\magnification \magstep1 \input amstex \nopagenumbers \documentstyle{amsppt} \hsize=6truein \vsize 9truein \hoffset .15truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \pageno=0 \centerline{\bf{THE DENSITY IN A ONE-DIMENSIONAL POTENTIAL}}\smallskip \centerline{by} \smallskip \bigskip\bigskip\bigskip \centerline{\bf{C. Fefferman\footnote"*"{partially supported by NSF grant \#DMS--9104455} $\quad$ and $\quad$ L. Seco}} \centerline{\bf Table of Contents} \medskip \line{\sl \hfil Pages} \smallskip \line{Introduction\hfil 1-14} \smallskip \line{Review of Earlier Results\hfil 15-26}\smallskip \line{A Reformulation of the Eigenvalue Theorem\hfil 27-45}\smallskip \line{Approximating Sums by Integrals\hfil 46-52}\smallskip \line{The Microlocalized Density in the Airey Region I\hfil 53-82}\smallskip \line{The Microlocalized Density in the Airey Region II\hfil 83-113}\smallskip \line{The Microlocalized Density in the Oscillatory Region\hfil 114-134} \smallskip \line{The Microlocalized Density Near the Minimum of the Potential \hfil 135-141}\smallskip \line{Combining the Microlocalized Results \hfil 142-173}\smallskip \line{The Density for a One-Dimensional Potential \hfil 174-183}\smallskip \line{The WKB Density Theorem in One Dimension \hfil 184-186}\smallskip \line{The Density for Degenerate One-Dimensional Potentials I\hfil 187-189} \smallskip \line{The Density for Degenerate One-Dimensional Potentials II\hfil 190-193} \smallskip \line{The Density for Degenerate One-Dimensional Potentials III\hfil 194-202} \smallskip \line{The Density for Degenerate One-Dimensional Potentials IV\hfil 203-206} \line{References\hfil 207} \smallskip \def\Icirc{\hbox{$\int$}\kern-6pt\raise 1pt\hbox{$\circ$}} \def\chm{\chi_{{}_-}} \def\chp{\chi_{{}_+}} \def\rnt{\rho_{{}_{NT}}} \def\kbar{\hbox{$k$}\kern-3pt\raise 5pt\hbox{$--$}} \def\chk{\chi_{{}_k}} \def\bib#1#2{ \hbox to \hsize{\hbox to 0.5 in{\hss #1}% \hskip 0.2 in \vtop {\advance \hsize -0.8 in\noindent #2}\hss}} %% \def\vr{\vrule height 0 pt depth 0.6 pt width 0.75in} %% \def\\{ \hfill \break} \vfill\eject \pageno=1 \heading{\bf{INTRODUCTION}} \endheading \medskip In [FS2] we proved precise estimates for the eigenvalues $E_k$ and normalized eigenfunctions $\varphi_k$ of an ordinary differential operator $H=-\frac{d^2}{dx^2}+V(x)$ on an interval $I_{\text{BVP}}$. In this paper, we apply the results of [FS2] to study the density $$\rho(x)=\sum\limits_{E_k\le 0}|\varphi_k(x)|^2\, .\tag 1$$ \noindent We will introduce a simple approximation $\rho_{sc}(x)$, and estimate $\rho-\rho_{sc}$. Our goal is to prove the asymptotic formula announced in [FS1] for the ground--state energy of an atom. This requires estimates for $\rho-\rho_{sc}$ in weighted Sobolev norms of order $-1$, which we will derive here. The complete proofs of the results of [FS1] are given here and in the papers [FS2$\ldots$7]. The potentials of interest to us are large and slowly varying. A basic example is $$V(x)=\lambda^2V_1(x)\, ,\tag 2$$ \noindent with $V_1(x)$ a fixed, smooth function, and $\lambda$ a large parameter. We suppose $V_1(x)$ is defined on $[-1,1]$ and satisfies $$V_1(0)<0,\,\, V_1^\prime(0)=0,\,\, V_1^{\prime\prime}>c>0\ \text{on}\ [-1,1], \,\, \{V_1<0\}\subset\subset [-1,1]\, .\tag 3$$ \noindent For such potentials, a standard approximate formula for the density is $$\rho(x)\approx \tfrac 1\pi(-V(x))_+^{1/2}\, ,\tag 4$$ \noindent where $t_+^s=\cases t^s &\text{if$t>0$}\\ 0 &\text{if$t\le 0$}\endcases$. Unfortunately, this is too crude for our purposes. In place of (4), we will use the sharper approximation $$\rho_{sc}(x)=\frac 1\pi(-V(x))_+^{1/2}-(-V(x))_+^{-1/2}(\int_{I_{\text{BVP}}} (-V(y))_+^{-1/2}\,dy)^{-1}\chi_{{}_-}(\frac 1\pi \phi)\, , \quad\text{where}\tag 5$$ $$\phi=\frac \pi2+\int_{I_{\text{BVP}}}(-V(x))_+^{1/2}\,dx\, ,\quad\text{and} \tag 6$$ $$\chi_{{}_-}(t)=t-k-1/2\quad\text{for}\quad k= (\text{greatest\ integer}\ \le t)\, . \tag 7$$ \noindent We call (5) the semiclassical approximation for $\rho$, even though the name usually refers to (4). The role of the last term in (5) is perhaps obscure, but we will explain it later in the introduction. For potentials of the form (2), (3) our basic result on the density is as follows. \vglue 1pc \proclaim{Theorem 1} Suppose $V$ is given by (2), (3), and let $\phi$ be defined by (6). \roster \item"{(A)}" If the distance from $\frac \phi\pi$ to the nearest integer is at least $C\lambda^{-1}$, then \break $\Vert \rho-\rho_{sc}\Vert_{H^{-1}}\le C_\varepsilon^\prime\, \lambda^{\varepsilon-\frac 2{43}}$ for any $\varepsilon>0$. \item"{(B)}" If the distance from $\frac \phi\pi$ to the nearest integer is less than $C\,\lambda^{-1}$, then \break $\Vert\rho-\rho_{sc}\Vert_{H^{-1}}\le C^\prime$. \endroster \noindent Here $C$, $C^\prime$, $C_\varepsilon^\prime$ may depend on $V_1$ but not on $\lambda$. \endproclaim \medskip Note that $\Vert\rho_{sc}\Vert_{H^{-1}}\sim \lambda$. The exponent $\varepsilon-\frac {2}{43}$ in (A) is surely not optimal. All we need is some negative power of $\lambda$. The distinction between cases (A) and (B) above is easily explained. Theorem 1 is closely related to the WKB approximation, which says that the eigenvalues $E_k$ are given approximately as the solutions to $$\int_{I_{\text{BVP}}}(E_k-V(x))_+^{1/2}\,dx\approx\pi(k-1/2),\quad k\ge 1 \qquad (\text{See\ Erd\'elyi\ [E]}.)\tag 8$$ \noindent In particular, there is an eigenvalue $E_{k_0}$ close to zero if $\frac\phi\pi$ is near an integer. We cannot predict the sign of $E_{k_0}$, hence we don't know whether $|\varphi_{k_0}(x)|^2$ occurs in the sum (1). If $\frac\phi\pi$ is not near an integer, we meet no such difficulty. To understand atoms, we need to deal with potentials more general than (2). Already the hydrogen atom leads to $$V(x)=\frac{\ell(\ell+1)}{x^2}-\frac 1x+E^0\quad\text{on}\quad (0,\infty) \tag 9$$ \noindent which is certainly not given by (2), (3). In [FS2] we studied potentials $V(x)$ satisfying estimates $$\vert(\frac {d}{dx})^\alpha V(x)|\le C_\alpha S(x)(B(x))^{-\alpha}\quad \text{when}\quad x\in I\, ,\tag 10$$ \noindent for suitable weight functions $S(x)$ and $B(x)$. Here $I$ is an interval containing \break $\{V<0\}$. Hypothesis (10) lets us treat all the potentials we need, simply by picking the proper weight functions $S(x)$, $B(x)$ and interval $I$. For instance, when $V$ is given by (2), (3), then (10) holds with $S(x)=\lambda^2$, $B(x)=1$, $I=[-1,1]$. If $V$ is given instead by (9) for suitable $E^0>0$, then we take $S(x)=\frac 1x$, $B(x)=x$, $I=[\frac{\ell(\ell+1)}{2},\frac{2}{E^0})$. The main result of this paper is a version of Theorem 1 that applies to potentials satisfying (10). To state our main theorem, we need the analogue of hypothesis (3) in the setting of (10). We assume $V$ takes its minimum at $x_0\in I$ and satisfies $$\gather V(x_0)<-cS(x_0);\,\, V^{\prime\prime}(x)>cS(x_0)(B(x_0))^{-2}\quad\text{if} \quad |x-x_0|\le c_1B(x_0);\\ |V^\prime(x)|>cS(x)(B(x))^{-1}\quad\text{if}\quad x\in I,|x-x_0|\ge c_1B (x_0).\tag 11\endgather$$ \noindent We need also a number $\Lambda$ that plays the role of $\lambda$ in (2). It is given by $$\Lambda=\Bigl(\int_{V(x)<0}\frac{dx}{(S(x))^{1/2}(B(x))^2}\Bigr)^{-1}\, ,\tag 12$$ \noindent as explained in [FS2]. \noindent Finally, we need an $H^{-1}$--norm adapted to weight functions $S(x)$, $B(x)$. For a function $f$ on $I_{\text{BVP}}=(a,b)$ and a point $y\in I$, we set $$\Cal N(f,y)=\Bigl(\frac{1}{B(y)}\int_{|x-y|<\hat cB(y)}\,\Big|\int_a^xf(t)\,dt \Big|^2\,dx\Bigr)^{1/2}\tag 13$$ \noindent for a small constant $\hat c$. This is an $H^{-1}$--norm, since it is an $L^2$--norm of a primitive of $f$. Now we can state our main result. \vglue 1pc \proclaim{WKB Density Theorem} Let $V(x)$ satisfy (10), (11) and various technical conditions, and suppose $\Lambda$ is large. \roster \item"{(A)}" If the distance from $\frac\phi\pi$ to the nearest integer is at least $C\Lambda^{-1}$, then $$\multline \Cal N(\rho-\rho_{sc},y)\le C_\varepsilon\Lambda^{\varepsilon-\frac{45} {43}}\int_a^{y+2\hat cB(y)}(-V(x))_+^{1/2}\,dx+C_N\Lambda^{-N}\, \\ \text{for\ any\ positive}\ N\ , \varepsilon\ .\endmultline$$ \item"{(B)}" If the distance from $\frac\phi\pi$ to the nearest integer is less than $C\Lambda^{-1}$, then \endroster $$\Cal N(\rho-\rho_{sc},y)\le C\Lambda^{-1}\int_a^{y+2\hat cB(y)}(-V(x))_+^{1/2} \,dx+C_N\Lambda^{-N}\, .$$\endproclaim \bigskip For additional conclusions and a careful statement of all the hypotheses, we refer the reader to the relevant section of this paper. We prepare to sketch the proofs of our results on the density. For simplicity, we confine ourselves to Theorem 1. We start by describing how the eigenfunctions behave. Let $\{x\in I_{\text{BVP}}\mid V(x)c\lambda^2$ in $\text{supp}\ g(x,E)$. Then \align \rho(x,g)&=\sum\limits_{E_k\le 0}|\varphi_k(x)|^2g(x,E_k)\\ &\qquad\approx \sum\limits_{E_k\le 0}\tfrac 1\pi (\phi^\prime(E_k))^{-1}g(x,E_k)(E_k-V(x))^{-1/2}\cos^2(\eta(x,E_k))\\ &=\frac 1{2\pi}\sum\limits_{E_k\le 0}\frac{(\phi^\prime(E_k))^{-1} g(x,E_k)}{(E_k-V(x))^{1/2}}\negthinspace +\negthinspace \frac 1{2\pi}\sum\limits_{E_k\le 0} \frac{(\phi^\prime(E_k))^{-1}g(x,E_k)}{(E_k-V(x))^{1/2}} \cos(2\eta(x,E_k))\\ &\equiv F(x)+G(x)\, .\tag 22\endalign \noindent Note that the negative powers of $(E_k-V(x))$ pose no problem, since $g(x,E)$ is supported in region (I). We prove that $F(x)\approx \rho_{sc}(x,g)$ and that $G$ has negligibly small norm in $H^{-1}$. To estimate $G$ in $H^{-1}$--norm, we compute its indefinite integral. In general, when $f(x)$ is smooth and $\eta^\prime(x)$ is large, the indefinite integral of $f(x)\cos(2\eta(x))$ is approximately $\frac{f(x)}{2\eta^\prime(x)} \sin(2\eta(x))$. Applying this remark to the summands in the definition of $G$, we obtain the approximate formula $$\multline \Cal G(x)\approx \frac{1}{2\pi}\,\sum\limits_{E_k\le 0}\frac{(\phi^\prime(E_k))^{-1} g(x,E_k)}{(E_k-V(x))^{1/2}}\,\, \frac{\sin(2\eta(x,E_k))}{(2\tfrac{\partial\eta} {\partial x}(x,E_k))}\\ =\frac{1}{2\pi}\,\sum\limits_{E_k\le 0}\frac{(\phi^\prime(E_k))^{-1}g(x,E_k)} {2(E_k-V(x))}\sin(2\eta(x,E_k))\endmultline$$ \noindent for a primitive of $G$. The terms on the right are nearly orthogonal because of the rapidly oscillating factors $\sin(2\eta(x,E_k))$, so that $$\Vert G\Vert_{H^{-1}}^2\approx \Vert\Cal G\Vert_{L^2}^2\sim \sum\limits_{E_k\le 0}\int_{I_{\text{BVP}}}\frac{(\phi^\prime(E_k))^{-2} |g(x,E_k)|^2}{4(E_k-V(x))^2}\,dx\, .$$ \noindent The sum on the right is easily dominated by a negative power of $\lambda$, so $G$ is negligibly small in $H^{-1}$. To analyze $F(x)$ in (22), we introduce the function $E(t)$ that solves the equation $\phi(E)=t$. As $t$ increases from $t_{\text{min}}=1/2$ to $t_{\text{max}}=\phi(0)=\tfrac 1\pi\phi$, $E(t)$ increases from $(\min V)$ to $0$. The WKB approximation (8) says that $\phi(E_k)\approx k$, hence $E_k\approx E(k)$. So $F(x)$ is given approximately by $$\multline F(x)\approx\frac{1}{2\pi}\sum\limits_{k\in [t_{\text{min}},t_{\text{max}}]} \frac{g(x,E(k))E^\prime(k)}{(E(k)-V(x))^{1/2}} =\tfrac{1}{2\pi}\sum\limits_{k\in [t_{\text{min}},t_{\text{max}}]}f(k,x) \endmultline\tag 23$$ \noindent with $$f(t,x)=\frac{g(x,E(t))E^\prime(t)}{(E(t)-V(x))^{1/2}}\, .\tag 24$$ For fixed $x$, $f(t,x)$ is nearly constant as $t$ varies over an interval of length $O(1)$. Hence it is reasonable to replace the sum in (23) by an integral. This yields $$\multline F(x)\approx\frac 1{2\pi}\int_{t_{\min}}^{t_{\max}} \frac{g(x,E(t))E^\prime(t)} {(E(t)-V(x))^{1/2}}\,dt=\frac{1}{2\pi}\int_{\text{min}\ V}^0 \frac{g(x,E)\,dE}{(E-V(x))^{1/2}}\\ =\frac{1}{2\pi}\int_{-\infty}^0g(x,E)(E-V(x))_+^{-1/2}\,dE\endmultline\tag25$$ \noindent since $g$ is supported in region (I). The right--hand side of (25) is the crude form of the semiclassical approximation to $\rho(x,g)$. The most serious error in the derivation of (25) comes from replacing the Riemann sum (23) by an integral. To remedy it, we study $\sum\limits_{k\in [a,b]}f(k)-\int_a^bf(t)\,dt$ for general slowly varying functions $f$. We find easily that $$\sum\limits_{k\in [a,b]} f(k)\approx \int_a^{b} f(t)dt-\chi_-(b)f(b)- \chi_+(a)f(a)\, ,\tag 26$$ \noindent with $\chi_{{}_-}$ as in (7), and $\chi_{{}_+}(t)=k-t-1/2$ for $k=\ (\text{least\ integer}\ \ge t)$. The last terms in (26) compensate for the fact that small changes in $a$ and $b$ typically change the integral in (26) but not the sum. Using (26) to evaluate the right--hand side of (23), we obtain \align F(x)&\approx \frac1{2\pi}\int_{t_{\text{min}}}^{t_{\text{max}}} \frac{g(x,E(t))E^\prime(t)}{(E(t)-V(x))^{1/2}}\,dt -\tfrac 1{2\pi} \chi_{{}_-}(t_{\text{max}})\frac{g(x,E(t_{\text{max}}))E^\prime(t_{\text{max}})} {(E(t_{\text{max}})-V(x))^{1/2}}\\ &\qquad\qquad -\frac {1}{2\pi}\chi_{{}_+}(t_{\text{min}})\frac {g(x,E(t_{\text{min}}))E^\prime(t_{\text{min}})} {(E(t_{\text{min}})-V(x))^{1/2}}\\ &=\frac{1}{2\pi}\int_{-\infty}^0g(x,E)(E-V(x))_+^{-1/2}dE\\ &\qquad\qquad-\chi_{{}_-}(\frac 1\pi\phi)g(x,0)(-V(x))_+^{-1/2} (\int_{I_{\text{BVP}}}(-V(y))_+^{-1/2}\,dy)^{-1}\, ,\tag27\endalign \noindent since $t_{\text{max}}=\frac1\pi\phi$, $E(t_{\text{max}})=0$, $E^\prime(t_{\text{max}})=(\phi^\prime(0))^{-1}= (\frac{1}{2\pi}\int_{I_{\text{BVP}}}(-V(y))_+^{-1/2}\,dy)^{-1}$ and $g$ is supported in region (I). Recalling that $G$ is negligibly small in $H^{-1}$, we conclude from (22) and (27) that $$\multline \rho(x,g)\approx F+G\approx \tfrac{1}{2\pi}\int_{-\infty}^0g(x,E) (E-V(x))_+^{-1/2}\,dE-\chi_{{}_-}(\tfrac\phi\pi)\frac{g(x,0)(-V(x))_+^{-1/2}} {\int_{I_{\text{BVP}}}(-V(y))_+^{-1/2}\,dy}\\ \equiv \rho_{\text{sc}}(x,g)\, .\endmultline\tag28$$ Thus we have shown that $\rho(x,g)\approx \rho_{sc}(x,g)$ when $g$ is supported in region (I). It is equation (28) that teaches us the corrected definitions (5), (21) for the semiclassical density. Next we study $\rho(x,g)$ when $g$ is supported in region (II). To deal with this case, we need to know something about the Airey function" $A(t)$. First of all, $$A(t)\sim t_+^{-1/4}\cos(\frac 23t^{3/2}-\frac\pi4 )\qquad \text{for}\quad |t|>>1\, .\tag 29$$ \noindent In particular, $A(t)$ oscillates more and more rapidly as $t\to +\infty$. At a crucial point we will use the following estimate from our previous paper [FS2]. $$X\equiv \int_{-\infty}^\infty\theta(\frac tR)[A^2(t)-\frac 12 t_+^{-1/2} ]\,dt=O(R^{-5/2})\quad\text{for}\quad R>>1\quad\text{and}\quad \theta\in C_0^\infty\, .\tag 30$$ \noindent Note that (29) yields only $X=O(1)$. Now we can start computing $\rho(x,g)$ in case (II). From (II) and the definition of the microlocalized density, we have \align \rho(x,g)&\approx \frac 1\pi \sum\limits_{E_k\le 0}g(x,E_k)(\phi^\prime (E_k))^{-1}\lambda^{-2/3}\Bigl(\frac{\partial y}{\partial x}(x,E_k)\Bigr)^{-1} A^2(\lambda^{2/3}y(x,E_k))\\ &\approx \frac 1\pi \sum\limits_{k\in [t_{\text{min}},t_{\text{max}}]} f(k,x)\, ,\qquad with\tag 31\endalign $$f(t,x)=g(x,E(t))E^\prime(t)\Bigl(\frac{\partial y}{\partial x}(x,E(t))\Bigr)^{-1} \lambda^{-2/3}A^2(\lambda^{2/3}y(x,E(t)))\, .\tag 32$$ \noindent Since $g(x,E)$ is supported in region (II), $|x-x_{\text{left}}(E(t))|$ is small in $\text{supp}\ f(t,x)$. Hence $|y(x,E(t))|$ is small also, as we see from (18). Consequently, $\lambda^{2/3}y(x,E(t))$ is not too large, so $A^2(\lambda^{2/3}y(x,E(t)))$ does not oscillate too rapidly in $\text{supp}\ f(t,x)$. It follows that $f(t,x)$ is nearly constant when $x$ stays fixed and $t$ varies by $O(1)$. Thus we may approximate the sum in (31) by appealing to our previous result (26) on general Riemann sums. We obtain \align \rho(x,g)&\approx \frac 1\pi\int_{t_{\text{min}}}^{t_{\text{max}}} \frac{g(x,E(t))E^\prime(t)\lambda^{-2/3}}{(\frac{\partial y}{\partial x} (x,E(t)))}\, A^2(\lambda^{2/3}y(x,E(t)))\,dt\\ &\qquad-\frac{2\chi_{{}_-}(\frac \phi\pi)g(x,0)(\int_{I_{\text{BVP}}} (-V(y))_+^{-1/2}\,dy)^{-1}} {\lambda^{2/3}(\frac{\partial y}{\partial x}(x,0))}\,A^2(\lambda^{2/3} y(x,0))\, ,\tag 33\endalign \noindent since $t_{\text{max}}=\frac \phi\pi$, $E(t_{\text{max}})=0$, $t_{\text{min}}=1/2$, $\chi_{{}_+}(t_{\text{min}})=\chi_{{}_+}(1/2)=0$. Changing variable in the integral in (33), we get \align \rho(x,g)&\approx\frac 1\pi\int_{\text{min}\ V}^{0} \frac{g(x,E)}{\lambda^{2/3}(\frac{\partial y}{\partial x}(x,E))}\,A^2 (\lambda^{2/3}y(x,E))dE\\ &\qquad -\frac{2\chi_-(\frac \phi\pi)g(x,0)}{\lambda^{2/3}(\frac{\partial y} {\partial x}(x,0))}\,\Bigl(\int_{I_{\text{BVP}}}(-V(y))_+^{-1/2} \,dy\Bigr)^{-1}\,A^2(\lambda^{2/3}y(x,0))\, .\tag34\endalign Next, we want to replace $A^2(\lambda^{2/3}y(x,E))$ by $\frac 12(\lambda^{2/3} y(x,E))_+^{-1/2}$ in (34). This is plausible, since (29) gives $$\multline A^2(\lambda^{2/3}y(x,E))\approx \frac 12(\lambda^{2/3}y(x,E))_+^{-1/2}\\ +\frac 12(\lambda^{2/3}y(x,E))_+^{-1/2}\cos(\frac 43\lambda y^{3/2} (x,E)-\frac \pi 2)\, .\endmultline$$ \noindent The last term oscillates rapidly about zero, so it should have small norm in $H^{-1}$ and a small effect on the integral in (34). To replace $A^2(\lambda^{2/3}y(x,E))$ by $\frac 12(\lambda^{2/3}y(x,E))_+^{-1/2}$ with a small error, we study integrals of the form $$F_R(x)=\int_{-\infty}^x\theta(\frac xR,\frac tR)[A^2(t)-\frac 12 t_+^{-1/2}]\,dt,\quad\text{for}\quad \theta\in C_0^\infty\quad\text{and} \quad R>>1\, .\tag 35$$ \noindent When $x$ is bounded or large negative, (35) is easily understood in terms of the asymptotics of $A(t)$. Unfortunately, when $x$ is large positive, the region of integration in (35) extends over $\{|t|\le C\}$, where $A^2(t)-\frac 12(t)_+^{-1/2}$ is neither small nor rapidly oscillating. To overcome this difficulty, we appeal to (30) and write $$F_R(x)=\int_{-\infty}^\infty\theta(\frac xR, \frac tR)[A^2(t)-\frac 12 t_+^{-1/2}]\,dt-\int_x^\infty\theta(\frac xR,\frac tR) [A^2(t)-\frac 12t_+^{-1/2}]\,dt\, .$$ \noindent The first term on the right is $O(R^{-5/2})$ by (30), and the second integral is easily analyzed in terms of the asymptotics of $A(t)$ when $x$ is large positive. Hence we can prove sharp bounds for $|F_R(x)|$ for all $x$. We are ready to replace $A^2(\lambda^{2/3}y(x,E))$ by $\frac 12(\lambda^{2/3} y(x,E))_+^{-1/2}$ in (34). This leads to errors \align G(x)&=\int_{E<0}\frac{g(x,E)}{\lambda^{2/3}(\frac{\partial y}{\partial x} (x,E))}\,[A^2(\lambda^{2/3}y(x,E))-\frac 12\,(\lambda^{2/3}y(x,E))_+^{-1/2}]\,dE \qquad\text{and}\\ H(x)&=\frac{g(x,0)}{\lambda^{2/3}(\frac{\partial y}{\partial x}(x,0))} \,[A^2(\lambda^{2/3}y(x,0))-\frac 12(\lambda^{2/3}y(x,0))_+^{-1/2}]\, ,\endalign \noindent which we have to estimate in the $H^{-1}$--norm. The primitives of $G$ and $H$ are $$\Cal G(z)=\operatornamewithlimits{\int} \Sb E<0\\ xx_+}\\ O(\sum\limits_k|g(x_0,E_k)|) &\text{for z\in J}.\endcases\tag 40$$ \noindent Equation (40) holds also for $\rho_{sc}(x,g)$, and therefore $$\Cal G(z)\equiv \int\limits_{xFrom (5), (6), (7) we obtain easily that \int_{I_{\text{BVP}}}\rho_{sc}(x)dx is the greatest integer in \frac\phi\pi. According to the WKB approximation (8), this is precisely the number of non--positive eigenvalues, provided \frac\phi\pi isn't too near to an integer. If we had used the standard formula (4) as our definition of \rho_{sc}, then \int_{I_{\text{BVP}}} \rho_{sc}(x)dx would differ from \int_{I_{\text{BVP}}}\rho(x)dx by an error of the order of magnitude 1. Thus the last term in (5) contributes to the accuracy of the approximation. We thank Maureen Schupsky for exceptional skill and patience in Texing our manuscript. \vfill\eject \head {\bf{REVIEW OF EARLIER RESULTS}} \endhead \medskip For the reader's convenience, we gather here those results from [FS2] which we require for our present work. \subhead The Airey Function \endsubhead We work with a variant of the standard Airey function. Our A(t) is a real--valued solution of (\frac{d^2}{dt^2}+t)A(t)=0, satisfying:$$ \Big\vert\Bigl(\frac{d}{dt}\Bigr)^m\,A(t)\Big|\le C_{mK}(1+|t|)^{-K}\quad \text{as}\quad t\to -\infty,\quad \text{for\ any}\ m\ \text{and}\ K\, ; \tag A1  A(t)\sim \text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4+\frac 23 t^{3/2}i}} {t^{1/4}} \Bigl(1+\sum\limits_{s=1}^\infty\frac{c_s}{t^{\frac {3}{2}s}}\Bigr)\Bigr] \quad\text{as}\quad t\to +\infty\ , \tag A2 $$\noindent in the following sense. Given m and K, we can find M so that$$ \Bigl(\frac{d}{dt}\Bigr)^\ell\Bigl\{A(t)-\text{Re}\Bigl[\frac {e^{\pm i\frac\pi 4+\frac 23t^{3/2}i}}{t^{1/4}}(1+\sum\limits_{s=1}^M \frac{c_s}{t^{\frac 32 s}})\Bigr]\Bigr\}=O(t^{-K}) $$\noindent for 0\le \ell\le m and t>1. We have c_1=\pm \frac{5i}{48}. \smallskip \noindent(A3) Suppose |(\frac{d}{dt})^m\theta(t)|\le C_m R^{-m} for all m, with R\ge 10. Assume also that \theta(t) has compact support. Then$$ \Big|\int_{-\infty}^\infty \theta(t)\bigl[A^2(t)-\frac 12 t_+^{-1/2}\bigr]\ dt\Bigr|\le C^\prime R^{-5/2}, $$\noindent where C^\prime depends only on finitely many of the C_m. \noindent (This is immediate from Lemmas 4 and 7 in the section on normalizing eigenfunctions in [FS2]. Recently, we have found a much simpler proof of this fact.) \vglue 1pc \subhead The WKB Theorems\endsubhead \smallskip \noindent\demo{Set-up} We are given the following: A potential V(x) defined on a (possibly unbounded) interval I_{\text{BVP}}; two positive functions S(x), B(x) defined on a subinterval I \subset I_{\text{BVP}}; two real numbers E_0\le E_\infty; positive numbers \varepsilon<\frac{1} {100}, K>1 and N>K\varepsilon^{-10}. We define N^\prime=[\varepsilon N/500] and N^{\prime\prime}=\frac 32 \varepsilon N^\prime-K-33.\enddemo Our goal is to understand the eigenvalues and eigenfunctions of the self--adjoint operator H=\frac{-d^2}{dx^2}+V(x) on L^2(I_{\text{BVP}}), with Dirichlet or Neumann conditions at the endpoints. \demo{Hypotheses} \noindent {\it Assumptions on V(x), S(x), B(x) in I\/} \roster \item"(Hyp0)" If x,y \in I and |x-y|cB(x_{\text{left}}), \text{dist}(x_{\text{rt}},\partial I)>cB(x_{\text{rt}}). \item"(Hyp3)" For x_{\text{left}}\le x\le x_{\text{left}}+c_1 B(x_{\text{left}}) we have -V^\prime(x)>cS(x_{\text{left}})B^{-1} (x_{\text{left}}), and for x_{\text{rt}}-c_1B(x_{\text{rt}})\le x\le x_{\text{rt}} we have +V^\prime(x)>cS(x_{\text{rt}})B^{-1}(x_{\text{rt}}). \item"(Hyp4)" For x_{\text{left}}+c_1B(x_{\text{left}})\le x\le x_{\text{rt}}-c_1B(x_{\text{rt}}) we have cS(x)E \quad for all\hfill\break x \in I_{\text{BVP}}\backslash [x_{\text{left}}(E), x_{\text{rt}}(E)]. \item"{(Hyp6)}" If x \in I_{\text{BVP}} satisfies xx_{\text{rt}}+\frac 12 \lambda_{\text{rt}} ^KB_{\text{rt}}, then V(x)\ge E_\infty+\frac{100}{|x-x_{\text{rt}}|^2}. \endroster\enddemo \medskip \noindent\demo{Technical Assumptions} \roster \item"{(Hyp7)}" \operatornamewithlimits{max}_{x \in I} S(x)\le \lambda_{\text{left}}^KS_{\text{left}} and \operatornamewithlimits {max}_{x \in I}S(x)\le \lambda_{\text{rt}}^KS_{\text{rt}} \item"{(Hyp8)}" \int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx} {S^{1/2}(x)}\le \Lambda^K\cdot\min(S_{\text{left}}^{-1/2}B_{\text{left}}, S_{\text{rt}}^{-1/2}B_{\text{rt}}) \item"{(Hyp9)}" [\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx} {S^{1/2}(x)B^{4}(x)}]\cdot[\int_{x_{\text{left}}}^{x_{\text{rt}}} \frac{dx}{S^{1/2}(x)}]\le \Lambda^K \endroster\enddemo \medskip \noindent \demo{WKB Condition} \roster \item"{(Hyp10)}" \Lambda is bounded below by a positive constant depending only on \varepsilon, K, N,\hfill\break and on c, C, c_1, c_2, C_\alpha in (Hyp0)\ldots(Hyp4). \endroster\enddemo \smallskip \noindent\demo{Definitions and Basic Properties of Phases} Assume hypotheses (Hyp0)\ldots(Hyp10). For |E-E_0|0}\frac {dx} {(E-V(x))^{1/2}}\Bigr)-1\Big|\le C_{\#}\Lambda^{3\varepsilon-2}\quad \text{and}$$ $$\Big|b_{\text{rt}}^2\cdot \Bigl(\frac 12 \int_{E-V(x)>0}\frac {dx} {(E-V(x))^{1/2}}\Bigr)-1\Big|\le C_{\#}\Lambda^{3\varepsilon-2},$$ \noindent with $C_{\#}$ depending only on $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$ in (Hyp 0)$\ldots$(Hyp 5).\endproclaim \vglue 1pc \subhead The WKB Theorem on Low Eigenvalues \endsubhead \smallskip Let $\varepsilon,K,N>0$ be given, with $\varepsilon N\ge 100$. Let $V(x)$ be a potential defined on a (possibly unbounded) interval $I_{\text{BVP}}$. Let $S$, $B$ be positive numbers, and let $x_0 \in I_{\text{BVP}}$ be given. Define $\lambda=S^{1/2}B$. Let $E_\infty$ be a given energy, with $E_\infty>V(x_0)$. We make the following assumptions. \roster \item"{(H0$^*$)}" $I=\{|x-x_0|\frac 12 \lambda^KB$, we have $V(x)\ge E_\infty +\frac{1000}{|x-x_0|^2}$. \item"{(H6$^*$)}" $\lambda$ is bounded below by a positive constant depending only on $c$, $c^\prime$, $c^{\prime \prime}$, $C_\alpha$ in (H0$^*$) and (H1$^*$), and on $\varepsilon$, $K$, $N$.\endroster Let $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_{\text{BVP}})$ with Dirichlet or Neumann conditions at the endpoints. For V(x_0)\delta \endSb V^{\prime\prime}(x)(E-V(x))^{-3/2}\,dx -q(E)\delta^{-1/2}\Bigr]\\ &=\lim_{\delta_{\text{left}},\delta_{\text{rt}}\to 0+} \Bigl[\int_{x_{\text{left}}(E)+\delta_{\text{left}}} ^{x_{\text{rt}}(E)-\delta_{\text{rt}}} V^{\prime\prime} (x)(E-V(x))^{-3/2}\,dx-q_{\text{left}}(E)\delta_{\text{ left}}^{-1/2}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\,\,\,\,\, -q_{\text{rt}}(E)\delta_{\text{rt}}^{-1/2} \Bigr]\endalign $$\noindent with q(E), q_{\text{left}}(E), q_{\text{ rt}}(E) uniquely specified by demanding the finiteness of the limits. \vglue 1pc \proclaim{Lemma 1} The phases \phi(E), \psi(E) satisfy the estimates$$ \Big|\Bigl(\frac{d}{dE}\Bigr)^\beta \phi(E)\Big|\le C_{\#}^\beta\lambda S^{-\beta}  \Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\psi(E)\Big|\le C_{\#}^\beta\lambda^{-1}S^{-\beta}  \frac{d}{dE}\phi(E)\ge c_{\#}\lambda S^{-1} $$\noindent for V(x_0)C_{\#}\lambda^{-2+4\varepsilon}, then k_{\text{max}}=\overline n. In any case, |k_{\text{max}}-\overline n|\le 1. \item"{(b)}" If 0\le kcB(x). \item"{(H\hat 1)}" For x \in I_{\text{center}} we have |(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x) B^{-\alpha}(x) and cS(x)0 defined on I. Set \lambda(x)=S^{1/2}(x)B(x). We are given an energy E_0. \enddemo We make the following assumptions. \demo{Hypotheses} \roster \item"{(H\overline 0)}" For x,y \in I with |x-y|cB(x). \item"{(H\overline 1)}" For x \in I we have |(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x)B^{-\alpha}(x). \item"{(H\overline 2)}" \{x \in I_{\text{BVP}}\mid V(x)cB(x_{\text{left}}), \text{dist}(x_{\text{right}},\partial I)>cB(x_{\text{right}}). \item"{(H\overline 3)}" In [x_{\text{left}},x_{\text{left}}+c_1 B(x_{\text{left}})] we have -V^\prime(x)\ge c S(x_{\text{left}})/ B(x_{\text{left}}), and in [x_{\text{right}}-c_1B(x_{\text{right}}), x_{\text{right}}] we have +V^\prime(x)\ge cS(x_{\text{right}})/B (x_{\text{right}}). \item"{(H\overline 4)}" In [x_{\text{left}}+c_1B(x_{\text{left}}), x_{\text{right}}-c_1B(x_{\text{right}})] we have cS(x)\xi+\tau. Then we have k=\ (\text{greatest\ integer}\ \le \xi+\tau). On the other hand, if k+1\le \xi+\tau then since \xi-\tau\le k+1\le \xi+\tau with 0<\tau<\frac {1}{10}, we have k=\ (\text{greatest\ integer}\ \le \xi-2\tau). Thus, (20) holds in either case. Now define a set$$\split \Cal K=\{k\in \Bbb Z\mid k_{\min}\le k\le k_{\max}\ \text{and}\ E_k\ \text{is\ an\ eigenvalue\ of}\ H\\ \text{belonging\ to}\ [E_{\ell o},E_{hi}]\}\ .\endsplit\tag 21 $$\noindent Note that \Cal K contains at least c_{\#}\Lambda integers, by (1), (10) and (19). (Here we take c_{\#}^1 in (1) to be smaller than \frac 14 c_{\#} in (10).) Also, note that$$ k_1k_{\ell o}+c_{\#}\Lambda\ .\tag 27 $$\noindent Also, comparing (21) with (26), we get$$ k_{\min}\le k_{\ell o}\le k_{hi}\le k_{\max}\ .