma_{1}\mid^{2r-2}}{(\mid x\mid^{2r}-\mid x-\gamma_{1}\mid^{2r})^{2}}=O(\rho^{2-2r-2\alpha_{1}+\alpha}),$ 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 (42) for $k=1$ follows from (4) and (43). Suppose that (42) holds for $k=s.$ Using this and (24), replacing $\mid x\mid^{2r}$ by $\mid x\mid^{2r}+F_{s}(x)$ in (43) and evaluating as above we obtain% $\frac{\partial}{\partial x_{i}}(\dfrac{1}{\mid x\mid^{2r}+F_{s}(x)-\mid x-\gamma_{1}\mid^{2r}})=$% $O(\rho^{2r-2-2\alpha_{1}+\alpha})+\dfrac{\frac{\partial F_{s}(x)}{\partial x_{i}}}{(\mid x\mid^{2r}+F_{s}-\mid x-\gamma_{1}\mid^{2r})^{2}}=$% $O(\rho^{2-2r-2\alpha_{1}+\alpha})+O(\rho^{2-2r-4\alpha_{1}+\alpha}% )=O(\rho^{2-2r-2\alpha_{1}+\alpha})$ This formula with the definition of $F_{k}$ implies (42) for $k=s+1.$ \end{remark} \section{Asymptotic Formulas for Bloch Functions} In this section using the asymptotic formulas for eigenvalues ( Theorem 1) and simplisity conditions (9), (10) we prove asymptotic formulas for Bloch functions with a quasimomenta of the simple set $B$. To use the simplicity conditions we write the asymptotic formulas in the following form. By Theorem 1(b) for $\gamma+t\in B\subset U^{1}(\rho^{\alpha_{1}},p)$ there exists an eigenvalue $\Lambda_{N}$ satisfying (5). Taking $k=k_{1}$ in (5), where $k_{1}=[\frac{d}{3\alpha}]+2,$ \ ( we can take $k=k_{1},$ since $k_{1}% \leq\frac{1}{3}(p-\frac{1}{2}q(d-1))$ (see the inequality 3. in the end of introduction)) using the relation $3k_{1}\alpha>d+2\alpha$ ( see the inequality 5. in the end of the introduction) and notations $F(\gamma +t)=\mid\gamma+t\mid^{2r}+F_{k_{1}-1}(\gamma+t)$, $\varepsilon_{1}% =\rho^{-d-2\alpha}$ ( see introduction) we obtain \begin{equation} \Lambda_{N}-F(\gamma+t)=o(\varepsilon_{1}) \end{equation} \begin{theorem} If $\gamma+t\in B$ and $\mid\gamma+t\mid\sim\rho$ then there exists a unique eigenvalue $\Lambda_{N}$ satisfying (5) for $k=1,2,...,n=[\frac{p}{3}]$. This is a simple eigenvalue and the corresponding eigenfunction $\Psi_{N}(x)$ satisfies \begin{equation} \Psi_{N}(x)=e^{i(\gamma+t,x)}+O(\rho^{-\alpha_{1}}), \end{equation} if $q(x)\in W_{2}^{s_{0}}(F),$ where $s_{0}=\frac{3d-1}{2}(3^{d}+d+2)+\frac {1}{4}d3^{d}+d+6.$ This eigenvalue is the $\Lambda(\gamma+t)$ ( see (6) for the definition of $\Lambda(\gamma+t)$ ). \end{theorem} % %TCIMACRO{\TeXButton{Proof}{\proof} }% %BeginExpansion \proof %EndExpansion We noted that ( see above) there exists an eigenvalue $\Lambda_{N}$ satisfying (44). Let $\Psi_{N}$ be any normalized eigenfunction corresponding to $\Lambda_{N}$. Since the normalized eigenfunction is defined up to constant of modulas $1$ we can suppose that $\arg b(N,\gamma)=0,$ where $b(N,\gamma )=(\Psi_{N},e^{i(\gamma+t,x)}).$ Therefore to prove (45) it suffices to verify the equality \begin{equation} \sum_{\gamma^{^{\prime}}\in\Gamma\backslash\{\gamma\}}\mid b(N,\gamma ^{^{\prime}})\mid^{2}=O(\rho^{-2\alpha_{1}}). \end{equation} Thus we need to show that (46) holds. First consider the case $\gamma ^{^{\prime}}\notin K$. Using (44), definition of $K$ (see introduction), and then (11), we get \begin{align} & \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2r}\mid>\frac{1}{4}% \rho^{\alpha_{1}},\gamma^{^{\prime}}\notin K,\\ \sum_{\gamma^{^{\prime}}\notin K} & \mid b(N,\gamma^{^{\prime}})\mid ^{2}=\parallel\Psi_{N}(x)q(x)\parallel^{2}O(\rho^{-2\alpha_{1}})=O(\rho ^{-2\alpha_{1}}).\nonumber \end{align} If $\gamma^{^{\prime}}\in K$ , then clearly \begin{equation} \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2r}\mid<\frac{1}{2}\rho ^{\alpha_{1}}, \end{equation} Now we prove that the simplicity conditions (9), (10) imply \begin{equation} \mid b(N,\gamma^{^{\prime}})\mid\leq c_{4}\rho^{-c\alpha},\forall \gamma^{^{\prime}}\in K\backslash\{\gamma\}, \end{equation} where $c=p-dq-\frac{1}{4}d3^{d}-3.$ If for $\gamma^{^{\prime}}+t\in U^{1}% (\rho^{\alpha_{1}},p)$ and $\gamma^{^{\prime}}\in K\backslash\{\gamma\}$ the inequality in (49) is not true then by Theorem 1(a) \ ( see Theorem 1(a), and inequalities (48), (49)) we have \begin{equation} \Lambda_{N}=\mid\gamma^{^{\prime}}+t\mid^{2r}+F_{k-1}(\gamma^{^{\prime}% }+t)+O(\rho^{-3k\alpha}), \end{equation} for $k=1,2,...,[\frac{1}{3}(p-c)]=[\frac{1}{3}(dq+\frac{1}{4}d3^{d}+3)].