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\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{$\circ$}}
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\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 $\{x\in I_{\text{BVP}}\mid V(x)<E\}=(x_{\text{left}}(E),x_{\text{rt}}
(E))$.  The endpoints $x_{\text{left}}(E)$, $x_{\text{rt}}(E)$
are called ``turning points".  Define
$$
\phi(E)=\frac 1\pi\int_{I_{\text{BVP}}}(E-V(x))_+^{1/2}\,dx+\frac 12,\quad
\text{and}\tag 14
$$
$$
\eta(x,E)=-\frac \pi 4+\int_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt
\quad\text{for}\quad x_{\text{left}}(E)<x<x_{\text{rt}}(E)\, .\tag 15
$$

Let $E_k$, $\varphi_k$ be an eigenvalue and its corresponding
normalized eigenfunction.  Then $\varphi_k$ is given on overlapping
intervals by the following approximate formulas.

\noindent (I) For $x\in (x_{\text{left}}(E_k),x_{\text{rt}}(E_k))$ not too near
the turning points,
$$
\varphi_k(x)\approx (\pi\phi^\prime(E_k))^{-1/2}(E_k-V(x))^{-1/4}
\text{cos}(\eta(x,E_k))\, .\tag 16
$$
\medskip

\noindent (II)  For $x$ near the turning point $x_{\text{left}}(E_k)$,
$$
\varphi_k(x)\approx\lambda^{-1/3}(\pi\phi^\prime(E_k))^{-1/2}
\Bigl(\frac{\partial y(x,E_k)}{\partial x}\Bigr)^{-1/2}\,A(\lambda^{2/3}
y(x,E_k))\, ,\tag17
$$

\noindent where $A(t)$ is a slight variant of the Airey function, and
$$
\lambda^2\Bigl(\frac{\partial y(x,E)}{\partial x}\Bigr)^2y(x,E)
\approx E-V(x)\, .\tag 18
$$

\noindent The auxiliary function $y(x,E)$ depends smoothly on 
$x,\frac{E}{\lambda^2}$.
\medskip

\noindent (III)  For $x$ near the turning point $x_{\text{rt}}(E_k)$,
$$
\varphi_k(x)\approx\pm \lambda^{-1/3}(\pi\phi^\prime(E_k))^{-1/2}
\Bigl(-\frac{\partial}{\partial x}\tilde y(x,E_k)\Bigr)^{-1/2}A(\lambda^{2/3}
\tilde y(x,E_k))\tag 19
$$
\noindent for an auxiliary function $\tilde y(x,E)$ analogous to $y(x,E)$.
\medskip

\noindent (IV) For $x$ outside $(x_{\text{left}}(E_k),x_{\text{rt}}(E_k))$
and not too near the turning points, $\varphi_k(x)$ is negligibly small.
\medskip

These formulas and (8) comprise the standard WKB approximations.  (Again,
see Erd\'elyi [E].)  Our previous paper [FS2] proves precise
theorems justifying and refining (8) and (I)$\ldots$(IV).  As $E_k$ gets
too near to the minimum of $V$, (I)$\ldots$(IV) lose accuracy.
\medskip

\noindent (V) For $E_k$ near the minimum of the potential, the 
approximation (8) remains highly accurate, and the eigenfunctions $\varphi_k
(x)$ are highly concentrated near the point $x_0$ where $V$ attains
its minimum. 

\noindent This concludes our discussion of the individual eigenvalues
and eigenfunctions of $\frac{-d^2}{dx^2}+V$.
\medskip

Our task in proving Theorem 1 is to use our precise knowledge of the
eigenfunctions $\varphi_k$ to understand the density (1).  Because the
$\varphi_k$ are described by different formulas in overlapping regions,
it is natural to introduce the {\it microlocalized density}
$$
\rho(x,g)=\sum\limits_{E_k\le 0}\,|\varphi_k(x)|^2g(x,E_k)\tag 20
$$

\noindent corresponding to a function $g(x,E)$.  We compare $\rho(x,g)$
with the {\it semiclassical microlocalized density\/}, defined by
$$
\rho_{sc}(x,g)=\frac 1{2\pi}\int_{-\infty}^0(E-V(x))_+^{-1/2}
g(x,E)\,dE-\frac{(-V(x))_+^{-1/2}g(x,0)}{(\int_{I_{\text{BVP}}}(-V(y))_+^{-1/2}\,
dy)}\, \chi_{{}_-}(\frac 1\pi\phi)\, .\tag 21
$$

\noindent When $g\equiv 1$, $\rho(x,g)$ and $\rho_{sc}(x,g)$ reduce to
the density (1) and its semiclassical approximation (5).  Theorem 1 is
a special case of the general assertion that $\rho_{sc}(x,g)$ approximates
$\rho(x,g)$ closely in $H^{-1}$ for all reasonable functions $g$.  By a
partition of unity, we may assume that $g$ is supported in one of
the regions (I)$\ldots$(V) above.

We deal with each case (I)$\ldots$(V) in turn.  Suppose first that
$g(x,E)$ is supported in region (I).  Thus, $E-V(x)$ is positive and
rather large in $\text{supp}\ g(x,E)$.  For simplicity, we assume here
that $E-V(x)>c\lambda^2$ 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
$$
\Cal G(z)=\operatornamewithlimits{\int} \Sb E<0\\ x<z \endSb 
\int\frac{g(x,E)}
{\lambda^{2/3}(\frac{\partial y}{\partial x}(x,E))}\,[A^2(\lambda^{2/3}
y(x,E))-\frac 12(\lambda^{2/3}y(x,E))_+^{-1/2}]\,dE\,dx
$$
\noindent and
$$
\Cal H(z)=\int_{x<z} \frac{g(x,0)}{\lambda^{2/3}(\frac{\partial y}{\partial x}
(x,0))}\, [A^2(\lambda^{2/3}y(x,0))-\frac 12(\lambda^{2/3}
y(x,0))_+^{-1/2}]\,dx\, .
$$

To estimate $G$ and $H$ in $H^{-1}$--norm, we bound the $L^2$--norms of
$\Cal G$ and $\Cal H$.  We can reduce $\Cal G$ and $\Cal H$ to the form (35)
by making simple changes of variable.  In fact, set $t=\lambda^{2/3}y(x,0)$
and we find that
$$\split
\Cal H(z)=\int_{-\infty}^{\lambda^{2/3}y(z,0)}\Theta(t)[A^2(t)-\frac 12
t_+^{-1/2}]\,dt,\qquad\text{with}\,\, \Theta\,\, \text{defined\ by}\\
\Theta(t)\,dt=\frac{g(x,0)}{\lambda^{2/3}(\frac{\partial y}{\partial x}
(x,0))}\,dx\, .\endsplit\tag 36
$$

\noindent Similarly, putting $t=\lambda^{2/3}y(x,E)$ gives
$$\split
\Cal G(z)=\operatornamewithlimits{\int}
\Sb x<z \\ t<\lambda^{2/3}y(x,0) \endSb \int h(x,t)[A^2(t)-\frac 12
t_+^{-1/2}]\,dt\,dx,\quad \text{with}\\
h(x,t)\,dt=\frac{g(x,E)}{\lambda^{2/3}(\frac{\partial y}{\partial x}
(x,E))}\,dE\qquad\text{for\ fixed}\ x\, .\endsplit\tag 37
$$

Then with $\tilde x(t)$ defined as the solution of $\lambda^{2/3}y(\tilde x,0)
=t$, we can rewrite (37) in the form
$$\split
\Cal G(z)=\int_{-\infty}^{\lambda^{2/3}y(z,0)}\tilde\Theta(z,t)[A^2(t)-\frac
12 t_+^{-1/2}]\,dt,\quad \text{where}\\
\tilde\Theta(z,t)=\int_{\tilde x(t)}^z h(x,t)\,dx\, .\endsplit\tag 38
$$

\noindent Equations (36) and (38) reduce $\Cal G(z)$ and $\Cal H(z)$
to the form (35).  Our pointwise bounds for integrals (35) therefore control
the $L^2$--norms of $\Cal G$ and $\Cal H$.  In other words, we can control the
$H^{-1}$--norms of the errors that arise when we replace
$A^2(\lambda^{2/3}y)$ by $\frac 12(\lambda^{2/3}y)_+^{-1/2}$ in (34).

Now we have 
$$\multline
\rho(x,g)\approx \frac 1{2\pi}\int_{E<0}\lambda^{-1}\Bigl(\frac{\partial y}
{\partial x}(x,E)\Bigr)^{-1}(y(x,E))_+^{-1/2}g(x,E)dE\\
-\frac{\chi_{{}_-}(\frac \phi\pi)\lambda^{-1}(\frac{\partial y}
{\partial x}(x,0))^{-1}(y(x,0))_+^{-1/2}g(x,0)}
{\int_{I_{\text{BVP}}}(-V(y))_+^{-1/2}\,dy}.\endmultline
\tag 39
$$

\noindent Equation (18) shows that $\lambda^{-1}(\frac{\partial y}{\partial x}
(x,E))^{-1}(y(x,E))_+^{-1/2}\approx (E-V(x))_+^{-1/2}$, with a small error
in $H^{-1}$.  Putting this into (39), we get
$$\multline
\rho(x,g)\approx \frac 1{2\pi}\int_{E<0}(E-V(x))_+^{-1/2}g(x,E)dE-
\frac{\chi_{{}_-}(\frac\phi\pi)(-V(x))_+^{-1/2}g(x,0)}{\int_{I_{\text{BVP}}}(-V(y))_+
^{-1/2}dy}\\
\equiv \rho_{sc}(x,g).\endmultline
$$

\noindent This completes the proof that $\rho(x,g)\approx\rho_{sc}(x,g)$
for $g$ suppported in region (II).  The proof for $g$ supported in region (III)
is the same as for region (II).

It is trivial to show that $\rho(x,g)\approx \rho_{sc}(x,g)$ when $g$
is supported in region (IV).  In fact, from (IV) and (20) we see that
$\rho(x,g)$ is negligibly small, while (21) gives $\rho_{sc}(x,g)\equiv 0$.

Finally, suppose $g(x,E)$ is supported in region (V), where $E\approx
\min\ V$.  A sharp analysis would compare $V(x)$ to the harmonic
oscillator, but the following crude argument is enough for our purposes.
According to (V), the relevant eigenfunctions $\varphi_k$ are highly
concentrated in a small interval $J=[x_-,x_+]$ about the point $x_0$
where $V$ takes its minimum.  Hence
$$
\rho(x,g)=\sum\limits_k|\varphi_k(x)|^2g(x,E_k)\approx \sum\limits_k
|\varphi_k(x)|^2g(x_0,E_k),\quad\text{and}
$$
$$\int\limits_{x<z}\rho(x,g)dx\approx \cases 0 &\text{for $z<x_-$}\\
\sum\limits_kg(x_0,E_k) &\text{for $z>x_+$}\\
O(\sum\limits_k|g(x_0,E_k)|) &\text{for $z\in J$}.\endcases\tag 40
$$

\noindent Equation (40) holds also for $\rho_{sc}(x,g)$, and therefore
$$
\Cal G(z)\equiv \int\limits_{x<z}(\rho(x,g)-\rho_{\text{sc}}(x,g))dx=
O(\sum\limits_k|g(x_0,E_k)|)\quad\text{for}\quad z\in J,
$$

\noindent $\Cal G(z)$ is negligibly small outside $J$.

\noindent Since also $J$ is small, it follows that $\Cal G$ has small 
$L^2$--norm.  Thus $\rho(x,g)-\rho_{sc}(x,g)$ has small norm in $H^{-1}$.
This concludes our sketch of case (V).  Later, when we fill in details, we will
restrict attention to $\rho(x,g)$ for $g(x,E)$ that depend on $E$ alone.
This is enough for our purposes.

We have shown that $\Vert\rho(x,g)-\rho_{sc}(x,g)\Vert_{H^{-1}}$ is small
in each of the cases\break  (I)$\ldots$(V), so the proof of Theorem 1 is
complete.  The proof of our general result, the WKB Density Theorem,
is essentially the same, although the details are more elaborate.  

We will need also crude variants of our density theorems in several
degenerate cases.  For instance, if $V(x)$ is as in (2), (3), then we
need a crude understanding of the density associated to $-\frac{d^2}
{dx^2}+V(x)+E_0$, where $-E_0$ is just slightly greater than the minimum
of $V$.  The relevant density results for degenerate cases are proved
in this paper, but we omit them from the introduction.

We next make a small remark to confirm that the extra
term in the definition of $\rho_{sc}$ improves the accuracy of the 
approximation.
If we calculate the integral of $\rho_{sc}$ over the whole interval
$I_{\text{BVP}}$, 
then we obtain an approximation to $\int_{I_{\text{BVP}}}\rho(x)dx$, which
is the number of non--positive eigenvalues of $-\frac{d^2}{dx^2}+V(x)$.
>From (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 $|x-y|<cB(x)$, then $c<\frac{B(y)}
{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$.
\item"(Hyp1)" For $x \in I$ and $\alpha \ge 0$ we have
$|(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$.
\item"(Hyp2)" The equation $V(x)=E_0$ has two solutions
$x_{\text{left}}<x_{\text{rt}}$ in $I$, and they satisfy
$\text{dist}(x_{\text{left}},\partial I)>cB(x_{\text{left}}),
\text{dist}(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$.
\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 $cS(x)<E_0-V(x)<CS(x)$.
\endroster\enddemo

To state the remaining hypotheses, we establish some notation.  Set
$\lambda(x)=S^{1/2}(x)B(x)$ for $x \in I$.  Then set
$$
B_{\text{left}}=B(x_{\text{left}}),\quad S_{\text{left}}=S(x_{\text{left}}),
\quad\lambda_{\text{left}}=\lambda(x_{\text{left}})\ .
$$
$$
B_{\text{rt}}=B(x_{\text{rt}}),\quad S_{\text{rt}}=S(x_{\text{rt}})
,\quad\lambda_{\text{rt}}=\lambda(x_{\text{rt}})\ .
$$

For $|E-E_0|<cS_{\text{left}}$, let $x_{\text{left}}(E)$ be the solution
of $V(x)=E$ nearest to $x_{\text{left}}$, and for $|E-E_0|<cS_{\text{rt}}$,
let $x_{\text{rt}}(E)$ be the solution of $V(x)=E$ nearest to $x_{\text{rt}}$.
Define $S_{\text{min}}=\operatornamewithlimits{inf}_{x_{\text{left}}<x<x_
{\text{rt}}}S(x)$ and $\Lambda=\Bigl(\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac
{dx}{S^{1/2}(x)B^2(x)}\Bigr)^{-1}$.

Our remaining hypotheses are as follows.

\noindent\demo{Assumptions on $V(x)$ in all of $I_{\text{BVP}}$}
\roster
\item"{(Hyp5)}" If $|E-E_0|<c_2S_{\text{min}}$ and $E\le E_\infty$, then
$V(x)>E$ \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
$x<x_{\text{left}}-\frac 12 \lambda_{\text{left}}^KB_{\text{left}}$ then
$V(x)\ge E_\infty+\frac{100}{|x-x_{\text{left}}|^2}$, and if
$x \in I_{\text{BVP}}$ satisfies $x>x_{\text{rt}}+\frac 12 \lambda_{\text{rt}}
^KB_{\text{rt}}$, 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 $|E-E_0|<c_{\#}S_{\text{min}}$,
define
$$\gather
\phi(E)=\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}(E-V(x))^{1/2}\,
dx\quad \text{and}\\
\psi(E)=\operatornamewithlimits{lim}_{\delta\to 0+} \Bigl[
\int_{x_{\text{left}}(E)+\delta}^{x_{\text{rt}}(E)-\delta}V^{\prime\prime}
(x)(E-V(x))^{-\frac 32}\,dx-q(E)\delta^{-1/2}\Bigr]\endgather
$$

\noindent with $q(E)$ uniquely specified by demanding the finiteness of
the limit.\enddemo

\vglue 1pc
\proclaim{Lemma 1}  For $|E-E_0|<c_{\#}S_{\text{min}}$ we have
$$\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\phi(E)\Big|\le C_{\#}^\beta
\int_{x_{\text{left}}}^{x_{\text{rt}}}S^{\frac 12-\beta}(x)\,dx
\quad\text{and}
$$
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\psi(E)\Big|\le C_{\#}^\beta
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{\frac 12+\beta}
(x)B^2(x)}\ .
$$
\noindent Also $c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{1/2}(x)}<\frac{d\phi(E)}{dE}<C_{\#}\int_{x_{\text{left}}}^{
x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}$.

