%% This document created by Scientific Word (R) Version 2.0 %% Starting shell: mathart1 \documentclass[12pt,thmsa]{article}% \usepackage{amssymb} \usepackage{sw20lart}% \usepackage{amsmath}% \setcounter{MaxMatrixCols}{30}% \usepackage{amsfonts}% \usepackage{graphicx} %TCIDATA{OutputFilter=3Dlatex2.dll} %TCIDATA{Version=3D4.00.0.2312} %TCIDATA{TCIstyle=3DArticle/art4.lat,lart,article} %TCIDATA{CSTFile=3Darticle.cst} %TCIDATA{LastRevised=3DWednesday, June 23, 2004 14:17:02} %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 $L(l,q(x))=3D-\Delta^{l}+q(x),$ = of arbitrary dimension $d$ with periodic, with respect to \ arbitrary = lattice, potential $q(x),$ where $l\geq1.$ Then we prove that the number of gaps = in the spectrum of the operator $L(l,q(x))$ is finite which is the = generalisation of the Bethe -Sommerfeld conjecture for this operator. In particular, = taking $l=3D1$ 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(l,q)=3D-\Delta^{l}+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 $s\geq3d(3^{d}+d+2),$ $F\equiv R^{d}/\Omega$ is a fundamental domain of $\Omega,$ and, without loss of generality, we assume that the measure $\mu(F)$ of $F$ is $1$. Let $L_{t}(l,q)$ be the operator generated in $F$ by (1) = and the quasiperiodic conditions:% \begin{equation} u(x+\omega)=3De^{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 $L_{t}(l,q)$ 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 $L$, that is, $Spec(L)=3D\left\{ \Lambda_{n}(t):t\in F^{\ast },n=3D1,2,...\right\} .$ The corresponding eigenfunctions $\Psi_{1,t}% (x),\ \Psi_{2,t}(x),...,$ of $L_{t}$ are known as Block functions. In = the case $q(x)=3D0$ these eigenvalues and eigenfunctions are = $\mid\gamma+t\mid^{2l}$ and $e^{i(\gamma+t,x)}$ respectively, for $\gamma\in\Gamma$. The paper consists of three sections. In section 1 the eigenvalues $\left\vert \gamma+t\right\vert ^{2l}$, 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 $\left\vert \gamma+t\right\vert ^{2l}$ of the order of $\rho^{2l}$ ( = written as $\left\vert \gamma+t\right\vert ^{2l}\sim\rho^{2l}$ 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 $\Lambda(a-b)<\rho^{2l},$ $\Lambda(a+b)>\rho^{2l}$\ and = $\Lambda(\gamma+t)$ \ is continuous on this intervals. Hence there exists $\gamma+t$ for = which $\Lambda(\gamma+t)=3D\rho^{2l}.$ It implies the validity of = Bethe-Sommerfeld conjecture for $L(l,q).$ 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 $L_{t}(l,q),$ lie in the = $\varepsilon_{1}% =3D\rho^{-d-2\alpha},$ neighborhood of a numbers = $F(\gamma+t)=3D\mid\gamma +t\mid^{2l}+F_{k_{1}-1}(\gamma+t)$, $\lambda_{j}(\gamma+t)$ ( see (5), = (7)), where $k_{1}=3D[\frac{d}{3\alpha}]+2.$ 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 $\gamma^{^{\prime}}\in K,\gamma^{^{\prime}}+t\in E_{k}^{1}\backslash E_{k+1}^{1},j=3D1,2,...,$ where $K$ is the set of $\gamma^{^{\prime}}\in\Gamma$ satisfying $\mid = F(\gamma+t)-\mid\gamma^{^{\prime}}+t\mid^{2l}\mid<\frac{1}% {3}\rho^{\alpha_{1}}$. 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 $L(1,q).$ 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 $\{\gamma+t\in = U^{1}(\rho^{\alpha_{1}},p):F(\gamma+t)=3D\rho^{2}\}.$ If $d=3D2,3$ then $F(\gamma+t)=3D\mid\gamma+t\mid^{2},$ and the matrix $C(\gamma+t)$ corresponds to the Schrodinger operator with directional potential $q_{\gamma_{1}}(x)=3D\sum_{n\in Z}q_{n\gamma_{1}}e^{i(n\gamma_{1},x)}$ ( = 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 $d=3D3$ = and in  for $d\geq2$ ( if $d=3D2$ 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. $2l>n;$ 2. $4l>n+1,$ $(2l\leq = n)$; 3. $n=3D3,l=3D1,$ 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 = $L_{t}(1,q)$ were obtained. In  asymptotic formulae for nonresonance eigenvalues of $L_{0}(1,q)$ 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 $d=3D2,3$ for arbitrary lattice, for dimension $d>3$ for rational lattice and for = the operator $L(l,q),$ for $2l>d.$ 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 the Bethe-Sommerfeld conjecture = for the Scrodinger operator for $d\leq4,$ for arbitrary lattice. Recendly Parnovski and Sobelev  proved this conjecture for the operator = $L(l,q),$ for $8l>d+3.$ 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 = $L(l,q),$ in order to avoid eclipsing the essence by technical details,\ we assume = that $l\geq1.$ It can be replaced by $l>n_{d,s},$ where $n_{d,s}<1$ and = depends on dimension $d$ of space and smoothness $s$ of potential $q(x)\in W_{2}^{s}(F).$ 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 $A$ 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 $L_{t}(l,q)$ lying in $(a^{2l}-\rho^{2l-1},a^{2l}+\rho^{2l-1})$ 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 $k=3D1,2,...,d,$ that can be easily verified by using the = definitions $p=3Ds-d,$ $\alpha_{k}=3D3^{k}\alpha,$ $\alpha=3D\frac{1}{q},$ = $q=3D3^{d}+d+2,$ $k_{1}=3D[\frac{d}{3\alpha}]+2,$ $p_{1}\equiv\lbrack\frac{p}{3}]+1$ 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^{2l})b(N,\gamma)=3D(\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^{2l})b(N,\gamma)=3D\sum_{\gamma_{1}\in\Gamm= a (\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}})=3D\dfrac{(\Psi_{N,t}q(x),e^{i(\gamma^{^{\prime}}+= t,x)} % )}{\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2l}}=3D% %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^{2l}}+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^{2l}\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^{2l}\mid<\frac{1}{2}\rho^{\alpha_{1}}, \end{equation} and $\mid\gamma+t\mid^{2l}$ is a non-resonance eigenvalue, i.e., $\gamma+t\in U^{l}(\rho^{\alpha_{1}},p)$ then \begin{equation} \mid\mid\gamma+t\mid^{2l}-\mid\gamma-\gamma_{1}+t\mid^{2l}\mid>\rho ^{\alpha_{1}},\mid\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2l}\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 $b(N,\gamma-\gamma_{1})=3D% %TCIMACRO{\dsum _{\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}{\Lambda_{N}-\mid \gamma-\gamma_{1}+t\mid^{2l}}+O(\rho^{-p\alpha}).