$ Now taking $k=k_{1}=[\frac{d}{3\alpha}]+2$ ( we can take $k=k_{1},$ since one can easily verify that $\frac{d}{3\alpha}+2\leq\frac{1}{3}(dq+\frac{1}{4}% d3^{d}+3)$ ), and arguing as in the prove of (44) we get% $\Lambda_{N}-F(\gamma^{^{\prime}}+t)=o(\varepsilon_{1})\text{ ,}%$ because $3k_{1}\alpha>d+2\alpha$ (see the inequality 5. in the end of introduction). This with (44) contradicts \ the simplicity condition (9). Similarly if, for $\gamma^{^{\prime}}+t\in(E_{k}^{1}\backslash E_{k+1}^{1})$ and $\gamma^{^{\prime}}\in K,$ the inequality in (49) does not hold then by Theorem 1(c) \begin{equation} \Lambda_{N}=\lambda_{j}(\gamma^{^{\prime}}+t)+O(\rho^{-(p-c-\frac{1}{4}% d3^{d})\alpha}). \end{equation} where $(p-c-\frac{1}{4}d3^{d})\alpha=(dq+3)\alpha>d+2\alpha$ . Therefore we have $\Lambda_{N}-\lambda_{j}(\gamma^{^{\prime}}+t)=o(\varepsilon_{1})$ This with (44) contradicts \ the simplicity condition (10). Hence the inequality in (49) holds. Therefore using $\mid K\mid=O(\rho^{q(d-1)\alpha})$ we have \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 (46). Thus we proved that for any normalized eigenfunction $\Psi_{N}$ corresponding to any eigenvalue $\Lambda_{N}$ \ satisfying (5) the equality (45) holds. If there exist two different eigenvalues or multiple eigenvalue satisfying (5) then there exist two orthogonal normalized eigenfunction satisfying (45) which is imposible. Therefore $\Lambda_{N}$\ is a simple eigenvalue. It follows from Theorem 1(a) that $\Lambda_{N}$ satisfies (5) for $k=1,2,...,n=[\frac{p}{3}],$ because the inequality (23) holds for $c=0$ ( see (45)). Since (45) holds it follows from the definition of $\Lambda(\gamma+t)$ ( see (6)) that if $\gamma+t\in B$ then the eigenvalue $\Lambda_{N}$ satisfying (5) is $\Lambda(\gamma+t).$ The theorem is proved$\blacksquare$ Now we prove the asymptotic formulas of arbitrary order for Bloch functions. \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 (8), for $k=1,2,...,n=[\frac{1}{6}(2p-(3d-1)q-\frac{1}{2}d3^{d}-6)]$, where, $\Phi_{0}(x)=0,$ $\Phi_{1}(x)=% %TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}e^{i(\gamma+t+\gamma_{1},x)}}{(\mid\gamma+t\mid^{2r}% -\mid\gamma+\gamma_{1}+t\mid^{2r})},$ and $\Phi_{k-1}(x)$ is a linear conbination of $e^{i(\gamma+t+\gamma^{^{\prime}},x)}$ for $\gamma^{^{\prime}% }\in\Gamma((k-1)\rho^{\alpha})\cup\{0\}$ with explicitly expressed coefficients (58), (59). \end{theorem} % %TCIMACRO{\TeXButton{Proof}{\proof} }% %BeginExpansion \proof %EndExpansion By Theorem 2 formula (8) for $k=1$ is proved. To prove formula (8) 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 accoding to (46). Assume that (53) is true for $k=m$ . Then by (54) for $k=m,$ and by (3) we have $\Psi_{N}(x)(q(x))=H(x)+O(\rho^{-m\alpha_{1}}),$ where $H(x)$ is linear conbination 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 $\gamma^{^{\prime}}% \in\Gamma^{c}(m).$ Therefore using (11) 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^{2r}}\mid^{2}=O(\rho^{-2(m+1)\alpha_{1}}), \end{equation} where the sum is taken under conditions $\gamma^{^{\prime}}\in\Gamma^{c}(m)$ and $\gamma+\gamma^{^{\prime}}\notin K.$ On the other hand, using $\alpha _{1}=3\alpha$ and definition of $n,$ from (52) we obtain $\sum_{\gamma^{^{\prime}}\in K\backslash\{\gamma\}}\mid b(N,\gamma^{^{\prime}% })\mid^{2}=O(\rho^{-2n\alpha_{1}}),$ This with (55) implies (53) for $k=m+1.$ Thus formula (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. In (12) replase $\gamma$ by $\gamma+\gamma^{^{\prime}}$. Iterate it $n$ times and every times isolate the terms with maltiplicant $b(N,\gamma).$ In other word apply (13) for $b(N,\gamma+\gamma^{^{\prime}}),$ where $\gamma^{^{\prime}}\in \Gamma((n-1)\rho^{\alpha}),$ and hence $\gamma^{^{\prime}}\neq0.$ Then apply (13) for $b(N,\gamma+\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i})$ when $\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}\neq0$ where $\gamma_{i}\in \Gamma(\rho^{\alpha}),$ $j=1,2,...,n-1.$ Then using (4) and the relation $\mid\Lambda_{N}-\mid\gamma+t+\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}% \mid^{2r}\mid>\frac{1}{2}\rho^{\alpha_{1}}$ ( see (16) and use the fact 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 $P(\gamma +t)=\mid\gamma+t\mid^{2r}+F_{[\frac{p}{3}]}(\gamma+t)$ ( see Theorem 2), 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 $A_{1}(\gamma^{^{\prime}})\equiv\dfrac{q_{\gamma^{^{\prime}}}}{P(\gamma +t)-\mid\gamma+\gamma^{^{\prime}}+t\mid^{2r}}=\dfrac{q_{\gamma^{^{\prime}}}% }{\mid\gamma+t\mid^{2r}-\mid\gamma+\gamma^{^{\prime}}+t\mid^{2r}}% +O(\rho^{-3\alpha_{1}})$ $A_{k}(\gamma^{^{\prime}})=% %TCIMACRO{\dsum _{\gamma_{1},...,\gamma_{k-1}}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},...,\gamma_{k-1}}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{k-1}}q_{\gamma^{^{\prime}% }-\gamma_{1}-\gamma_{2}-...