The constants $c_{\#}$, $C_{\#}$, $C_{\#}^\beta$ depend only on
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$,  $c_2$, $C_\alpha$ in the
hypotheses (Hyp 0)$\ldots$(Hyp 4).
\endproclaim

\vfill\eject
\proclaim{WKB Eigenvalue Theorem}  If (Hyp0)$\ldots$(Hyp10) hold, then
there is a function $\Phi(E)$ on $[E_0-c_{\#}S_{\text{min}},E_0+c_{\#}
S_{\text{min}}]$ and there are numbers $E_{k_{\text{min}}},E_{k_{\text{min}}+1},
\ldots,E_{k_{\text{max}}}\hfill\break \le E_\infty$ with the following 
properties.
\roster
\item"{(A)}" $\Phi(E)=\pm\frac \pi 2+\phi(E)+\frac{1}{48}\psi(E)+\phi_
{\text{error}}(E)$, with
$$
\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\phi_{\text{error}}\Big|\le C_{\#}
^\beta\Lambda^{-1}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{\frac 12+\beta}(x)B^2(x)}, \quad\text{all}\ \beta\ge 0\ .
$$
\item"{(B)}" $\{k_{\text{min}},k_{\text{min}}+1,\ldots,k_{\text{max}}\}$
is exactly the set of integers $k$ for which $|\Phi(E)-\pi k|<C_{\#}
\Lambda^{-N^{\prime\prime}}$ for some $E \in [E_0-\frac 14
c_{\#}S_{\text{min}},E_0+\frac 14c_{\#}S_{\text{min}}]\cap (-\infty,E_{\infty}]$.
\item"{(C)}" If $k_{\text{min}}\le k<k_{\text{max}}$, then $E_k$ is an 
eigenvalue of $H$.
\item"{(D)}" If $k=k_{\text{max}}$, then $E_k$ is either an eigenvalue of $H$
or equal to $E_\infty$.
\item"{(E)}" For $k_{\text{min}}\le k\le k_{\text{max}}$ we have
$|E_k-E_0|<c_{\#}S_{\text{min}}$ and $|\Phi(E_k)-\pi k|<C_{\#}
\Lambda^{-N^{\prime\prime}}$.
\item"{(F)}" Every eigenvalue of $H$ in the interval $[E_0-\frac 14
c_{\#}S_{\text{min}},E_0+\frac 14c_{\#}S_{\text{min}}]\hfill\break
 \cap (-\infty,E_\infty]$
is one of the $E_k$ $(k_{\text{min}}\le k\le k_{\text{max}})$.\endroster
\endproclaim

\noindent The constants $c_{\#}$, $C_{\#}^\beta$ depend only on $\varepsilon$,
$K$, $N$, $c$, $C$, $c_1$ $c_2$, $C_\alpha$ in (Hyp0)$\ldots$(Hyp5).

\demo{Remark} Perhaps $E_{k_{\text{min}}}$ or $E_{k_{\text{max}}}$ or both lie
slightly outside 
$$
[E_0-\frac 14c_{\#}S_{\text{min}},E_0+\frac 14 c_{\#}
S_{\text{min}}]\ .$$ \enddemo

We prepare to describe the eigenfunctions of $H$.  Given an energy $E$
with $|E-E_0|<c_{\#}S_{\text{min}}$, define intervals:
$$
I_{\text{far\ left}}^E=I_{\text{BVP}}\cap (-\infty,x_{\text{left}}(E)
-\lambda_{\text{left}}^{\varepsilon-2/3}B_{\text{left}}]
$$
$$ 
I_{\text{Airey\ left}}^E=[x_{\text{left}}(E)-\lambda_{\text{left}}^
{-\varepsilon}B_{\text{left}},x_{\text{left}}(E)+\lambda_{\text{left}}^
{-\varepsilon} B_{\text{left}}]
$$
$$
I_{\text{medium\ left}}^E=[x_{\text{left}}(E)+\lambda_{\text{left}}^{
\varepsilon-\frac 23}B_{\text{left}}, x_{\text{left}}(E)+c_{\#}
B_{\text{left}}]
$$
$$
I_{\text{center}}^E=[x_{\text{left}}(E)+\frac 12 c_{\#}B_{\text{left}},
x_{\text{rt}}(E)-\frac 12 c_{\#}B_{\text{rt}}]
$$
$$
I_{\text{medium\ rt}}^E=[x_{\text{rt}}(E)-c_{\#}B_{\text{rt}},
x_{\text{rt}}(E)-\lambda_{\text{rt}}^{\varepsilon-2/3}B_{\text{rt}}]
$$
$$
I_{\text{Airey\ rt}}^E=[x_{\text{rt}}(E)-\lambda_{\text{rt}}^{-\varepsilon}
B_{\text{rt}},x_{\text{rt}}(E)+\lambda_{\text{rt}}^{-\varepsilon}B_{\text{rt}}]
$$
$$
I_{\text{far\ rt}}^E=[x_{\text{rt}}(E)+\lambda_{\text{rt}}^{\varepsilon-2/3}
B_{\text{rt}},\infty)\cap I_{\text{BVP}}.
$$
\noindent Also define regions
$$
U_{\text{Airey\ left}}=\{(x,E)\Bigm| |E-E_0|<c_{\#}S_{\text{min}},
|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}\},
$$
$$
U_{\text{Airey\ rt}}=\{(x,E)\Bigm| |E-E_0|<c_{\#}S_{\text{min}},|x-x_{\text{rt}}
(E)|<\lambda_{\text{rt}}^{-\varepsilon}B_{\text{rt}}\}
$$
$$
U_{\text{oscil}}=\{(x,E)\Bigm| |E-E_0|<c_{\#}S_{\text{min}},x_{\text{left}}(E)<
x<x_{\text{rt}}(E)\}\ .
$$
\noindent If $Y(x,E)$ is any smooth function with $\frac{\partial Y}
{\partial x}\ne 0$, then define $\{Y,x\}\equiv \frac 12 \frac
{\partial_x^3Y}{\partial_xY}-\frac 34 \bigl(\frac{\partial_x^2Y}
{\partial_xY}\bigr)^2$.
\vglue 1pc
\proclaim{WKB Eigenfunction Theorem}  Assume (Hyp0)$\ldots$(Hyp10).  Then
there exist real--valued functions $Y_{\text{left}}(x,E)$ on
$U_{\text{Airey\ left}}$, $Y_{\text{rt}}(x,E)$ on $U_{\text{Airey\ rt}}$,
and complex--valued functions $u_{\text{left}}(x,E)$, $u_{\text{rt}}(x,E)$
on $U_{\text{oscil}}$, for which the following hold.\endproclaim
\medskip

\subhead Properties of the Auxiliary Functions\endsubhead
\roster
\item"{(1)}" On $U_{\text{Airey\ left}}$\ we can write
$Y_{\text{left}}=Y_0^{\text{left}}+\lambda_{\text{left}}^{-2}
Y_1^{\text{left}}$, with $|\partial_x^\alpha\partial_E^\beta Y_i^{\text{left}}|
\le C_{\#}^{\alpha\beta}B_{\text{left}}^{-\alpha}S_{\text{left}}^{-\beta}$
and $\lambda_{\text{left}}^2(\frac{\partial Y_0^{\text{left}}}{\partial x})^2Y_0^
{\text{left}}=E-V(x)$ to order at least $N^\prime$ at $x=x_{\text{left}}(E)$.
More precisely, $|\lambda_{\text{left}}^2(\frac{\partial Y_{\text{left}}}
{\partial x})^2\cdot Y_{\text{left}}+\{Y_{\text{left}},x\}-
(E-V(x))|\le C_{\#}\lambda_{\text{left}}^{-N^\prime}
S_{\text{left}}$ on $U_{\text{Airey}}^{\text{left}}$. 
Similarly, on
$U_{\text{Airey\ rt}}$ we can write $Y_{\text{rt}}=Y_0^{\text{rt}}+
\lambda_{\text{rt}}^{-2}Y_1^{\text{rt}}$, with $|\partial_x^\alpha
\partial_E^\beta Y_i^{\text{rt}}|\le C_{\#}^{\alpha\beta}
B_{\text{rt}}^{-\alpha}S_{\text{rt}}^{-\beta}$ and $\lambda_{\text{rt}}^2
(\frac{\partial Y_0^{\text{rt}}}{\partial x})^2Y_0^{\text{rt}}=E-V(x)$
to order at least $N^\prime$ at $x=x_{\text{rt}}(E)$.  More precisely,
$|\lambda_{\text{rt}}^2
(\frac{\partial Y_{\text{rt}}}{\partial x})^2Y_{\text{rt}}+\{Y_{\text{rt}},x\}-
(E-V(x))|\le C_{\#}\lambda_{\text{rt}}^{-N^\prime}S_{\text{rt}}$ on
$U_{\text{Airey}}^{\text{rt}}$.

\item"{(2)}" For $|E-E_0|<c_{\#}S_{\text{min}}$ and $x \in I_{\text{medium\
left}}^E$ we have $|\partial_x^\alpha u_{\text{left}}|\le \hfill\break
C_{\#}^\alpha
\lambda_{\text{left}}^{-1}(\frac{x-x_{\text{left}}(E)}{B_{\text{left}}})
^{-\frac 32}\cdot(x-x_{\text{left}}(E))^{-\alpha}$, and
$|\text{Re}\ u_{\text{left}}|\le \hfill\break
C_{\#}\lambda_{\text{left}}^{-2}
(\frac{x-x_{\text{left}}(E)}{B_{\text{left}}})^{-3}$.  Similarly, for
$|E-E_0|<c_{\#}S_{\text{min}}$ and $x \in I_{\text{medium\ rt}}^E$ we have
$|\partial_x^\alpha u_{\text{rt}}|\le C_{\#}^\alpha\lambda_{\text{rt}}^{-1}
(\frac{x_{\text{rt}}(E)-x}{B_{\text{rt}}})^{-\frac 32}\cdot (x_{\text{rt}}
(E)-x)^{-\alpha}$ and $|\text{Re}\ u_{\text{rt}}|\le C_{\#}
\lambda_{\text{rt}}^{-2}(\frac{x_{\text{rt}}(E)-x}{B_{\text{rt}}})^{-3}$.
\item"{(3)}" For $|E-E_0|<c_{\#}S_{\text{min}}$ and $x \in I_{\text{center}}^E$
we have
$$
|\partial_x^\alpha u_{\text{left}}|\le C_{\#}^\alpha\Lambda^{-1}B^{-\alpha}
(x)\quad\text{and}\quad |\text{Re}\ u_{\text{left}}|\le C_{\#}\Lambda^{-2}\ .
$$
Similarly, on the same region we have $|\partial_x^\alpha u_{\text{rt}}|\le
C_{\#}^\alpha\Lambda^{-1}B^{-\alpha}(x)$ and $|\text{Re}\ u_{\text{rt}}|
\hfill\break\le C_{\#}\Lambda^{-2}$.
\endroster

\smallskip
\subhead Description of Eigenfunctions\endsubhead

Let $F(x)$ be a real--valued eigenfunction of $H$ with eigenvalue $E$ and
norm $1$.  Assume $|E-E_0|<c_{\#}S_{\text{min}}$ and $E\le E_\infty$.  Then
there exist real constants $b_{\text{left}}$, $b_{\text{rt}}$ for which we
have:
\roster
\item"{(4)}" $\int_{I_{\text{far\ left}}^E}|F(x)|^2\,dx\le\Lambda^
{-N^{\prime\prime}}$ and $\int\limits_{I_{\text{far\ rt}}^E}|F(x)|^2\,dx\le
\Lambda^{-N^{\prime\prime}}$
\item"{(5)}" $\int_{I_{\text{Airey\ left}}^E}|F(x)-b_{\text{left}}
\lambda_{\text{left}}^{-1/3}\Bigl(\frac{\partial Y_{\text{left}}(x,E)}
{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{left}}^{2/3}Y_{\text{left}}(x,E))|^2\,dx
\le \hfill\break
\Lambda^{-N^{\prime\prime}}$ and $\int\limits_{I_{\text{Airey\ rt}}^E}
|F(x)-b_{\text{rt}}\lambda_{\text{rt}}^{-1/3}\Bigl(\frac{-\partial Y_{\text{rt}}
(x,E)}{\partial x}\Bigr)^{-1/2}A(\lambda_{\text{rt}}^{2/3}
Y_{\text{rt}}(x,E))|^2\,dx\hfill\break\le
\Lambda^{-N^{\prime\prime}}$
\item"{(6)}"$$\split
\int_{I_{\text{medium\ left}}^E\cup I_{\text{center}}^E}
\Big|F(x)-b_{\text{left}}\text{Re}\Bigl[\frac{e^{\pm i\frac \pi 4}\,e^{i\int_
{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\\
\cdot(1+u_{\text{left}}(x,E))\Bigr]\Big|^2\,dx\le \Lambda^{-N^{\prime\prime}}
\endsplit
$$
\noindent and
$$\split
\int_{I_{\text{medium\ rt}}^E\cup I_{\text{center}}^E}
|F(x)-b_{\text{rt}}\text{Re}\Bigl[\frac{e^{\mp i\frac \pi 4}\,
e^{-i\int_x^{x_{\text{rt}}(E)}(E-V(t))^{1/2}\,dt}}{(E-V(x))^{1/4}}\\
\cdot (1+u_{\text{rt}}(x,E))\Bigr]|^2\,dx\le \Lambda^{-N^{\prime\prime}}
\endsplit
$$
\item"{(7)}" $c_{\#}<(|b_{\text{left}}|^2+|b_{\text{rt}}|^2)
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}<C_{\#}$, and
\item"{(8)}" $\Bigl||b_{\text{left}}|\cdot|b_{\text{rt}}|^{-1}-1\Bigr|
\le C_{\#}\Lambda^{-2}$.
\endroster
\medskip

\subhead What the Constants may Depend on\endsubhead
\smallskip

The constants $c_{\#}$, $C_{\#}$, $C_{\#}^\alpha$ above depend only on 
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$ in
(Hyp0)$\ldots$(Hyp5).

\vglue 1pc
\proclaim{WKB Normalization Theorem}  The constants $b_{\text{left}}$,
$b_{\text{rt}}$ in the WKB Eigenfunction Theorem satisfy
$$
\Big|b_{\text{left}}^2\cdot \Bigl(\frac 12 \int_{E-V(x)>0}\frac {dx}
{(E-V(x))^{1/2}}\Bigr)-1\Big|\le C_{\#}\Lambda^{3\varepsilon-2}\quad
\text{and}
$$
$$
\Big|b_{\text{rt}}^2\cdot \Bigl(\frac 12 \int_{E-V(x)>0}\frac {dx}
{(E-V(x))^{1/2}}\Bigr)-1\Big|\le C_{\#}\Lambda^{3\varepsilon-2},
$$
\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$^*$)}" $I=\{|x-x_0|<cB\}\subset I_{\text{BVP}}$
\item"{(H1$^*$)}" $|(\frac{d}{dx})^\alpha V(x)|\le
C_\alpha SB^{-\alpha}$ in $I$
\item"{(H2$^*$)}" $\frac{d^2}{dx^2}V\ge c^\prime SB^{-2}$
in $I$
\item"{(H3$^*$)}" $V^\prime(x_0)=0$
\item"{(H4$^*$)}" For $x \in I_{\text{BVP}}\backslash I$
we have $V(x)\ge \text{min}\{E_\infty,V(x_0)+c^{\prime
\prime}\lambda^{-2\varepsilon} S\}$.
\item"{(H5$^*$)}" For $x \in I_{\text{BVP}}$ with
$|x-x_0|>\frac 12 \lambda^KB$, 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 $V(x_0)<E<V(x_0)+\lambda^{-2\varepsilon}S$, define
$x_{\text{left}}(E)<x_{\text{rt}}(E)$ to be the
two values of $x \in I$ at which $V(x)=E$.  Then define
$$
\phi(E)=\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
(E-V(x))^{1/2}\,dx
$$
$$\align
\psi(E)&=\lim_{\delta\to 0+}\Bigl[\operatornamewithlimits{\int} \Sb x \in I\\
E-V(x)>\delta  \endSb V^{\prime\prime}(x)(E-V(x))^{-3/2}\,dx
-q(E)\delta^{-1/2}\Bigr]\\
&=\lim_{\delta_{\text{left}},\delta_{\text{rt}}\to 0+}
\Bigl[\int_{x_{\text{left}}(E)+\delta_{\text{left}}}
^{x_{\text{rt}}(E)-\delta_{\text{rt}}} V^{\prime\prime}
(x)(E-V(x))^{-3/2}\,dx-q_{\text{left}}(E)\delta_{\text{
left}}^{-1/2}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\qquad\qquad\,\,\,\,\,
-q_{\text{rt}}(E)\delta_{\text{rt}}^{-1/2}
\Bigr]\endalign
$$
\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 $V(x_0)<E<V(x_0)+\lambda^{-2\varepsilon}S$.

