Content-Type: multipart/mixed; boundary="-------------0810141016556" This is a multi-part message in MIME format. ---------------0810141016556 Content-Type: text/plain; name="08-187.comments" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="08-187.comments" This is a revision of a paper posted in novembre 2006. The paper in the present form is due to appear con Comm.PDE's ---------------0810141016556 Content-Type: text/plain; name="08-187.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="08-187.keywords" dispersion, Bloch functions ---------------0810141016556 Content-Type: application/x-tex; name="revised1.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="revised1.tex" \input amstex.tex \input amsppt.sty \TagsAsMath \NoRunningHeads \magnification=1200 \hsize=5.0in\vsize=7.0in \hoffset=0.2in\voffset=0cm \nonstopmode \document \document \topmatter \title{Dispersion for Schr\"odinger equation with periodic potential in 1D} \endtitle \author Scipio Cuccagna \endauthor \address DISMI University of Modena and Reggio Emilia, Via Amendola 2, Padiglione Morselli, Reggio Emilia 42100 Italy \endaddress \email cuccagna.scipio\@unimore.it\endemail \abstract We extend a result on dispersion for solutions of the linear Schr\"odinger equation, proved by Firsova for operators with finitely many energy bands only, to the case of smooth potentials in 1D with infinitely many bands. The proof consists in an application of the method of stationary phase. Estimates for the phases, essentially the band functions, follow from work by Korotyaev. Most of the paper is devoted to bounds for the Bloch functions. For these bounds we need a detailed analysis of the quasimomentum function and the uniformization of the inverse of the quasimomentum function. \endabstract \endtopmatter \head \S 1 Introduction \endhead We consider an operator $H_0= -\frac {d^2}{dx^2} +P(x)$ with $P(x)$ a smooth periodic potential of period 1. We will prove: \proclaim{Main Theorem } There is a $C>0$ such that for any $p\ge 2$ and $t>0$ we have $ \| e^{itH_0} \colon L^{\frac{p}{p-1} }(\Bbb R) \to L^{p }(\Bbb R) \| \le C \max \left \{ t^{-\frac 12}, t ^{-\frac 13} \right \} ^{(1-\frac 2p)} . $ \endproclaim We recall that the spectrum $\Sigma (H_0)$ is a union of closed intervals (bands) and that the theorem in the case of finitely many bands is in Firsova \cite{F1}, for alternative proofs in the two bands case see \cite{Cai,Cu}. Here we will consider the case when $\Sigma (H_0)$ is a union of infinitely many bands. Just to simplify the notation, we will formulate the proof only in the generic case when all the gaps are nonempty. This generic case contains all the essential difficulties, and a proof for cases when some gaps are empty goes through similarly, only with more complicated notation. We are motivated by nonlinear problems, see \cite{Cu}. Indeed dispersion for linear operators is central to nonlinear equations, see for instance \cite{Str}. For $P\equiv 0$ we have $ e^{-it {d^2}/{dx^2}} (x,y)= (4\pi it)^{-\frac{1}{2}} e^ {-i {(x-y)^2}/{4t} }$. For no nonconstant $P(x)$ a similar explicit formula seems to be available. We outline the proof, with terminology and formulas introduced rigourously later. The proof of the Main Theorem reduces to a pointwise bound on the kernel $e^{itH_0}(x,y).$ The kernel is expressed by means of the distorted Fourier transform, written using Bloch functions and is a sum of oscillatory integrals, one term per energy band. In each integral the phase involves the band functions, which express energy $E$ with respect to quasimomentum $k$. $k$ varies in $\Bbb C$ cut along some slits which correspond to the gaps. We derive various estimates for the derivatives of the band function $E(k)$ from the representation of the quasimomentum function $k=k(w)$, $w=\sqrt{E}$, exploiting formulas in Korotyaev \cite{K1}. We have essentially $E(k)=k^2$, except for very narrow regions near the edges of the spectral bands where, in particular, the third derivative of $ E(k)$ is very large. In this region very close the edges our bounds on the Bloch functions are large, but in the stationary phase formula are more than offset by the very small upper bound for the inverse of the third derivative of $ E(k)$. Away from the edges, the contribution to $e^{itH_0}(x,y) $ is essentially the same of the constant coefficients case and there are good bounds for the Bloch functions. There is an intermediate region, close but not very close, in a relative sense, to the edges, where our bounds for the Bloch functions are large and where the third derivative of $ E(k)$ is not large. Yet, combining the narrowness of this region with good enough bounds for the Bloch functions, we control the corresponding contribution to $e^{itH_0}(x,y) $. Since estimates for $E(k)$ follow by fairly direct elaboration on material by Korotyaev \cite{K1}, our most serious effort is for bounds on the Bloch functions and their first derivative in $k$. It is probably because of these that \cite{F1} considers finitely many bands only. The Bloch functions can be expressed as a linear combination of a nice fundamental set of solutions of the equation $H_0u=Eu$. The coefficient in this linear combination is the Weyl-Titchmarsh function, which near the edges of the bands of higher energy is hard to bound, because it is the ratio of two very small quantities. In the finite bands case, Firsova \cite{F1} uses the fact that the Bloch functions are analytic on the uniformization of the function $w(k)$. Since there are only finitely many bands, this gives a uniform bound. Bounds for the derivatives follow from the Cauchy integral formula. In the case of infinitely many bands, near each edge the Bloch functions are bounded, but there is no obvious uniform bound over an infinite number of bands. Furthermore, the distance of the edges from the boundary of the domain of analyticity goes to 0 as we take higher energies. Hence, for large energy, the Cauchy integral formula gives bad estimates on the $k$ derivative of the Bloch functions, near the edges of the band. This problem for the derivative is the main source of trouble in the paper. Taking Fourier expansion, if one differentiates in $k$ one gets a small divisors problem, with small divisors for modes $n=0$ and for $\pm n\pi +k\approx 0$. Is seems convenient to use the Cauchy integral formula with bounds for complex $k$. The Bloch functions are of the form $e^{\pm ikx}m_{\pm }(x,k)$. Thanks to estimates on a nice pair of fundamental solutions of $H_0u=Eu$ and relations between the various terms in the Weyl-Titchmarsh function and a certain normalization term denoted by $N(k)$, we see that for large energies in the Fourier series expansion $m_{\pm }(x,k)=\sum \widehat{ m}_{\pm }(n,k)e^{2\pi i nx}$ most terms are small compared to $\widehat{ m}_{\pm}(0,k)$ and $\widehat{ m}_{\pm}(\pm n,k)$ with $| \pi n +k|\ll 1$ and $\| m_{\pm }(\cdot ,k)\| _{L^\infty}\approx \| m_{\pm }(\cdot ,k)\| _{L^2}$. For $k\in \Bbb R$ the Bloch functions are normalized, $\| m_{\pm }(\cdot ,k)\| _{L^2}=1$. However for $k\not \in \Bbb R$ and near the boundary of the domain of analyticity, that is near the slits, it is problematic to bound $\| m_{\pm}(\cdot ,k)\| _{L^2}.$ In correspondence to the interior of the band $|\widehat{ m}_{\pm }(0,k)|\gg |\widehat{ m}_{\pm }(\mp n,k)|$. Using the normalization of the Bloch functions we get $$1\approx \widehat{ m}_{+}(0,k)\widehat{ m}_{-}(0,k)+\widehat{ m}_{+}(-n,k)\widehat{ m}_{-}(n,k),\tag 1.1$$ so $\| m_{+ }(\cdot ,k)\| _{L^2} \| m_{-}(\cdot ,k)\| _{L^2} \lesssim 1$. Near the slits however $|\widehat{ m}_{\pm }(0,k)|\approx |\widehat{ m}_{\pm }(\mp n,k)|$ and in the right in (1.1) we could have a cancelation. However from some explicit formula and thanks to (9.1) we get $\widehat{ m}_{+}(0,k)\widehat{ m}_{-}(0,k)-\widehat{ m}_{+}(-n,k)\widehat{ m}_{-}(n,k)\approx \dot E/(2k)$. By the Schwartz Christoffel formula we get a good bound for this quantity for $k$ close enough to the real axis. Hence for $k$ close enough to the real axis in (1.1) there is no cancelation and we get a uniform bound on $\| m_{+ }(\cdot ,k)\| _{L^2} \| m_{-}(\cdot ,k)\| _{L^2} $. The argument shows however that near the extremes of the slits the two terms in (1.1) are in fact unbounded. Away from the edges of the bands we improve the estimates significantly thanks to more information on $E(k)$. One can relax significantly the regularity requirements on $P(x)$ maintaining the proof. The proof goes from \S 4 to \S 10. In particular, the estimates on the Bloch functions are in \S 9 and \S 10. Here the spectrum is denoted by $\Sigma (H_0)$, with the usual notation $\sigma (H_0)$ reserved for something else. In a statement or in a proof, notation $a_n\lesssim b _n$ means that there is a fixed constant $C>0$ independent of $n$, with $a_n\le C b _n$. If we write $a_n\ll b _n$ we mean $a_n\le C b _n$ for a very small fixed constant $C>0$. If we write $a_n\approx b _n$ we mean $(1/C)b_n \le a_n\le C b _n$ for a fixed constant $C>0$ independent of $n$. If we write $a_n=O(b_n)$ (resp. $a_n=o(b_n)$) we mean $a_n\lesssim b _n$ (resp. $a_n\ll b_n$). For $p\in [1,\infty ]$, by $\| f \|_p$ we mean the usual $L^p$ norm of $f(x)$, where the $x$ varies in a set indicated in the context. \head \S 2 Band function, Bloch function, quasimomentum and uniformization \endhead The spectrum is of the form $\Sigma (H_0)=\cup _{n=0}^\infty \Sigma _n$, with each two compact intervals $\Sigma _n$ and $\Sigma _{n+1}$ separated by an open gap $G_n$. We assume $\inf \Sigma _0=0.