-\gamma_{k-1}}}{\prod_{j=0}^{k-1}(P(\gamma +t)-\mid\gamma+t+\gamma^{^{\prime}}-\sum_{i=1}^{j}\gamma_{i}\mid^{2r})}% =O(\rho^{-k\alpha_{1}}),$ \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 $k=n$ and (56), we obtain $\Psi_{N}=b(N,\gamma)e^{i(\gamma+t,x)}+\sum_{\gamma^{\ast}\in\Gamma ((n-1)\rho^{\alpha})}\sum_{k=1}^{n-1}(A_{k}(\gamma^{\ast})b(N,\gamma )+O(\rho^{-n\alpha_{1}}))e^{i(\gamma+t+\gamma^{\ast},x)})+H_{n}%$ Since $\parallel\Psi_{N}\parallel=1,\arg b(N,\gamma)=0,$ 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 and $\parallel H_{n}\parallel=O(\rho^{-n\alpha_{1}})$ we have $1=\mid b(N,\gamma)\mid^{2}+\sum_{k=1}^{n-1}(\sum_{\gamma^{\ast}\in \Gamma((n-1)\rho^{\alpha})}\mid A_{k}(\gamma^{\ast})b(N,\gamma)\mid^{2}% +O(\rho^{-n\alpha_{1}})),$ \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).) Now 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 $b(N,\gamma )=1+O(\rho^{-2\alpha_{1}}),$ $b(N,\gamma+\gamma^{^{\prime}})=A_{1}(\gamma^{^{\prime}})+O(\rho^{-2\alpha _{1}})=\dfrac{q_{\gamma^{^{\prime}}}}{\mid\gamma+t\mid^{2r}-\mid\gamma +\gamma^{^{\prime}}+t\mid^{2r}}+O(\rho^{-2\alpha_{1}})$ for all $\gamma ^{^{\prime}}\in\Gamma(\rho^{\alpha}).$ These and (54) for $k=2$ imply the formula for $\Phi_{1}\diamondsuit$ \section{Simple Sets and Bethe-Sommerfeld conjecture} In this section we construct a part of the simple set in neighbourhood of the surface $S_{\rho}=\{x\in U^{1}(2\rho^{\alpha_{1}},p):F(x)=\rho^{2r}\}$. Due to (5) ( replace $\gamma+t$ by $x$ in $F(\gamma+t)$ ) it is natural to call $S_{\rho}$ the approximated isoenergetic surfaces in the nonresonance domain. As we noted in introduction in order that the nonresonance eigenvalue $\Lambda(\gamma+t)$ can not coincide with any other nonresonance eigenvalue $\Lambda(\gamma+t+b)$ we choose the $\gamma+t$ such that $\mid F(\gamma+t)-F(\gamma+t+b)\mid>2\varepsilon_{1}$ for $\gamma+t+b\in U^{1}(\rho^{\alpha_{1}},p)$ and $b\in\Gamma\backslash\{0\}$ (see (9)). Therefore we eliminate \begin{equation} P_{b}=\{x:x,x+b\in U^{1}(\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 $\varepsilon=\frac{\varepsilon_{1}}{7r\rho^{2r-1}},$ $U_{\varepsilon }(a)=\{x\in R^{d}:\mid x-a\mid<\varepsilon\}.$ In this set the first simplicity condition (9) holds (see Lemma 4(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 (10) holds, we discart from $U_{\varepsilon}(S_{\rho}^{^{\prime}})$ the translation $Tr(A(\rho))$ of \begin{equation} A(\rho)=\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 $A_{k,i}(\gamma_{1},...,\gamma_{k})=\{x\in(\cap_{i=1}^{k}V_{\gamma_{i}}% ^{1}(\rho^{\alpha_{k}})\backslash E_{k+1}^{1})\cap K_{\rho}:\lambda_{i}% (x)\in(\rho^{2r}-3\varepsilon_{1},\rho^{2r}+3\varepsilon_{1})\},$ and \begin{equation} K_{\rho}=\{x\in R^{d}:\mid\mid x\mid^{2r}-\rho^{2r}\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 intervals $\{a+sb:s\in\lbrack-1,1]\}$ such that $\Lambda (a-b)<\rho^{2r},$ $\Lambda(a+b)>\rho^{2r}$\ and $\Lambda(\gamma+t)$ \ is continuous on this intervals. Hence there exists $\gamma+t$ for which $\Lambda(\gamma+t)=\rho^{2r}.$ It implies the validity of Bethe-Sommerfeld conjecture for $L(r,q)$. For this we need the following lemma. \begin{lemma} $(a)$ If $x\in U_{\varepsilon}(S_{\rho}^{^{\prime}})$ and $x+b\in U^{1}% (\rho^{\alpha_{1}},p),$ where $b\in\Gamma,$ then \begin{equation} \mid F(x)-F(x+b)\mid>2\varepsilon_{1}% \end{equation} Hence for $\gamma+t\in U_{\varepsilon}(S_{\rho}^{^{\prime}})$ the first simplicity condition (9) 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 $E\subset R^{d}$ is bounded set 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} % %TCIMACRO{\TeXButton{Proof}{\proof} }% %BeginExpansion \proof %EndExpansion $(a)$ If $x\in U_{\varepsilon}(S_{\rho}^{^{\prime}})$ then there is a point $a$ in $S_{\rho}^{^{\prime}}$ such that $x\in U_{\varepsilon}(a)$. Since $S_{\rho}^{^{\prime}}$ does not contain the set $P_{b}$ ( see (60)) we have \begin{equation} \mid F(a)-F(a+b)\mid\geq3\varepsilon_{1}% \end{equation} On the other hand using (42) and the relations $\mid x\mid<\rho+1,$ $\mid x-a\mid<\varepsilon,$ $\mid x+b-a-b\mid <\varepsilon$ we obtain \begin{align} & \mid F(x)-F(a)\mid<3r\rho^{2r-1}\varepsilon,\\ & \mid F(x+b)-F(a+b)\mid<3r\rho^{2r-1}\varepsilon.