\noindent The constants $c_{\#}$, $C_{\#}^\beta$ depend
only on $c$, $c^\prime$,$ c^{\prime\prime}$, $C_{\alpha}$,
$\varepsilon$, $K$, $N$ in hypotheses (H0$^*$)$\ldots$
(H6$^*$).\endproclaim

\vglue 1pc
\proclaim{WKB Theorem on Low Eigenvalues} Assume
(H0$^*$)$\ldots$(H6$^*$).  Then there is a finite
sequence $E_0,E_1,\ldots,E_{k_{\max}}$ with the
following properties.
\roster
\item"{(a)}" Let $w=\phi(E_*)+\frac{1}{48}\psi(E_*)$
with $E_*=\text{min}\{E_\infty,V(x_0)+c_{\#}\lambda^{-2\varepsilon}S\}$,
and let  $\overline n$ be the largest integer
with $\pi(\overline n+1/2)\le  w$.  If
$\operatornamewithlimits{min}_{k \in \Bbb Z}|w-
\pi(k+1/2)|>C_{\#}\lambda^{-2+4\varepsilon}$, then
$k_{\text{max}}=\overline n$.  In any case,
$|k_{\text{max}}-\overline n|\le 1$.
\item"{(b)}" If $0\le k<k_{\text{max}}$, then $E_k$
is an eigenvalue of $H$.
\item"{(c)}" Either $E_{k_{\text{max}}}=E_\infty$ or
else $E_{k_{\text{max}}}$ is an eigenvalue of $H$.
\item"{(d)}" Every eigenvalue $E$ of $H$ satisfying
$E\le E_\infty$,
$|E-V(x_0)|<c_{\#}\lambda^{-2\varepsilon}S$ is one of
the $E_k$.
\item"{(e)}" For $0 \le k\le k_{\text{max}}$ we have
$V(x_0)<E_k<V(x_0)+2c_{\#}\lambda^{-2\varepsilon}S$
and $|\phi(E_k)+\frac{1}{48}\psi(E_k)-\pi(k+1/2)|\le
C_{\#}\lambda^{-2+4\varepsilon}$.
\endroster
\noindent The constants $c_{\#}$, $C_{\#}$ depend only on
$\varepsilon$, $K$, $N$, $c$, $c^\prime$,
$c^{\prime\prime}$, $C_\alpha$ in hypotheses
(H0$^*$)$\ldots$(H6$^*$).\endproclaim

\vglue 1pc
\proclaim{Lemma 2}  Assume (H0$^*$)$\ldots$(H6$^*$).
Let $F(x)$ be an eigenfunction of $H$ whose eigenvalue
$E$ satisfies
$V(x_0)<E<V(x_0)+c_{\#}\lambda^{-2\varepsilon}S$ and $E\le
E_\infty$.  Then
$$
\int_{I_{\text{BVP}}\backslash\{|x-x_0|
<\lambda^{-\varepsilon}B\}}|F(x)|^2\,dx\le
C_{\#}\lambda^{-N}\int_{\{|x-x_0|<\lambda^{-\varepsilon}
B\}}|F(x)|^2\,dx.
$$
\noindent The constants $c_{\#}$, $C_{\#}$ depend only 
on $\varepsilon$, $K$, $N$, $c$, $c^\prime$,
$c^{\prime\prime}$, $C_\alpha$ in the hypotheses
(H0$^*$)$\ldots$(H6$^*$).\endproclaim 
\subhead WKB Theory with Weak Turning Points\endsubhead
\smallskip

\noindent\demo{Set--up}  We are given an energy
$E_0$ and a potential $V(x)$ defined on a (possibly
unbounded) interval $I_{\text{BVP}}$.  The interval $I_{\text{BVP}}$
is partitioned into subintervals
 $I_{\text{far\ left}}$, $I_{\text{left}}$,
$I_{\text{center}}$, $I_{\text{rt}}$, $I_{\text{far\ rt}}$
with $I_{\text{far\ left}}$ to the left of
$I_{\text{left}}$, $I_{\text{left}}$ to the left of
$I_{\text{center}}$, etc. Here, $I_{\text{far\ left}}$ and 
$I_{\text{far\ right}}$ may be empty.  On $I_{\text{center}}$ we
are given positive weight functions $S(x)$, $B(x)$.  Set
$\lambda(x)=S^{1/2}(x)B(x)$ and
$\Lambda=(\int_{I_{\text{center}}}\frac{dx}
{\lambda(x)B(x)})^{-1}$.  We make the following
assumptions.\enddemo

\noindent\demo{Hypotheses}
\roster
\item"{(H$\hat 0$)}" $I_{\text{center}}$ is non--empty,
and for $x,y \in I_{\text{center}}$ with $|x-y|<cB(x)$
we have $c<B(y)/B(x)<C$ and $c<S(y)/S(x)<C$, and
$|I_{\text{center}}|>cB(x)$.

\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 $cS(x)<E_0-V(x)<CS(x)$.

\item"{(H$\hat 2$)}" $\Lambda$ is bounded below by a
large number depending only on $c$, $C$, $C_\alpha$ in
(H$\hat0$), (H$\hat1$).

\item"{(H$\hat 3$)}" $I_{\text{left}}$, $I_{\text{rt}}$
are non--empty.  If $I_{\text{center}}=[x_{\text{left}},
x_{\text{rt}}]$, then we have $|I_{\text{left}}|\le
\underline{C}B(x_{\text{left}})$, $|I_{\text{rt}}|\le
\underline{C}B(x_{\text{rt}})$, $\lambda(x_{\text{left}})
\le \underline{C}$, $\lambda(x_{\text{rt}})\le 
\underline{C}$.

\item"{(H$\hat 4$)}" If $I_{\text{left}}=[x_{\text{far\
left}},x_{\text{left}}]$, then we have $|V(x)-E_0|\le
\underline{C}|I_{\text{left}}|^{-1}(x-\hat x_{\text{far\
left}})^{-1}$ in $I_{\text{left}}$.  Here
$\hat x_{\text{far\ left}}\le x_{\text{far\ left}}$ with
strict inequality unless $I_{\text{far\ left}}=\emptyset$.

\item"{(H$\hat 5$)}" For $x \in I_{\text{rt}}$ we have
$|V(x)-E_0|\le \underline{C}|I_{\text{rt}}|^{-2}$.

\item"{(H$\hat 6$)}" For $x \in I_{\text{far\ left}}$
we have $V(x)-E_0\ge \underline{c}|I_{\text{left}}|^{-2}$.
$V(x)$ is $C^\infty$ in the interior of $I_{\text{far\
left}}$.

\item"{(H$\hat 7$)}" For $x \in I_{\text{far\ rt}}$
we have $V(x)-E_0\ge
\frac{-10^{-9}}{(x-x_{\text{rt}})^2}$.\endroster
\enddemo

Our goal is to understand the eigenfunctions and
eigenvalues of $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_
{\text{BVP}})$ with Dirichlet boundary conditions.

\noindent Denote by $C_{\#}$ a constant depending on $c$,
$C$, $C_\alpha$ in (H$\hat 0$), (H$\hat 1$).
\noindent Denote by $C_*$, $c_*$ etc. constants
depending only on $c$, $C$, $C_\alpha$, $\underline{c}$,
$\underline{C}$ in (H$\hat 0$)$\ldots$(H$\hat 7$).

Note that $I_{\text{left}}$ and $I_{\text{rt}}$ don't
play completely analogous r$\hat{\text{o}}$les in our
hypotheses.

\vglue 1pc
\proclaim{Theorem 1}  Under the assumptions (H$\hat 0$)$\ldots$(H$\hat 7$)
we have
$$
|(\text{Number\ of\ eigenvalues\ of}\ H<E_0)-\frac 1\pi 
\int_{I_{\text{center}}}(E_0-V(t))^{1/2}\,dt|\le C_\ast\ .
$$
\noindent Suppose $E_0$ is an eigenvalue of $H$, and suppose $u$ is the
corresponding eigenfunction, with $u$ real--valued and having $L^2$--norm
$1$ on $I_{\text{BVP}}$.  Then
$$|u(x)|^2\le C_*\Bigl(\int_{I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x))^{-1/2}\ \text{for}\ x \in I_{\text{center}}
$$
$$\split
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x_{\text{left}}))^{-1/2}\exp\Bigl(\frac{-c_*(x_{\text{left}}-x)}
{B(x_{\text{left}})}\Bigr)\\
\text{for}\ x \in I_{\text{BVP}}, x \le x_{\text{left}}=
\text{min}\ I_{\text{center}}\ .\endsplit
$$
$$
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{center}}}(E_0-V(t))^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x_{\text{right}}))^{-1/2}\quad\text{for}\ x \in I_{\text{right}}\ .
$$\endproclaim

As an application of this theorem, we study the following situation.
\smallskip
\demo{Set--up} $V(x)$ is a potential defined on a (possibly
unbounded) interval $I_{\text{BVP}}$.  We are given a subinterval
$I \subset I_{\text{BVP}}$ and weight functions $S(x)$, $B(x)>0$ 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 $|x-y|<cB(x)$ we have
$c<\frac{B(y)}{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$, and $|I|>cB(x)$.
\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$\overline 2$)}" $\{x \in I_{\text{BVP}}\mid V(x)<E_0\}
=(x_{\text{left}},x_{\text{right}})\subset I$ with
$\text{dist}(x_{\text{left}},\partial I)>cB(x_{\text{left}})$,
$\text{dist}(x_{\text{right}},\partial I)>cB(x_{\text{right}})$.
\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 $cS(x)<E_0-V(x)<CS(x)$.
\item"{(H$\overline 5$)}" In $I_{\text{BVP}}^{\text{interior}}\cap
(-\infty,x_{\text{left}})$, $V(x)$ is decreasing and $C^\infty$.
\item"{(H$\overline 6$)}" $\Lambda=(\int_{x_{\text{left}}}^{x_{\text{right}}}
\frac{dx}{\lambda(x)B(x)})^{-1}$ is bigger than a large positive number
depending only on $c$, $C$, $c_1$, $C_\alpha$ in (H$\overline 0$)$\ldots$
(H$\overline 4$).
\endroster\enddemo

Let $H=-\frac{d^2}{dx^2}+V(x)$ on $L^2(I_{\text{BVP}})$ with Dirichlet
boundary conditions.

\vglue 1pc
\proclaim{Theorem 2}  Under assumptions (H$\overline 0$)$\ldots$
(H$\overline 6$) above, we have
$$
|\text{(Number\ of\ eigenvalues\ of}\ H<E_0)-\frac 1\pi \int_{I_{\text{BVP}}}
(E_0-V(t))_+^{1/2}\,dt|\le C_*.
$$
\noindent If $E_0$ is an eigenvalue of $H$, and if $u(x)$ (real--valued,
with norm $1$ in $L^2(I_{\text{BVP}})$) is the corresponding eigenfunction,
then
$$
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{BVP}}}(E_0-V(t))_+^{-1/2}\,dt\Bigr)^{-1}
(E_0-V(x))^{-1/2}\quad\text{for}\ x_{\text{left}}<x<x_{\text{right}}
$$
$$\split
|u(x)|^2\le C_*\Bigl(\int_{I_{\text{BVP}}}(E_0-V(t))_+^{-1/2}\,dt\Bigr)^{-1}
\Bigl[S(x_{\text{left}})\lambda^{-2/3}(x_{\text{left}})\Bigr]^{-1/2}\cdot\\
\cdot\exp\Bigl(-c_*\lambda^{2/3}(x_{\text{left}})\frac{(x_{\text{left}}-x)}
{B(x_{\text{left}})}\Bigr)\quad
\roman{for}\ x \in I_{\text{BVP}}\cap (-\infty,x_{\text{left}}]\ .\endsplit
$$
\noindent The constants $c_*$, $C_*$ depend only on $c$, $C$, $c_1$,
$C_\alpha$, in (H$\overline 0$)$\ldots$(H$\overline 4$).\endproclaim

\vfill\eject












\heading{\bf{A REFORMULATION OF THE EIGENVALUE THEOREM}}\endheading
\medskip

In this section, we restate the WKB Eigenvalue Theorem in the form in which
we will use it.

\vglue 1pc
\proclaim{Reformulated Eigenvalue Theorem}  Assume the hypotheses of
the WKB\break Eigenvalue Theorem, with $E_\infty=0$.  Let $\phi(E)$ and
$\psi(E)$ be defined as in that theorem.  Then the eigenvalues of $H$ that
lie in the interval
$$
[E_{\ell o},E_{hi}]\equiv (-\infty,0]\cap \{E\mid |E-E_0|\le c_{\#}^1
S_{\min}\}\tag 1
$$
\noindent may be written as $\{E_k\mid k_{\ell o}\le k\le k_{hi}\}$, where the
following properties are satisfied:
$$k_{\ell o}<k_{hi}\ .\tag 2
$$
$$
\text{For}\ \ k_{\ell o}\le k\le k_{hi}\ \ \text{we\ have}\ \
|\phi(E_k)+\frac 1{48}\psi(E_k)-\pi(k+1/2)|\le C_{\#}\Lambda^{-2}\ .\tag 3
$$
$$\multline
\text{The\ set}\ \ \{k\in \Bbb Z\mid k_{\ell o}\le k\le k_{hi}\}\ \
\text{consists\ exactly}\\ 
\text{of\ all\ those\ integers}\ \ k\ \
\text{that\ lie\ in\ the\ interval}\\
[a,b]\equiv [\frac 1\pi \phi(E_{\ell o})+\frac{1}{48\pi}\psi(E_{\ell o})
-\frac 12+\omega_{\ell o}, \frac 1\pi \phi(E_{hi})+\frac {1}{48\pi}
\psi(E_{hi})-\frac 12 +\omega_{hi}]\\
\text{with}\ |\omega_{\ell o}|, |\omega_{hi}|\le C_{\#}\Lambda^{-2}\ .
\endmultline\tag 4
$$
$$\multline
\text{If}\ \ \min_{k\in \Bbb Z}|\frac 1 \pi \phi(E_{hi})+\frac{1}
{48\pi}\psi(E_{hi})-\frac 12-k|\ge C_{\#}\Lambda^{-2}\ ,\\
\text{then\ we\ can\ take}\ \ \omega_{hi}=0\ \ \text{in\ (4)}\ .\endmultline
\tag 5
$$
\noindent Here, $c_{\#}^1$ and $C_{\#}$ depend only on the constants
$\varepsilon$, $K$, $N$, etc. appearing in the hypotheses of the
WKB Eigenvalue Theorem.\endproclaim

To derive the result from the WKB Eigenvalue Theorem is merely a
tedious exercise.  We provide details for the reader's convenience.