\tag 28 $$\noindent From (10) and (1), we know that every eigenvalue of H that lies in [E_{\ell o}, E_{hi}] is equal to E_{\tilde k} for some \tilde k (k_{\min}\le \tilde k\le k_{\max}). The definition (21) then shows that \tilde k\in \Cal K. So the set of eigenvalues of H in [E_{\ell o},E_{hi}] is contained in \{E_k\mid k\in \Cal K\}. The converse inclusion is immediate from (21), so the eigenvalues of H that lie in [E_{\ell o}, E_{hi}] are precisely the E_k for k\in \Cal K. In view of (26), this means that the eigenvalues of H that lie in [E_{\ell o},E_{hi}] are precisely the E_k for k_{\ell o}\le k\le k_{hi}, as in the statement of the Reformulated Eigenvalue Theorem. Moreover, (2) follows from (27); and (3) follows from (9), (11) and the estimate$$ |\phi_{\text{error}}(E)|\le C_{\#}\Lambda^{-2}\ , \text{valid\ for}\ E\in [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]\ .\tag 29 $$\noindent Estimate (29) is merely the special case \beta=0 of (12). So the only assertions of the Reformulated Eigenvalue Theorem that remain to be proved are (4) and (5). To establish (4) and (5), the main step is to show that$$ k_{hi}= (\, \text{greatest\ integer}\ \le \frac 1\pi\phi (E_{hi})+\frac{1}{48\pi}\psi(E_{hi})-\frac 12+\tilde\omega)\tag 30 $$\noindent for an \tilde\omega with$$ |\tilde\omega|\le C_{\#}\Lambda^{-2}\ .\tag 31 $$\noindent To prove (30) and (31), we distinguish three cases. \smallskip \demo{Case A} k_{hi}=k_{\max}\enddemo \smallskip \demo{Case B} k_{hi}From (6) we see that |\Phi(E)-\pi(k_{\max}+1)|\ge C_{\#} \Lambda^{-N^{\prime\prime}} for all E\in (-\infty,0]\cap [E_0-\frac 14 c_{\#}S_{\min}, E_0+\frac 14 c_{\#}S_{\min}]\equiv [E_-,E_+]. That is,$$ \pi(k_{\max}+1)\ \ \text{has\ distance\ at\ least}\ \ C_{\#}\Lambda^{-N^{\prime \prime}}\ \ \text{from}\ \ [\Phi(E_-),\Phi(E_+)\ ] \ .\tag 32 $$\noindent (Recall that \Phi is increasing on an interval containing [E_-,E_+], so [\Phi(E_-), \Phi(E_+)] is the image of [E_-,E_+] under \Phi.) On the other hand, another application of (6) shows similarly that$$ \pi k_{\max}\ \text{has\ distance\ less\ than}\ \ C_{\#}\Lambda^{-N^{\prime \prime}}\ \text{from}\ \ [\Phi(E_-), \Phi(E_+)]\ .\tag 33 $$\noindent The constant C_{\#} is the same in (32) and (33). Hence (32) and (33) imply$$ \pi(k_{\max}+1)\ge \Phi(E_+)+C_{\#}\Lambda^{-N^{\prime\prime}}\ .\tag 34 $$\noindent Comparing (1) with the definition of [E_-, E_+] and taking c_{\#}^1 small enough, we see that E_{hi}\le E_+, and therefore \Phi(E_{hi})\le \Phi(E_+). So (34) implies$$ \pi(k_{hi}+1)\ge \Phi(E_{hi})+C_{\#}\Lambda^{-N^{\prime\prime}}\ ,\text{ since\ we're\ assuming}\ \ k_{\max}=k_{hi}\ .\tag 35 $$\noindent On the other hand, (21) and (26) show that E_{k_{hi}}\in [E_{\ell o},E_{hi}]. In particular, E_{k_{hi}}\le E_{hi}, so \Phi(E_{k_{hi}})\le \Phi (E_{hi}). Hence (9) implies$$ \pi k_{hi}\le \Phi(E_{hi})+C_{\#}\Lambda^{-N^{\prime\prime}}\ .\tag 36 $$In view of (11) and (29) estimates (35) and (36) imply$$ k_{hi}+1\ge \frac 1\pi \phi(E_{hi})+\frac{1}{48\pi}\psi(E_{hi})-\frac 12 -C_{\#}^\prime\Lambda^{-2}\ \ \text{and} \tag 37  k_{hi}\le \frac 1\pi \phi(E_{hi})+\frac 1{48\pi}\psi(E_{hi})-\frac 12+ C_{\#}^\prime\Lambda^{-2}\ . \tag 38 $$The desired conclusions (30) and (31) follow at once from (37), (38) and (20). So we have proven (30) and (31) in Case A. Next we verify (30) and (31) in Case B. Thus we assume k_{hi}E_{hi}. Therefore \Phi(E_{k_{hi}+1})> \Phi(E_{hi}) since \Phi is strictly increasing on [E_0-c_{\#} S_{\min}, E_0+c_{\#}S_{\min}]. Applying (9), we obtain$$ \pi(k_{hi}+1)>\Phi(E_{hi})-C_{\#}\Lambda^{-N^{\prime\prime}}\ .\tag 42 $$On the other hand, (41) yields \Phi(E_{k_{hi}})\le \Phi(E_{hi}) since \Phi is increasing, and therefore (9) implies$$ \pi k_{hi}\le \Phi(E_{hi})+C_{\#}\Lambda^{-N^{\prime\prime}}\ .\tag 43 $$\noindent From (42), (43) and (29), we conclude that$$ k_{hi}+1\ge \frac 1\pi \phi(E_{hi})+\frac {1}{48\pi} \psi(E_{hi})-\frac 12 -C_{\#}^\prime\Lambda^{-2}\ \ \text{and}  k_{hi}\le \frac 1\pi \phi(E_{hi})+\frac{1}{48\pi}\psi(E_{hi})-\frac 12 +C_{\#}^\prime\Lambda^{-2}\ . $$\noindent These two inequalities and (20) show that (30), (31), hold in Case C. Thus, we have verified (30), (31) in all three cases A, B, C, so we know that (30), (31) hold always. Note that under the additional assumption$$\split \min_{k\in \Bbb Z}\Big|\bigl(\frac 1\pi \phi(E_{hi})+\frac 1{48\pi} \psi(E_{hi})-\frac 12\bigr)-k\Big|>C_{\#}\Lambda^{-2}\\ \text{with}\ \ C_{\#}\ \ \text{as\ in (31)}\ ,\endsplit $$\noindent we obtain$$\split k_{hi}= (\, \text{greatest\ integer}\ \le \frac 1\pi \phi(E_{hi})+\frac{1}{48\pi}\psi(E_{hi})-\frac 12)\\ \text{as\ a\ consequence\ of\ (30), (31)}\ .\endsplit $$\noindent Therefore, we have proven (30), (31) in the following stronger form:$$ k_{hi}=(\, \text{greatest\ integer}\ \le \frac 1\pi \phi(E_{hi})+ \frac{1}{48\pi}\psi(E_{hi})-\frac 12 +\omega_{hi})\ , \text{with} \tag 44  |\omega_{hi}|\le C_{\#}^\prime\Lambda^{-2}\ ,\ \ \text{and}\tag 45  \omega_{hi}=0\ \ \text{if}\ \ \min_{k\in \Bbb Z} \big|\bigl(\frac 1\pi \phi(E_{hi})+\frac{1}{48\pi}\psi(E_{hi})-\frac 12\bigr) -k\Big|\ge C_{\#}^\prime \Lambda^{-2}\ . \tag 46 $$We turn our attention to k_{\ell o}. Applying (20) to -\xi, -k, \tau in place of \xi, k, \tau, we get the following trivial observation.$$\multline \text{Suppose}\ k \in \Bbb Z, \xi\in \Bbb R\ \text{and}\ 0<\tau\le \frac {1} {10}.\ \text{If}\ k-1\le \xi+\tau\ \text{and}\ k\ge \xi-\tau\ ,\\ \text{then}\ k=(\, \text{least\ integer}\ \ge \xi+\omega)\ \text{for\ some}\ \omega\ \text{with}\ |\omega|\le 2\tau\ .\endmultline\tag 47 $$\noindent We shall use (47) to prove$$ k_{\ell o}=(\, \text{least\ integer}\ \ge \frac 1\pi \phi(E_{\ell o})+ \frac{1}{48\pi}\psi(E_{\ell o})-\frac 12+\omega_{\ell o})\ ,\tag 48 $$\noindent with$$ |\omega_{\ell o}|\le C_{\#}^\prime \Lambda^{-2}\ .\tag 49 $$To prove (48), (49), we first note that$$ k_{\min}c_{\#}^\prime S_{\min}), we find that there are eigenvalues $\tilde E$ of $H$ that belong to $J$. >From (10) we get $\tilde E=E_{\tilde k}$ for $k_{\min}\le \tilde k \le k_{\max}$. Thus, $E_{\tilde k}\in J$. On the other hand, $E_{k_{\ell o}} \in [E_{\ell o},E_{hi}]$ by (21), (26). Since $J$ lies to the left of $[E_{\ell o}, E_{hi}]$, it follows that $E_{\tilde k}From (71) and (72) we get$\phi(E_{\ell o})<\phi(\overline E)<\phi(E_{hi})$. Thus, $$\int_{I_{\text{BVP}}}(E_{\ell o}-V(x))_+^{1/2}\, dx <\int_{I_{\text{BVP}}}(\overline E-V(x))_+^{1/2}\ dx< \int_{I_{\text{BVP}}}(E_{hi}-V(x))_+^{1/2}\ dx\ ,$$ \noindent which implies$\overline E\in [E_{\ell o},E_{hi}]$, completing the proof of {\it CLAIM A\/}. The next step is to prove the following. \medskip \noindent{\it CLAIM B\/}: If$E$,$E^\prime$are distinct eigenvalues of$H$belonging to$[E_{\ell o},E_{hi}]$, then$|\phi(E)-\phi(E^\prime)|>\frac {1}{100}$.\smallskip To see this suppose instead$|\phi(E)-\phi(E^\prime)|\le \frac{1}{100}$. We may assume$E^\primeFrom (87), (88) we see that $$\split \Cal K\subset \Bbb Z\cap [\frac 1\pi \phi(E_{\ell o})-\frac 12 +\tilde\omega_{\ell o}, \frac 1\pi \phi(E_{hi})-\frac 12+\tilde\omega_{hi}]\ ,\\ \text{which\ is\ half\ of\ (86)}\ .\endsplit\tag 89$$ \noindent Now suppose $\kbar \in \Bbb Z\cap [\frac 1\pi\phi(E_{\ell o}) -\frac 12+\tilde\omega_{\ell o}, \frac 1\pi\phi(E_{hi})-\frac 12+\tilde\omega _{hi}]$. Then either $$\frac 1\pi\phi(E_{\ell o})-\frac 12+\tilde\omega_{\ell o}\le \kbar<\frac 1\pi\phi(E_{\ell o})-\frac 12+C_{\#}^\prime\Lambda_{\min}^{-1},\ \text{or} \tag 90$$ $$\frac 1\pi \phi(E_{\ell o})-\frac 12+C_{\#}^\prime\Lambda_{\min}^{-1}\le \kbar \le \frac 1\pi \phi(E_{hi})-\frac 12-C_{\#}^\prime\Lambda_{\min}^{-1},\ \text{or} \tag 91$$ $$\frac 1\pi \phi(E_{hi})-\frac 12-C_{\#}^\prime\Lambda_{\min}^{-1}<\kbar \le \frac 1\pi \phi(E_{hi})-\frac 12+\tilde\omega_{hi}\ .\tag 92$$ \noindent If (90) holds, then $\tilde\omega_{\ell o}\ne C_{\#}^\prime\Lambda_{\min}^{-1}$, so (81), (82) show that $\tilde\omega_{\ell o}=-C_{\#}\Lambda_ {\min}^{-1}$ and that there is some $k\in \Cal K$ satisfying $$\frac 1\pi \phi(E_{\ell o})-\frac 12-C_{\#}\Lambda_{\min}^{-1}\le k\le \frac 1\pi \phi(E_{\ell o})-\frac 12+C_{\#}^\prime\Lambda_{\min}^{-1}\ .\tag 93$$ \noindent From (90), (93) we get $|k-\kbar|\le (C_{\#}+C_{\#}^\prime)\Lambda _{\min}^{-1}<<1$. Since $k$ and $\kbar$ are both integers, it follows that $\kbar=k$. Hence $\kbar$ belongs to $\Cal K$. If (91) holds, then $\kbar$ belongs to $\Cal K$, by (79). If (92) holds, then $\tilde\omega_{hi}\ne -C_{\#}^\prime\Lambda_{\min}^{-1}$, so (83), (84) show that $\tilde\omega_{hi}=C_{\#}\Lambda_{\min}^{-1}$, and there is some $k\in \Cal K$ satisfying $$\frac 1\pi \phi(E_{hi})-\frac 12-C_{\#}^\prime\Lambda_{\min}^{-1}\le k\le \frac 1\pi \phi(E_{hi})-\frac 12+C_{\#}\Lambda_{\min}^{-1}\ .$$ \noindent This and (92) yield $|\kbar-k|\le (C_{\#}+C_{\#}^\prime)\Lambda _{\min}^{-1}<<1$. Since $k$ and $\kbar$ are both integers, it follows that $\kbar=k$. Hence $\kbar\in \Cal K$. Thus, in all three cases (90), (91), (92) we have $\kbar \in \Cal K$. So we have proven $$\Bbb Z\cap [\frac 1\pi \phi(E_{\ell o})-\frac 12+\tilde\omega_{\ell o}, \frac 1\pi \phi(E_{hi})-\frac 12+\tilde\omega_{hi}]\subset \Cal K\ .$$ \noindent This together with (89) completes the proof of (86). Now define $\omega_{\ell o}$, $\omega_{hi}$ by setting $$\split \omega_{\ell o}=\tilde\omega_{\ell o},\ \omega_{hi}=0\ \text{if}\ \min_{k\in \Bbb Z}|\frac 1\pi \phi(E_{hi})-\frac 12-k|\ge 2C_{\#}^\prime \Lambda_{\min}^{-1}\ ,\\ \omega_{hi}=\tilde\omega_{hi}\ \text{otherwise}\ .\endsplit$$ \noindent Evidently, $$\multline \Bbb Z\cap\ [\frac 1\pi \phi(E_{\ell o})-\frac 12+\omega_{\ell o},\frac 1\pi \phi(E_{hi})-\frac 12+\omega_{hi}]\\ =\Bbb Z\cap\ [\frac 1\pi \phi(E_{\ell o}) -\frac 12+\tilde\omega_{\ell o},\frac 1\pi \phi(E_{hi})-\frac 12+ \tilde\omega_{hi}]=\Cal K\quad\text{by\ (86)}\ .\endmultline\tag 94$$ \noindent Also, $$|\omega_{\ell o}|, |\omega_{hi}|\le C_{\#}^\prime\Lambda_{\min}^{-1}\ , \ \text{by\ (85);\ and}\tag 95$$ $$\omega_{hi}=0\quad\text{if}\quad \min_{k\in \Bbb Z} |\frac 1\pi \phi(E_{hi})-\frac 12-k|\ge 2C_{\#}^\prime \Lambda_{\min}^{-1}\ .\tag 96$$ >From (94) and the hypothesis $\phi(E_{hi})-\phi(E_{\ell o})\ge 100$, we see that $\Cal K$ has the form $$\Cal K=\{k\in \Bbb Z\mid k_{\ell o}\le k\le k_{hi}\}\ , \quad\text{with} \tag 97$$ $$k_{\ell o}0, \tau(t)\ge 1 in [a,b]; and assume that whenever t_1,t_2\in [a,b] with |t_1-t_2|From these estimates our desired conclusion is obvious.\qquad\qquad\blacksquare \medskip \demo{Trivial Remarks} There are obvious variants of the Lemma on Riemann sums with the interval [a,b] replaced by an open or half--open interval. These results follow from the case of a closed interval by a simple limiting argument. For instance, applying the above lemma to [a+\delta,b-\delta] and letting \delta \to 0+, we obtain$$\multline \sum\limits_{k\in \Bbb Z\cap (a,b)}f(k)=\int_a^b f(t)dt-f(b)\chm(b-)-f(a)\chp(a+)+\frac 12f^\prime(b)\tilde\chi (b)\\ -\frac 12 f^\prime (a)\tilde\chi(a)+\ \text{Error}\ ,\endmultline $$\noindent with \chp(a+)\equiv \operatornamewithlimits{\lim}_{\delta \to 0+} \chp(a+\delta), \chm(b-)\equiv \operatornamewithlimits{\lim}_{\delta \to 0+} \chm(b-\delta), and |\text{Error}| estimated as before. Note that \tilde\chi is a continuous function, so we needn't write \tilde\chi(b-) or \tilde\chi(a+).\enddemo Note also that \frac 12 f^\prime(b)\tilde\chi(b) is dominated by \frac{\sigma(b)}{\tau(b)}, and similarly for \frac 12 f^\prime (a) \tilde\chi(a). Hence in the lemma on Riemann Sums, we may omit the terms \frac 12 f^\prime(b)\tilde\chi(b) and \frac 12 f^\prime(a)\tilde\chi(a), provided we weaken our estimates on the error to$$ |\text{Error}|\le C^\prime \frac{\sigma(a)}{\tau(a)}+C^\prime \frac{\sigma(b)}{\tau(b)}+C_N^\prime \int_a^b\frac{\sigma(t)} {\tau^N(t)}\ dt\ . $$\noindent Again we may omit \frac{\sigma(a)}{\tau(a)} here if f vanishes to infinite order at a, and similarly for b. \vfill\eject \heading{\bf{THE MICROLOCALIZED DENSITY IN THE AIREY REGION I}}\endheading \medskip In this section we adopt the notation and hypotheses of the WKB Theorems, with E_\infty=0. Our goal is to understand the microlocalized density \rho(x,g) when g(x,E) is supported in the region where eigenfunctions are closely approximated in terms of the Airey function. More precisely, we make the following assumptions on g(x,E). \medskip \noindent {\it ASSUMPTION\/} 1. \text{supp}\ g(x,E)\subset \{|E-E_0|<\hat c\ S_{\min}, |x-x_{\text{left}}(E)| <\delta x\}.\smallskip \noindent {\it ASSUMPTION\/} 2. |\partial_E^\beta g(x,E)|\le \hat C_\beta (\delta E)^{-\beta} for all x, E.\smallskip \noindent {\it ASSUMPTION\/} 3. \delta x and \delta E are positive numbers with \lambda_{\text{left}}^{-2/3+\varepsilon}B_{\text{left}} <\delta x<\frac 1{20}\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}} and \delta E=\min\{S_{\text{min}},\frac{S_{\text{left}}}{B_{\text{left}}} (\delta x)\}.\smallskip \noindent {\it ASSUMPTION\/} 4. The constant \hat c in {\it ASSUMPTION\/} 1 is bounded above by a certain small positive number, determined by the constants \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in the hypotheses of the WKB Theorems.\smallskip \noindent {\it ASSUMPTION\/} 5. \Lambda is bounded below by a certain large positive number, determined by \hat c and the \hat C_\beta in {\it ASSUMPTIONS\/} 1 and 2, and by \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in the hypotheses of the WKB Theorems. \medskip \noindent Here, {\it ASSUMPTION\/} 5 strengthens the hypothesis on the hugeness of \Lambda in the WKB Theorems. We denote by c_{\#}, C_{\#}, C_{\#}^\alpha etc\. constants that depend only on \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in the hypotheses of the WKB Theorems. Constants denoted by c_\ast, C_\ast, C_\ast^\alpha etc\. are allowed to depend on \varepsilon, K, N, c, C, c_1, c_2, C_\alpha and also on \hat c, \hat C_\beta in {\it ASSUMPTIONS\/} 1 and 2. The microlocalized density \rho(x,g) is defined as a weighted sum of squares of eigenfunctions. The WKB Eigenfunction Theorem lets us approximate the eigenfunctions by explicit formulas involving the Airey function. This is possible by virtue of our assumption on the support of g(x,E). Hence \rho(x,g) is given approximately as a weighted sum of squares of Airey functions. The Lemma on Riemann sums then lets us approximate this sum by an integral involving the square of the Airey function. In this section we carry out the details, and show that \rho(x,g) is closely approximated by an integral involving the square of the Airey function. In the next section, we will relate that integral to the semiclassical density \rho_{sc}(x,g). We begin by applying the WKB Eigenvalue Theorem to understand the eigenvalues in the interval\smallskip$$ [E_{\ell o},E_{hi}]=(-\infty,0]\cap \{E\colon |E-E_0|\le c_{\#}^1 S_{\text{min}}\}.\tag 1$$\noindent These eigenvalues may be written as E_{k_{\ell o}},E_{k_{\ell o}+1} \ldots, E_{k_{hi}}, with the following properties.$$ \text{For}\ k_{\ell 0}\le k\le k_{hi}\ \text{we\ have}\ |\phi(E_k)+\frac 1{48}\psi(E_k)-\pi(k+1/2)|\le C_{\#}\Lambda^{-2}\ .\tag 2 \multline \text{The\ integers}\ k_{\ell o}\le k\le k_{hi}\ \text{are\ precisely\ those\ integers}\ k\\ \text{that\ lie\ in\ the\ interval}\ [a,b]=[\frac{1}{\pi} \phi(E_{\ell o})+\frac {1}{48\pi}\psi(E_{\ell o})-\frac 12+\omega_{\ell o}\ , \frac 1\pi \phi(E_{hi})+\frac {1}{48\pi}\psi(E_{hi})\\ -\frac 12+\omega_{hi}]\, \text{with}\ |\omega_{\ell o}|, |\omega_{hi}|\le C_{\#} \Lambda^{-2}\ .\endmultline\tag 3 \multline \text{If}\ \operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(E_{hi}) +\frac 1{48}\psi(E_{hi})-\pi(k+1/2)|\ge C_{\#}\Lambda^{-2}\ ,\\ \text{then\ we\ can\ take}\ \omega_{hi}=0\ .\endmultline\tag 4 $$\noindent These properties are the conclusions of the reformulated WKB Eigenvalue Theorem. \noindent Next we recall the basic estimates for the phase \phi(E), \psi(E), from the section on the WKB Theorems. They are as follows$$\multline \Big|\Bigl(\frac{d}{dE}\Bigr)^m\phi(E)\Big|\le C_{\#}^m\int_{x_{\text{left}}} ^{x_{\text{rt}}}S^{\frac {1}{2}-m}(x)\,dx \quad \text{and}\\ \Big|\Bigl(\frac{d}{dE}\Bigr)^m \psi(E)\Big|\le C_{\#}^m\int_{x_{\text{left}}}^{x_{\text{rt}}} S^{-\frac {1}{2}-m}(x)B^{-2}(x)\,dx\quad(m\ge 0)\\ \text{for}\quad |E-E_0|\le 33\ c_{\#}^1S_{\text{min}}\ .\endmultline\tag 5 $$\noindent Also,$$ \frac{d\phi}{dE}\ge c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}S^{-1/2}(x)dx\quad \text{for}\quad |E-E_0|\le 33\ c_{\#}^1S_{\min}\ .\tag 6 $$\noindent (See the remark just after the proof of the Reformulated Eigenvalue Theorem.)\smallskip \noindent Since S(x)\ge S_{\min} and S(x)B^2(x)=\lambda^2(x)\ge c_{\#} \Lambda^2 in (x_{\text{left}},x_{\text{rt}}), it follows that$$ \Big|\bigl(\frac{d}{dE}\bigr)^m\phi\Big|\le C_{\#}^m\ \Gamma S_{\min}^{-m} \ \ (m\ge 1)\quad \text{for}\quad |E-E_0|\le 33\ c_{\#}^1S_{\min} \tag 7  \Big|\bigl(\frac{d}{dE}\bigr)^m\psi\Big|\le C_{\#}^m\Lambda^{-2} \Gamma S_{\min}^{-m}\ \ (m\ge 1)\quad\text{for}\quad |E-E_0|\le 33\ c_{\#}^1S_{\min}\tag 8  \frac{d\phi}{dE}\ge c_{\#}\Gamma S_{\min}^{-1}\quad\text{for}\quad |E-E_0|\le 33\ c_{\#}^1S_{\min}\ ,\quad \text{with}\tag 9  \Gamma=S_{\min}\int_{x_{\text{left}}}^{x_{\text{rt}}}S^{-1/2}(x)dx\ .\tag 10 $$We note two useful lower bounds for \Gamma. Restricting the region of integration to [x_{\text{left}},x_{\text{left}}+c_{\#}B_{\text{left}}] in (10), we get$$ \Gamma S_{\min}^{-1}\ge c_{\#}S_{\text{left}}^{-1/2}B_{\text{left}}\ . \tag 11 $$\noindent Also, we can find x_{\min}\in (x_{\text{left}},x_{\text{rt}}) with S(x_{\min})\sim S_{\min} and \text{dist}(x_{\min},x_{\text{left}}), \text{dist}(x_{\min},x_{\text{rt}})>c_{\#}B(x_{\min}). Restricting the region of integration in (10) to [x_{\min}-c_{\#}B(x_{\min}), x_{\min}+c_{\#}B(x_{\min})], we get$$ \Gamma\ge c_{\#}S_{\min}\cdot S^{-1/2}(x_{\min})B(x_{\min})\ge c_{\#}^\prime S^{1/2}(x_{\min})B(x_{\min})=c_{\#}^\prime\lambda(x_{\min})\ge c_{\#} \Lambda\ .\tag 12 $$We use (7), (8), (9) to solve the equation \phi(E)+\frac{1}{48}\psi(E)= \pi(t+1/2). In fact, those estimates show that$$ \Gamma^{-1}\Bigl[\Bigl\{\frac 1\pi\phi(E)+\frac{1}{48\pi}\psi(E)-\frac 12 \Bigr\}-\Bigl\{\frac 1\pi \phi(E_0)+\frac{1}{48\pi}\psi(E_0)-\frac 12\Bigr\} \Bigr]=f_1\bigl(\frac{E-E_0}{S_{\min}}\bigr)\ , $$\noindent with a--priori estimates on the C^\infty seminorms of f_1 and an a--priori lower bound for (f_1)^\prime. Here f_1 is defined on [-33\ c_{\#}^1,+33\ c_{\#}^1]. Hence on the image f_1([-33 c_{\#}^1,+33 c_{\#}^1]), the inverse function f_1^{-1} is well--defined, and we have a--priori bounds on its C^\infty seminorms. In other words, we have the following results. Let \Cal J=\{\frac 1\pi\phi(E)+\frac{1}{48\pi}\psi(E)-\frac 12| |E-E_0|\le 33\ c_{\#}^1S_{\text{min}}\}. For t\in \Cal J, the equation \frac 1\pi\phi(E)+\frac{1}{48\pi}\psi(E)-\frac 12=t has a unique solution E=E(t) with |E-E_0|\le 33\ c_{\#}^1S_{\min}, and we have$$ \Big|\Bigl(\frac{d}{dt}\Bigr)^mE(t)\Big|\le C_{\#}^mS_{\min} \Gamma^{-m}\quad (m\ge 1) \quad\text{for}\quad t\in \Cal J\tag 13  \frac{d}{dt}E(t)\ge c_{\#}S_{\min}\Gamma^{-1}\quad\text{for}\quad t\in \Cal J\ .\tag 14 $$>From (8), (9) we see that \Cal J is an interval and that$$ \align \min\ \Cal J&= \Bigl\{\frac{1}{\pi}\phi(E_0-33\ c_{\#}^1S_{\min})+\frac{1} {48\pi}\psi(E_0-33\ c_{\#}^1S_{\min})-\frac 12\Bigr\}\\ &\le \Bigl\{\frac{1}{\pi}\phi(E_0-c_{\#}^1S_{\min})+\frac{1}{48\pi} \psi(E_0-c_{\#}^1S_{\min})-\frac 12\Bigr\} -c_{\#}\Gamma\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad =a-\omega_{\ell o}-c_{\#}\Gamma\ .\endalign $$\noindent Here we use the definitions of a and E_{\ell o}. By (12) and |\omega_{\ell o}|\le C_{\#}\Lambda^{-2}, we obtain$$ \min \Cal J\le a-c_{\#}^\prime\Gamma\ .\tag 15 $$\noindent Similarly, (9) implies$$\align \max\Cal J&=\Bigl\{\frac 1\pi\phi(E_0+33\ c_{\#}^1S_{\min})+\frac{1}{48\pi} \psi(E_0+33\ c_{\#}^1S_{\min})-\frac 12 \Bigr\}\\ &\ge \Bigl\{\frac 1\pi \phi(E_{hi})+\frac {1}{48\pi}\psi(E_{hi})-\frac 12\Bigr\} +c_{\#}\Gamma=b-\omega_{hi}+c_{\#}\Gamma\ ,\endalign $$\noindent so that$$ \max \Cal J\ge b+c_{\#}^\prime\Gamma\ .\tag 16 $$\noindent This tells us in particular that$$ [a,b]\subset \Cal J\ .\tag 17 $$\noindent For k_{\ell o}\le k\le k_{hi} we have k\in [a,b] by (3), so E(k) is well--defined and lies in \{|E_0-E(k)|\le 33\ c_{\#}^1 S_{\text{min}}\}. Also |E_k-E_0|\le c_{\#}^1S_{\min} by definition of the E_k. Hence$$\align |E_k-E(k)|&\le \max_{t\in \Cal J}\Big|\frac{dE(t)} {dt}\Big|\cdot\Bigl|\bigl\{\frac 1\pi\phi(E_k)+\frac{1}{48\pi} \psi(E_k)-\frac 12\bigr\}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad -\{\frac 1\pi\phi(E(k))+\frac{1}{48\pi}\psi(E(k))-\frac 12\}\Bigr|\\ &=\max_{t\in \Cal J}\Big|\frac{dE(t)}{dt}\Big|\cdot \Big|\frac 1\pi\phi(E_k)+\frac{1}{48\pi}\psi(E_k)-\frac 12-k\Big|\le C_{\#}S_{\min}\Gamma^{-1}\Lambda^{-2}\ ,\tag 18\endalign $$\noindent by estimates (2) and (13). Similarly, define E_{\max}=E(b) and E_{\min}=E(a), and we have$$\align |E(b)-E_{hi}|&\le \max_{t\in \Cal J}\Big|\frac{dE(t)}{dt}\Big|\cdot \Bigl|b-\bigl\{\frac 1\pi \phi(E_{hi})+\frac{1}{48\pi}\psi(E_{hi})-\frac 12 \bigr\}\Bigr|\\ &\le C_{\#}S_{\min}\Gamma^{-1}|\omega_{hi}|\le C_{\#}S_{\min}\Gamma^{-1} \Lambda^{-2}\ .\tag 19\endalign $$\noindent The same reasoning gives also$$ |E(a)-E_{\ell o}|\le C_{\#}S_{\min}\Gamma^{-1}\Lambda^{-2}\ .\tag 20 $$\noindent We know that g(x,E)=0 for |E-E_0|>\hat cS_{\min}. In particular, if E\le E_{\min}=E(a)\le E_{\ell o}+C_{\#}S_{\min} \Gamma^{-1}\Lambda^{-2} (by (20)) =(E_0-c_{\#}^1S_{\min})+ C_{\#}S_{\min}\Gamma^{-1}\Lambda^{-2} (by definition of E_{\ell o}), then evidently E\le E_0-\hat cS_{\min} if \Lambda and \Gamma are big enough and \hat c is small enough. {\it ASSUMPTIONS\/} 4 and 5 and estimate (12) show that \Lambda and \Gamma are big enough and \hat c is small enough, so we get$$ g(x,E)=0\quad\text{if}\quad E\le E_{\min}\ .\tag 21 $$\noindent Similarly, suppose E\ge E_{\max} and |E_0|\ge 2\hat cS_{\min}. Then E_{hi}=\min(0,E_0+c_{\#}^1S_{\min})\ge E_0+2\hat c\ S_{\min}, so$$\align E&\ge E_{\max}=E(b)\ge E_{hi}-C_{\#}S_{\min}\Gamma^{-1}\Lambda^{-2} \quad \text{((by\ (19))}\\ &\ge E_0+2\hat c\ S_{\min}-C_{\#}S_{\min}\Gamma^{-1}\Lambda^{-2}\ge E_0+\hat c\ S_{\min}\quad \text{by\ {\it ASSUMPTION\/} 5}\ .\endalign $$\noindent Hence we get$$ g(x,E)=0\quad \text{if}\quad E\ge E_{\max}\quad\text{and}\quad |E_0|\ge 2\hat cS_{\min}\ .\tag 22 $$\noindent If |E_0|<2\hat c\ S_{\min}, then perhaps g(x,E_{\max}) will not vanish. Next we apply the WKB Eigenfunction Theorem to approximate closely the (normalized) eigenfunction u_k(x) associated with E_k (k_{\ell o} \le k\le k_{hi}). (See also the remark just after the proof of the Reformulated Eigenvalue Theorem.) We obtain$$\split \int\limits_{|x-x_{\text{left}}(E_k)|<\lambda_{\text{left}}^{-\varepsilon} B_{\text{left}}}\ |u_k(x)-b_{\text{left}}F(x,E_k)|^2\, dx\le\Lambda^{-N^{\prime\prime}}\\ \text{for}\quad k_{\ell o}\le k\le k_{hi}\ ,\endsplit\tag 23 $$\noindent with$$ F(x,E)=\lambda_{\text{left}}^{-1/3}\Bigl(\frac{\partial Y_{\text{left}} (x,E)}{\partial x}\Bigr)^{-1/2}\ A(\lambda_{\text{left}}^{2/3}Y_{\text{left}} (x,E))\ .\tag 24 $$\noindent The WKB Normalization Theorem gives$$ \Big|b_{\text{left}}^2\cdot\Bigl(\frac 12\int\limits_{E_k-V(x)>0}(E_k-V(x))^{-1/2}\ dx\Bigr)-1\Big|\le \Lambda^{4\varepsilon-2}\ ,\quad\text{i.e.}  |b_{\text{left}}^2\phi^\prime(E_k)-1|\le \Lambda^{4\varepsilon-2}\ .\tag 25 $$\noindent In (23) and (25) we want to replace E_k by E(k). To do this for (25), note that$$\align &|\phi^\prime(E_k)-\phi^\prime(E(k))|\le \operatornamewithlimits{\max}_ {|E-E_0|\le 33\ c_{\#}^1S_{\text{min}}}\Big|\frac{d^2\phi}{dE^2}\Big| \cdot |E_k-E(k)|\\ &\ \ \le C_{\#}\Gamma S_{\min}^{-2}\cdot C_{\#}S_{\min}\Gamma^{-1}\Lambda^{-2} \quad \text{(by\ (7)\ and\ (18))}\\ &\ \ =C_{\#}(\Gamma S_{\min}^{-1})\cdot \Gamma^{-1}\Lambda^{-2}\le C_{\#}^\prime \phi^\prime(E_k)\cdot\Gamma^{-1}\Lambda^{-2}\\ &\ \ \qquad\qquad\qquad\qquad\qquad\qquad \text{(by\ (9))}\quad\le C_{\#}^{\prime\prime}\phi^\prime(E_k)\cdot \Lambda^{-3}\quad\text{(by\ (12))}\ . \endalign $$\noindent This and (25) show that$$ |b_{\text{left}}^2\phi^\prime(E(k))-1|\le C_{\#}\Lambda^{4\varepsilon-2} \ , \quad\text{as\ desired}\ .\tag 26 $$The argument for changing E_k to E(k) in (23) is more involved. First of all, we want an interval I_k\subset \{|x-x_{\text{left}}(E_k)|< \lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\} that contains the support of x\mapsto g(x,E) for all E between E_k and E(k). We take$$ I_k=\{|x-x_{\text{left}}(E_k)|<2(\delta x)\}\quad \text{ and\ check\ that\ it\ works.}\tag 27 $$\noindent Certainly I_k\subset\{|x-x_{\text{left}}(E_k)|< \lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\} by {\it ASSUMPTION\/} 3. For E between E_k and E(k) we have$$\align &|x_{\text{left}}(E)-x_{\text{left}}(E_k)|\le \frac{C_{\#} B_{\text{left}}}{S_{\text{left}}}|E-E_k|\le \frac{C_{\#} B_{\text{left}}}{S_{\text{left}}}|E(k)-E_k|\\ &\qquad\le \frac{C_{\#} B_{\text{left}}}{S_{\text{left}}}\cdot S_{\min}\Gamma^{-1} \Lambda^{-2}\le \frac{C_{\#}B_{\text{left}}}{S_{\text{left}}} S_{\text{left}}^{1/2}B_{\text{left}}^{-1}\Lambda^{-2}\quad\text{(by\ (11))}\\ & \qquad =\frac{C_{\#}\Lambda^{-2}B_{\text{left}}}{S_{\text{left}}^{1/2} B_{\text{left}}} =\frac{C_{\#}\Lambda^{-2}B_{\text{left}}}{\lambda_{\text{left}}}< \lambda_{\text{left}}^{-\frac 23+\varepsilon}B_{\text{left}}\le (\delta x) \ \ \ \text{by\ {\it ASSUMPTION\/} 3}\ .\tag 28\endalign $$\noindent For fixed E between E_k and E(k), x\mapsto g(x,E) is supported in \{|x-x_{\text{left}}(E)|<(\delta x)\}\subset I_k by {\it ASSUMPTION\/} 1 and the previous chain of inequalities. So I_k does what we said it does. Next for x\in I_k and E between E_k and E(k) we estimate$$\align \Big|\frac{\partial F(x,E)}{\partial E}\Big|&\le C_{\#}\lambda_{\text{left}}^{-1/3}\Big|\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\Big|^{-3/2}\ \Big|\frac{\partial^2Y_{\text{left}}(x,E)} {\partial x\ \partial E}\Big|\ |A(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))|\\ &\ + C_{\#}\lambda_{\text{left}}^{-1/3}\Big|\frac{\partial Y_{\text{left}} (x,E)}{\partial x}\Big|^{-1/2}\ \lambda_{\text{left}}^{2/3}\ \Big|\frac{\partial Y_{\text{left}}(x,E)}{\partial E}\Big|\ | A^\prime(\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E))|\ .\endalign $$\noindent We know that |\frac{\partial Y_{\text{left}}}{\partial x}| \sim B_{\text{left}}^{-1}, |\frac{\partial Y_{\text{left}}}{\partial E}| \sim S_{\text{left}}^{-1}, |\frac{\partial^2Y_{\text{left}}}{\partial x\ \partial E}|\le C_{\#}B_{\text{left}}^{-1}S_{\text{left}}^{-1}, by the properties of Y_{\text{left}} enumerated in the WKB Eigenfunction Theorem. Also, |A(\xi)|\le C(1+|\xi|)^{-1/4} and |A^\prime(\xi)|\le C(1+|\xi|)^{+1/4} for a universal constant C. Hence,$$\align \Big|\frac{\partial F(x,E)}{\partial E}\Big|&\le C_{\#}\lambda_{\text{left}}^{-1/3}B_{\text{left}}^{1/2}S_{\text{left}} ^{-1}\ (1+\lambda_{\text{left}}^{2/3}|Y|)^{-1/4}\\ &\ \ + C_{\#}\lambda_{\text{left}}^{+1/3}B_{\text{left}}^{1/2} S_{\text{left}}^{-1}\ (1+\lambda_{\text{left}}^{2/3}|Y|)^{+1/4}\ .\endalign $$\noindent The second term on the right clearly dominates the first term, so$$\split \Big|\frac{\partial F(x,E)}{\partial E}\Big|\le C_{\#} \lambda_{\text{left}}^{1/3}B_{\text{left}}^{1/2}S_{\text{left}}^{-1} (1+\lambda_{\text{left}}^{2/3}|\ Y_{\text{left}}(x,E)|)^{+\frac 14}\\ \text{for}\ \ x\in I_k,\ E\ \ \text{between}\ \ E_k\ \ \text{and}\ \ E(k)\ .