\nonumber \end{align} These inequalities together with (64) give (63), since $6r\rho^{2r-1}% \varepsilon<\varepsilon_{1}.$ $(b)$ This is a simple consequence of $(a).$ Indeed if $x$ and $x+b$ lie in $U_{\varepsilon}(S_{\rho}^{^{\prime}})$ then there are 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 $\mid F(c)-F(x+b)\mid <3r\rho^{2r-1}\varepsilon.$ This, the first ineguality 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 are only finite number of the vectors $\gamma_{1},\gamma_{2},...,\gamma_{s}$ such that $E(k)\equiv (E+\gamma_{k})\cap U_{\varepsilon}(S_{\rho}^{^{\prime}})\neq\emptyset ,k=1,2,...,s$ 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 by devided into finite number of the pairwise disjoint sets $E_{1},E_{2},...,E_{n}$ such that there are 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 $k\neq j$ and $k,j=1,2,...,n.$ Using $\mu(E_{k}+\gamma_{k})=\mu(E_{k})$ we get the proof of $(d),$ becouse $Tr_{F^{\star}}(E)$ and $E$ are union of piarwise disjoint sets $E_{k}+\gamma_{k}$ and $E_{k}$ for $k=1,2,...,n$ respectively$\diamondsuit$ \begin{theorem} $(a)$ The set $U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho))$ is a subset of $B.$ Hence if $\gamma+t$ lies in this subset then Theorem 2 and Theorem 3 hold. For every connected open subset $E$ of $U_{\varepsilon }(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho)$ there exists a unique index $N$ such that $\Lambda(\gamma+t)=\Lambda_{N}(t)$ for all $\gamma+t\in E.$ $(b)$ For the part $V_{\rho}\equiv S_{\rho}^{^{\prime}}\backslash U_{\varepsilon}(Tr(A(\rho)))$ of the approximated isoenergetic surface the following holds \begin{equation} \mu(V_{\rho})>(1-c_{8}\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 number $\rho^{2r}$ for $\rho\gg1$ lies in the spectrum of $L(r,q(x)),$ that is, the number of the gaps in the spectrum of $L(r,q(x))$ is finite, where $r\geq1,$ $q(x)\in W_{2}^{s_{0}}(R^{d}/\Omega),$ $d\geq2,$ and $\Omega$ is an arbitrary lattice. \end{theorem} % %TCIMACRO{\TeXButton{Proof}{\proof}}% %BeginExpansion \proof %EndExpansion $(a)$ Suppose $\gamma+t\equiv x\in U_{\varepsilon}(S_{\rho}^{^{\prime}% })\backslash Tr(A(\rho)),$ that is , there is $a\in S_{\rho}^{^{\prime}}$ such that $x\in U_{\varepsilon}(a).$ Then, by lemma 4(a), the simplisity condition (9) holds. Now we prove that (10) holds too. The first inequality in (65) and the relation $F(a)=\rho^{2r}$ give \begin{equation} F(\gamma+t)\in(\rho^{2r}-3r\rho^{2(r-1)}\varepsilon,\rho^{2r}+3r\rho ^{2(r-1)}\varepsilon)\subset(\rho^{2r}-\varepsilon_{1},\rho^{2r}% +\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 $\gamma^{^{\prime}}+t\in E_{k}^{1}\backslash E_{k+1}^{1}$ then by definition of $K$ ( see introduction) the inequality $\mid F(\gamma+t)-\mid\gamma^{^{\prime}}+t\mid^{2r}\mid<\frac{1}{3}% \rho^{\alpha_{1}}$ holds. This (24) and (67) imply that $\gamma^{^{\prime}% }+t\in(E_{k}^{1}\backslash E_{k+1}^{1})\cap K_{\rho}$ ( see (62) for the definition of $K_{\rho}$). Sinse $\gamma^{^{\prime}}+t\notin A(\rho)$ we have $\lambda_{i}(\gamma^{^{\prime}}+t)\notin(\rho^{2r}-3\varepsilon_{1},\rho ^{2r}+3\varepsilon_{1})$ for $\gamma^{^{\prime}}\in K$ and $\gamma^{^{\prime}% }+t\in E_{k}\backslash E_{k+1}.$ Therefore (10) follows from (67). 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 2 for $a\in E\subset U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho)$ the eigenvalue $\Lambda(a)$ is simple and the corresponding eigenfunction $\Psi_{a}(x)$ satisfies $\mid(\Psi_{a}(x),e^{i(a,x)})\mid^{2}>\frac{1}{2}$. So for $y\in E$ there exists a unique $N(y)$ such that $\Lambda(y)=\Lambda_{N(y)}(y)$% ,$\Psi_{y}(x)=\Psi_{N(y),y}(x)$, \begin{equation} \mid(\Psi_{N(y),y}(x),e^{i(y,x)})\mid^{2}>\frac{1}{2}. \end{equation} 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 $t$ if $\Lambda_{N}(t)$ is a simple eigenvalue. Therefore for $a\in E$ 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 every $y\in U(a).$ This and (68) yield that $N(y)=N(a)$ for $y\in U(a),$ since there is a unique integer $N$ satisfying (68). Hence we proved the Proposition 1 which is $\forall a\in E,\exists U(a)\subset E:N(y)=N(a),\forall y\in U(a).