\vglue 1pc
\demo{Proof of the Reformulated Eigenvalue Theorem} We use $c_{\#}$,
$c_{\#}^\prime$ $C_{\#}$ etc. for constants depending only on the
constants $\varepsilon$, $K$, $N$ etc. appearing in the hypotheses of the
WKB Eigenvalue Theorem.  From that theorem, we get a function $\Phi(E)$
on $[E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]$ and a finite sequence
$E_{k_{\min}}, E_{k_{\min}+1},\ldots,E_{k_{\max}}\le 0$ with the following
properties.
$$\multline
\{k\in \Bbb Z\mid k_{\min}\le k\le k_{\max}\}\ \ \text{is\ exactly\
the\ set\ of\ those\ integers}\ \ k\ \ \text{for\ which}\\
|\Phi(E)-\pi k|<C_{\#}\Lambda^{-N^{\prime\prime}}\ \  \text{for\ some}\\
E\in (-\infty,0]\cap [E_0-\frac 14 c_{\#}S_{\min}, E_0+\frac 14 c_{\#}
S_{\min}]\ .\endmultline\tag 6
$$
$$
\text{If}\ \ k_{\min}\le k<k_{\max}\ , \ \text{then}\ \
E_k\ \ \text{is\ an\ eigenvalue\ of}\ \ H\ .\tag 7
$$
$$
E_{k_{\max}}\ \ \text{is either}\ \ 0\ \ \text{or\ an\ eigenvalue\ of}\ \
H\ .\tag 8
$$
$$\multline
\text{For}\ \ k_{\min}\le k\le k_{\max}\ \ \text{we\ have}\ \
E_k\in [E_0-c_{\#}S_{\min}, E_0+c_{\#} S_{\min}]\\
\text{and}\ \ |\Phi(E_k)-\pi k|<C_{\#}\Lambda^{-N^{\prime\prime}}\ .
\endmultline\tag 9
$$
$$\multline
\text{Every\ eigenvalue\ of}\ \ H\ \ \text{in\ the\ interval}\ \
(-\infty,0]\cap [E_0-\frac 14 c_{\#}S_{\min}, E_0+\frac 14 c_{\#}
S_{\min}]\\
\text{is\ one\ of\ the}\ \ E_k\ \ (k_{\min}\le k\le k_{\max})\ .
\endmultline\tag 10
$$
\noindent Moreover, either
$$
\Phi(E)=-\frac \pi 2+\phi(E)+\frac 1{48}\psi(E)+\phi_{\roman{error}}(E)\ \
\text{with}\tag 11
$$
$$\split
\Big|\bigl(\frac{d}{dE}\bigr)^\beta \phi_{\roman{error}}(E)\Big|\le
C_{\#}^\beta\Lambda^{-1}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{\frac 12+\beta}(x)B^2(x)}\ \ \text{for\ all}\ \ \beta \ge 0\ ,\\
\text{and\ all}\ \ E\in [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]\endsplit
\tag 12
$$
\noindent or else
$$
\Phi(E)=+\frac \pi 2+\phi(E)+\frac{1}{48}\psi(E)+\phi_{\roman{error}}
(E)\ \ \text{with}\ \ \phi_{\roman{error}}\ \text{satisfying\ (12)}\ .
\tag 13
$$\enddemo

We can assume (11) holds instead of (13).  For, if $\Phi$ satisfies (13),
then in place of $\Phi$, $E_k$, $k_{\min}$, $k_{\max}$ we use
$\tilde\Phi(E)=\Phi(E)-\pi$, $\tilde E_k=E_{k+1}$, $\tilde k_{\min}=k_{\min}-1$,
$\tilde k_{\max}=k_{\max}-1$, and we see that
$\tilde\Phi$, $\tilde E_k$, $\tilde k_{\min}$, $\tilde k_{\max}$ satisfy
(6)$\ldots$(12).

Next we estimate the derivative of the phase $\Phi$.  Lemma 1 preceding
the WKB Eigenvalue Theorem gives
$$\multline
\Big|\frac{d}{dE}\psi(E)\Big|\le C_{\#}
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{3/2}(x)B^2(x)}\\
\le C_{\#}\bigl(\operatornamewithlimits{inf}_{
x\in [x_{\text{left}},x_{\text{rt}}]}S(x)B^2(x)\bigr)^{-1}
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}
\le C_{\#}\Lambda^{-2}\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{1/2}(x)}\ \ \text{and}\endmultline\tag 14
$$
$$
c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}
{S^{1/2}(x)}<\frac{d}{dE}\phi(E)<C_{\#}\int_{x_{\text{left}}}
^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}\tag 15
$$
\noindent for $E\in [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]$ .
\noindent  From (12) we get
$$
\Big|\frac{d}{dE}\phi_{\text{error}}(E)\Big|\le C_{\#}\Lambda^{-1}
\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{3/2}(x)B^2(x)}
\le C_{\#}\Lambda^{-3}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{1/2}(x)}\tag 16
$$
\noindent for 
$E\in [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]$,
where the last inequality follows as in (14).  Putting (14), (15),
(16) into (11), we find that
$$\multline
c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx}{S^{1/2}(x)}
<\frac{d}{dE}\Phi(E)<C_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{1/2}(x)}\\
\text{for}\ E\in [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]\ .\endmultline
\tag 17
$$
\noindent In particular, $\Phi$ is strictly increasing on $[E_0-c_{\#}
S_{\min}, E_0+c_{\#}S_{\min}]$. Moreover if $J\subset (-\infty,0]\cap
[E_0-\frac{c_{\#}}{4}S_{\min},E_0+\frac{c_{\#}}{4}S_{\min}]$ is an
interval of length
$\ge c_{\#}^\prime S_{\min}$, then the image $\Phi(J)$ is
an interval with length at least
$$c_{\#}^\prime S_{\min}\cdot c_{\#}\int_{x_{\text{left}}}^{x_{\text{rt}}}
\frac{dx}{S^{1/2}(x)}\ge
c_{\#}^{\prime\prime}\int_{[x_{\text{left}},x_{\text{rt}}]\cap
[\tilde x-c_{\#}B(\tilde x),\tilde x+c_{\#}B(\tilde x)]}
S_{\min}\frac{dx}{S^{1/2}(x)}
$$
\noindent for any $\tilde x\in [x_{\text{left}},x_{\text{rt}}]$.  Picking
$\tilde x$ so that $S(\tilde x)\sim S_{\min}\equiv \operatornamewithlimits
{\inf}_{x\in [x_{\text{left}},x_{\text{rt}}]}S(x)$, we conclude that
$$
\text{length}\ \Phi(J)\ge c_{\#}S^{1/2}(\tilde x)B(\tilde x)\ge
 c_{\#}^{\prime\prime}\Lambda\ .
$$
\noindent Therefore, $\Phi(J)$ contains at least $c_{\#}\Lambda$
distinct intervals of the form
$$
I_k=[\pi k-C_{\#}\Lambda^{-N^{\prime\prime}}, \pi k+C_{\#}\Lambda^{-N^{\prime
\prime}}]\quad (k\in \Bbb Z)\ .
$$
\noindent However, suppose $I_k\subset \Phi(J)$.  Then
$$
\pi k\in I_k\subset \Phi(J)\subset \Phi((-\infty,0]\cap [E_0-\frac{c_{\#}}{4}
S_{\min}, E_0+\frac{c_{\#}}{4}S_{\min}])\ ,
$$
\noindent so that $k_{\min}\le k\le k_{\max}$ by (6) and
$$
\Phi(E_k)\in I_k\subset \Phi(J)\quad \text{by (9)}\ .\tag 18
$$
\noindent Since $\Phi$ is strictly increasing on $[E_0-c_{\#}S_{\min},
E_0+c_{\#}S_{\min}]$, and since
$J\subset [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]$ while 
$E_k\in [E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]$ by (9), it
follows from (18) that $E_k$ belongs to $J$.  Consequently, $J$
contains at least $c_{\#}\Lambda$ of the $E_k$ $(k_{\min}\le
k\le k_{\max})$.  In view of (7), all but at most one of these
$E_k$ are eigenvalues of $H$.  Thus we have proven the following
fact.
$$\multline
\text{If}\ \ J\subset (-\infty,0]\cap [E_0-\frac 14 c_{\#}S_{\min},E_0+\frac 14
c_{\#}S_{\min}]\ \ \text{is\ an\ interval\ of\ length\ at\ least}\\ 
c_{\#}^\prime S_{\min}\ ,\ \ 
\text{then\ at\ least}\ \ c_{\#}^{\prime\prime}\Lambda\ \
\text{distinct\ eigenvalues\ of}\ \ H\ \ \text{lie\ in}\ \ J\ .\endmultline
\tag 19
$$
\noindent Next we make a trivial observation that will be used
repeatedly.
$$\multline
\text{Suppose}\ \ k\in \Bbb Z\ , \xi \in \Bbb R\ \ \text{and}\ \
0<\tau<\frac {1}{10} \ .\quad \text{If}\ \ k+1\ge \xi-\tau\ \ \text{and}\ \
k\le \xi+\tau\ ,\\
\text{then\ for\ some}\ \  \omega \ \ \text{with}\ \ |\omega|\le 2\tau\ \
\text{we\ have}\ \ k=\ (\text{greatest\ integer}\ \le \xi+\omega).\endmultline
\tag 20
$$

\noindent To see this, assume first that $k+1>\xi+\tau$.  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
$$
k_1<k_2<k_3\ , \ k_1\in \Cal K\ \text{and}\ k_3\in \Cal K\
\text{imply}\ k_2\in \Cal K\ .\tag 22
$$
\noindent To see this, note that the $E_k$ increase with $k$, as follows
from (9) and the fact that $\Phi$ is strictly increasing on $[E_0-c_{\#}
S_{\min},E_0+c_{\#}S_{\min}]$.  Suppose $k_1<k_2<k_3$ with
$k_1,k_3\in \Cal K$.  Then
$$
k_{\min}\le k_1< k_2<k_3\le k_{\max}\ ,\ \text{and}\tag 24
$$
$$
E_{k_1},E_{k_3}\in [E_{\ell o},E_{hi}]\ .\tag 25
$$

\noindent From (25) and the fact that the $E_k$ increase, we see that
$E_{k_2}\in [E_{\ell o},E_{hi}]$.  From (24) and (7), we see
that $E_{k_2}$ is an eigenvalue of $H$.  Hence $k_2\in \Cal K$, completing
the proof of (22).

Since $\Cal K$ is a non--empty finite set of integers satisfying (22), it
follows that
$$
\Cal K=\{k \in \Bbb Z\mid k_{\ell o}\le k\le k_{hi}\}\tag 26
$$
\noindent for integers $k_{\ell o}\le k_{hi}$.  Recalling that $\Cal K$
contains at least $c_{\#}\Lambda$ integers, we conclude from (26) that
$$
k_{hi}>k_{\ell o}+c_{\#}\Lambda\ .\tag 27
$$
\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}  $k_{hi}<k_{\max}$ and $E_{k_{hi}+1}\in [E_{\ell o},
E_{hi}]$ but $E_{k_{hi}+1}$ isn't an eigenvalue of $H$.\enddemo
\smallskip
\demo{Case C}  $k_{hi}<k_{\max}$ and $E_{k_{hi}+1}\notin [E_{\ell o},
E_{hi}]$.\enddemo
\smallskip

Let us check that we are always in one of these three cases.  If we're
not in Case A, then (28) gives $k_{\min}\le k_{hi}+1\le k_{\max}$, yet
$k_{hi}+1\notin\Cal K$ by (26).  Hence, (21) shows that
$E_{k_{hi}+1}\notin [E_{\ell o},E_{hi}]$ or else
$E_{k_{hi}+1}$ isn't an eigenvalue of $H$.  Consequently, we're either
in Case B or Case C.  This shows we are always in one of the three 
cases $A$, $B$, $C$.

Next we prove (30), (31) in case A.  Thus we assume $k_{hi}=k_{\max}$.
>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 $k_{hi}<k_{max}$
and
$$
E_{k_{hi}+1}\in [E_{\ell o},E_{hi}]\ ,\tag 39
$$
\noindent but $E_{k_{hi}+1}$ isn't an eigenvalue of $H$.  From (7) and (8)
we see that $k_{hi}+1=k_{\max}$ and $E_{k_{hi}+1}=0$.  Then from (39) we
get $0\in [E_{\ell o},E_{hi}]\subset (-\infty,0]$, so that $E_{hi}=0$.
Since $E_{hi}=E_{k_{hi}+1}=0$, (9) shows that $|\Phi(E_{hi})-\pi
(k_{hi}+1)|\le C_{\#}\Lambda^{-N^{\prime\prime}}$.  In view of (11)
and (29), this implies $|(\frac 1\pi \phi(E_{hi})+\frac{1}{48\pi}
\psi(E_{hi})-\frac 12)-(k_{hi}+1)|\le C_{\#}^\prime\Lambda^{-2}$.
Therefore, $k_{hi}=(\, \text{greatest\ integer}\ \le \frac 1\pi
\phi(E_{hi})+\frac 1{48\pi}\psi(E_{hi})-\frac 12-2C_{\#}^\prime
\Lambda^{-2})$, which immediately yields (30), (31).  So we have proven
(30), (31) in Case B.

Next we prove (30), (31) in Case C.  Thus we assume $k_{hi}<k_{max}$ and
$$
E_{k_{hi}+1}\notin [E_{\ell o},E_{hi}] \ .\tag 40
$$
\noindent From (21) and (26) we get
$$
E_{k_{hi}}\in [E_{\ell o},E_{hi}]\ .\tag 41
$$

\noindent Since we noted that $E_k$ increases with $k$, (40) and (41)
imply $E_{k_{hi}+1}>E_{hi}$.  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
$$
k_{\min}<k_{\ell o}\ .\tag 50
$$
\noindent In fact, applying (19) to $J=[E_0-\frac 14 c_{\#}
S_{\min}, E_0-2c_{\#}^1S_{\min}]$ (and taking $c_{\#}^1<
\frac{c_{\#}}{20}$ so that $\text{length}\ J>c_{\#}^\prime S_{\min}$),
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 $E_{\tilde k}<E_{k_{\ell o}}$.
Since the $E_k$ increase with $k$, we obtain $\tilde k<k_{\ell o}$,
and therefore $k_{\min}\le \tilde k<k_{\ell o}$, proving (50).

In view of (50), $k_{\min}\le k_{\ell o}-1\le k_{\max}$, so
$E_{k_{\ell o}-1}$ is well--defined.  Also, $k_{\ell o}-1<k_{\ell o}\le
k_{\max}$, so $E_{k_{\ell o}-1}$ is an eigenvalue of $H$, by (7).  If
$E_{k_{\ell o}-1}$ belonged to $[E_{\ell o},E_{hi}]$, then (21) would
give $k_{\ell o}-1\in \Cal K$, contradicting (26).  So
$E_{k_{\ell o}-1}$ doesn't belong to $[E_{\ell o},E_{hi}]$.  On the other
hand, $E_{k_{\ell o}}$ does belong to $[E_{\ell o},E_{hi}]$, as we noted
above.  Since the $E_k$ increase with $k$, it follows that $E_{k_{\ell o}-1}<
E_{\ell o}\le E_{k_{\ell o}}$.  Hence $\Phi(E_{k_{\ell o}-1})<\Phi
(E_{\ell o})\le \Phi(E_{k_{\ell o}})$ since $\Phi$ is increasing on
$[E_0-c_{\#}S_{\min}, E_0+c_{\#}S_{\min}]$.  Applying (9), we find that
$$
\pi(k_{\ell o}-1)-C_{\#}\Lambda^{-N^{\prime\prime}}<\Phi(E_{\ell o})\le
\pi k_{\ell o}+C_{\#}\Lambda^{-N^{\prime\prime}}\ .
$$
\noindent Together with (11) and (29), this shows that
$$
k_{\ell o}-1-C_{\#}\Lambda^{-2} <\frac 1\pi \phi(E_{\ell o})+\frac 1{48\pi}
\psi(E_{\ell o})-\frac 12\le k_{\ell o}+C_{\#}\Lambda^{-2}\ .\tag 51
$$
\noindent From (51) and (47), we deduce the desired inequalities
(48), (49).

Now it is trivial to complete the proof of the Reformulated Eigenvalue
Theorem \vfill\eject
\noindent by checking (4) and (5).  In fact, (4) is immediate from 
(44), (48) and the estimates (45), (49).  Finally, (5) is contained in (46).
The proof is complete.$\qquad\blacksquare$

\vglue 1pc
\demo{Remark}  When we picked $c_{\#}^1$ in the proof of the Reformulated
Eigenvalue Theorem, we needed only to make sure that $c_{\#}^1$ is small
enough.  Hence we can take $c_{\#}^1<\frac{1}{33}c_{\#}$, where $c_{\#}$
is the small constant appearing in the statements of the WKB Theorems
(and Lemma 1 preceding them).  This ensures that we have precise information
on the phases $\phi(E)$, $\psi(E)$ for $E\in [E_0-33c_{\#}^1S_{\min},
E_0+33 c_{\#}^1S_{\min}]$, and on eigenfunctions corresponding to
eigenvalues in $[E_{\ell o}, E_{hi}]$.\enddemo

The Reformulated Eigenvalue Theorem identifies conveniently those
eigenvalues of $H=-\frac{d^2}{dx^2}+V(x)$ that lie in an interval
$[E_0-c_{\#}^1S_{\min},E_0+c_{\#}^1S_{\min}]\cap (-\infty,0]$.

This makes it trivial to identify those eigenvalues of $H$ that lie in a
union of such intervals.  Again we provide details for the reader's
convenience.