$ Putting it into right side of (12) and isolating the terms with the coefficient $b(N,\gamma)$, i.e., isolating the case $\gamma_{1}+\gamma_{2}=3D0$ we get $(\Lambda_{N}-\mid\gamma+t\mid^{2l})b(N,\gamma)=3D% %TCIMACRO{\dsum _{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}% {\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2l}}+O(\rho^{-p\alpha})=3D$% $%TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{\mid q_{\gamma_{1}}\mid^{2}b(N,\gamma)}{\Lambda_{N}-\mid\gamma -\gamma_{1}+t\mid^{2l}}+% %TCIMACRO{\dsum _{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},\gamma_{2}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}b(N,\gamma-\gamma_{1}-\gamma_{2})}% {\Lambda_{N}-\mid\gamma-\gamma_{1}+t\mid^{2l}}+O(\rho^{-p\alpha}).$ 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 last formula replacing $b(N,\gamma^{^{\prime}})$ for $\gamma^{^{\prime}}=3D\gamma-\gamma_{1}-\gamma_{2},$ and then for = $\gamma ^{^{\prime}}=3D\gamma-\gamma_{1}-\gamma_{2}-\gamma_{3},...,$ 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^{2l})b(N,\gamma)=3DA_{p_{1}}(\Lambda_{N}% ,\gamma+t)b(N,\gamma)+C_{p_{1}}+O(\rho^{-p\alpha}), \end{equation} where = $A_{p_{1}}(\Lambda_{N},\gamma+t)=3D\sum_{k=3D1}^{p_{1}}S_{k}(\Lambda _{N},\gamma+t)$ , $S_{k}(\Lambda_{N},\gamma+t)=3D% %TCIMACRO{\dsum _{\gamma_{1},...,\gamma_{k}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},...,\gamma_{k}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{k}}q_{-\gamma_{1}-\gamma= _{2}-...-\gamma_{k}}}{\prod_{j=3D1}^{k}(\Lambda_{N}-\mid\gamma+t-\sum_{i=3D= 1}% ^{j}\gamma_{i}\mid^{2l})},$% $C_{p_{1}}=3D\sum_{\gamma_{1},...,\gamma_{p_{1}+1}\in\Gamma(\rho^{\alpha})= }% \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{p_{1}+1}}b(N,\gamma -\gamma_{1}-\gamma_{2}-...-\gamma_{p_{1}+1})}{\prod_{j=3D1}^{p_{1}}(\Lamb= da _{N}-\mid\gamma+t-\sum_{i=3D1}^{j}\gamma_{i}\mid^{2l})}.$ 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 $s=3D1,2,...,k$ and for $s=3D1,2,...,p_{1}$ respectively. These = conditions and the inclusion $\gamma_{i}\in\Gamma(\rho^{\alpha}),$ for $i=3D1,2,...,p_{1}$ = imply the relation $\sum_{i=3D1}^{j}\gamma_{i}\in\Gamma(p\rho^{\alpha})$ 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}}=3DO(\rho^{-p_{1}\alpha_{1}})=3DO(\rho^{-p\alpha}),\text{ = }S_{k}% (\Lambda_{N},\gamma+t)=3DO(\rho^{-k\alpha_{1}}),\forall = k=3D1,2,...,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)=3DO(\rho^{-k\alpha_{1}}),\forall k=3D1,2,...,p_{1}% \end{equation} for all $a$ satisfying $\mid a-\mid\gamma+t\mid^{2l}\mid<\frac{1}{2}% \rho^{\alpha_{1}}.$ Thus finding the number $N$ such that $\Lambda_{N}$ = is close to $\mid\gamma+t\mid^{2l}$ 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 = $L_{t}(l,q(x))$ is close to the eigenvalue of the unperturbed operator $L_{t}(l,0).$ = However in Theorem 1(c) we prove that if $\mid\gamma+t\mid^{2l}$ is resonance eigenvalue, say, if = $\gamma+t\in(\cap_{i=3D1}^{k}V_{\gamma_{i}}^{l}(\rho^{\alpha_{k}}))$ 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 $L_{t}(l,q(x))$ is close to the = eigenvalue of the matrix constructed as follows. Introduse the sets \begin{align*} B_{k}(\gamma_{1},\gamma_{2},...,\gamma_{k}) & = =3D\{b:b=3D\sum_{i=3D1}^{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) & =3D\gamma+t+B_{k}\equiv\{\gamma+t+b:b\in B_{k},\},\\ B_{k}(\gamma+t,p_{1}) & =3DB_{k}(\gamma+t)+\Gamma(p_{1}\rho^{\alpha}). \end{align*} Denote by $h_{i}+t$ for $i=3D1,2,...,b_{k}$ 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 $C(\gamma+t,\gamma _{1},\gamma_{2},...,\gamma_{k})=3D(c_{i,j})$ by the formula $c_{i,j}=3Dq_{h_{i}-h_{j}},\forall i\neq j,c_{i,i}=3D\mid = h_{i}+t\mid^{2l},$ $i,j=3D1,2,...,b_{k}.$ 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^{2l}-\mid x+\gamma_{1}\mid^{2l}=3Da^{2(l-1)}(\mid = x\mid^{2}-\mid x+\gamma_{1}\mid^{2}) \end{equation} where $a\sim\rho.$ Therefore = $V_{\gamma_{1}}^{l}(\rho^{\alpha_{1}})\subset V_{\gamma_{1}}^{1}(\rho^{\alpha_{1}}),$% \begin{equation} (\cap_{i=3D1}^{k}V_{\gamma_{i}}^{l}(\rho^{\alpha_{k}}))\subset\cap_{i=3D1= }% ^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}})), \end{equation}% \begin{equation} U^{1}(\rho^{\alpha_{1}},p)\subset U^{l}(\rho^{\alpha_{1}},p) \end{equation} \ for $l\geq1$ and $k=3D1,2,....$Taking into account these inclusions we consider the resonance eigenvalue \ $\mid\gamma+t\mid^{2l}$ for $\gamma +t\in(\cap_{i=3D1}^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}})).$ \begin{theorem} $(a)$ Suppose $\gamma+t\in U^{l}(\rho^{\alpha_{1}},p),$ $\mid\gamma\mid \sim\rho.$ 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 = $k=3D1,2,...,[\frac{1}% {3}(p-c)],$ where \begin{equation} F_{s}=3DO(\rho^{-\alpha_{1}}),\forall s=3D0,1,..., \end{equation} and $F_{0}=3D0,$ = $F_{s}=3DA_{s}(\mid\gamma+t\mid^{2l}+F_{s-1},\gamma+t),$ for $s=3D1,2,....$ $(b)$ For $\gamma+t\in U^{l}(\rho^{\alpha_{1}},p),$ = $\mid\gamma\mid\sim\rho$ there exists an eigenvalue $\Lambda_{N}$ of $L_{t}(q(x))$ satisfying = (5). $(c)$Suppose $\mid\gamma\mid\sim\rho,$ $\gamma+t\in(\cap_{i=3D1}^{k}% V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}}))\backslash E_{k+1}^{1},$ where = $1\leq k\leq d-1.