\endsplit $$\noindent Using again the fact that \frac{\partial Y}{\partial x} \sim B_{\text{left}}^{-1}, we conclude that$$\multline \int\limits_{I_k}\Big|\frac{\partial F(x,E)}{\partial E}\Big|^2\ dx\le C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2}\int\limits_{I_k} (1+\lambda_{\text{left}}^{2/3}|\ Y_{\text{left}}(x,E)|)^{+\ 1/2}\\ \cdot\lambda_{\text{left}}^{2/3}\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\ dx =C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2}\int\limits_{\hat I_k} (1+|\xi|)^{+\frac 12}\ d\xi\ ,\endmultline\tag 29 $$\noindent where \hat I_k is the image of I_k under x\mapsto \lambda_{\text{left}}^{2/3}\ Y_{\text{left}}(x,E). Estimate (28) shows that x_{\text{left}}(E)\in I_k, hence \lambda_{\text{left}}^{2/3}Y_{\text{left}}(x_{\text{left}}(E),E)\in \hat I_k. On the other hand, |Y_{\text{left}}(x_{\text{left}}(E),E)|=|Y_{\text{left}} (x_{\text{left}}(E),E)-Y_0^{\text{left}}(x_{\text{left}}(E),E)|\le C_{\#} \lambda_{\text{left}}^{-2}. Hence \hat I_k contains a point \xi with |\xi|\le C_{\#}\lambda_{\text{left}}^{\frac 23-2}<<1. The length of \hat I_k is\hfill\break \sim \lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1} |I_k|\sim \lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}(\delta x), since \frac{\partial Y_{\text{left}}}{\partial x}\sim B_{\text{left}}^{-1}. For an interval \hat I_k containing \xi with absolute value <1 and having length \sim \lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1} (\delta x)>1 by {\it ASSUMPTION\/} 3, we have$$ \int\limits_{\hat I_k}(1+|\xi|)^{1/2}\ d\xi\sim |\hat I_k|^{3/2}\sim \lambda_ {\text{left}}\Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{3/2}. $$\noindent Putting this into (29), we get$$\split \int\limits_{I_k}\Big|\frac{\partial F(x,E)}{\partial E}\Big|^2\ dx\le C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2}\lambda_{\text{left}} \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{3/2}\\ \text{for}\ E\ \text{between} \ E_k\ \text{and}\ E(k)\ .\endsplit\tag 30 $$\noindent Since (\int_{I_k}|F(x,E_k)-F(x,E(k))|^2dx)^{1/2}\le \int_{\min(E_k,E(k))}^{\max(E_k,E(k))} (\int_{I_k}|\frac{\partial F(x,E)}{\partial E}|^2dx)^{1/2}dE, we conclude that$$\align \int\limits_{I_k}|F(x,E_k)-F(x,E(k))|^2dx&\le C_{\#}B_{\text{left}}^{2}S_{\text{left}}^{-2} \lambda_{\text{left}}\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{3/2}|E_k-E(k)|^2\\ &\le C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2} \lambda_{\text{left}}\Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{3/2}(S_{\min}\Gamma^{-1}\Lambda^{-2})^2\ .\tag 31\endalign $$\noindent Also (7), (9), (25) show that |b_{\text{left}}^2|\sim (S_{\min}\Gamma^{-1}). This and (31) show that$$\multline \int\limits_{I_k}|b_{\text{left}}F(x,E_k)-b_{\text{left}}F(x,E(k)) |^2dx\\ \le C_{\#}(S_{\min}\Gamma^{-1})^3\Lambda^{-4} \lambda_{\text{left}}\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{3/2}B_{\text{left}}^2 S_{\text{left}}^{-2}\ . \endmultline $$\noindent Combining this with (23) and recalling that I_k\subset \{|x-x_{\text{left}}(E_k)|< \lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\}, we get$$ \int\limits_{I_k}|u_k(x)-b_{\text{left}}F(x,E(k))|^2dx\le C_{\#}(S_{\min}\Gamma^{-1})^3\Lambda^{-4}\lambda_ {\text{left}}\Bigl(\frac{\delta x}{B_{\text{left}}} \Bigr)^{3/2}B_{\text{left}}^2S_{\text{left}}^{-2} +\Lambda^{-N^{\prime\prime}}\ .\tag 32 $$Estimate (32) is the desired analogue of (23) with E_k replaced by E(k). It doesn't matter that the region of integration has been cut down to I_k, since the support of x\mapsto g(x,E) is contained in I_k for all E between E_k and E(k). We next use (26) to replace b_{\text{left}} by \pm[\phi^\prime(E(k))]^{-1/2} in (32). After possibly replacing u_k(x) by -u_k(x), we may suppose b_{\text{left}}>0. Estimate (26) may then be written as |b_{\text{left}}-\bigl(\phi^\prime(E(k))\bigr)^{-1/2}|\le C_{\#} \Lambda^{4\varepsilon-2}|\phi^\prime(E(k))|^{-1/2}, which implies$$\multline \int\limits_{I_k}|b_{\text{left}}F(x,E(k))-[\phi^\prime (E(k))]^{-1/2}F(x,E(k))|^2dx\le\\ C_{\#}\Lambda^{8\varepsilon-4}[\phi^\prime(E(k))]^{-1} \int\limits_{I_k}|F(x,E(k))|^2dx\le C_{\#} \Lambda^{8\varepsilon-4}S_{\min}\Gamma^{-1} \int\limits_{I_k}|F(x,E(k))|^2dx\\\qquad \text{by\ (9)}\ .\endmultline\tag 33 $$To estimate the right--hand side of (33), we again use the estimates \frac{\partial Y_{\text{left}}} {\partial x}\sim B_{\text{left}}^{-1}, |A(\xi)|\le C(1+|\xi|)^{-1/4} to conclude that$$\align &\int\limits_{I_k}|F(x,E(k))|^2dx=\int\limits_{I_k} \lambda_{\text{left}}^{-2/3}\Bigl(\frac{\partial Y_ {\text{left}}}{\partial x}\Bigr)^{-1}A^2 (\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E(k)))\ dx\\ &\ \ < C_{\#}\int\limits_{I_k}\lambda_{\text{left}} ^{-4/3}\Bigl(\frac{\partial Y_{\text{left}}} {\partial x}\Bigr)^{-2}\bigl(1+\lambda_{\text{left}}^{2/3}\ |Y_{\text{left}}(x,E(k))|\bigr)^{-1/2}\cdot\lambda_{\text{left}}^{2/3} \frac{\partial Y_{\text{left}}}{\partial x} \ dx\\ &\ \ \le C_{\#}\lambda_{\text{left}}^{-4/3} B_{\text{left}}^2\int\limits_{I_k}\bigl(1+\lambda_{\text{left}} ^{2/3}\ |Y_{\text{left}}(x,E(k))|\bigr)^{-1/2}\cdot \lambda_{\text{left}}^{2/3}\frac{\partial Y_{\text{left}}(x,E(k))} {\partial x} \ dx\\ &\ \ = C_{\#}\lambda_{\text{left}}^{-4/3}B_{\text{left}}^2 \int\limits_{\hat I_k}(1+|\xi|)^{-1/2}d\xi\quad \text{with}\quad \hat I_k \quad \text{as\ in\ (29)}\ . \tag 34\endalign $$\noindent Recall from the paragraph following (29) that \hat I_k is an interval of length\hfill\break \sim \lambda_{\text{left}}^{+2/3}(\frac{\delta x} {B_{\text{left}}}) >1, containing a point \xi with |\xi|<1. Therefore, \int\limits_{\hat I_k} (1+|\xi|)^{-1/2}d\xi\sim |\hat I_k|^{1/2}\sim \lambda_{\text{left}}^{+\,1/3} (\frac{\delta x} {B_{\text{left}}})^{1/2}, so that (34) implies$$ \int\limits_{I_k}|F(x,E(k))|^2\ dx\le C_{\#} \lambda_{\text{left}}^{-1}B_{\text{left}}^2 \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2} \ . \tag 35 $$\noindent From (33) and (35), we see that$$\multline \int\limits_{I_k}|b_{\text{left}}F(x,E(k))-[\phi^\prime (E(k))]^{-1/2}F(x,E(k))|^2\ dx\\ \le C_{\#}\Lambda^{8\varepsilon-4}S_{\min} \Gamma^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{1/2}\ .\endmultline\tag 36 $$\noindent This and (32) imply$$\multline \int\limits_{I_k}|u_k(x)-[\phi^\prime(E(k))]^{-1/2}F (x,E(k))|^2\ dx\\ \le C_{\#}\Lambda^{8\varepsilon-4}(S_{\min}\Gamma^{-1}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2\Bigl( \frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}\\ +C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2} \lambda_{\text{left}} \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{3/2} \Lambda^{-4}(S_{\min}\Gamma^{-1})^3+\Lambda^{-N^{\prime \prime}} \ . \endmultline\tag 37 $$Fortunately, the right--hand side of (37) simplifies. In fact, (11) shows that$$\multline C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2} \lambda_{\text{left}}\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{3/2}\Lambda^{-4} (S_{\min}\Gamma^{-1})^3\\ \le C_{\#}B_{\text{left}}^2S_{\text{left}}^{-2} \lambda_{\text{left}}\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{3/2}\Lambda^{-4} (S_{\min}\Gamma^{-1})(S_{\text{left}}^{1/2} B_{\text{left}}^{-1})^2\\ = C_{\#}B_{\text{left}}^2\lambda_{\text{left}} \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{3/2} \Lambda^{-4}(S_{\min}\Gamma^{-1})\cdot (S_{\text{left}} ^{-1}B_{\text{left}}^{-2})\\ =C_{\#}B_{\text{left}}^2\lambda_{\text{left}}^{-1} \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{3/2} \Lambda^{-4}(S_{\min}\Gamma^{-1})\\ \le C_{\#}\Lambda^{8\varepsilon-4}(S_{\min}\Gamma^{-1}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2\Bigl( \frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}\ .\endmultline $$\noindent This means that the second term on the right of (37) is dominated by the first term. So (37) simplifies to$$\multline \int\limits_{I_k}|u_k(x)-[\phi^\prime (E(k))]^{-1/2}F(x,E(k))|^2\ dx\\ \le C_{\#}\Lambda^{8\varepsilon -4}(S_{\min} \Gamma^{-1})\lambda_{\text{left}}^{-1}B_{\text{left}}^2 \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}+\Lambda^ {-N^{\prime\prime}}\ .\endmultline\tag 38 $$\noindent Next we use the inequality$$ \multline \int\limits_{I_k}|u^2(x)-v^2(x)|\ dx\le \bigl( \int\limits_{I_k}|u(x)-v(x)|^2\ dx\bigr)^{1/2}\bigl(\int\limits_{I_k} |u(x)+v(x)|^2 \ dx\bigr)^{1/2}\\ \le \bigl(\int\limits_{I_k}|u(x)-v(x)|^2\ dx\bigr)^{1/2} \Bigl[2\bigl(\int\limits_{I_k}|v(x)|^2\ dx\bigr)^{1/2}+ \bigl(\int\limits_{I_k}|u(x)-v(x)|^2\ dx\bigr)^{1/2}\Bigr]\ , \endmultline\tag 39 $$\noindent as follows from \Vert u+v\Vert=\Vert 2v-(v-u)\Vert \le 2 \Vert v\Vert+\Vert u-v\Vert. \noindent We take u=u_k(x) and v(x)=[\phi^\prime (E(k))]^{-1/2}F(x,E(k)). By (35) and (9) we have$$\multline \int\limits_{I_k}|v(x)|^2\ dx\le C_{\#} S_{\min}\Gamma^{-1}\int\limits_{I_k} |F(x,E(k))|^2\ dx\\ \le C_{\#}S_{\min}\Gamma^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2\Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}\ .\endmultline\tag 40 $$\noindent Hence (38) and (39) show that$$\multline \int\limits_{I_k}|u_k^2(x)-[\phi^\prime (E(k))]^{-1}F^2(x,E(k))|\ dx\\ \le \Bigl[C_{\#}\Lambda^{8\varepsilon-4}(S_{\min} \Gamma^{-1})\lambda_{\text{left}}^{-1} B_{\text{left}}^2\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{1/2}+\Lambda^{-N^{\prime\prime}} \Bigr]^{1/2}\\ \cdot\Bigl[C_{\#}(S_{\min}\Gamma^{-1}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2 \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2} +\Lambda^{-N^{\prime\prime}}\Bigr]^{1/2}\\ \le C_{\#}\Lambda^{4\varepsilon-2}(S_{\min} \Gamma^{-1})\lambda_{\text{left}}^{-1} B_{\text{left}}^2\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{1/2}\\ +C_{\#}\Lambda^{-\frac {N^{\prime\prime}}{2}}\Bigl[(S_{\min} \Gamma^{-1})\lambda_{\text{left}}^{-1} B_{\text{left}}^2\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{1/2}\Bigr]^{1/2} +\Lambda^{-N^{\prime\prime}}\\ \le C_{\#}\Lambda^{4\varepsilon-2}(S_{\min} \Gamma^{-1})\lambda_{\text{left}}^{-1} B_{\text{left}}^2\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{1/2}+C_{\#} \Lambda^{10-N^{\prime\prime}}\endmultline\tag 41 $$\noindent (we see the last inequality by using \Lambda^{-\frac {N^{\prime\prime}}{2}} X^{1/2}\le \Lambda^{10-N^{\prime\prime}}+\Lambda^{-2}X) . Since I_k contains the supports of both g(x,E_k) and g(x,E(k)), we have$$\multline \int\limits_{I_{\text{BVP}}}|g(x,E_k)u_k^2(x)-g(x,E(k))[\phi ^{\prime}(E(k))]^{-1}F^2(x,E(k))|\ dx\\ \le \Bigl[\operatornamewithlimits{\max}_{x\in I_{\text{BVP}}}|g(x,E_k)|\Bigr]\int\limits_{I_k} |u_k^2(x)-[\phi^\prime(E(k))]^{-1}F^2(x,E(k))|\ dx\\ +\Bigl[\operatornamewithlimits{\max}_{x\in I_{\text{BVP}}} |g(x,E_k)-g(x,E(k))|\Bigr] \int\limits_{I_k}\ \Bigl|[\phi^\prime(E(k))]^{-1}F^2(x,E(k))\Bigr|\ dx\ .\endmultline\tag 42 $$We know that \max_{x\in I_{ \text{BVP}}}|g(x,E_k)|\le C_{\ast} and$$\multline \Bigl[\operatornamewithlimits{\max}_{ x\in I_{\text{BVP}}}|g(x,E_k)-g(x,E(k))|\Bigr]\le C_\ast (\delta E)^{-1}|E_k-E(k)|\\ \le C_\ast(\delta E)^{-1}S_{\min}\Gamma^{-1}\Lambda^{-2} \endmultline $$\noindent (by (18)) \le C_\ast[S_{\text{left}}^{-1} \bigl(\frac{B_{\text{left}}}{\delta x}\bigr)+ S_{\min}^{-1}]S_{\min}\Gamma^{-1}\Lambda^{-2}. (See {\it ASSUMPTION\/} 3.) The integrals on the right of (42) are controlled by (40) and (41). Substituting these estimates into (42), we get the following.$$\multline \int_{I_{\text{BVP}}}|g(x,E_k)u_k^2(x)-g(x,E(k))[\phi^\prime (E(k))]^{-1}F^2(x,E(k))|\ dx\\ \le C_\ast\Lambda^{4\varepsilon-2}(S_{\min}\Gamma^{-1})\lambda_{\text{left}} ^{-1}B_{\text{left}}^2\Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}+ C_\ast\Lambda^{10-N^{\prime\prime}}\\ +\Bigl\{C_\ast\Bigl[S_{\text{left}}^{-1}\Bigl(\frac{B_{\text{left}}} {\delta x}\Bigr)+S_{\min}^{-1}\Bigr](S_{\min}\Gamma^{-1})\Lambda^{-2}\\ \cdot (S_{\min}\Gamma^{-1})\lambda_{\text{left}}^{-1}B_{\text{left}}^2 \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}\Bigr\}\ .\endmultline\tag 43 $$\noindent The right--hand side simplifies. We shall check that the term in curly brackets on the right is dominated by the first term on the right. This amounts to checking that \Bigl[S_{\text{left}}^{-1}\Bigl(\frac{B_{\text{left}}}{\delta x} \Bigr)+S_{\min}^{-1}\Bigr](S_{\min}\Gamma^{-1})\le \Lambda^{4\varepsilon}, i.e.$$ S_{\text{left}}^{-1}(B_{\text{left}}/\delta x)\ (S_{\min}\Gamma^{-1})+ \Gamma^{-1}\le \Lambda^{4\varepsilon} \ .\tag 44 $$\noindent We have \Gamma>>1 by (12), so (44) will follow from$$ S_{\text{left}}^{-1}\Bigl(\frac{B_{\text{left}}}{\delta x}\Bigr) (S_{\min}\Gamma^{-1})\le \frac 12 \Lambda^{4\varepsilon}\ . \tag 45 $$\noindent Since \frac{B_{\text{left}}}{\delta x}\le \lambda_{\text{left}}^ {2/3} by {\it ASSUMPTION\/} 3, estimate (11) shows that$$\multline S_{\text{left}}^{-1}\Bigl(\frac{B_{\text{left}}}{\delta x}\Bigr)(S_{\min} \Gamma^{-1})\le C_{\#}S_{\text{left}}^{-1}\Bigl(\frac{B_{\text{left}}} {\delta x}\Bigr)(S_{\text{left}}^{1/2}B_{\text{left}}^{-1})\\ \le C_{\#}\lambda_{\text{left}}^{2/3}(S_{\text{left}}^{-1/2} B_{\text{left}}^{-1})=C_{\#}\lambda_{\text{left}}^{-1/3}<<1\ .\endmultline $$\noindent So (45) holds, and the term in curly brackets in (43) is dominated by the other terms on the right--hand side. Thus (43) may be rewritten as$$\multline \int\limits_{I_{\text{BVP}}}|g(x,E_k)u_k^2(x)-g(x,E(k))[\phi^\prime (E(k))]^{-1}F^2(x,E(k))|\ dx\\ \le C_\ast\Lambda^{4\varepsilon-2}(S_{\min}\Gamma^{-1})\lambda_{\text{left}} ^{-1}B_{\text{left}}^2\Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2} +C_\ast\Lambda^{10-N^{\prime\prime}}\ . \endmultline\tag 46 $$\noindent We have proven (46) for k_{\ell o}\le k\le k_{hi}. Next recall that g(x,E)\equiv 0 outside \{|E-E_0|\le c_{\#}^1S_{\min}\}, and that E_{k_{\ell o}}\ldots E_{k_{hi}} are precisely the eigenvalues of -\frac{d^2}{dx^2}+V(x) belonging to \{|E-E_0|\le c_{\#}^1S_{\min}\}\hfill\break\cap (-\infty,0]. The definition \rho(x,g)=\sum\limits_{E\ \text{eigenvalue}\ \le 0} g(x,E)u^2(x,E) (with u(x,E)= the real normalized eigenfunction corresponding to E) therefore shows that \rho(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}} g(x,E_k)u_k^2(x). So (46) implies:$$\multline \int\limits_{I_{\text{BVP}}}|\rho(x,g)-\sum\limits_{k=k_{\ell o}} ^{k_{hi}}g(x,E(k))[\phi^\prime(E(k))]^{-1}F^2(x,E(k))| \ dx\\ \le [C_\ast\Lambda^{4\varepsilon-2}(S_{\min}\Gamma^{-1}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2\Bigl(\frac{\delta x} {B_{\text{left}}}\Bigr)^{1/2}+C_\ast\Lambda^{10-N^{\prime\prime}}] \cdot (k_{hi}-k_{\ell o}+1) \ . \endmultline\tag 47 $$\noindent It is easy to estimate (k_{hi}-k_{\ell o}+1), since \{k_{hi}\le k\le k_{\ell o}\}=\Bbb Z\cap [a,b] by (3), and thus k_{hi}-k_{\ell o}+1\le b-a+10\le \frac 1\pi \phi(E_{hi})-\frac 1\pi \phi(E_{\ell o})+20. (To see the last estimate, recall that b=\frac 1\pi\phi(E_{hi})+\frac 1{48\pi}\psi(E_{hi})-\frac 12 +\omega_{hi}, a=\frac 1\pi\phi(E_{\ell o})+\frac {1}{48\pi}\psi(E_{\ell o})-\frac 12 +\omega_{\ell o}, with |\psi(E_{hi})|, |\psi(E_{\ell o})|, |\omega_{hi}|, |\omega_{\ell o}|<1.) From (7) we get \phi(E_{hi})-\phi(E_{\ell o})\le C_{\#}\Gamma S_{\min}^{-1} |E_{hi}-E_{\ell o}|\le C_{\#}\Gamma by definition (1) of E_{hi}, E_{\ell o}. Hence$$ k_{hi}-k_{\ell o}+1\le C_{\#}\Gamma+20\le C_{\#}\Gamma\qquad \text{by\ (12)}\ . \tag 48$$\noindent Putting this into (47) and using (3), we get$$\multline \int\limits_{I_{\text{BVP}}}|\rho(x,g)-\sum\limits_{k\in \Bbb Z\cap [a,b]} g(x,E(k))[\phi^\prime(E(k))]^{-1}F^2(x,E(k))| \ dx\\ \le C_\ast\Lambda^{4\varepsilon-2}S_{\min} \lambda_{\text{left}}^{-1}B_{\text{left}}^2 \Bigl(\frac{\delta x}{B_{\text{left}}}\Bigr)^{1/2}+C_\ast\Lambda^{10-N^{\prime \prime}}\Gamma\ .\endmultline\tag 49 $$We want to approximate the sum in (49) by using the Lemma on Riemann sums. This requires that we bound the derivatives of t\mapsto g(x,E(t))[\phi^\prime(E(t))]^{-1}F^2(x,E(t)). To produce the bounds, we write$$ g(x,E(t))[\phi^\prime(E(t))]^{-1}F^2(x,E(t))=f_{\text{Airey}}(x,t) f_{\text{other}}(x,t)\ , \qquad \text{with}\tag 50  f_{\text{Airey}}(x,t)=A^2\bigl(\lambda_{\text{left}}^{2/3}Y_{\text{left}} (x,E(t))\bigr)\qquad\text{and}\tag 51  f_{\text{other}}(x,t)=g(x,E(t))\lambda_{\text{left}}^{-2/3} \Bigl(\frac{\partial Y_{\text{left}}(x,E(t))}{\partial x}\Bigr)^{-1} [\phi^\prime(E(t))]^{-1}\ . \tag 52 $$\noindent (See (24).) To handle f_{\text{Airey}}, we note that$$ \Bigl|\bigl(\frac{d}{d\xi}\bigr)^mA^2(\xi)\Bigr|\le C_m(1+|\xi|)^{-\frac 12+\frac m2}\ \ (m\ge 0)\quad \text{for\ a\ universal\ constant}\ C_m\ .\tag 53 $$\noindent The estimates for Y_{\text{left}}(x,E) in the WKB Eigenfunction Theorem include$$ \Bigl|\bigl(\frac{\partial}{\partial E}\bigr)^\beta Y_{\text{left}}(x,E)\Bigr| \le C_{\#}^\beta S_{\text{left}}^{-\beta}\quad \text{for}\ (x,E)\in\ \text{supp}\ g\ .\tag 54 $$\noindent The derivative (\frac{\partial}{\partial E})^\beta A^2 (\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E)) is a sum of terms of the form$$ \Bigl[\bigl(\frac{d}{d\xi}\bigr)^mA^2(\xi)\mid_{\xi=\lambda_{\text{left}} ^{2/3}Y_{\text{left}}(x,E)}\Bigr]\cdot \prod\limits_{\nu=1}^m\Bigl[\bigl(\frac{\partial}{\partial E}\bigr)^{\beta\nu} \{\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E)\}\Bigr]\tag 55 $$\noindent with \beta_\nu\ge 1 and \beta_1+\ldots+\beta_m=\beta. In particular, 0\le m\le \beta. By (53) and (54), the term (55) is dominated by$$ C_{\#}^\beta(1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E)|)^{-\frac 12 +\frac m2}\lambda_{\text{left}}^{\frac 23 m}S_{\text{left}}^{-\beta}\ ; $$\noindent and since 0\le m\le \beta, this is in turn dominated by$$ C_{\#}^\beta(1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E)|)^{-\frac 12} \Bigl[(\lambda_{\text{left}}^{2/3}+\lambda_{\text{left}}|Y_{\text{left}} (x,E)|^{1/2})S_{\text{left}}^{-1}\Bigr]^\beta\ . $$\noindent Hence,$$\split \Bigl|\bigl(\frac{\partial}{\partial E}\bigr)^\beta A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E))\Bigr|\le C_{\#}^\beta (1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E)|)^{-1/2}\\ \cdot \Bigl[ \frac{\lambda_{\text{left}}^{2/3}+\lambda_{\text{left}} |Y_{\text{left}}(x,E)|^{1/2}}{S_{\text{left}}}\Bigr]^\beta \text{for}\quad (x,E)\in\ \text{supp}\ g\ .\endsplit\tag 56 $$\noindent The derivative (\frac{d}{dt})^mf_{\text{Airey}}(x,t) is a sum of terms of the form$$ \Bigl[\bigl(\frac{\partial}{\partial E})^\beta A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))\mid_{E=E(t)}\Bigr]\ \cdot \prod\limits_{\nu=1}^\beta \Bigl[\bigl(\frac{d}{dt}\bigr)^{m_\nu}E(t)\Bigr]\tag 57 $$\noindent with m_\nu\ge 1 and m_1+\ldots+m_\beta=m. In particular, 0\le \beta\le m. \noindent For t\in \Cal J and (x,E(t))\in\ \text{supp}\ g we have (56) and (13), so that the term (57) is dominated by$$ C_{\#}^m\bigl(1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E(t))|\bigr)^{-1/2} \Bigl[\frac{\lambda_{\text{left}}^{2/3}+\lambda_{\text{left}}|Y_{\text{left}} (x,E(t))|^{1/2}}{S_{\text{left}}}\Bigr]^\beta S_{\min}^\beta\Gamma^{-m}\ .\tag 58 $$\noindent We rewrite (58) in the form$$\split C_{\#}^m\bigl(1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E(t))|\bigr) ^{-1/2}\Bigl[\frac{(\lambda_{\text{left}}^{2/3}+\lambda_{\text{left}} |Y_{\text{left}}(x,E(t))|^{1/2})}{S_{\text{left}}}\\ \cdot(S_{\min}\Gamma^{-1})\Bigr]^\beta\ \Gamma^{\beta-m}\ . \endsplit\tag 59 $$\noindent Hence |(\frac{d}{dt})^mf_{\text{Airey}}(x,t)| is dominated by a sum of terms (59) for t\in \Cal J, (x,E(t))\in\ \text{supp}\ g, with 0\le \beta\le m. Estimate (11) shows that the quantity in square brackets in (59) is dominated by$$\multline \frac{(\lambda_{\text{left}}^{2/3}+\lambda_{\text{left}}|Y_{\text{left}} (x,E(t))|^{1/2})} {S_{\text{left}}}\ (S_{\text{left}}^{1/2}B_{\text{left}}^{-1})\\ =(\lambda_{\text{left}}^{-1/3}+|Y_{\text{left}}(x,E(t))|^{1/2})\le C_{\#} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ \text{for}\quad (x,E(t))\in\ \text{supp}\ g\ ,\endmultline $$\noindent since then |x-x_{\text{left}}(E(t))|<(\delta x) so$$ |Y_{\text{left}}(x,E(t))|\le |Y_0^{\text{left}}(x,E(t))|+C_{\#} \lambda_{\text{left}}^{-2}\le C_{\#}\bigl(\frac{\delta x}{B_{\text{left}}}) +C_{\#}\lambda_{\text{left}}^{-2} $$\noindent by the WKB Eigenfunction Theorem. Hence, (59) is dominated by$$ C_{\#}^m(1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E(t))|)^{-1/2} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{\frac 12 \beta} \Gamma^{\beta-m}\quad (0\le \beta\le m)\ , $$\noindent so$$\split \Big|\bigl(\frac{d}{dt}\bigr)^m f_{\text{Airey}}(x,t)\Big| \le C_{\#}^m(1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E(t))|)^{-1/2}\\ \cdot\ [\min\{\Gamma, (\frac{B_{\text{left}}}{\delta x})^{1/2}\}]^{-m}\ , \quad\text{for}\quad t \in \Cal J, (x,E(t))\in\ \text{supp}\ g \ .\endsplit\tag 60 $$Next we study f_{\text{other}}(x,t). The WKB Eigenfunction Theorem gives$$\split \Big|\bigl(\frac{\partial}{\partial E}\bigr)^\beta\frac{\partial Y_{\text{left}}(x,E)}{\partial x}\Big|\le C_{\#}^\beta B_{\text{left}}^{-1}S_{\text{left}}^{-\beta}\\ \text{and}\quad \Big|\frac{\partial Y_{\text{left}}(x,E)}{\partial x}\Big|> c_{\#}B_{\text{left}}^{-1}\endsplit $$\noindent in \text{supp}\ g, and therefore$$ \Big|\bigl(\frac{\partial}{\partial E}\bigr)^\beta\bigl(\frac{\partial Y_ {\text{left}}(x,E)}{\partial x}\bigr)^{-1}\Bigr|\le C_{\#}^\beta B_{\text{left}}S_{\text{left}}^{-\beta}\quad\text{in}\quad \text{supp}\ g\ . \tag 61 $$\noindent {\it ASSUMPTION\/} 2 and the inequalities (\delta E)\le S_{\min}\le S_{\text{left}}, together with (61), imply$$\split \Big|\bigl(\frac{\partial}{\partial E}\bigr)^\beta\bigl\{g(x,E) (\frac{\partial Y_{\text{left}}(x,E)}{\partial x})^{-1}\bigr\}\Big|\le C_\ast^\beta B_{\text{left}}(\delta E)^{-\beta}\quad \text{for\ all}\ x\ , E\ . \endsplit\tag 62 $$\noindent From (7) and (9) we see that$$ \Big|\bigl(\frac{d}{dE}\bigr)^\beta\bigl[\frac{d\phi}{dE}\bigr]^{-1}\Big| \le C_{\#}^\beta(S_{\min}\Gamma^{-1})S_{\min}^{-\beta}\quad \text{for}\quad |E-E_0|\le 33\ c_{\#}^1S_{\min}\ . $$\noindent Combining this with (62) and recalling that g(x,E)=0 for |E-E_0|>c_{\#}^1S_{\min}, we get$$\split \Big|\bigl(\frac{\partial}{\partial E}\bigr)^\beta \bigl\{\lambda_{\text{left}}^{-2/3}g(x,E)\bigl(\frac{\partial Y_{\text{left}} (x,E)}{\partial x}\bigr)^{-1}[\phi^\prime(E)]^{-1}\bigr\}\Big|\\ \le C_\ast^\beta\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1}) (\delta E)^{-\beta}\ ,\ \text{all}\ x, E\ . \endsplit\tag 63 $$\noindent Here again we use \delta E\le S_{\min}. The derivative (\frac{d}{dt})^mf_{\text{other}}(x,t) is a sum of terms$$\split \Bigl[\bigl(\frac{\partial}{\partial E}\bigr)^\beta\bigl\{\lambda_{\text{left}} ^{-2/3}g(x,E)\bigl(\frac{\partial Y_{\text{left}}(x,E)}{\partial x}\bigr)^{-1}\\ \cdot [\phi^\prime(E)]^{-1}\bigr\}\bigm|_{E=E(t)}\Bigr] \cdot \prod\limits_{\nu=1}^\beta \Bigl[\bigl(\frac{d}{dt}\bigr)^{m_\nu} E(t)\Bigr]\endsplit\tag 64 $$\noindent with m_\nu\ge 1 and m_1+\ldots+m_\beta=m. (See (52).) In particular, 0\le \beta\le m. For t\in \Cal J, estimates (63) and (13) show that the term (64) is dominated by$$ C_\ast^m\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min} \Gamma^{-1})(\delta E)^{-\beta}\cdot S_{\min}^\beta\Gamma^{-m}\ .\tag 65 $$\noindent Since 0\le \beta\le m and \delta E\le S_{\min}, the term (65) is in turn dominated by$$ C_\ast^m\lambda_{\text{left}}^{-2/3} B_{\text{left}}(S_{\min}\Gamma^{-1}) \bigl[(\delta E)^{-1}S_{\min}\Gamma^{-1}\bigr]^m, $$\noindent and therefore$$\split \Big|\bigl(\frac{d}{dt}\bigr)^mf_{\text{other}}(x,t)\Big|\le C_\ast^m \lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1})[(\delta E) ^{-1}S_{\min}\Gamma^{-1}]^m\\ \text{for}\quad t\in \Cal J\ .\endsplit\tag 66 $$\noindent {\it ASSUMPTION\/} 3 shows that the expression in square brackets in (66) has the order of magnitude$$\align &(S_{\min}^{-1}+S_{\text{left}}^{-1}B_{\text{left}}(\delta x)^{-1}) S_{\min}\Gamma^{-1}\\ &=\Gamma^{-1}+S_{\text{left}}^{-1}(B_{\text{left}}/(\delta x))(S_{\min} \Gamma^{-1})\le \Gamma^{-1}+\bigl(\frac{B_{\text{left}}}{\delta x}\bigr) S_{\text{left}}^{-1}(S_{\text{left}}^{1/2}B_{\text{left}}^{-1})\endalign $$\noindent (by (11)) = \Gamma^{-1}+(\frac{B_{\text{left}}}{\delta x})\lambda_ {\text{left}}^{-1}. Therefore, (66) implies$$\split \Big|\bigl(\frac{d}{dt}\bigr)^mf_{\text{other}}(x,t)\Big|\le C_\ast^m\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1}) \bigl[\min\bigl\{\Gamma,\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)\lambda_ {\text{left}}\bigr\}\bigr]^{-m}\\ \text{for}\quad t\in \Cal J\ .\endsplit\tag 67 $$\noindent From (60) and (67), we learn that$$\multline \Big|\bigl(\frac{d}{dt}\bigr)^m\bigl\{f_{\text{Airey}}(x,t)f_{\text{other}} (x,t)\bigr\}\Big|\\ \le C_\ast^m\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1}) \cdot (1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E(t))|)^{-1/2}\\ \cdot \bigl[\min\{\Gamma,\lambda_{\text{left}}(\frac{\delta x}{B_{\text{left}}}), (\frac{B_{\text{left}}}{\delta x})^{1/2}\bigr\}\bigr]^{-m}\ ,\\ \text{for}\quad t\in \Cal J, (x,E(t))\in\ \text{supp}\ g\ .\endmultline\tag 68 $$\noindent The right--hand side of (68) simplifies. First of all, the WKB Eigenfunction Theorem gives$$ |Y_{\text{left}}(x,E)-Y_0^{\text{left}}(x,E)|\le C_{\#}\lambda_{\text{left}}^{-2} \ ,\ |Y_0^{\text{left}}(x,E)|\sim \frac{|x-x_{\text{left}}(E)|}{B_{\text{left}}} $$\noindent in (\text{supp}\ g), and therefore$$ 1+\lambda_{\text{left}}^{2/3}|Y_{\text{left}}(x,E(t))|\sim 1+\lambda_{\text{left}} ^{2/3}B_{\text{left}}^{-1}|x-x_{\text{left}}(E(t))| \quad\text{for}\quad (x,E(t))\in\ \text{supp}\ g\ . $$\noindent Also, {\it ASSUMPTION\/} 3 gives \frac{\delta x}{B_{\text{left}}}>\lambda^{-2/3}, so that (\frac{B_{\text{left}}}{\delta x})^{1/2}\le \lambda_{\text{left}}^{+1/3} \le \lambda_{\text{left}}(\frac{\delta x}{B_{\text{left}}}). Hence, \min\{\Gamma,\lambda_{\text{left}}(\frac{\delta x}{B_{\text{left}}}), (\frac{B_{\text{left}}}{\delta x})^{1/2}\}=\min\{\Gamma,(\frac{B_{\text{left}}} {\delta x})^{1/2}\}, so that (68) yields$$ \Big|\bigl(\frac{d}{dt}\bigr)^m\bigl\{f_{\text{Airey}}(x,t)f_{\text{other}} (x,t)\bigr\}\Big|\le C_\ast^m\sigma_x(t)\tau^{-m}\quad \text{for}\quad t\in \Cal J\ , x\in I_{\text{BVP}}\tag 69 $$\noindent with$$ \tau=\min\{\Gamma,(\frac{B_{\text{left}}}{\delta x})^{1/2}\}\quad\text{and} \tag 70  \sigma_x(t)=\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min} \Gamma^{-1})\cdot (1+\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}\ |x-x_{\text{left}}(E(t))|)^{-1/2}\ . \tag 71 $$Estimate (69) is the main hypothesis in the Lemma on Riemann sums. The other hypotheses are \tau\ge 1 and c_\ast<\sigma_x(t_2)/\sigma_x(t_1) \lambda_{\text{left}}^\varepsilon and \Gamma\ge c_{\#}\Lambda, (70) shows that \tau \ge 1. That c_\ast<\sigma_x(t_2)/\sigma_x(t_1)\lambda_{\text{left}}^{-2/3})=C_{\#}S_{\text{left}}\cdot \lambda_{\text{left}}^{-2/3}, which is the desired hypothesis for the Lemma on Riemann sums. So we have verified all the hypotheses of the Lemma on Riemann sums. To know which variant of the lemma to apply, we must still check whether the integrand \{f_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\} vanishes to infinite order, both at t=a and at t=b. From (21) we see that g(x,E(t))=0 for t\le a, so the integrand vanishes to infinite order at t=a, in all cases. Similarly, (22) shows that the integrand vanishes to infinite order at t=b, provided |E_0|\ge 2\hat cS_{\min}. If |E_0|<2\hat cS_{\min}, then we still know that \text{supp}\ g(x,E)\subset \{|x-x_{\text{left}}(E)|<\delta x\}, and therefore g(x,E(t)) vanishes to infinite order at t=b if |x-x_{\text{left}}(E_{\max})|>(\delta x). To summarize,$$\multline f_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\ \text{vanishes\ to\ infinite\ order\ at}\ t=a\ \text{always}\ ,\\ \text{and\ at}\ t=b\ \text{unless}\ |E_0|<2\hat c\ S_{\min}\\ \text{and}\ |x-x_{\text{left}}(E_{\max})|\le(\delta x)\ .\endmultline \tag"{(71\text{bis})}" $$\noindent Therefore, the lemma on Riemann sums yields the following conclusions.$$\multline \sum\limits_{k\in \Bbb Z\cap [a,b]} f_{\text{Airey}}(x,k)\cdot f_{\text{other}}(x,k)\\ =\int_a^bf_{\text{Airey}}(x,t)f_{\text{other}}(x,t)dt-f_{\text{Airey}} (x,b)f_{\text{other}}(x,b)\chm(b)+\text{Error}_1(x)\\ \text{if}\quad |E_0|\le 2\hat c\ S_{\min}\quad\text{and}\quad |x-x_{\text{left}}(E_{\max})|\le \delta x\ , \quad\text{where}\endmultline \tag 72  |\text{Error}_1(x)|\le C_\ast\sigma_x(b)\tau^{-1}+C_\ast^{\overline N} \int_a^b\sigma_x(t)\tau^{-\overline N}\ dt \qquad (\overline N\ \text{ will\ be\ picked\ later})\ .