\vglue 1pc
\proclaim{Corollary to the Reformulated Eigenvalue Theorem} Let
$[{E_{\ell o},E_{hi}}]\hfill\break \subset (-\infty,0]$.  Assume for each 
$E_0\in [E_{\ell o},E_{hi}]$ that the hypotheses of the WKB Eigenvalue Theorem
are satisfied, with $E_\infty=0$, 
and with constants $\varepsilon$, $K$, $N$, $C$, $c$, $c_1$, $c_2$, $C_\alpha$
independent of $E_0$.  Let $S_{\min}(E_0)$, $\Lambda(E_0)$ denote the
quantities called $S_{\min}$, $\Lambda$ in the statement of the WKB
Eigenvalue Theorem, corresponding to the given $E_0$.  Define
$\Lambda_{\min}={\roman{inf}}_{E_0\in [E_{\ell o},E_{hi}]}\Lambda(E_0)$.
Assume that
$\phi(E_{hi})-\phi(E_{\ell o})\ge 100$, with $\phi(E)\equiv
\int_{I_{\text{BVP}}}(E-V(x))_+^{1/2}dx$ for $E\in [E_{\ell o},E_{hi}]$.
Then the eigenvalues of $H$ that lie in $[E_{\ell o},E_{hi}]$
may be written in the form $(E_k)_{k_{\ell o}\le k\le k_{hi}}$, so that
the following properties hold.
$$
k_{\ell o}<k_{hi}\tag 52
$$
$$
|\phi(E_k)-\pi(k+1/2)|\le C_{\#}\Lambda_{\min}^{-1}\quad{\roman{for}}\quad
k_{\ell o}\le k\le k_{hi}\ .\tag 53
$$
$$
\{k\in \Bbb Z\mid k_{\ell o}\le k\le k_{hi}\}=\Bbb Z\cap [\frac 1\pi
\phi(E_{\ell o})-\frac 12 +\omega_{\ell o},\frac 1\pi \phi(E_{hi})-\frac 12
+\omega_{hi}]\ ,\tag 54
$$
\noindent with
$$
|\omega_{\ell o}|, |\omega_{hi}|\le C_{\#}\Lambda_{\min}^{-1}\, \quad{\roman{and}}
\tag 55
$$
$$\text{If}\quad \min_{k\in \Bbb Z}|\frac 1\pi \phi(E_{hi})-\frac 12 -k|\ge
C_{\#}\Lambda_{\min}^{-1}\ , \quad\text{then}\quad \omega_{hi}=0\ .
\tag 56
$$

\noindent Here $C_{\#}$ depends only on the constants $\varepsilon$, $K$,
$N$, $c$, $c_1$, $c_2$, $C$, $C_\alpha$ in the hypotheses of the WKB
Theorems.\endproclaim

\vglue 1pc

\demo{Proof}  Let $c_{\#}$, $C_{\#}$ etc. denote constants determined by
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$.  We begin
with a few remarks on the phase $\phi$.  Recall that in the WKB Theorems
we defined $\phi(E)=\int_I
(E-V(x))_+^{1/2}\, dx$ for a certain subinterval $I\subset I_{\text{BVP}}$.
\enddemo\medskip

If $E_0\in [E_{\ell o},E_{hi}]$, then the hypotheses of the WKB Theorems
hold with $E_\infty=0$.  It follows that $\{x\in I_{\text{BVP}}\mid
V(x)<E\}\subset I$ for $E\in [E_0-c_{\#}^1S_{\min}(E_0)$,
$E_0+c_{\#}^1S_{\min}(E_0)]\cap (-\infty,0]\equiv [E_{\ell o}(E_0),
E_{hi}(E_0)]$.  Hence
$$\split
\phi(E)=\int\limits_{I_{\text{BVP}}}(E-V(x))_+^{1/2}\, dx\quad\text{for}\quad
E_0\in [E_{\ell o},E_{hi}]\quad\text{and}\\
E\in [E_{\ell o}(E_0),E_{hi}(E_0)]\ .\endsplit\tag 57
$$

\noindent We recall an important lower bound for $\phi^\prime
(E)=\frac 12 \int_{I}(E-V(x))_+^{-1/2}\ dx$, when $E$ is as in (57).
In fact, we can find an interval of the form $J=[\tilde x-c_{\#}
B(\tilde x), \tilde x+c_{\#}B(\tilde x)]$ contained in
$I\cap \{c_{\#}S(\tilde x)<E-V(x)<C_{\#}S(\tilde x)\}$ and with
$S(\tilde x)\sim S_{\min}(E_0)$.  Restricting the region of integration
to $J$ in the formula for $\phi^\prime(E)$, we get
$\phi^\prime(E)\ge c_{\#}(S(\tilde x))^{-1/2}B(\tilde x)=c_{\#}
[(S(\tilde x))^{1/2} B(\tilde x)](S(\tilde x))^{-1}=c_{\#}
\lambda(\tilde x)(S(\tilde x))^{-1}\ge c_{\#}(\Lambda(E_0))(S_{\min}
(E_0))^{-1}$.  This holds for $E\in [E_{\ell o}(E_0),E_{hi}(E_0)]$.  Therefore,
$$\align
\phi(E_0)&\ge \phi(E_{\ell o}(E_0))+\Bigl(\frac{c_{\#}\Lambda(E_0)}
{S_{\min}(E_0)}\Bigr)\cdot (E_0-E_{\ell o}(E_0))\\
&=\phi(E_{\ell o}(E_0))+\frac{c_{\#}\Lambda(E_0)}{S_{\min}(E_0)}
\cdot c_{\#}^1S_{\min}(E_0)\ge \phi(E_{\ell o}(E_0))+c_{\#}^\prime
\Lambda_{\min}\tag 58\endalign
$$
\noindent for $E_0\in [E_{\ell o},E_{hi}]$.  This concludes our small
discussion of $\phi$.

For each $E_0\in [E_{\ell o},E_{hi}]$, we apply the Reformulated Eigenvalue
Theorem, to obtain the following results.
$$
\split
\text{The\ eigenvalues\ of}\ H\ \text{that\ lie\ in}\ [E_{\ell o}(E_0),
E_{hi}(E_0)]\ \text{may\ be\ written\ as}\\
\{\Cal E(k,E_0)\colon k_{\ell o}(E_0)\le k\le k_{hi}(E_0)\}\endsplit\tag 59
$$
$$
k_{\ell o}(E_0)<k_{hi}(E_0)\tag 60
$$
$$
|\phi(\Cal E(k,E_0))+\frac{1}{48}\psi(\Cal E(k,E_0))-\pi(k+1/2)|\le
C_{\#}(\Lambda(E_0))^{-2}\tag 61
$$
$$
\{k\in \Bbb Z\colon k_{\ell o}(E_0)\le k\le k_{hi}(E_0)\}=\Bbb Z\cap
[a(E_0),b(E_0)]\ ,\quad\text{where}\tag 62
$$
$$
a(E_0)=\frac 1\pi \phi(E_{\ell o}(E_0))+\frac{1}{48\pi}\psi(E_{\ell o}
(E_0))-\frac 12 +\omega_{\ell o}(E_0)\ ,\tag 63
$$
$$
b(E_0)=\frac 1\pi \phi(E_{hi}(E_0))+\frac{1}{48\pi}\psi(E_{hi}(E_0))-\frac 12
+\omega_{hi}(E_0)\ , \quad\text{and}\ \tag 64
$$
$$
|\omega_{\ell o}(E_0)|, |\omega_{hi}(E_0)|\le C_{\#}(\Lambda(E_0))^{-2}\ .\tag 65
$$

\noindent Lemma 1 immediately preceding the WKB Eigenvalue Theorem gives
$|\psi(E)|\le C_{\#}(\Lambda(E_0))^{-1}$ for $E\in [E_{\ell o}(E_0),
E_{hi}(E_0)]$, $E_0\in [E_{\ell o},E_{hi}]$.  Applying this to
$E=E_{\ell o}(E_0)$, $E=E_{hi}(E_0)$ and $E=\Cal E(k,E_0)$, and setting
$$
\tilde\omega_{\ell o}(E_0)=\frac{1}{48\pi}\psi(E_{\ell o}(E_0))+
\omega_{\ell o}(E_0)\ , \quad\tilde \omega_{hi}(E_0)=\frac {1}{48\pi}
\psi(E_{hi}(E_0))+\omega_{hi}(E_0)\ ,
$$
\noindent we obtain the following from (61)$\ldots$(65).
$$
|\phi(\Cal E(k,E_0))-\pi(k+1/2)|\le C_{\#}(\Lambda(E_0))^{-1}\quad\text{for}
\quad k_{\ell o}(E_0)\le k\le k_{hi}(E_0)\ ;\tag 66
$$
$$
\{k \in \Bbb Z\colon k_{\ell o}(E_0)\le k\le k_{hi}(E_0)\}=
\Bbb Z\cap [a(E_0),b(E_0)]\ ;
\tag 67
$$
$$
a(E_0)=\frac 1\pi \phi(E_{\ell o}(E_0))-\frac 12 +\tilde\omega_{\ell o}(E_0)\ ;
\tag 68
$$
$$
b(E_0)=\frac 1\pi \phi(E_{hi}(E_0))-\frac 12+\tilde\omega_{hi}(E_0)\ ;
\tag 69
$$
$$
|\tilde\omega_{\ell o}(E_0)|, |\tilde\omega_{hi}(E_0)|\le C_{\#}(\Lambda(E_0))
^{-1}\ .\tag 70
$$

\noindent Here (66)$\ldots$(70) hold for all $E_0\in [E_{\ell o},E_{hi}]$.

We shall prove the following for a constant $C_{\#}^\prime$ larger than
$C_{\#}$ in (66)$\ldots$(70).  

\bigskip
\noindent{\it Claim A\/}:

Suppose $\kbar\in \Bbb Z$ satisfies
$$
\frac 1\pi \phi(E_{\ell o})-\frac 12 +C_{\#}^\prime\Lambda_{\min}^{-1}\le
\kbar \le \frac 1\pi \phi(E_{hi})-\frac 12-C_{\#}^\prime\Lambda_{\min}^{-1}\ .
\tag 71
$$

\noindent Then there is an eigenvalue $\overline E$ of $H$ that lies in
$[E_{\ell o},E_{hi}]$ and satisfies
$$
|\phi(\overline E)-\pi(\kbar+1/2)|\le C_{\#}\Lambda_{\min}^{-1}\ .\tag 72
$$
\smallskip

\noindent To see this, note that $\frac 1\pi\phi(E_{\ell o})-\frac 12
\le \kbar+C_{\#}^\prime\Lambda_{\min}^{-1}\le \frac 1\pi \phi(E_{hi})-\frac 12$,
so we can find an $E_0\in [E_{\ell o},E_{hi}]$ with $\frac 1\pi \phi(E_0)-
\frac 12=\kbar+C_{\#}^\prime\Lambda_{\min}^{-1}$.  We will check that $a(E_0)\le
\kbar \le b(E_0)$.
\smallskip

\noindent In fact, $\kbar+C_{\#}^\prime\Lambda_{\min}^{-1}=\frac 1\pi
\phi(E_0)-\frac 12\le \frac 1\pi\phi(E_{hi}(E_0))-\frac 12=b(E_0)
-\tilde\omega_{hi}(E_0)\le b(E_0)+C_{\#}\Lambda_{\min}^{-1}$ by (69), (70).
Hence $\kbar\le b(E_0)$.  On the other hand, (58) gives
$\frac 1\pi \phi(E_{\ell o}(E_0))-\frac 12\le \frac 1\pi \phi(E_0)-\frac 12
-c_{\#}\Lambda_{\min}=\kbar +C_{\#}^\prime\Lambda_{\min}^{-1}-c_{\#}\Lambda_{\min}
\le \kbar -\frac 12 c_{\#}\Lambda_{\min}$.  This, together with (68), (70),
gives $a(E_0)\le \kbar$.  Thus $a(E_0)\le \kbar \le b(E_0)$, as asserted.
Hence $k_{\ell o}(E_0)\le \kbar \le k_{hi}(E_0)$ by (67), so that
$\overline E=\Cal E(\kbar,E_0)$ is an eigenvalue of $H$ lying in
$[E_{\ell o}(E_0),E_{hi}(E_0)]$ by (59).  We have $|\phi(\overline E)
-\pi(\kbar+1/2)|\le C_{\#}\Lambda_{\min}^{-1}$ by (66), which proves (72).
It remains to check that $\overline E\in [E_{\ell o},E_{hi}]$.  Since
$\overline E\in [E_{\ell o}(E_0),E_{hi}(E_0)]$, we have $\phi(\overline E)
=\int_{I_{\text{BVP}}}(\overline E-V(x))_+^{1/2}\ dx$ by (57).  Equation (57)
yields also $\phi(E_{\ell o})=\int_{I_{\text{BVP}}}(E_{\ell o}-V(x))_+^{1/2}\ dx$
and $\phi(E_{hi})=\int_{I_{\text{BVP}}}(E_{hi}-V(x))_+^{1/2}\ dx$.

>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 $E^\prime<E$.  If $E^\prime$, $E$ both belonged to
$[E_{\ell o}(E),E_{hi}(E)]$, then we would get a contradiction from
(59), (66).  Since $E\in [E_{\ell o}(E),E_{hi}(E)]$, we must have
$E^\prime\not\in [E_{\ell o}(E),E_{hi}(E)]$.  Since also
$E^\prime<E\in [E_{\ell o}(E),E_{hi}(E)]$, we must have
$$
E^\prime<E_{\ell o}(E)\le E\ .\tag 73
$$
\noindent From (73) we see that $E^\prime$, $E_{\ell o}(E)$, $E$ all belong to
$[E_{\ell o},E_{hi}]$, so that (73) implies
$$
\phi(E^\prime)\le \phi(E_{\ell o}(E))\ .
$$
\noindent Therefore, $\phi(E^\prime)\le \phi(E)-c_{\#}\Lambda_{\min}$ by
(58), which contradicts our original assumption $|\phi(E)-\phi(E^\prime)|
\le \frac{1}{100}$.  This completes the proof of {\it CLAIM B}.

Next we prove

\noindent{\it CLAIM C\/}:  Let $E\in [E_{\ell o},E_{hi}]$ be an eigenvalue of $H$.
Then
$$
|\phi(E)-\pi(k+1/2)|\le C_{\#}\Lambda_{\min}^{-1}\tag 74
$$
\noindent for an integer $k$ which satisfies
$$
\frac 1\pi \phi(E_{\ell o})-\frac 12-C_{\#}\Lambda_{\min}^{-1}
\le k\le \frac 1\pi \phi(E_{hi})-\frac 12 +C_{\#}\Lambda_{\min}^{-1}\ .
\tag 75
$$

\noindent In fact, since $E\in [E_{\ell o}(E),E_{hi}(E)]$, (59) gives
$E=\Cal E(k,E)$ for an integer $k$ with $k_{\ell o}(E)\le k\le k_{hi}(E)$.
Then from (66) we get (74).  To prove (75), we recall that $\phi$ is
increasing on $[E_{\ell o},E_{hi}]$, so that $\frac 1\pi \phi(E_{\ell o})
-\frac 12\le \frac 1\pi \phi(E)-\frac 12\le \frac 1\pi\phi(E_{hi})-\frac 12$.
These inequalities and (74) imply (75).\medskip

Let $\Cal K$ be the set of all integers $k$ for which there is an eigenvalue
$E\in [E_{\ell o},E_{hi}]$ satisfying
$$
|\phi(E)-\pi(k+1/2)|\le C_{\#}\Lambda_{\min}^{-1}\ .\tag 76
$$

\noindent {\it Claim B\/} shows that for each $k\in \Cal K$ there is only one
eigenvalue $E\in [E_{\ell o},E_{hi}]$ satisfying (76); we call that eigenvalue
$E_k$.  Thus, $E_k$ is well-defined and belongs to $[E_{\ell o},E_{hi}]$
for $k\in \Cal K$.  Moreover, $E_k$ is an eigenvalue of $H$ and satisfies
$$
|\phi(E_k)-\pi(k+1/2)|\le C_{\#}\Lambda_{\min}^{-1}\quad\text{for}\quad
k\in \Cal K\ ;\quad\text{and}\tag 77
$$
$$
\split
\text{If}\ k\in \Cal K\ \text{and}\ E\ \text{is\ an\ eigenvalue\ of}\ H\
\text{belonging\ to}\ [E_{\ell o},E_{hi}]\\
\text{and\ satisfying}\ |\phi(E)-\pi(k+1/2)|\le C_{\#}\Lambda_{\min}^{-1},\ 
\text{then}\ E=E_k\ . \endsplit\tag 78
$$
\medskip

{\it CLAIM C\/}, (78) and the definition of $\Cal K$ show that the eigenvalues
of $H$ that belong to $[E_{\ell o},E_{hi}]$ are precisely the
$(E_k)_{k\in \Cal K}$.