$ We set now $ \sigma = \cup _{n=-\infty}^\infty \sigma _n$, with each two compact intervals $\sigma _n$ and $\sigma _{n+1}$ separated by an open gap $g_n$, with $g_0$ empty, and with $\sigma _{-n}=-\sigma _n$ and $\sigma ^2_n=\Sigma _n$. We will assume each $g_n$ non empty for $n\neq 0$. We recall now the following standard result, see \cite{Ea} ch. 4: \proclaim{Theorem 2.1} Let $P(x)$ be smooth. Set $\sigma _n=[a_n^+, a_{n+1}^-]$ and $g_n=]a_n^-, a_{n }^+[$. Then there exist a strictly increasing sequence $\{ \ell _n \in \Bbb Z \} _{n\in \Bbb Z}$ and a fixed constant $C $ such that $$|a_n^- -\ell _n\pi |+ |a_n^+ -\ell _n\pi |\le C \langle \ell _n \rangle ^{-1}. $$ For any $N$ there exists a fixed constant $C_N $ such that the length $|g_n|$ of the gap $g_n$ is $|g_n|\le C _N \langle \ell _n \rangle ^{-N}. $ \endproclaim To simplify notation we will assume in the rest of the paper that $\ell _n\equiv n$, the case when all spectral gaps, from a certain one on, are not empty, which is generic, but essentially the same proof goes through in general. For any $w \in {\Bbb C_+}$ (the open upper half plane) we consider the fundamental solutions $\theta (x,w)$ and $\varphi (x,w)$ of $H_0u=w^2u$ which satisfy the initial conditions $$\varphi (0,w)= \theta ^\prime (0,w) =0\, , \quad \varphi ^\prime (0,w) =\theta (0,w)=1.\tag 2.1$$ The Floquet determinant $D(w)$ is defined by $2D(w)={\varphi ^\prime (w) +\theta (w)}$ where $\varphi ^\prime (w) =\varphi ^\prime (1,w)$ and $\theta (w) =\theta (1,w)$. For any $w \in {\Bbb C_+}$ there is a unique $ k\in {\Bbb C_+}$, called quasimomentum, and a unique choice of constants $m^\pm (w)$ such that the functions $$\tilde \phi _\pm (x,w)=\theta (x,w)+m^\pm (w) \varphi (x,w)\tag 2.2$$ are of the form $\tilde \phi _\pm (x,w)=e^{\pm ikx} \xi _\pm (x,w)$ with $\xi _\pm (x,w)$ periodic of period $1$ in $x$. We have $$m^\pm (w)=\frac{\varphi ^\prime (w) - \theta (w) }{2\varphi (w)} \pm i \frac{\sin k}{\varphi (w)}.\tag 2.3$$ We have the relation $D(w)=\cos k$. The correspondence between $w$ and the corresponding quasimomentum $k$ is a conformal mapping between ${\Bbb C_+}$ and a "comb" $K$, that is a set $K={\Bbb C_+}\backslash \cup _{n\neq 0} [ n\pi , n\pi +ih _n ]$ where the $[ n\pi , n\pi +ih _n ]$ are vertical slits with $h_n\ge 0$. In particular, $|g_n|\le 2 h_n \le (1+C n^{-2}) |g_n|$ for a fixed $C$, see Theorem 1.2 \cite{KK}. Now we will use that all gaps are nonempty, but the following standard discussion extends easily. The map $k(w)$ is called quasimomentum map and extends into a continuous map in $\overline{\Bbb C_+}$ with $k(\sigma _n)=[ n\pi , (n +1)\pi ]$, with $k(w)$ a one to one and onto map between $\sigma _n$ and $[ n\pi , (n +1)\pi ]$, and with $k(g _n)=] n\pi , n\pi +ih_n ]$. We have $k(-\bar w) =- \overline{k(w)}$. By the Schwartz reflection principle, $k(w)$ extends into a conformal map from $\Bbb C \backslash \cup _{n\neq 0}\overline{g _n}$ into $\Cal K=\Bbb C\backslash \cup _n \gamma _n $ with $\gamma _n= [n\pi -ih_n, n\pi +ih_n]$. So we have $k( \bar w) = \overline{k(w)}$ and $k(w)=-k(-w)$. Hence also $ w( \bar k)= \overline{w(k)}$ and $w(k)=-w(-k)$. Then for $|t|0$ for $w\in \sigma $ (so that we define $N(w)>0$ for $w\in \sigma $). $N^2(w)$ is well defined and analytic in $\Bbb C - \cup _{\neq 0}g _n$. We have $N^2(w)=N^2(-w)=\overline{N^2(\overline{w})}$, the first equality by (2.6) and the second by (2.5). From formula (1.4) \cite{F2} we have $D' (w)=-4w \varphi (w)N^2(w)$, for a sketch of proof see \S 3 \cite{Cu}. From $D(w)=\cos k$ we get $D' (w)= - \frac{dk}{dw}\sin k$. Since $k(w)$ is a conformal map, $\frac{dk}{dw} \neq 0$ for $\Im w >0$ and so for $k\in K$. Hence $N^2(w)\neq 0$ for any $w\in \Bbb C \backslash \cup _{n\neq 0}\overline{g _n}$. We set now $$ e^{ik (x-y)}m _+ ^0(x,w)m _- ^0 (y,w)=\frac{\tilde \phi _+(x,w)\tilde \phi _-(y,w)}{N^2 (w)} .\tag 2.7$$ We express $w=w(k)$ for $k\in K$ and with an abuse of notation we write $m^0 _\pm (x,k)$ for $m^0 _\pm (x,w(k))$. Then the product $m _+ ^0(x,k)m _- ^0 (y,k)$ extends analytically for $k\in \Cal K$ and to $\Cal R$. Fix a square root $N$ of $N^2$ with $N(k)>0$ for $k\in \Bbb R$, ignoring problems of monodromy (here we need the product $m _+ ^0 m _- ^0$ to be analytic). We set $m _\pm ^0(x,k)=\xi _\pm ^0(x,k)/N(k)$. \bigskip \head \S 3 Fourier transform \endhead From Theorem XIII.98 \cite{RS} it is possible to conclude: \proclaim{Lemma 3.1} Let $N(k)=\sqrt{N^2(k)}>0$ for $k\in \Bbb R$. Set $\phi _\pm (y,k)=\widetilde{\phi} _\pm (y,k)/N(k)$ and $\hat f (k)=\int _\Bbb R {\phi _+ (y,k)} f(y)dy$. Then: $$\align & \int _\Bbb R| f(y)|^2dy= \int _{\Bbb R}|\hat f (k)|^2dk \tag a \\& f(x)= \int _{\Bbb R} {\phi _- (x,k)} \hat f(k) dk \tag b \\& \widehat{H_0f }(k)= E (k)\hat f(k) . \tag c \endalign $$ \endproclaim Lemma 3.1 implies: \proclaim{Lemma 3.2} We have $ e^{itH_0} (x,y)= K (t,x,y) =\sum _{n\in \Bbb Z} K^n(t,x,y)$ with $$ K^n (t,x,y)= \int _{n\pi }^{(n+1) \pi} e^{i(t E (k) -(x-y)k) } {m _- ^0(x,k)}{ m _+ ^0(y,k)} dk . \tag 3.1 $$ \endproclaim Then the Main Theorem follows from: \proclaim{Theorem 3.3} There is $C$ fixed such that $ | K(t,x,y)|\le C \max \{ t ^{-\frac 13}, t ^{-\frac 12} \} . $ \endproclaim \noindent Theorem 3.3 follows by the method of stationary phase. Notice that the phase in $K^n(t,x,y)$ satisfies the following result by Korotyaev \cite{K2}: \proclaim{Theorem 3.4} Consider $E(k)$ for $k\in [ \pi n, \pi (n+1)]$. Then $ E ' (k)=0$ for $k=n\pi , (n+1)\pi $ and $ E ' (k)>0 $ in $]\pi n, \pi (n+1) [$ for $n\ge 0$ ($E(k)$ is even). In $[ \pi n, \pi (n+1)] $ the equation $E '' (k)=0$ admits exactly one solution $k_n$. We have $k_n\in ] \pi n, \pi (n+1) [$ and $E ^{\prime \prime \prime } (k_n)\neq 0$. \endproclaim Naively the proof of Theorem 3.3 would go as follows. Theorem 3.4 and estimates on $m _\pm ^0(x,k)$ for $k\in [ \pi n, \pi (n+1)]$ lead to an estimate $|K^n(t,x,y)|\le D_n \langle t \rangle ^{-\frac 13}$ thanks to the method of stationary phase, which we quote from p. 334 \cite{Ste}: \proclaim{Lemma 3.5} Suppose $\phi (x)$ is real valued and smooth in $[a,b]$ with $|\phi ^{(m)}(x) |\ge c_m>0$ in $]a,b[$ for $m\ge 1$ . For $m=1$ assume furthermore that $\phi ^\prime (x)$ is monotonic in $]a,b[$. Then we have for $C_m= 5 \cdot 2^{m-1} -2 $: $$ \big |\int _a^b e^{i\mu \phi (x)} \psi (x) dx \big | \le C_m (c _m\mu )^{-\frac 1m} \left [ \min \{ |\psi (a) |, |\psi (b) | \}+\int _a^b |\psi ' (x) | dx \right ] . $$ \endproclaim Since we need to add up over all the $K^n$ we have to control the constants $D_n$. In the next section we state a list of estimates on the band function $E(k)$ and on the Bloch functions $\phi _\pm (x,k)$ which are sufficient to the purpose of the present paper and then prove Theorem 3.3. In the subsequent sections we prove the estimates on $E(k)$ and $\phi _\pm (x,k)$. \bigskip \head \S 4 Proof of Theorem 3.3 \endhead We state estimates for the first three derivatives of $E(k)$. These are proved later using material in Korotyaev \cite{K1}. Notice that $E(k)$ is even in $k$, so we consider only $k\ge 0$. We start with the first derivative $\dot E (k)$: \proclaim{Lemma 4.1} For all $k\ge 0$ we have $\dot E(k) \ge 0$. $\dot E(k) =0$ implies $k=n\pi $ for some $n\in \Bbb Z$. There are fixed constants $C>0$ and $c>0$ such that for any $n$ we have $$\aligned & \frac{k}{C} \frac {\sqrt{ w-a^+_n } }{\sqrt{|g_n|} } \le \dot E(k) \le C k \frac {\sqrt{ w-a^+_n } }{\sqrt{|g_n|} } \quad \text{for} \quad a^+_n \le w \le a^+_n + c|g_n| \\& |\dot E(k) - 2k| \le \frac{C}{\langle k \rangle } \quad \text{for} \quad a^+_n + c|g_n| \le w \le a^-_{n+1}- c|g_{n+1}| \\& \frac{k}{C}\frac {\sqrt{ a^-_{n+1} -w } }{\sqrt{|g_{n+1}|} } \le \dot E(k) \le k C \frac {\sqrt{ a^-_{n+1} -w } }{\sqrt{|g_{n+1}|} } \quad \text{for} \quad a^-_{n+1} - c|g_{n+1}|\le w \le a^-_{n+1} .\endaligned $$ \endproclaim Now we consider the second derivative $\ddot E (k)$: \proclaim{Lemma 4.2} There are fixed constants $C>0,$ $C_1> C_2>0$ and $c>0$ such that for any $n$ and any $k\in [n\pi , (n+1)\pi ] $, that is for any $w\in [a^+_n , a^-_{n+1}] $, we have: $$\aligned & w \le a^+_n + c|g_n| \Rightarrow \big |\ddot{E}-\frac{n}{|g_{n }|} \big | \le C \\& w \ge a^-_{n+1} - c|g_{n+1}| \Rightarrow \big |\ddot{E}+\frac{{n+1}}{|g_{{n+1} }|} \big | \le C\\& a^+_n +c|g_n| \le w\le a^-_{n+1}-C _1|n+1|^{\frac{1}{3}}|g_{n+1 }|^{\frac{2}{3}} \Rightarrow \ddot{E} \approx \frac 12 +\frac{n|g_n|^2}{ |w-a_n^+|^{3}} . \endaligned $$ \endproclaim Finally we have (see Lemma 7.5): \proclaim{Lemma 4.3} There are $c>0$ and $c_1>0$ such that $\forall$ $\alpha \in (1/2, 1)$ and for any $w\in \sigma _{n}\cup \sigma _{n-1}$ with $c|g_{n }|\le |w - a^\pm _{n }| \le |g_{n }|^{ \alpha}$ we have $ |\dddot{E} |\ge c_1 |n| \, |g_n|^{4(\frac{1}{2}- \alpha )}.$ \endproclaim Next we need estimates for the Bloch functions. First of all we have, see in \S 9: \proclaim{Lemma 4.4} There are fixed constants $C>0$, $C_3>0$, $\delta >0$, $\Gamma >0$ and $c>0$ such that for all $x$, all $n$ we have : {\item{ (1)}} $ \forall \, w \in [a^+_{n } +C_3n^5 |g_n| ,a^-_{n+1 }-C_3(n+1)^5 |g _{n+1}| ]$ we have $$\aligned & \big | m ^0_+(x,k) m ^0_-(y,k) -1 \big | \le C \langle k\rangle ^{-1}; \endaligned$$ {\item{ (2)}} for all $k=p+iq$ with $ k\in \Cal K$, $\pi n < p< \pi (n+1)$, $k$ in $\{ k: |q|<\delta |g_n|\} \cup \{ k: |\pi n - p|>\Gamma |g_n|\} \cup \{ k: |\pi (n+1) - p|>\Gamma |g_{n+1}|\} $ and $|q|\le 1$ we have $$\big | m ^0_+(x,k) m ^0_-(y,k) \big | \le C . $$ \endproclaim The proof of (1) Lemma 4.4 is elementary and that of (2) if $k\in \Bbb R$, that is the case near the edges not covered by (1), is relatively easy. We will state later more estimates for $k\not \in \Bbb R$, needed to bound $\dot m ^0_\pm (x,k)= {\partial _k} m ^0_\pm (x,k)$. We have: \proclaim{Lemma 4.5} There are fixed constants $C>0$ and $C_4>0$, with $C_40$ there is a fixed $C_\epsilon $ such that $ |K^n_1| \le C_\epsilon t^{-\frac{1}{2}}| g_{n}|^{\frac{1}{2}- \epsilon}$ and $ |K^n_5| \le C_\epsilon t^{-\frac{1}{2}}| g_{n+1}|^{\frac{1}{2}- \epsilon}$. \endproclaim {\it Proof of $\ell =1$.} By Lemma 3.5, by $\ddot E \ge \frac{|n|}{2| g_{n}|}$, Lemma 4.1, and by Lemmas 4.4 and 4.5, $$|K^n_1(t,x,y)| \le \frac{C\sqrt{| g_{n}|}}{\sqrt{|n| t}} \int _{a^{+}_{n}} ^{a^{+}_{n}+c|g_n|} \frac{ \frac{dk}{dw} dw}{ |k-\pi n| +|g_n| } . $$ By Lemma 4.1, $\frac{dk}{dw}\approx \frac{\sqrt{| g_{n}|}}{\sqrt{w-a^{+}_{n}}}$ and $|k-\pi n| \approx \sqrt{|g_n|} \sqrt{w-a^+_n} $. Hence $$|K^n_1(t,x,y)| \le \frac{C }{\sqrt{ |n| t}} \int _{a^{+}_{n}} ^{a^{+}_{n}+c|g_n|} \frac{ dw}{\sqrt{w-a^{+}_{n}}} \le \frac{C_ 1 \sqrt{|g_n| } }{\sqrt{ |n| t}}. $$ With a similar argument we get the estimate for $K_5$. \bigskip \proclaim{Claim $\ell =2$ } There is $C>0 $ such that $ |K^n_2| \le C \min\{ t^{-\frac{1}{2}}\log (1/|g_n|), | g_{n}|^{\frac{1}{4}} \}$.