$ If (15) and (23) hold then there is an indeks $j$ such that \begin{equation} \Lambda_{N}=3D\lambda_{j}(\gamma+t)+O(\rho^{-(p-c-\frac{1}{4}d3^{d})\alph= a}), \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 $\Lambda(\gamma+t)\sim\rho^{2l}$ ( see (6) for = the definition of $\Lambda(\gamma+t)$) of the operator $L_{t}(l,q(x))$ = satisfies either (5) or (25) for $c=3D\frac{q(d-1)}{2},$ namely = $\Lambda(\gamma+t)$ satisfies (5) for $\gamma+t\in U^{l}(\rho^{\alpha_{1}},p)$ and satisfies (25) for $\gamma+t\in E_{k}^{1}\backslash E_{k+1}^{1},$ where = $k=3D1,2,...,d-1.$ \end{theorem}% %TCIMACRO{\TeXButton{Proof}{\proof}}% %BeginExpansion \proof %EndExpansion $(a)$ To prove (5) in case $k=3D1$ 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^{2l}=3D\mid\gamma+t\mid^{2l}+O(\rho^{-\alpha= _{1}} ) \end{equation} This and $\alpha_{1}=3D3\alpha$ ( see the end of the introduction) imply = that formula (5) for $k=3D1$ holds and $F_{0}=3D0.$ Hence (24) for $s=3D0$ is = also proved. Moreover by (19) we have $S_{k}(\mid\gamma+t\mid^{2l}+O(\rho^{-\alpha_{1}}),\gamma+t)=3DO(\rho ^{-\alpha_{1}})$ for $k=3D1,2,....$ 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 $k=3Dj$, then putting the expression from = (5) (for $k=3Dj)$, 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} & =3D\mid\gamma+t\mid^{2l}+A_{p_{1}}(\mid\gamma+t\mid^{2l}% +F_{j-1}+O(\rho^{-j\alpha_{1}}),\gamma+t)+O(\rho^{-(p-c)\alpha})=3D\\ \{A_{p_{1}}( & \mid\gamma+t\mid^{2l}+F_{j-1}+O(\rho^{-j\alpha_{1}}% ),\gamma+t)-A_{p_{1}}(\mid\gamma+t\mid^{2l}+F_{j-1},\gamma+t)\}+\\ A_{p_{1}}( & = \mid\gamma+t\mid^{2l}+F_{j-1},\gamma+t)+O(\rho^{-(p-c)\alpha}) \end{align*} To prove $(a)$ for $k=3Dj+1$ 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=3D1}^{s}(\mid\gamma+t\mid^{2l}+F_{j-1}+O(\rho^{-j\alpha= _{1}% })-\mid\gamma+t-\sum_{i=3D1}^{s}\gamma_{i}\mid^{2l})}-\\ & \frac{1}{\prod_{j=3D1}^{s}(\mid\gamma+t\mid^{2l}+F_{j-1}-\mid\gamma +t-\sum_{i=3D1}^{s}\gamma_{i}\mid^{2l})}\\ & =3D\frac{1}{\prod_{j=3D1}^{s}(\mid\gamma+t\mid^{2l}+F_{j-1}-\mid\gamma +t-\sum_{i=3D1}^{s}\gamma_{i}\mid^{2l})}(\frac{1}{1-O(\rho^{-(j+1)\alpha_= {1}}% )}-1) \end{align*} $=3DO(\rho^{-(j+1)\alpha_{1}})$, $\forall s=3D1,2,...,p_{1}.$ $(b)$ Let $A$ be the set of indices $N$ satisfying (15). Using (11) and Bessel inequality, we obtain% $\sum_{N\notin A}\mid b(N,\gamma)\mid^{2}=3D\sum_{N\notin = A}\mid\dfrac{(\Psi _{N}(x),(q(x)e^{i(\gamma+t,x)})}{\Lambda_{N}-\mid\gamma+t\mid^{2l}}\mid ^{2}=3DO(\rho^{-2\alpha_{1}})$ Hence by the Parseval equality we have $\sum_{N\in A}\mid = b(N,\gamma)\mid ^{2}=3D1-O(\rho^{-2\alpha_{1}}).$ This together with $\mid = A\mid\frac{1}{2}(c_{\Gamma})^{-1}\rho^{-\frac{(d-1)q}{2}\alpha}$ , that is, = (23) holds, for $c=3D\frac{(d-1)q}{2}$ . 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^{2l})b(N,h_{i})=3D\sum_{\gamma^{^{\prime}}% \in\Gamma(\rho^{\alpha})}q_{\gamma^{^{\prime}}}b(N,h_{i}-\gamma^{^{\prime= }% })+O(\rho^{-p\alpha}), \end{equation} for $i=3D1,2,...,b_{k}.$ 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}})=3D\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^{^{\pr= ime }% }-\sum_{i=3D1}^{p_{1}}\gamma_{i})}{\prod_{j=3D0}^{p_{1}-1}(\Lambda_{N}-\m= id h_{i}-\gamma^{^{\prime}}+t-\sum_{i=3D1}^{j}\gamma_{i}\mid^{2l})}+ \end{equation}% $+O(\rho^{-p\alpha})=3DO(\rho^{-p\alpha}).$ 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^{2l}\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 $j=3D1,2,...,s$ and $s=3D0,1,...,p_{1}-1.$ 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 $s=3D0,1,...,p_{1}-1.$ Since $B_{k}(\gamma+t,p)=3D\gamma+t+B_{k}+\Gamma (p\rho^{\alpha})$ we have a decomposition $a_{s}=3D\gamma+t+b+a,$ 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 $l=3D1.$ To prove (31) for $l=3D1$ 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 $\gamma+t=3Dx+v$ 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=3DO(\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), $\gamma+t=3Dx+v,$ $(x,v)=3D(x,a)=3D(x,b)=3D0,$% \begin{equation} \mid\gamma+t+a+b\mid^{2}-\mid\gamma+t\mid^{2}=3D\mid a+b+v\mid^{2}-\mid v\mid^{2}% \end{equation} give us the estimation (31) for $l=3D1.$ 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 $\gamma+t\in\cap_{i=3D1}^{k}% V_{\gamma_{i}}^{1}(\rho^{\alpha_{k+1}}).$ 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 $d_{1}=3D\mid\gamma+t+a\mid^{2}-\mid\gamma+t\mid^{2}+2\langle a,b\rangle,$ it follows from the inequalities (34), (30) that $\mid = d_{1}% \mid>\frac{2}{3}$ $\rho^{\alpha_{k+1}}$. Taking $a=3D0$ in (33) we have $d_{2}=3D\mid b+v\mid^{2}-\mid v\mid^{2}.$ Therefore (32) and (30) 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% $(\Lambda_{N}-\mid h_{i}+t\mid^{2})b(N,h_{i})=3D\sum_{\gamma^{^{\prime}}% }q_{\gamma^{^{\prime}}}b(N,h_{i}-\gamma^{^{\prime}})+O(\rho^{-p\alpha }),i=3D1,2,...,b_{k},$ 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 $(C-\Lambda_{N}I)(b(N,h_{1}),b(N,h_{2}),...b(N,h_{b_{k}}))=3DO(\rho^{-p\al= pha}) ,$ where the rigth side of this system is a vector having the norm $\mid\mid O(\rho^{-p\alpha})\mid\mid=3DO(\sqrt{b_{k}}\rho^{-p\alpha})$. = 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\alp= ha },\\ \max_{i=3D1,2,...,b_{k}} & = \mid\Lambda_{N}-\lambda_{i}\mid^{-1}=3D\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 $\mid B_{k}\mid=3DO(\rho^{\frac{k}{2}\alpha_{k+1}}),$ $\mid \Gamma(p_{1}\rho^{\alpha})\mid=3DO(\rho^{d\alpha})$ and = $d\alpha<\frac{1}% {2}\alpha_{d}$ ( see the end of introduction) we get \begin{equation} b_{k}=3DO(\rho^{d\alpha+\frac{k}{2}\alpha_{k+1}})=3DO(\rho^{\frac{d}{2}\a= lpha_{d }% })=3DO(\rho^{\frac{d}{2}3^{d}\alpha}),\forall k=3D1,2,...,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 $\rho^{2l}$ of the operator $L_{t}(l,q(x)).$ Denote by $D$ the set = of all vectors $\gamma\in\Gamma$ satisfying (15). From (11), arguing as in = the proof of ($b$), we obtain $\sum_{\gamma\in D}\mid b(N,\gamma)\mid^{2}% =3D1-O(\rho^{-2\alpha_{1}}).