\tag 73 $$\noindent Also$$ \multline \sum\limits_{k\in \Bbb Z\cap [a,b]}f_{\text{Airey}}(x,k)\cdot f_{\text{other}} (x,k)=\int_a^bf_{\text{Airey}}(x,t)\cdot f_{\text{other}}(x,t)dt+\ \text{Error}_2 (x)\\ \text{if}\quad |E_0|\ge 2\hat c\ S_{\min}\quad\text{or}\quad |x-x_{\text{left}} (E_{\max})|>(\delta x)\ , \text{where}\endmultline\tag 74  |\text{Error}_2(x)|\le C_\ast^{\overline N}\int_a^b\sigma_x(t)\tau^ {-\overline N}\ dt\ .\tag 75 $$We estimate the L^1--norms of \text{Error}_1(x) and \text{Error}_2(x). >From the definitions of \sigma_x(t) and \tau we have$$\align &\int\limits_{|x-x_{\text{left}}(E_{\max})|<\delta x}\,\sigma_x(b)\tau^{-1}dx\\ &\le C_\ast(\Gamma^{-1}+(\frac{\delta x}{B_{\text{left}}})^{1/2}) \int\limits_{|x-x_{\text{left}}(E_{\max})|<\delta x}\sigma_x(b)dx\\ &\le C_\ast (\Gamma^{-1}+(\frac{\delta x}{B_{\text{left}}})^{1/2}) \int\limits_{|x-x_{\text{left}}(E_{\max})|<\delta x} \lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1})\\ &\qquad\qquad\qquad\qquad\qquad \cdot (1+\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1} |x-x_{\text{left}}(E_{\max})|)^{-1/2}\ dx\\ &\le C_\ast (\Gamma^{-1}+(\frac{\delta x}{B_{\text{left}}})^{1/2}) \lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1})\cdot \lambda_{\text{left}}^{-1/3}B_{\text{left}}^{1/2}(\delta x)^{1/2}\\ &=C_\ast(\Gamma^{-1}+(\frac{\delta x}{B_{\text{left}}})^{1/2}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2(S_{\min}\Gamma^{-1}) (\frac{\delta x}{B_{\text{left}}})^{1/2}\\ &=C_\ast(\Gamma^{-2}+(\frac{\delta x}{B_{\text{left}}})^{1/2} \Gamma^{-1})\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} (\frac{\delta x}{B_{\text{left}}})^{1/2}\\ &\le C_\ast (\Lambda^{-2}+(\frac{\delta x}{B_{\text{left}}})^{1/2} \Lambda^{-1})\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} (\frac{\delta x}{B_{\text{left}}})^{1/2}\quad\text{by\ (12)}\ . \tag 76\endalign $$\noindent Similarly,$$\multline \int\limits_{|x-x_{\text{left}}(E_0)|\delta x\ ,\endmultline\tag 80 $$\noindent since we know in that case that f_{\text{Airey}}(x,t) f_{\text{other}}(x,t) vanishes to infinite order at t=b. From (72) and (80), we have$$\split \sum\limits_{k\in \Bbb Z\cap [a,b]}f_{\text{Airey}}(x,k)f_{\text{other}} (x,k)=\int_a^bf_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\ dt\\ -f_{\text{Airey}}(x,b)f_{\text{other}}(x,b)\chm(b)+\ \text{Error}_3(x)\ , \endsplit\tag 81 $$\noindent with \text{Error}_3(x)=\text{Error}_1(x) for |E_0|\le 2\hat cS_{\min}, |x-x_{\text{left}}(E_{\max})|\le \delta x, and \text{Error}_3(x)=\ \text{Error}_2(x) otherwise. \noindent Equations (78) and (79) yield$$\align \int\limits_{I_{\text{BVP}}}|\text{Error}_3(x)|\ dx &\le C_\ast(\Lambda^{-2}+(\frac{\delta x}{B_{\text{left}}})^{1/2} \Lambda^{-1})\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} (\frac{\delta x}{B_{\text{left}}})^{1/2}\\ &\quad+C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\quad\text{if}\quad|E_0|\le 2\hat cS_{\min}\ ,\tag 82\endalign $$\noindent and$$ \int\limits_{I_{\text{BVP}}}|\text{Error}_3(x)| dx\le C_\ast\Lambda^{-N^\prime} \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min},\quad\text{if}\quad |E_0|>2\hat cS_{\min}\ .\tag 83 $$\noindent Here we used the fact that \text{Error}_3(x) is supported in \{|x-x_{\text{left}}(E_0)|2\hat cS_{\min}\ .\endsplit\tag 91$$ \noindent In the integral in (89), we want to change variable from $t$ to $E=E(t)$. This introduces a Jacobian factor $dt=[\frac 1\pi\phi^\prime(E)+\frac{1} {48\pi}\psi^\prime(E)]\ dE$, in place of the desired factor $\frac 1\pi\phi^\prime(E)$. To control the resulting error terms, we again use (84) and argue as follows. For $t\in \Cal J$, estimates (8), (9) give $\big|\frac{\psi^\prime(E)}{\phi^\prime(E)}\big|\le C_{\#}\Lambda^{-2}$ with $E=E(t)$, hence $|[1+ \frac{1}{48}\frac{\psi^\prime(E)}{\phi^\prime(E)} ]^{-1}-1|\le C_{\#}\Lambda^{-2}$. Therefore, $$\multline \int\limits_{I_{\text{BVP}}}\Big|\int_{a_0}^{b_0} f_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\ dt\\ -\int_{a_0}^{b_0} f_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\bigl[1+\frac{1} {48}\frac{\psi^\prime(E(t))}{\phi^\prime(E(t))}\bigr]^{-1} dt\Big|\ dx\\ \le C_{\#}\Lambda^{-2}\int_{a_0}^{b_0}\int_{I_{\text{BVP}}} |f_{\text{Airey}}f_{\text{other}}(x,t)|\ dxdt\\ \le C_\ast\Lambda^{-2}(b_0-a_0)\lambda_{\text{left}}^{-1} B_{\text{left}}^2(S_{\min}\Gamma^{-1})(\frac{\delta x} {B_{\text{left}}})^{1/2}\quad \text{(by\ (84))}\\ \le C_\ast\Lambda^{-2}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}(\frac{\delta x}{B_{\text{left}}} )^{1/2}\ , \text{by\ (86)\ and\ the\ fact\ that}\ b-a\le C_{\#}\Gamma\ .\endmultline\tag 92$$ \noindent On the other hand, changing variable from $t$ to $E=E(t)$ gives $$\multline \int_{a_0}^{b_0}f_{\text{other}}(x,t)f_{\text{Airey}} (x,t)\bigl[1+\frac{1}{48}\frac{\psi^\prime(E(t))} {\phi^\prime(E(t))}\bigr]^{-1}\ dt\\ =\frac 1\pi \int_{E_{\ell 0}}^{E_{hi}}g(x,E)\lambda_{\text{left}} ^{-2/3}\Bigl(\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\Bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))\ dE\ .\endmultline\tag 93$$ \noindent Here we used (51), (52) and $dt=\frac 1\pi (\phi^\prime(E)+\frac{1}{48}\psi^\prime(E))\ dE$ for $E=E(t)$ to write down the integrand on the right, and we used (85) for the limits of integration. Next we show that the region of integration on the right of (93) can be changed from $[E_{\ell o},E_{hi}]$ to $(-\infty,0]$ without affecting the value of the integral. Since $E_{\ell o}=E_0-c_{\#}^1S_{\min}$, {\it ASSUMPTION\/} 1 shows that $g(x,E)=0$ for $E2\hat cS_{\min}$, then $E_{hi}=\min\{0,E_0+c_{\#}^1S_{\min}\}> E_0+\hat cS_{\min}$, so $E_{hi}\le0$, and $g(x,E)=0$ for $E_{hi}\le E\le 0$. Thus the upper limit of integration in (93) may be changed from $E_{hi}$ to $0$ without changing the integral, provided $|E_0|>2\hat cS_{\min}$. On the other hand, if $|E_0|\le 2\hat cS_{\min}$, then already $E_{hi}=0$ so there is nothing to prove. Hence, (93) may be rewritten as $$\multline \int_{a_0}^{b_0}f_{\text{other}}(x,t)f_{\text{Airey}} (x,t)\bigl[1+\frac{1}{48}\frac{\psi^\prime(E(t))} {\phi^\prime(E(t))}\bigr]^{-1}\ dt\\ =\frac{1}{\pi}\int_{-\infty}^0 g(x,E)\lambda_{\text{left}} ^{-2/3}\bigl(\frac{\partial Y_{\text{left}}(x,E)} {\partial x} \bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}} (x,E))\ dE\ .\endmultline\tag 94$$ \noindent From (92) and (94) we get $$\multline \int_{a_0}^{b_0}f_{\text{other}}(x,t)f_{\text{Airey}} (x,t)\ dt=\\ \frac 1\pi\int_{-\infty}^0g(x,E)\lambda_{\text{ left}}^{-2/3}\bigl(\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))\ dE+\ \text{Error}_5(x) \endmultline\tag 95$$ \noindent with $$\int_{I_{\text{BVP}}}|\text{Error}_5(x)| dx \le C_\ast \Lambda^{-2}\lambda_{\text{left}}^{-1}B_{\text{left}}^2 S_{\min}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\ .\tag 96$$ \noindent Equations (89)$\ldots$(91) and (95), (96) imply $$\multline \sum\limits_{k\in \Bbb Z\cap [a,b]} f_{\text{Airey}}(x,k)f_{\text{other}}(x,k)=\\ \frac 1\pi \int_{-\infty}^0 g(x,E)\lambda_{\text{left}} ^{-2/3}\bigl(\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}} ^{2/3}Y_{\text{left}}(x,E))\ dE\\ -f_{\text{Airey}}(x,b)f_{\text{other}}(x,b)\chm (b)+\ \text{Error}_6(x)\endmultline\tag 97$$ \noindent with $$\split \int_{I_{\text{BVP}}}|\text{Error}_6(x)|\ dx\le C_\ast (\Lambda^{-2}+\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\Lambda^{-1}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\quad\text{if}\quad |E_0|\le 2\hat cS_{\min}\ ,\quad\text{and}\endsplit \tag 98$$ $$\split \int_{I_{\text{BVP}}}|\text{Error}_6(x)|\ dx\le C_\ast \Lambda^{-2}\lambda_{\text{left}}^{-1}B_{\text{left}}^2 S_{\min}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\quad\text{if}\quad |E_0|>2\hat cS_{\min}\ .\endsplit\tag 99$$ \noindent In equation (97) we want to change $f_{\text{Airey}}(x,b)f_{\text{other}}(x,b)\chm(b)$ to $f_{\text{Airey}}(x,b_0)f_{\text{other}}(x,b_0)\hfill\break\cdot \chm(b)$. To do so, we estimate for fixed $t\in \Cal J$: $$\multline \int_{I_{\text{BVP}}}\Big|\frac{\partial}{\partial t} \bigl\{f_{\text{Airey}}(x,t)f_{\text{other}} (x,t)\bigr\}\Big|\ dx\le C_\ast\int\limits_{|x-x_{\text{ left}}(E(t))|<\delta x}\sigma_x(t)\tau^{-1}\ dx\\ \le C_\ast(\Lambda^{-2}+\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\Lambda^{-1})\lambda_ {\text{left}}^{-1}B_{\text{left}}^2S_{\min} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ \text{by\ the\ proof\ of\ (76)\, with}\ E(t)\ \text{in\ place\ of}\ E_{\max}\ .\endmultline$$ \noindent Since $|b-b_0|2\hat cS_{\min}$, then $f_{\text{other}}(x,b)f_{\text{Airey}}(x,b)\hfill\break \equiv 0$ by (71 bis). Also, $|E_0|>2\hat cS_{\min}$ implies $E(b_0)=E_{hi}= \min(0,E_0+c_{\#}^1S_{\min})\hfill\break >E_0+\hat cS_{\min}$, so that $g(x,E(b_0))\equiv 0$ by {\it ASSUMPTION\/} 1, and hence $f_{\text{other}} (x,b_0)\equiv 0$. Thus $$\split f_{\text{Airey}}(x,b)f_{\text{other}}(x,b)=f_{\text{Airey}} (x,b_0)f_{\text{other}}(x,b_0)=0\\ \text{for\ all}\ x \in I_{\text{BVP}},\ \text{if}\ |E_0|>2\hat cS_{\min}\ .\endsplit\tag 101$$ \noindent Putting (100) and (101) into (97)$\ldots$(99) and noting that $E(b_0)=E_{hi}=0$ if $|E_0|\le 2\hat c S_{\min}$, we derive the following conclusion $$\multline \sum\limits_{k\in \Bbb Z\cap [a,b]}f_{\text{Airey}} (x,k)f_{\text{other}}(x,k)\\ =\frac 1\pi \int_{-\infty}^0 g(x,E)\lambda_{\text{left}} ^{-2/3}\bigl(\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))\ dE\\ -g(x,0)\lambda_{\text{left}}^{-2/3}\bigl(\frac {\partial Y_{\text{left}}(x,0)}{\partial x}\bigr)^{-1} A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}} (x,0))[\phi^\prime(0)]^{-1}\chm(b)\\ +\ \text{Error}_7(x)\ , \text{with}\endmultline\tag102$$ $$\split \int_{I_{\text{BVP}}}|\text{Error}_7(x)|\ dx\le C_\ast\Bigl(\Lambda^{-2}+\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\Lambda^{-1}\Bigr) \lambda_{\text{left}}^{-1}B_{\text{left}}^2 S_{\min}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast \Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\quad\text{if}\quad |E_0|\le 2\hat cS_{\min}\ ,\quad\text{and}\endsplit \tag 103$$ $$\split \int_{I_{\text{BVP}}}|\text{Error}_7(x)|\ dx\le C_\ast\Lambda^{-2}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\quad\text{if}\quad |E_0|>2\hat cS_{\min}\ .\endsplit\tag 104$$ Here we used also the fact that $g(x,0)\equiv 0$ if $|E_0|>2\hat cS_{\min}$. In particular, the expression $g(x,0)\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y_{\text{left}} (x,0)}{\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,0))[\phi^\prime(0)]^{-1}\chm(b)$ is defined to be zero whenever $g(x,0)\equiv 0$. We make this remark because eg. $Y_{\text{left}}(x,0)$ needn't be well-defined when $|E_0|>c_{\#}S_{\min}$. In (102), we want to replace $\chm(b)$ by $\chm(\frac 1\pi\phi(0)-\frac 12)$. If $|E_0|\le 2\hat cS_{\min}$, then $E(b_0)=E_{hi}=0$ as we just noted, and therefore $b_0=\frac 1\pi\phi(0)+\frac 1{48\pi}\psi(0)-\frac 12$ by (85), so the estimates for $\psi(E)$ in the WKB Theorems imply |b_0-(\frac 1\pi \phi(0)-\frac 12)| \overline C_{\#}\Lambda^{-1}\ \text{and}\ |E_0|\le 2\hat cS_{\min}\ .\endmultline\tag 105 $$\noindent If \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}, then we know only that$$ |\chm(b)-\chm(\frac 1\pi\phi(0)-\frac 12)|\le C_{\#}\ . \tag"(105 bis)" $$\noindent Under the assumptions of (105), we have$$\multline \int_{I_{\text{BVP}}}\Big|\bigl[\chm(b)-\chm(\frac 1\pi \phi(0)-\frac 12 )\bigr]g(x,0)\lambda_{\text{left}}^{-2/3} \bigl(\frac{\partial Y_{\text{left}}(x,0)} {\partial x}\bigr)^{-1}\\ \cdot A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}} (x,0))[\phi^\prime(0)]^{-1}\Big|\ dx\\ =|\chm(b)-\chm(\frac 1\pi\phi(0)-\frac 12)| \int_{I_{\text{BVP}}}|f_{\text{Airey}}(x,b_0) f_{\text{other}}(x,b_0)|\ dx\\ \le C_\ast \Lambda^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2(S_{\min}\Gamma^{-1}) (\frac{\delta x}{B_{\text{left}}})^{1/2}\quad \text{(by\ (84)\ and\ (105))}\\ \le C_\ast \Lambda^{-2}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\quad\text{by\ (12)}\ . \endmultline\tag 106 $$\noindent Estimate (106) holds when |E_0|\le 2\hat c S_{\min} and \min_{k \in \Bbb Z}|\phi(0)-\pi (k+1/2)|>\overline C_{\#}\Lambda^{-1}. If instead |E_0|\le 2\hat cS_{\min} and \min_{k\in \Bbb Z} |\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}, then instead of (105) we can use merely (105 bis), so the proof of (106) gives only$$\split \int_{I_{\text{BVP}}}\Big|\bigl[\chm(b)- \chm(\frac 1\pi \phi(0)-\frac 12)\bigr]g(x,0) \lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y_{\text{left}} (x,0)}{\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^ {2/3}Y_{\text{left}}(x,0))[\phi^\prime(0)]^{-1}\Big|\ dx\\ \le C_\ast \Lambda^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\ .\endsplit\tag 107 $$Finally, if |E_0|>2\hat cS_{\min}, then g(x,0)\equiv 0 so obviously we can replace \chm(b) by \chm(\frac 1\pi\phi(0)-\frac 12) in (102). Combining these observations with (102)\ldots(104), we obtain the following.$$\multline \sum\limits_{k\in \Bbb Z\cap [a,b]}f_{\text{Airey}} (x,k)f_{\text{other}}(x,k)\\ =\frac 1\pi\int_{-\infty}^0 g(x,E)\lambda_{\text{left}} ^{-2/3}\bigl(\frac{\partial Y_{\text{left}}(x,E)} {\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))\ dE\\ -g(x,0)\lambda_{\text{left}}^{-2/3} \bigl(\frac{\partial Y_{\text{left}}(x,0)} {\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,0))[\phi^\prime(0)]^{-1} \chm(\frac 1\pi \phi(0)-\frac 12)\\ +\ \text{Error}_8(x)\ ,\quad\text{where}:\endmultline\tag 108 $$If |E_0|\le 2\hat cS_{\min} and \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#} \Lambda^{-1} then$$\split \int_{I_{\text{BVP}}}|\text{Error}_8(x)|\ dx\le C_\ast(\Lambda^{-2}+\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\Lambda^{-1})\lambda_{\text{ left}}^{-1}B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\ .\endsplit\tag 109 $$\noindent If |E_0|\le 2\hat cS_{\min} and \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}, then$$\split \int_{I_{\text{BVP}}}|\text{Error}_8(x)|\ dx\le C_\ast\Lambda^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\ .\endsplit\tag 110 $$\noindent If |E_0|\ge2\hat cS_{\min}, then$$\split \int_{I_{\text{BVP}}}|\text{Error}_8(x)|\ dx \le C_\ast\Lambda^{-2}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast \Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\ .\endsplit\tag 111 $$\noindent The left--hand side of (108) is equal to the sum in (49), as one sees from (50). Therefore, (49) and (108)\ldots(111) combine to prove the following estimates.$$\multline \rho(x,g)=\frac 1\pi \int_{-\infty}^0 g(x,E)\lambda_ {\text{left}}^{-2/3}\bigl(\frac{\partial Y_{\text{left}}(x,E)}{\partial x}\bigr)^{-1} A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E))\ dE\\ -g(x,0)\lambda_{\text{left}}^{-2/3}\bigl(\frac{ \partial Y_{\text{left}}(x,0)}{\partial x}\bigr)^{-1} A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,0)) [\phi^\prime(0)]^{-1}\chm(\frac 1\pi \phi(0)-\frac 12)\\ +\ \text{Error}_9(x) , \quad\text{where}: \endmultline\tag 112 $$\noindent If |E_0|\le 2\hat cS_{\min} and \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#} \Lambda^{-1}, then$$\split \int_{I_{\text{BVP}}}|\text{Error}_9(x)|\ dx\le C_\ast(\Lambda^{4\varepsilon-2}+(\frac{\delta x} {B_{\text{left}}})^{1/2}\Lambda^{-1})\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x}{B_{ \text{left}}}\bigr)^{1/2}\\ + C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}+ C_\ast\Lambda^{10-N^{\prime\prime}}\Gamma\ .\endsplit \tag 113 $$\noindent If |E_0|\le 2\hat cS_{\min} and \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}, then$$\split \int_{I_{\text{BVP}}}|\text{Error}_9(x)|\ dx\le C_\ast\Lambda^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}+C_\ast\Lambda^{10-N^{\prime\prime}} \Gamma \ .\endsplit\tag 114 $$\noindent If |E_0|>2\hat cS_{\min}, then$$\split \int_{I_{\text{BVP}}}|\text{Error}_9(x)|\ dx\le C_\ast\Lambda^{4\varepsilon-2}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}+ C_\ast\Lambda^{10-N^{\prime \prime}}\Gamma \ .\endsplit\tag 115 $$\noindent Equations (113)\ldots(115) simplify, because$$\split \Lambda^{10-N^{\prime\prime}}\Gamma\ge c_{\#} \Lambda^{10-N^{\prime\prime}}S_{\min}S_{\text{left}} ^{-1/2}B_{\text{left}}\quad \text{(by\ (11))}\quad\ = c_{\#}\Lambda^{10-N^{\prime\prime}}\lambda_{\text{left}} ^{-1}B_{\text{left}}^2S_{\min}\\ >\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\ . \endsplit $$\noindent Hence the term C_\ast\Lambda^{-N^\prime} \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} may be deleted from the right--hand side of (113)\ldots(115). Also, it will be convenient to note that \Gamma\le C_\ast\int_{E_0-\hat cS_{\min}}^{E_0} \phi^\prime(E)\ dE (by (9)), and to substitute this in (113) \ldots(115). Thus we arrive at the main result of this section. \vglue 1pc \proclaim{Lemma} Suppose V(x), S(x), B(x), g(x,E) satisfy the hypotheses of the WKB Theorems with E_\infty=0, as well as {\it ASSUMPTIONS\/} 1\ldots 5 above. Then the microlocalized density for g(x,E) is given by$$\multline \rho(x,g)=\frac 1\pi\int_{-\infty}^0 g(x,E)\lambda_{\text{left}}^{-2/3}\bigl(\frac {\partial Y_{\text{left}}(x,E)}{\partial x} \bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3} Y_{\text{left}}(x,E))\ dE\\ -g(x,0)\lambda_{\text{left}}^{-2/3}\bigl( \frac{\partial Y_{\text{left}}(x,0)}{\partial x}\bigr)^ {-1}A^2(\lambda_{\text{left}}^{2/3}Y_{\text{left}} (x,0))[\phi^\prime(0)]^{-1}\chm(\frac 1\pi \phi(0)-\frac 12)\\ +\ \roman{Error}(x)\ ,\endmultline $$\noindent where \lambda_{\text{left}}, Y_{\text{left}}(x,E), \phi(E) are as in the WKB Theorems, and \roman{Error}(x) satisfies the following estimates. \roster \item"(a)" If |E_0|>2\hat cS_{\min}, then$$\split \int_{I_{\text{BVP}}}|\roman{Error}(x)|\ dx\le C_\ast\Lambda^{4\varepsilon-2}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\min}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{10-N^{\prime\prime}}\int_{E_0-\hat c S_{\min}}^{E_0}\phi^\prime(E)\ dE\ .\endsplit $$\item"(b)" If |E_0|\le 2\hat cS_{\min} and \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#} \Lambda^{-1}, then$$\split \int_{I_{\text{BVP}}}|\roman{Error}(x)| \ dx\le C_\ast(\Lambda^{4\varepsilon-2}+\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\Lambda^{-1}) \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda^{10-N^{\prime\prime}}\int_{E_0-\hat c S_{\min}}^{E_0}\phi^\prime(E)\ dE\ .\endsplit $$\item"(c)" If |E_0|\le 2\hat cS_{\min}, and \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#} \Lambda^{-1}, then$$\split \int_{I_{\text{BVP}}}|\roman{Error}(x)|\ dx\le C_\ast \Lambda^{-1}\lambda_{\text{left}}^{-1}B_{\text{left}}^2 S_{\min}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2} \\ +C_\ast\Lambda^{10-N^{\prime\prime}}\int_{E_0-\hat c S_{\min}}^{E_0}\phi^\prime(E)\ dE\ .\endsplit $$\endroster \noindent For the definition of S_{\min}, see the section on the WKB Theorems. The constant \overline C_{\#} depends only on \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in the hypotheses of the WKB Theorems. The constant C_\ast depends only on \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in the hypotheses of the WKB Theorems, and on \hat c, \hat C_{\beta} in {\it ASSUMPTIONS\/} 1 and 2.\endproclaim \vglue 1pc \demo{Remarks} \roster \item"{1.}" The term containing g(x,0) in the above lemma is defined to be identically zero whenever |E_0|\ge 2\hat cS_{\min}. \item"{2.}" For typical g(x,E) satisfying {\it ASSUMPTIONS\/} 1\ldots5, we have$$\split \int_{I_{\text{BVP}}}\Big|\int_{-\infty}^0 g(x,E)(E-V(x))_+^{-1/2}\ dE\Big|\ dx\sim \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\ . \endsplit $$\endroster\enddemo \smallskip \noindent Thus in inequality (a) above, the \text{Error} has L^1--norm roughly \Lambda^{4\varepsilon-2} as large as the L^1--norm of the main term. In (b) the factor is \Lambda^{4\varepsilon-2}+(\frac{\delta x} {B_{\text{left}}})^{1/2}\Lambda^{-1}, and in (c) the factor is \Lambda^{-1}. \vfill\eject \heading{\bf{THE MICROLOCALIZED DENSITY IN THE AIREY REGION II}}\endheading \medskip \noindent{\it Set-Up\/}. We are given a potential V(x) defined on a (possibly unbounded) interval I_{\text{BVP}}. On a subinterval I\subset I_{\text{BVP}} we are given positive functions S(x), B(x). We are given positive numbers \varepsilon, K, N with \varepsilon<\frac{1} {100} and K>100, N>K/\varepsilon^{20}. Set N^\prime=[\varepsilon N/500] and N^{\prime\prime}=\frac 32\varepsilon N^\prime-100 K-33. We are given a function g(x,E) defined on all of \Bbb R^2. We are given numbers E_0 and \delta E with E_0\le 0 and \delta E>0. \smallskip Define H=-\frac{d^2}{dx^2}+V(x) on I_{\text{BVP}}, with Dirichlet or Neumann boundary conditions. We make the following assumptions. \medskip \subhead{Assumptions on V(x), S(x), B(x) on I}\endsubhead \smallskip \roster \item"(X0)" If x,y\in I and |x-y|cB(x_{\text{left}}(E)), \text{dist}(x_{\text{rt}}(E),\partial I)> cB(x_{\text{rt}}(E)). \item"(X3)" For E\in \Cal J and x\in [x_{\text{left}}(E), x_{\text{left}} (E)+c_1B(x_{\text{left}}(E))] we have -V^\prime(x)>cS(x)B^{-1}(x). Similarly, for E\in \Cal J and x\in [x_{\text{rt}}(E)-c_1B(x_{\text{rt}}(E)), x_{\text{rt}}(E)] we have +V^\prime(x)>cS(x)B^{-1}(x). \item"(X4)" For E\in \Cal J and x\in [x_{\text{left}}(E)+c_1B (x_{\text{left}}(E)), x_{\text{rt}}(E)-c_1B(x_{\text{rt}}(E))] we have cS(x)E. \item"(X7)" For E\in \Cal J and x\in I_{\text{BVP}}, we have$$ \align V(x)&\ge \frac{100}{|x-x_{\text{left}}(E)|^2}\quad\text{if}\ xx_{\text{rt}}(E)+\frac 12\lambda_{\text{rt}}^K(E)B_{\text{rt}}(E)\ . \endalign $$\endroster \vglue 1pc \subhead Polynomial Growth Conditions on S(x), B(x)\endsubhead \smallskip \roster \item"(X8)" We have$$\align \max_{x\in I}B(x)&<\Lambda_{\min}^K\min_{x\in I}B(x)\ ,\\ \max_{x\in I}S(x)&<\Lambda_{\min}^K\min_{x\in I}S(x)\ ,\\ |I|&\le \Lambda_{\min}^K\min_{x\in I}B(x)\ .\endalign $$\endroster \vglue 1pc \subhead Assumptions on g(x,E)\endsubhead \smallskip \roster \item"(X9)" \text{supp}\ g(x,E)\subset \{|E-E_0|<\hat c(\delta E), |x-x_{\text{left}}(E)|\le (\delta x)\},\hfill\break with \lambda_{\text{left}}^{-\frac 23+2\varepsilon}(E_0)B_{\text{left}}(E_0)< (\delta x)<\frac{1}{20}\lambda_{\text{left}}^{-2\varepsilon}(E_0) B_{\text{left}}(E_0). \item"(X10)" |\partial_x^\alpha\partial_E^\beta g(x,E)|\le \hat C_{\alpha\beta}(\delta x)^{-\alpha}\Bigl[\frac{S_{\text{left}}(E_0) (\delta x)}{B_{\text{left}}(E_0)}\Bigr]^{-\beta} for all x, E. \item"(X11)" The constant \hat c in (X9) is bounded above by a certain small\hfill\break positive number determined by \varepsilon, K, N, c, c_1, C_\alpha, C in\hfill\break (X0)\ldots(X8) above. \endroster \vfill\eject \subhead The WKB Hypothesis\endsubhead \roster \item"(X12)" \Lambda_{\min} is bounded below by a certain large positive number determined by \varepsilon, K, N, c, c_1, C_\alpha, C, \hat c, \hat C_{\alpha\beta} in (X0)\ldots(X10) above. \endroster \smallskip Under these hypotheses, we will use the results of the previous section to compare the microlocalized density \rho(x,g) with its semiclassical approximation. We write c_{\#}, C_{\#}, C_{\#}^\alpha etc. for constants that depend only on \varepsilon, K, N, c, C, c_1, C_\alpha; and we write C_\ast, c_\ast, C_\ast^\alpha etc. for constants that depend also on \hat c, \hat C_{\alpha\beta}. Define \Cal J^0=[E_0-c_{\#}\delta E,E_0+c_{\#}\delta E]\cap (-\infty,0] for a small constant c_{\#}. Our present assumptions easily imply that for any E_0^\prime\in \Cal J^0, the hypotheses of the WKB Theorems are satisfied, with E_0^\prime in place of E_0, with E_\infty=0, and with 100K in place of K. However, our present assumptions on g(x,E) are different from the assumptions imposed on g in the previous section. Our current g(x,E) is smoother and has larger support than the g considered there. The hypotheses in the previous section were appropriate for applying the WKB Theorems. Our present hypotheses are natural for the study of integrals of squares of Airey functions, such as that in the Lemma of the previous section. The first step in our analysis will be to write g(x,E) as a sum of pieces g_\nu(x,E) with small supports, to which we can apply the Lemma of the previous section. This allows us to approximate \rho(x,E) closely by an integral involving the square of the Airey function. Most of the work in this section is devoted to comparing that integral with the semiclassical density \rho_{sc}(x,g). Note that x_{\text{left}}(E) and x_{\text{rt}}(E) play different r\^oles in the hypotheses. This is caused by our desire to study g(x,E) supported as in (X9), which is inherently asymmetrical. Let us start by cutting g(x,E) into suitable g_\nu(x,E). We use the following result. \vfill\eject \proclaim{Lemma 1} For all \hat c^\prime\frac 12 \hat c^\prime S_{\min}(E_\nu^\prime). Here, E_\nu^\prime is any point in \Cal J_\nu^\prime\subset \Cal J^0. Also \big|\bigl(\frac{d}{dE}\bigr)^\beta\theta_\nu(E)\big|\le C_{\#}^\beta (\text{length}\ \Cal J_\nu^\prime)^{-\beta}\le C_{\#}^\beta([\hat c^\prime]^2 S_{\min}(E_\nu^\prime))^{-\beta} for E\in \Cal J^0. Since E=0 belongs to at most C_{\#} of the doubles of the \Cal J_\mu, and length \Cal J_\mu\sim S_{\min}(E_\nu^\prime) for E_\nu^\prime \in \Cal J_\mu, it follows that |E_\nu^\prime|\le c_{\#}S_{\min}(E_\nu^\prime) for at most C_{\#}M of the E_\nu^\prime. Extend \theta_\nu(E) from \Cal J^0 to \Bbb R^1, preserving the properties$$ \Big|\bigl(\frac{d}{dE}\bigr)^\beta\theta_\nu(E)\Big|\le C_{\#}^\beta ([\hat c^\prime]^2 S_{\min}(E_\nu^\prime))^{-\beta}\quad\text{and}  \theta_\nu(E)=0\quad\text{for}\quad |E-E_\nu^\prime|>\hat c^\prime S_{\min}(E_\nu^\prime)\ . $$\noindent Properties (a)\ldots(e) are then obvious.\qquad\blacksquare \smallskip Now we can write g(x,E)=\sum\limits_\nu g_\nu(x,E) for E\le 0. We take g_\nu(x,E)= \theta_\nu(E)g(x,E), where \theta_\nu(E) and E_\nu^\prime come from Lemma 1, with \hat c^\prime equal to a small enough constant of the form c_\ast. The hypotheses of the WKB Theorems, and the assumptions of the preceding section are satisfied, with: E_\nu^\prime in place of E_0; 0 in place of E_\infty; g_\nu(x,E) in place of g(x,E); 100K in place of K; and \hat c^\prime in place of \hat c. (Note that our present \delta E is not the same as the \delta E in the previous section. Also note that B_{\text{left}}(E)\sim B_{\text{left}}(E_0) and S_{\text{left}}(E)\sim S_{\text{left}}(E_0) for E\in \Cal J^0, since |E-E_0|2\hat c^\prime S_{\min}(E_\nu^\prime)\ . \endmultline\tag 2$$ $$\multline \int_{I_{\text{BVP}}}|\Cal E_\nu^1(x)|\, dx\le C_\ast\Bigl(\Lambda_{\min} ^{4\varepsilon-2}+\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\Lambda_{\min}^{-1}\Bigr)\lambda_{\text{left}} ^{-1}B_{\text{left}}^2S_{\min}(E_\nu^\prime)\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)\, dE\\ \text{if}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)\ \text{and}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#} \Lambda_{\min}^{-1}\ .\endmultline\tag 3$$ $$\multline \int_{I_{\text{BVP}}}|\Cal E_\nu^1(x)|\, dx\le C_\ast\Lambda_{\min}^{-1}\lambda_{\text{left}}^{-1}B_{\text{left}}^2 S_{\min}(E_\nu^\prime)\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)\, dE\\ \text{if}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)\quad \text{and}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#} \Lambda_{\min}^{-1}\ .