Moreover, {\it CLAIMS A\/} and {\it C\/} show that
$$
\Cal K\supset \Bbb Z\cap [\frac 1\pi \phi(E_{\ell o})-\frac 12
+C_{\#}^\prime\Lambda_{\min}^{-1},\frac 1\pi \phi(E_{hi})-\frac 12-
C_{\#}^\prime\Lambda_{\min}^{-1}]\ \text{and}\tag 79
$$
$$
\Cal K\subset \Bbb Z\cap [\frac 1\pi \phi(E_{\ell o})-\frac 12-C_{\#}
\Lambda_{\min}^{-1}, \frac 1\pi \phi(E_{hi})-\frac 12+C_{\#}\Lambda_{\min}^{-1}
]\ .\tag 80
$$

Define $\tilde\omega_{\ell o}$ and $\tilde\omega_{hi}$ as follows.
$$\multline
\tilde\omega_{\ell o}=-C_{\#}\Lambda_{\min}^{-1}\ \text{if\ there\ is\ some}\\
k\in \Cal K\cap [\frac 1\pi \phi(E_{\ell o})-\frac 12-C_{\#}\Lambda_{\min}^{-1}
,\frac 1\pi \phi(E_{\ell o})-\frac 12+C_{\#}^\prime\Lambda_{\min}^{-1}]\ ,
\endmultline\tag 81
$$
$$
\tilde\omega_{\ell o}=+C_{\#}^\prime\Lambda_{\min}^{-1}\quad\text{otherwise}\ .
\tag 82
$$
$$\multline
\tilde\omega_{hi}=+C_{\#}\Lambda_{\min}^{-1}\quad \text{if\ there\ is\ some}\\
k\in \Cal K\ \cap [\frac 1\pi \phi(E_{hi})-
\frac 12-C_{\#}^\prime\Lambda_{\min}^{-1},
\frac 1\pi \phi(E_{hi})-\frac 12 +C_{\#}\Lambda_{\min}^{-1}]\ ,\endmultline\tag 83
$$
$$
\tilde\omega_{hi}=-C_{\#}^\prime\Lambda_{\min}^{-1}\quad\text{otherwise}\ .
\tag 84
$$

Evidently,
$$
|\tilde\omega_{\ell o}|, |\tilde\omega_{hi}|\le C_{\#}^\prime\Lambda_{\min}^{-1}
\ .\tag 85
$$

Let us check that
$$
\Cal K=\Bbb Z\cap [\frac 1\pi \phi(E_{\ell o})-\frac 12+\tilde\omega_{\ell o},
\frac 1\pi \phi(E_{hi})-\frac 12+\tilde\omega_{hi}]\ .\tag 86
$$

\noindent Suppose $\kbar \in \Cal K$.  From (80) we get $\kbar\le
\frac 1\pi \phi(E_{hi})-\frac 12+C_{\#}\Lambda_{\min}^{-1}$.
If $\kbar \ge \frac 1\pi \phi(E_{hi})-\frac 12-C_{\#}^\prime\Lambda_{\min}^{-1}$,
then the definition (83) gives $\tilde\omega_{hi}=C_{\#}\Lambda_{\min}^{-1}$
so that
$$
\kbar \le \frac 1\pi\phi(E_{hi})-\frac 12+\tilde\omega_{hi}\ .\tag 87
$$
\noindent On the other hand, if $\kbar\le \frac 1\pi \phi(E_{hi})-\frac 12
-C_{\#}^\prime\Lambda_{\min}^{-1}$, then (87) follows from the definition
(83), (84) of $\tilde\omega_{hi}$.  Hence (87) holds in all cases.

Similarly, (80) yields $\kbar \ge \frac 1\pi \phi(E_{\ell o})-\frac 12-C_{\#}
\Lambda_{\min}^{-1}$.  If $\kbar \le \frac 1\pi\phi(E_{\ell o})-\frac 12+
C_{\#}^\prime\Lambda_{\min}^{-1}$, then (81) gives $\tilde\omega_{\ell o}=
-C_{\#}\Lambda_{\min}^{-1}$, so that
$$
\kbar \ge \frac 1\pi \phi(E_{\ell o})-\frac 12+\tilde\omega_{\ell o}\ .\tag 88
$$

On the other hand, if $\kbar \ge \frac 1\pi \phi(E_{\ell o})-\frac 12
+C_{\#}^\prime\Lambda_{\min}^{-1}$, then (88) follows from the definition
(81), (82) of $\tilde\omega_{\ell o}$.  Hence (88) holds in all cases. 

>From (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
$$\Cal K=\{k\in \Bbb Z\mid k_{\ell o}\le k\le k_{hi}\}\ , \quad\text{with}
\tag 97
$$
$$
k_{\ell o}<k_{hi}\ .\tag 98
$$

\noindent Since we saw already that the eigenvalues of $H$ belonging to
$[E_{\ell o},E_{hi}]$ are precisely the $(E_k)_{k\in \Cal K}$, we now
have from (97) that these eigenvalues are $(E_k)_{k_{\ell o}\le k\le
k_{hi}}$.  The conclusions of the present Corollary are now obvious.  In
fact, (52) is the same as (98); (53) follows from (77) and (97); (54)
follows from (94) and (97); and (55), (56) with an enlarged constant
$C_{\#}$ are contained in (95), (96).  The proof of the Corollary is complete.

\line{\hss$\blacksquare$}


\vfill\eject


\heading{\bf{APPROXIMATING SUMS BY INTEGRALS}}\endheading
\medskip


In this section we study the error that arises when we approximate
a Riemann sum by an integral.  Our formulas involve the following
elementary functions.
$$\gather
\chp(x)=k-x-\frac 12\quad \text{for}\quad  k \quad\text{the\ smallest\
integer}\ \ge x\ ;\\
\chm(x)=x-k-\frac 12\qquad \text{for}\quad k\quad\text{the largest\
integer}\ \le x\ ;\\
\tilde\chi(x)=|x-k-\frac 12|^2-\frac{1}{12}\quad \text{for\ an\ integer}\quad
k\quad \text{that\ minimizes}\ |x-k-\frac 12|\ .
\endgather
$$

\vglue 1pc
\proclaim{Lemma 1}  Assume $|(\frac{d}{dx})^mf(x)|\le C\, R^{-m}$ for
$0\le m\le 4$ with $R\ge 1$.  If $a\le b$, then
$$\split
\sum\limits_{k\in \Bbb Z\cap [a,b]}f(k)=\int\limits_a^bf(t)\, dt
-f(b)\chm(b)-f(a)\chp(a)+\frac 12 f^\prime(b)
\tilde\chi(b)\\
-\frac 12 f^\prime(a)\tilde\chi(a)+\ \text{Error}\, ,\endsplit
$$
\noindent with $|\roman {Error}|\le C^\prime R^{-2}+C^\prime R^{-4}
|b-a|$ and $C^\prime$ determined entirely by $C$.\endproclaim

\vglue 1pc
\demo{Proof}  First assume $\Bbb Z\cap [a,b]\ne \emptyset$.  Thus for
suitable $m_0,m_1\in \Bbb Z$ we have $m_0-1<a\le m_0\le m_1\le b<m_1+1$.  Then
$\chp(a)=m_0-a-1/2$, $\chm(b)=b-m_1-1/2$,
$\tilde\chi(a)=|a-m_0+\frac 12|^2-\frac{1}{12}$,
$\tilde\chi(b)=|b-m_1-\frac 12|^2-\frac{1}{12}$.  For $m_0\le k\le m_1$
we have the following:
$$\gather
\int\limits_{k-1/2}^{k+1/2}f(t)\,dt=f(k)+\frac{1}{24}f^{\prime\prime}(k)+\ \text
{Error}_k, \,\,\, |\text{Error}_k|\le C^\prime R^{-4}\, ,\\
\int\limits_{k-1/2}^{k+1/2}f^{\prime\prime}
(t)\,dt=f^{\prime\prime}(k)+\ \text{Error}_k
^{\prime\prime},\ |\text{Error}_k^{\prime\prime}|\le C^\prime R^{-4}\, .
\endgather
$$
\noindent These formulas follow by Taylor--expanding $f$ to order $3$ with
remainder, and $f^{\prime\prime}$ to order $1$ with remainder.  Summing
over $k$ from $m_0$ to $m_1$, we get the following equations:
$$\gather
\int\limits_{m_0-1/2}^{m_1+1/2}\, f(t)dt=\sum\limits_{m_0\le k\le m_1}
f(k)+\frac{1}{24}\sum\limits_{m_0\le k\le m_1}f^{\prime\prime}(k)+\ 
\text{Error}\, ,\\
f^\prime(m_1+\frac 12)-f^\prime(m_0-1/2)=\int\limits_{m_0-1/2}^{m_1+1/2}
f^{\prime\prime}(t)dt=\sum\limits_{m_0\le k\le m_1} f^{\prime\prime}(k)+\
\text{Error}^{\prime\prime}\endgather
$$
\noindent with $|\text{Error}|$, $|\text{Error}^{\prime\prime}|\le
C^\prime R^{-4}(b-a+1)$.  Therefore
$$
\sum\limits_{m_0\le k\le m_1} f(k)=\int\limits_{m_0-1/2}^{m_1+1/2}
f(t)dt-\frac{1}{24}\Bigl(f^\prime(m_1+\frac 12)-f^\prime(m_0-\frac 12)\Bigr)
+\ \text{Error}\, ,
$$
\noindent with $|\text{Error}|\le C^\prime R^{-4}(b-a+1)$.\enddemo

We simplify slightly, by noting that $|f^\prime(m_1+1/2)-f^\prime(b)|$,
$|f^\prime(m_0-1/2)-f^\prime(a)|$ are dominated by $C^\prime R^{-2}$.  Thus,
$$\split
\sum\limits_{m_0\le k\le m_1}f(k)=\int\limits_{m_0-1/2}^{m_1+1/2}f(t)dt
-\frac{1}{24}(f^\prime(b)-f^\prime(a))+\ \text{Error}\, ,\\
|\text{Error}|\le C^\prime R^{-2}+C^\prime R^{-4}(b-a)\ .\endsplit\tag 1
$$

Next, Taylor--expanding $f(t)$ about $t=b$ to order $1$ with remainder,
we get
$$
\split
\int\limits_{m_1+1/2}^{b}f(t)dt=f(b)\cdot(b-m_1-1/2)-\frac 12
f^\prime(b)\cdot (b-m_1-1/2)^2+\ \text{Error}\ ,\\
|\text{Error}|\le C^\prime R^{-2}\ .\endsplit
$$
\noindent Similarly,
$$\split
\int\limits_a^{m_0-1/2}f(t)dt=f(a)\cdot (m_0-a-1/2)+\frac 12
f^\prime (a)\cdot (m_0-a-1/2)^2+\ \text{Error}\, ,\\
|\text{Error}|\le C^\prime R^{-2}\ .\endsplit
$$
\noindent Combining these equations with (1), we get
$$\multline
\sum\limits_{m_0\le k\le m_1}f(k)=\int\limits_a^bf(t)dt-f(b)\cdot
(b-m_1-1/2)-f(a)\cdot (m_0-1/2-a)+\frac 12 f^\prime(b)\\
\cdot \{(b-m_1-1/2)^2-\frac{1}{12}\}- \frac 12f^\prime(a)\cdot
\{(m_0-a-1/2)^2-\frac {1}{12}\}+\ \text{Error}\ ,\\
|\text{Error}|\le C^\prime R^{-2}+C^\prime R^{-4}(b-a)\ .\endmultline
$$
\noindent That is the conclusion of Lemma 1.  It remains to check the
case $[a,b]\cap \Bbb Z= \emptyset$.  Thus $m<a\le b<m+1$ for an
integer $m$.  We then have $\chp(a)=m+\frac 12-a$, $\chm(b)=b-m-\frac 12$,
$\tilde\chi(a)=(a-m-\frac 12)^2-\frac{1}{12}$, $\tilde\chi(b)=
(b-m-\frac 12)^2-\frac{1}{12}$, so our lemma asserts that
$$\multline
0=\int\limits_a^bf(t)dt-f(b)(b-m-\frac 12)-f(a)(m+\frac 12-a)\\
+\frac 12 f^\prime(b)\cdot \{\bigl(b-m-\frac 12\bigr)^2-\frac{1}{12}\}-\frac 12
f^\prime (a)\cdot \{(a-m-\frac 12)^2-\frac{1}{12}\}+\ \text{Error}\ ,\\
|\text{Error}|\le C^\prime R^{-2}\ .\endmultline\tag 2
$$

Taylor--expanding $f(t)$ about $t=a$ yields
$$\multline
\int\limits_a^{m+\frac 12}f(t)\,dt=f(a)\cdot(m+\frac 12-a)+\frac 12
f^\prime(a)\cdot(m+\frac 12-a)^2+\ \text{Error}\ ,\\
|\text{Error}|\le C^\prime R^{-2}\ .\endmultline
$$
\noindent Taylor--expanding $f(t)$ about $t=b$ yields
$$\multline
\int\limits_{m+1/2}^bf(t)\, dt=f(b)\cdot(b-m-\frac 12)-\frac 12 f^\prime
(b)\cdot(m+\frac 12-b)^2+\ \text{Error}\  ,\\
|\text{Error}|\le C^\prime R^{-2}\, .\endmultline
$$
\noindent Adding these and noting that $|f^\prime(b)-f^\prime(a)|\le
C^\prime R^{-2}$, we obtain (2).  So the lemma holds also
when $\Bbb Z\cap [a,b]= \emptyset$. $\qquad\blacksquare$


\vglue 1pc
\proclaim{Lemma 2} Assume $|(\frac {d}{dx})^mf(x)|\le C_mR^{-m} (m\ge 0)$
with $R\ge 1$.  Assume also that $f$ is supported in an interval of length
$CR$.  Then
$$
\sum\limits_{k\in \Bbb Z}f(k)=\int\limits_{-\infty}^\infty f(t)\, dt+\ 
\roman{ Error},\quad  \text{and}\quad |\roman{ Error}|\le C_N^\prime R^{-N}\quad
\text{with}\quad N\quad \text{as\ large}
$$
\noindent as we please.   The constant $C_N^\prime$ depends only on $N$,
on $C$, and on finitely many of the $C_m$.\endproclaim
\vfill\eject

\demo{Proof}  We may assume that $\text{supp}\ f$ is contained in an
interval of length $CR$ centered at $0$.  Since $x\mapsto f(xR)$ is a
Schwartz function, the Fourier transform $\hat f(\xi)$ satisfies the
estimate $|\hat f(\xi)|\le C_N^\prime R(1+R|\xi|)^{-N}$ .  Therefore,
$$
\Big\vert\sum\limits_{k\in \Bbb Z\backslash\{0\}}\hat f(k)\Big|
\le C_N^\prime R\sum\limits_{k\in \Bbb Z\backslash \{0\}} (R|k|)^{-N}
\le C_N^\prime R^{1-N}\, ,
$$
\noindent so the lemma follows at once from the Poisson summation formula.
$\qquad\blacksquare$\enddemo
\medskip

The interested reader can easily supply a more direct proof of Lemma 2.
The main result of this section is as follows.

\vglue 1pc
\proclaim{Lemma on Riemann Sums} Let $f(t)$, $\sigma(t)$, $\tau(t)$ be
defined on a non--empty interval $[a,b]$.  Suppose $\sigma(t)>0$,
$\tau(t)\ge 1$ in $[a,b]$; and assume that whenever $t_1,t_2\in [a,b]$ with
$|t_1-t_2|<c\tau(t_1)$, we have 
$c<\frac{\tau(t_2)}{\tau(t_1)}<C$ and $c<\frac{\sigma(t_2)}{\sigma(t_1)}
<C$.  Finally assume $|(\frac{d}{dt})^mf(t)|\le C_m\sigma(t)\tau^{-m}
(t)$ for $t\in [a,b]$.  Then $\sum\limits_{k\in \Bbb Z\cap [a,b]}
f(k)=\int\limits_a^b f(t)dt-f(b)\chm(b)-f(a)\chp(a)+\frac 12f^\prime (b)
\tilde\chi(b)
-\frac 12f^\prime (a)\tilde\chi(a)+\ \roman{Error}$ with
$|\roman{ Error}|\le C^\prime\sigma(a)\tau^{-2}(a)+C^\prime\sigma(b)\tau^{-2}
(b)+C_N^\prime\int_a^b\sigma(t)\tau^{-N}(t)dt$.  Here, $C^\prime$
depends only on $c$, $C$, $C_m$; and $C_N^\prime$ depends only on $c$,
$C$, $C_m$, $N$.  If $f(t)=0$ to infinite order at $t=a$, then we have
the sharper estimate $|\roman{Error}|\le C^\prime \sigma(b)\tau^{-2}
(b)+C_N^\prime\int_a^b\sigma(t)\tau^{-N}(t)dt$, with $C^\prime$,
$C_N^\prime$ as before.  Similarly, if $f(t)=0$ to infinite order
at $t=b$, then $|\roman{Error}|\le C^\prime\sigma(a)\tau^{-2}(a)+C_N^\prime
\int_a^b\sigma(t)\tau^{-N}(t)dt$.  If $f(t)=0$ to infinite order
at both $t=a$ and $t=b$, then
$|\roman{Error}|\le
C_N^\prime\int\limits_a^b\sigma(t)\tau^{-N}(t)dt$.\endproclaim
\vglue 1pc

\demo{Proof} We distinguish several cases.\enddemo

\medskip

\demo{Case 1}  $b-a\ge c\tau(a)$.\enddemo

Then we can find a partition of unity $\theta_0(x)\ldots\theta_M(x)$ with
the following properties\smallskip
\roster
\item"{($\cdot$)}" Each $\theta_\nu(x)$ is supported in an interval $I_\nu$
of length $\sim \tau(x_\nu)$, $x_\nu\in I_\nu\cap [a,b]$.