$ Since $\mid D\mid=3DO(\rho^{d-1})$ ( see = the end of the introduction), from definition of $\Lambda(\gamma+t)$ we get $\mid b(N,\gamma)\mid>c_{6}\rho^{-\frac{(d-1)}{2}}=3Dc_{6}\rho^{-\frac {(d-1)q}{2}\alpha}$, that is, condition (23) for $c=3D\frac{(d-1)q}{2}$ = 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 $k=3D1,2,...,d-1$ ( see (41) in Remark 1). The Theorem is proved.$\diamondsuit$ \begin{remark} Here we note that the nonresonance domain = $U^{l}(c_{7}\rho^{\alpha_{1}},p)$ 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 $B(\rho)=3D\{x\in R^{d}:\mid x\mid=3D\rho\}.$ By (22) it is enough = to prove this for $l=3D1.$ 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 $\{x:\mid x\mid^{2}-\mid x+b\mid^{2}=3D-c_{7}\rho^{\alpha_{1}}\}$and = $\{x:\mid x\mid^{2}-\mid x+b\mid^{2}=3Dc_{7}\rho^{\alpha_{1}}\}.$ The distance of = these hyperplanes from origin are $\frac{c_{7}\rho^{\alpha_{1}}}{\mid b\mid}.$ This, the relation $\mid\Gamma(p\rho^{\alpha})\mid=3DO(\rho^{d\alpha}),$ 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))=3DO(\rho ^{d-1-\alpha})\\ \mu(U^{1}(c_{7}\rho^{\alpha_{1}},p)\cap B(\rho)) & = =3D(1+O(\rho^{-\alpha}% ))\mu(B(\rho)). \end{align} Now we consider = $E\equiv(\cap_{i=3D1}^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}% }))\cap B(\rho).$ We turn the coordinate axis so that = $Span\{\gamma_{1,}% \gamma_{2},...,\gamma_{k}\}$ coincides with the span of the vectors $e_{1}=3D(1,0,0,...,0)$, $e_{2}=3D(0,1,0,...,0),...,$ $e_{k}$. Then = $\gamma _{s}=3D\sum_{i=3D1}^{k}\gamma_{s,i}e_{i}$ for $s=3D1,2,...,k$ . = Therefore the relation $x\in\cap_{i=3D1}^{k}V_{\gamma_{i}}^{1}(\rho^{\alpha_{k}})$ = implies that $\sum_{i=3D1}^{k}\gamma_{s,i}x_{i}=3DO(\rho^{\alpha_{k}}),s=3D1,2,...,k;\t= ext{ }% x_{n}=3D\frac{\det(b_{j,i}^{n})}{\det(\gamma_{j,i})}\text{ = ,}n=3D1,2,...,k,$ where $x=3D(x_{1},x_{2},...,x_{d}),\gamma_{j}=3D(\gamma_{j,1},\gamma _{j,2},...,\gamma_{j,k},0,0,...,0),$ $b_{j,i}^{n}=3D\gamma_{j,i}$ for = $n\neq j$ and $b_{j,i}^{n}=3DO(\rho^{\alpha_{k}})$ for $n=3Dj.$ Taking into = account that the determinant $\det(\gamma_{j,i})$ of the system is a volume of the parallelepiped $\{\sum_{i=3D1}^{k}b_{i}\gamma_{i}:b_{i}\in\lbrack0,1],i=3D1,2,...,k\}$ = , hence greater than some constant $c$ and using that $\mid\gamma_{j,i}\mid From this for projection$v$of$x\in\cap_{i=3D1}^{k}V_{\gamma_{i}}(\rho ^{\alpha_{k}})$onto$Span\{\gamma_{1,}\gamma_{2},...,\gamma_{k}\}$we obtain estimation (32). Moreover, note that if$k=3Dd,$then$v=3Dx$and by = (40) we have$\mid x\mid=3DO(\rho^{\alpha_{d}+(d-1)\alpha})$which is imposible, = becouse$\alpha_{d}+(d-1)\alpha<1$( see the inequality 6. in the end of the introduction) and$x\in B(\rho).$It means that the intersection of$d$resonance domains lying in$\{\mid x\mid>\rho_{0}\}$for$\rho_{0}\gg1$= is empty set. Therefore we have \begin{equation} R^{d}\cap\{\mid x\mid>\rho_{0}\}=3D(U^{1}\cup(\cup_{s=3D1}^{d-1}(E_{s}% ^{1}\backslash E_{s+1}^{1})))\cap\{\mid x\mid>\rho_{0}\}. \end{equation} \end{remark} \begin{remark} Here we note some properties of the known part$\mid\gamma+t\mid^{2l}+F_{k}(\gamma+t)$( for$k=3D1,2,...$see Theorem = 1) of the nonresonance eigenvalues of$L_{t}(l,q(x))$. Denoting$\gamma+t$by =$x$, where$\mid\gamma+t\mid\sim\rho,\gamma+t\in = U^{1}(\rho^{\alpha_{1}},p),$we prove \begin{equation} \frac{\partial F_{k}(x)}{\partial = x_{i}}=3DO(\rho^{2-2l-2\alpha_{1}+\alpha }),\forall i=3D1,2,...,d;\forall k=3D1,2,... \end{equation} We prove (42) by induction with respect to$k.Using (20) one can = easily verify that% \begin{align*} & \mid x\mid^{2l}-\mid x-\gamma_{1}\mid^{2l}\sim\rho^{2l-2}(\mid x\mid ^{2}-\mid x-\gamma_{1}\mid^{2}),\\ & \mid x\mid^{2l-2}-\mid x-\gamma_{1}\mid^{2l-2}\sim\rho^{2l-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% $\frac{\partial}{\partial x_{i}}(\dfrac{1}{\mid x\mid^{2l}-\mid x-\gamma _{1}\mid^{2l}})=3D$% \begin{equation} \dfrac{2lx_{1}(\mid x\mid^{2l-2}-\mid x-\gamma_{1}\mid^{2l-2})}{(\mid x\mid^{2l}-\mid x-\gamma_{1}\mid^{2l})^{2}})+ \end{equation}% $\dfrac{-2\gamma_{1}(i)\mid x-\gamma_{1}\mid^{2l-2}}{(\mid = x\mid^{2l}-\mid x-\gamma_{1}\mid^{2l})^{2}}=3DO(\rho^{2-2l-2\alpha_{1}+\alpha}),$ 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$k=3D1$follows from (4) and (43). Suppose that (42) holds for$k=3Ds.$= Using this and (24), replacing$\mid x\mid^{2l}$by$\mid x\mid^{2l}+F_{s}(x)$in = (43) and evaluating as above we obtain% $\frac{\partial}{\partial x_{i}}(\dfrac{1}{\mid x\mid^{2l}+F_{s}(x)-\mid x-\gamma_{1}\mid^{2l}})=3D$% $O(\rho^{2l-2-2\alpha_{1}+\alpha})+\dfrac{\frac{\partial = F_{s}(x)}{\partial x_{i}}}{(\mid x\mid^{2l}+F_{s}-\mid x-\gamma_{1}\mid^{2l})^{2}}=3D$% $O(\rho^{2-2l-2\alpha_{1}+\alpha})+O(\rho^{2-2l-4\alpha_{1}+\alpha}% )=3DO(\rho^{2-2l-2\alpha_{1}+\alpha})$ This formula with the definition of$F_{k}$implies (42) for$k=3Ds+1.$\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$k=3Dk_{1}$in (5), = where$k_{1}=3D[\frac{d}{3\alpha}]+2,$\ ( we can take$k=3Dk_{1},$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$F(\gamma +t)=3D\mid\gamma+t\mid^{2l}+F_{k_{1}-1}(\gamma+t)$,$\varepsilon_{1}% =3D\rho^{-d-2\alpha}$( see introduction) we obtain \begin{equation} \Lambda_{N}-F(\gamma+t)=3Do(\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 =$k=3D1,2,...,n=3D[\frac{p}{3}]$. This is a simple eigenvalue and the corresponding eigenfunction$\Psi_{N}(x)$satisfies \begin{equation} \Psi_{N}(x)=3De^{i(\gamma+t,x)}+O(\rho^{-\alpha_{1}}). \end{equation} 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$\Lambda_{N}$. The formula \ (45) is equivalent to =$b(N,\gamma)=3D1+O(\rho ^{-\alpha_{1}}),$where$b(N,\gamma)=3D(\Psi_{N},e^{i(\gamma+t,x)}).$= Since the normalized eigenfunction is defined up to constant of modulas$1$we can suppose that$\arg b(N,\gamma)=3D0.