\endmultline\tag 4$$ \noindent Here we set\lambda_{\text{left}}=\lambda_{\text{left}}(E_0)$,$B_{\text{left}}=B_{\text{left}}(E_0)$,$S_{\text{left}}=S_{\text{left}}(E_0)$, and we use the fact that$\lambda_\nu\sim\lambda_{\text{left}}$,$B_{\text{left}}(E_\nu^\prime)\sim B_{\text{left}}$,$S_{\text{left}} (E_\nu^\prime)\sim S_{\text{left}}$,$\Lambda(E_\nu^\prime)\ge \Lambda_{\min}$to deduce (2), (3), (4) from the lemma of the preceding section. Throughout this section, we make the convention that terms such as the$g_\nu(x,0)$-term in (1) are defined to be identically zero whenever$g_\nu(x,0)\equiv 0$. (Recall that there are cases in which$g_\nu(x,0)\equiv 0$but$Y_\nu(x,0)$isn't well-defined.) Our next task is to sum (1) over$\nu$to get a formula for$\rho(x,g)$. To do so , we need to know that the various$\lambda_\nu^{2/3}Y_\nu(x,E)$agree closely with one another as$\nu$varies. This follows from the proof of the WKB Theorems, but we prefer to give a more self-contained argument. We use the fact that$\lambda_\nu\sim \lambda_{\text{left}}, and that $$\multline \big|\lambda_\nu^2(\partial_xY_\nu)^2Y_\nu+\{Y_\nu,x\}-(E-V(x))\big|< C_{\#}\lambda_{\text{left}}^{-N^{\prime}}S_{\text{left}}\\ \text{for}\quad |E-E_\nu^\prime|c_{\#} B_{\text{left}}^{-1}\quad \text{and}\quad\Big|\frac{\partial Y}{\partial E} (x,E)\Big|>c_{\#}S_{\text{left}}^{-1}\\ \text{for}\quad |E-E_0|c>0, |y_i(0)|0\}. Hence |\{y_1,x\}-\{y_2,x\}| \le \overline C\lambda^{-m+3\varepsilon} for |x|<\lambda^{-\varepsilon}, and so by hypothesis,$$ \Big|\bigl(\frac{dy_1}{dx}\bigr)^2y_1-\bigl(\frac{dy_2}{dx}\bigr)^2y_2\Big| \le \overline C\lambda^{-m+3\varepsilon-2}\quad\text{for}\quad |x|< \lambda^{-\varepsilon}\ .\tag 10 $$\noindent Let x_0 be the zero of y_1 with |x_0|\le \overline C \lambda^{-2}. Estimate (10) and \frac{dy_2}{dx}>c show that$$ |y_2(x_0)|<\overline C\lambda^{-m+3\varepsilon-2}\ .\tag 11 $$\noindent Set y_3(x)=y_2(x)-y_2(x_0), so that y_1(x) and y_3(x) both vanish at x_0, \frac{d}{dx}y_3>c>0, and$$ \Big|\bigl(\frac{dy_1}{dx}\bigr)^2y_1-\bigl(\frac{dy_3}{dx}\bigr)^2 y_3\Big|\le \overline C\lambda^{-m+3\varepsilon-2}\quad\text{for}\quad |x|<\lambda^{-\varepsilon}\ .\tag 12 $$\noindent Both y_1(x) and y_3(x) are N^{\text{th}} order polynomials, bounded a-priori on \{|x|\le c\}. If x_02\hat c^\prime S_{\min}(E_\nu^\prime) we have g_\nu(x,0)\equiv 0, so the left side of (20) is equal to zero. Combining (19) and (20) with (1)\ldots(4), we obtain the following results.$$\multline \rho(x,g_\nu)=\frac 1\pi\int_{-\infty}^0 g_\nu(x,E)\lambda_{\text{left}} ^{-2/3}\bigl(\frac{\partial Y(x,E)}{\partial x}\bigr)^{-1}A^2 (\lambda_{\text{left}}^{2/3}Y(x,E))\, dE\\ -g_\nu(x,0)\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y(x,0)} {\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3}Y(x,0))[\phi^\prime (0)]^{-1}\chm(\frac 1\pi \phi(0)-\frac 12)+\Cal E_\nu^2(x)\ .\endmultline \tag 21 \multline \text{If}\quad |E_\nu^\prime|>2\hat c^\prime S_{\min}(E_\nu^\prime)\ ,\ \text{then}\\ \int_{I_{\text{BVP}}}|\Cal E_\nu^2(x)|\, dx\le C_\ast \Lambda_{\min}^{4\varepsilon-2}\lambda_{\text{left}}^{-1}B_{\text{left}} ^2S_{\min}(E_\nu^\prime)\cdot \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)\, dE +C_\ast S_{\min}(E_\nu^\prime)\lambda_{\text{left}}^{-N^{\prime\prime}} B_{\text{left}}^2\ .\endmultline\tag 22 \multline \text{If}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime) \ \text{and}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#} \Lambda_{\min}^{-1}\ ,\ \text{then}\\ \int_{I_{\text{BVP}}}|\Cal E_\nu^2(x)|\, dx\le C_\ast\bigl(\Lambda_{\min} ^{4\varepsilon-2}+\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2} \Lambda_{\min}^{-1}\bigr)\lambda_{\text{left}}^{-1}B_{\text{left}} ^2S_{\min}(E_\nu^\prime)\cdot \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2} \\ +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)\, dE+ C_\ast S_{\min}(E_\nu^\prime)\lambda_{\text{left}}^{-N^{\prime\prime}} B_{\text{left}}^2 +C_\ast\lambda_{\text{left}}^{1-N^{\prime\prime}}\ .\endmultline\tag 23 \multline \text{If}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)\ \text{and}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#} \Lambda_{\min}^{-1}\ ,\ \text{then}\\ \int_{I_{\text{BVP}}}|\Cal E_\nu^2(x)| dx\le C_\ast\Lambda_{\min}^{-1} \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}(E_\nu^\prime)\cdot \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\ +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)dE+C_\ast S_{\min}(E_\nu^\prime)\lambda_{\text{left}}^{-N^{\prime\prime}}B_{\text{left}}^2 +C_\ast\lambda_{\text{left}}^{1-N^{\prime\prime}}\ .\endmultline\tag 24 $$\noindent We now sum (21)\ldots(24) over all \nu, to get a formula for \rho(x,g). To sum the error terms on the right of (22)\ldots(24), we use the following observations. First of all, \sum_\nu S_{\min} (E_\nu^\prime)\le \sum_\nu C_\ast|\Cal J_\nu^{\prime\prime}|\le C_{\ast}|\bigcup_{\nu}\Cal J_\nu^{\prime\prime}|\le C_\ast S_{\text{left}} and \sum_\nu\int_{E_\nu^\prime-\hat c^\prime S_{\min}(E_\nu^\prime)} ^{E_\nu^\prime}\phi^\prime(E)\, dE\hfill\break \le \sum_\nu\int_{\Cal J_\nu^{\prime\prime} \cap \Cal J}\phi^\prime(E)\, dE\le C_{\ast}\int_\Cal J\phi^\prime(E)\, dE\le C_{\ast}\phi(\text{max}\ \Cal J) by Lemma 1. Secondly, there are at most C_\ast distinct \nu for which (23) or (24) applies. This also follows from Lemma 1. If |E_0|>2\hat c(\delta E), then (23), (24) will not be needed at all, provided we take \hat c^\prime small enough depending on \hat c. From these remarks, we get the following results by summing (21)\ldots(24) over \nu.$$\multline \rho(x,g)=\frac 1\pi \int_{-\infty}^{0}g(x,E)\lambda_{\text{left}}^{-2/3} \bigl(\frac{\partial Y(x,E)}{\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}} ^{2/3}Y(x,E))\, dE\\ -g(x,0)\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y(x,0)}{\partial x} \bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3}Y(x,0))[\phi^\prime(0)]^{-1} \chm(\frac 1\pi \phi(0)-\frac 12)+\Cal E_3(x)\ .\endmultline\tag 25 \multline \text{If}\quad |E_0|>2\hat c(\delta E)\ ,\ \text{then}\\ \int_{I_{\text{BVP}}}|\Cal E_3(x)| dx\le C_\ast\Lambda_{\min}^{4\varepsilon-2} \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\text{left}}\cdot \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}+C_\ast\Lambda_{\min} ^{10-N^{\prime\prime}}\phi(\text{max}\ \Cal J)\\ +C_\ast S_{\text{left}}\lambda_{\text{left}} ^{-N^{\prime\prime}}B_{\text{left}}^2 \le C_\ast\Lambda_{\min}^{4\varepsilon-2}\lambda_{\text{left}} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}+C_\ast \Lambda_{\min}^{10-N^{\prime\prime}}\phi(\text{max}\ \Cal J)\\ +C_\ast\Lambda_{\min}^{2-N^{\prime \prime}}\equiv\Omega\ .\endmultline\tag 26 \multline \text{If}\quad |E_0|\le 2\hat c(\delta E)\ \text{and}\ \min_{k\in \Bbb Z} |\phi(0)-\pi(k+1/2)|>\overline C_{\#}\Lambda_{\min}^{-1}\ ,\ \text{then}\\ \int_{I_{\text{BVP}}}|\Cal E_3(x)| dx\le \Omega+C_\ast\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{1/2}\Lambda_{\min}^{-1}\lambda_{\text{left}}^{-1} B_{\text{left}}^2S_{\text{left}}\bigl(\frac{\delta x}{B_{\text{left}}} \big)^{1/2}+C_\ast\lambda_{\text{left}}^{1-N^{\prime\prime}}\\ \le C_\ast\Omega+C_\ast\Lambda_{\min}^{-1}\lambda_{\text{left}}\bigl( \frac{\delta x}{B_{\text{left}}}\bigr)\ .\endmultline\tag 27 \multline \text{If}\ |E_0|\le 2\hat c(\delta E)\ \text{and}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda_{\min}^{-1} \ ,\ \text{then}\\ \int_{I_{\text{BVP}}}|\Cal E_3(x)| dx\le C_\ast\Lambda_{\min}^{-1} \lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\text{left}}\bigl(\frac{ \delta x}{B_{\text{left}}}\bigr)^{1/2} +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\phi(\text{max}\ \Cal J)\\ +C_\ast\lambda_{\text{left}}^{-N^{\prime\prime}}S_{\text{left}} B_{\text{left}}^2 +C_\ast\lambda_{\text{left}}^{1-N^{\prime\prime}}\le C_\ast\Lambda_{\min}^{-1} \lambda_{\text{left}}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2} +C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\phi(\text{max}\ \Cal J)\\ +C_\ast\Lambda_{\min}^{2-N^{\prime\prime}}\ .\endmultline\tag 28 $$Equations (25)\ldots(28) reduce the study of \rho(x,g) to that of an integral involving the Airey function. It remains to relate (25) to the semiclassical density \rho_{sc}(x,g). Our goal is to control \rho(x,g)-\rho_{sc}(x,g) in the Sobolev norm H^{-1}. The main step is to understand the integrals$$\multline \Cal F(x)=\int_{-\infty}^x\int_{-\infty}^0 g(x^\prime,E)\lambda_{\text{left}} ^{-2/3}\bigl(\frac{\partial Y(x^\prime,E)}{\partial x^\prime}\bigr)^{-1} [A^2(\lambda_{\text{left}}^{2/3}Y(x^\prime,E))\\ -\frac 12 \lambda_{\text{left}} ^{-1/3}(Y(x^\prime,E))_+^{-1/2}]\, dE\, dx^\prime\endmultline\tag 29 $$\noindent and$$\multline \Cal G(x)=\int_{-\infty}^{x}g(x^\prime,0)\lambda_{\text{left}}^{-2/3} \bigl(\frac{\partial Y(x^\prime,0)}{\partial x^\prime}\bigr)^{-1} [A^2(\lambda_{\text{left}}^{2/3}Y(x^\prime,0))\\ -\frac 12\lambda_{\text{left}}^{-1/3}(Y(x^\prime,0))_+^{-1/2}]\, dx^\prime \ . \endmultline\tag 30 $$\noindent We begin with \Cal F, and suppose that |E_0|\le 2\hat c (\delta E). Set y=Y(x^\prime,E) for fixed x^\prime, and define h(y,x^\prime) by h(y,x^\prime)=g(x^\prime,E)\bigl(\frac{\partial Y(x^\prime, E)}{\partial x^\prime}\bigr)^{-1}\bigl(\frac{\partial Y(x^\prime,E)} {\partial E}\bigr)^{-1} for y=Y(x^\prime,E). Thus g(x^\prime,E)\bigl(\frac{\partial Y(x^\prime,E)}{\partial x^\prime}\bigr)^{-1} dE=h(y,x^\prime)dy for fixed x^\prime. Also note that E\to y=Y(x^\prime,E) is increasing for |E-E_0|<\hat c(\delta E), |x-x_{\text{left}}(E)|0. So at x=x_{\text{left}}(E) we have \frac{\partial Y}{\partial E}, \frac{\partial Y}{\partial x}>0.) Let us estimate the inner integral in (31). We know that$$\align |\partial_{x^\prime}^\alpha\partial_E^\beta g(x^\prime,E)|&\le \hat C_{\alpha\beta}(\delta x)^{-\alpha}\bigl(\frac{S_{\text{left}}} {B_{\text{left}}}\delta x\bigr)^{-\beta}\\ |\partial_{x^\prime}^\alpha\partial_E^\beta(\frac{ \partial Y}{\partial x^\prime}(x^\prime,E))^{-1}|&\le C_{\#}^{\alpha\beta}B_{\text{left}}\cdot B_{\text{left}}^{-\alpha} S_{\text{left}}^{-\beta}\le C_{\#}^{\alpha\beta} B_{\text{left}}(\delta x)^{-\alpha}\bigl(\frac{S_{\text{left}}} {B_{\text{left}}}\delta x\bigr)^{-\beta}\\ |\partial_{x^\prime}^\alpha\partial_E^\beta(\frac{\partial Y}{\partial E} (x^\prime,E))^{-1}|&\le C_{\#}^{\alpha\beta}S_{\text{left}}\cdot B_{\text{left}}^{-\alpha}S_{\text{left}}^{-\beta}\le C_{\#} ^{\alpha\beta}S_{\text{left}}(\delta x)^{-\alpha}\bigl(\frac{S_{\text{left}}} {B_{\text{left}}}\delta x\bigr)^{-\beta}\ , \endalign $$\noindent so f(x^\prime,E)\equiv g(x^\prime,E)\bigl(\frac{\partial Y} {\partial x^\prime}(x^\prime,E))^{-1}\bigl(\frac{\partial Y}{\partial E} (x^\prime,E))^{-1} satisfies$$ |\partial_{x^\prime}^\alpha\partial_E^\beta f(x^\prime,E)|\le C_\ast^{\alpha \beta}S_{\text{left}}B_{\text{left}}(\delta x)^{-\alpha} \bigl(\frac{S_{\text{left}}}{B_{\text{left}}}\delta x\bigr)^{-\beta}\ . \tag 32 $$Now y=Y(x^\prime,E) is a smooth function of \frac{x^\prime-x_{\text{left}} (0)}{B_{\text{left}}} and \frac{E}{S_{\text{left}}}. That smooth function has its C^\infty seminorms bounded a-priori, and the derivative with respect to the second argument (\frac{E}{S_{\text{left}}}) is bounded a-priori away from zero. Hence \frac{E}{S_{\text{left}}} is a smooth function of \frac{x^\prime-x_{\text{left}}(0)}{B_{\text{left}}} and y, i.e.$$ |\partial_{x^\prime}^\alpha\partial_y^\beta E(y,x^\prime)|\le C_{\#} ^{\alpha\beta}S_{\text{left}}B_{\text{left}}^{-\alpha}\,\ \text{with}\ E(y,x^\prime)\ \text{the\ solution\ of}\ Y(x^\prime,E)=y\ . \tag 33 $$\noindent By definition, h(y,x^\prime)=f(x^\prime,E(y,x^\prime)). Hence the derivative \partial_{x^\prime}^\alpha\partial_y^\beta h(y,x^\prime) is a sum of terms$$ \partial_{x^\prime}^{\alpha_0}\partial_E^\gamma f(x^\prime,E) \bigm|_{E=E(y,x^\prime)} \cdot \prod\limits_{\nu=1}^{\gamma}[\partial_{x^\prime}^{\alpha_\nu} \partial_y^{\beta_\nu}E(y,x^\prime)]\tag 34 $$\noindent with \alpha_0+\alpha_1+\ldots+\alpha_\gamma=\alpha, \beta_1+ \ldots+\beta_\gamma=\beta, \alpha_\nu+\beta_\nu\ge 1. (In particular, 0\le \gamma\le \beta+(\alpha_1+\ldots+\alpha_\gamma).) The term (34) is dominated by\hfill\break C_\ast^{\alpha\beta}S_{\text{left}}B_{\text{left}} (\delta x)^{-\alpha_0}\bigl(\frac{S_{\text{left}}}{B_{\text{left}}} \delta x\bigr)^{-\gamma}\cdot S_{\text{left}}^\gamma B_{\text{left}}^ {-(\alpha_1+\ldots+\alpha_\gamma)} (by (32) and (33)), which is dominated by C_\ast^{\alpha\beta}S_{\text{left}}B_{\text{left}} (\delta x)^{-\alpha}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-\beta} since 0\le \gamma\le \beta+(\alpha_1+\ldots+\alpha_\gamma). Thus,$$ |\partial_{x^\prime}^\alpha\partial_y^\beta h(y,x^\prime)| \le C_\ast^{\alpha\beta} S_{\text{left}}B_{\text{left}}(\delta x)^{-\alpha}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{-\beta}\ .\tag 35 $$\noindent For y10\ . \tag 51$$ We look separately at the casesx>R$,$10R\ ,\quad \text{and\ that} \tag 53  F(x)=-\int_x^\infty \theta(x,y)[A^2(y)-\frac 12 y_+^{-1/2}]\, dy+O(R^{-5/2}) \quad\text{for}\quad 1010$. Hence $$A^2(y)-\frac 12 y_+^{-1/2}=\ \text{Re}\, \Bigl[\frac{e^{\pm i\frac \pi 2}\, e^{\frac 43 iy^{3/2}}}{2y^{1/2}}\Bigr]+\ \text{Re}\Bigl[\frac {\text{(const)}e^{\frac 43iy^{3/2}}}{y^2}\Bigr]+O(y^{-7/2})\ .$$ \noindent Here the$O(y^{-7/2})$error term contributes to the right-hand side of (54) at most$\int_x^\infty(R^{-1}|x-y|)y^{-7/2}dy\le C^\prime R^{-1}x^{-3/2}$, since$|\theta(x,y)|\le CR^{-1}|x-y|$. Therefore, (54) implies $$\multline F(x)=\sum\limits_{k=\pm 1}c_k\int_x^\infty \theta(x,y)\frac{ e^{\frac 43 iky^{3/2}}}{y^{1/2}}\, dy +\sum\limits_{k=\pm 1}\tilde c_k\int_x^\infty\theta(x,y)\frac{e^{\frac 43 iky^{3/2}}}{y^2}dy+O(R^{-1}x^{-3/2})\\ \text{for}\quad 10100\ .\endmultline$$ \noindent Since$\partial_y^m\theta(x,y)\mid_{y=x}=\cases O(R^{-1}) &\text{for $m\ge 1$}\\ 0&\text{for $m=0$}\endcases, this yields \Big|\int_x^\infty \frac{e^{\frac 43 iky^{3/2}}}{y^{q}} \theta(x,y)dy\Big|\le C^\prime R^{-1}x^{-1}\quad\text{for}\quad 1010\ . \endalign \noindent Estimates (46 bis)\ldots$(49) show that$\theta_{\text{in}}$,$\theta_{\text{out}}$both satisfy hypothesis (51). Applying (50) and (61), (61 bis) for$\theta=\theta_{\text{in}}, \theta_{\text{out}}, and comparing with (45), we learn that \align |\tilde \Cal F(\tilde x)|&\le C_\ast^\prime\Bigl[\lambda_{\text{left}}^{4/3} \frac{\delta x}{B_{\text{left}}}\Bigr]\bigl[C_{\#} \lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\Bigr]^{-1} (1+|\tilde x|)^{-1}\\ &=C_\ast\lambda_{\text{left}}^{2/3}(1+|\tilde x|)^{-1}\quad\text{for}\quad |\tilde x|\le C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}} \ ,\endalign \noindent and $$|\tilde \Cal F(\tilde x)|\le C_\ast^\prime[\lambda_{\text{left}}^{4/3}] \bigl[\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\Bigr]^{-5/2} \quad\text{for}\quad C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x} {B_{\text{left}}}<|\tilde x|c_{\#} \lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}, so$$\multline \Big|\bigl(\frac{d}{d\xi}\bigr)^\gamma x^\prime\Big|\le C_{\#}^\gamma B_{\text{left}}\lambda_{\text{left}}^{-\frac 23\gamma}\ (\gamma\ge 1)\ ,\\ \text{where}\quad x^\prime=x^\prime(\xi)\quad \text{is\ the\ solution\ of}\ \lambda_{\text{left}}^{2/3}Y(x^\prime,0)=\xi\ .\endmultline\tag 73 $$\noindent The derivative \lambda_{\text{left}}^{4/3}\bigl(\frac{d}{d\xi}\bigr)^m\tilde g(\xi) is a sum of terms$$\multline \Bigl[\partial_{x^\prime}^\alpha\bigl\{g(x^\prime,0)\bigl(\frac{\partial Y} {\partial x^\prime}(x^\prime,0)\bigr)^{-2}\bigr\}\Big]\bigm|_{x^\prime=x^\prime(\xi)}\cdot \prod\limits_{\nu=1}^\alpha\Bigl[\bigl(\frac{d}{d\xi}\bigr)^{m_\nu} x^\prime\Bigr]\ ,\\ \text{with}\quad m_\nu\ge 1\quad\text{and}\quad m_1+\ldots+m_\alpha=m \ . \endmultline\tag 74 $$\noindent By (72) and (73), the term (74) is dominated by C_\ast^mB_{\text{left}}^2(\delta x)^{-\alpha}\cdot B_{\text{left}}^\alpha \lambda_{\text{left}}^{-\frac 23 m}, which in turn is at most C_\ast^mB_{\text{left}}^2 \bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\bigr)^{-m}, since 0\le \alpha\le m. Hence,$$ \Big|\bigl(\frac{d}{d\xi}\bigr)^m\tilde g(\xi)\Big|\le C_\ast^m\lambda_{\text{left}}^{-4/3}B_{\text{left}}^2\bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\bigr)^{-m}\ , \quad m\ge 0\ .\tag 75 $$\noindent Also, since g(x^\prime,0) is supported in |x^\prime-x_{\text{left}}(0)|<\delta x, it follows that \tilde g(\xi) is supported in \{|\xi|\le C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x} {B_{\text{left}}}\}. \noindent We investigate in general integrals of the form$$\multline G(x)=\int_{-\infty}^x\theta(\xi)[A^2(\xi)-\frac 12\xi_+^{-1/2}]\, d\xi\ ,\quad \text{with}\quad \big|\bigl(\frac{d}{d\xi}\bigr)^m\theta\big|\le C_mR^{-m}\ ,\\ \text{supp}\ \theta\subset \{|\xi|10\ .\endmultline\tag 76 $$Taking \theta(\xi)=\frac{\tilde g(\xi)}{\lambda_{\text{left}}^{-4/3} B_{\text{left}}^2} and R=C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x} {B_{\text{left}}}, we satisfy the assumptions of (76); and (71), (76) together give$$ \Cal G(x)=\lambda_{\text{left}}^{-4/3}B_{\text{left}}^2G(\lambda_{\text{left}} ^{2/3}Y(x,0))\quad\text{for}\quad |x-x_{\text{left}}(0)|R, $10R$ we have $G(x)=\int_{-\infty}^\infty \theta(\xi)[A^2(\xi)-\frac 12\xi_+^{-1/2}]\, d\xi=O(R^{-5/2})$, by (A3) from the section Review of Earlier Results". For $1010$, we can throw away information from (82) to get $$\multline \int_{|x-x_{\text{left}}(E_0)|2\hat c(\delta E)\ \text{then}\\ \frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|\overline C_{\#} \Lambda_{\min}^{-1}\ , \quad\text{then}\\ \frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|c>0 and |X(0,0)|<<1. Let \varphi(t) be supported in \{|t|\le \delta \} and satisfy |\varphi^{(m)}(t)|\le C_m\delta^{-m}. Set F(s,\tau)=\int_{-\infty} ^s\varphi(t)(X(t,\tau))_+^{-1/2}dt. Then \int_{|s|<1}|F(s,0)- F(s,\tau)|^2ds\le C\tau^2(\delta^{-1}+\ell \text{n}\ \frac{1}{|\tau|}) for |\tau|<<1.\endproclaim \medskip \demo{Proof} Set \xi=X(t,\tau), and define t=T(\xi,\tau) to be the solution of X(t,\tau)=\xi. Changing variable from t to \xi in the definition of F gives F(s,\tau)=\break \int_{\{0<\xic>0 for |s|\le 1, it follows that$$\align \int_{|s|\le 1}|F(s,\tau)-F(s,0)|^2ds&\le C\tau^2\int_{|s|\le 1} (|X(s,0)|+|\tau|)^{-1}ds+C\delta^{-1}\tau^2\\ &\le C\tau^2(\delta^{-1}+\ell\text{n} \frac{1}{|\tau|})\ .\endalign $$\line{\hss\blacksquare}\enddemo \medskip To apply the lemma, fix E with |E-E_0|<2\hat c(\delta E), and set X(t,\tau)\hfill\break =B_{\text{left}}^2\bigl[\bigl(\frac{\partial Y(x,E)}{\partial x}\bigr)^2 Y(x,E)+\tau\{Y(x,E),x\}\bigr] with x=x_{\text{left}}(E)+B_{\text{left}}t. Also, we set \varphi(t)=g(x_{\text{left}}(E)+B_{\text{left}}t,E) and \delta=\bigl(\frac{\delta x}{B_{\text{left}}}\bigr). The hypotheses of Lemma 4 are then satisfied, and we use the conclusion of Lemma 4 with \tau=\lambda_{\text{left}}^{-2}. Note that \delta^{-1}+\ell\text{n} \frac{1}{|\tau|}\sim \delta^{-1}=\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{-1} since we take \frac{\delta x}{B_{\text{left}}}< \lambda_{\text{left}}^{-\varepsilon}. Hence, Lemma 4 gives$$ C_\ast\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1} \lambda_{\text{left}}^{-4} \ge \int_{|s|\le 1}|F(s,\lambda_{\text{left}}^{-2})-F(s,0)|^2ds\ , \quad\text{with}\tag 91 \align F(s,\tau)&=\int_{-\infty}^s g(x_{\text{left}}(E)+B_{\text{left}}t,E)\cdot \bigl[B_{\text{left}}^2((\partial_xY)^2Y+\tau\{Y,x\})\bigr]_+^{-1/2}dt\\ &=\int_{-\infty}^{x_{\text{left}}(E)+B_{\text{left}}s}g(x,E)B_{\text{left}} ^{-1}\bigl[(\partial_xY)^2Y+\tau\{Y,x\}]_+^{-1/2}B_{\text{left}}^{-1}dx\ .\\ \tag 92\endalign $$\noindent Hence, making the change of variable x^\prime=x_{\text{left}} (E)+B_{\text{left}}s in (91), (92), we get$$\multline \frac{1}{B_{\text{left}}}\int_{|x^\prime-x_{\text{left}}(E)|2\hat c(\delta E)$, then (97) holds trivially, since$g(x,0)\equiv 0$. The right-hand sides of (95) and (97) are dominated by the right-hand sides of (88), (89), (90), since the latter all contain$\lambda_{\text{left}}^{-2} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}$. Hence in (84), we may replace$\lambda_{\text{left}}^2(\partial_xY)^2Y$by$\lambda_{\text{left}}^2(\partial_xY)^2Y+\{Y,x\}$on the right, without affecting the error estimates (88)$\ldots$(90). That is, $$\multline \rho(x,g)=\frac{1}{2\pi}\int_{-\infty}^0 g(x,E)\bigl(\lambda_{\text{left}}^2 \bigl(\frac{\partial Y(x,E)}{\partial x}\bigr)^2Y(x,E)+\{Y(x,E),x\}\bigr)_+ ^{-1/2}dE\\ -\frac 12 g(x,0)\bigl(\lambda_{\text{left}}^2\bigl(\frac{\partial Y(x,0)} {\partial x}\bigr)^2Y(x,0)+\{Y(x,0),x\}\bigr)_+^{-1/2}[\phi^\prime(0)]^{-1} \chm(\frac 1\pi \phi(0)-\frac 12)\\ +\frac{d}{dx}\Cal H(x)\ , \quad\text{with}\quad \Cal H(x)\quad \text{satisfying\ the\ following}\ .\endmultline \tag 98$$ $$\multline \text{If}\quad |E_0|>2\hat c(\delta E)\ , \text{then}\ \frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|\overline C_{\#}\Lambda_{\min}^{-1}\ , \ \text{then}\\ \frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)||x-x_{\text{left}}(E)|>\lambda_{\text{left}} ^{-\frac 23 N^\prime}B_{\text{left}}}C_\ast\lambda_{\text{left}}^{-\frac 13 N^\prime}\bigl(\frac{S_{\text{left}}}{B_{\text{left}}}|x-x_{\text{left}} (E)|\bigr)^{-1/2}\, dx\\ +\int_{|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\frac 23 N^\prime} B_{\text{left}}}\Bigl\{C_\ast\bigl(\frac{S_{\text{left}}}{B_{\text{left}}} |x-x_{\text{left}}(E)|\bigr)^{-1/2}\\ +C_\ast\bigl(\frac{S_{\text{left}}} {B_{\text{left}}}|x-x_{\text{perturbed}}(E)|\bigr)^{-1/2}\Bigr\}dx\\ \le C_\ast\lambda_{\text{left}}^{-\frac 13 N^\prime}S_{\text{left}}^{-1/2} B_{\text{left}}\ .\endmultline\tag 102$$ \noindent Since$g(x,E)$is supported inside$\{|E-E_0|<\hat c\delta E\}$, we may integrate (102) to obtain $$\multline \int_{x\in I_{\text{BVP}}}\Big|\int_{-\infty}^0 g(x,E)\bigl(E-V(x)+f(x,E)\bigr) _+^{-1/2}dE\\ -\int_{-\infty}^0 g(x,E)(E-V(x))_+^{-1/2}dE| dx\\ \le C_\ast\lambda_{\text{left}}^{-\frac 13 N^\prime}S_{\text{left}}^{-1/2} B_{\text{left}}(\delta E)\le C_\ast\lambda_{\text{left}}^{-\frac 13 N^\prime} S_{\text{left}}^{+1/2} B_{\text{left}}=C_\ast\lambda_{\text{left}}^{1-\frac 13 N^{\prime}}\ .\endmultline\tag 103$$ \noindent Also, (19 bis) and (102) imply $$\multline \int_{x\in I_{\text{BVP}}}|g(x,0)\bigl(-V(x)+f(x,0)\bigr)_+^{-1/2} [\phi^\prime(0)]^{-1}-g(x,0)\bigl(-V(x)\bigr)_+^{-1/2}[\phi^\prime(0)]^{-1} |dx\\ \le C_\ast\lambda_{\text{left}}^{-\frac 13 N^\prime}\quad\text{for}\quad |E_0|<2\hat c(\delta E)\ .\endmultline \tag 104$$ \noindent If$|E_0|\ge 2\hat c(\delta E)$, then (104) holds trivially, since$g(x,0)\equiv 0$in that case. Putting (103), (104) into (98) and recalling the definition of$\rho_{sc}(x,g)$, we get the main result of this section. \vglue 1pc \proclaim{Airey Density Lemma} Suppose the potential$V(x)$, the weight functions$S(x)$,$B(x)$, and the cutoff function$g(x,E)$satisfy assumptions (X0)$\ldots$(X12). Then the microlocalized density is given by$\rho(x,g)=\rho_{sc}(x,g)+\frac{d}{dx}H(x)$, where$\rho_{sc}(x,g)$is the semiclassical approximation, and$H(x)$satisfies the following estimates on$I_{\text{left}}=\{|x-x_{\text{left}}(E_0)|2\hat c(\delta E)$, then $$\multline \frac{1}{B_{\text{left}}}\int_{I_{\text{left}}}|H(x)|^2dx\le C_\ast\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x}{B_{\text{left}}} \bigr)^{-5} +C_\ast\Lambda_{\min}^{8\varepsilon-4}\lambda_{\text{left}}^2 \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)\\ +C_\ast\Lambda_{\min}^{-N^{\prime\prime}}(\phi(\text{max}\ \Cal J))^2 +C_\ast\Lambda_{\min}^{-N^{\prime\prime}}\ .\endmultline$$ \noindent (B) If$|E_0|\le 2\hat c(\delta E)$and$\min_{k\in \Bbb Z} |\phi(0)-\pi(k+1/2)|>\overline C_{\#}\Lambda_{\min}^{-1}$, then $$\multline \frac{1}{B_{\text{left}}}\int_{I_{\text{left}}}|H(x)|^2dx\le C_\ast \lambda_{\text{left}}^{-2/3}+C_\ast\lambda_{\text{left}}^{-2} \bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}+C_\ast\Lambda_{\min} ^{8\varepsilon-4}\lambda_{\text{left}}^2\bigl(\frac{\delta x} {B_{\text{left}}})\\ +C_\ast\Lambda_{\min}^{-2}\lambda_{\text{left}}^2\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^2+C_\ast\Lambda_{\min}^{-N^{\prime\prime}} (\phi(\text{max}\ \Cal J))^2+C_\ast\Lambda_{\min}^{-N^{\prime\prime}}\ . \endmultline$$ \noindent (C) If$|E_0|\le 2\hat c(\delta E)$and$\min_{k\in \Bbb Z} |\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda_{\min}^{-1}$, then $$\multline \frac{1}{B_{\text{left}}}\int_{I_{\text{left}}}|H(x)|^2dx \le C_\ast\lambda_{\text{left}}^{-2/3} +C_\ast\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)^{-5}\\ +C_\ast\Lambda_{\min}^{-2}\lambda_{\text{left}}^2\bigl(\frac{\delta x} {B_{\text{left}}}\bigr)+C_\ast\Lambda_{\min}^{-N^{\prime\prime}} (\phi(\text{max}\ \Cal J))^2+C_\ast\Lambda_{\min}^{-N^{\prime\prime}}\ . \endmultline$$ \noindent Here, the constants$c_{\#}$and$\overline C_{\#}$are determined by$\varepsilon$,$K$,$N$,$c$,$C$,$c_1$,$C_\alpha$in (X0)$\ldots$(X12), while$C_\ast$is determined by$\varepsilon$,$K$,$N$,$c$,$C$,$c_1$,$C_\alpha$,$\hat c$,$\hat C_{\alpha\beta}$in\hfill\break (X0)$\ldots$(X12).\endproclaim \vglue 1pc \demo{Remark} Since$E_{\text{upper}}\equiv \max\Cal J\le 0$, we have$\phi(\max\ \Cal J)=\int_I(E_{\text{upper}}-V(x))_+^{1/2}dx\le \int_{I_{\text{BVP}}}(-V(x))_+^{1/2}dx$.\enddemo \vfill\eject \head {\bf THE MICROLOCALIZED DENSITY IN THE OSCILLATORY REGION}\endhead \medskip In this section we study$\rho(x,g)$for$g(x,E)$supported in the region where eigenfunctions are given approximately by $$u(x)\sim \text{Re}\Bigl[e^{\pm i\frac \pi 4}\exp\bigl(i\int_{x_{\text{left}}(E)} ^x (E-V(t))^{1/2}\ dt\bigr)\cdot \frac{(1+u_{\text{left}}(x))} {(E-V(x))^{1/4}}\Bigr]\ ,$$ \noindent or by $$u(x)\sim \text{Re}\Bigl[e^{\mp i\frac \pi 4}\exp\bigl(-i \int_x^{x_{\roman{rt}}(E)}(E-V(t))^{1/2}\ dt\bigr)\cdot \frac {(1+u_{\roman{rt}}(x))}{(E-V(x))^{1/4}}\Bigr]\ .$$ \noindent Our set-up and assumptions on$V(x)$and$g(x,E)$are as follows. \medskip \noindent{\it Set-Up}: We are given the following data. A potential$V(x)$defined on an interval$I_{\text{BVP}}$; Weight functions$S(x)$,$B(x)>0$defined on a subinterval$I\subset I_{\text{BVP}}$; An interval$[E_{\ell o},E_{hi}]\subset (-\infty,0]$; A point$x_0\in I$; A function$g(x,E)$defined on$I_{\text{BVP}}\times \Bbb R^1$; Positive constants$\varepsilon$,$K$,$N$,$c$,$C$,$c_1$,$c_2$,$c_3$,$C_\alpha$,$\hat c$,$\hat c_1$,$\hat C$,$\hat C_{\alpha\beta}$. \medskip \noindent{\it Assumptions\/}: \noindent (A1)$\quad$For all$E_0\in [E_{\ell o},E_{hi}]$, the hypotheses of the WKB Theorems are satisfied, with$E_\infty=0$, and with$\varepsilon$,$K$,$N$,$c$,$C$,$c_1$,$c_2$,$C_\alpha$as in our present {\it set-up\/}. Denote by$\Lambda(E_0)$,$S_{\min}(E_0)$the quantities called$\Lambda$,$S_{\min}$in the section on the WKB Theorems. Set$\Lambda_{\min}=\text{inf}\{\Lambda(E_0)\colon E_0\in [E_{\ell o},E_{hi}]\}$. \noindent (A2)$\quad$There is a point$\tilde x_0\in I$with$|\tilde x_0-x_0|c_3S(\tilde x_0)$. \noindent (A3)$\quad$Let$x\in I_{\text{BVP}}$and$E\in \Bbb R^1$be given. Then$g(x,E)=0$unless the following conditions are satisfied:$|x-x_0|<\hat cB(x_0)$; either$E_{\ell o}+\hat c_1S_{\min}(E_{\ell o}) \le E\le E_{hi}$or else$E>0; $$\hat c\tau S(x_0)0 are bounded above by a small, positive number determined by \varepsilon, K, N, c, C, c_1, c_2, c_3, C_\alpha. \noindent (A4)\quad \Lambda_{\min}^{\varepsilon-\frac 23}<\tau\le 1. \noindent (A5)\quad |\partial_x^\alpha\partial_E^\beta g(x,E)|\le \hat C_{\alpha\beta}(\tau B(x_0))^{-\alpha}(\tau S(x_0))^{-\beta}. \noindent (A6)\quad \Lambda_{\min} is bounded below by a large positive number determined by \varepsilon, K, N, c, C, c_1, c_2, c_3, C_\alpha, \hat c, \hat c_1, \hat C, \hat C_{\alpha\beta}. Let c_{\#}, C_{\#}, C_{\#}^\prime etc. denote constants determined by \varepsilon, K, N, c, C, c_1, c_2, c_3, C_\alpha. Let c_\ast, C_\ast, C_\ast^\prime etc. denote constants determined by \varepsilon, K, N, c, C, c_1, c_2, c_3, C_\alpha, \hat c, \hat c_1, \hat C, \hat C_{\alpha\beta}. \noindent (That is, C_{\#} is determined by the constants appearing in our hypotheses on V(x), S(x), B(x) and [E_{\ell o},E_{hi}]. Constants C_\ast may depend also on the constants appearing in our assumptions on the function g(x,E).) We may assume that the set G=\{(x,E)\mid E_{\ell o}+\hat c_1S_{\min} (E_{\ell o})\le E\le E_{hi}, |x-x_0|<\hat cB(x_0), \hat c\tau S(x_0)0. (ii) G\subset \{(x,E)\mid |x-x_0|<\hat c B(x_0),\ x-x_{\text{left}}(E)\sim \tau B(x_0)\} (iii) G\subset \{(x,E)\mid |x-x_0|<\hat c B(x_0), \ x_{\text{rt}}(E)- x\sim \tau B(x_0)\}. Here A\sim B means c_{\ast}100.) To deduce (5)\ldots(10) from the WKB Eigenfunction Theorem, we use the fact that |x-x_{\text{left}}(E)|>c_{\ast}\tau B(x_{\text{left}}(E)), |x-x_{\text{rt}}(E)|>c_{\ast}\tau B(x_{\text{rt}}(E)) for (x,E)\in \text{supp}\ g, by virtue of our assumption (A3) on the support of g. In (11) we take S_{\min}(E)=\ \text{inf}_{x_{\text{left}}(E)From (13) and (15), we get \phi^\prime(E^\prime)\sim\phi^\prime (E) for E^\prime \in J(E), and so the number of E_k \in J(E) is at most C_{\#}, by (1). If |\tilde E-E|\le c_{\#}^\prime(\phi^\prime (E))^{-1}, then E\in J(\tilde E) because \phi^\prime(\tilde E)\sim \phi^\prime(E). Another consequence of (15) is that \phi(E_{\min})\le \phi(E_{\ell o}-c_{\#} S_{\min}(E_{\ell o}))\le \phi(E_{\ell o})-c_{\#}\Lambda_{\min}. Similarly, \phi(E_{\max})\ge \phi(E_{hi})+c_{\#}\Lambda_{\min}. These inequalities and (3), (4) show that a_0, b_0, a, b all belong to [a_{\min},b_{\max}]. Note that \frac{dE(t)}{dt} has constant order of magnitude for t between a_0 and a, and thus |E(a)-E_{\ell o}|=|E(a)-E(a_0)|\le C_{\#}\frac{dE(t)}{dt}\mid_{t=a_0}\cdot |a-a_0|\le C_{\#}(\phi^\prime(E_{\ell o}))^{-1}\Lambda_{\min}^{-1}\le C_{\#}S_{\min}(E_{\ell o})\Lambda_{\min}^{-2} by (15). So E(a)\le E_{\ell o}+\hat c_1S_{\min}(E_{\ell o}). Therefore g(x,E)=0 whenever E\le E(a), by virtue of (A3). In particular, g(x,E(t))=0 to infinite order at t=a. We use the above information to estimate$$ \rho(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}}g(x,E_k)u_k^2(x) \ . $$\noindent First of all,$$\split\int\limits_{(x,E_k)\in\ \text{supp}\ g}\Big|u_k(x)-b_k\text{Re} \Bigl[\frac{e^{i\eta(x,E_k)}}{(E_k-V(x))^{1/4}}(1+w_k(x))\Bigr]\Big|^2\ dx\le \Vert\Cal E_k(x)\Vert_{L^2(I_{\text{BVP}})}^2\\ \le \Lambda_{\min}^{-N^{\prime\prime}}\ .