\endproclaim {\it Proof.} For $k\in [a^{+}_{n}+ c|g_n|, a^+_{n }+ |g_{n }|^{\frac{1}{4}}]$ we have $|\ddot E| \gtrsim 1$ by Lemma 4.2 and $\frac{dk}{dw}\approx 1$ and $w-a^+_n \approx k-\pi n$ by Lemma 4.1. Hence $ |K^n_2(t,x,y)| \le $ $$\le \frac{C _1}{\sqrt{ t}} \int _{a^{+}_{n}+c|g_n|} ^{a^{+}_{n}+ |g_{n }|^{\frac{1}{4}}} \frac{ \frac{dk}{dw} dw}{ |k-\pi n| +|g_n| } \le C t^{-\frac{1}{2}} \int _ {a^+_{n }+ c|g_{n }| } ^{ a^+_{n }+ |g_{n }|^{\frac{1}{4}}} \frac{ dw}{ w-a^+_{n } } \le A t^{-\frac{1}{2}} \log \frac 1{|g_n|} . $$ On the other hand, taking absolute value in the integral defining $K^n_2$, thanks to Lemma 4.4 we get $|K^n_2|\le C |g_n|^{\frac{1}{4}}.$ \bigskip \proclaim{Claim $\ell =4$} There is $C $ such that $ |K^n_4| \le C t^{-\frac{1}{3}} | g_{n+1}|^{ \frac 1 {31}}$. \endproclaim {\it Proof.} $K^n_4$ is an integral on an interval where $|\dddot E| \gtrsim |n+1| |g_{n+1 }|^{-\frac{1}{10}}$ by Lemma 4.3. $\frac{dk}{dw}\approx 1$ and $ a^-_{n+1}-w \approx \pi (n+1)-k$ by Lemma 4.1. So, by Lemmas 3.5, 4.4 and 4.5, we get a contribution bounded by $$ t^{-\frac{1}{3}} |n+1|^{\frac{1}{3}} |g_{n +1}|^{ \frac{1}{30}} \int ^ {a^-_{n +1}- c|g_{n+1 }| } _{ a^-_{n +1}- |g_{n+1 }|^{\frac{3}{5}}} \frac{ dw}{ a^-_{n+1}-w } \lesssim t^{-\frac{1}{3}} |n+1|^{\frac{1 }{3} } |g_{n+1 }|^{ \frac{1}{30}} \log \frac 1{|g_{n+1 }|} . $$ \bigskip We get $ \sum _n \sum _{\ell \neq 3,\ell =1}^5 |K^n_\ell (t,x,y)| \le C \max \{ t^{-\frac{1}{3}}, t^{-\frac{1}{2}}\} $ by the above claims and by Lemma 2.1. Finally we consider $K^n_3.$ Set $K_3=\sum _n K _3^n $. \proclaim{Lemma 4.6 } There is a fixed $C$ such that $| K _3 (t,x,y)| \le C \max \{ t^{-\frac{1}{2}}, t^{-\frac{1}{3}} \} .$ \endproclaim {\it Proof.} Set $\chi _{int} (k)=\sum _n \chi _1 (\frac{w - a_{n}^+ }{ |g_n|^{ \frac{1}{4}}} ) \chi _1 (\frac{ a_{n+1}^--w }{ |g _{n+1}|^{ \frac{3}{5}}} )$. Then $$K _3 (t,x,y)=\int _{\Bbb R } e^{i(t E (k) -(x-y)k) } \chi _{int} (k) {m ^0_-(x,k)}{ m ^0_+(y,k)} dk .$$ We will use the following lemma: \proclaim{Lemma 4.7} In the support of $\chi _{int} $ we have $w\approx k$, $\dot w= 1+O(k^{-2})$, $\ddot w= O(k^{-3})$. Furthermore, we can extend $w$ from the support of $\chi _{int} $ to the whole of $\Bbb R$ so that the extension (which we denote again with $w$) satisfies the same relations \endproclaim $w\approx k$ by Lemma 5.2, $\dot w= 1+O(k^{-2})$ by Lemma 7.1, $\ddot w= O(k^{-3})$ by Lemma 7.4 and the last statement is Lemma 7.6. Now assume Lemma 4.7. We have $E=w^2$, $\dot E =2w\dot w$ and $\ddot E=2\dot w^2-2w\ddot w$. So by Lemma 4.7 we can extend $E(k)$ from the support of $\chi _{int}$ into a function defined on all $\Bbb R$ convex with $\dot E\approx k$, $\ddot E \approx 1$. We express $$ m _- ^0(x,k) m _+ ^0(y,k) = \left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) +1.\tag 1$$ Let $k_0$ be the unique solution of $\dot \Phi (k) =0$. Then set $q^2/2=\Phi (k) -\Phi (k_0)= \frac{1}{2} \ddot \Phi (\tilde k ) (k-k_0)^2$. Since $1/C \le \ddot \Phi \le C$ then $ \frac{1}{\sqrt{C}} \le \frac{q}{k-k_0} \le \sqrt{C} $. From $q\dot q=\dot \Phi (k)= \ddot \Phi (k_1) (k-k_0)$ we conclude that for some $C$ we have $1/C\le \dot q \le C$. Now we insert (1) in the definition of $K _3$ obtaining $ K _3 =H_1+H_2 $ with $$\aligned & H_1 (t,x,y)=e^{i (tE(k_0)-(x-y) k_0) } \int _{\Bbb R} e^{itq^2 } \chi _{int}(k)\left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) \frac{dk}{dq} dq\\& H_2 (t,x,y)= e^{i (tE(k_0)-(x-y) k_0) } \int _{\Bbb R} e^{itq^2 } \chi _{int}(k)\frac{dk}{dq} dq. \endaligned $$ By $\| \widehat{\rho} \| _1\le \tilde C_\varepsilon \| \rho \| _{H^{ \frac{1}{2} + \varepsilon}}$, in what follows we can use $$ \big | \int _{\Bbb R} e^{itq^2 }\rho (q) dq\big | \le C_\varepsilon t ^{-\frac{1}{2}} \| \rho \| _{H^{ \frac{1}{2} + \varepsilon}}. \tag 4.1$$ \proclaim{Lemma 4.8 } For a fixed $C$ we have $\big | H_2 (t,x,y) \big | \le C t ^{-\frac{1}{2}}$. \endproclaim { \it Proof . } We write $\chi _{int}(k)= 1-( 1- \chi _{int}(k)).$ Let $\zeta (k)$ be either 1 or $-( 1- \chi _{int}(k)).$ For $\chi (t)$ a cutoff supported near $t=0$, we insert the partition of unity $\chi (k-k_0)+(1-\chi (k-k_0))$inside $ \int e^{i (tE(k )-(x-y) k ) }\zeta (k) dk$. Then by $\| \zeta (k)\| _{ \infty } + \| \zeta ^\prime (k)\| _{1 }\lesssim 1$ and by Lemmas 3.5 and 4.2, for a fixed $C$ $$\big | \int e^{i (tE(k )-(x-y) k ) } \zeta (k) \chi (k-k_0) dk\big | \le C t ^{-\frac{1}{2}}.$$ Next we want to bound $$\big | \int e^{i t\Phi (k) }\zeta (k) (1- \chi (k-k_0)) dk\big | =\big | \int e^{i \frac{t}{2}q^2 } \zeta (k) (1- \chi (k-k_0)) \frac{dk}{dq} dq\big | .\tag 2 $$ Let first $\zeta (k)\equiv 1.$ Then $ \int e^{i \frac{t}{2}q^2 } \zeta (k) (1- \chi (k-k_0)) \frac{dk}{dq} dq =$ $$= \int e^{i \frac{t}{2}q^2 } (1- \chi (k-k_0)) dq -\int e^{i \frac{t}{2}q^2 } (1- \chi (k-k_0)) \left (1- \frac{dk}{dq} \right ) dq.$$ By standard arguments the first term in $O(t^{-\frac{1}{2}})$. By (4.1), with $\varepsilon =1/2$, the second term is also $O(t^{-\frac{1}{2}})$ thanks to the following lemma: \proclaim{Lemma 4.9 } $\forall \, a>0$ $\exists$ a fixed constant $C_a$ such that $\| 1- \frac{dk}{dq}\| _{H^1 (\{ |q|\ge a \} }\le C_a$. \endproclaim {\it Proof.} We set $\dot q=\frac{dq}{dk}$. We have $2q\dot q=\dot E(k)- \dot E (k_0) =2(k-k_0) + O(\langle k \rangle ^{-1}) + O(\langle k _0\rangle ^{-1}) $ by Lemma 4.1. Hence after integration $$ \frac{q^2}{(k-k_0)^2}= 1+ \frac{O(\langle k _0\rangle ^{-1}) }{k-k_0 } +\frac{O( \log ( \langle k/k_0 \rangle ))}{ (k-k_0)^2} . \tag 3$$ We know from $q\approx k-k_0$ that (3) is uniformly bounded for $|k-k_0|\le 1$. Hence we conclude $$ 1- \frac{q }{ k-k_0 }=\left ( 1+ \frac{q }{ k-k_0 } \right ) ^{-1} \left ( 1- \frac{q ^2}{ (k-k_0)^2} \right ) $$ is in $L^2(\Bbb R)$ with norm independent from $k_0$. By $ q\dot q=k-k_0+ O(\langle k \rangle ^{-1})+O(\langle k _0\rangle ^{-1}) $, $$\dot q =1 +\left ( 1- \frac{q }{ k-k_0 } \right ) \dot q + \frac{O(\langle k \rangle ^{-1}) + O(\langle k _0\rangle ^{-1})}{k-k_0}. $$ Since we know $\dot q $ is uniformly bounded, we conclude that for a fixed constant $C$ we have $\| 1- \dot q \| _{2}\le C$. Next, since in $\ddot q= \frac{\ddot E -\dot q^{2}}{q} $ the numerator is bounded, we see that $\ddot q \in L^2(\{ |q|\ge 1 \} ) .$ By $\frac{d^2k}{dq^2}= - (\dot q) ^{-3} \ddot q$ we obtain the desired result. \bigskip To complete the proof of Lemma 4.8, we have to show that (2) is $O(t^{-\frac{1}{2}})$ when $ \zeta (k)=-( 1- \chi _{int}(k)).$ We have $\| \zeta \| _{1}\lesssim 1$, so for $t\lesssim 1$ we have $(2)=O(1)$. We suppose now $t\gg 1$. In (2) split $\int _\Bbb R= \int _{|k|\le \sqrt{t}}+\int _{|k|\ge \sqrt{t}}$. By Lemma 2.1, for any $N$ $\int _{|k|\ge \sqrt{t}}=O( \sum _{|n|\gtrsim \sqrt{t}} |g_n|^{\frac{1}{4}}) \ll t^{-N}.$ We have $$\int _{|k|\le \sqrt{t}}e^{i t\Phi (k) }\zeta (k) (1- \chi (k-k_0)) dk=O( t^{-1} \sum _{|n|\lesssim \sqrt{t}} (1+1))\approx t^{-\frac{1}{2}},$$ where the terms 1 are of the form $ |g_n|^{-\frac{1}{4}} \int _{a^{+} _{n}} ^ {a^{+} _{n}+ |g_n|^{ \frac{1}{4}} } dk $ with $ |g_n|^{-\frac{1}{4}}\gtrsim |\zeta ^\prime | $ in $[a^{+} _{n}, a^{+} _{n}+ |g_n|^{ \frac{1}{4}}]$ and $ |g_{n+1}|^{-\frac{3}{5}} \int ^{a^{-} _{n+1}} _ {a^{-} _{n+1}- |g_{n+1}|^{ \frac{3}{5}} } dk $ with $ |g_{n+1}|^{-\frac{3}{5}}\gtrsim |\zeta ^\prime | $ in $[a^{-} _{n+1}-|g_{n+1}|^{ \frac{3}{5}}, a^{-} _{n+1}]$. \bigskip \proclaim{Lemma 4.10 } There is a fixed $C$ such that $| H_1 (t,x,y)| \le C \max \{ t^{-\frac{1}{2}}, t^{-\frac{1}{3}} \} .$ \endproclaim \noindent {\it Proof}. We split $ H_1 (t,x,y)= H _{11} (t,x,y)+H _{12} (t,x,y)$ with $H _{11} (t,x,y) $ defined inserting the additional factor $\widetilde{\chi } _{int}(k)=\sum _n\chi _1 \left (\frac{k - \pi n }{ {n}^{ \varepsilon -\frac{1}{2}} } \right ) \chi _1 \left (\frac{(n+1) \pi -k }{{(n+1)}^{ \varepsilon -\frac{1}{2}}} \right )$ in the definition $H _{1 } (t,x,y) $, and with $H _{12} (t,x,y) $ defined inserting $\widetilde{\chi } _{edge}=1- \widetilde{\chi } _{int}$. The integrals defining $H _{11} (t,x,y) $ are supported in $\pi n +c{n}^{ \varepsilon -\frac{1}{2}}\le k \le (n+1) \pi -c(n+1)^{ \varepsilon -\frac{1}{2}}$, that is in the interior of the bands, while the integrals defining $H _{12} (t,x,y) $ are supported near the edges. \proclaim{Claim} There is a fixed $C$ such that $|H _{11} (t,x,y) |\le C t^{-\frac{1}{2}}$. \endproclaim {\it Proof.} We have $ {\chi }_{int} \widetilde{\chi }_{int}=\widetilde{\chi }_{int}$ since $\widetilde{\chi }_{int}$ is the characteristic function of the union of $\pi n +c{n}^{ \varepsilon -\frac{1}{2}}\le k \le (n+1) \pi -c(n+1)^{ \varepsilon -\frac{1}{2}}$ smoothed and $ {\chi }_{int}$ is the characteristic function of the union of $\pi n + |g_n|^{\frac{1}{4}} \le k \le (n+1) \pi - |g _{n+1}|^{\frac{3}{5}}$ smoothed. We split $H _{11} (t,x,y) $ in two pieces. For the first piece we have for $\chi (t)$ a cutoff supported near 0, by Lemma 3.5, by $\| \widetilde{\chi }_{int} \|_\infty + \| \widetilde{\chi }_{int} ^\prime \chi (k-k_0)\|_1\lesssim 1$, $ m _- ^0(x,k) m _+ ^0(y,k)-1= O( k^{-1})$ and $\widetilde{\chi } _{int} (k) \partial _k \left ( m _- ^0(x,k) m _+ ^0(y,k)\right ) = O( k^{-\frac{1}{2} -\varepsilon}) $, $$\big | \int e^{i t\Phi (k) }\widetilde{\chi } _{int} (k) \chi (k-k_0) \left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) dk\big | \le C t^{-\frac{1}{2}} .$$ Next we consider $$ \int e^{i tq^2/2 } \widetilde{\chi } _{int} (k) (1-\chi (k-k_0)) \left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) \frac{dk}{dq}dq.