$ Now let $a_{1}$ and $a_{2}$ be two points of $E$ and $C\subset E$ be the arc which joins these points. (Note that the open connected subset of $R^{d}$ is arcwise connected). Let $U(y_{1}),U(y_{2}),...,U(y_{k})$ be 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 Proposition 1. By Proposition 1 $N(y)=N(y_{i}% )=N_{i}$ for $y\in U(y_{i}).$ Clearly if $U(y_{i})\cap U(y_{j})\neq\emptyset$ then $N_{i}=N_{j}$ since for $a\in U(y_{i})\cap U(y_{j})$ there is a unique $N(a)$. Thus $N_{1}=N_{2}=...=N_{k}$ and $N(a_{1})=N(a_{2}).$ $(b)$ First we note that if $a\in V_{\rho}$ then $U_{\varepsilon}(a)\in U_{\varepsilon}(S_{\rho}^{^{\prime}})\backslash Tr(A(\rho)),$ because the relation $a\in S_{\rho}^{^{\prime}}$ implies that $U_{\varepsilon}(a)\in U_{\varepsilon}(S_{\rho}^{^{\prime}})$ and the relation $a\notin U_{\varepsilon}(Tr(A(\rho)))$ implies that $U_{\varepsilon}(a)\cap Tr(A(\rho))=\emptyset.$ To prove (66) 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_{9}\rho^{-\alpha})\mu(B(\rho)),\\ \mu(S_{\rho}^{^{\prime}}) & >(1-c_{10}\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 (64) of the measure of set $V_{\rho}$ is done in Estimation 4 by using Estimations 1, 2, 3. $(c)$ Since $F(a)=\rho^{2r}$ for $a=(a_{1},a_{2},...,a_{d})=\sum_{i=1}% ^{d}a_{i}e_{i}\in V_{\rho}\subset S_{\rho}$ it follows from (24) that $\rho-1<\mid a\mid<\rho+1,$ and there is an indeks $i$ such that $\mid a_{i}\mid>\frac{1}{d}\rho$ . Without loss of generality, assume that $\ a_{i}>0.$ Then (42) and (44) imply that $F(a-\varepsilon e_{i})<\rho ^{2r}-c_{11}\varepsilon_{1},$ $F(a+\varepsilon e_{i})>\rho^{2r}+c_{11}% \varepsilon_{1}$ and $\Lambda(a-\varepsilon e_{i})<\rho^{2r},$ $\Lambda (a+\varepsilon e_{i})>\rho^{2r}.$ Since $\Lambda(\gamma+t)$ \ is continuous there exists a point $y(a)\in\lbrack a-\varepsilon e_{i},a+\varepsilon e_{i}]$ such that $\Lambda(y(a,i))=\rho^{2r}.$ The Theorem is proved $\diamondsuit$ In Estimations 1-4 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 $U_{\varepsilon}(S_{\rho }^{^{\prime}})\cup U_{2\varepsilon}(A(\rho))$ , hence 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,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 $F(x)=\rho^{2r}$ and by (24) we have $\mid x\mid=\rho+O(\rho^{-1-\alpha_{1}}).$ Hence the inequalities in (72) hold for $x\in U_{\varepsilon}(S_{\rho}^{^{\prime}}).$ If $x\in$ $A(\rho)$ then by definition of $A(\rho)$ we have $x\in K_{\rho},$ and hence $\mid x\mid=\rho+O(\rho^{-1+\alpha_{1}})$. Therefore the inequalities in (72) hold for $x\in U_{2\varepsilon}(A(\rho))$ too. The inclusion in (72) follows from these inequalities. If $G$ $\subset S_{\rho}$ then for $x\in G(+k,\rho^{-\alpha})$ by (42) we have $\frac{\partial F(x)}{\partial x_{k}}>0,$ . 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 $R^{m}$ then we use \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 (69) by using (73). During this estimation the set $S_{\rho}$ is redenoted by $G.$ If $x\in G$ then $x\notin V_{b}^{1}% (\rho^{\alpha_{1}})),$ 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 some $b\in\Gamma(p\rho^{\alpha})$ coinsides with the direction of $(1,0,0,...,0),$ that is, $b=(b_{1},0,0,...,0),b_{1}>0.$ Then the relation $x\notin V_{b}^{1}(\rho^{\alpha_{1}}))$ implies that $\mid x_{1}\mid>a,$ where $a=(\rho^{\alpha_{1}}-b_{1}^{2})(2b_{1})^{-1}>\rho^{\alpha}$ because $\mid b\mid\rho^{\alpha},\text{ }(\frac{\partial F}{\partial x_{1}})^{-1}\mid grad(F)\mid=\frac{\mid x\mid}{x_{1}}% +O(\rho^{-2\alpha}), \end{equation}% \begin{equation} \Pr(G(+1,a))\supset\Pr(A(+1,2a)), \end{equation} where$x\in G(+1,a),A=B(\rho)\cap U^{1}(3\rho^{\alpha_{1}},p).$Here (75) follows from (42) and (76) follows from (75). Now we prove the inclusion in (77). If$(x_{2},...,x_{d})\in\Pr(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}.$This and (24) give$F(x_{1}+h,x_{2},...,x_{d})>\rho^{2r}$,$F(x_{1}-h,x_{2},...,x_{d})<\rho^{2r}% $. Since$F$is a continuous function there is$y_{1}\in(x_{1}-h,x_{1}+h)$such that (see (78)) \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>2\rho^{\alpha_{1}}, \end{equation} because the expression under the absolute value in (79) differ from the expression under the absolute value in (78) by$2(y_{1}-x_{1})b_{1},$whose absolute value is less than$\rho^{\alpha_{1}}.