$Hence to prove (45) it suffices to = prove the equality$\mid b(N,\gamma)\mid=3D1+O(\rho^{-\alpha_{1}})$which = equavalent to \begin{equation} \sum_{\gamma^{^{\prime}}\in\Gamma\backslash\{\gamma\}}\mid b(N,\gamma ^{^{\prime}})\mid^{2}=3DO(\rho^{-\alpha_{1}}) \end{equation} Thus we need to show that (46) holds. First consider the case$\gamma ^{^{\prime}}\notin K$( for the definition of$K$see introduction). = Using (44), definition of$K, and then (11), we get \begin{align} & \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2l}\mid>\frac{1}{4}% \rho^{\alpha_{1}},\gamma^{^{\prime}}\notin K,\\ \sum_{\gamma^{^{\prime}}\notin K} & \mid b(N,\gamma^{^{\prime}})\mid ^{2}=3D\parallel\Psi_{N}(x)q(x)\parallel^{2}O(\rho^{-2\alpha_{1}})=3DO(\r= ho ^{-2\alpha_{1}}).\nonumber \end{align} If\gamma^{^{\prime}}\in K$, then clearly \begin{equation} \mid\Lambda_{N}-\mid\gamma^{^{\prime}}+t\mid^{2l}\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} for$c=3D2dq-\frac{1}{4}d3^{d}-d-3=3D\frac{7}{4}d3^{d}+2d^{2}+3d-3.$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}=3D\mid\gamma^{^{\prime}}+t\mid^{2l}+F_{k-1}(\gamma^{^{\prime}= % }+t)+O(\rho^{-3k\alpha}), \end{equation} for$k=3D1,2,...,[\frac{1}{3}(p-c)].$Now taking =$k=3Dk_{1}=3D[\frac{d}{3\alpha}]$( we can take$k=3Dk_{1},$since one can easily verify that =$\frac{d}{3\alpha }+2\leq\frac{1}{3}(p-c)$), and arguing as in the prove of (44) we get% $\Lambda_{N}-F(\gamma^{^{\prime}}+t)=3Do(\varepsilon_{1})\text{ ,}$ 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}=3D\lambda_{j}(\gamma^{^{\prime}}+t)+O(\rho^{-(p-c-\frac{1}{4}= % d3^{d})\alpha}). \end{equation} It is not hard to verify that$(p-c-\frac{1}{4}d3^{d})\alpha>d+2\alpha$= . Therefore we have $\Lambda_{N}-\lambda_{j}(\gamma^{^{\prime}}+t)=3Do(\varepsilon_{1})$ This with (44) contradicts \ the simplicity condition (10). Hence the inequality in (49) holds. Therefore using$\mid K\mid=3DO(\rho^{q(d-1)\alpha})$we have \begin{equation} \sum_{\gamma^{^{\prime}}\in K\backslash\{\gamma\}}\mid b(N,\gamma^{^{\prime}% })\mid^{2}=3DO(\rho^{-(2c-q(d-1))\alpha}), \end{equation} for$c=3D2dq-\frac{1}{4}d3^{d}-d-3.$Since =$c-q(d-1)>3,\alpha_{1}=3D3\alpha$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$k=3D1,2,...,n=3D[\frac{p}{3}],$= because the inequality (23) holds for$c=3D0$( 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$k=3D1,2,...,n=3D[\frac{1}{3}(c-\frac{1}{2}q(d-1))]$, where =$c=3D2dq-\frac{1}% {4}d3^{d}-d-3,\Phi_{0}(x)=3D0,\Phi_{1}(x)=3D% %TCIMACRO{\dsum _{\gamma_{1}\in\Gamma(\rho^{\alpha})}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1}\in\Gamma(\rho^{\alpha})}} %EndExpansion \dfrac{q_{\gamma_{1}}e^{i(\gamma+t+\gamma_{1},x)}}{(\mid\gamma+t\mid^{2l}= % -\mid\gamma+\gamma_{1}+t\mid^{2l})},$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$k=3D1$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}=3DO(\rho^{-2k\alpha_{1}}), \end{equation}% \begin{equation} \Psi_{N}=3Db(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$\parallel H_{k}(x)\parallel=3DO(\rho^{-k\alpha_{1}}).$The case =$k=3D1$is proved accoding to (46). Assume that (53) is true for$k=3Dm$. Then by = (54) for$k=3Dm,$and by (3) we have =$\Psi_{N}(x)(q(x))=3DH(x)+O(\rho^{-m\alpha_{1}}),$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$(H(x),e^{i(\gamma+t+\gamma^{^{\prime}},x)})=3D0$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}% =3D\sum_{\gamma^{^{\prime}}}\mid\dfrac{(O(\rho^{-m\alpha_{1}}),e^{i(\gamm= a +t+\gamma^{^{\prime}},x)})}{\Lambda_{N}-\mid\gamma+\gamma^{^{\prime}}% +t\mid^{2l}}\mid^{2}=3DO(\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 =$\alpha _{1}=3D3\alpha$and definition of$n,$from (52) we obtain $\sum_{\gamma^{^{\prime}}\in K\backslash\{\gamma\}}\mid b(N,\gamma^{^{\prime}% })\mid^{2}=3DO(\rho^{-2n\alpha_{1}}),$ This with (55) implies (53) for$k=3Dm+1.$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$b(N,\gamma+\gamma^{^{\prime}}-\sum_{i=3D1}^{j}\gamma_{i})$= when$\gamma^{^{\prime}}-\sum_{i=3D1}^{j}\gamma_{i}\neq0$where =$\gamma_{i}\in \Gamma(\rho^{\alpha}),j=3D1,2,...,n-1.$Then using (4) and the = relation$\mid\Lambda_{N}-\mid\gamma+t+\gamma^{^{\prime}}-\sum_{i=3D1}^{j}\gamma_{= i}% \mid^{2l}\mid>\frac{1}{2}\rho^{\alpha_{1}}$( see (16) and use the fact = that$\gamma^{^{\prime}}-\sum_{i=3D1}^{j}\gamma_{i}\in\Gamma(p\rho^{\alpha})$= since$p>2n$),$\Lambda_{N}=3DP(\gamma+t)+O(\rho^{-n\alpha_{1}}),$where$P(\gamma +t)=3D\mid\gamma+t\mid^{2l}+F_{[\frac{p}{3}]}(\gamma+t)$( see Theorem = 2), we obtain% \begin{equation} b(N,\gamma+\gamma^{^{\prime}})=3D\sum_{k=3D1}^{n-1}A_{k}(\gamma^{^{\prime= }% })b(N,\gamma)+O(\rho^{-n\alpha_{1}}), \end{equation} where$A_{1}(\gamma^{^{\prime}})\equiv\dfrac{q_{\gamma^{^{\prime}}}}{P(\gamma +t)-\mid\gamma+\gamma^{^{\prime}}+t\mid^{2l}}=3D\dfrac{q_{\gamma^{^{\prim= e}}}% }{\mid\gamma+t\mid^{2l}-\mid\gamma+\gamma^{^{\prime}}+t\mid^{2l}}% +O(\rho^{-3\alpha_{1}})A_{k}(\gamma^{^{\prime}})=3D% %TCIMACRO{\dsum _{\gamma_{1},...,\gamma_{k-1}}}% %BeginExpansion {\displaystyle\sum_{\gamma_{1},...,\gamma_{k-1}}} %EndExpansion \dfrac{q_{\gamma_{1}}q_{\gamma_{2}}...q_{\gamma_{k-1}}q_{\gamma^{^{\prime= }% }-\gamma_{1}-\gamma_{2}-...-\gamma_{k-1}}}{\prod_{j=3D0}^{k-1}(P(\gamma +t)-\mid\gamma+t+\gamma^{^{\prime}}-\sum_{i=3D1}^{j}\gamma_{i}\mid^{2l})}= % =3DO(\rho^{-k\alpha_{1}}),$\begin{equation} \sum_{\gamma^{\ast}\in\Gamma((n-1)\rho^{\alpha})}\mid = A_{1}(\gamma^{\ast}% )\mid^{2}=3DO(\rho^{-2\alpha_{1}}),\sum_{\gamma^{\ast}\in\Gamma((n-1)\rho= ^{\alpha})}\mid A_{k}(\gamma^{\ast})\mid=3DO(\rho^{-k\alpha_{1}}), \end{equation} for$k>1.