\tag 1$$ By Lemmas 4.5 and 4.9 $$\big \| \widetilde{\chi } _{int} (k)(1-\chi (k-k_0))\left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) \frac{dk}{dq} \big \| _{H^{1}} \le C \| \{ {\langle n\rangle ^{ -\frac{1}{2}-\varepsilon}} \} \| _{ l^2(\Bbb N )} . $$ We can apply (4.1) and bound (1) by $C t^{-\frac{1}{2}}$. \bigskip \noindent We consider $H _{12} (t,x,y)=\sum _n H _{12} ^{n}(t,x,y)$, $ H _{12}^{n} (t,x,y) =\int _{n\pi}^{(n+1) \pi } e^{it \Phi ( k)} f(k) dk$ $$\aligned &\text{with } f(k)=\Psi _n(k) \left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) \text{ where}\\& \Psi _n(k)= \chi _1 (\frac{w - a_{n}^+ }{ |g_n|^{ \frac{1}{4}}} ) \chi _1 (\frac{ a_{n+1}^--w }{ |g _{n+1}|^{ \frac{3}{5}}} )\chi _0 \left (\frac{k - \pi n }{ {n}^{ \varepsilon -\frac{1}{2}} } \right ) \chi _0 \left (\frac{(n+1) \pi -k }{{(n+1)}^{ \varepsilon -\frac{1}{2}}} \right )\endaligned $$ and with $ \Phi ( k)= E(k)-E(k_0) - t^{-1} (x-y) (k-k_0)$. Observe that $ \Psi _n(k)= \Psi _{n 1}(k) +\Psi _{n 2}(k)$ with $\Psi _{n 1}(k)$ supported in $ |g_n|^{ \frac{1}{4}}\lesssim k-\pi n\lesssim n^{ \varepsilon -\frac{1}{2}}$ and with $\Psi _{n 2}(k)$ supported in $ |g _{n+1}|^{ \frac{3}{5 }} \gtrsim \pi (n+1)-k\gtrsim (n+1)^{ \varepsilon -\frac{1}{2}}.$ Correspondingly write $f=f_1+f_2$ and $H _{12}^{n}= H _{12}^{n1}+ H _{12}^{n2} .$ \proclaim{Lemma 4.11} For a fixed $C$ and for $j=1,2$: $|H _{12 }^{nj}(t,x,y)|\le C \langle t \rangle ^{-\frac{1}{2}} |\log t |^2 .$ \endproclaim {\it Proof.} We focus on $H _{12}^{n1}$, the proof for $H _{12}^{n2}$ being almost the same. We have $$ H _{12}^{n1} (t,x,y) =\int _{n\pi}^{(n+\frac{1}{2}) \pi } F^\prime (k) f_1(k)dk \text{ with } F(k)= \int _{n\pi}^{k}e^{it \Phi ( k^\prime )} dk^\prime .$$ For $H _{12}^{n2}$ the proof is the same but with $F(k)= \int _{(n+1) \pi}^{k}e^{it \Phi ( k^\prime )} dk^\prime $. We get $$\aligned & H _{12}^{n1} (t,x,y) =-H _{121}^{n1} (t,x,y)-H _{122}^{n1} (t,x,y) \text{ with}\\& H _{121}^{n1} (t,x,y)=\int _{n\pi}^{(n+\frac{1}{2}) \pi } \left ( \int _{n\pi}^{k}e^{it \Phi ( k^\prime )} dk^\prime \right ) \Psi _{n1} ^\prime (k) \left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) dk\\& H _{122}^{n1} (t,x,y)=\int _{n\pi}^{(n+\frac{1}{2}) \pi } \left ( \int _{n\pi}^{k}e^{it \Phi ( k^\prime )} dk^\prime \right ) \Psi _{n1} (k) \partial _k\left ( m _- ^0(x,k) m _+ ^0(y,k)-1 \right ) dk .\endaligned $$ \proclaim{Claim} For $ | k_0-n\pi |\ge 2 \pi $ for a fixed $C$ we have $$|H _{121}^{n1} (t,x,y) |\le C \langle t(\pi n-k_0) \rangle ^{ -1} \langle n \rangle ^{ -1}.$$ \endproclaim {\it Proof.} Indeed we have $|F(k)|\le C \langle t(\pi n-k_0) \rangle ^{ -1} $, $|m _- ^0(x,k) m _+ ^0(y,k)-1| \le C \langle k \rangle ^{ -1}$ and $\| \Psi _{n1} ^\prime (k) \| _{L^1(\pi n , \pi (n+1))} \le C$. In a similar fashion we obtain \proclaim{Claim} For $ | k_0-n\pi |\le 2 \pi $ for a fixed $C$ we have $$|H _{121}^{n1} (t,x,y) |\le C \langle t \rangle ^{ -\frac{1}{2}} \langle n \rangle ^{ -1}.$$ \endproclaim Next we use that there is a fixed $C$ such that for any $x_0$ and any $t>0$, $$\int _{|x-x_0|\ge 1} \frac{dx}{\langle x \rangle \langle t(x-x_0) \rangle } \le C \min \left \{ t^{-1} , |\log t|\right \} $$ to conclude that for a fixed $C$ $$\sum _n |H _{121}^{n1} (t,x,y) |\le C \min \left \{ \langle t \rangle ^{ -\frac{1}{2}} , |\log t|\right \} .$$ \bigskip We now consider $H _{122}^{n1} (t,x,y)$. We start by assuming $ | k_0-n\pi |\ge 2 \pi $. Then notice that for a fixed $C$ $$\big | \int _{n\pi}^{k}e^{it \Phi ( k^\prime )} dk^\prime \big | \le \min \{ C \langle t(\pi n-k_0) \rangle ^{ -1} , |k- \pi n| \} .$$ Next we split $$H _{122}^{n} (t,x,y)=\int _{n\pi}^{n \pi +\langle t(\pi n-k_0) \rangle ^{ -1} } \dots +\int ^{(n+\frac{1}{2})\pi}_{n \pi +\langle t(\pi n-k_0) \rangle ^{ -1} } \dots $$ But now $$ \big |\int _{n\pi}^{n \pi +\langle t(\pi n-k_0) \rangle ^{ -1} } \dots \big | \le C \int _{n\pi}^{n \pi +\langle t(\pi n-k_0)\rangle ^{ -1} } |k- \pi n| |k- \pi n| ^{-1} \langle n \rangle ^{ -1} $$ and $$ \big |\int ^{(n+\frac{1}{2})\pi}_{n \pi +\langle t(\pi n-k_0) \rangle ^{ -1} } \dots \big | \le \int ^{(n+\frac{1}{2})\pi}_{n \pi +\langle t(\pi n-k_0) \rangle ^{ -1} } C \langle t(\pi n-k_0)\rangle ^{ -1}|k- \pi n| ^{-1} \langle n \rangle ^{ -1} $$ and so $$|H _{122}^{n} (t,x,y)| \le C\langle n \rangle ^{ -1}\langle t(\pi n-k_0)\rangle ^{ -1} \log \left ( \langle t(\pi n-k_0)\rangle \right ) .$$ Similarly for $ | k_0-n\pi |\le 2 \pi $ we get $$|H _{122}^{n} (t,x,y)| \le C\langle t \rangle ^{ -\frac{1}{2}} \log \left ( \langle t \rangle \right ) .$$ We use that there is a fixed $C$ such that for any $x_0$ and any $t>0$, $$\int _{|x-x_0|\ge 1} \frac{\log \left ( \langle t(x-x_0) \rangle \right ) dx}{\langle x \rangle \langle t(x-x_0) \rangle } \le C \min \left \{ t^{-1} |\log t| , |\log t|^2 \right \}$$ to conclude that for a fixed $C$ $$\sum _n |H _{122}^{n1} (t,x,y) |\le C \langle t \rangle ^{ -\frac{1}{2}} |\log t| ^2.$$ \bigskip \head \S 5 Asymptotic expansion for $ w-k$ \endhead \noindent We consider $w=u+iv$ and $k=p+iq$. We set $Q_\ell =\frac 1\pi \int _\Bbb R u^\ell q(u) \,du .$ Then $Q_{2\ell +1}=0$. In particular, see \cite{KK} p. 601, we have $Q_0=\frac 12 \int _0^1 P(t) \, dt $ and $ Q_2=\frac 1 8 \int _0^1 P ^2(t) \, dt.$ For $u\in \sigma $ we have $q(u)=0.$ For $u\in g_n$ we have formula (4.12) \cite{K1}: $$q(u)=\sqrt{ (u-a_n^-) ( a_n^+-u )} \left ( 1+\frac 1\pi \sum _{m\neq n}\int _{g_m} \frac{q(t) \, dt }{|t-u| \sqrt{ (t-a_n^-) ( t-a_n^+ )}} \right ) .\tag 5.1$$ By Lemma 2.1 there is a fixed constant $C>0$ such that for $|\sigma _m|$ the length of $ \sigma _m $ we have $ |\sigma _m| \ge C $ for all $ m.$ For $u\in g_n$ and $ C_0= 1+ \frac{Q_0}{\min \{ |\sigma _m| \colon m\in \Bbb Z \} }$ by \cite{K1} p.16 $$\aligned & \sqrt{ (u-a_n^-) ( a_n^+-u )} \le q(u)\le C_0\sqrt{ (u-a_n^-) ( a_n^+-u )}. \endaligned \tag 5.2$$ \proclaim{Lemma 5.1} 1) $\forall$ $N$ there is a $C_N>0$ such that $|q(u)| \le C_N \langle u \rangle ^{-N}$ $\forall$ $u\in \Bbb R$. \noindent 2) The distributional derivative $q^\prime (u)$ satisfies $\langle u \rangle ^Nq^\prime \in L^{ r}(\Bbb R)$ for any $1\le r<2 $ and any $N$. \endproclaim {\it{Proof.}} We have $ 0\le q(u)\le C_0|g_n| \le C_N \langle u \rangle ^{-N}$ for $u\in g_n$ by (5.2) and Lemma 2.1. Turning to the second claim, by (5.1) the pointwise derivative $q^\prime (u)$ is well defined except at the points $a_n^\pm $ for $n\in \Bbb Z$. Obviously $q^\prime (u)=0$ for any $a_n^+0$, we conclude there is a fixed $C$ such that $$|q^\prime (u)|< C\left ( \frac{a_{ n } ^{+}-u}{u-a_{ n } ^{-}}\right ) ^{\frac{1 }{2}} + C \left ( \frac {u-a_{ n } ^{-}} {a_{ n } ^{+}-u}\right ) ^{\frac{1}{2}} . $$ From this we conclude that the pointwise $q^\prime (u)$ coincides with the distributional derivative and that $\int _{ g_n}\langle u \rangle ^N|q^\prime (u)|^rdu\le \frac{C\langle n \rangle ^N}{2-r} |g_n|$ for some fixed $C$. By Lemma 2.1 we conclude $\| \langle u \rangle ^Nq^\prime \| _r \le \frac{C_N}{2-r}$ for some $C_N$. \bigskip From the second claim in Lemma 5.1 we obtain: \proclaim{Lemma 5.2} For any integer $N$ there is a constant $C_N>0$ such that, for any $w=u+iv$ with $ v\ge 0$ and $|u|>1$, we have $$w-k(w) = \sum _{\ell =0} ^{N} \frac{ Q_\ell }{w^{\ell +1}}+ R_N(w)\, , \quad \big | R_N(w) \big | \le \frac{C _N}{\langle w \rangle ^{N+1} }.$$ \endproclaim {\it{Proof}}. We have for $v>0$ by (4.1) \cite{K1} $k(w)-w= \frac 1\pi \int _{\cup g_n} \frac{ q(t) }{t-w } dt$ and so $$ w-k(w)= \frac 1\pi \sum _{\ell =0} ^N \frac{1}{ w ^{\ell+1}} \int _{\cup g_n} t^\ell q(t) dt + \frac 1{ w^{N+1}} \left ( Q_v - i P_v \right ) \ast (t^{N+1 } q(t) ) (u) $$ with in the last term the convolution of $t^{N+1 } q(t)$ with the Poisson kernels $$P_v(x)= \frac{1}{\pi}\frac{v}{x^2+v^2} \, , \quad Q_v(x)=\frac{1}{\pi} \frac{x}{x^2+v^2}.$$ Since for $1 0$, see p. 121 \cite{Ste}, $$\| \left ( Q_v - i P_v \right ) \ast (t^{N+1 } q(t) ) \| _{W^{1, r}} \le C_r\| t^{N+1 } q(t) \| _{W^{1, r}}\, , $$ by Lemma 5.1 and by the Sobolev embedding theorem there is a fixed constant $C_N>0$ such that $|\left ( Q_v - i P_v \right ) \ast (t^{N+1 } q(t) ) (u)|\le C_N. $ \bigskip As an immediate corollary of Lemma 5.2 we obtain: \proclaim{Lemma 5.3} There are two constants $C_1$ and $C_2$ such that, for $w=u+iv$ with $1\ge v\ge 0$ and $|u|>C_1$, we have $$\big | u-p- \frac{ Q_0 }{u}+ \frac{ Q_0 v^2}{u^3}-\frac{ Q_2 }{u^3}\big | \le \frac{C _2}{|u| ^{4} }.$$ \endproclaim \noindent By $w-k= Q_0 \frac{ \overline{w}}{|w|^2} + Q_2\frac{ \overline{w}^3}{|w|^6}+ O(w^{-4})$, we get $u-p=Q_0 \frac{ u}{u^2+v^2} + Q_2\frac{ u^3}{(u^2+v^2)^3}+ O(u^{-4})$. Then by Taylor series $\frac{ 1}{u^2+v^2} = u^{-2}- v^2 u^{-4}+\dots ,$ we get the desired result. \bigskip \head \S 6 Relation between $v$ and $q(u+iv)$ \endhead In the proof of Lemma 4.4 we will need to use the relative size of the coordinates of $w=u+iv$ and $k(w)=p(w)+iq(w).$ Lemma 5.3 gives some information on $u-p(u+iv)$. We now consider the relation between $v$ and $q(u+iv)$. Recall that $$q(u+iv)=v+\frac{v}{\pi}\int _\Bbb R\frac{q(t)\, dt}{(t-u)^2+v^2}. \tag 6.1 $$ The unnatural restriction on $u$ in the following lemma is not sharp and is only justified by estimates needed later. \proclaim{Lemma 6.1} Let $a^{+}_{n}+ |g_n|^{\frac{1}{4}}\le u \le a^{-}_{n+1}- |g _{n+1}|^{\frac{3}{5}}$ for $n \gg 1$ large, and let $1\ge v \ge 0$ Then there is a constant $C $ such that $$|q(u+iv)-v-\frac{v}{u^2}Q_0|\le C v u^{-4}.$$ \endproclaim {\it Proof.} By (6.1) write $$q=v+\frac{v }{\pi u^2} \int _\Bbb R q(t) \left ( 1+\frac{2t}{u}- \frac{t^2}{u^2}-\frac{v^2}{u^2} \right ) dt +\frac{v }{\pi } \int _\Bbb R \frac{q(t) \left ( \frac{2t}{u}- \frac{t^2}{u^2}-\frac{v^2}{u^2} \right )^2}{(t-u)^2+v^2} dt $$ and use the formulas for $Q_\ell $ above (5.1) to express the second term on the rhs as $\frac{v }{ u^2} ( (1-\frac{v^2}{u^2})Q_0-\frac{Q_2 }{u^2}).$ In the second integral we expand the square on the numerator, treating the resulting terms separately. For example we write $$\aligned \frac{4v }{\pi u^2} \int _\Bbb R \frac{t^2q(t) }{(t-u)^2+v^2} dt& = \frac{4v }{\pi u^4} \int _\Bbb R t^2q(t) \left ( 1+\frac{2t}{u}- \frac{t^2}{u^2}-\frac{v^2}{u^2} \right ) dt+\\& + \frac{4v }{\pi u^4} \int _\Bbb R \frac{t^2q(t) \left ( 2t - \frac{t^2}{u }-\frac{v^2}{u } \right )^2 }{(t-u)^2+v^2} dt.\endaligned $$ The first term on the rhs is $O(vu^{-4})$. To show that the second term is $O(vu^{-4})$, we need to show bounds of the form $ \int _\Bbb R \frac{|t|^N q(t) }{(t-u)^2+v^2} dt \le C_N.