\item"{($\cdot$)}" If $\nu=0$ then $x_\nu=a$; if $\nu=M$ then $x_\nu=b$; if
$0<\nu<M$ then $I_\nu\subset [a,b]$.
\item"{($\cdot$)}" If $\nu=0$, then $\theta_\nu=1$ in a neighborhood of $a$;
and if $\nu=M$, then $\theta_\nu=1$ in a neighborhood of $b$.

\item"{($\cdot$)}" Each $\theta_\nu(x)$ satisfies
$|(\frac{d}{dx})^m\theta_\nu(x)|\le C_m^\prime\tau^{-m}(x_\nu)$.

\item"{($\cdot$)}" $\theta_0(x)+\theta_1(x)+\ldots+\theta_M(x)=1$ on
$[a,b$].

\item"{($\cdot$)}" No point belongs to more than $C^\prime$ of the intervals
$I_\nu$.
\endroster
\smallskip

\noindent Using the $\theta_\nu$ we write
$f=\sum\limits_{\nu=0}^M\theta_\nu f\equiv \sum\limits_{\nu=0}^Mf_\nu$.
\smallskip

\noindent For $0<\nu<M$, we have $|(\frac{d}{dx})^mf_\nu|\le C_m^{\prime\prime}
\sigma(x_\nu)\tau^{-m}(x_\nu)$ and diam supp $f_\nu\le C\tau(x_\nu),
\text{supp} f_\nu\subset [a,b]$.  For
$\nu=0$ or $M$, we can extend $f_\nu$ to a $C_0^\infty$
function satisfying again $|(\frac{d}{dx})^mf_\nu|\le C_m^{\prime\prime}
\sigma(x_\nu)\tau^{-m}(x_\nu)$ and diam supp $f_\nu\le C\tau(x_\nu)$, but we
don't have $\text{supp}\ f_\nu\subset [a,b]$.  However, if $\nu=0$
and $f(t)$ vanishes to infinite order at $a$, then we can take
$f_\nu\equiv 0$ outside $[a,b]$ and thus $\text{supp}\ f_\nu\subset
[a,b]$.  Similarly for $b$.  Now we compute $\sum\limits_{k\in \Bbb Z\cap
[a,b]} f(k)=\sum\limits_{\nu=0}^M \bigl(\sum\limits_{k\in \Bbb Z\cap
[a,b]}f_\nu(k)\bigr)$ by applying lemmas 1 and 2.  If $\text{supp}\ f_\nu
\subset [a,b]$, then $\sum\limits_{k\in \Bbb Z\cap [a,b]}f_\nu(k)
=\sum\limits_{k\in \Bbb Z}f_\nu(k)$, so lemma 2 applies.  If
$\text{supp}\ f_\nu\not\subset [a,b]$, then we must apply lemma 1.  Since
$\sum\limits_{\nu=0}^M\sigma(x_\nu)\tau^{-(N-1)}(x_\nu)\sim
\sum\limits_{\nu=0}^M \int\limits_{I_\nu}\sigma(t)\tau^{-N}
(t)dt\sim\int\limits_a^b\frac{\sigma(t)}{\tau^N(t)}\,dt$,
$f_0(a)=f(a)$, $f_0^\prime(a)=f^\prime(a)$, $f_M(b)=f(b)$, $f_M^\prime(b)
=f^\prime(b)$, the result of our computation is the conclusion of the
present lemma.  This settles the case $b-a\ge c\tau(a)$.

\medskip
\demo{Case 2} $b-a<c\tau(a)$, and $f(t)$ vanishes to only finite order
at $a$ or $b$.  Then the conclusion of the present lemma follows from
lemma 1.\enddemo
\medskip

\demo{Case 3}  $c(\tau(a))^{1/2}<b-a<c\tau(a)$, and $f(t)$ vanishes to infinite
order at $a$ and $b$.  Replacing $\tau(t)$ by $\tilde\tau=\tau^{1/2}(t)$,
we find that $\sigma$, $\tilde\tau$, $f$ satisfy the hypotheses of
the present lemma, and we find ourselves in Case 1.  So our desired conclusion
follows by the reasoning of Case 1.\enddemo

\medskip
\demo{Case 4} $c\le b-a\le c(\tau(a))^{1/2}$ and $f(t)$ vanishes to
infinite order at $a$ and $b$.  Then in $[a,b]$ we have $|f(t)|\le 
\max_{[a,b]} |(\frac{d}{dt^\prime})^{2N}
f(t^\prime)|\cdot (b-a)^{2N}$ by Taylor's theorem, hence
$|f(t)|\le C_N\frac{\sigma(t)}{\tau^{2N}(t)}(b-a)^{2N}\le 
C_N\frac{\sigma(t)}{\tau^N(t)}\le C_N^\prime \frac{\sigma(a)}
{\tau^N(a)}$ for $t\in [a,b]$.  Consequently,
$$\multline
\Big|\sum\limits_{k\in \Bbb Z\cap [a,b]} f(k)\Big|\le C_N^\prime
\frac{\sigma(a)}{\tau^N(a)}\cdot (b-a+1)\\
\le C_N^{\prime\prime}\frac{\sigma(a)}{\tau^N(a)}\cdot
(b-a)\le C_N^{\prime\prime}\le C_N^{\prime\prime}\int_a^b
\frac{\sigma(t)}{\tau^N(t)}\ dt\endmultline
$$
\noindent and
$$
\Big|\int_a^bf(t)dt\Big|\le C_N^\prime \frac{\sigma(a)}{\tau^N(a)}
\cdot (b-a)\le C_N^{\prime\prime}\int_a^b\frac{\sigma(t)}
{\tau^N(t)}\ dt\ .
$$
\noindent From these estimates, the conclusion of the lemma is obvious.
\enddemo

\medskip
\demo{Case 5} $b-a\le c$ and $f(t)$ vanishes to infinite order at $a$ and
$b$.  Then Taylor's theorem tells us that $|f(\overline t)|\le \max_{[a,b]}
|(\frac{d}{dx})^{2N}f(x)|\cdot (b-a)^{2N}\le
C_N^\prime \frac{\sigma(a)}{\tau^{2N}(a)}(b-a)^{2N}\le C_N^\prime
\frac{\sigma(a)}{\tau^N(a)}(b-a)\le C_N^{\prime\prime}
\int_a^b\frac{\sigma(t)}{\tau^N(t)}\ dt$ for $\overline t\in [a,b]$.  Therefore
$$
\Big|\int_a^bf(t)\ dt\Big|\le C_N^{\prime\prime}\int_a^b\frac{\sigma(t)}
{\tau^N(t)}\ dt\qquad \text{since}\ b-a<1\ ,
$$
\noindent and $|\sum\limits_{k\in \Bbb Z\cap [a,b]}f(k)|\le 
C_N^{\prime\prime}\int_a^b\frac{\sigma(t)}{\tau^N(t)}\ dt$ since
the sum has at most one term.\enddemo

>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 $c_\ast<\sigma_x(t_2)/\sigma_x(t_1)
<C_\ast$ for $t_1,t_2\in\Cal J$, $|t_1-t_2|<c_\ast\tau$.  Since
$\frac{B_{\text{left}}}{\delta x}>\lambda_{\text{left}}^\varepsilon$
and $\Gamma\ge c_{\#}\Lambda$, (70) shows that $\tau \ge 1$.  That
$c_\ast<\sigma_x(t_2)/\sigma_x(t_1)<C_\ast$ amounts to saying that
$|x_{\text{left}}(E(t_2))-x_{\text{left}}(E(t_1))|\le C_\ast
\lambda_{\text{left}}^{-2/3}B_{\text{left}}$ for $|t_1-t_2|<c_\ast\tau$;
which in turn amounts to $|E(t_2)-E(t_1)|\le C_\ast\lambda_{\text{left}}^{-2/3}
S_{\text{left}}$ for $t_1,t_2\in \Cal J$, $|t_1-t_2|<c_\ast\tau$.  We know
from (13) that $|E(t_2)-E(t_1)|\le C_{\#}S_{\min}\Gamma^{-1}|t_1-t_2|\le
C_{\#}S_{\min}\Gamma^{-1}\tau\le C_{\#}(S_{\text{left}}^{1/2}
B_{\text{left}}^{-1})(\frac{B_{\text{left}}}{\delta x})^{1/2}$ (by (11) and
(70)) $\le C_{\#}S_{\text{left}}(S_{\text{left}}^{-1/2}B_{\text{left}}^{-1})
\lambda_{\text{left}}^{1/3}$ (since $\frac{\delta x}{B_{\text{left}}}
>\lambda_{\text{left}}^{-2/3})=C_{\#}S_{\text{left}}\cdot
\lambda_{\text{left}}^{-2/3}$, 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,
$$\multline
\int\limits_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
\int_a^b\sigma_x(t)\tau^{-\overline{N}}dt\ dx\le \int_a^b
\Bigl\{\int\limits_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
\sigma_x(t)\ dx\Bigr\}dt\cdot \tau^{-\overline{N}}\\
\le C_\ast \tau^{-\overline{N}}(b-a)\cdot \operatornamewithlimits{max}_
{t\in (a,b)} \Bigl\{\int\limits_{|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}}\frac{\lambda_{\text{left}}^{-2/3}B_{\text{left}}
(S_{\min}\Gamma^{-1})\ dx}
{(1+\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}
|x-x_{\text{left}}(E(t))|)^{1/2}}\Bigr\}\\
\le C_\ast \tau^{-\overline{N}}(b-a)\cdot \frac{\lambda_{\text{left}}^{-2/3}
B_{\text{left}}^2(S_{\min}\Gamma^{-1})}{\lambda_{\text{left}}^{1/3}}
\le C_\ast\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}\tau^{-\overline 
{N}}\endmultline\tag 77
$$

\noindent (since we see from the proof of (48) that $b-a\le C_{\#}\Gamma$)
$\le C_\ast (\Gamma^{-\overline N}+(\frac{\delta x}{B_{\text{left}}})^{\frac
{\overline N}{2}})\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}$
(by (70)) $\le C_\ast (\Lambda^{-N}+(\frac{\delta x}{B_{\text{left}}})^N)
\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}$.  From (73), (75),
(76), (77) we get
$$
\int\limits_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
|\text{Error}_2(x)|\ dx\le C_\ast(\Lambda^{-N}
+(\frac{\delta x}{B_{\text{left}}})^N)\lambda_{\text{left}}^{-1}
B_{\text{left}}^2S_{\min}\ , \quad\text{and}\tag 78
$$
$$\split
\int\limits_{|x-x_{\text{left}}(E_{\max})|<(\delta x)}|\text{Error}_1(x)|
dx\le C_\ast(\Lambda^{-2}+(\frac{\delta x}{B_{\text{left}}})^{1/2}
\Lambda^{-1})\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}
(\frac{\delta x}{B_{\text{left}}})^{1/2}\\
+C_\ast (\Lambda^{-N}+(\frac{\delta x}{B_{\text{left}}})^N)\lambda_{\text{
left}}^{-1}B_{\text{left}}^2S_{\min}\ .\endsplit\tag 79
$$

\noindent Note that $(\frac{\delta x}{B_{\text{left}}})^N\le
\lambda_{\text{left}}^{-\varepsilon N}\le \Lambda^{-N^\prime}$ and
$\Lambda^{-N}\le \Lambda^{-N^\prime}$.  Equation (74) may be rewritten as
$$\multline
\sum\limits_{k\in \Bbb Z\cap [a,b]}f_{\text{Airey}}(x,k)\cdot
f_{\text{other}}(x,k)=\int_a^bf_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\ dt\\
-f_{\text{Airey}}(x,b)f_{\text{other}}(x,b)\chm(b)+\text{Error}_2(x)\\
\text{for}\quad |E_0|\ge 2\hat cS_{\min}\quad\text{or}\quad
|x-x_{\text{left}}(E_{\max})|>\delta x\ ,\endmultline\tag 80
$$

\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 $\{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}\}$, which follows
from (52), (81) and {\it ASSUMPTIONS\/} 1 and 3.

To make further progress, we need to estimate $\int\limits_{I_{\text{BVP}}}
|f_{\text{Airey}}(x,t)f_{\text{other}}(x,t)|dx$ for fixed $t\in \Cal J$.
Using (69) and the fact that the integrand is supported in
$\{|x-x_{\text{left}}(E(t))|<\delta x\}$, we get
$$\align
&\int\limits_{I_{\text{BVP}}}|f_{\text{Airey}}(x,t)f_{\text{other}}
(x,t)|\ dx\le C_\ast\int\limits_{|x-x_{\text{left}}(E(t))|<\delta x}
\sigma_x(t)\ dx\\
&\le C_\ast\int\limits_{|x-x_{\text{left}}(E(t))|<\delta x}
\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1})\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad \cdot
(1+\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}
|x-x_{\text{left}}(E(t))|)^{-1/2}\ dx\\
&\le C_\ast\lambda_{\text{left}}^{-2/3}B_{\text{left}}(S_{\min}\Gamma^{-1})
\lambda_{\text{left}}^{-1/3}B_{\text{left}}^{1/2}(\delta x)^{1/2}\\
&=C_\ast\lambda_{\text{left}}^{-1}B_{\text{left}}^2(S_{\min}\Gamma^{-1})
(\frac{\delta x}{B_{\text{left}}})^{1/2}\quad\text{for}\quad t\in \Cal J\ .
\tag 84\endalign
$$

Our first use of (84) is to change the region of integration in (81)
from $[a,b]$ to
$$
[a_0,b_0]=\bigl[\frac1\pi \phi(E_{\ell o})+\frac{1}{48\pi}
\psi(E_{\ell o})-\frac 12\ ,\ \frac{1}{\pi}\phi(E_{hi})+\frac{1}{48\pi}
\psi(E_{hi})-\frac 12\bigr]\ .\tag 85
$$
\noindent From (3) and (85) we get
$$
|a-a_0|\ ,\ |b-b_0|\le C_{\#}\Lambda^{-2}\ ,\tag 86
$$
\noindent and therefore by (84) we have
$$\multline
\int\limits_{I_{\text{BVP}}}\Big|\int_{b_0}^b f_{\text{Airey}}
(x,t)f_{\text{other}}(x,t)\ dt\Big|\ dx\le |b_0-b|\cdot C_\ast
\lambda_{\text{left}}^{-1}B_{\text{left}}^2(S_{\min}\Gamma^{-1})(\frac{\delta x}
{B_{\text{left}}})^{1/2}\\
\le C_\ast\Lambda^{-2}\lambda_{\text{left}}^{-1}B_{\text{left}}^2(S_{\min}
\Gamma^{-1})(\frac{\delta x}{B_{\text{left}}})^{1/2}\le
C_\ast\Lambda^{-3}\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}
(\frac{\delta x}{B_{\text{left}}})^{1/2}\\
\text{by\ (12)}\ .\endmultline\tag 87
$$

\noindent Similarly
$$
\int\limits_{I_{\text{BVP}}}\Big|\int_{a_0}^af_{\text{Airey}}
(x,t)f_{\text{other}}(x,t)\ dt\Big|\ dx\le C_\ast
\Lambda^{-3}\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}
(\frac{\delta x}{B_{\text{left}}})^{1/2}\ .\tag 88
$$
\noindent (In fact, the left--hand side is zero, but never mind.)
\smallskip

\noindent Equations (81)$\ldots$(83) and (87), (88) imply
$$\split
\sum\limits_{k\in \Bbb Z\cap [a,b]}f_{\text{Airey}}(x,k)f_{\text{other}}
(x,k)=\int_{a_0}^{b_0}f_{\text{Airey}}(x,t)f_{\text{other}}(x,t)\ dt\\
-f_{\text{Airey}}(x,b)f_{\text{other}}(x,b)\chm(b)+\ \text{Error}_4(x)
\endsplit\tag 89
$$
\noindent with
$$\split
\int\limits_{I_{\text{BVP}}}|{\roman {Error}}_4(x)|\ dx\le C_\ast
(\Lambda^{-2}+(\frac{\delta x}{B_{\text{left}}})^{1/2}\Lambda^{-1})
\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}(\frac{\delta x}
{B_{\text{left}}})^{1/2}\\
+C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1}B_{\text{left}}^2
S_{\min}\quad\text{if}\quad |E_0|\le 2\hat cS_{\min}\ ,\quad\text{and}
\endsplit\tag 90
$$
$$\split
\int\limits_{I_{\text{BVP}}}|\text{Error}_4(x)|\ dx\le C_\ast
\Lambda^{-3}\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}
(\frac{\delta x}{B_{\text{left}}})^{1/2}\\
+C_\ast\Lambda^{-N^\prime}\lambda_{\text{left}}^{-1}B_{\text{left}}^2
S_{\min}\quad\text{if}\quad |E_0|>2\hat cS_{\min}\ .\endsplit\tag 91
$$

\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
$E<E_{\ell o}$.  So changing the region of integration
from $[E_{\ell o},E_{hi}]$ to $(-\infty,E_{hi}]$
leaves the integral unchanged on the right--hand side
of (93).  Similarly, if $|E_0|>2\hat cS_{\min}$,
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 $|b-b_0|<C_{\#}\Lambda^{-2}$ by (86),
this implies
$$\multline
\int_{I_{\text{BVP}}}\Big|f_{\text{Airey}}
(x,b)f_{\text{other}}(x,b)-f_{\text{Airey}}(x,b_0)
f_{\text{other}}(x,b_0)\Big|\ dx\\
\le C_\ast\Bigl(\Lambda^{-4}+\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{1/2}\Lambda^{-3}\Bigr)
\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\ .
\endmultline\tag 100
$$

\noindent We use (100) if $|E_0|\le 2\hat c
S_{\min}$.  If instead $|E_0|>2\hat cS_{\min}$,
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 $|b_0-(\frac 1\pi \phi(0)-\frac 12)|
<C_{\#}\Lambda^{-1}$.  By (86) we have also $|b-
(\frac 1\pi\phi(0)-\frac 12)|\le C_{\#}\Lambda^{-1}$.
This implies
$$\multline
\big|\chm(b)-\chm(\frac 1\pi \phi(0)-\frac 12)\big|\le
C_{\#}\Lambda^{-1}\\
\text{if}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|
>\overline C_{\#}\Lambda^{-1}\ \text{and}\
|E_0|\le 2\hat cS_{\min}\ .\endmultline\tag 105
$$

\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$\ldots$5, 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 $|x-y|<cB(x)$, then $c<\frac{B(y)}{B(x)}<C$
and $c<\frac{S(y)}{S(x)}<C$.