$Now from (54) for$k=3Dn$and (56), we obtain $\Psi_{N}=3Db(N,\gamma)e^{i(\gamma+t,x)}+\sum_{\gamma^{\ast}\in\Gamma ((n-1)\rho^{\alpha})}\sum_{k=3D1}^{n-1}(A_{k}(\gamma^{\ast})b(N,\gamma )+O(\rho^{-n\alpha_{1}}))e^{i(\gamma+t+\gamma^{\ast},x)})+H_{n}$ Since$\parallel\Psi_{N}\parallel=3D1,\arg b(N,\gamma)=3D0,$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 =$\parallel H_{n}\parallel=3DO(\rho^{-n\alpha_{1}})$we have$1=3D\mid b(N,\gamma)\mid^{2}+\sum_{k=3D1}^{n-1}(\sum_{\gamma^{\ast}\in \Gamma((n-1)\rho^{\alpha})}\mid A_{k}(\gamma^{\ast})b(N,\gamma)\mid^{2}% +O(\rho^{-n\alpha_{1}})),$\begin{equation} b(N,\gamma)=3D(1+\sum_{k=3D1}^{n-1}(\sum_{\gamma^{\ast}\in\Gamma((n-1)\rh= o ^{\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}})=3D(\sum_{k=3D1}^{n-1}A_{k}(\gamma^{^{\prim= e}% }))(1+\sum_{k=3D1}^{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$n=3D2.$By (58), (57), (59) we have$b(N,\gamma )=3D1+O(\rho^{-2\alpha_{1}}),b(N,\gamma+\gamma^{^{\prime}})=3DA_{1}(\gamma^{^{\prime}})+O(\rho^{-2\al= pha _{1}})=3D\dfrac{q_{\gamma^{^{\prime}}}}{\mid\gamma+t\mid^{2l}-\mid\gamma +\gamma^{^{\prime}}+t\mid^{2l}}+O(\rho^{-2\alpha_{1}}) $for all$\gamma^{^{\prime}}\in\Gamma(\rho^{\alpha}).$These and (54) for$k=3D2$= imply the formula for$\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}=3D\{x\in = U^{1}(2\rho^{\alpha_{1}},p):F(x)=3D\rho^{2l}\} $. Due to (5) ( replace$\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\}$(see (9)). Therefore = we eliminate \begin{equation} P_{b}=3D\{x:x,x+b\in U^{1}(\rho^{\alpha_{1}},p),\mid F(x)-F(x+b)\mid <3\varepsilon_{1}\} \end{equation} for$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}})=3D\cup_{a\in = S_{\rho }^{^{\prime}}}U_{\varepsilon}(a)\}$of$S_{\rho}^{^{\prime}}$, where$\varepsilon=3D\frac{\varepsilon_{1}}{7l\rho^{2l-1}},U_{\varepsilon }(a)=3D\{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)=3D\{\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))$of \begin{equation} A(\rho)=3D\cup_{k=3D1}^{d-1}(\cup_{\gamma_{1},\gamma_{2},...,\gamma_{k}\i= n \Gamma(p\rho^{\alpha})}(\cup_{i=3D1}^{b_{k}}A_{k,i}(\gamma_{1},\gamma _{2},...,\gamma_{k}))), \end{equation} where$A_{k,i}(\gamma_{1},...,\gamma_{k})=3D\{x\in(\cap_{i=3D1}^{k}V_{\gamma_{i= }}% ^{1}(\rho^{\alpha_{k}})\backslash E_{k+1}^{1})\cap K_{\rho}:\lambda_{i}% (x)\in(\rho^{2l}-3\varepsilon_{1},\rho^{2l}+3\varepsilon_{1})\},$and \begin{equation} K_{\rho}=3D\{x\in R^{d}:\mid\mid = x\mid^{2l}-\rho^{2l}\mid<\rho^{\alpha_{1}}\}. \end{equation} As a result we construct the part$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^{2l},\Lambda(a+b)>\rho^{2l}$\ and$\Lambda(\gamma+t)$\ is continuous on this intervals. Hence there exists$\gamma+t$for which$\Lambda(\gamma+t)=3D\rho^{2l}.$It implies the validity of = Bethe-Sommerfeld conjecture for$L(l,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))=3D\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 <\varepsilonwe obtain \begin{align} & \mid F(x)-F(a)\mid<3l\rho^{2l-1}\varepsilon,\\ & \mid F(x+b)-F(a+b)\mid<3l\rho^{2l-1}\varepsilon.\nonumber \end{align} These inequalities together with (64) give (63), since6l\rho^{2l-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 <3l\rho^{2l-1}\varepsilon.$This, the first ineguality in (65), and the relations$F(a)=3D\rho^{2},F(c)=3D\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=3D1,2,...,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})=3D\mu(E(k)),E(k)-\gamma _{k}\subset E.$Moreover by$(b)(E(k)-\gamma_{k})\cap(E(j)-\gamma_{j})=3D\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=3D1,2,...,n.$Using$\mu(E_{k}+\gamma_{k})=3D\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=3D1,2,...,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$lies in this subset then Theorem 2 holds. For every connected open subset$E$of =$U_{\varepsilon}(S_{\rho}^{^{\prime}% })\backslash Tr(A(\rho)$there exists a unique index$N$such that$\Lambda(\gamma+t)=3D\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^{2l}$for$\rho\gg1$lies in the spectrum of$L(l,q),$that is, the number of the gaps in the spectrum of$L(l,q)$is finite. \end{theorem}% %TCIMACRO{\TeXButton{Proof}{\proof}}% %BeginExpansion \proof %EndExpansion$(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)=3D\rho^{2l}$give \begin{equation} F(\gamma+t)\in(\rho^{2l}-3l\rho^{2(l-1)}\varepsilon,\rho^{2l}+3l\rho ^{2(l-1)}\varepsilon)\subset(\rho^{2l}-\varepsilon_{1},\rho^{2l}% +\varepsilon_{1}) \end{equation} for$\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^{2l}\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+t)\notin(\rho^{2l}-3\varepsilon_{1},\rho^{2l}% +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)=3D\Lambda_{N(y)}(y)$% ,$\Psi_{y}(x)=3D\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)=3DN(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)=3DN(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)=3DN(y_{i})=3DN_{i}$for =$y\in U(y_{i}).$Clearly if$U(y_{i})\cap U(y_{j})\neq\emptyset$then$N_{i}=3DN_{j}$since for$a\in U(y_{i})\cap U(y_{j})$there is a unique$N(a)$. Thus$N_{1}=3DN_{2}=3D...=3DN_{k}$and$N(a_{1})=3DN(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))=3D\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}}) & =3D(1-c_{10}\rho^{-\alpha})\mu(B(\rho)),\\ \mu(U_{2\varepsilon}(A(\rho))) & = =3DO(\rho^{-\alpha})\mu(B(\rho))\varepsilon. \end{align} ( See below, Estimations 1, 2, 3). The estimation (64) of the measure of = setV_{\rho}$is done in Estimation 4 by using Estimations 1, 2, 3.$(c)$Since$F(a)=3D\rho^{2l}$for =$a=3D(a_{1},a_{2},...,a_{d})=3D\sum_{i=3D1}% ^{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 ^{2l}-c_{11}\varepsilon_{1},F(a+\varepsilon e_{i})>\rho^{2l}+c_{11}% \varepsilon_{1}$and$\Lambda(a-\varepsilon e_{i})<\rho^{2l},\Lambda (a+\varepsilon e_{i})>\rho^{2l}.$Since$\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))=3D\rho^{2l}.