$ Say that $u\le \frac{a^{+}_{n}+a^{-}_{n+1}}{2}$, with the case $u\ge \frac{a^{+}_{n}+a^{-}_{n+1}}{2}$ treated similarly. Then $$ \int _\Bbb R \frac{|t|^N q(t) }{(t-u)^2+v^2} dt= \left ( \sum _{\ell \neq n } \int _{g_\ell }\frac{|t|^N q(t) }{(t-u)^2+v^2} dt \right ) + \int _{a^{-}_{n} } ^{a^{+}_{n} }\frac{|t|^N q(t) }{(t-u)^2+v^2} dt $$ where the first term in the rhs is bounded by a $C_N\langle u \rangle ^{-2}$ thanks to $|t-u|\approx |\ell -n |$, Lemma 5.1 and $ \langle u \rangle ^{-2}\ast \langle u \rangle ^{-N}\lesssim \langle u \rangle ^{-2}.$ Next we write $$\int _{a^{-}_{n} } ^{a^{+}_{n} }\frac{|t|^N q(t) }{(t-u)^2+v^2} dt \le \frac{C|n|^N |g_n|^2}{(a^{+}_{n}-u)^2}\le C_N $$ by $q(t)\le (1+C n^{-2}) |g_n|$, see \cite{KK}, by $a^{+}_{n} -a^{-}_{n}=|g_n|$, by our restriction on $u$ and by Lemma 2.1. \bigskip We have the following corollary: \proclaim{Lemma 6.2} In $\{ 1\ge v \ge 0\} $ for any preassigned $C_1>0$ in $a^{+}_{n}+ |g_n|^{\frac{1}{4}}\le u \le a^{-}_{n+1}- |g _{n+1}|^{\frac{3}{5}}$ there is a $C$ such that $$ \big | \Im (E-k^2) \big | \le C \frac{ v}{u^3} .$$ \endproclaim \noindent {\it Proof.} Write $\Im (E-k^2) =(v-q)(u+p)+(u-p)(v+q) $ with $0\le v \le q$. By Lemma 5.3 we have $u-p=\frac{Q_0}{u}-\frac{Q_0v^{2}}{u^3}+\frac{Q_2}{u^3} + O(u^{-4} ) $ and $u+p=2u- \frac{Q_0}{u}+\frac{Q_0v^{2}}{u^3} -\frac{Q_2}{u^3} + O(u^{-4} ). $ By Lemma 6.1 $v-q=-vu^{-2}Q_0+O( vu^{-4})$ and $q+v=2v+vu^{-2}Q_0+O( vu^{-4})$. The $O(vu^{-1 })$ term in $\Im (E-k^2)$ cancels and the following one is $O(vu^{-3 })$. \bigskip \noindent Now we extend Lemma 6.2 without the restriction on $u$. From $q(t)\ge 0$ and $q(t)\not \equiv 0$ we get $ q(u+iv)> v$ by (6.1). The following is an elementary consequence of Lemmas 2.1 and 5.1: \proclaim{Lemma 6.3} For any $u\in \Bbb R$ there is at most one $g_n$ such that $\text{dist}(u,g_n)\ll 1$. For such an $n$ we have $$ 0\le \sum _{\ell \neq n}\frac{v}{\pi}\int _{g_\ell }\frac{q(t)\, dt}{(t-u)^2+v^2} \frac{|g_m|}{4}$ or if $v\ge |g_m|$ we have $$\frac 1C \frac{|g_m|^2 v}{(u-a _{m})^2+v^2}< I_n(u,v )< C \frac{|g_m|^2 v}{(u-a _{m})^2+v^2} ; $$ \medskip \item if $u\in g_n=[a^{-} _{n}, a^{+} _{n}]$ for $v\ge \frac{|g_n|}{2}$ we have $$ \frac 1C \frac{|g_n|^2 }{v} v$ we get $I_n(u,v)\approx \sqrt{|g _{n}|\tilde u}$ and hence claim 3 in Lemma 6.4. \bigskip \head \S 7 Estimates on the band function \endhead We will need to bound $\frac{d E}{dk }$, $\frac{d^2E}{dk^2}$ and $\frac{d^3E}{dk^3}$. \proclaim{Lemma 7.1} There are constants $C_1>C_2>0$ such $\forall m$ and $\forall u\in \sigma _m=[a^+_m,a^-_{m+1}]$ and if $v=0$, we have for $A(u)= \frac{ |g_m|^2 }{( u-a_m^+ )^{\frac{1}{2}} ( u-a_m^++|g_m|)^{\frac{3}{2}} } + \frac{ |g_{m+1}|^2 }{( a_{m+1}^- -u )^{\frac{1}{2}} ( a_{m+1}^--u+|g _{m+1}|)^{\frac{3}{2}} }$ $$ \aligned & 1+ C_2\left ( A(u) + \frac{1}{\langle u \rangle ^2} \right ) \ge p^\prime (u) \ge 1+C_1 A(u) .\endaligned \tag 1$$ Correspondingly for $p\in [m\pi , (m+1) \pi ]$ we have $$ \align &\frac 1{1+C_2 \left ( A(u) + \frac{1}{\langle u \rangle ^2 } \right )}\le \frac{du}{dp}\le \frac 1{1+C_1 A(u)} \tag 2\\& \frac {2\big |u \big | }{1+C_2 \left ( A(u) + \frac{1}{\langle u \rangle ^2} \right )} \le \big | \frac{dE}{dp}\big | \le \frac{2\big |u \big | }{1+C_1 A(u)} .\tag 3\endalign$$ If $a^{+}_{n}+ |n| ^{3} |g_n| \le u \le a^{-}_{n+1}- |{n}+1|^{3} |g _{n+1}|$ with $n \gg 1$ large and $v=0$, then there is a fixed $C$ such that $$|\dot w-1|+|\dot E (k)-2k | \le C \langle k \rangle ^{-2}. \tag 4$$ There is $n_0$ such that if for $n\ge n_0$, $ |a_{n}^{+}-u| +|v|\ge |k| |g_n| $ with $a^{+}_{n} \le u \le \frac{a^{+}_{n}+a^{-}_{n+1}}{2} $ and $1\ge v\ge 0 $ then there is a $C$ such that for the corresponding $k=p+iq$ we have $$ \big | \frac{\dot E}{2k}-1 \big | \le C |k|^{-1} .\tag 5$$ (5) holds also in $ |a_{n+1}^{-}-u| +|v|\ge |k| |g_{n+1}| $ and in particular for $a^{+}_{n}+ |g_n|^{\frac{1}{4}}\le u \le a^{-}_{n+1} -|g_{n+1}|^{\frac{3}{5}} $ and $1\ge v\ge 0 $. \endproclaim Estimate (5) is used in the proof of Lemma 4.4. {\it Proof of Lemma 7.1. } By \cite{K1}, $ p^\prime (u)=1+ \frac 1\pi \sum _n \int _{ g_n} \frac{ q(t) }{(t-u)^2} dt\ge 1+I(u) $ with $$I(u)= \frac 12 \sum _n \frac{|g_n|^2 }{\sqrt{ |(u-a_n^-) ( a_n^+-u )| } \left ( \sqrt{ | u-a_n^- |} + \sqrt{ | a_n^+-u | } \right ) ^2} .$$ For $u\in \sigma _m$, ignoring all the terms in the sum defining $I(u)$ except for $n=m,m+1$, we get the lower bound for $p^\prime (u)$ in (1) by $ \sum _{n = m, m+1 }\cdots \approx A(u)$. Turning to the upper bound, by (5.2) we have $p^\prime (u)\le 1 + C_0I(u)$. We split now $2I(u)=$ $$ \left ( \sum _{n \neq m, m+1 } + \sum _{n = m, m+1 } \right ) \frac{|g_n|^2 }{\sqrt{ |(u-a_n^-) ( a_n^+-u )| } \left ( \sqrt{ | u-a_n^- |} + \sqrt{ | a_n^+-u | } \right )^2} .$$ Since there is a fixed $c>0$ such that for $n \neq m, m+1$ and for $u\in \sigma _m$ we have $| a_n^\pm -u |\ge c \langle n-m \rangle $ then $$\sum _{n \neq m, m+1 }\cdots \le C \sum _{n } \frac{|g_n|^2 }{\langle n-m \rangle ^2} \le C_2 \frac{1}{\langle u \rangle ^2} .$$ This gives the upper bound for $p^\prime (u)$ in (1). (2) is obtained taking the inverses in (1) and (3) follows from $\frac{d E}{dk }= 2w\frac{d w}{dk }$. To prove (4) we claim \proclaim{Claim} For $a^{+}_{n}+C_1 |n| ^{3} |g_n| \le u \le a^{-}_{n+1}-C_1 |{n}+1|^{3} |g _{n+1}|$ for any fixed $C_1$, $n \gg 1$ large and $v=0$, we have $ p^\prime (u)=1+ \frac{Q_0}{u^2} +O(u^{-3})$. \endproclaim \noindent Assume for a moment the Claim. Then $\dot w =1- \frac{Q_0}{u^2}+O(u^{-3})$ and by Lemma 5.3 $$\dot E= 2u \dot w= 2 u-2\frac{Q_0}{u}+O(u^{-2})= 2p +2\frac{Q_0}{u}- 2\frac{Q_0}{u} +O(u^{-2}).$$ \noindent{\it Proof of the Claim.} Is suggested by formal differentiation of $p(u)=u -\frac{Q_0}{u}-\frac{Q_2}{u^3}+\dots $, but for a proof we return to $ p^\prime (u)=1+ \frac 1\pi \sum _n \int _{ g_n} \frac{ q(t) }{(t-u)^2} dt$. By the argument in the proof of Lemma 6.1, simply setting $v=0$ in the appropriate integral, $$p^\prime (u)=1+ \frac{Q_0}{u^2}- \frac{Q_2}{u^4}+\frac{1}{\pi u^2}\int _\Bbb R \frac{q(t) \left ( 2t - \frac{t^2}{u} \right )^2}{(t-u)^2 } dt . $$ Then $$ \int _\Bbb R \frac{q(t) \left ( 2t - \frac{t^2}{u} \right )^2}{(t-u)^2}= \left [ \left ( \sum _{\ell \neq n } \int _{g_\ell } \right ) + \int _{a^{-}_{n} } ^{a^{+}_{n} } \right ] \frac{q(t) \left ( 2t - \frac{t^2}{u} \right )^2}{(t-u)^2} dt $$ where the first term in the rhs is bounded by a $C \langle u \rangle ^{-2}$ thanks to $|t-u|\approx |\ell -n |$ and where $$\int _{a^{-}_{n} } ^{a^{+}_{n} }\frac{q(t) \left ( 2t - \frac{t^2}{u} \right )^2}{(t-u)^2} dt \lesssim \int _{a^{-}_{n} } ^{a^{+}_{n} }\frac{q(t) t ^2}{(t-u)^2} dt \le \frac{C |n|^2|g_n| ^2}{(a^{+}_{n}-u)^2} \lesssim n^{-4} $$ by $t\approx u\approx n$ i.e. Lemma 2.1, $q(t)\le (1+C n^{-2}) |g_n|$, see \cite{KK}, $a^{+}_{n} -a^{-}_{n}=|g_n|$, see definition of $|g_n|,$ and by our restriction on $u$, that is $u-a^{+}_{n}\gtrsim n^3|g_n| $. To prove (5) we write $$ k^\prime (w)=1+ \frac 1\pi \sum _{\ell \neq n} \int _{ g_\ell } \frac{ q(t) }{(t-w)^2} dt+ \frac 1\pi \int _{a_n^-}^ {a_n^+} \frac{ q(t) }{(t-w)^2} dt . $$ The second term in the rhs is $O(n ^{-2})$. For $|a_{n}^{+}-u|+|a_{n}^{-}-u| +|v|\ge |k| |g_n|, $ so in particular for $a^{+}_{n}+c {n} ^{\frac{1}{3}} |g_n|^{\frac{2}{3}}\le u \le \frac{a^{+}_{n}+a^{-}_{n+1}}{2} $ and $1\ge v\ge 0 $, the third term has absolute value less than $$ \frac 1\pi \int _{a_n^-}^ {a_n^+} \frac{ q(t) }{(t-u)^2+v^2} dt\le C \frac{ |g_n|^2} {|k|^2 |g_n| ^{2} } = O(k^{-2}).$$ So $\dot w = 1+O(k^{-2})$ and $\dot E=2w\, \dot w=2w+O(k^{-1})= 2k+O(k^{-1})$ by Lemma 5.2. \bigskip We will need the following formulas, see (6.1) \cite{K1}:\proclaim{Lemma 7.2} For any $u\not \in [a,b]$ we have $$\align & \int _{a} ^b \frac{ \sqrt{(t-a)(b-t)} }{(t-u)^3} dt=\frac{\pi }{8}\text{sign} (u-a) \frac{(b-a)^2}{|u-a|^{\frac 32}|u-b|^{\frac 32}} \tag 1 \\& \int _{a} ^b \frac{ \sqrt{(t-a)(b-t)} }{(t-u)^4} dt=\frac{ \pi }{16} \frac{(b-a)^2}{|u-a|^{\frac 32}|u-b|^{\frac 32}} \left ( \frac{1}{|u-a| } + \frac{1}{|u-b| } \right ) . \tag 2 \endalign $$ \endproclaim The proof is as Lemma 6.1 \cite{K1}. Let us suppose $u>b$. Then setting $\ell =b-a$, $\ell s= t-a$, $\ell h=a-u$, $q = \sqrt{1+ \frac{1}{h }}$ and introducing a new variable $x$ defined by $x\, s = \sqrt{s(1-s)}$ and so $s=1/(1+x^2),$ $ds=-2xs^2dx$, we obtain $$\frac{1}{\ell }\int _0^1 \frac{\sqrt{s(1-s)}}{(s+h)^3}ds=\frac{1}{\ell h^3}\int _0^1 \frac{\sqrt{s(1-s)}}{s^3(\frac 1s+\frac 1h)^3}ds =\frac{1}{\ell h^3}\int _\Bbb R \frac{x^2}{(x^2+q^2)^3}dx$$ which by the Residue Theorem is equal to $ \frac{ \pi i }{\ell h^3} \left [ \frac{z^2}{(z+iq)^3} \right ] ^{\prime\prime} _{z=iq } = \frac{ \pi }{8\ell h^3 q^3} . $ Proceeding similarly we get (1) also for $u m$) are negative (resp. positive). We have $$\aligned &p^{\prime \prime }(u) \le \frac 2\pi \sum _{n\le m} \int _{ g_n} \frac{ \sqrt{ (t-a_n^-) ( a_n^+-t ) } }{(t-u)^3} dt +\frac {2C_0}\pi \sum _{n> m}\int _{ g_n} \frac{ \sqrt{ (t-a_n^-) ( a_n^+-t ) } }{(t-u)^3} dt \\& = -\frac 14 \sum _{n\le m} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}} +\frac {C_0}4 \sum _{n> m} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}}. \endaligned \tag 6 $$ Similarly we have $$\aligned &p^{\prime \prime }(u) \ge -\frac {C_0}4 \sum _{n\le m} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}} +\frac {1}4 \sum _{n> m} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}}. \endaligned \tag 7 $$ We are considering $m\gg 1$. Observe that by Lemma 2.1 for $u\in \sigma _m$ there are fixed constants such that for arbitrary $N$ $$\aligned & \sum _{n\ge m+1} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}}\approx \sum _{n> m+1} \frac{|g_n|^2}{\langle n-m \rangle ^{3}}\le \frac{C_N }{\langle u \rangle ^{N}} \\& \sum _{n\le m-1} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}} \approx \sum _{n\le m-1} \frac{|g_n|^2}{\langle n-m \rangle ^{3}}\approx \langle u \rangle ^{-3} . \endaligned \tag 8$$ Hence for $u\in [a_m^++ \alpha ,a_{m+1}^--\alpha ] $ with $\alpha >0$, (6)-(8) and $|g _{n}|\lesssim \langle n \rangle ^{-N}$ imply (1). Assume now $u\in (a^+_m , a^+_m+\alpha ]$. From (6)-(8), $|g _{m+1}|\lesssim \langle m+1 \rangle ^{-N}$ and the signs, we get (2) and (3) $$ \aligned &p^{\prime \prime }(u) \approx -\langle u \rangle ^{-3} - \frac{(a_m^+-a_m^-)^2}{|u-a_m^-|^{\frac 32}|u-a_m^+|^{\frac 