$The relations in (79) means that$(x_{2},...,x_{d})\in\Pr G(+1,a).$Hence the inclusion in (77) is proved. Now it follows from (73), (76) and the obvious relation$\mu(\Pr G(+1,a))=O(\rho^{d-1})$( see the inequality in (72))% $\mu(G(+1,a))=\int\limits_{\Pr(G(+1,a))}\frac{\mid x\mid}{x_{1}}dx_{2}% dx_{3}...dx_{d}+O(\rho^{-\alpha})\mu(B(\rho))\geq$% $\int\limits_{\Pr(A(+1,2a))}\frac{\mid x\mid}{x_{1}}dx_{2}dx_{3}...dx_{d}% -c_{11}\rho^{-\alpha}\mu(B(\rho))=$% $\mu(A(+1,2a))-c_{12}\rho^{-\alpha}\mu(B(\rho)).$ Similarly$\mu(G(+1,a))\geq\mu(A(-1,2a))-c_{13}\rho^{-\alpha}\mu(B(\rho)).$Hence$\mu(G)>\mu(A)-c_{14}\rho^{-\alpha}\mu(B(\rho)).$This inequality implies (69) since$\mu(A))=(1+O(\rho^{-\alpha}))\mu(B(\rho))$(see (39) ). ESTIMATION 2 Here we prove (70). For this we estimate the measure of the set$S_{\rho}\cap P_{b}$( see (60)) by using (73). During this estimation the set$S_{\rho}\cap P_{b}$is redenoted by$G$. We choose the coordinate axis so that the direction of$b$coincides with the direction of$(1,0,0,...,0),$i.e.,$b=(b_{1},0,0,...,0)$and$b_{1}>0$. Clearly if$(x_{1},x_{2}% ,...,x_{d})\in G$then ( see definition of$S_{\rho},G$and$F(x)) \begin{align} (x_{1}^{2}+x_{2}^{2}+...+x_{d}^{2})^{r}+F_{k_{1}-1}(x) & =\rho^{2r},\\ ((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})^{r}+F_{k_{1}-1}(x+b) & =\rho^{2r}+h \end{align} whereh\in(-3\varepsilon_{1},3\varepsilon_{1}).$It follows from (80), (81), and (24) that \begin{equation} (x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})=\rho^{2}+O(\rho^{-\alpha_{1}}) \end{equation}% \begin{equation} ((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})=\rho^{2}+O(\rho ^{-\alpha_{1}}) \end{equation} Subtrracting (82) from (83) we get% \begin{equation} (2x_{1}-b_{1})b_{1}=O(\rho^{-\alpha_{1}}), \end{equation} This with inequalities in (72) implies \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} Conside two cases. Case 1:$b\in\Gamma_{1}\equiv\{b\in\Gamma:\mid\rho ^{2}-(\frac{b_{1}}{2})^{2}\mid<3d\rho^{-2\alpha}\}.$In this case from (85), and (82) we obtain \begin{equation} x_{1}^{2}=\rho^{2}+O(\rho^{-2\alpha}),\mid x_{1}\mid=\rho+O(\rho^{-2\alpha -1}),\text{ }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 (73), the obvious relation$\mu(\Pr_{1}(G(+1,a))=O(\rho^{-(d-1)\alpha})$(see the lsat equality in (86)) and the fact that the expression under the integral (73) for$k=1$is equal to$1+O(\rho^{-\alpha})$(see (76) and (86)) and doing the same for$G(-1,a)$we get$\mu(G)=O(\rho^{-(d-1)\alpha}).$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:$b\notin\Gamma_{1},$i.e.,$\mid\rho^{2}-(\frac{b_{1}}{2})^{2}\mid \geq3d\rho^{-2\alpha}.$By (85) and (82) we have \begin{equation} \mid x_{1}^{2}-\rho^{2}\mid>2d\rho^{-2\alpha},x_{2}^{2}+x_{3}^{2}% +...+x_{d}^{2}>d\rho^{-2\alpha},\mid x_{k}\mid>\rho^{-\alpha}, \end{equation} for some$k=2,3,...,d.$Therefore$G\subset\cup_{k\geq2}(G(+k,\rho^{-\alpha })\cup G(-k,\rho^{-\alpha})).$We calculate$\mu(G(+d,\rho^{-\alpha}))$by (73). Redenote by$D$the set$\Pr_{d}G(+d,\rho^{-\alpha}).$If$x\in G(+d,\rho^{-\alpha})$then by (82) and (42) the under integral expression in (73) 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 (89). To prove the first equality of (89) we use (74) for$m=d-1;k=1$and the relations$\mu(\Pr_{1}D)=O(\rho^{d-2})$, \begin{equation} \mu(D(x_{2},x_{3},...,x_{d-1}))<6\varepsilon_{1}\mid b\mid^{-1},\text{ }\forall(x_{2},x_{3},...,x_{d-1})\in\Pr D. \end{equation} First relation follows from the fact that$\mid x\mid<\rho+1$for$x\in G(+d,\rho^{-\alpha})$(see (72)). So we need to prove (90). If$x_{1}\in D(x_{2},x_{3},...,x_{d-1})$then (80) and (81) holds. Subtructing (80) from (81), we get% $((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})^{r}-((x_{1}^{2}% +x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})^{r}+$% \begin{equation} 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 (80) and (91) with respect two unknown$x_{1}$and$x_{d}$. Using (42), the implicit function theorem, and the inequalities$\mid x_{d}\mid>\rho^{-\alpha },\alpha_{1}>2\alpha$from (80), we obtain \begin{equation} x_{d}=f(x_{1}),\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 (91), we get $((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+f^{2}(x_{1}))^{r}-((x_{1}^{2}% +x_{2}^{2}+x_{3}^{2}+...+f^{2}(x_{1}))^{r}+$% \begin{equation} 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 (42), (92), the first equality in (85) and$\mid x_{d}\mid>\rho ^{-\alpha}$we see the derivative (w.r.t.