\endsplit $$\noindent So$$\align &\int\limits_{(x,E_k)\in\ \text{supp}\ g}|u_k^2(x)-b_k^2\Bigl(\text{Re} \bigl[\frac{e^{i\eta(x,E_k)}}{(E_k-V(x))^{1/4}}\ (1+w_k(x))\bigr]\Bigr)^2| \ dx\\ &\le \Bigl(\int\limits_{(x,E_k)\in\ \text{supp}\ g} |u_k(x)-b_k\ \text{Re}[\text{etc}]|^2\ dx\Bigr)^{1/2}\\ &\qquad\qquad \cdot\Bigl(\int\limits_{(x,E_k)\in\ \text{supp}\ g_k}|u_k(x)+b_k\ \text{Re}[\text{etc}]|^2\ dx\Bigr)^{1/2}\\ &\le \Lambda_{\min}^{-\frac 12 N^{\prime\prime}}\Bigl\{\Bigl(\int\limits_ {(x,E_k)\ \text{supp}\ g_k}|u_k(x)-b_k\text{Re}[\text{etc}]|^2\ dx\Bigr)^{1/2} +2\Vert u_k\Vert_{L^2(I_{\text{BVP}})}\Bigr\}\\ &\le C_{\#}\Lambda_{\min}^{-\frac 12 N^{\prime\prime}}\ . \endalign$$\noindent Summing over k, we get$$\multline \int_{I_{\text{BVP}}} \Big|\sum\limits_kg(x,E_k)u_k^2(x)-\sum\limits_k g(x,E_k)b_k^2\Bigl(\text{Re}\bigl[\frac{e^{i\eta(x,E_k)}}{(E_k-V(x))^{1/4}} (1+w_k(x))\bigr]\Bigr)^2\Big|\ dx\\ \le C_{\ast}\Lambda_{\min}^{-\frac 12 N^{\prime\prime}}\cdot (\text{Number\ of}\ k)\le C_{\ast}\Lambda_{\min}^{-\frac 12 N^{\prime\prime}}(\phi(0)+1) \quad\text{by (1)}\ .\endmultline $$\noindent That is,$$ \int_{I_{\text{BVP}}}|\rho(x,g)-\rho_0(x,g)-\sum\limits_\pm \rho_\pm (x,g)|\ dx \le C_{\ast}\Lambda_{\min}^{-\frac 12 N^{\prime\prime}}(\phi(0)+1)\tag 16 $$\noindent with$$ \rho_0(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}} g(x,E_k)b_k^2 \frac{|1+w_k(x)|^2}{2(E_k-V(x))^{1/2}}\tag 17 $$\noindent and$$ \rho_+(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}} g(x,E_k)\frac{b_k^2e^{2i\eta(x,E_k)}}{4(E_k-V(x))^{1/2}}(1+w_k(x))^2 \tag 18  \rho_-(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}} g(x,E_k)b_k^2\,e^{-2i\eta(x,E_k)}(1+\overline{w_k(x)})^2 /\bigl\{4(E_k-V(x))^{1/2}\bigr\}\ .\tag 19 $$\noindent We will see that \rho_{\pm}(x,g) are small in H^{-1}--norm, while \rho_0(x,g) is closely approximated by \rho_{sc}(x,g) in L^1. We begin with \rho_0. In \text{supp}\ g(x,E_k) we have |1+w_k(x)|^2=1+O(\Lambda_{\min}^{-2}\tau^{-3}) by (7); and b_k^2=[\phi^\prime(E_k)]^{-1}\cdot (1+O(\Lambda_{\min}^{4\varepsilon-2})) by (9). Hence \frac{b_k^2|1+w_k(x)|^2}{2(E_k-V(x))^{1/2}}= \frac{[\phi^\prime(E_k)]^{-1}}{2(E_k-V(x))^{1/2}}\cdot (1+O(\Lambda_{\min} ^{4\varepsilon-2}\tau^{-3})), so that$$ |\rho_0(x,g)-\rho_1(x,g)|\le C_{\ast}\sum\limits_k|g(x,E_k)| \frac{[\phi^\prime(E_k)]^{-1}}{(E_k-V(x))^{1/2}}\ \Lambda_{\min}^{4\varepsilon -2}\tau^{-3}\tag 20 $$\noindent with$$ \rho_1(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}} g(x,E_k)\frac{[\phi^\prime(E_k)]^{-1}}{2(E_k-V(x))^{1/2}}\ .\tag 21 $$To control the right--hand side of (20), we use the properties of the energy intervals J(E). For each E_k we have$$ 1\le C_{\#}\int\limits_{E_{\min}}^{E_{\max}}\chi_{{}_{E_k\in J(E)}} \frac{dE}{|J(E)|}\quad\text{since}\quad |E-E_k|From (13), (15) we conclude that\phi^\prime(E(t))$has constant order of magnitude when$t$varies over an interval of length$c_{\#} \Lambda_{\min}$inside$[a_{\min},b_{\max}]$. Note also that$\tau\Lambda_{\min}>\Lambda_{\min}^{1/3}>1$by (A4) and (A5). These remarks and (33) show that the hypotheses of the lemma on Riemann sums are satisfied. Applying that lemma, and noting that$g(x,E(a))\equiv 0$, we obtain $$\rho_2(x,g)=\sum\limits_{k\in \Bbb Z\cap [a,b]}F_x(E(k))=\\ \int_a^bF_x(E(t))\ dt-F_x(E(b))\chm(b)+\Cal E(x)\ ,\tag 34$$ \noindent with $$\multline |\Cal E(x)|\le C_\ast\frac{[\phi^\prime(E(b))]^{-1}}{(\tau S(x_0))^{1/2}} (\tau\Lambda_{\min})^{-1}+\frac{C_\ast[\phi^\prime(E(a))]^{-1}} {(\tau S(x_0))^{1/2}}(\tau\Lambda_{\min})^{-1}\\ + C_\ast^{\overline N}\int_a^b\frac{[\phi^\prime(E(t))]^{-1}} {(\tau S(x_0))^{1/2}}(\tau\Lambda_{\min})^{-\overline N}\ dt\endmultline\tag 35$$ \noindent ($\overline N$as large as we please.) Since$[\phi^\prime(E(a))]^{-1}$,$[\phi^\prime(E(b))]^{-1}\le C_{\#} S^{1/2}(x_0)B^{-1}(x_0)$, the first two terms on the right of (35) are at most$\frac{C_\ast}{\tau^{3/2}}\Lambda_{\min}^{-1}B^{-1}(x_0)$. The last term on right of (35) has the order of magnitude$\frac{(\tau\Lambda_{\min})^{-\overline N}(E(b)-E(a))}{(\tau S(x_0))^{1/2}}\le \frac{C_\ast(\Lambda_{\min})^{-\overline N/3}}{\tau^{1/2}} (S(x_0))^{1/2}$. Hence,$|\Cal E(x)|\le \frac{C_\ast^{\overline N} \Lambda_{\min}^{-\overline N/3}}{\tau^{1/2}}(S(x_0))^{1/2}+\frac {C_\ast\Lambda_{\min}^{-1}}{\tau^{3/2}}(B(x_0))^{-1}, so that $$\int\limits_{|x-x_0|\Lambda_{\min}^{\varepsilon-2/3}\ge \Lambda_{\min}^\varepsilon (\lambda(x_0))^{-2/3}. We make repeated use of this observation, starting with h_k=g_k, to construct by successive approximation a function \sigma_k(x) with the following properties.$$ \text{supp}\ \sigma_k(x)\subset\ \text{supp}\ g_k\subset\{|x-x_0|<\hat c B(x_0), E_k-V(x)\sim \tau S (x_0)\}\tag 54  |(\frac{d}{dx})^\alpha\sigma_k(x)|\le C_\ast^\alpha\frac{[\phi^\prime(E_k)]^{-1}} {\tau S(x_0)}(\tau B(x_0))^{-\alpha}\tag 55  \frac{d}{dx}\bigl\{\sigma_k(x)e^{2i\eta(x,E_k)}\bigr\} =g_k(x)e^{2i\eta(x,E_k)}+\ \text{err}_k(x)\ ,\ \text{with}\tag 56  |\text{err}_k|\le C_\ast\Lambda_{\min}^{-N}\frac{[\phi^\prime(E_k)]^{-1}} {(\tau S(x_0))^{1/2}}\ \chi_{{}_{E_k-V(x)\sim \tau S(x_0)}}\ . \tag 57 $$Thus,$$ \rho_+(x,g)=\frac{d}{dx}\bigl\{\sum\limits_{k=k_{\ell o}} ^{k_{hi}}\sigma_k(x)e^{2i\eta(x,E_k)}\bigr\}-\sum\limits_{k=k_{\ell o}} ^{k_{hi}}\ \text{err}_k(x)\ .\tag 58 $$\noindent We have$$ |\sum\limits_k\ \text{err}_k(x)|\le C_\ast\sum\limits_k\frac {[\phi^\prime(E_k)]^{-1}\Lambda_{\min}^{-N}} {(\tau S(x_0))^{1/2}}\chi_{{}_{E_k-V(x)\sim \tau S(x_0)}} \le \frac{C_\ast\Lambda_{\min}^{-N}}{(\tau S(x_0))^{1/2}}\cdot C_{\ast} \tau S(x_0) $$\noindent by (27 bis), and therefore \int\limits_{|x-x_0|\lambda^{\varepsilon-\frac 12}). We apply these observations to study the sum \sum\limits_{k=0}^{k_{hi}} g(E_k). In fact,$$ |g(E_k)-g(E(k))|\le |E(k)-E_k|\cdot \max |g^\prime|\le C_{\#} S\lambda^{-2}\cdot C_\ast(\tau^2S)^{-1}=C_\ast(\tau^2\lambda^2)^{-1}\ . $$\noindent Hence,$$\split \Big|\sum\limits_{k=0}^{k_{hi}}g(E_k)-\sum\limits_{k=0}^{k_{hi}} g(E(k))\Big| \le (k_{hi}+1)\cdot C_\ast(\tau^2\lambda^2)^{-1}\\ \le C_{\#}\tau^2\lambda\cdot C_\ast(\tau^2\lambda^2)^{-1} &\le C_\ast\lambda^{-1}\ . \endsplit\tag 2 $$\noindent We want to compute \sum\limits_{k=0}^{k_{hi}}g(E(k)) using the lemma on Riemann sums. This requires bounds for \bigl(\frac{d}{dt}\bigr)^m g(E(t)), which is a sum of terms of the form$$\split \bigl(\frac{d}{dE}\bigr)^\beta g(E)\mid_{E=E(t)}\cdot\prod\limits_{\nu=1}^{\beta}\Bigl\{\bigl(\frac{d}{dt} \bigr)^{m_\nu}E(t)\Bigr\}\quad\text{with}\quad m_\nu\ge 1, m_1+\ldots+m_\beta=m\ ,\\ \text{hence}\quad 0\le \beta \le m\ .\endsplit $$\noindent This term is dominated by C_\ast^m(\tau^2S)^{-\beta}\cdot S^\beta \lambda^{-m}=C_\ast^m(\tau^2)^{-\beta}\lambda^{-m}\le C_\ast^m (\tau^2\lambda)^{-m}. Therefore,$$ \Big|\bigl(\frac{d}{dt}\bigr)^mg(E(t))\Big|\le C_\ast^m(\tau^2\lambda)^{-m} \quad\text{for}\quad t\in \Cal J, m\ge 0\ . $$\noindent Note that \tau^2\lambda\ge 10, as required in the hypothesis of the lemma on Riemann sums. That lemma now implies$$ \sum\limits_{k=0}^{k_{hi}}g(E(k))=\sum\limits_{k\in \Bbb Z \cap [-\frac 12, b]}g(E(k))=\int\limits_{-\frac 12}^{b}g(E(t))dt -g(E(b))\chm(b)+\ \text{Err}_0\ , $$\noindent with$$|\text{Err}_0|\le C_\ast(\tau^2\lambda)^{-1}+\int\limits_{-\frac 12} ^{b}C_{\ast}^{\overline N}(\tau^2\lambda)^{-\overline N}\ dt \le C_\ast(\tau^2\lambda)^{-1},\quad\text{since}\quad b\le C_{\#}\tau^2\lambda\ . $$\noindent Combining this with (2), we get$$\split \sum\limits_{k=0}^{k_{hi}}g(E_k)=\int\limits_{-\frac 12}^{b} g(E(t))\ dt-g(E(b))\chm(b)+\ \text{Err}_1\ ,\quad\text{with}\\ |\text{Err}_1|\le C_\ast(\tau^2\lambda)^{-1}\ .\endsplit\tag 3 $$\noindent In (3), we want to use b_0 in place of b. Since |b_0-b|\le C_{\#}\lambda^{-1}, |g(E(t))|\le C_\ast, |\frac{d}{dt}g(E(t))|\le C_\ast(\tau^2\lambda)^{-1}, we have$$ \Big|\int\limits_b^{b_0}g(E(t))dt\Big|\le C_\ast\lambda^{-1}\ , $$\noindent and$$ |g(E(b))-g(E(b_0))|\le C_{\#}\lambda^{-1}\cdot C_\ast(\tau^2\lambda)^{-1} \ . $$\noindent Putting these estimates into (3), we find that$$\split \sum\limits_{k=0}^{k_{hi}}g(E_k)=\int\limits_{-\frac 12}^{b_0}g(E(t))dt-g(E(b_0))\chm(b)+\ \text{Err}_2\ ,\quad\text{with}\\ |\text{Err}_2|\le C_\ast(\tau^2\lambda)^{-1}\ .\endsplit\tag 4 $$We have E(b_0)=E_{hi} by definition of b_0 and E(t). Either g(E_{hi})=g(0)=0 or else E_{hi}=0 and b=\frac 1\pi\phi(0)-\frac 12 +\omega_{hi}. In either case, we have$$ -g(E(b_0))\chm(b)=-g(0)\chm\bigl(\frac 1\pi\phi(0)-\frac 12+\omega_{hi} \bigr)\ .\tag 5 $$\noindent Also,$$\align \int\limits_{-1/2}^{b_0}g(E(t))dt&=\int\limits_{V(x_0)}^{E_{hi}} g(E)\cdot \frac 1\pi\frac{d\phi(E)}{dE}\ dE\\ &=\int\limits_{E\in [V(x_0),E_{hi}]}g(E)\Bigl\{\frac{1}{2\pi} \int\limits_{x\in I_{\text{BVP}}}(E-V(x))_+^{-\frac 12}\ dx\Bigr\}\ dE\\ &=\int\limits_{x\in I_{\text{BVP}}}\Bigl\{\frac {1}{2\pi} \int\limits_{-\infty}^0 g(E)\cdot (E-V(x))_+^{-\frac 12}\ dE\Bigr\}\ dx\ ,\endalign $$\noindent in view of our assumption on the support of g. Putting this and (5) into (4), and writing -g(0)[\phi^\prime(0)]^{-1}\cdot \int\limits_{x\in I_{\text{BVP}}}\frac 12 (-V(x))_+^{-\frac 12} \ dx\chm(\frac 1\pi \phi(0)-\frac 12+\omega_{hi}) for the right-hand side of (5),we get$$\split \sum\limits_{k=0}^{k_{hi}}g(E_k)=\int\limits_{x\in I_{\text{BVP}}} \ dx\Bigl\{\frac 1{2\pi}\int\limits_{-\infty}^{0}g(E)\cdot (E-V(x))_+^{-\frac 12}\ dx-g(0)[\phi^\prime(0)]^{-1}\\ \cdot\frac 12(-V(x))_+^{-\frac 12}\chm(\frac 1\pi\phi(0)-\frac 12+\omega _{hi})\Bigr\}+\ \text{Err}_3\ ,\quad\text{with}\quad |\text{Err}_3|\le C_\ast(\tau^2\lambda)^{-1}\ .\endsplit\tag 6 $$Since |\omega_{hi}|\le C_{\#}\lambda^{-1}, we have$$\multline \Big|\chm(\frac1\pi \phi(0)-\frac 12+\omega_{hi}\bigr)-\chm\bigl(\frac 1\pi \phi(0)-\frac 12\bigr)\Big|\\ \le \cases C_{\#}\lambda^{-1} &\text{if\min_{k\in \Bbb Z}|\ \phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\lambda^{-1}$}\\ C_{\#} &\text{if$\min_{k\in \Bbb Z}|\ \phi(0)-\pi(k+1/2)|\le \overline C_{\#}\lambda^{-1}}\ ,\endcases\endmultline $$\noindent So (6) implies$$ \sum\limits_{k=0}^{k_{hi}}g(E_k)=\int\limits_{x\in I_{\text{BVP}}} \rho_{sc}(x,g)dx+\ \text{Err}_4\ ,\qquad\text{with}\tag 7  |\text{Err}_4|\le C_\ast(\tau^2\lambda)^{-1}\quad\text{if}\quad \operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\lambda^{-1}\tag 8  |\text{Err}_4|\le C_\ast\quad\text{if}\quad \operatornamewithlimits{\min}_ {k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\lambda^{-1}\ .\tag 9 $$Now let u_k(x) be the real, normalized eigenfunction corresponding to E_k. Thus,$$ \rho(x,g)=\sum\limits^{k_{hi}}_{k=0}g(E_k)u_k^2(x)\ .\tag 10 $$\noindent We know that$$ \int\limits_{|x-x_0|>\lambda^\varepsilon\tau B}|u_k(x)|^2\ dx\le C_{\#} \lambda^{-N^\prime}\ .\tag 11 $$\noindent This follows by applying lemma 2 in the section on the WKB Theorem for low eigenvalues, to the potential \tilde V(x)=V(x)-\min(0,V(x_0)+\tilde S), with \tilde S=\lambda^{2\varepsilon}\tau^2S, \tilde B= \lambda^\varepsilon\tau B in place of S, B. \noindent Let$$\gather {\align H_-(x)&= \int\limits_{x^\prime \in I_{\text{BVP}}\cap (-\infty,x]} [\rho(x^\prime,g)-\rho_{sc}(x^\prime,g)]\ dx^\prime\\ H_+(x)&=\int\limits_{x^\prime \in I_{\text{BVP}}\cap [x,\infty)}[\rho (x^\prime,g)-\rho_{sc}(x^\prime,g)]\ dx^\prime \ .\endalign}\\ \endgather $$\noindent Note that for |x^\prime-x_0|>\lambda^\varepsilon\tau B we have E-V(x^\prime)<0 for E\in \text{supp}\ g\ \cap (-\infty,0]. Hence \rho_{sc}(x^\prime,g) is supported in \{|x^\prime-x_0|<\lambda^\varepsilon \tau B\}. Hence for x\in I_{\text{BVP}}, xx_0+\lambda^\varepsilon\tau B we have$$ |H_+(x)|\le C_\ast\lambda^{1-N^\prime}\ .\tag 13 $$\noindent Also, H_-(x)+H_+(x)=\int\limits_{x^\prime\in I_{\text{BVP}}} [\rho(x^\prime,g)-\rho_{sc}(x^\prime,g)]\ dx^\prime= \sum\limits_{k=0}^{k_{hi}}g(E_k)-\hfill\break \int\limits_{I_{\text{BVP}}}\rho_{sc} (x^\prime,g)\ dx^\prime, so (7), (8), (9), (13) yield the estimates:$$\multline |H_-(x)|\le C_\ast(\tau^2\lambda)^{-1}\quad \text{for}\quad x>x_0+ \lambda^\varepsilon\tau B,\\ \text{if}\quad \operatornamewithlimits{\min}_{k\in \Bbb Z}| \phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\lambda^{-1}\endmultline\tag 14 \multline |H_-(x)|\le C_\ast\quad\text{for}\quad x>x_0+\lambda^\varepsilon\tau B\ ,\\ \text{if}\quad \operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(0)-\pi (k+1/2)|\le \overline C_{\#}\lambda^{-1}\ .\endmultline\tag 15 $$We estimate H_-(x) in \{|x-x_0|<\lambda^\varepsilon\tau B\}, contenting ourselves with the crudest result. Trivially,$$\multline \Big|\int\limits_{x^\prime \in I_{\text{BVP}}\cap (-\infty,x]}\rho(x^\prime,g) \ dx^\prime\Big|\le \sum\limits_{k=0}^{k_{hi}}|g(E_k)|\int\limits_{x^\prime \in I_{\text{BVP}}}|u_k(x^\prime)|^2\ dx^\prime\\ \le C_\ast(k_{hi}+1)\le C_\ast\tau^2\lambda\ ,\quad\text{by\ our\ estimate\ for}\ k_{hi}\ .\endmultline\tag 16 $$\noindent Also,$$\multline \Big|\int\limits_{x^\prime\in I_{\text{BVP}}\cap (-\infty,x]}\Bigl\{ \int_{-\infty}^0 g(E)(E-V(x^\prime))_+^{-1/2}\ dE\Bigr\}\ dx^\prime\Big|\le\\ \qquad\qquad \int\Sb V(x^\prime)x_0+\lambda^\varepsilon\tau B\ ,\\ \operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\lambda^{-1}\endsplit\tag"(c)" \split |H(x)|\le C_\ast\quad\text{if}\quad x\in I_{\text{BVP}}\ ,\ x>x_0+\lambda^\varepsilon\tau B\ ,\\ \operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\lambda^{-1}\ .\endsplit\tag"(d)" $$\noindent The constant \overline C_{\#} depends only on the constants in the hypotheses of the WKB Theorem on low eigenvalues. The constant C_\ast depends also on the \hat C_{\beta} above.\endproclaim \vfill\eject \heading{\bf{COMBINING THE MICROLOCALIZED RESULTS}}\endheading \medskip In this section, we combine our previous results on microlocalized densities by using a partition of unity. Our goal is to control \rho(x,\varphi) where \varphi(E) is a function of energy alone, that vanishes near the minimum of the potential. In the next section, we will remove the assumption that \varphi vanishes near the minimum of the potential, and take \varphi\equiv 1 to control the density \rho(x). Our present set-up is as follows. We are given a potential V(x) defined on a (possibly unbounded) interval I_{\text{BVP}}. On a subinterval I\subset I_{\text{BVP}} we are given positive functions S(x), B(x). We are given positive numbers K, \varepsilon, N with K>100, \varepsilon<\frac{1}{1000 K}, N>\frac{K}{\varepsilon^{50}}. We set N^\prime=[\varepsilon N/500] and N^{\prime\prime}=\frac 32 \varepsilon N^\prime-30000 K-33. We are given a point x_0\in I. Define H=-\frac{d^2}{dx^2}+V(x) on I_{\text{BVP}}, with Dirichlet or Neumann boundary conditions. When we speak of eigenvalues or eigenfunctions, we mean those of H. We are given a function \varphi(E) and a constant \hat c>0. We make the following hypotheses. \bigskip \subhead Assumptions concerning V(x), S(x), B(x) on I\endsubhead \smallskip \roster \item"(Y0)" If x,y\in I and |x-y|cB(x_{\text{left}}(0)), \text{dist}(x_{\text{rt}}(0),\partial I)>cB(x_{\text{rt}}(0)). \item"(Y3)" We have V(x_0)<-c_2S(x_0), V^\prime(x_0)=0; and for |x-x_0|\le c_1B(x_0) we have V^{\prime\prime}(x)\ge cS(x_0)B^{-2} (x_0). \item"(Y4)" For x_{\text{left}}(0)\le x\le x_0-c_1B(x_0) we have -V^\prime(x)\ge cS(x)B^{-1}(x), and for x_0+c_1B(x_0)\le x\le x_{\text{rt}}(0) we have +V^\prime(x)\ge cS(x)B^{-1}(x). \endroster \noindent As usual, define \lambda(x)=S^{1/2}(x)B(x). Set \Lambda=\bigl(\int_{x_{\text{left}}(0)}^{x_{\text{rt}}(0)}\frac{dx}{\lambda(x) B(x)}\bigr)^{-1}. \bigskip \subhead Assumptions concerning V(x) on all of I_{\text{BVP}}\endsubhead \smallskip \roster \item"(Y5)" For all x\in I_{\text{BVP}}\backslash [x_{\text{left}}(0), x_{\text{rt}}(0)] we have V(x)>0. \item"(Y6)" For all x\in I_{\text{BVP}} with xx_{\text{rt}}(0)+\Lambda^KB(x_{\text{rt}}(0)), we have V(x)\ge \frac{1000}{|x-x_{\text{rt}}(0)|^2}. \endroster \bigskip \subhead Polynomial growth conditions on S(x), B(x)\endsubhead \smallskip \roster \item"(Y7)" We have \operatornamewithlimits{\max}_{x\in I}B(x)<\Lambda^K \cdot \operatornamewithlimits{\min}_{x\in I}B(x),\hfill\break \operatornamewithlimits{\max}_{x\in I}S(x)<\Lambda^K\cdot \operatornamewithlimits{\min}_{x\in I}S(x), and |I|<\Lambda^K \cdot\operatornamewithlimits{\min}_{x\in I}B(x). \endroster \bigskip \subhead Assumptions on the function \varphi(E) and the constant \hat c \endsubhead \smallskip \roster \item"(Y8)" We have |(\frac{d}{dE})^\beta\varphi(E)|\le \hat C_{\beta} (S(x_0))^{-\beta} for all E. \item"(Y9)" For -\infty c_{\#}S(x)B^{-1}(x)\\ \text{for}\quad x\in \tilde I\cap (-\infty, x_{\text{left}}(0)]\ ;\quad\text{and}\endmultline$$ $$\multline S(x)\le C_{\#}S(x_{\text{rt}}(0))\quad\text{and}\quad +V^\prime(x)>c_{\#}S(x)B^{-1}(x)\\ \text{for}\quad x\in \tilde I\cap [x_{\text{rt}}(0),+\infty)\ . \endmultline$$ \noindent Then set $$\check I=[x_{\text{left}}(0)-\Lambda^{-\varepsilon}B(x_{\text{left}}(0)), x_{\text{rt}}(0)+\Lambda^{-\varepsilon}B(x_{\text{rt}}(0))]\subset \tilde I \ .$$ We can produce a partition of unity\{\theta_\nu(x)\}$with the following properties. $$\sum\limits_\nu\theta_\nu(x)=1\quad\text{for}\quad x\in \check I\tag 1$$ $$\text{supp}\ \theta_\nu(x)\subset \{|x-x_\nu|<\hat cB(x_\nu)\}\subset \tilde I\tag 2$$ $$\Big|\Bigl(\frac{d}{dx}\Bigr)^\alpha\theta_\nu(x)\Big|\le C_\ast^\alpha (B(x_\nu))^{-\alpha}\tag 3$$ $$\text{Each\ point}\ \tilde x\ \text{belongs\ to\ at\ most}\ C_{\#}\ \text{of\ the\ intervals}\ \{|x-x_\nu|<10\, \hat cB(x_\nu)\}\tag 4$$ $$\theta_\nu\ge 0\ \text{everywhere,\ and}\ \theta_\nu(x_\nu)\ge c_{\#}>0\ .\tag 5$$ \noindent We can also achieve $$\multline \text{supp}\ \theta_\nu\ \text{meets}\ \check I\ \text{for\ each}\ \nu,\\ \text{simply\ by\ deleting\ the}\ \theta_\nu\ \text{whose\ supports\ fail\ to\ meet}\ \check I\ .\endmultline\tag"(5a)"$$ \noindent (To avoid doing violence to the notation, we may suppose$x_0$in (Y0)$\ldots$(Y11) is equal to our present$x_\nu$with$\nu=0$.) Next, pick$k_{\max}$so that$\frac 12\Lambda^{-2/7}<2^{-k_{\max}}\le \Lambda^ {-2/7}$, and take functions\hfill\break$\chi_{{}_{0}}(t),\chi_{{}_{1}}(t),\ldots,\chi_{{}_{k_{\max}}}(t)with the following properties. \chi_{{}_{k}}(t)\ \text{is\ supported\ in}\ \{2^{-k}\le t\le C_\ast2^{-k}\}\ \text{for}\ 0\le kE in [x_{\text{left}}(0),x_{\text{rt}}(0)]\backslash [x_{\text{left}} (E),x_{\text{rt}}(E)]. Hypotheses (Y2), (Y5) show that in fact \{x\in I_{\text{BVP}}\mid V(x)E for x\in I_{\text{BVP}}\backslash [x_{\text{left}}(E),x_{\text{rt}}(E)]. If V(x_0)E\quad &\text{for}\quad x\in I_{\text{BVP}}\backslash [x_{\text{left}} (E),x_{\text{rt}}(E)]\ .\endalign \noindent In particular,\text{dist}(x_{\text{left}}(E),\partial I)\hfill\break>c_{\#}B(x_{\text{left}}(E))$and$\text{dist}(x_{\text{rt}}(E),\partial I)> c_{\#}B(x_{\text{rt}}(E))$by virtue of (Y2). Taking$E=E_0$, we get (Hyp 2). To check (Hyp 3) we note that$x_{\text{left}}(E_0)\not\in [x_0-c_{\#}B(x_0),x_0+c_{\#}B(x_0)]$. (Otherwise, we could not have$E_0=V(x_{\text{left}}(E_0))\ge V(x_0)+\frac{1}{10}c_2S(x_0)$.) Hypotheses (Y3), (Y4) then imply$-V^\prime(x_{\text{left}}(E_0))>c_{\#}^\prime S(x_{\text{left}}(E_0))B^{-1}(x_{\text{left}}(E_0))$, since$x_{\text{left}}(E_0)c_{\#}^\prime S(x_{\text{rt}}(E_0))B^{-1}(x_{\text{rt}}(E_0))$. Our estimates on$|V^{\prime\prime}|$imply therefore $$\multline -V^\prime(x)>c_{\#}^{\prime\prime}S(x_{\text{left}}(E_0))B^{-1} (x_{\text{left}}(E_0))\\ \text{for}\quad x_{\text{left}}(E_0)\le x\le x_{\text{left}}(E_0)+ 2c_{\#}^{\prime\prime}B(x_{\text{left}}(E_0))\endmultline$$ \noindent and $$\multline +V^\prime(x)>c_{\#}^{\prime\prime}S(x_{\text{rt}}(E_0))B^{-1} (x_{\text{rt}}(E_0))\\ \text{for}\quad x_{\text{rt}}(E_0)-2c_{\#}^{\prime\prime}B(x_{\text{rt}} (E_0))\le x\le x_{\text{rt}}(E_0)\ .\endmultline$$\enddemo \noindent These estimates imply (Hyp 3), with$c_{\#}^{\prime\prime}$in place of$c_1$. To check (Hyp 4), we first note that$E_0\le 0$, so$E_0-V(x)\le -V(x)\le |V(x)|\le C_{\#}S(x)$. Hence to establish (Hyp 4), we need only check that$E_0-V(x)\ge c_{\#}^{\prime\prime\prime} S(x)$for$x\in [x_{\text{left}}(E_0)+c_{\#}^{\prime\prime}B(x_{\text{left}} (E_0))$,$x_{\text{rt}}(E_0)-c_{\#}^{\prime\prime}B(x_{\text{rt}} (E_0))]$. We verify this as follows. First of all,$V(x)\le V(x_0)+\frac{1}{20}c_2S(x_0)$if$|x-x_0|\tilde c_{\#}S(x)$if$|x-x_0|x+\tilde c_{\#}^\prime B(x)>x>x_0+c_{\#}B(x_0)\ . $$\noindent Here we discuss only the first case since the second is analogous. For x_{\text{left}}(E_0)<\hfill\break x^\prime0, so E_0-V(x)=V(x_{\text{left}}(E_0))-V(x)=\int_{x_{\text{left}}(E_0)}^{x} (-V^\prime(x^\prime))dx^\prime\ge\int_{x-\tilde c_{\#}^\prime B(x)}^x \tilde c_{\#}^{\prime\prime}S(x^\prime)B^{-1}(x^\prime)dx^\prime\ge c_{\#}^{\prime\prime\prime}S(x) as needed. The proof of (Hyp 4) is complete. To check (Hyp 5), we take 10^{-5}c_2 in place of c_2. Since S_{\min}\hfill\break =\operatornamewithlimits{\min}_{x\in [x_{\text{left}}(E_0),x_{\text{rt}}(E_0)]}S(x)\le S(x_0) and E_0\ge V(x_0)+\frac{1}{10}c_2S(x_0), the assumptions on E in (Hyp 5) imply 0\ge E\ge E_0-10^{-5}c_2S_{\min}\ge (V(x_0)+\frac{1}{10} c_2S(x_0))-10^{-5}c_2S(x_0)>V(x_0). As we saw before and noted in the discussion of Hyp(2), 0\ge E>V(x_0) implies V(x)>E for all x\in I_{\text{BVP}}\backslash[x_{\text{left}}(E),x_{\text{rt}} (E)]. This proves (Hyp 5). Next we note that the quantity called \Lambda in (Hyp 7)\ldots(Hyp 10) is in fact \Lambda(E_0)=\bigl(\int_{x_{\text{left}}(E_0)}^{x_{\text{rt}}(E_0)} \frac{dx}{\lambda(x)B(x)}\bigr)^{-1}\ge \bigl(\int_{x_{\text{left}}(0)} ^{x_{\text{rt}}(0)}\frac{dx}{\lambda(x)B(x)}\bigr)^{-1}=\Lambda. So the quantity called \Lambda in \hfill\break (Hyp 7)\ldots(Hyp 10) is greater than or equal to our present \Lambda from (Y0)\ldots(Y11). Hence in proving (Hyp 7)\ldots(Hyp 10), it is enough to use our present \Lambda in place of \Lambda(E_0). To check (Hyp 6) with 300K in place of K, suppose x\in I_{\text{BVP}}, with xx_{\text{rt}}(E_0)+\frac 12(\lambda(x_{\text{rt}}(E_0)))^{300K} B(x_{\text{rt}}(E_0)). This completes the proof of (Hyp 6). (Hyp 7) with 300K in place of K follows trivially from (Y7) and the fact that \lambda(x_{\text{left}}(E_0)), \lambda(x_{\text{rt}}(E_0)) \ge c_{\#}\Lambda. To check (Hyp 8) with 300K in place of K let \tilde x\in I be given, and note that S(x)\ge \Lambda^{-K}S(\tilde x) for x\in I by (Y7). Hence \int_I\frac{dx}{S^{1/2}(x)}\le \Lambda^{\frac 12 K}S^{-1/2}(\tilde x) |I|\le \Lambda^{\frac 12 K}S^{-1/2}(\tilde x)\cdot\Lambda^KB(\tilde x)\ ,\ \text{again\ by (Y7)} . Taking \tilde x=x_{\text{left}}(E_0), x_{\text{rt}}(E_0), we obtain (Hyp 8). To check (Hyp 9) with 300K in place of K we again fix \tilde x\in I and apply (Y7), to deduce$$\multline \Bigl[\int_I\frac{dx}{S^{1/2}(x)B^4(x)}\Bigr]\Bigl[\int_I\frac{dx} {S^{1/2}(x)}\Bigr]\\ \le \Bigl[\Lambda^{1/2K}S^{-1/2}(\tilde x)\Lambda^{4K}B^{-4} (\tilde x)|I|\Bigr]\Bigl[\Lambda^{1/2K}S^{-1/2}(\tilde x)|I|\Bigr]\\ =\Lambda^{5K}S^{-1}(\tilde x)B^{-4}(\tilde x)|I|^2\le \Lambda^{5K} S^{-1}(\tilde x)B^{-4}(\tilde x)[\Lambda^KB(\tilde x)]^2\\ =\frac{\Lambda^{7K}}{S(\tilde x)B^2(\tilde x)}=\frac{\Lambda^{7K}} {\lambda^2(\tilde x)}\le C_{\#}\Lambda^{7K-2}\ .\endmultline $$This trivially implies (Hyp 9). By reviewing our proofs of (Hyp 0)\ldots(Hyp 9), the reader will see that the constants called \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in (Hyp 0)\ldots(Hyp 9) depend only on \varepsilon, K, c, C, c_1, c_2, C_\alpha in (Y0)\ldots(Y11). Hence, (Hyp 10) follows from the WKB hypothesis (Y11). The verification of (Hyp 0)\ldots(Hyp 10) is complete. We already checked that the quantity called \Lambda in (Hyp 0)\ldots(Hyp 10) is greater than or equal to the number \Lambda in (Y0)\ldots(Y11). Thus, we have verified all the statements in Lemma 1.\qquad\blacksquare \vglue 1pc \demo{Proof of Lemma 2} We may assume g_{\nu k}(x,E)\neq 0 for some (x,E)\in I_{\text{BVP}}\times (-\infty,0]. Our first step is to pick \tilde x_0, E_{\ell o}, E_{hi}. We distinguish two cases: (i) V(x_\nu)\le V(x_0)+\frac 12 c_2S(x_0). (ii) V(x_\nu)>V(x_0)+\frac 12 c_2S(x_0).\enddemo \smallskip \noindent In case (i), we take \tilde x_0=x_\nu, E_{\ell o}=V(x_0)+ \frac 34 c_2S(x_0), E_{hi}=0, and note that E_{\ell o}<0. In case (ii), we first note that |x_\nu-x_0|>c_{\#}B(x_0). Hence, with (\text{sgn})=\pm 1, we have$$\multline c_{\#}S(x_\nu)B^{-1}(x_\nu)< (\text{sgn})\cdot V^\prime(x)0$for$|x-x_\nu|<\hat cB(x_\nu)$, which implies easily that$g_{\nu k}(x,E)=0$for all$(x,E)\in I_{\text{BVP}}\times (-\infty,0]$, in contradiction to our initial assumption. Hence$[E_{\ell o},E_{hi}]$is a non-empty subinterval of$(-\infty,0]$in case (ii). This completes our selection of$\tilde x_0$,$E_{\ell o}$,$E_{hi}$. The next step is to show that $$E_{\ell o}\ge V(x_0)+\frac{1}{10}c_2S(x_0)\ .\tag"(20.6)"$$ \noindent This is obvious in case (i). In case (ii) we use (20.5), (ii) and (18) to write $$\multline E_{\ell o}=V(\tilde x_0)+\frac 12 c_{\#}^{\prime\prime}bS(x_\nu)> V(\tilde x_0)>V(x_\nu)-C_{\#}^{\prime\prime}bS(x_\nu)\\ >[V(x_0)+\frac 12 c_2S(x_0)]-C_{\#}^{\prime\prime}bS(x_\nu)\\ >V(x_0)+\frac 12 c_2S(x_0)-C_{\#}^{\prime\prime\prime}bS(x_0)\ ,\endmultline \tag"(20.7)"$$ \noindent which yields (20.6) provided we take $$b<\frac{1}{10}c_2(C_{\#}^{\prime\prime\prime})^{-1}\ , \quad \text{with}\quad C_{\#}^{\prime\prime\prime}\quad \text{as\ in\ (20.7)}\ .\tag"(20.8)"$$ \noindent The proof of (20.6) is complete. \medskip Applying (20.6) and Lemma 1, we draw the following conclusions. The hypotheses (Hyp 0)$\ldots$(Hyp 10) hold for any$E_0$in$[E_{\ell o},E_{hi}]$, with$E_\infty=0$and with$300 K$in place of$K$. The constants called$c$,$C$,$c_1$,$c_2$,$C_\alpha$in (Hyp 0)$\ldots$(Hyp 10) may all be taken to have the form$c_{\#}$. In particular, let$c_{\#}^{\text{HYP}}$denote the constant called$c$in (Hyp 0)$\ldots$(Hyp 10). Finally,$\Lambda(E_0)\ge \Lambda$for$E_0\in [E_{\ell o},E_{hi}]$. These conclusions imply hypothesis (A1) of the oscillatory density lemma. Next we check (A2), with$x_\nu$in place of$x_0$, and with$c_{\#}^{\text{HYP}}$in place of$c$. In case (i) we have$|x_\nu-\tilde x_0|0$. That is, we must show that $$E>E_{\ell o}+\hat c_1S_{\min}(E_{\ell o})\ .\tag"(20.13)"$$ \noindent We check (20.13) in case (i). Since$E_{\ell o}<0$and$V(x_0)V(x)>V(\tilde x_0)+c_{\#}^{\prime\prime}bS(x_\nu)=E_{\ell o}+\frac 12 c_{\#}^{\prime\prime}bS(x_\nu)\\ \ge E_{\ell o}+\frac 12 c_{\#}^{\prime\prime}c_{\#}bS(\tilde x_0)\ .