32}} . \endaligned $$ Now we consider $u$ close to $a^{-}_{n+1}$. We now prove (4). From (7) we get $$ \aligned &p^{\prime \prime }(u) \ge -\frac {C_0}4 \sum _{n\le m} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}} +\frac {1}4 \sum _{n> m+1} \frac{(a_n^+-a_n^-)^2}{|u-a_n^-|^{\frac 32}|u-a_n^+|^{\frac 32}} +\\& +\frac {1}4 \frac{(a_{m+1}^+-a_{m+1}^-)^2}{|u-a_{m+1}^-|^{\frac 32}|u-a_{m+1}^+|^{\frac 32}} . \endaligned $$ We absorb the first two terms in the right hand side inside the term $\frac{-C_2}{\langle u \rangle ^3}$ of (4) and we get (4). The proof of (5) proceeds similarly starting from (6). \bigskip \noindent In the following two lemmas the symbols $\approx $, $\lesssim $ and $\ll $ involve fixed constants. We remark that $E(k)$ is even in $k$ so for this reason we will assume now only $k\gg 1$. \proclaim{Lemma 7.4} There are fixed $C_1,C_2,c>0$, with $C_1>C_2$, and $m_0>0$ such that for any $m\ge m_0>0$ and any $u\in \sigma _m$ we have $ \big |\frac{d^2E}{du^2}\big | \approx $ $$\aligned \frac{m}{|g_{m }|} \quad &\text{for}\quad u-a_m^+ \le c|g_{m }|; \\ 1+ \frac{m|g_m|^2}{ |u-a_m^+|^3} \quad &\text{for}\quad a_m^++c |g_{m }|\le u \le a_{m+1 }^--C _1|m+1|^{\frac{1}{3}}|g_{m+1 }|^{\frac{2}{3}}; \\ \frac{-m|g_{m+1 }|^2}{|u-a_{m+1 }^-|^{\frac 32}|u-a_{m+1 }^+|^{\frac 32}} \quad &\text{for}\quad c|g_{m+1 }|\le a_{m+1 }^--u \le C_2 |m+1|^{\frac{1}{3}}|g_{m+1 }|^{\frac{2}{3}}; \\ \frac{-m}{|g_{m+1 }|} \quad &\text{for}\quad a_{m+1 }^--u\le c |g_{m+1 }|. \endaligned $$ For $a^{+}_m+|g_{m}|^{\frac{1}{4}}\le w \le a^{-}_{m+1}-|g_{m+1}|^{\frac{3}{5}}$ we have $\ddot w =O(k^{-3})$. \endproclaim We start by the last statement. We have $\ddot w =-(\dot w)^3 k''$ with $\dot w\approx 1$ for $a^{+}_m+|g_{m}|^{\frac{1}{4}}\le w \le a^{-}_{m+1}-|g_{m+1}|^{\frac{3}{5}}$. In this interval we have $ |g_m|^2|u-a_{m}^+|^{-\frac{3}{2}}|u-a_{m}^-|^{-\frac{3}{2}}\le |g_m|^{\frac{5 }{4}}\ll k^{-3}$ and $ |g_{m+1 }|^2|u-a_{m+1 }^+|^{-\frac{3}{2}}|u-a_{m+1 }^-|^{-\frac{3}{2}}\le |g_{m+1 }|^{\frac{1 }{5}}\ll k^{-3}$. We have $$ {\ddot E} =2\left ( {\dot u} \right )^2-2 u \left (\dot u \right )^3 p''. \tag 1$$ Let us first assume $u\le a^-_{m +1} -\alpha $. Then $A(u) \approx \sqrt{\frac{ |g_m|}{u-a^+_m}} +|g_{m+1}|^2$. By (1-3) Lemma 7.3 and by Lemma 7.1 we have $$ \frac{d^2E}{dp^2}\approx \frac{1}{(1+ \sqrt{\frac{ |g_m|}{u-a^+_m}} )^2} + \frac{m}{(1+ \sqrt{\frac{ |g_m|}{u-a^+_m}})^3}\left ( \frac{(a_m^+-a_m^-)^2}{|u-a_m^-|^{\frac 32}|u-a_m^+|^{\frac 32}} + \frac{1}{\langle m \rangle ^3 }\right ) . $$ For $|u-a_m^+|\lesssim |g_m|$ we have $ \frac{d^2E}{dp^2}\approx \frac{m}{|g_m|}.$ For $a^+_{m }+|g_m| \lesssim u\le a^-_{m +1} -\alpha $ we have $$\frac{d^2E}{dp^2}\approx 1+ \frac{m|g_m|^2}{ |u-a_m^+|^{3}} . $$ Now we consider $u\ge a^-_{m +1} -\alpha $. Then $A(u) \approx \sqrt{\frac{ |g _{m+1}|}{u-a^- _{m+1}}} +|g_{m }|^2$. The two terms in the right hand side of (1) can be equal for $|u- a^- _{m+1}|\approx |m+1|^{\frac{1}{3}}|g_{m+1 }|^{\frac{2}{3}}$. For $|u- a^- _{m+1}|\gg |m+1|^{\frac{1}{3}}|g_{m+1 }|^{\frac{2}{3}}$ we claim that $ \frac{d^2E}{dp^2}\approx 1$. Indeed $A(u)\ll 1$, $ \frac{du}{dp} \approx 1$, and $$ \big | \frac{d^2p}{du^2} \big | \lesssim \langle u \rangle ^{-3}+\frac{ m|g_{m+1 }|^2}{|u-a_{m+1 }^-|^{\frac 32}|u-a_{m+1 }^+|^{\frac 32}}\ll 1.$$ For $|u- a^- _{m+1}|\ll |m+1|^{\frac{1}{3}}|g_{m+1 }|^{\frac{2}{3}}$ we distinguish between $|u- a^- _{m+1}|\gtrsim |g_{m+1 }|$ and $|u- a^- _{m+1}|\lesssim |g_{m+1 }|$. If $|g_{m+1 }|\lesssim |u- a^- _{m+1}| \ll |m+1|^{\frac{1}{3}}|g_{m+1 }|^{\frac{2}{3}}$ then $A(u) \lesssim 1$, $ \frac{du}{dp} \approx 1$ and by (4-5) Lemma 7.3 $$\frac{d^2E}{dp^2}\approx \frac{-m|g_{m+1 }|^2}{|u-a_{m+1 }^-|^{\frac 32}|u-a_{m+1 }^+|^{\frac 32}}. $$ For $|g_{m+1 }|\gtrsim |u- a^- _{m+1}|$, as $u\nearrow a_{m+1 }^-$ then $A(u)$ starts getting larger and $ \frac{du}{dp}$ starts getting smaller without however matching $p''$ which is very large, and we have $$ u \left (\frac{d u}{dp } \right )^3 \frac{d^2p}{du^2}\approx (m+1)\frac{|u- a^- _{m+1}|^{\frac{3}{2} }}{|g_{m+1 }|^{\frac{3}{2} } } \frac{(a_m^+-a_m^-)^2}{|u-a _{m+1}^-|^{\frac 32}|u-a _{m+1}^+|^{\frac 32}}\approx \frac{m}{|g_{m+1 }|} $$ and hence $\frac{d^2E}{dp^2}\approx -\frac{m+1}{|g_{m+1 }|} $. \bigskip \proclaim{Lemma 7.5} Let here $w=u+i0$. There are fixed $c >0$, $c_1> 0$ and $n_0$ such that for $ m \ge n_0$ and for $\alpha \in (1/2, 1)$ then $a_{m+1}^- - |g _{m+1}|^{\alpha}u>a_{m }^++ c |g _{m }| $ implies $ |\dddot{E} |\ge c_1 |{m }| \, |g_{m }|^{4(\frac{1}{2}- \alpha )}.$ \endproclaim We prove the $m+1$ case, the other being similar. We have $$\align & \frac{d^3E}{dp^3}=-6\left (\frac{d u}{dp } \right )^4 \frac{d^2p}{du^2}+ 6u\left (\frac{d u}{dp } \right )^5\left ( \frac{d^2p}{du^2} \right )^2- 2 u \left (\frac{d u}{dp } \right )^4 \frac{d^3p}{du^3} .\tag 1\endalign$$ If $c $ is large, in our domain $\frac{du}{dp}\approx 1$ and $ \frac{d^2p}{du^2} \approx \frac{|g _{m+1}|^{2}}{ |u -a^+ _{m+1}|^{\frac{3}{2}} |u -a^- _{m+1}|^{\frac{3}{2}} } . $ We claim that the dominating term in the rhs of (1) is the third. We write $$p^{\prime \prime \prime }(u)= \frac 2\pi \sum _n \int _{ g_n} \frac{ q(t) }{(t-u)^4} dt .$$ For $u$ as in the statement, by (2) Lemma 7.2 $$\big | \frac{d^3p}{du^3} \big | \approx \frac{|g _{m+1}|^{2}}{ |u -a^+ _{m+1}|^{\frac{3}{2}} |u -a^- _{m+1}|^{\frac{5}{2}} } \gtrsim |g_{m+1}|^{4(\frac{1}{2}- \alpha )}. $$ Since $\frac{du}{dp}\approx 1$ for $|g _{m+1}|\ll |u -a^- _{m+1}|$, we have $ (p^{\prime \prime }) ^2 \ll |p^{\prime \prime \prime}| $ and the last term in the rhs of (1) is the dominating one. \bigskip In Lemma 4.6 we have defined $\chi _{int} (k)=\sum _{n\ge n_0} \chi _1 (\frac{w - a_{n}^+ }{ |g_n|^{ \frac{1}{3}}} ) \chi _1 (\frac{ a_{n+1}^--w }{ |g _{n+1}|^{ \frac{3}{5}}} )$ for $n_0\gg 1$. In the support of $\chi _{int} $ we have $w\approx k$, $ k'= 1+O(\langle w\rangle ^{-2})$, $k''= O(\langle w\rangle^{-3})$. Now, to complete the proof of Lemma 4.7 it is enough to prove: \proclaim{Lemma 7.6} We can extend the restriction of $p(u)$ on the support of $\chi _{int} (p(u)) $ to the whole of $\Bbb R$ so that the extension $\widetilde{p}(u)$ satisfies the same relations $w\approx \widetilde{p}$, $ \widetilde{p}'= 1+O(\langle u\rangle ^{-2})$, $\widetilde{p}''= O(\langle u\rangle^{-3})$. \endproclaim Recall $p(u)=u+ \frac 1\pi \sum _{n\in \Bbb Z}\int _{ g_n} \frac{ q(t) }{t-u } dt$ for $u\in \sigma =\cup \sigma _n$. For $u\in supp \chi _{int} (p(u))$ we have $p(u)=\widetilde{p}(u)$ with $$\widetilde{p}(u):=u+ \frac 1\pi \sum _{na^{+}_{n_0}$ we obtain that $\widetilde{p}(u)$ satisfies the relations stated in in the statement. This follows from the fact that near a gap $g _{m}$ we have $$\aligned \widetilde{p}'&= 1+O(\langle u\rangle ^{-2}) +\frac 1\pi \left ( \chi _{1} (\frac{u - a_{m}^+ }{ |g_m|^{ \frac{1}{3}}} ) \chi _{1} (\frac{ a_{m }^--u }{ |g _{m }|^{ \frac{3}{5}}} ) \int _{ g_m} \frac{ q(t) }{t-u } dt \right ) '\\& =1+O(\langle u\rangle ^{-2})+O(|g_m| ^{\frac{4}{5}})\\ \widetilde{p}'' & = O(\langle u\rangle^{-3})+\frac 1\pi \left ( \chi _{1} (\frac{u - a_{m}^+ }{ |g_m|^{ \frac{1}{3}}} ) \chi _{1} (\frac{ a_{m }^--u }{ |g _{m }|^{ \frac{3}{5}}} ) \int _{ g_m} \frac{ q(t) }{t-u } dt \right ) ''\\& =O(\langle u\rangle^{-3})+ O(|g_m| ^{\frac{1}{5}}). \endaligned$$ \bigskip \head \S 8 Estimates on fundamental solutions \endhead In what follows $\dot f =\frac{d}{dk}f$ and $ f^\prime =\frac{d}{dx}f$. Referring to formulas (2.1) and (2.2), we write $\theta (x,k)$ and $\varphi (x,k)$ for $\theta (x,w(k))$ and $\varphi (x,w(k))$. Then we have: \proclaim{Lemma 8.1} For $x\in [0,1]$ we have $$\align & \big | \theta (x,k) -\cos (kx) \big | \le \frac 1{|k|} e^{\frac{x}{|k|}(\|P \| _\infty +|k^2-E(k)|) }\tag 1 \\& \big | \varphi (x,k) -\frac{\sin (kx)}{k} \big | \le \frac{1}{|k|^2}\, e^{\frac{x}{|k|}(\|P \| _\infty +|k^2-E(k)|) } . \tag 2\endalign $$ Furthermore there is a fixed constant $C$ such that for $x\in [0,1]$ $$\align &\big | \dot \theta (x,k) +x \sin (kx)- \int _0^x \theta (s,k) \frac{\partial}{\partial k} \frac{\sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ]}{k} ds \big | \le \frac{C}{\langle k \rangle} \\ & \big |\dot \varphi (x,k) - \frac{\partial}{\partial k} \frac{\sin (kx)}{k}- \int _0^x \varphi (s,k) \frac{\partial}{\partial k} \frac{\sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ]}{k} ds \big | \le \frac{C}{\langle k \rangle ^2} . \endalign $$ \endproclaim {\it Proof.} The argument is routine. $\theta (x,k)$ and $ \varphi (x,k)$ satisfy the following integral equations: $$\align & \theta (x,k) =\cos (kx)+\frac 1k \int _0^x \sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ] \theta (s,k) ds \tag 3 \\& \varphi (x,k) =\frac{\sin (kx)}{k}+\frac 1k \int _0^x \sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ] \varphi (s,k) ds . \tag 4 \endalign $$ Now we write $$\align & \theta (x,k) =\sum _{n=0}^\infty \theta _n(x,k) \, , \quad \varphi (x,k) =\sum _{n=0}^\infty \varphi _n(x,k)\\& \theta _0(x,k)=\cos (kx)\, , \quad \varphi _0(x,k) =\frac{\sin (kx)}{k} \\& \theta _{n+1}(x,k) = \frac 1k \int _0^x \sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ] \theta _{n }(s,k) ds\\& \varphi _{n+1}(x,k) = \frac 1k \int _0^x \sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ] \varphi _{n }(s,k) ds. \endalign $$ Singling out $\theta (x,k)$, we have for $x_{n+1}=x$: $ \theta _{n+1}(x,k)=$ $$ =\frac 1{k^{n+1}} \int _{0\le x_1 \le \dots \le x_n\le x} \prod _{j=1}^{n}\left \{ \sin \left ( k(x_{j+1}-x_{j}) \right ) \left [ P(x_j)+k^2-E(k) \right ] dx_j\right \} \cos(kx_1). $$ This implies the following estimate which implies (1): $$ \big | \theta _{n+1}(x,k) \big | \le \frac{\left ( \int _0^x(|P(s)|+|k^2-E(k)|) ds \right ) ^n}{k^{n+1}\, n!}\le \frac{x^n}{k^{n+1} \, n!} (\|P \| _\infty +|k^2-E(k)|)^n. $$ Proceeding similarly we obtain the following inequality, which gives us (2): $$ \big | \varphi _{n+1}(x,k) \big | \le \frac{ \left ( \int _0^x(|P(s)|+|k^2-E(k)|) ds \right ) ^n}{k^{n+2}\, n!}\le \frac{x^{n }}{k^{n+2} \, n!