$x_{1}$) of the rigth side$a_{r}(x)$of (93) is% \begin{equation} \frac{\partial a_{r}(x)}{\partial x_{1}}=r\mid x-b\mid^{2(r-1)}(2(x_{1}% -b_{1})+2f(x_{1})f^{^{\prime}}(x_{1})) \end{equation}$-r\mid x\mid^{2(r-1)}(2x_{1}+2f(x_{1})f^{^{\prime}}(x_{1}))+O(\rho ^{-2\alpha_{1}+\alpha})(1+\frac{x_{1}}{x_{d}})+O(\rho^{-3\alpha_{1}+\alpha}) .$If$r=1$then using the first equality in (85) and$\mid x_{d}\mid >\rho^{-\alpha}$\ we have% \begin{equation} \mid\frac{\partial a_{r}(x)}{\partial x_{1}}\mid>b_{1}. \end{equation} If$r>1then it follows from (80), (81), and (24) that $\mid x\mid^{2r}=\rho^{2r}(1+O(\rho^{-2r-\alpha_{1}})),\text{ }\mid x-b\mid^{2r}=\rho^{2r}(1+O(\rho^{-2r-\alpha_{1}}))$% \begin{align*} & \mid x\mid^{2(r-1)}=\rho^{2(r-1)}(1+O(\rho^{-2r-\alpha_{1}}))^{\frac {r-1}{r}}=\rho^{2(r-1)}+O(\rho^{-2-\alpha_{1}})\text{ }\\ & \mid x-b\mid^{2(r-1)}=\rho^{2(r-1)}(1+O(\rho^{-2r-\alpha_{1}}))^{\frac {r-1}{r}}=\rho^{2(r-1)}+O(\rho^{-2-\alpha_{1}})\text{ }% \end{align*} Using these in (94) and arguing as in proof of (95), forr=1,$we get the proof of (95) for$r>1.$Thus in any case (95) holds. Therefore from (93) by implicit function theorem, we get$\mid\frac{dx_{1}}{dh}\mid<\frac{1}{\mid b\mid}.$This inequality and relation$h\in(-3\varepsilon_{1},3\varepsilon _{1})$imply (90). Hence (89) is proved. In the same way we get the same estimation for$G(+k,\rho^{-\alpha})$and$G(-k,\rho^{-\alpha})$for$k\geq2$. Thus$\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$( see (85)) and$\varepsilon_{1}=\rho^{-d-2\alpha}$, we infer$\mu(\cup_{b\notin\Gamma _{1}}(S_{\rho}\cap P_{b}))=O(\rho^{2d-1+\alpha}\varepsilon_{1})=O(\rho ^{-\alpha})\mu(B(\rho)).$This, (87) and (69) give the proof of (70). ESTIMATION 3. Here we prove (71). Let$G\equiv U_{2\varepsilon}(A_{k,j}% (\gamma_{1,}\gamma_{2},...,\gamma_{k})),$where$\gamma_{1,}\gamma _{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha}),k=\leq d-1.$We turn the coordinate axis so that$Span\{\gamma_{1,}\gamma_{2},...,\gamma_{k}\}=\{x=(x_{1},x_{2},...,x_{k}% ,0,0,...,0):x_{1},x_{2},...,x_{k}\in R\}$. Then by (40) we have$x_{n}% =O(\rho^{\alpha_{k}+(k-1)\alpha}),$for$n\leq k,x\in G$. This, (72), and$\alpha_{k}+(k-1)\alpha<1$( see the inequality 6. in the end of the introduction) give$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.$Now using this and (74) for$m=d$we prove \begin{equation} \mu(G(+i,\rho d^{-1}))=O(\varepsilon\rho^{k(\alpha_{k}+(k-1)\alpha )+(d-1-k)}),\forall i>k. \end{equation} For this we redenote by$D$the set$G(+i,\rho d^{-1})$and prove \begin{equation} \mu((D(x_{1},x_{2},...x_{i-1},x_{i+1},...x_{d}))\leq(42d^{2}+4)\varepsilon ,\forall i>k, \end{equation} for all$(x_{1},x_{2},...x_{i-1},x_{i+1},...x_{d})\in\Pr_{i}(D).$To prove (97) it is sufficient to show that if both$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$. We may assume that$\rho d^{-1}42d^{2}\varepsilon,(a_{i}^{^{\prime}})^{2}-(a_{i})^{2}>2(\rho d^{-1}-2\varepsilon)(a_{i}^{^{\prime}}-a_{i}),\mid\mid a_{s}\mid-\mid a_{s}^{^{\prime}}\mid\mid<4\varepsilon,\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<\frac27\rho d^{-1}(a_{i}-a_{i}^{^{\prime}}),$% \begin{equation} \mid\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}\mid>\frac{3}{2}\rho d^{-1}\mid a_{i}^{^{\prime}}-a_{i}\mid,\mid a_{i}^{^{\prime}}-a_{i}\mid>42d^{2}% \varepsilon \end{equation} Moreover using mean value theorem and the relations$\mid a\mid=\rho+O(1),\mid a^{^{\prime}}\mid=\rho+O(1),$we get \begin{equation} \mid a\mid^{2r}-\mid a^{^{\prime}}\mid^{2r}=r(\rho+O(1))^{2(r-1)})(\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}) \end{equation} Let$r_{i}(x)=\lambda_{i}(x)-\mid x\mid^{2r}.$Hence$r_{1}(x)\leq r_{2}(x)\leq...