\endmultline \tag"(20.14)" $$\noindent Also V(\tilde x_0)V(x_0)+\frac 12 c_2S(x_0)\ ,\tag 21$$ \noindent which implies $|x_\nu-x_0|>c_{\#}B(x_0)$. Hence either $x_\nux_0+c_{\#}B(x_0)$. We assume we are in the first case, and show we are in alternative (A) or (B). An analogous argument (which we omit) applies to the second case, to derive alternative (A) or (C). So for the rest of the proof of Lemma 3, we may assume $$x_\nu\in \tilde I\cap (-\infty,x_0-c_{\#}B(x_0)]\ .\tag 22$$ \noindent (We know that $x_\nu\in \tilde I$ by (2).) \noindent The defining properties of $\tilde I$, together with hypotheses (Y3), (Y4), show that (22) implies $$-V^\prime(x_\nu)\ge c_{\#}S(x_\nu)B^{-1}(x_\nu)\ .\tag 23$$ Assume for a moment that $V(x)\ge \Lambda^{\varepsilon-2/7}S(x_\nu)$ for all $x\in\ \text{supp}\ \theta_\nu$. Then the function $\theta_\nu(x)\chi_{{}_{k_{\max}}}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)$ would be supported in the set $\{V(x)\ge \Lambda^{\varepsilon-2/7} S(x_\nu)$, $|E-V(x)|\le C_\ast\Lambda^{-2/7}S(x_\nu)\}\subset\{ E>0\}$. Therefore, $g_{\nu k_{\max}}(x,E)=0$ whenever $E\le 0$, and we are in alternative (A). So for the rest of the proof of Lemma 3, we may assume $$V(x_\nu^\prime)\le \Lambda^{\varepsilon-2/7}S(x_\nu)\ \text{for\ some}\ x_\nu^\prime\in\ \text{supp}\ \theta_\nu\subset \{|x-x_\nu|< \hat cB(x_\nu)\}\ .\tag 24$$ \noindent By (23), (24), and the estimates (Y1) for $|V^{\prime\prime} (x)|$, we conclude that $$V(x_\nu^{\prime\prime})<0\ \text{for\ some}\ x_\nu^{\prime\prime}\ \text{with}\ |x_\nu^{\prime\prime}-x_\nu|<2\hat cB(x_\nu)\ .\tag 25$$ \noindent (To deduce (25) we used also the WKB hypothesis (Y11).) Now recall $E_0=\min(V(x_\nu),0)$. If $V(x_\nu)>0$, then by (25) there is a point $\overline x_\nu$ satisfying $$V(\overline x_\nu)=E_0\quad\text{and}\quad |\overline x_\nu-x_\nu| <2\hat cB(x_\nu)\ .\tag 26$$ \noindent If instead $V(x_\nu)\le 0$, then we take $\overline x_\nu=x_\nu$, and (26) still holds. Hence, we can always take $\overline x_\nu$ to satisfy (26). Note that $V^\prime(\overline x_\nu)<0$ by (23), (26) and (Y10), so (26) implies $$\overline x_\nu=x_{\text{left}}(E_0)\ .\tag 27$$ \noindent Also, from (21) and the inequality $V(x_0)+c_2S(x_0)<0$ (contained in (Y3)), we get $$V(x_0)+\frac 12 c_2S(x_0)\le \min(V(x_\nu),0)=E_0\le 0\ .\tag 28$$ We prepare to pick $\delta E$ for use in hypotheses (X0)$\ldots$(X12) of the Airey density lemma. By (28) and (23), we have $|x_{\text{left}}(E)-x_{\text{left}}(E_0)|\le C_{\#}B(x_\nu)\frac{|E-E_0|}{S(x_\nu)}$ for $|E-E_0|From (26), (32) we get also$|V(x_\nu)-E_0|c_\ast|I_\nu|$, we have $$|c_{\nu k}^-|=(Av_{I_\nu^-}|H_{\nu k}|^2)^{1/2}\le C_\ast (Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\ ,\tag 44$$ \noindent and similarly $$|c_{\nu k}^+|\le C_\ast(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\ .\tag 45$$ \noindent Instead of$H_{\nu k}(x)$, we are really interested in $$\tilde H_{\nu k}(x)=\int_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime, g_{\nu k})-\rho_{sc}(x^\prime,g_{\nu k}))\ dx^\prime \ .$$ \noindent From (36) and (43), together with the fundamental theorem of calculus and the fact that$\rho(x,g_{\nu k})$,$\rho_{sc}(x,g_{\nu k})$are supported in$I_\nu^0$, we see that $$\tilde H_{\nu k}(x)=0\quad\text{to\ the\ left\ of}\ I_\nu^0\tag 46$$ $$\tilde H_{\nu k}(x)=H_{\nu k}(x)-c_{\nu k}^-\quad\text{for}\ x\in I_\nu\ (\text{and\ in\ particular\ for}\ x\in I_\nu^0)\tag 47$$ $$\tilde H_{\nu k}(x)=c_{\nu k}^+-c_{\nu k}^-\quad\text{to\ the\ right\ of}\ I_\nu^0\ .\tag 48$$ We use our estimates to bound the average size of $$\tilde H(x)=\sum\limits_{\nu k}\tilde H_{\nu k}(x)=\int\limits_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime, \sum\limits_{\nu k}g_{\nu k}) -\rho_{sc}(x^\prime,\sum\limits_{\nu k}g_{\nu k}))\, dx^\prime$$ \noindent on an interval$J(\tilde x)=\{|x-\tilde x|\le \hat c B(\tilde x)\}\subset I$. Note that$I_\nu^0=J(x_\nu)$. For fixed$\tilde x$, we divide the set of all$\nuinto three classes: \align \nu\in \Bbb N_{\text{nil}}(\tilde x)\quad&\text{if}\ I_\nu^0\ \text{lies\ entirely\ to\ the\ right\ of}\ J(\tilde x)\\ \nu\in \Bbb N_{\text{const}}(\tilde x)\quad &\text{if}\ I_\nu^0\ \text{lies\ entirely\ to\ the\ left\ of}\ J(\tilde x)\\ \nu\in \Bbb N_{\text{var}}(\tilde x)\quad &\text{if}\ I_\nu^0\ \text{intersects}\ J(\tilde x)\ .\endalign \align \text{If}\ \nu \in \Bbb N_{\text{nil}}(\tilde x)\ , &\text{then}\ \tilde H_{\nu k}=0\ \text{throughout}\ J(\tilde x)\ ,\ \text{by\ (46)}\ .\\ \text{If}\ \nu\in \Bbb N_{\text{const}}(\tilde x)\ , &\text{then}\ \tilde H_{\nu k}=c_{\nu k}^+-c_{\nu k}^-\ \text{throughout}\ J(\tilde x), \text{by\ (48)}\ .\\ \text{If}\ \nu \in \Bbb N_{\text{var}}(\tilde x)\ , &\text{then}\ J(\tilde x)\subset I_\nu\ \text{(by\ (Y10)),\ and}\ |J(\tilde x)|>c_{\ast}|I_\nu|\ .\tag 49\endalign \noindent Hence for\nu\in \Bbb N_{\text{var}}(\tilde x)$we have$\tilde H_{\nu k}=H_{\nu k}-c_{\nu k}^-$on$J(\tilde x)$, and$Av_{J(\tilde x)}|\tilde H_{\nu k}|^2\le C_\ast Av_{I_\nu} |H_{\nu k}-c_{\nu k}^-|^2\le C_\ast^\prime Av_{I_\nu}|H_{\nu k}|^2$by (44). Since also$|c_{\nu k}^+-c_{\nu k}^-|^2\le C_\ast Av_{I_\nu} |H_{\nu k}|^2$by (44), (45), we have shown that $$\bigl(Av_{J(\tilde x)}|\tilde H_{\nu k}|^2\bigr)^{1/2}\le C_\ast(Av_{I_\nu}|H_{\nu k}|^2\bigr)^{1/2}\quad\text{if}\ \nu\in \Bbb N_{\text{const}}(\tilde x)\cup \Bbb N_{\text{var}}(\tilde x) \tag 50$$ \noindent Equations (49), (50) and the definition of$\tilde H(x)$imply $$\bigl(Av_{x\in J(\tilde x)}|\tilde H(x)|^2\bigr)^{1/2}\le C_\ast\sum\limits_{\nu\in \Bbb N_{\text{const}}(\tilde x)\cup \Bbb N_{\text{var}}(\tilde x)}\Bigl(\sum\limits_{0\le k\le k_{\max}}\bigl(Av_{I_\nu}|H_{\nu k}|^2 \bigr)^{1/2}\Bigr)\ .$$ Applying (41) and (42) to estimate the right-hand side, we get $$\multline \bigl(Av_{J(\tilde x)}|\tilde H|^2\bigr)^{1/2}\le C_\ast\Lambda^{4\varepsilon-9/7}\sum\limits_{\nu\in \Bbb N_{\text{const}} (\tilde x)\cup \Bbb N_{\text{var}}(\tilde x)}\lambda(x_\nu) +\text{Junk}\\ \text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\Lambda^{-1}\ ;\endmultline\tag 51$$ $$\multline \bigl(Av_{J(\tilde x)}|\tilde H|^2\bigr)^{1/2}\le C_\ast\Lambda^{-1} \sum\limits_{\nu\in \Bbb N_{\text{const}}(\tilde x)\cup \Bbb N_{\text{var}} (\tilde x)}\lambda(x_\nu)+\ \text{Junk}\\ \text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}\ .\endmultline\tag 52$$ \noindent Here, $$\text{Junk}\ = C_\ast\Lambda^{-\frac{N^{\prime\prime}}{3}}(\phi(0)+1)\cdot (\text{Number\ of\ distinct}\ \nu)\ .\tag 53$$ \noindent To interpret these estimates, we bound$\sum\limits_{\nu\in \Bbb N_{\text{const}}(\tilde x)\cup \Bbb N_{\text{var}}(\tilde x)} \lambda(x_\nu)$,\hfill\break$(\text{Number\ of\ distinct}\ \nu)$, and$(\phi(0)+1)$. From the properties (1)$\ldots$(5) of the$\theta_\nu$we can argue as follows:$\int\limits_I\theta_\nu\, dx\ge c_\ast|I_\nu|$for each$\nu$, and$0\le \sum\limits_\nu\theta_\nu\le C_\ast$everywhere so, $$C_\ast|I|\ge \sum\limits_\nu\int\limits_{I}\theta_\nu\, dx\ge c_\ast\sum\limits_{\nu}|I_\nu|\ .$$ \noindent Also,$|I_\nu|>c_\ast B(x_\nu)>c_\ast\Lambda^{-K}|I|$by (Y7), so $$C_\ast|I|\ge c_\ast\sum\limits_{\nu}|I_{\nu}|\ge c_\ast\Lambda^{-K} |I|\cdot(\text{Number\ of\ distinct}\ \nu)\ .$$ \noindent Thus, $$(\text{Number\ of\ distinct}\ \nu)\le C_\ast\Lambda^K\ .\tag 54$$ \noindent Recall that (Y7) tells us that$\lambda(x)>\Lambda^{-\frac 32 K} \lambda(\overline x)$for$x,\overline x\in I$, hence$\Lambda^{-1}\le \int_I\frac{dx}{\lambda(x)B(x)}\le \Lambda^{\frac 32 K} \lambda^{-1}(\overline x)\int_I\frac{dx}{B(x)}<\Lambda^{\frac 32 K} \lambda^{-1}(\overline x)\int_I\frac{dx}{\Lambda^{-K}|I|}=\Lambda^{\frac 52 K} \lambda^{-1}(\overline x)$, i.e.$\lambda(\overline x)\negthinspace\le \Lambda^{\frac 52 K+1}. Hence \align \phi(0)&\le C_{\#}\int_IS^{1/2}(x)\, dx=C_{\#}\int_I\frac{\lambda(x)dx} {B(x)}\le C_{\#}\Lambda^{\frac 52 K+1}\int_I\frac{dx}{B(x)}\\ &\le C_{\#}\Lambda^{\frac 52 K+1}\int_I\frac{dx}{\Lambda^{-K}|I|} =C_{\#}\Lambda^{\frac 72 K+1}\ ,\ \text{again\ by\ (Y7)}\ . \endalign \noindent Thus, $$(\phi(0)+1)<\Lambda^{5K}\ .\tag 55$$ \noindent Combining (53), (54), (55) and recalling thatN^{\prime\prime} >1000K$, we get $$\text{Junk}\ <\Lambda^{-\frac{N^{\prime\prime}}{4}}\ .\tag 56$$ \noindent To estimate$\sum\limits_{\nu\in \Bbb N_{\text{const}} (\tilde x)\cup \Bbb N_{\text{var}}(\tilde x)}\lambda(x_\nu)$, note that each$x_\nu$satisfies $$\text{dist}(x_\nu,[x_{\text{left}}(0),x_{\text{rt}}(0)])<2\hat cB(x_\nu)\ .\tag 57$$ \noindent In fact, (2) and (5a) show that$\text{dist}(x_\nu,\check I)\le \hat cB(x_\nu)$, i.e. $$|x_\nu-\check x_\nu|\le \hat c B(x_\nu)\quad\text{for\ an}\ \check x_\nu \in \check I\ .\tag 58$$ \noindent Evidently,$B(\check x_\nu)\le C_{\#}B(x_\nu)$. Moreover, by definition of$\check I$, any point$\check x\in \check I$satisfies$\text{dist} (\check x,[x_{\text{left}}(0),x_{\text{rt}}(0)])c_{\#}S(x_{\text{left}}(0))B^{-1}(x_{\text{left}}(0))\cdot (x-x_{\text{left}}(0))\\ \text{for}\quad x\in [x_{\text{left}}(0),x_{\text{left}}(0)+\overline c_{\#} B(x_{\text{left}}(0))]\ .\endmultline\tag 59 $$\noindent Similarly,$$\multline -V(x)>c_{\#}S(x_{\text{rt}}(0))B^{-1}(x_{\text{rt}}(0))\cdot (x_{\text{rt}} (0)-x)\\ \text{for}\quad x\in [x_{\text{rt}}(0)-\overline c_{\#}B(x_{\text{rt}} (0)),x_{\text{rt}}(0)]\ .\endmultline\tag 60 $$\noindent Also from Lemma 1 with E=0 we get$$ -V(x)>c_{\#}S(x)\quad\text{for}\quad x\in [x_{\text{left}} (0)+\overline c_{\#}B(x_{\text{left}}(0)),x_{\text{rt}}(0)- \overline c_{\#}B(x_{\text{rt}}(0))]\ .\tag 61 $$\noindent From (57), (59), (60), (61) we will show that$$ \int\limits_{|x-x_\nu|<3\hat cB(x_\nu)}\bigl(-V(x)\bigr)_+^{1/2} \ge c_\ast\lambda(x_\nu)\ .\tag 62 $$\noindent To see (62), we shrink the region of integration on the left to \{|x-x_\nu|<3\hat cB(x_\nu)\}\hfill\break\cap [x_{\text{left}}(0)+(\hat c)^2B (x_{\text{left}}(0)),x_{\text{rt}}(0)-(\hat c)^2B(x_{\text{rt}}(0))]\equiv \Cal E. Estimates (59), (60), (61) show that -V(x)>c_\ast S(x_\nu) on \Cal E, and (57) shows that \Cal E is an interval of length at least \frac 12 \hat c B(x_\nu). Hence, the left-hand side of (62) is at least c_\ast S^{1/2}(x_\nu)B(x_\nu)=c_\ast\lambda(x_\nu), which proves (62). Next we note that \nu\in \Bbb N_{\text{const}}(\tilde x)\cup \Bbb N_{\text{var}}(\tilde x) implies x_\nu<\tilde x+C_{\#} \hat cB(\tilde x) and therefore also \{|x-x_\nu|<3\hat c B(x_\nu)\}\subset (-\infty,\tilde x+C_{\#}^\prime\hat cB(\tilde x)]. Hence (62) implies$$ \lambda(x_\nu)\le C_\ast\int\limits_{I_{\text{BVP}}\cap (-\infty,\tilde x+C_{\#}^\prime\hat cB(\tilde x)]}\chi_{{}_{|x-x_\nu| <3\hat cB(x_\nu)}}\bigl(-V(x)\bigr)_+^{1/2}\, dx $$\noindent for \nu\in \Bbb N_{\text{const}}(\tilde x)\cup\Bbb N_{\text{var}} (\tilde x). Summing over \nu and using (4), we get$$ \sum\limits_{\nu\in \Bbb N_{\text{const}}(\tilde x)\cup \Bbb N_{\text{var}} (\tilde x)}\lambda(x_\nu)\le C_\ast\int\limits_{I_{\text{BVP}}\cap (-\infty,\tilde x+C_{\#}^\prime\hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2} \, dx\ .\tag 63 $$\noindent Putting (56) and (63) into (51), (52) yields:$$\multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|\tilde H(x)|^2\Bigr)^{1/2} \le C_\ast\Lambda^{4\varepsilon-9/7}\int\limits_{I_{\text{BVP}} \cap (-\infty,\tilde x+C_{\#}\hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}\, dx\\ +\Lambda^{-\frac{N^{\prime\prime}}{4}}\\ \text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\Lambda^{-1}\ ;\endmultline\tag 64 \multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|\tilde H(x)|^2\Bigr)^{1/2} \le C_\ast\Lambda^{-1}\int\limits_{I_{\text{BVP}}\cap (-\infty, \tilde x+C_{\#}\hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}\, dx +\Lambda^{-\frac{N^{\prime\prime}}{4}}\\ \text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#} \Lambda^{-1} .\endmultline\tag 65 $$Now set$$\align g_{\text{err}}(x,E)=\varphi-\sum\limits_{\nu k}g_{\nu k}&= \varphi(E)\cdot\bigl(1-\sum\limits_{\nu k}h_{\nu k}(x,E)\bigr)\\ &\equiv \varphi(E)\cdot h_{\text{err}}(x,E)\ .\endalign $$\noindent From (12) and (14) we get$$\multline |h_{\text{err}}(x,E)|\le C_\ast\chi_{{}_{x\in I_{\text{BVP}}\backslash [x_{\text{left}}(E)-\Lambda^{-3/7}B(x_{\text{left}}(E)),x_{\text{rt}}(E)+ \Lambda^{-3/7}B(x_{\text{rt}}(E))]}}\ ,\\ \text{for}\quad V(x_0) \lambda^{\varepsilon-2/3}(x_{\text{left}}(E)), \lambda^{\varepsilon-2/3} (x_{\text{rt}}(E))$. Hence, Lemma 1 and the WKB Eigenfunction Theorem show that the integral on the right in (67) is at most$C_{\#}\Lambda^{-N^{\prime\prime}}$for$V(x_0)+c_2S(x_0)\le E_k\le 0$. For$E_k\le V(x_0)+c_2S(x_0)$we have$\varphi(E_k)=0$by hypothesis (Y9). Hence (67) implies $$\int\limits_{I_{\text{BVP}}}|\rho(x,g_{\text{err}})|\, dx\le C_{\#}\Lambda^{-N^{\prime\prime}}\cdot\ \text{(Number\ of\ eigenvalues}\ E_k\ \text{in}\ [V(x_0)+c_2S(x_0),0])\ .\tag 68$$ \noindent We estimate the number of eigenvalues on the right in (68). If$E_k\in [V(x_0)\hfill\break +c_2S(x_0),0]$is an eigenvalue, then by applying Lemma 1 with$E=E_k$and the WKB Eigenvalue Theorem with$E_0=E_k$, we learn the following: $$|\phi(E_k)-\pi(k^\prime+1/2)|\le C_{\#}\Lambda^{-1}\quad \text{for\ an\ integer}\ k^\prime\ .\tag 69$$ $$\multline E_k\ \text{is\ the\ only\ eigenvalue}\ E\ \text{satisfying}\ |E_k-E|\le c_{\#}S_{\min}(E_k)\ ,\\ E\le 0,\ \text{and}\ |\phi(E)-\pi(k^\prime+1/2)|\le C_{\#}\Lambda^{-1}\ . \endmultline\tag 70$$ \noindent Here,$S_{\min}(E_k)=\operatornamewithlimits{\inf}_{ x\in [x_{\text{left}}(E_k),x_{\text{rt}}(E_k)]}S(x)$as in the WKB Theorems. For$|E-E_k|\pi(k^\prime+1/2)+c_{\#}^{\prime\prime}\Lambda $$\noindent by (69). In particular, we cannot have |\phi(E)-\pi(k^\prime+1/2)| \le C_{\#}\Lambda^{-1}. Similarly, if V(x_0)0, hence \rho_{sc}(x,\varphi)=0. \noindent For any x in (-\infty,x_{\text{left}}(0)-\hat cB(x_{\text{left}} (0))] we have therefore$$\multline |H(x)|=\Big|\int\limits_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime, \varphi)-\rho_{sc}(x^\prime,\varphi))\, dx^\prime\Big|\\ =\Big|\int\limits_{I_{\text{BVP}}\cap (-\infty,x]}\rho(x^\prime, \varphi)\, dx^\prime\Big|\le \Lambda^{-\frac{N^{\prime\prime}}{4}}\ . \endmultline\tag 77 $$\noindent Similarly, for any x, y in [x_{\text{rt}}(0)+\hat c B(x_{\text{rt}}(0)),+\infty) we have$$\multline |H(x)-H(y)|=\Big|\int\limits_{I_{\text{BVP}}\cap [\min\{x,y\},\max \{x,y\}]}(\rho(x^\prime,\varphi)-\rho_{sc}(x^\prime,\varphi))\, dx^\prime\Big|\\ =\Big|\int\limits_{I_{\text{BVP}}\cap [\min\{x,y\},\max\{x,y\}]} \rho(x^\prime,\varphi)\, dx^\prime\Big|\le \Lambda^{-\frac{N^{\prime\prime}} {4}}\ .\endmultline\tag 78 $$\noindent Take \tilde x=x_{\text{rt}}(0)+\sqrt{\hat c}\,B(x_{\text{rt}}(0)), so that \{|x-\tilde x|<\hat cB(\tilde x)\}\subset I, and |x-\tilde x|<\hat cB(\tilde x) implies x\ge x_{\text{rt}}(0)+ \hat cB(x_{\text{rt}}(0)). Using (74), (75), (78) and the obvious estimate |H(y)|\le Av_{\{|x-\tilde x|<\hat cB(\tilde x)\}}\,|H(x)-H(y)|+(Av_{\{| x-\tilde x|<\hat cB(\tilde x)\}}|H(x)|^2\bigr)^{1/2}, we find that$$\multline |H(y)|\le C_\ast\Lambda^{4\varepsilon-9/7}\int\limits_{I_{\text{BVP}}} \bigl(-V(x)\bigr)_+^{1/2}\, dx+10\Lambda^{-\frac{N^{\prime\prime}}{4}}\\ \text{when}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\Lambda^{-1}\ \text{and}\ y\ge x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ ;\endmultline\tag 79 $$\noindent and$$\multline |H(y)|\le C_{\ast}\Lambda^{-1}\int\limits_{I_{\text{BVP}}} \bigl(-V(x)\bigr)_+^{1/2}\, dx+10\Lambda^{-\frac{N^{\prime\prime}}{4}}\\ \text{when}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#} \Lambda^{-1}\ \text{and}\ y\ge x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 80 $$\noindent Our basic results on \rho(x,\varphi) are (74), (75), (77), (79), (80). We collect these results in the following. \vglue 1pc \proclaim{Lemma} Assume hypotheses (Y0)\ldots(Y11), and define H(x)=\hfill\break \int\limits_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime,\varphi) -\rho_{sc}(x^\prime,\varphi))\, dx^\prime for arbitrary real x. Then we have the following estimates for H(x). \medskip \noindent{\it CASE I\/}: Suppose \min_{k\in \Bbb Z} |\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}\Lambda^{-1}. Then for -\infty100, \varepsilon<\frac{1}{1000 K}, N>\frac{K}{\varepsilon^{50}}; and we set N^\prime=[\varepsilon N/500], N^{\prime\prime}=\frac 32 \varepsilon N^\prime-30000 K-33. Define H=-\frac{d^2}{dx^2}+V(x) on I, with Dirichlet or Neumann conditions. When we speak of eigenfunctions or eigenvalues, we mean those of H. We suppose we are given a positive constant \hat c. Our hypotheses are as follows. We assume conditions (Y0)\ldots(Y7) from the previous section. In addition, we assume (Y10) and the following variant of (Y11). \roster \item"(Y11^\ast)" \Lambda, defined as in the previous section, is bounded below by a certain large, positive number determined by \varepsilon, K, N, c, C, c_1, c_2, C_\alpha, \hat c. \endroster \noindent Our present assumptions differ from those of the previous section in that we have dropped the function \varphi(E) and the constants \hat C_\beta used to control it. We use c_{\#}, C_{\#} etc. to denote constants that depend only on \varepsilon, K, N, c, C, c_1, c_2, C_\alpha in (Y0)\ldots(Y7) and (Y10), while constants denoted c_\ast, C_\ast etc. depend also on \hat c. Our plan is to make a partition of unity 1=\sum\limits_{0\le k\le k_{\text{last}}}\varphi_k(E), and write$$ H(x)=\int_{I_{\text{BVP}}\cap(-\infty,x]} (\rho(x^\prime,1)-\rho_{sc}(x^\prime,1))dx^\prime $$\noindent as a sum$$\align H(x)&=\sum\limits_{0\le k\le k_{\text{last}}}H_k(x)\ ,\quad\text{with}\\ H_k(x)&=\int_{I_{\text{BVP}}\cap(-\infty,x]} (\rho(x^\prime,\varphi_k)-\rho_{sc} (x^\prime,\varphi_k))dx^\prime \ .\endalign $$\noindent We will use the results of the previous section to control H_k(x) for 0\le k0 to be picked later.$$ \varphi_0(E)\ \quad\text{is\ supported\ in}\quad \{E\ge V(x_0)+c_{\#}^2 S(x_0)\}\tag 1  \Bigl|\bigl(\frac{d}{dE}\bigr)^\beta\varphi_0(E)\Big|\le C_{\#}^\beta (S(x_0))^{-\beta}\quad\text{for\ all}\quad E\ , \beta\ .\tag 2 $$\noindent For 1\le kc_{\#}\Lambda^{7/43} \hfill\break >>1 when 1\le k\le k_{\max}. \noindent Therefore, for 1\le k\le k_{\max}, we can find an energy \tilde E_k\le -c_{\#}S(x_0) (\underbar{not} an eigenvalue -- the eigenvalues are called E_k) with$$ \tilde E_k-V(x_0)\ ,\quad \tilde E_k-\max(\text{supp}\ \varphi_k)\sim c_{\#}^2 2^{-2k}S(x_0)\tag 11  \min_{k^\prime\in \Bbb Z}\ |\phi(\tilde E_k)-\pi(k^\prime+1/2)|\ge \frac{1} {20}\ .\tag 12 $$\noindent We can take the \tilde E_k to decrease with k, so that x_{\text{left}}(\tilde E_k) increases with k, and x_{\text{rt}}(\tilde E_k) decreases with k. \noindent Now for 1\le k\le k_{\text{last}} we set S_k=2^{-2k}S(x_0)\ , B_k=2^{-k}B(x_0), \lambda_k=S_k^{1/2}B_k=2^{-2k}\lambda (x_0), V_k(x)=V(x)-\tilde E_k, and I_k=\{|x-x_0|c_{\#}\Lambda^{7/43}. To check (Y6), note that V(x)=\tilde E_k at the point x_{k,\text{left}} that plays the r\hat ole of x_{\text{left}}(0) for V_k. We have (x_0-x_{k,\text{left}})\sim B_k and -V^\prime\sim S_kB_k^{-1} whenever (x_0-x)\sim B_k. Hence there is a point x_k^- with (x_{k,\text{left}} -x_{k}^-)\sim B_k and 0\ge V(x_k^-)\ge \tilde E_k+c_{\#}S_k. Our hypotheses on V(x) show that V(x) is decreasing in [x_{\text{left}} (0),x_k^-] and V(x)>0 for xV(x_k^-)\ge \tilde E_k+c_{\#}S_k, so V_k(x)=V(x)-\tilde E_k\ge c_{\#}S_k=c_{\#}\lambda_k^2B_k^{-2}. \noindent If xx_{k,\text{rt}}+C_{\#}B_k, then V_k(x)\ge c_{\#} \lambda_k^2\cdot (x-x_{k,\text{rt}})^{-2}. We have proven more than required for (Y6). Property (Y7) is trivial, and (Y8), (Y9) for \varphi_k(E) follow at once from (3), (4). Properties (Y10) and (Y11) for V_k(x) follow at once from our present hypotheses (Y10), (Y11^\ast), and the fact that \hat C_\beta has here the form C_{\#}^\beta. Thus, part (B) of the Lemma is verified. To prove part (C) we first check (H0^\ast)\ldots(H6^\ast). Note that the quantity that plays the r\hat ole of \lambda for V_{k_{\text{last}}} is \lambda(x_0)\ge c_{\#}\Lambda. Properties (H0^\ast)\ldots(H6^\ast) are trivial, except (perhaps) for (H5^\ast), which follows from the same argument just used above for (Y6) in part B. The properties of \varphi_{k_{\text{last}}}(E), which we assumed for g(E) in the section on the microlocalized density near the minimum of the potential, simply amount to (5) and (6). The proof of Lemma 1 is complete.\qquad\blacksquare\enddemo \medskip Define \tilde\varphi_k=\varphi_k(E-\tilde E_k) for 1\le k\le k_{\text{last}}. The functions \rho(x,\varphi_k) and \rho_{sc}(x,\varphi_k) arising from the potential V(x) are equal to the functions \rho(x,\tilde\varphi_k) and \rho_{sc}(x,\tilde\varphi_k) respectively, arising from V_k(x). (This fact is trivially verified, using the fact that \varphi_k(E)=0 for E\ge \tilde E_k.) Therefore, Lemma 1(B) and the lemma of the previous section allow us to control H_k(x) for 1\le k\le k_{\text{last}}. Note that the phase \phi(0) corresponding to V_k(x) is equal to the phase \phi(\tilde E_k) arising from V(x), which is not too near any \pi(k^\prime+1/2), in view of (12). So the lemma of the previous section gives us the following estimates on H_k(x) (1\le k\le k_{\text{last}}):$$ |H_k(x)|\le C_\ast\lambda_k^{-\frac{N^{\prime\prime}}{4}}\quad\text{for}\quad xx_{\text{rt}}(\tilde E_k)+\hat cB_k\ .\tag 15 $$\noindent These estimates in turn trivially imply the following.$$ |H_k(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\quad\text{for} \quad xx_{\text{rt}}(\tilde E_1) +\hat cB(x_0)\quad (1\le kx_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 20 $$\medskip \noindent{\it{Case II\/}}: Suppose \min_{k\in \Bbb Z}\ |\phi(0) -\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}. Then$$ |H_0(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\quad\text{for} \quad xx_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 23 $$Applying Lemma 1(C) and the lemma from the section on the microlocalized density near the minimum of the potential, and recalling (12), we obtain$$ |H_{k_{\text{last}}}(x)|\le C_\ast(\lambda(x_0))^{1-N^\prime} \quad\text{if}\quad xx_0+(\lambda(x_0))^{\varepsilon-\frac{18}{43}}B(x_0)\ . \tag 26 $$\noindent From (24) we get at once$$ |H_{k_{\text{last}}}(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}} {1000}}\quad\text{if}\quad xx_0+\hat cB(x_0)\ .\tag 29 $$\noindent We use (16), (17), (17 bis) and (27), (28), (29) to control \sum\limits_{1\le k\le k_{\text{last}}}H_k(x).$$\multline \text{If}\quad x\le x_{\text{left}}(\tilde E_1) -\hat cB(x_0)\ ,\quad\text{then}\quad \Big|\sum\limits_{1\le k\le k_{\text{last}}}H_k(x)\Big|\le \Lambda^{-10^{-4}N^{\prime\prime}}\\ \text{by\ (16), (27)\ and}\quad 2^{-k_{\text{last}}}\sim (\lambda(x_0))^{-\frac{18} {43}}\ge \Lambda^{-1000 K}\ .\endmultline\tag 30 $$\noindent (The last estimate follows from (Y7)). \noindent If x\ge x_{\text{rt}}(\tilde E_1) +\hat cB(x_0), then \Big|\sum\limits_{ 1\le k\le k_{\text{last}}}H_k(x)\Bigr|\le \sum\limits_{1\le kx_{\text{rt}}(\tilde E_1)+\hat cB(x_0)\ .\tag 31$$ \noindent From (17) and (28), we get for any $\tilde x\in \Bbb R$ that $$\Bigl(Av_{|x-\tilde x|<\hat cB(x_0)}\Big|\sum\limits_{1\le k\le k_{\text{last}}}H_k(x)\Big|^2\Bigr)^{1/2}\\ \le \sum\limits_{1\le kx_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ .\tag 36$$ \noindent If $x_{\text{left}}(\tilde E_1)-C_{\#}\hat cB(x_0)\le \tilde x$, then $\lambda(x_0)\le C_\ast\int_{x_{\text{left}}(\tilde E_1)-2C_{\#}\hat cB(x_0)}^{x_{\text{left}}(\tilde E_1)-C_{\#} \hat cB(x_0)}\bigl(-V(x)\bigr)_+^{1/2}dx$, so \align \lambda^{\varepsilon-\frac{2}{43}}(x_0)&\le C_\ast\bigl(\lambda(x_0)\bigr)^ {\varepsilon-\frac{45}{43}}\int_{I_{\text{BVP}}\cap (-\infty,\tilde x]} \bigl(-V(x)\bigr)_+^{1/2}dx\\ &\le C_\ast\Lambda^{\varepsilon-\frac{45}{43}}\int_{I_{\text{BVP}} \cap (-\infty,\tilde x]}\bigl(-V(x)\bigr)_+^{1/2}dx\ .\endalign \noindent Therefore (33)$\ldots$(36) imply: $$\Big|\sum\limits_{1\le k\le k_{\text{last}}}H_k(x)\Big|\le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}\quad\text{if}\quad x\le x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\ .\tag 37$$ $$\multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\Big|\sum\limits_{1\le k\le k_{ \text{last}}}H_k(x)\Big|^2\Bigr)^{1/2}\\ \le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}+C_\ast\Lambda^{\varepsilon- \frac{45}{43}}\int_{I_{\text{BVP}}\cap (-\infty,\tilde x+C_{\#} \hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}dx\\ \text{if}\quad x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le \tilde x\le x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ .\endmultline\tag 38$$ $$\multline \Big|\sum\limits_{1\le k\le k_{\text{last}}}H_k(x)\Big|\le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}+C_\ast\Lambda^{\varepsilon- \frac{45}{43}}\int_{I_{\text{BVP}}}\bigl(-V(x^\prime)\bigr)_+^{1/2}dx^\prime\\ \text{if}\quad x>x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ .\endmultline \tag 39$$ \noindent Since $H(x)=H_0(x)+\bigl(\sum\limits_{1\le k\le k_{\text{last}}} H_k(x)\bigr)$, estimates (18)$\ldots$(23) and\hfill\break (37)$\ldots$(39) yield the following results. \medskip \noindent{\it{Case I\/}}: Suppose $\min_{k\in \Bbb Z} |\phi(0)-\pi(k+1/2)|\ge \overline C_{\#} \Lambda^{-1}$. Then $$|H(x)|\le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}\quad\text{for} \quad x\le x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\tag 40$$ $$\multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|H(x)|^2\Bigr)^{1/2}\\ \le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}+C_\ast\Lambda^{\varepsilon- \frac{45}{43}}\int_{I_{\text{BVP}}\cap(-\infty,\tilde x+C_{\#} \hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}dx\\ \text{for}\quad x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le \tilde x\le x_{\text {rt}}(0)+\hat cB(x_{\text{rt}}(0))\endmultline\tag 41$$ $$\multline |H(x)|\le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}+C_\ast\Lambda^{\varepsilon -\frac{45}{43}}\int_{I_{\text{BVP}}}\bigl(-V(x^\prime)\bigr)_+^{1/2}dx^\prime\\ \text{for}\quad x\ge x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 42$$ \medskip \noindent{\it{Case II\/}}: Suppose $\min_{k\in \Bbb Z}|\phi(0)- \pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}$. Then $$|H(x)|\le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}\quad\text{for}\quad x\le x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\tag 43$$ $$\multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|H(x)|^2\Bigr)^{1/2}\\ \le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}+C_\ast\Lambda^{-1} \int_{I_{\text{BVP}}\cap(-\infty,\tilde x+C_{\#}\hat cB(\tilde x)]} \bigl(-V(x)\bigr)_+^{1/2}dx\\ \text{for}\quad x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le \tilde x \le x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ \endmultline\tag 44$$ $$\multline |H(x)|\le C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}+C_\ast\Lambda^{-1} \int_{I_{\text{BVP}}}\bigl(-V(x^\prime)\bigr)_+^{1/2}dx^\prime\\ \text{for}\quad x\ge x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 45$$ After a change of notation, we may drop the assumptions $K>100$, $\varepsilon<\frac{1}{1000 K}$, $N>\frac{K}{\varepsilon^{50}}$; and we may replace $10^{-4}N^{\prime\prime}$ in (40)$\ldots$(45) by $N$. To do this, we merely pick new $K$, $\varepsilon$, $N$ by setting $K_\ast=K+100$, $\varepsilon_\ast=\min(\varepsilon,\frac{1}{2000 K_\ast})$, $N_\ast=(N+1)\cdot\frac{K_\ast}{(\varepsilon_\ast)^{50}}$. Then $K_\ast$, $\varepsilon_\ast$, $N_\ast$ satisfy $K_\ast>100$, $\varepsilon_\ast<\frac{1} {1000 K_\ast}$, $N_\ast>\frac{K_\ast}{(\varepsilon_\ast)^{50}}$. The hypotheses of this section hold with $K_\ast$, $\varepsilon_\ast$, $N_\ast$ in place of $K$, $\varepsilon$, $N$. The quantity $10^{-4}N^{\prime\prime}$ arising from $N_\ast$ is greater than $N$. Thus, we obtain the result of the next section, which is our main theorem on the density for a one-dimensional potential. \vfill\eject \heading{\bf{THE WKB DENSITY THEOREM IN ONE DIMENSION}}\endheading \medskip \noindent{\it{Set-Up\/}}: We are given positive numbers $\varepsilon$, $K$, $N$, $\hat c$; two intervals $I\subset I_{\text{BVP}}$ (possibly unbounded); a point $x_0\in I$; a potential $V(x)$ defined on $I_{\text{BVP}}$; and two positive functions $S(x)$, $B(x)$ defined on $I$. Our assumptions are as follows. \subhead Assumptions Concerning $V(x)$, $S(x)$, $B(x)$ on $I$\endsubhead \smallskip \roster \item"(Z0)" If $x,y\in I$ and $|x-y|cB(x_{\text{left}})$ and $\text{dist}\,(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$. \item"(Z3)" We have $V(x_0)<-cS(x_0)$, $V^\prime(x_0)=0$; and for $|x-x_0|\le c_1B(x_0)$ we have $V^{\prime\prime}(x)\ge cS(x_0)B^{-2} (x_0)$. \item"(Z4)" For $x_{\text{left}}\le x\le x_0-c_1B(x_0)$ we have $-V^\prime(x)>cS(x)B^{-1}(x)$; and for $x_0+c_1B(x_0)\le x\le x_{\text{rt}}$ we have $+V^\prime(x)>cS(x)B^{-1}(x)$. \endroster \noindent Define $\lambda(x)=S^{1/2}(x)B(x)$ for $x\in I$, and set $$\Lambda=\Bigl(\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx} {\lambda(x)B(x)}\Bigr)^{-1}\ .