} (\|P \| _\infty +|k^2-E(k)|)^n. $$ \bigskip Next we consider $$\aligned & \dot \theta (x,k) =-x \sin (kx)+ \int _0^x \theta (s,k) \frac{\partial}{\partial k} \frac{\sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ]}{k} ds+\\& +\frac 1k \int _0^x \sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ] \dot \theta (s,k) ds \endaligned $$ and $$\aligned & \dot \varphi (x,k) = \frac{\partial}{\partial k} \frac{\sin (kx)}{k}+ \int _0^x \varphi (s,k) \frac{\partial}{\partial k} \frac{\sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ]}{k} ds+\\& +\frac 1k \int _0^x \sin \left ( k(x-s) \right ) \left [ P(s)+k^2-E(k) \right ] \dot \varphi (s,k) ds. \endaligned $$ We have for a fixed $C_1$ $$ \aligned & \big | {\partial _k} \left ( k^{-1} \sin \left ( k(x-s) \right ) P(s) \right ) +\left ( k^2-E(k)\right ) {\partial _k} \left ( k^{-1} \sin \left ( k(x-s) \right ) \right ) + \\& + \left (2k-\dot E (k)\right ) k^{-1} \sin \left ( k(x-s) \right ) \big | \le C \langle k\rangle ^{-1} +C \langle k\rangle ^{-1}\big |2k-\dot E (k)\big | \le C_1. \endaligned \tag 8.1 $$ By this estimate, by (1) and (2) and by the above arguments we obtain the last two inequalities of lemma 8.1. \bigskip \proclaim{Lemma 8.2} There is a fixed $C$ such that $$ \align & \big | \int_0^{1} \theta ^2(x,k) dx-\frac{1}{2}-\frac{\sin (2k)}{4k } \big |\le C k^{-2} ; \tag 1 \\& \big | k^2 \int_0^{1} \varphi ^2(x,k) dx- \frac{1}{2}+\frac{\sin (2k)}{4k } \big |\le C k^{-2};\tag 2 \\& \big | 2k \int_0^{1} \theta (x,k) \varphi (x,k)dx-\frac{1-\cos (2k)}{2k} \big |\le C k^{-2}.\tag 3 \endalign $$ \endproclaim {\it Proof.} We use the notation in the proof of Lemma 8.1. To prove (1) is enough to show that $ \int_0^{1} \cos (x k)\theta _1 (x,k) dx=O(k^{-2}).$ By its definition and elementary computation $$\theta _1 (x,k)=\frac{\sin (x k)}{2k} \left [ \int _0^xP(s) ds+(k^2-E )x\right ] +O(k^{-2}).\tag 4$$ By $k^2-E=-2Q_0+o(1)$ and elementary integration, (1) follows. To prove (2) is enough to show that $ \int_0^{1} \sin (x k)k\varphi _1 (x,k) dx=O(k^{-2}).$ By its definition and elementary computation $$k\, \varphi _1 (x,k)=-\frac{\cos (x k)}{2k} \left [ \int _0^xP(s) ds+(k^2-E )x\right ] +O(k^{-2}) .\tag 5$$ Elementary integration gives (2). To prove (3) is enough $ \int_0^{1} \left ( \cos (x k)k\varphi _1 (x,k) + \sin (x k)\theta _1 (x,k) \right ) dx=O(k^{-2}).$ So (3) follows from $ \int_0^{1} \left ( \cos (x k)k\varphi _1 (x,k) + \sin (x k)\theta _1 (x,k) \right ) dx=$ $$= - \frac{1}{2k} \int_0^{1} \cos (2kx) \left [ \int _0^xP(s) ds+(k^2-E )x\right ] dx+O(k^{-2}). $$ \bigskip \head \S 9 Proof of Lemma 4.4 \endhead Formulas (2.2), (2.3), (2.7) and Lemma 8.1 imply with Lemma 9.1 below gives the first claim of Lemma 4.4. \proclaim{Lemma 9.1} There is a fixed constant $C>0$ such that for $w=u+iv$ with $v=0$ and for $a_n^+ +Cn^5|g_n|\le u\le a_{n+1}^{-} -Cn^5|g_{n+1}|$, we have $$ \big | N^2(k)-1 \big |+ \big | \frac{ \sin k }{k \varphi (k) } -1\big |+ \big |\frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \big |\le \frac{C}{\langle k \rangle }. $$ \endproclaim {\it Proof}. Expanding in the definition of $N^2(k)$ we write $$\aligned &N^2(k)= \int_0^{1} \left ( \theta (x,k) + \frac{\varphi ^\prime (k) - \theta (k) }{2\varphi (k)} \varphi (x,k) \right ) ^2 dx + \int_0^{1} \frac{ \sin ^2k }{ \varphi ^2(k) } \varphi ^2 (x,k) dx=\\& = \int_0^{1} \left [ \theta ^2(x,k) +2\theta (x,k) \varphi (x,k) \frac{\varphi ^\prime (k) - \theta (k) }{2\varphi (k)}+ \left (\frac{\varphi ^\prime (k) - \theta (k) }{2\varphi (k)} \varphi (x,k)\right )^2 \right ] +\\& +\int_0^{1} \frac{ \sin ^2k }{ \varphi ^2(k) } \varphi ^2 (x,k) dx .\endaligned$$ We recall now from formulas (1.4) and (3.1) in \cite{F2}: $$\dot E(k) =\frac{2\sin k }{ \varphi (k) N^2(k)}. \tag 9.1$$ For $a_n^+ +Cn^5|g_n|\le u \le a_{n+1}^{-} -Cn^5|g_{n+1}|$ we have $\big | \dot E(k)-2k\big | \lesssim \langle n \rangle ^{-2} $ by claim 4 in the statement of Lemma 7.1. So $ N^2(k) =\frac{\sin k }{k \varphi (k) } \left ( 1+O_2(\frac 1{k^2}) \right ) $ with $O_2$ a big $O$. Hence by Lemma 8.2 we obtain, for a certain number of big O's, $$ \aligned & \left ( \frac{ \sin k }{k \varphi (k) } \right ) ^2-2\frac{\sin k }{k \varphi (k) } \frac{1+O_2(k^{-2})}{1 -\frac{\sin (2k)}{2k} +O _1( k^{-2}) } +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right ) ^2 +\\& +\frac{1 +\frac{\sin (2k)}{2k} +O _3( k^{-2}) }{1 -\frac{\sin (2k)}{2k} +O_1 ( k^{-2}) } +O _4( k^{-1}) \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} =0\endaligned $$ which implies $ \frac{ \sin k }{ k\varphi (k) } = \frac{1+O_2(k^{-2})}{1 -\frac{\sin (2k)}{2k} +O _1( k^{-2}) }\pm \sqrt{\Delta}$, for $$ \Delta= { \frac{\frac{\sin ^2(2k)}{4k^2}}{ \left ( 1- \frac{\sin (2k)}{2k} \right )^2}+ O(k^{-2})- \left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right ) ^2 -O _4( k^{-1}) \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)}}. $$ So $ \frac{ \sin k }{k \varphi (k) } =1+O(\frac 1k)$ and $ \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} =O(\frac 1{ {k}}) $ when $k\in \Bbb R$ and under the restriction $a_n^+ +Cn^5|g_n|\le u\le a_{n+1}^{-} -Cn^5|g_{n+1}|$. \bigskip \noindent The proof of the second claim of Lemma 4.4 is trickier and proceeds in several steps. We set $\|m_\pm ^0( k)\| _2=(\int _0^1|m_\pm ^0(x,k)| ^2dx) ^{\frac{1}{2}}$. First of all we consider the Fourier series expansion $m_{\pm }^{0}(x,k)=\sum \widehat{m}_{\pm}(\ell ,k)e^{2\pi i \ell x}$ and show that the $L^2$ is concentrated in two harmonics. One harmonic is $|\widehat{m_\pm }(0,k )|\approx \| {m_\pm }(k )\| _2 $. If there is an $n$ such that $|n\pi +k|\ll 1$ then also $ \widehat{m_\pm }(\mp n,k )$ can be significant. We then bound $N^ {-1}(k)$, $m^{\pm }(k)k^{-1}N^ {-1}(k)$ in terms $\| {m_\pm }(k )\| _2 $. Next, we express $m_\pm ^0(x,k)$ in terms of $\varphi (x,k)$ and $\theta (x,k)$, we expand the latter in terms of $\sin(kx)$ and $\cos(kx)$ and a reminder, and we conclude that in $L^\infty$ sense $m_\pm ^0(x,k)$ can be approximated by the two terms of the Fourier expansion discussed above. Next we look at the normalization of the Bloch functions. Slightly off the slits we have $|\widehat{m_\pm }(0,k )|\gg |\widehat{m_\pm }(\mp n,k )|$ and so $1\approx \widehat{m_+ }(0,k ) \widehat{m_- }(0,k )$. From this we get the desired bound on $|{m_+ }(x,k ) {m_- }(y,k )|\lesssim 1$ off the slits. Near the slits we have $|\widehat{m_\pm }(0,k )|\approx |\widehat{m_\pm }(\mp n,k )|$ so to exploit the normalization we have to exclude a significant cancelation in a certain formula. Let us start with the first step, and show that there are at most two significant harmonics. \proclaim{Lemma 9.2} For all $n$ except possibly for one $ n_0$, we have $|n\pi +k|\gtrsim 1$. Then for $n\neq 0, n_0$ we have $ \sum _{n \neq 0,n_0 } | \widehat{m_+}(n ) |^2 \le C |k| ^{-2}\| m_+ ^0( k)\| _2^2$ for a fixed $C$. Furthermore for a fixed $C$ we have $ \sum _{n \neq 0,n_0 } n^2 | \widehat{m_+}(n ) |^2 \le C \| m_+ ^0( k)\| _2^2$. The same statement holds for $m_- ^0$ with $ n_0$ replaced by $-n_0$. If for all $n\neq 0$ we have $|n\pi +k|\gtrsim 1$ we can extend the above inequalities to the sum on all $n \neq 0$. \endproclaim {\it Proof.} The first sentence is straightforward. We will assume there is $n_0$ with $|n_0\pi +k|\ll 1$. If such $n_0$ does not exist, the proof is almost the same. Set for $n\neq 0, n_0$ $$ \aligned & \widehat{m_+}(n)+\sum _{\ell \neq 0, n_0 } T(n,\ell ) \widehat{m_+}(\ell ) =-\frac{ \widehat{P}( n) \widehat{m_+}(0 ) + \widehat{P}( n-n_0) \widehat{m_+}(n_0 ) }{ 4\pi n (n\pi +k) } \\& T(n,\ell ) =\frac{\widehat{P}(0) +k^2-E }{4\pi n (n\pi +k)} \delta (n-\ell )+\frac {\widehat{P}(n-\ell ) }{ 4\pi n (n\pi +k) } .\endaligned \tag 1 $$ We have $\widehat{P}(0) +k^2-E=O(k^{-2}) $ by Lemma 5.2 and by $\widehat{P}(0)=2Q_0$. Equation (1) is of the form $(I+T)u=f$ where $\| f\| _{l^2} \le C |k|^{-1} \| m_\pm ^0( k)\| _2 $ and $\| f\| _{l^2_1} \le C \| m_\pm ^0( k)\| _2 $ where $\| f\| _{l^2_1}^2= \sum _{n \neq 0, n_0 } n^2|f (n)|^2.$ By $$ \sup _\ell \sum _n \left ( |k T(n,\ell ) | +|n T(n,\ell ) |\right ) +\sup _n \sum _\ell \left ( |k T(n,\ell ) | +|n T(n,\ell ) |\right ) \le C $$ for $u\in l^2 (\Bbb N)$ we have $\sum _n |(Tu)(n)|^2 \le C |k|^{-2} \| u\| _{l^2}^{2}$ and $\sum _n n^2 |(Tu)(n)|^2 \le C \| u\| _{l^2}^{2}$. So inverting and after a Neumann expansion, we see that for $n\neq 0, n_0$ we have $$\widehat{m_+}(n)=-\frac{ \widehat{P}( n) \widehat{m_+}(0 ) + \widehat{P}( n-n_0) \widehat{m_+}(n_0 ) }{ 4\pi n (n\pi +k) } + \widehat{e} (n)\tag 2$$ where $\widehat{e}=\sum _{m=1}^\infty (-)^mT^mf$ satisfies $ \widehat{e} (0)=\widehat{e}(n_0)=0$ and $$\| \widehat{e} \| _{l^2 } \le \| \widehat{e} \| _{l^2_1} \le C \| \sum _{m=0}^\infty (-)^mT^mf\| _{l^2} \le C |k|^{-1} \sum _{m=0}^\infty (C|k|^{-1})^m \| m_+ ^0( k)\| _2 .$$ Hence by (2), $ \sum _{n \neq 0, n_0 } |\widehat{m_+} (n)|^2\le C |k|^{-2}\| m_+ ^0( k)\| _2^2$ and $ \sum _{n \neq 0, n_0 }n^2 |\widehat{m_+} (n)|^2\le C \| m_+ ^0( k)\| _2^2$. The proof for $m_-$ is similar. \bigskip We express now the Bloch functions in terms of the fundamental solutions as in \S 2. Using the notation in the proof of Lemma 8.1 and for $m^\pm (k)= \frac{\varphi ^\prime (k) - \theta (k) }{2\varphi (k)} \pm i \frac{\sin k}{\varphi (k)} $, $$\aligned m_\pm ^0(x ,k) &=e^{\mp ikx} \left ( \frac{\cos(x k)}{N} + \frac{m^\pm (k)}{kN} \sin(x k)\right ) +\\& + \frac{1}{N }e^{\mp ikx}\sum _{j=1}^\infty \theta _j(x,k) + \frac{m^\pm (k)}{kN} e^{\mp ikx}\sum _{j=1}^\infty k\varphi _j (x,k) . \endaligned \tag 9.2$$ By the proof of Lemma 8.1 $ \big | e^{\mp ikx}\sum _{j=1}^\infty \theta _j(x,k)\big | +\big | e^{\mp ikx}\sum _{j=1}^\infty k\, \varphi _j(x,k)\big | =O(k^{-1})$ for $x\in [0,1]$. Then we have $$\aligned &m_\pm ^0(x ,k) =A_\pm (k) + B_\pm (k) e^{\mp i 2kx} + O(N^{-1}k^{-1}) + O(m^\pm (k)N^{-1}k^{-2}), \\& A_\pm (k) =\frac{1 -i\, k^{-1} m^\pm (k) }{2N(k)}=\frac{1+ \frac{\sin ( k)}{ k\varphi (k) } \mp i \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} }{2N(k)}\\& B_\pm (k)= \frac{1+i\, k^{-1} m^\pm (k) }{2N(k)}= \frac{1- \frac{\sin ( k)}{ k\varphi (k) } \pm i \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} }{2N(k)} .\endaligned $$ We have $$\widehat{m_\pm ^0(\cdot ,k)}(0)=A_\pm (k)+ O\left (k^{-1} B_\pm (k) \right )+ O(N^{-1}k^{-1}) + O(m^\pm (k)N^{-1}k^{-2}).