\leqr_{b_{k}}(x)$be the eigenvalue of the matrix$C(x)-\mid x\mid^{2r}I\equiv C^{^{\prime}}(x),$where$C(x)$is defined in Section 1 (see Theorem 1(c)). By definition of$C^{^{\prime}}(x)$only diagonal elements of the matrix$C^{^{\prime}}(x)$depend on$x$and they are$\mid x-d_{i}% \mid^{2r}-\mid x\mid^{2r},$where$d_{i}=h_{i}+t-\gamma-t\in B_{k}% +\Gamma(p_{1}\rho^{\alpha})$. Clearly% \begin{equation} \mid d_{i}\mid<\rho^{\frac{1}{2}\alpha_{d}},\text{ }\mid r_{j}(a^{^{\prime}% })-r_{j}(a)\mid\leq\parallel C^{^{\prime}}(a^{^{\prime}})-C^{^{\prime}% }(a)\parallel=\max_{i}\mid a_{i,i}\mid. \end{equation} where$C^{^{\prime}}(a^{^{\prime}})-C^{^{\prime}}(a)=(a_{i,j}),a_{i,i}=\mid a\mid^{2r}-\mid a-d_{i}\mid^{2r}-\mid a^{^{\prime}}\mid^{2r}+\mid a^{^{\prime }}-d_{i}\mid^{2r},a_{i,j}=0$for$i\neq j,$that is$C^{^{\prime}% }(x)-C^{^{\prime}}(x^{^{\prime}})$is diagonal matrix. To estimate$\mid a_{i,i}\mid$using mean value theorem and the relations$\mid a\mid =\rho+O(1),\mid a^{^{\prime}}\mid=\rho+O(1),\mid a-d_{i}\mid=\rho +O(\rho^{\frac{1}{2}\alpha_{d}}),\mid a^{^{\prime}}-d_{i}\mid=\rho +O(\rho^{\frac{1}{2}\alpha_{d}}),$we obtain $a_{i,i}=r(\rho+O(\rho^{\frac{1}{2}\alpha_{d}}))^{2(r-1)}(\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2})-$% $r(\rho+O(\rho^{\frac{1}{2}\alpha_{d}}))^{2(r-1)}(\mid a-d_{i}\mid^{2}-\mid a^{^{\prime}}-d_{i}\mid^{2})=$% $r(\rho+O(\rho^{\frac{1}{2}\alpha_{d}}))^{2(r-1)}(\mid a\mid^{2}-\mid a-d_{i}\mid^{2}-\mid a^{^{\prime}}\mid^{2}+\mid a^{^{\prime}}-d_{i}\mid^{2})+$% $O(\rho^{\frac{1}{2}\alpha_{d}+2r-3})(\mid a\mid^{2}-\mid a^{^{\prime}}\mid ^{2}).$ Since$\mid a\mid^{2}-\mid a-d_{i}\mid^{2}-\mid a^{^{\prime}}\mid^{2}+\mid a^{^{\prime}}-d_{i}\mid^{2}=2(a-a^{^{\prime}},d_{i})$we have ( see (100))$\mid r_{j}(a)-r_{j}(a^{^{\prime}})\mid\leq3r\rho^{\frac{1}{2}\alpha_{d}% +2r-2}\mid a-a^{^{\prime}}\mid-c_{15}\rho^{\frac{1}{2}\alpha_{d}+2r-3}\mid\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}\mid.$Therefore using$\frac{1}{2}% \alpha_{d}<1,$(99), (98) and$\lambda_{i}(x)=r_{i}(x)+\mid x\mid^{2r}$we obtain$\mid\lambda_{j}(a)-\lambda_{j}(a^{^{\prime}})\mid\geq\mid\mid a\mid^{2r}-\mid a^{^{\prime}}\mid^{2r}\mid-\mid r_{j}(a)-r_{j}(a^{^{\prime}})\mid>r\rho ^{2r-1}d^{-1}\mid a_{i}^{^{\prime}}-a_{i}\mid>r\rho^{2r-1}42d\varepsilon>6\varepsilon_{1}$which controdicts the fact that both$\lambda_{j}(a)$and$\lambda_{j}(a^{^{\prime}})$lie in$(\rho ^{2}-3\varepsilon_{1},\rho^{2}+3\varepsilon_{1}),$becouse$a$and$a^{^{\prime}}$lie in$A_{k,j}$( see definition$A_{k,j}$). Thus (97), hence (96) is proved. In the same way we get the same formula for$G(-i,\frac{\rho }{d}).$So$\mu(U_{2\varepsilon}(A_{k,j}(\gamma_{1,}\gamma_{2},...,\gamma _{k})))=O(\varepsilon\rho^{k(\alpha_{k}+(k-1)\alpha)+d-1-k}),$where$j=1,2,...,b_{k}(\gamma_{1},\gamma_{2},...,\gamma_{k}),$and$\gamma _{1},\gamma_{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha}).$From this using that$b_{k}=O(\rho^{d\alpha+\frac{k}{2}\alpha_{k+1}})$( see (37)) 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 (71) if$d\alpha+\frac{k}{2}\alpha_{k+1}+dk\alpha+k(\alpha_{k}+(k-1)\alpha)+d-1-k\le q d-1-\alpha$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 (101) by$k\alpha$and using$\alpha_{k}=3^{k}\alpha,\alpha=\frac{1}{q},q=3^{d}+d+2$we see that (101) is equivalent to$\frac{d+1}{k}+\frac{3^{k+1}}{2}+3^{k}+k-1\leq3^{d}+2$The right-hand side of this inequality gets its maximum value 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 \$\frac{d+1}{d-1}\leq3,d<\frac{1}{6}3^{d}+1$for$d\geq2.$ESTIMATION 4. Here we prove (66). During this estimation we denote by$G$the set$S_{\rho}^{^{\prime}}\cap U_{\varepsilon}(Tr(A(\rho))$. Since$V_{\rho }\equiv S_{\rho}^{^{\prime}}\backslash G$, by (70) and (72), it is enough to prove$\mu(G(+i,\rho d^{-1}))=O(\rho^{-\alpha})\mu(B(\rho)),$for$i=1,2,...,d.$(The same estimation for$G(-i,\rho d^{-1})$can be proved in the same way.) We prove this by using (73). By (42) for$x\in G(+i,\rho d^{-1})$the under integral expression in (73), for$k=i;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$and by (74), \begin{equation} \mu(U_{\varepsilon}(G))\geq2\varepsilon\mu(\Pr(G(+i,\rho d^{-1})). \end{equation} Hence to prove (102) we need to estimate$\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$). Now we prove the third inclusion in (104). If$x\in U_{\varepsilon}(G)$then by the second inclusion of (104) there exists$b$such that$b\in Tr(A(\rho)),\mid x-b\mid<2\varepsilon.$Then by the definition of$Tr(A(\rho))$there are$\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}})$(see the first inclusion of (104)) give$x\in Tr(U_{2\varepsilon}(A(\rho)))$, i.e., the third inclusion in (104) is proved. The third inclusion, Lemma 4 (c) and (71) imply that$\mu(U_{\varepsilon }(G))=O(\rho^{-\alpha})\mu(B(\rho))\varepsilon.$This and (103) give the proof of (102)$\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}J. Feldman, H. 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