$The Theorem is proved =$\diamondsuit$In Estimations 1-4 we use the notations:$G(+i,a)=3D\{x\in G,x_{i}>a\},G(-i,a)=3D\{x\in G,x_{i}<-a\},$where =$x=3D(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$the followings hold \begin{equation} \rho-1<\mid x\mid<\rho+1,G\subset(\cup_{i=3D1}^{d}(G(+i,\rho d^{-1})\cup G(-i,\rho d^{-1})) \end{equation} Indeed if$x\in S_{\rho}^{^{\prime}}$then$F(x)=3D\rho^{2l}$and by = (24) we have$\mid x\mid=3D\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=3D\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}$we use the formula \begin{equation} \mu(G(+k,a))=3D\int\limits_{\Pr_{k}(G(+k,a))}(\frac{\partial F}{\partial x_{k}% })^{-1}\mid grad(F)\mid dx_{1}...dx_{k-1}dx_{k+1}...dx_{d}, \end{equation} where$\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}=3D0.$Instead of$\Pr_{k}(G)$we write$\Pr(G)$if$k$is unambiguous. If$D$= is$m-$dimensional subset of$R^{m}$then we use \begin{equation} \mu(D)=3D\int\limits_{\Pr_{k}(D)}\mu(D(x_{1},...x_{k-1},x_{k+1},...,x_{m}= % ))dx_{1}...dx_{k-1}dx_{k+1}...dx_{m}, \end{equation} where$D(x_{1},...x_{k-1},x_{k+1},...,x_{m})=3D\{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=3D(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=3D(\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=3D\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 $x\in G(+1,a),$ $A=3DB(\rho)\cap U^{1}(3\rho^{\alpha_{1}},p).$ = 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}=3D\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 $h=3D\rho^{-\alpha}$ 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 $F(x_{1}+h,x_{2},...,x_{d})>\rho^{2l}$, $F(x_{1}-h,x_{2},...,x_{d})<\rho^{2l}%$. 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})=3D\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 $\mu(\Pr G(+1,a))=3DO(\rho^{d-1})$ ( see the inequality in (72))% $\mu(G(+1,a))=3D\int\limits_{\Pr(G(+1,a))}\frac{\mid x\mid}{x_{1}}dx_{2}% dx_{3}...dx_{d}+O(\rho^{-\alpha})\mu(B(\rho))\geq$% $\int\limits_{\Pr(A(+1,2a))}\frac{\mid = x\mid}{x_{1}}dx_{2}dx_{3}...dx_{d}% -c_{10}\rho^{-\alpha}\mu(B(\rho))=3D$% $\mu(A(+1,2a))-c_{12}\rho^{-\alpha}\mu(B(\rho)).$ 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 $\mu(A))=3D(1+O(\rho^{-\alpha}))\mu(B(\rho))$ (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., $b=3D(b_{1},0,0,...,0)$ 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})^{l}+F_{k_{1}-1}(x) & =3D\rho^{2l},\\ ((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})^{l}+F_{k_{1}-1}(x+b= ) & =3D\rho^{2l}+h \end{align} where $h\in(-3\varepsilon_{1},3\varepsilon_{1}).$ It follows from (80), (81), and (24) that \begin{equation} (x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})=3D\rho^{2}+O(\rho^{-\alpha_= {1}}) \end{equation} \begin{equation} ((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})=3D\rho^{2}+O(\rho ^{-\alpha_{1}}) \end{equation} Subtrracting (82) from (83) we get% \begin{equation} (2x_{1}-b_{1})b_{1}=3DO(\rho^{-\alpha_{1}}), \end{equation} This with inequalities in (72) implies \begin{equation} \mid b_{1}\mid<2\rho+3,\text{ = }x_{1}=3D\frac{b_{1}}{2}+O(\rho^{-2\alpha_{1}% }b_{1}^{-1}),\mid = x_{1}^{2}-(\frac{b_{1}}{2})^{2}\mid=3DO(\rho^{-2\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}=3D\rho^{2}+O(\rho^{-2\alpha}),\mid = x_{1}\mid=3D\rho+O(\rho^{-2\alpha -1}),\text{ }x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2}=3DO(\rho^{-2\alpha}). \end{equation} Therefore $G\subset G(+1,a)\cup G(-1,a),$ where $a=3D\rho-\rho^{-1}$. = Using (73), the obvious relation = $\mu(\Pr_{1}(G(+1,a))=3DO(\rho^{-(d-1)\alpha})$ (see (82)) and the fact that the expression under the integral (73) for = $k=3D1$ is equal to $1+O(\rho^{-\alpha})$ (see (76) and (86)) and doing the same = for $G(-1,a)$ we get $\mu(G)=3DO(\rho^{-(d-1)\alpha}).$ Since = $\mid\Gamma_{1}% \mid=3DO(\rho^{d-1})$ we have \begin{equation} \text{ }\mu(\cup_{b\in\Gamma_{1}}(S_{\rho}\cap = P_{b})=3DO(\rho^{-(d-1)\alpha +d-1})=3DO(\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 $k=3D2,3,...,d.$ 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 $k=3Dd$ is $O(\rho^{1+\alpha}).$ Therefore the first equality = in \begin{equation} \mu(D)=3DO(\varepsilon_{1}\mid b\mid^{-1}\rho^{d-2}),\text{ = }\mu(G(+d,\rho ^{-\alpha}))=3DO(\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 $m=3Dd-1;$ $k=3D1$ and the relations = $\mu(\Pr_{1}D)=3DO(\rho^{d-2})$, \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% $((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})^{l}-((x_{1}^{2}% +x_{2}^{2}+x_{3}^{2}+...+x_{d}^{2})^{l}-$% \begin{equation} F_{k_{1}-1}(x-b)-F_{k_{1}-1}(x)=3Dh \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}=3Df(x_{1}),\frac{df}{dx_{1}}=3D\frac{2x_{1}+O(\rho^{-2\alpha_{1}+\a= lpha}% )}{2x_{d}+O(\rho^{-2\alpha_{1}+\alpha})}=3D\frac{x_{1}}{x_{d}}+O(\rho ^{-\alpha_{1}}) \end{equation} Substituting this in (91), we get $((x_{1}-b_{1})^{2}+x_{2}^{2}+x_{3}^{2}+...+f^{2}(x_{1}))^{l}-((x_{1}^{2}%= +x_{2}^{2}+x_{3}^{2}+...+f^{2}(x_{1}))^{l}+$% \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)=3Dh \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 $a_{l}(x)$ of (93) is% \begin{equation} \frac{\partial a_{l}(x)}{\partial x_{1}}=3Dl\mid = x-b\mid^{2(l-1)}(2(x_{1}% -b_{1})+2f(x_{1})f^{^{\prime}}(x_{1})) \end{equation} $-l\mid x\mid^{2(l-1)}(2x_{1}+2f(x_{1})f^{^{\prime}}(x_{1}))+O(\rho ^{-2\alpha_{1}+\alpha})(1+\frac{x_{1}}{x_{d}})+O(\rho^{-3\alpha_{1}+\alph= a}) .