\tag 9.3 $$ For the $n_0$ of Lemma 9.2 we have $$\widehat{m_\pm ^0(\cdot ,k)}(\pm n_0)= B_\pm (k) \frac{ e^{\pm 2i ( \pi n_0-k)} -1}{\pm 2i ( \pi n_0-k)} + O(N^{-1}k^{-1}) + O(m^\pm (k)N^{-1}k^{-2}).\tag 9.4 $$ \proclaim{Lemma 9.3} For a fixed $C>0$ $$\align & |1/N(k)| + |\frac{ m^\pm (k) }{ kN(k)} | +|A_{\pm }| + |B_{\pm }|\le C \| m_\pm ^0( k)\| _2 . \endalign$$ \endproclaim \noindent {\it Proof.} Take $$A_\pm (k)=\frac{1}{2N}+i\frac{ m^\pm (k)}{2kN}\, , \quad B_\pm (k)=\frac{1}{2N}-i\frac{ m^\pm (k)}{2kN}. \tag 1 $$ Then by (9,3-4) $$ \align & |A_\pm (k)| \lesssim \| m_\pm ^0( k)\| _2+ O( N^{-1}k^{-1})+O(m^\pm (k)N^{-1}k^{-2}) \tag 2\\& |B_\pm (k)| \lesssim \| m_\pm ^0( k)\| _2+ O( N^{-1}k^{-1})+O(m^\pm (k)N^{-1}k^{-2}) .\tag 3 \endalign$$ By the triangular inequality, $$|1/N(k)| \le |A_\pm (k)|+|B_\pm (k)| \lesssim \| m_\pm ^0( k)\| _2+ O(m^\pm (k)N^{-1}k^{-2}).\tag 4$$ By (1-3) if one of $|1/N(k)|$ and $|m^\pm (k)N^{-1}k^{-1}| $ is $\gg \| m_\pm ^0( k)\| _2$, then $|1/N(k)|\approx |m^+ (k)N^{-1}k^{-1}| $ . Then $|1/N(k)|\lesssim \| m_\pm ^0( k)\| _2$ by (4) and $|m^\pm (k)N^{-1}k^{-1}|\lesssim \| m_\pm ^0( k)\| _2 .$ So these last two formulas hold. So $|A_\pm (k)|+|B_\pm (k)| \lesssim \| m_\pm ^0( k)\| _2 $ by (2-3) . \bigskip We now use the normalization of Bloch functions $1=\int _0^1m_+ ^0(x, k)m_- ^0(x, k)dx.$ We will denote $n_0$ by $n$. By (9,3-4) and Lemma 9.3 $$1=\left [ A_+ (k)A_- (k)+B_+ (k)B_- (k)\right ]+ O(k^{-1}\| m_+ ^0( k)\| _2\| m_- ^0( k)\| _2).\tag 9.5$$ We can also write by Lemma 9.2 $$1= \widehat{m} _{+} (0,k ) \widehat{m} _{-} (0,k )+ \widehat{m} _{+} (- n ,k )\widehat{m} _{-} ( n ,k )+ O(k^{-2}\| m_+ ^0( k)\| _2\| m_- ^0( k)\| _2).\tag 9.6$$ \proclaim{Lemma 9.4} Suppose that $k$ is in a region such that $|\widehat{m} _{\pm } (\mp n ,k )|\le \frac{1}{2}|\widehat{m} _{\pm } (0,k )|$. Then for a fixed constant $C$ we have $ \| m_+ ^0( k)\| _2 \| m_-^0( k)\| _2\le C $. As a consequence also $| m_+ ^0(x, k) m_- ^0(y, k)|0$. Set $m=m_\pm ^{0}$. Multiply by $\overline{m }$ the equation $ m'' \pm 2ikm' -P(x)m+(E-k^2)m=0 $, integrate in $[0,1]$ and take imaginary part to obtain $$ 2q \, \Im \int_0^1 \overline{m} \, m' dx = \Im (E-k^2) \| m \| _2^2. \tag 9.8 $$ Then by Lemma 9.2, $\int_0^1 \overline{m} \, m' dx= -2\pi n i |\widehat{m} (\mp n,k)|^2+O(k^{-1}\| m\| _2^2)$ and $\Im (E-k^2) =(v-q)(u+p)+(u-p)(v+q) =-(2u +O(u^{-1})) I_n+O((|v|+|q|) u^{-1}).$ Substitute in (9.8), divide by $4\pi q|n|$ and use $0 \le v \le q$ and $ I_n\le q$. \bigskip \proclaim{Lemma 9.6} There is a fixed $\Gamma >0$ such that for $|p-n\pi |>\Gamma |g_n|$ we have $|\widehat{m} _{\pm } (\mp n ,k )|\le \frac{1}{2}|\widehat{m} _{\pm } (0,k )|$. \endproclaim {\it Proof.} For $\Gamma \gg 1$ we have $I_n/q\ll 1$ in (9.7) by Lemma 6.2. Furthermore $ {|u|}/{ (|n |\pi )}= 1+O(1/|u|).$ \bigskip \proclaim{Lemma 9.7} There is a fixed $C$ such that for any fixed $\Gamma >0$ there are $N,\delta >0$ such that for $|n|\ge N$ in the region $|p-n\pi |<\Gamma |g_n|$, and $0\le q \le \delta |g_n| $ we have $\| m_+ ^0( k)\| _2\| m_- ^0( k)\| _2\le C.$ \endproclaim {\it Proof}. First of all, by Lemma 9.5 for $k$ near the slit we have $| \widehat{m}_\pm ^0(j, k) |^2 \approx \| m_+ ^0( k)\| _2^2/2$ for $j=0,\mp n$. These harmonics could be large with a large cancelation in (9.6). We have $$\aligned & A_+ (k)A_- (k)+B_+ (k)B_- (k) =\frac{\left (1+ \frac{\sin ( k)}{ k\varphi (k) } \right )^2 +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right )^2 }{4N^2(k)} +\\& +\frac{\left (1- \frac{\sin ( k)}{ k\varphi (k) } \right )^2 +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right )^2 }{4N^2(k)} =\frac{1+ \left ( \frac{\sin ( k)}{ k\varphi (k) } \right )^2 +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right )^2 }{2N^2(k)} . \endaligned $$ By (9.1) we have $$A_+ (k)A_- (k)-B_+ (k)B_- (k)=\frac{\sin ( k)}{ k\varphi (k) N^2}= \frac{\dot E}{2k}= \frac{ w}{ k \frac{dk}{dw}}\approx \frac{ 1}{ \frac{dk}{dw}}.\tag 9.9$$ We have $ \| m_+ ^0( k)\| _2 \| m_- ^0( k)\| _2\approx $ $$ \approx |A_+ (k)A_- (k)|\le 1+2/|k'(w)|+O(k^{-1}\| m_+ ^0( k)\| _2 \| m_- ^0( k)\| _2).$$ By Lemma 9.8 below we obtain $ \| m_+ ^0( k)\| _2 \| m_- ^0( k)\| _2\lesssim 1.$ \proclaim{Lemma 9.8} There is a fixed $C>0$ such that for any fixed $\Gamma >0$ there are $N,\delta >0$ such that for $|n|\ge N$ in the region $|p-n\pi |<\Gamma |g_n|$, and $0\le q \le \delta |g_n| $ we have $|k'(w) | > C.$ \endproclaim {\it Proof.} We recall by Schwartz Christoffel formula, for $c_m \in g_m $ with $k'(c_m )=0$, $$k'(w)= a\frac{1-w/c_n }{\sqrt{ (1-w/a^{-}_n ) (1-w/a^{+}_n) } } \prod _{\ell \neq n} \frac{1-w/c_\ell }{\sqrt{ (1-w/a^{-}_\ell ) (1-w/a^{+}_\ell) } } , \tag 1$$ see (1.10) \cite{MO}. By taking the derivative $dq/du$ in (5.1) we see that $q'(u)>0$ for $u\in ( a_m ^-, a_m ^-+ \varepsilon |g_m |] $ and $q'(u)<0$ for $u\in [ a_m ^+- \varepsilon |g_m |, a_m ^+) $ for a fixed $\varepsilon >0$. So $c_m \in [ a_m ^-+ \varepsilon |g_m|,a_m ^+- \varepsilon |g_m | ]$ for a fixed sufficiently small $\varepsilon >0$. Since the infinite product in (1) has value approximately 1 for $w$ near $\pi n$, because of the second factor there is a fixed $C>0$ such that $|k'(w) | > C $ holds for $ w=u+iv$ with $|v|\le \delta |g_n|$ and either $a^{-}_n -\Gamma |g_n|\le u\le a_n^-+ \varepsilon |g_n|/2$ or $a_n^+- \varepsilon |g_n|/2\le u \le a^{+}_n +\Gamma |g_n| $ with $\delta >0$ fixed and sufficiently small. Now we need to show that the values of $k=p+iq$ in the statement are inside this region in the $w$ plane. First of all $0\le q \le \delta |g_n| $ implies $0\le v\le q \le \delta |g_n| $ by (6.1). For $0\le v \le \delta |g_n| $ and $a_n^-+ \delta |g_n|\ll u \ll a_n^+- \delta |g_n|$ by Lemma 6.3 we have $q\approx I_n(u,v)$ and by Lemma 6.4 we have $I_n(u,v)\approx |g_n|^{ \frac{1}{2}} \sqrt{\min \{ |u- a^{-} _{n}|, |u- a^{+} _{n}| \}} \gg \delta |g_n|$. Obviously the latter is incompatible with $q\le \delta |g_n|$. So for small $\delta >0$ we have either $a^{-}_n -\Gamma |g_n|\lesssim u\le a_n^-+ \varepsilon |g_n|/2$ or $a_n^+- \varepsilon |g_n|/2\le u \lesssim a^{+}_n +\Gamma |g_n| $. \bigskip Notice that by (9.9) and (1) in Lemma 9.8 we can see that near the tip of the slit the product $\| m_+^0(k)\| _2\| m_-^0(k)\| _2$ is unbounded. \bigskip \head \S 10 Proof of Lemma 4.5: case $u\in [a^{+}_{n}+ |g_n|^{\frac{1}{4}}, a^{-}_{n+1}- |g _{n+1}|^{\frac{3}{5}}]$\endhead Lemma 4.5 consists in 3 claims. The third one follows immediately from Lemma 4.4 by the Cauchy integral formula. The first two claims follow immediately from the Cauchy integral formula from Lemma 10.1 which is an improvement of the second claim of Lemma 4.4 in the case when $a^{+}_{n}+ |g_n|^{\frac{1}{4}}\le u \le a^{-}_{n+1}- |g _{n+1}|^{\frac{3}{5}}$. We will assume this restriction on $u$ everywhere below in this section: \proclaim{Lemma 10.1} For $a^{+}_{n}+ |g_n|^{\frac{1}{4}}\le u \le a^{-}_{n+1}- |g _{n+1}|^{\frac{3}{5}}$ with $n\ge n_0$ and for $1\ge v\ge 0$, there is a $C$ such that $ \big | m_- ^0(x ,k)m_+ ^0(y ,k)-1 \big | \le \frac{C}{|k|} .$ \endproclaim Set $m_\pm (x)= m_\pm ^0(x ,k)$ and consider the expansion $m_\pm (x)= \sum _n e^{2\pi i n x}\widehat{m_\pm }(n).$ For $a^{+}_{n}+ |g_n|^{\frac{1}{4}}\le u \le a^{-}_{n+1}- |g _{n+1}|^{\frac{3}{5}}$ we have the following strengthening of Lemma 9.3: \proclaim{Lemma 10.2 } We have for a fixed $C$ $$||\widehat{m_\pm }(0 ) |^2 -\| m_\pm ^0( k)\| _2^2| +\sum _{n \neq 0 } | \widehat{m_\pm}(n ) |^2 \le C |k| ^{-2}\| m_\pm ^0( k)\| _2^2.$$ \endproclaim \noindent {\it Proof .} By (9.8) and Lemma 6.2 we get $ \Im \int _0^1 \bar m _\pm m^\prime _\pm dx = O( u^{-2} )\| m_\pm ^0( k)\| _2 ^2.$ By Lemma 9.2 $\Im \int _0^1 \bar m _\pm m^\prime _\pm dx =\mp 2\pi i n_{0}|\widehat{m}_\pm (\mp n_{0},k)|^2+O(k^{-1}\| m_\pm ^0( k)\| _2 ^2)$. Hence $| \widehat{m_\pm}(\pm n_0 ) | \le C |k| ^{-1}\| m_\pm ^0( k)\| _2 $. The latter and Lemma 9.2 imply the inequality $||\widehat{m_\pm }(0 ) |^2 -\| m_\pm ^0( k)\| _2^2 |\le C |k| ^{-2}\| m_\pm ^0( k)\| _2^2.$ \bigskip Now we have the following lemma: \proclaim{Lemma 10.3} For fixed constants we have $| N ^{-1}|\approx |m^\pm (k)N^{-1}k^{-1}|\approx |A_{\pm }| \approx \| m_\pm ^0( k)\| _2 $ and $|B_{\pm }| \lesssim |k|^{-1} \| m_\pm ^0( k)\| _2$. \endproclaim \noindent {\it Proof of Lemma 10.3.} We get $|B_{\pm }| \lesssim |k|^{-1} \| m_\pm ^0( k)\| _2$ by Lemmas 10.2 \& 9.3 and by (9.4) . By $|\widehat{m}_\pm (0,k)|\approx \| m_\pm ^0( k)\| _2 $ and by (9.3) we get $|A_{\pm }| \approx \| m_\pm ^0( k)\| _2 $. By definition of $A_{\pm }$ and of $B_{\pm }$, estimates $|B_{\pm }| \lesssim |k|^{-1} \| m_\pm ^0( k)\| _2$ and $|A_{\pm }|\approx \| m_\pm ^0( k)\| _2 $ imply $|1/N(k)|\approx |m^\pm (k)N^{-1}k^{-1}|\approx \| m_\pm ^0( k)\| _2$. \bigskip \proclaim{Lemma 10.4} We have for fixed constants $\| m_\pm ^0( k)\| _2 \approx 1$.\endproclaim \noindent {\it Proof of Lemma 10.4.} By Lemma 10.3 $|A_\pm (k)|\approx |1/N|\approx \| m_\pm ^0( k)\| _2 $. By Lemma 10.2 and the formulas immediately above (9.5) $ m_\pm ^0( x,k) \approx A_\pm (k)$. In particular by normalization of Bloch functions we have the following which gives us Lemma 10.4: $$1=\int _0^1m_+ ^0( x,k)m_- ^0( x,k)dx\approx A_+ (k) A_- (k).$$ \bigskip {\it Proof of Lemma 10.1. } We write $$ m_- ^0(x ,k)m_+ ^0(y ,k)=\frac{\left (1+ \frac{\sin ( k)}{ k\varphi (k) } \right )^2 +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right )^2 }{4N^2(k)} + O(k^{-1}) .$$ By the expansion of $N^2(k)$ in Lemma 9.1 and by Lemmas 8.2, 10.3 and 10.4, we have $$2N^2(k)=1+\left ( \frac{\sin ( k)}{ k\varphi (k) } \right )^2 +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right )^2+O(k^{-1}).$$ This, formula (9.1) and $\frac{\dot E}{2k} = 1+ O(k^{-1})$, Lemma 7.1 (5), imply the following, $$\frac{1+ \left ( \frac{\sin ( k)}{ k\varphi (k) } \right )^2 +\left ( \frac{\varphi ^\prime (k) - \theta (k) }{2k\varphi (k)} \right )^2 }{4N^2(k)} + \frac{\frac{\sin ( k)}{ k\varphi (k) }}{2N^2(k)} =\frac{1}{2}+\frac{\dot E}{4k}+ O(k^{-1})= 1+ O(k^{-1}), $$ which ends the proof of Lemma 10.1. 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