$$ \subhead Assumptions Concerning $V(x)$ on all of $I_{\text{BVP}}$\endsubhead \smallskip \roster \item"(Z5)" We have $V(x)>0$ for all $x\in I_{\text{BVP}}\backslash [x_{\text{left}},x_{\text{rt}}]$. \item"(Z6)" For all $x\in I_{\text{BVP}}$ with $xx_{\text{rt}}+\Lambda^KB(x_{\text{rt}})$, we have $V(x)\ge \frac{1000} {|x-x_{\text{rt}}|^2}$. \endroster \smallskip \subhead Polynomial Growth Assumptions on $S(x)$, $B(x)$, $I$\endsubhead \smallskip \roster \item"(Z7)" We have $\max_{x\in I}B(x)<\Lambda^K\min_{x\in I}B(x)$; $\max_{x\in I}S(x)<\Lambda^K\min_{x\in I}S(x)$; and $|I|<\Lambda^K\cdot \min_{x\in I}B(x)$. \endroster \smallskip \subhead Smallness of the Constant $\hat c$\endsubhead \smallskip \roster \item"(Z8)" The constant $\hat c$ is bounded above by a certain small, positive number determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$.\endroster \smallskip \subhead The WKB Hypothesis\endsubhead \smallskip \roster \item"(Z9)" $\Lambda$ is bounded below by a certain large, positive number determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $\hat c$, $C_\alpha$. \endroster Let $E_k$ and $u_k(x)$ be the eigenvalues and (normalized) eigenfunctions of $-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{BVP}}$, with Dirichlet or Neumann boundary conditions. Then define the density $\rho(x)$ and its semiclassical approximation $\rho_{sc}(x)$ on $I_{\text{BVP}}$ by setting: \align \rho(x)&=\sum\limits_{E_k\le 0}|u_k(x)|^2\quad (x\in I_{\text{BVP}})\ ;\\ \rho_{sc}(x)&=\frac 1\pi \bigl(-V(x)\bigr)_+^{1/2}- \frac{\bigl(-V(x)\bigr)_+^{-1/2}}{\int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{-1/2} dy}\chm\Bigl(\frac 1\pi \int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{1/2}dy-\frac 12\Bigr)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad(x\in I_{\text{BVP}})\ . \endalign \noindent Recall that $t_+^a=t^a$ for $t>0$, $t_+^a=0$ for $t\le 0$, and that $\chm(t)=x-k-\frac 12$ for $k=\ (\text{largest\ integer}\ \le x)$. For any $x\in \Bbb R$, define $$H(x)=\int_{I_{\text{BVP}}\cap (-\infty,x]}\ (\rho(\overline x)-\rho_{sc} (\overline x))d\overline x \ .$$ \vglue 1pc \proclaim{WKB Density Theorem} Assume (Z0)$\ldots$(Z9). \smallskip \noindent{\it{Case I\/}}: Suppose $\min_{k\in \Bbb Z}\ \Big|\int_{I_{\text{BVP}}} \bigl(-V(y)\bigr)_+^{1/2}dy-\pi(k+1/2)\Big|>\overline C\Lambda^{-1}$. Then $$|H(x)|\le \Lambda^{-N}\quad\text{for}\quad x\le x_{\text{left}}-\hat c B(x_{\text{left}})\ .\tag"(A)"$$ $$\multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|H(x)|^2\Bigr)^{1/2}\\ \le\Lambda^{-N}+C_\ast\Lambda^{\varepsilon-\frac{45}{43}} \int_{I_{\text{BVP}}\cap (-\infty,\tilde x+ \overline C\hat cB(\tilde x)]}\bigl(-V(y)\bigr)_+^{1/2}dy\\ \text{for}\quad x_{\text{left}}-\hat cB(x_{\text{left}})\le \tilde x\le x_{\text{rt}} +\hat cB(x_{\text{rt}}) \ .\endmultline\tag"(B)"$$ $$|H(x)|\le \Lambda^{-N}+C_\ast\Lambda^{\varepsilon-\frac{45}{43}} \int_{I_{\text{BVP}}}\bigl(-V(y)\big)_+^{1/2}dy\quad \text{for}\quad x\ge x_{\text{rt}}+\hat cB(x_{\text{rt}})\ . \tag"(C)"$$ \medskip \noindent{\it{Case II\/}}: Suppose instead $\min_{k\in \Bbb Z}\ \Big|\int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{1/2}dy-\pi(k+1/2)\Big| \le \overline C\Lambda^{-1}$. Then $$|H(x)|\le \Lambda^{-N}\quad\text{for}\quad x\le x_{\text{left}}- \hat cB(x_{\text{left}})\ .\tag"(A)"$$ $$\multline \Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|H(x)|^2\Bigr)^{1/2}\\ \le \Lambda^{-N}+C_\ast\Lambda^{-1}\int_{I_{\text{BVP}}\cap(-\infty, \tilde x+\overline C\hat cB(\tilde x)]}\bigl(-V(y)\bigr)_+^{1/2}dy\\ \text{for}\quad x_{\text{left}}-\hat cB(x_{\text{left}})\le\tilde x\le x_{\text{rt}} +\hat cB(x_{\text{rt}})\ .\endmultline\tag"(B)"$$ $$|H(x)|\le \Lambda^{-N}+C_\ast\Lambda^{-1}\int_{I_{\text{BVP}}} \bigl(-V(y)\bigr)_+^{1/2}dy\quad\text{for}\quad x\ge x_{\text{rt}}+\hat cB(x_{\text{rt}})\ .\tag"(C)"$$ \noindent Here $\overline C$ depends only on $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$; and $C_\ast$ depends only on $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $\hat c$, $C_\alpha$.\endproclaim \vglue 1pc \demo{Remarks} The exponent $\varepsilon-\frac{45}{43}$ in {\it Case I\/} is not important to us, and surely not optimal. Any exponent strictly less than $-1$ would serve our purpose.\enddemo \smallskip \vfill\eject \heading{\bf{THE DENSITY FOR DEGENERATE ONE-DIMENSIONAL POTENTIALS I}} \endheading \medskip In the next sections, we prove crude results on the density $\rho(x)$ in various degenerate cases in which the hypotheses in the preceding section break down. We begin by treating a potential $V(x)$ whose minimum $V(x_0)$ is negative but has relatively small absolute value. \medskip \noindent{\it Set-up\/}. We are given positive numbers $\varepsilon$, $K$, $N$, $S$, $B$; a potential $V(x)$ defined on a (possibly unbounded) interval $I_{\text{BVP}}$; and a point $x_0\in I_{\text{BVP}}$. Our assumptions are as follows. \medskip \subhead Hypotheses\endsubhead \roster \item"(Z0$^\ast$)" $I=\{x\colon |x-x_0|0$. \item"(Z5$^\ast$)" For $x\in I_{\text{BVP}}$ with $|x-x_0|>\frac 12\lambda^KB$, we have $V(x)\ge \frac{1000}{|x-x_0|^2}$. \item"(Z6$^\ast$)" $\lambda$ is bounded below by a certain large, positive number determined by $\varepsilon$, $K$, $N$, $c$, $c^\prime$, $C_{\alpha}$. \endroster \noindent Let $E_k$, $u_k(x)$ be the eigenvalues and (normalized) eigenfunctions for $-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{BVP}}$, with Dirichlet or Neumann boundary conditions. As in the previous section, define the density $\rho(x)$ and its semiclassical approximation $\rho_{sc}(x)$ on $I_{\text{BVP}}$, by the formulas \align \rho(x)&=\sum\limits_{E_k\le 0}|u_k(x)|^2\\ \rho_{sc}(x)&=\frac 1\pi\bigl(-V(x)\bigr)_+^{1/2}- \frac{\bigl(V(x)\bigr)_+^{-1/2}} {\int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{-1/2}dy}\ \chm\bigl(\frac 1\pi \int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{1/2}dy-\frac 12\bigr)\endalign \noindent Then set $$H(x)=\int_{I_{\text{BVP}}\cap (-\infty,x]}\bigl(\rho(\overline x)-\rho_{sc} (\overline x)\bigr)\, d\overline x \ .$$ \vglue 1pc \proclaim{First Degenerate Density Lemma} Assume (Z0$^\ast$)$\ldots$(Z6$^\ast$). \noindent {\sl CASE I\/}: Suppose $min_{k\in \Bbb Z}\Big|\int_{I_{\text{BVP}}} \bigl(-V(y)\bigr)_+^{1/2}dy-\pi(k+1/2)\Big|\ge \overline C\lambda^{-1}$. Then \noindent (A) $|H(x)|\le \lambda^{-N}$ if $x$ lies to the left of $I$. \noindent (B) $\bigl(Av_I|H|^2\bigr)^{1/2}\le C_\ast\lambda^{\varepsilon-2/43}$. \noindent (C) $|H(x)|\le C_\ast\lambda^{\varepsilon-2/43}$ if $x$ lies to the right of $I$. \bigskip \noindent{\sl CASE II\/}: Suppose instead $\min_{k\in \Bbb Z}\Big|\int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{1/2}dy-\pi(k+1/2)\Big|\le \overline C\lambda^{-1}$. Then \noindent (A) $|H(x)|\le \lambda^{-N}$ if $x$ lies to the left of $I$. \noindent (B) $(Av_I|H|^2)^{-1/2}\le C_\ast$. \noindent (C) $|H(x)|\le C_\ast$ if $x$ lies to the right of $I$. The constants $C_\ast$, $\overline C$ depend only on $\varepsilon$, $K$, $N$, $c$, $c^\prime$, $C_\alpha$.\endproclaim \demo{Proof} Set $\tau\sim \bigl(\lambda(x_0)\bigr)^{-\frac{18}{43}}$, and pick $g(E)$ as in the section on the microlocalized density near the minimum of the potential, with $g(E)=1$ for $V(x_0)\le E\le 0$. We can take $g(E)$ so that the constants $\hat C_\beta$ used to control it depend only on $\varepsilon$, $K$, $N$, $c$, $c^\prime$, $C_\alpha$. Then $\rho(x,g)=\rho(x)$ because $g(E_k)=1$ for all the eigenvalues $E_k\le 0$. Also, $\rho_{sc}(x,g)=\rho_{sc}(x)$ because $g(E)=1$ for $V(x_0)\le E\le 0$. Hence our present function $H(x)$ is the same as the $H(x)$ in the Lemma in the section on the microlocalized density near the minimum of the potential. That lemma yields: (a) $|H(x)|\le C_\ast\lambda^{1-N^\prime}\quad\text{if}\ x \in I_{\text{BVP}}\ , xx_0+\lambda^{\varepsilon-\frac{18}{43}}B$ in {\it Case I\/}. (d) $|H(x)|\le C_\ast\quad x\in I_{\text{BVP}}\ , x>x_0+\lambda^{\varepsilon -\frac{18}{43}}B$ in {\it Case II\/}. \noindent The restriction to $x\in I_{\text{BVP}}$ is irrelevant, since $H(x)=H(\text{inf}\ I_{\text{BVP}})$ for $x<\text{inf}\ I_{\text{BVP}}$ and $H(x)=H(\text{sup}\ I_{\text{BVP}})$ for $x>\ \text{sup}\ I_{\text{BVP}}$. >From (a) we get Case I (A) and Case II (A), provided we enlarge the $N$ in the Lemma on the microlocalized density near the minimum of the potential. >From (c) we get Case I (C), and from (d) we get Case II (C). To handle (B), we write $$\multline \int_I|H(x)|^2dx\le \int_{|x-x_0|<\lambda^{\varepsilon-\frac{18}{43}}B} |H(x)|^2dx\\ +\int_{I\backslash \{|x-x_0|<\lambda^{\varepsilon-\frac{18}{43}}B\}} |H(x)|^2dx\ .\endmultline\tag 1$$ \noindent In Case I this yields $$\int_I|H(x)|^2dx\le C_\ast\lambda^{\frac{14}{43}}\cdot \lambda^{\varepsilon -\frac{18}{43}}B+C_\ast\lambda^{-\frac{14}{43}}\cdot B\le C_\ast\lambda^ {\varepsilon-\frac{4}{43}}B\ ,$$ \noindent proving the assertion of Case I (B). In Case II, (1) yields instead $$\int_I|H(x)|^2dx\le C_\ast\lambda^{\frac{14}{43}}\cdot \lambda^{\varepsilon -\frac{18}{43}}B+C_\ast B\le C_\ast^\prime B \ ,$$ \noindent proving the assertion of Case II (B).$\qquad\blacksquare$\enddemo \smallskip \demo{Remark} Again, the precise exponents in Case I (B), (C) are irrelevant to us, and are surely not optimal.\enddemo \vfill\eject \heading{\bf{THE DENSITY FOR DEGENERATE ONE-DIMENSIONAL POTENTIALS II}}\endheading \medskip In this section we derive (very) crude results for the density without making any polynomial growth assumptions on the weight functions $S(x)$, $B(x)$ or the interval $I$. These results will be used later for ODE arising from three-dimensional problems, with angular momentum $\ell$ in the range $[\text{Large\ Constant,}\ Z^\varepsilon]$. The precise formulation is as follows. \medskip \noindent{\it Set-Up\/}. We are given a potential $V(x)$ defined on a (possibly unbounded) interval $I_{\text{BVP}}$; positive functions $S(x)$, $B(x)$, defined on a subinterval $I\subset I_{\text{BVP}}$; a point $x_{\text{crit}}\in I_{\text{BVP}}$; an energy $E_{\text{crit}}\le 0$; and a number $\delta$ strictly between $0$ and $1$. \subhead Assumptions\endsubhead \smallskip \roster \item"(Z$\overline 0$)" For $x,y \in I$ with $|x-y|cB(x)$. \item"(Z$\overline 1$)" For $x\in I$ and $\alpha\ge 0$ we have $\big|\bigl(\frac{d}{dx}\bigr)^\alpha V(x)\big|\le C_\alpha S(x)B^{-\alpha}(x)$. \item"(Z$\overline 2$)" For $E_{\text{crit}}\le E\le 0$, the set $\{x\in I_{\text{BVP}}\mid V(x)\le E\}$ is a non-empty interval $(x_{\text{left}}(E), x_{\text{rt}}(E))$ contained in $I$, with $\text{dist}(x_{\text{left}}(E),\partial I)>cB(x_{\text{left}}(E))$ and $\text{dist}\,(x_{\text{rt}}(E),\partial I)>cB(x_{\text{rt}} (E))$. \item"(Z$\overline 3$)" For $E_{\text{crit}}\le E\le 0$, we have $-V^\prime(x)\ge cS(x)B^{-1}(x)$ for \hfill\break $x\in [x_{\text{left}}(E), x_{\text{left}}(E)+c_1B(x_{\text{left}}(E))]$ and $+V^\prime(x)\ge cS(x)B^{-1}(x)$ for $x\in [x_{\text{rt}}(E)- c_1B(x_{\text{rt}}(E)), x_{\text{rt}}(E)]$. \item"(Z$\overline 4$)" For $E_{\text{crit}}\le E\le 0$, we have $cS(x)cS(x)B^{-1} (x)$. \item"(Z$\hat 3$)" $\Lambda=\bigl(\int_I\frac{dx}{\lambda(x)B(x)}\bigr)^{-1}$ is greater than a certain large, positive number determined by $c$, $C$, $C_\alpha$ in (Z$\hat 0$)$\ldots$(Z$\hat 2$). \item"(Z$\hat 4$)" For $x\in (0,x_{\text{small}}]$ we have $V(x)\ge \underline cx_0^{-2}$ \item"(Z$\hat 5$)" For $x\in [x_{\text{small}},x_0]$ we have $|V(x)| \le \underline Cx_0^{-2}$ \item"(Z$\hat 6$)" We have $x_{\text{big}}<\underline Cx_1$ and $V(x)$ is increasing in $[x_1,x_{\text{big}}]$. \item"(Z$\hat 7$)" For $x\in \bigl[\frac{x_1}{8},x_{\text{big}}]$, we have $|V(x)|\le \underline Cx_1^{-2}$. \item"(Z$\hat 8$)" For $x\in [x_{\text{big}},\infty)$, we have $V(x)\ge 0$. \item"(Z$\hat 9$)" For $E\in [V(x_\ast),0]$ we have\hfill\break $\int_{x_0}^{x_{\text{crit}}}\bigl(E-V(x)\bigr)^{-1/2}dx\le \delta\cdot\int_{x_0}^{\frac 12 x_\ast}\bigl(E-V(x)\bigr)^{-1/2}dx$. \endroster When $\delta<<1$ hypothesis (Z$\hat 9$) shows that in the semiclassical approximation, most of the $L^2$-norm of an eigenfunction $u_k(x)$ with $E_k\in [V(x_\ast),0]$ is concentrated outside of $(0,x_{\text{crit}}]$. We denote by $c_{\#}$, $C_{\#}$, etc. constants that depend only on $c$, $C$, $C_\alpha$; while $c_\ast$, $C_\ast$ etc. denote constants that depend also on $\underline c$, $\underline C$. \vglue 1pc \proclaim{Lemma 1} Set $E_0=0$, $I_{\text{center}}=[x_0,x_1]$, $I_{\text{left}}=[x_{\text{small}},x_0]$, $I_{\text{rt}}=[x_1,x_{\text{big}}]$, $I_{\text{far\ left}}=(0,x_{\text{small}}]$, $I_{\text{far\ rt}}\ = [x_{\text{big}},\infty)$. Then the hypotheses (H$\hat 0$)$\ldots$(H$\hat 7$) of Theorem 1 in the section on WKB Theory with Weak Turning Points are satisfied. The constants called $c$, $C$, $C_\alpha$ in (H$\hat 0$)$\ldots$ (H$\hat 7$) may be taken to be of the form $C_{\#}$. The constants called $\underline c$, $\underline C$ in (H$\hat 0$)$\ldots$(H$\hat 7$) may be taken to be of the form $C_\ast$.\endproclaim \vglue 1pc \demo{Proof} (H$\hat 0$) follows from (Z$\hat 0$), and from $B(x)\equiv x$, $2x_0c_{\#}\tilde x$. Also, for $\alpha\ge 1$ we have $\big|\bigl(\frac{d}{dx}\bigr)^\alpha\tilde V(x)\big|=\big| \bigl(\frac{d}{dx}\bigr)^\alpha V(x)\big|\le C_{\#}^\alpha S(x)B^{-\alpha} (x)\le C_{\#}^\alpha\tilde S(x)\tilde B^{-\alpha}(x)$ since $\tilde S(x)=S(x)\cdot\bigl(\frac{\tilde x-x}{\tilde x}\bigr)$ and $\tilde B(x)\sim B(x)\cdot \bigl(\frac{\tilde x-x}{\tilde x}\bigr)$. Hence $|(\frac{d}{dx})^\alpha\tilde V(x)|\le C_{\#}^\alpha\tilde S(x)\tilde B ^{-\alpha}(x)$ is verified, except for $\alpha=0$, $|x-\tilde x|c_{\#}\tilde S(x)$ for $x\in I_{\text{center}}$. If $|\tilde x-x|\ge c_{\#}\tilde x$, then $V$ is increasing on $[x,\tilde x]$, and $V^\prime>c_{\#}S(x)x^{-1}$ on $[x,(1+c_{\#})x]\subset [x,\tilde x]$, as we see from (Z$\hat 2$). Hence, $-\tilde V(x)=V(\tilde x)-V(x)\ge \int_x^{(1+c_{\#})x}V^\prime \ge c_{\#}S(x)\ge c_{\#} \tilde S(x)$, as required. On the other hand, if $|\tilde x-x|\le c_{\#}\tilde x$, then (Z$\hat 2$) gives $V^\prime>c_{\#}S(\tilde x)\tilde x^{-1}$ on $[x,\tilde x]$. Hence $$-\tilde V(x)=V(\tilde x)-V(x)=\int_x^{\tilde x}V^\prime \ge c_{\#} S(\tilde x)\bigl(\frac{\tilde x-x}{x}\bigr)\ge c_{\#}\tilde S(x)\ .$$ \noindent Thus, in all cases, we know that $-\tilde V(x)\ge c_{\#}\tilde S(x)$ for $x\in \tilde I_{\text{center}}$, completing the proof of (H$\hat 1$). \noindent Note that the constants called $c$, $C$, $C_\alpha$ in (H$\hat 0$) and (H$\hat 1$) do not depend on our choice of $\hat C_{\#}$. To prove (H$\hat 2$), we compute $\tilde\lambda(x)=\tilde S^{1/2}(x) \tilde B(x)$ and $\tilde\Lambda^{-1}=\int_{\tilde I_{\text{center}}}\frac{dx} {\tilde\lambda(x)\tilde B(x)}$. For $x\in \tilde I_{\text{center}}$ with $|x-\tilde x|>c_{\#}\tilde x$ we have $\tilde S(x)\sim S(x)$, $\tilde B(x) \sim B(x)$, hence $\tilde\lambda(x)\sim \lambda(x)$. For $x\in \tilde I_{\text{center}}$ with $|x-\tilde x|\le c_{\#}\tilde x$, we have $\tilde S(x)=S(x) \cdot \bigl(\frac{\tilde x-x}{\tilde x}\bigr)$, $\tilde B(x)=(\tilde x-x)\sim B(x)\cdot \bigl(\frac{\tilde x-x}{\tilde x}\bigr)$, hence $$\multline \tilde\lambda(x)\sim \bigl[S(x)\cdot\bigl(\frac{\tilde x-x}{\tilde x}\bigr) \bigr]^{1/2}\bigl[B(x)\cdot \bigl(\frac{\tilde x-x}{\tilde x}\bigr)\bigr]= \lambda(x)\cdot \bigl(\frac{\tilde x-x}{\tilde x}\bigr)^{3/2}\\ \sim\lambda(\tilde x)\cdot\bigl(\frac{\tilde x-x}{\tilde x}\bigr)^{3/2}\ . \endmultline$$ \noindent So $$\multline \tilde \Lambda^{-1}\sim\int_{\tilde I_{\text{center}} \backslash\{|x-\tilde x|\le c_{\#}\tilde x\}}\, \frac{dx}{\lambda(x)B(x)}\\ +\int_{\tilde I_{\text{center}}\cap\{|x-\tilde x|\le c_{\#}\tilde x\}}\ \frac{dx}{\lambda(\tilde x)\cdot(\frac{\tilde x-x}{\tilde x})^{3/2} \cdot (\tilde x-x)}\\ \le \int_I\frac{dx}{\lambda(x)B(x)}+\int_{|x-\tilde x|>\hat C_{\#} \lambda^{-2/3}(\tilde x)\cdot\tilde x}\frac{dx} {\lambda(\tilde x)\big|\frac{\tilde x-x}{\tilde x}\big|^{3/2} \cdot|\tilde x-x|}\\ =\Lambda^{-1}+C_{\#}(\hat C_{\#})^{-3/2}\ .\endmultline$$ \noindent Hence (H$\hat 2$) follows from (Z$\hat 3$) provided we take $\hat C_{\#}$ large enough. This completes the proof of (H$\hat 2$). To prove (H$\hat 3$) we argue as follows. We have $\tilde I_{\text{left}}, \tilde I_{\text{rt}}\neq \emptyset$ by definition. We have $|\tilde I_{\text{left}}|=x_0-x_{\text{small}}\le x_0\sim \tilde B(x_{\text{left}})$, since $\tilde B(x_{\text{left}})=\min (x_0,\tilde x-x_0)$ and $\tilde x>2x_0$. Next, note that $x_{\text{rt}}=\tilde x-\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot\tilde x\ge \frac{9}{10}\tilde x$ since $\tilde I_{\text{center}}$ has been seen to contain $[x,\frac{9}{10}\tilde x]$. Therefore, $\tilde B(x_{\text{rt}}) =\min(\tilde x-x_{\text{rt}},x_{\text{rt}})=\tilde x-x_{\text{rt}}=\hat C_{\#} \lambda^{-2/3}(\tilde x)\cdot\tilde x$. Hence, $|\tilde I_{\text{rt}}|=2\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot\tilde x =2\tilde B(x_{\text{rt}})$, which proves the required bounds for $|\tilde I_{\text{left}}|$, $|\tilde I_{\text{rt}}|$. Regarding $\tilde\lambda(x_{\text{left}})$, $\tilde\lambda(x_{\text{rt}})$, we note that $\tilde\lambda(x_{\text{left}})=\tilde\lambda(x_0)\sim \lambda(x_0)\le C_\ast$, as we saw in the proof of Lemma 1; and $\tilde \lambda(x_{\text{rt}})\sim\lambda(\tilde x)\cdot\bigl (\frac{\tilde x-x_{\text{rt}}}{x}\bigr)^{3/2}=(\hat C_{\#})^{3/2}$. The right-hand side has the form $C_\ast$, completing the proof of (H$\hat 3$). To prove (H$\hat 4$), we need to show $|V(x)-V(\tilde x)|\le C_\ast x_0^{-2}$ for $x\in [x_{\text{small}},x_0]$. Hypothesis (Z$\hat 5$) reduces this to $|V(\tilde x)|\le C_\ast x_0^{-2}$. From (Z$\hat 2$) we get $V(x_0)0$. Hence $\int_{x_0} ^{\frac 32 x_0}\bigl(E_k-V(x)\bigr)^{-1/2}dx\ge \int_{x_0}^{\frac 32 x_0} \bigl[c_{\#}S^{-1/2}(x_0)\bigr]dx=c_{\#}S^{-1/2}(x_0)\cdot x_0$, so $\int_0^{x_0}|u_k(x)|^2dx\le C_\ast\Bigl(\int_{\tilde I_{\text{center}}} \bigl(E_k-V(y)\bigr)^{-1/2}dy\Bigr)^{-1}\int_{x_0}^{\frac 32 x_0} \bigl(E_k-V(x)\bigr)^{-1/2}dx$. Recall from the Remark following the statement of Lemma 2 that $\tilde I_{\text{center}}$ contains $[x_0,\frac{9}{10}\tilde x]$, which in turn contains $[x_0,\frac 12 x_\ast]$. Hence $$\Bigl(\int_{\tilde I_{\text{center}}}\bigl(E_k-V(y)\bigr)^{-1/2}dy\Bigr)^{-1} \le \Bigl(\int_{x_0}^{\frac 12 x_\ast}\bigl(E_k-V(y)\bigr)^{-1/2}dy \Bigr)^{-1}\ .\tag 3$$ \noindent Also $\frac 32 x_0\le x_{\text{crit}}$, so we get $$\int_0^{x_0}|u_k(x)|^2dx\le C_\ast\Bigl(\int_{x_0}^{x_{\text{crit}}} \bigl(E_k-V(y)\bigr)^{-1/2}dy\Bigr)\Bigl(\int_{x_0}^{\frac 12 x_\ast} \bigl(E_k-V(y)\bigr)^{-1/2}dy\Bigr)^{-1}\ .$$ \noindent Applying (Z$\hat 9$), we conclude that $$\int_0^{x_0}|u_k(x)|^2dx\le C_\ast\delta\quad\text{for}\quad E_k\in [\tilde E_{\text{low}},\tilde E_{\text{hi}}]\ .\tag 4$$ \noindent Similarly, (2), (3) imply $$\int_{x_0}^{x_{\text{crit}}}|u_k(x)|^2dx\le C_\ast\Bigl(\int_{x_0} ^{x_{\text{crit}}}\bigl(E_k-V(x)\bigr)^{-1/2}dx\Bigr)\Bigl( \int_{x_0}^{\frac 12 x_\ast}\bigl(E_k-V(y)\bigr)^{-1/2}dy\Bigr)^{-1}\ ,$$ \noindent so another application of (Z$\hat 9$) gives $$\int_{x_0}^{x_{\text{crit}}}|u_k(x)|^2dx\le C_\ast\delta \quad\text{for}\quad E_k\in [\tilde E_{\text{low}},\tilde E_{\text{hi}}]\ .$$ \noindent This and (4) imply $$\int_0^{x_{\text{crit}}}|u_k(x)|^2dx\le C_\ast\delta\quad\text{for} \quad E_k\in [\tilde E_{\text{low}},\tilde E_{\text{hi}}]\ .$$ \noindent Of course, for arbitrary $E_k$ we have $\int_0^{x_{\text{crit}}} |u_k(x)|^2dx\le 1$, since $u_k(x)$ is normalized. \noindent Hence, \align \int_0^{x_{\text{crit}}}\rho(x)dx&\le C_\ast\delta\cdot(\,\text{Number\ of}\ E_k\in [\tilde E_{\text{low}},\tilde E_{\text{hi}}])+C_\ast\cdot \bigl(\,\text{Number\ of}\ E_k<\tilde E_{\text{low}}\bigr)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad +C_\ast\cdot(\,\text{ Number\ of}\ E_k\in (\tilde E_{hi},0])\\ &\le C_\ast\delta\cdot(\,\text{Number\ of}\ E_k\le 0)+C_\ast(\,\text{Number\ of}\ E_k\le \tilde E_{\text{low}})\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad +C_\ast\cdot(\,\text{Number\ of}\ E_k\in (\tilde E_{hi},0])\ . \tag 5\endalign \noindent Again applying Theorem 1 (on WKB with weak turning points) and Lemma 2, with $\tilde E=\tilde E_{\text{hi}}$, we obtain $$\big|\big(\text{Number\ of}\ E_k\le \tilde E_{\text{hi}})-\frac 1\pi \int_{\tilde I_{\text{center}}}\bigl(\tilde E_{\text{hi}}-V(x)\bigr) ^{1/2}dx\big|\le C_\ast\ ,$$ \noindent with $\tilde I_{\text{center}}$ containing $[x_0,\frac{9}{10} \cdot \frac 14 x_1]$ by the remark following the statement of Lemma 2. Hence $$(\,\text{Number\ of}\ E_k\le \tilde E_{\text{hi}})\ge \frac 1\pi \int_{x_0}^{\frac 18x_1}\bigl(\tilde E_{\text{hi}}-V(x)\bigr)^{1/2} dx-C_\ast\ .\tag 6$$ \noindent Similarly, applying Theorem 1 and Lemma 2, with $\tilde E=\tilde E_{\text{low}}$, and noting that $\tilde I_{\text{center}}\subset I$, we get $$\big|\bigl(\text{Number\ of}\ E_k\le \tilde E_{\text{low}})- \frac 1\pi\int_{\tilde I_{\text{center}}}\bigl(\tilde E_{\text{low}} -V(x)\bigr)^{1/2}dx\big|\le C_\ast\ ,$$ \noindent so $$(\text{Number\ of}\ E_k\le \tilde E_{\text{low}})\le \frac 1\pi \int_{x_0}^{x_1}\bigl(\tilde E_{\text{low}}-V(x)\bigr)_+^{1/2}dx+C_\ast\ . \tag 7$$ Lemma 1 and Theorem 1 (on WKB with weak turning points) imply $$(\,\text{Number\ of}\ E_k\le 0)\le \frac 1\pi\int_{x_0}^{x_1}\bigl(-V(x)\bigr)^{1/2}dx+C_\ast\ .\tag 8$$ \noindent From (6) and (8) we get $$\multline (\,\text{Number of}\ E_k\in (\tilde E_{\text{hi}},0])\le C_\ast+\frac 1\pi\int_{x_0}^{\frac{x_1}{8}}\bigl\{\bigl(-V(x)\bigr)^{1/2} -(\tilde E_{\text{hi}}-V(x)\bigr)^{1/2}\bigr\}dx\\ +\frac 1\pi\int_{\frac{x_1}{8}}^{x_1}\bigl(-V(x)\bigr)^{1/2}dx\ . \endmultline\tag 9$$ We estimate the right-hand side of (9). From (Z$\hat 7$), we see that the last term on the right is dominated by $C_\ast$. Moreover, in $[x_0,\frac{x_1}{8}]$, we have $V(x)cS(x)B^{-1} (x)$. \item"(Z3$^\dag$)" $\Lambda=\bigl(\int_I\frac{dx}{\lambda(x)B(x)}\bigr)^{-1}$ is greater than a certain large, positive number determined by $c$, $C$, $C_\alpha$ in (Z0$^\dag$)$\ldots$(Z2$^\dag$). \item"(Z4$^\dag$)" $|V(x)|\le \underline C/(x_0x)$ for $x\in (0,x_0]$. \item"(Z5$^\dag$)" $V(x)$ is increasing and negative in $[\frac{x_1}{8}, x_{\text{big}}]$, and satisfies there $|V(x)|<\underline{C} x_1^{-2}$. Also, $x_{\text{big}}<\underline Cx_1$. \item"(Z6$^\dag$)" $V(x)\ge -10^{-9}x^{-2}$ for $x\in [x_{\text{big}},\infty)$. \item"(Z7$^\dag$)" For $E\in [V(x_\ast),0]$, we have\hfill\break $\int_{x_0}^{x_{\text{crit}}}\bigl(E-V(x)\bigr)^{-1/2}dx\le \delta \cdot \int_{x_0}^{\frac 12 x_\ast}\bigl(E-V(x)\bigr)^{-1/2}dx$. \endroster \noindent As usual, (Z7$^\dag$) says that certain eigenfunctions live mostly outside of $[x_0,x_{\text{crit}}]$ in the semiclassical approximation. \noindent We denote by $c_{\#}$, $C_{\#}$, etc. constants that depend only on $c$, $C$, $C_\alpha$ in (Z0$^\dag$)$\ldots$(Z7$^\dag$); while $c_\ast$, $C_{\ast}$, etc. denote constants that depend also on $\underline C$. \vglue 1pc \proclaim{Lemma 1} Set $I_{\text{far\ left}}=\emptyset$, $I_{\text{left}}= (0,x_0]$, $I_{\text{center}}=[x_0,x_1]$, $I_{\text{rt}}=[x_1,x_{\text{big}}]$, $I_{\text{far\ rt}}=[x_{\text{big}},\infty)$, $E_0=0$. Then the hypotheses (H$\hat 0$)$\ldots$(H$\hat 7$), from the section on WKB with weak turning points, are satisfied. The constants called $c$, $C$, $C_\alpha$ in (H$\hat 0$)$\ldots$(H$\hat 7$) may be taken of the form $C_{\#}$. The constants called $\underline c$, $\underline C$ in (H$\hat 0$)$\ldots$(H$\hat 7$) may be taken of the form $C_\ast$. \endproclaim \vglue 1pc \demo{Proof} (H$\hat 0$) follows from (Z0$^\dag$) and from $B(x)\equiv x$, $x_1>2x_0$. (H$\hat 1$) follows from (Z1$^\dag$) and (Z2$^\dag$). (H$\hat 2$) follows from (Z3$^\dag$). (H$\hat 3$) is proven as follows. $I_{\text{left}}$, $I_{\text{rt}}\neq \emptyset$ by definition. $|I_{\text{left}}|=x_0=B(x_{\text{left}})$, $|I_{\text{rt}}|=x_{\text{big}}-x_1\le \underline Cx_1$ (by (Z5$^\dag$)) $=\underline CB(x_{\text{rt}})$. Also $\lambda(x_{\text{left}})=S^{1/2} (x_0)\cdot x_0\le C_\ast$, since $-\underline Cx_0^{-2}\le V(x_0)\le -cS(x_0)<0$ by (Z2$^\dag$) and (Z4$^\dag$). Similarly, $\lambda(x_{\text{rt}})=S^{1/2}(x_1)\cdot x_1\le C_\ast$, since $-\underline Cx_1^{-2}\le V(x_1)\le -cS(x_1)<0$ by (Z2$^\dag$) and (Z5$^\dag$). We have proven all the assertions in (H$\hat 3$). (H$\hat 4$) is (Z4$^\dag$). (H$\hat 5$) follows from (Z5$^\dag$). (H$\hat 6$) holds vacuously, since $I_{\text{far\ left}}=\emptyset$. (H$\hat 7$) follows from (Z6$^\dag$).$\qquad\blacksquare$\enddemo \vglue1pc \proclaim{Lemma 2} Suppose $\tilde E=V(\tilde x)$ with $\tilde x\in [x_\ast,\frac 14 x_1]$. Take $\tilde V(x)=V(x)-\tilde E$, $\tilde S(x)= S(x)\cdot \bigl(\frac{\tilde x-x}{\tilde x}\bigr)$, $\tilde B(x)=\min (x,\tilde x-x)$, $\tilde E_0=0$, $\tilde I_{\text{far\ left}}=\emptyset$, $\tilde I_{\text{left}}=(0,x_0]$, $\tilde I_{\text{center}}=[x_0, \tilde x-\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot \tilde x]$ with $\hat C_{\#}$ to be picked large enough, $\tilde I_{\text{rt}}= [\tilde x-\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot \tilde x,\tilde x+\hat C_{\#} \lambda^{-2/3}(\tilde x)\cdot \tilde x]$, $\tilde I_{\text{far\ rt}}= [\tilde x+\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot \tilde x,\infty)$. Then the hypotheses (H$\hat 0$)$\ldots$(H$\hat 7$), in the section on WKB with weak turning points, are satisfied. The constants called $c$, $C$, $C_\alpha$ in (H$\hat 0$)$\ldots$(H$\hat 7$) may be taken of the form $C_{\#}$. The constants called $\underline c$, $\underline C$ in (H$\hat 0$)$\ldots$(H$\hat 7$) may be taken of the form $C_\ast$. \endproclaim \vglue 1pc \demo{Sketch of Proof} (H$\hat 0$), (H$\hat 1$), (H$\hat 2$) are proven exactly as in the proof of Lemma 2 in the preceding section. Regarding (H$\hat 3$), the assertions $|\tilde I_{\text{rt}}|\le C_\ast \tilde B(x_{\text{rt}})$ and $\tilde\lambda(x_{\text{rt}})\le C_\ast$ are proven just as in the discussion of (H$\hat 3$) in the proof of Lemma 2 in the preceding section. We have also $|\tilde I_{\text{left}}|=x_0 =\tilde B(x_{\text{left}})$, since $x_{\text{left}}=x_0<\frac 12 \tilde x$. In addition, $\tilde\lambda(x_{\text{left}})\sim \lambda(x_0)$ (since $x_{\text{left}}=x_0<\frac 12 \tilde x)=S^{1/2}(x_0)\cdot x_0\le C_\ast$ (as we saw in the proof of Lemma 1 above.) The intervals $\tilde I_{\text{left}}$, $\tilde I_{\text{rt}}$ are non-empty by definition. We have thus proven all the assertions in (H$\hat 3$). To check (H$\hat 4$), let $x\in \tilde I_{\text{left}}. (Z4^\dag)$ gives $|V(x)|\le \frac{C_\ast}{x_0x}$, $|V(x_0)|\le \frac{C_\ast}{x_0^2}$; and (Z2$^\dag$) gives \$V(x_0)