$ If $l=3D1$ then using the first equality in (85) and $\mid x_{d}\mid >\rho^{-\alpha}$ \ we have% \begin{equation} \mid\frac{\partial a_{l}(x)}{\partial x_{1}}\mid>b_{1}. \end{equation} If $l>1$ then it follows from (80), (81), and (24) that $\mid x\mid^{2l}=3D\rho^{2l}(1+O(\rho^{-2l-\alpha_{1}})),\text{ }\mid x-b\mid^{2l}=3D\rho^{2l}(1+O(\rho^{-2l-\alpha_{1}}))$% \begin{align*} & \mid = x\mid^{2(l-1)}=3D\rho^{2(l-1)}(1+O(\rho^{-2l-\alpha_{1}}))^{\frac{l-1}% {l}}=3D\rho^{2(l-1)}+O(\rho^{-2-\alpha_{1}})\text{ }\\ & \mid = x-b\mid^{2(l-1)}=3D\rho^{2(l-1)}(1+O(\rho^{-2l-\alpha_{1}}))^{\frac {l-1}{l}}=3D\rho^{2(l-1)}+O(\rho^{-2-\alpha_{1}})\text{ }% \end{align*} Using these in (94) and arguing as in proof of (95), for $l=3D1,$ we get = the proof of (95) for $l>1$ 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 $\mu(S_{\rho}\cap P_{b})=3DO(\rho^{d-1+\alpha}\varepsilon_{1}\mid b\mid^{-1}),$ for $b\notin\Gamma_{1}.$ Since $\mid b\mid<2\rho+3$ ( see (85)) and $\varepsilon_{1}=3D\rho^{-d-2\alpha}$, we infer = $\mu(\cup_{b\notin\Gamma _{1}}(S_{\rho}\cap = P_{b}))=3DO(\rho^{2d-1+\alpha}\varepsilon_{1})=3DO(\rho ^{-\alpha})\mu(B(\rho)).$ 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 $\gamma_{1,}\gamma _{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha}),k=3D\leq d-1.$ We turn the coordinate axis so that $Span\{\gamma_{1,}\gamma_{2},...,\gamma_{k}\}=3D\{x=3D(x_{1},x_{2},...,x_= {k}% ,0,0,...,0):x_{1},x_{2},...,x_{k}\in R\}$. Then by (40) we have $x_{n}% =3DO(\rho^{\alpha_{k}+(k-1)\alpha}),$ for $n\leq k,$ $x\in G$. This, = (72), and $\alpha_{k}+(k-1)\alpha<1$ ( see the end of the introduction) give $G\subset(\cup_{i>k}(G(+i,\rho d^{-1})\cup G(-i,\rho d^{-1})),$ $\mu(\Pr_{i}(G(+i,\rho d^{-1})))=3DO(\rho^{k(\alpha_{k}+(k-1)\alpha)+(d-1-k)}),$ for $i>k.$ Now using this and (74) for $m=3Dd$ we prove \begin{equation} \mu(G(+i,\rho d^{-1}))=3DO(\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)\varepsilo= n ,\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 $x=3D(x_{1},x_{2},...,x_{i}% ,...x_{d})$ and $x^{^{\prime}}=3D(x_{1},x_{2},...,x_{i}^{^{\prime}},...,x_{d})$ 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 $\mid a\mid=3D\rho+O(1),\mid a^{^{\prime}}\mid=3D\rho+O(1),$ we get \begin{equation} \mid a\mid^{2l}-\mid = a^{^{\prime}}\mid^{2l}=3Dl(\rho+O(1))^{2(l-1)})(\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}) \end{equation} Let $r_{i}(x)=3D\lambda_{i}(x)-\mid x\mid^{2l}.$ Hence $r_{1}(x)\leq r_{2}(x)\leq...\leq$ $r_{b_{k}}(x)$ be the eigenvalue of the matrix $C(x)-\mid x\mid^{2l}I\equiv C^{^{\prime}}(x),$ 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 $\mid x-b_{i}% \mid^{2l}-\mid x\mid^{2l},$ where $b_{i}=3Dh_{i}+t-\gamma-t\in B_{k}% +\Gamma(p_{1}\rho^{\alpha}).$ Hence $\mid b_{i}\mid<\rho^{\frac{1}{2}% \alpha_{d}}$ . Clearly% \begin{equation} \mid r_{j}(a^{^{\prime}})-r_{j}(a)\mid\leq\parallel C^{^{\prime}}(a^{^{\prime }})-C^{^{\prime}}(a)\parallel=3D\max_{i}\mid a_{i,i}\mid. \end{equation} where $C^{^{\prime}}(a^{^{\prime}})-C^{^{\prime}}(a)=3D(a_{i,j}),$ $a_{i,i}=3D\mid a\mid^{2l}-\mid a-b_{i}\mid^{2l}-\mid a^{^{\prime}}\mid^{2l}+\mid a^{^{\prime }}-b_{i}\mid^{2l},$ $a_{i,j}=3D0$ 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 $\mid a\mid =3D\rho+O(1),\mid a^{^{\prime}}\mid=3D\rho+O(1),$ $\mid = a-b_{i}\mid=3D\rho +O(\rho^{\frac{1}{2}\alpha_{d}}),\mid a^{^{\prime}}-b_{i}\mid=3D\rho +O(\rho^{\frac{1}{2}\alpha_{d}}),$ we obtain $a_{i,i}=3D(\rho+O(\rho^{\frac{1}{2}\alpha_{d}}))^{2(l-1)}(\mid = a\mid^{2}-\mid a^{^{\prime}}\mid^{2})-$% $(\rho+O(\rho^{\frac{1}{2}\alpha_{d}}))^{2(l-1)}(\mid = a-b_{i}\mid^{2}-\mid a^{^{\prime}}-b_{i}\mid^{2})=3D$% $(\rho+O(\rho^{\frac{1}{2}\alpha_{d}}))^{2(l-1)}(\mid a\mid^{2}-\mid a-b_{i}\mid^{2}-\mid a^{^{\prime}}\mid^{2}+\mid a^{^{\prime}}-b_{i}\mid^{2})+$% $O(\rho^{\frac{1}{2}\alpha_{d}+2l-3})(\mid a\mid^{2}-\mid = a^{^{\prime}}\mid ^{2}).$ Since $\mid a\mid^{2}-\mid a-b_{i}\mid^{2}-\mid = a^{^{\prime}}\mid^{2}+\mid a^{^{\prime}}-b_{i}\mid^{2}=3D2(a-a^{^{\prime}},b_{i})$ we have ( see = (100)) $\mid r_{j}(a)-r_{j}(a^{^{\prime}})\mid\leq3\rho^{\frac{1}{2}\alpha_{d}% +2l-2}\mid a-a^{^{\prime}}\mid-c_{15}\rho^{\frac{1}{2}\alpha_{d}+2l-3}\mid\mid a\mid^{2}-\mid a^{^{\prime}}\mid^{2}\mid$ Therefore using (99), (100) = and (98) we obtain $\mid\lambda_{j}(a)-\lambda_{j}(a^{^{\prime}})\mid\geq\mid\mid a\mid^{2l}-\mid a^{^{\prime}}\mid^{2l}\mid-\mid r_{j}(a)-r_{j}(a^{^{\prime}})\mid>l\rho ^{2l-1}d^{-1}\mid a_{i}^{^{\prime}}-a_{i}\mid>$ $l\rho^{2l-1}42d\varepsilon>6\varepsilon_{1}$ 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 = $\mu(U_{2\varepsilon}(A_{k,j}(\gamma_{1,}\gamma_{2},...,\gamma _{k})))=3DO(\varepsilon\rho^{k(\alpha_{k}+(k-1)\alpha)+d-1-k}),$ where $j=3D1,2,...,b_{k}(\gamma_{1},\gamma_{2},...,\gamma_{k}),$ and $\gamma _{1},\gamma_{2},...,\gamma_{k}\in\Gamma(p\rho^{\alpha}).$ From this = using that $b_{k}=3DO(\rho^{d\alpha+\frac{k}{2}\alpha_{k+1}})$ 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% \begin{equation} d\alpha+\frac{k}{2}\alpha_{k+1}+dk\alpha+k(\alpha_{k}+(k-1)\alpha)+d-1-k\= leq d-1-\alpha \end{equation} for $1\leq k\leq d-1$. Using $\alpha_{k}=3D3^{k}\alpha,$ = $\alpha=3D\frac{1}{q},$ $q=3D3^{d}+d+2$ 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 = $k=3Dd-1.$ 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 $\mu(G(+i,\rho d^{-1}))=3DO(\rho^{-\alpha})\mu(B(\rho)),$ for $i=3D1,2,...,d.$ (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 $k=3Di;a=3D\rho = d^{-1},$ is less than $d+1.$ Therefore it is sufficient to prove \begin{equation} \mu(\Pr(G(+i,\rho d^{-1}))=3DO(\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 $b=3D\gamma+c$. Therefore $\mid x-\gamma-c\mid=3D\mid = x-b\mid<2\varepsilon,$ $x-\gamma\in U_{2\varepsilon}(c)\subset U_{2\varepsilon}(A(\rho)).$ 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 $\mu(U_{\varepsilon }(G))=3DO(\rho^{-\alpha})\mu(B(\rho))\varepsilon.$ 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. 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