\item"(X1)" If $x\in I$ and $\alpha\ge 0$, then $|(\frac{d}{dx})^\alpha
V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$.

\item"(X2)" For $E\in \Cal J=[E_0-\delta E,E_0+\delta E]\cap (-\infty,0]$,
the equation $V(x)=E$ has two solutions $x_{\text{left}}(E)<x_{\text{rt}}
(E)$ in $I$, and they satisfy $\text{dist}(x_{\text{left}}(E),\partial I)
>cB(x_{\text{left}}(E))$, $\text{dist}(x_{\text{rt}}(E),\partial I)>
cB(x_{\text{rt}}(E))$.

\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 $cS(x)<E-V(x)<CS(x)$.

\item"(X5)" For $E\in \Cal J$ we have $c(\delta E)<S(x_{\text{left}}(E))<
C(\delta E)$
\endroster

To state the remaining hypotheses, we set up some notation.  Set
$\lambda(x)=\hfill\break S^{1/2}(x)B(x)$ as usual, and let
$$
B_{\text{left}}(E)=
B(x_{\text{left}}(E))\ ,\ B_{\text{rt}}(E)=B(x_{\text{rt}}(E))\ ,
$$
$$
S_{\text{left}}(E)=S(x_{\text{left}}(E))\ ,\ S_{\text{rt}}
(E)=S(x_{\text{rt}}(E))\ ,
$$
$$
\lambda_{\text{left}}(E)=\lambda(x_{\text{left}}
(E))\ ,\ \lambda_{\text{rt}}(E)=\lambda(x_{\text{rt}}(E))
\quad\text{for}\ E\in \Cal J\ .
$$

\noindent Then define $S_{\min}(E)=\operatornamewithlimits{\text{inf}}_
{x_{\text{left}}(E)\le x\le
x_{\text{rt}}(E)}S(x)$ and
$$
\Lambda(E)=\Bigl(\int_{x_{\text{left}}(E)}^{x_{\text{rt}}(E)}
\frac{dx}{S^{1/2}(x)B^2(x)}\Bigr)^{-1}\ , \quad\text{for}\ E\in \Cal J\ .
$$

\noindent Set $\Lambda_{\min}=\text{inf}_{E\in \Cal J}\Lambda(E)$.  
\smallskip

\noindent Now we can state the rest of our assumptions.
\medskip

\subhead Assumptions on $V(x)$ in all of $I_{\text{BVP}}$\endsubhead
\smallskip

\roster
\item"(X6)" For $E\in \Cal J$ and $x\in I_{\text{BVP}}\backslash
[x_{\text{left}}(E),x_{\text{rt}}(E)]$ we have $V(x)>E$.

\item"(X7)" For $E\in \Cal J$ and $x\in I_{\text{BVP}}$, we have
$$
\align
V(x)&\ge \frac{100}{|x-x_{\text{left}}(E)|^2}\quad\text{if}\
x<x_{\text{left}}(E)-\frac 12 \lambda_{\text{left}}^K(E)B_{\text{left}}(E)\ ,
\ \text{and}\\
V(x)&\ge \frac{100}{|x-x_{\text{rt}}(E)|^2}\quad\text{if}\
x>x_{\text{rt}}(E)+\frac 12\lambda_{\text{rt}}^K(E)B_{\text{rt}}(E)\ .
\endalign
$$
\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 $\hat c^\prime<c_{\#}$ we can produce a sequence
of functions $\theta_\nu(E)$ and energies $E_\nu^\prime$ with the
following properties.

\roster
\item"(a)" $\sum\limits_{\nu}\theta_\nu(E)=1$\quad for\quad 
$E\in \Cal J^0$.

\item"(b)" Each $\theta_\nu(E)$ is supported in $\{|E-E_\nu^\prime|\le
\hat c^\prime S_{\min}(E_\nu^\prime)\}\equiv \Cal J_\nu^{\prime\prime}$,
and each $E\in \Bbb R^1$ belongs to at most $C_{\#}[\hat c^\prime]^{-2}$
of the $\Cal J_\nu^{\prime\prime}$.

\item"(c)" Each $\theta_\nu(E)$ satisfies $\big|\bigl(\frac{d}{dE}\bigr)^\beta
\theta_\nu(E)\big|\le C_{\#}^\beta([\hat c^\prime]^2
S_{\min}(E_\nu^\prime))^{-\beta}$.

\item"(d)" We have $|E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)
$ for at most $C_{\#}[\hat c^\prime]^{-2}$ of the $E_\nu^\prime$.

\item"(e)" Each $E_\nu^\prime$ belongs to $\Cal J^0$.
\endroster
\endproclaim

\medskip

\noindent{\it Sketch of Proof\/}.  The point is that if $E_1, E_0\in \Cal J^0$
with $|E_1-E_2|<c_{\#}S_{\min}(E_1)$ then $c_{\#}<
\frac{S_{\min}(E_2)}{S_{\min}(E_1)}<C_{\#}$.  This implies that
$\Cal J^0$ can be partitioned into subintervals $\Cal J_\mu$ with the
following properties.  If $E\in \Cal J^0$ belongs to the double of
$\Cal J_\mu$, then $\text{length}(\Cal J_\mu)\sim S_{\min}(E)$.  Each
point $E\in \Bbb R^1$ belongs to at most $C_{\#}$ of the doubles of the
$\Cal J_\mu$.

Next, pick an integer $M\sim (\hat c^\prime)^{-2}$.  Then cut each
$\Cal J_\mu$ into $M$ equal subintervals, to partition $\Cal J^0$ into
smaller subintervals $\Cal J_\nu^\prime$, with the following properties.

If $E\in \Cal J^0$ belongs to the double of $\Cal J_\nu^\prime$, then
$\text{length}\,(\Cal J_\nu^\prime)\sim \frac{S_{\min}(E)}{M}$.
Consequently, if the doubles of $\Cal J_{\nu_1}^\prime$ and $\Cal
J_{\nu_2}^\prime$ meet, then $\text{length}\ \Cal J_{\nu_1}
^\prime\sim\
\text{length}\ \Cal J_{\nu_2}^\prime$.
Also, each point $E\in \Bbb R^1$ belongs to at most $C_{\#}$ of the doubles
of the $\Cal J_\nu^\prime$.

Now take $\chi_{{}_\nu}(E)$ supported in the double of $\Cal J_\nu^\prime$,
equal to $1$ on $\Cal J_\nu^\prime$, and satisfying $\big|\bigl(\frac{d}
{dE}\bigr)^\beta\chi_{{}_{\nu}}(E)\big|\le C_{\#}^\beta\bigl[\text{length}
(\Cal J_\nu^\prime)\bigr]^{-\beta}$.  For $E\in \Cal J^0$, define
$\theta_\nu(E)=\hfill\break \chi_{{}_{\nu}}^2(E)\bigl[\sum\limits_{\nu^\prime}
\chi_{{}_{\nu^\prime}}^2(E)\bigr]^{-1}$.  Certainly $\sum\limits_\nu
\theta_\nu(E)=1$ on $\Cal J^0$; and $\theta_\nu(E)=0$ if $E\in \Cal J^0$,
$|E-E_\nu^\prime|>\frac 12 \hat c^\prime S_{\min}(E_\nu^\prime)$.  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 $|E-E_0|<c_{\#}
S_{\text{left}}(E_0)$ implies $|x_{\text{left}}(E)-x_{\text{left}}
(E_0)|<c_{\#}B_{\text{left}}(E_0)$.  The last remark is needed to verify
{\it ASSUMPTION 3\/} in the previous section.)  The constants $\hat C_\beta$
in the previous section now have the form $C_\ast^\beta$.  Hence any
constant of the form $C_\ast$ in that section still has the form $C_\ast$
here.  The lemma of the previous section therefore shows that
$$\multline
\rho(x,g_\nu)=\frac 1\pi \int_{-\infty}^0 g_\nu(x,E)\lambda_\nu^{-2/3}
\Bigl(\frac{\partial Y_\nu(x,E)}{\partial x}\Bigr)^{-1}A^2(\lambda_\nu^{2/3}
Y_\nu(x,E))\, dE\\
-g_\nu(x,0)\lambda_\nu^{-2/3}\Bigl(\frac{\partial Y_\nu}{\partial x}
(x,0)\Bigr)^{-1}A^2\bigl(\lambda_\nu^{2/3}Y_\nu(x,0)\bigr)[\phi^\prime(0)]^{-1}
\chm(\frac 1\pi \phi(0)-\frac 12)+\Cal E_\nu^1(x)\endmultline\tag 1
$$

\noindent where $\lambda_\nu=\lambda_{\text{left}}(E_\nu^\prime)$, and
$Y_\nu(x,E)$ is the function $Y_{\text{left}}(x,E)$ that comes from
applying the WKB Theorems with $E_\nu^\prime$ in place of $E_0$ etc.,
as above.  For the error $\Cal E_\nu^1(x)$ we have the following
estimates from the lemma of the previous section.
$$\multline
\int_{I_{\text{BVP}}}|\Cal E_\nu^1(x)|\, dx\le C_\ast\Lambda_{\min}
^{4\varepsilon-2}\lambda_{\text{left}}^{-1}B_{\text{left}}^2S_{\min}
(E_\nu^\prime)\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\
+C_{\ast}\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-
\hat c^\prime S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}
\phi^\prime(E)\, dE\\
\text{if}\quad |E_\nu^\prime|>2\hat c^\prime S_{\min}(E_\nu^\prime)\ .
\endmultline\tag 2
$$
$$\multline
\int_{I_{\text{BVP}}}|\Cal E_\nu^1(x)|\, dx\le C_\ast\Bigl(\Lambda_{\min}
^{4\varepsilon-2}+\bigl(\frac{\delta
x}{B_{\text{left}}}\bigr)^{1/2}\Lambda_{\min}^{-1}\Bigr)\lambda_{\text{left}}
^{-1}B_{\text{left}}^2S_{\min}(E_\nu^\prime)\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{1/2}\\
+C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime
S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)\, dE\\
\text{if}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)\
\text{and}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#}
\Lambda_{\min}^{-1}\ .\endmultline\tag 3
$$
$$\multline
\int_{I_{\text{BVP}}}|\Cal E_\nu^1(x)|\, dx\le
C_\ast\Lambda_{\min}^{-1}\lambda_{\text{left}}^{-1}B_{\text{left}}^2
S_{\min}(E_\nu^\prime)\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{1/2}\\
+C_\ast\Lambda_{\min}^{10-N^{\prime\prime}}\int_{E_\nu^\prime-\hat c^\prime
S_{\min}(E_\nu^\prime)}^{E_\nu^\prime}\phi^\prime(E)\, dE\\
\text{if}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)\quad
\text{and}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}
\Lambda_{\min}^{-1}\ .\endmultline\tag 4
$$

\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
$$\multline
\big|\lambda_\nu^2(\partial_xY_\nu)^2Y_\nu+\{Y_\nu,x\}-(E-V(x))\big|<
C_{\#}\lambda_{\text{left}}^{-N^{\prime}}S_{\text{left}}\\
\text{for}\quad |E-E_\nu^\prime|<c_{\#}^\prime S_{\min}(E_\nu^\prime)\ ,
|x-x_{\text{left}}(E)|<c_{\#}\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}
\endmultline\tag 5
$$
$$\multline
\big|\partial_x^\alpha Y_\nu(x,E)\big|\le C_{\#}^\alpha B_{\text{left}}
^{-\alpha}\quad\text{for}\
|E-E_\nu^\prime|<\hat c^\prime S_{\min}(E_\nu^\prime)\ , |x-x_{\text{left}}
(E)|<c_{\#}B_{\text{left}}\ .\endmultline\tag 6
$$

\noindent Estimates (5) and (6) are part of the conclusion of the WKB
Eigenfunction Theorem.  We also use the following result.


\vglue 1pc
\proclaim{Lemma 2}  There is a function $Y(x,E)$ satisfying the estimates
$$\multline
\big|\lambda_{\text{left}}^2(\partial_xY)^2Y+\{Y,x\}-(E-V(x))\big|\le
C_{\#}\lambda_{\text{left}}^{-N^{\prime}}S_{\text{left}}\\
\text{for}\quad |E-E_0|<c_{\#}(\delta E)\sim c_{\#}
S_{\text{left}}\quad\text{and}\quad
|x-x_{\text{left}}(E)|\le c_{\#}\lambda_{\text{left}}^{-\varepsilon}
B_{\text{left}}\endmultline\tag 7
$$
\noindent and
$$\multline
\big|\partial_x^\alpha\partial_E^\beta Y(x,E)\big|\le C_{\#}^{\alpha\beta}
B_{\text{left}}^{-\alpha}S_{\text{left}}^{-\beta}\\
\text{for}\quad |E-E_0|<c_{\#}(\delta E)\quad\text{and}\quad
|x-x_{\text{left}}(E)|<c_{\#}B_{\text{left}}\ . \endmultline\tag 8
$$

\noindent Also,
$$\multline
\Big|\frac{\partial}{\partial x}Y(x,E)\big|>c_{\#}
B_{\text{left}}^{-1}\quad \text{and}\quad\Big|\frac{\partial Y}{\partial E}
(x,E)\Big|>c_{\#}S_{\text{left}}^{-1}\\
\text{for}\quad |E-E_0|<c_{\#}(\delta E)\quad\text{and}\quad
|x-x_{\text{left}}(E)|<c_{\#}B_{\text{left}}\ .\endmultline\tag 9
$$
\endproclaim
\smallskip

\noindent This is contained in Lemma 1 from the section on approximate global
solutions of ODE's in [FS2].  We compare $Y(x,E)$ with $Y_\nu(x,E)$, using the
following.

\vglue 1pc
\proclaim{Lemma 3}  Suppose $y_1(x)$ and $y_2(x)$ are defined
on $\{|x|<c\}$ and satisfy there $\big|\bigl(\frac{d}{dx}\bigr)^\alpha
y_i(x)\big|\le C_\alpha$, $\frac{d}{dx}y_i(x)>c>0$, $|y_i(0)|<C\lambda^{-2}$.
Suppose also $(\frac{d}{dx}y_1)^2y_1+\lambda^{-2}\{y_1,x\}=\bigl(\frac{d}{dx}
y_2\bigr)^2y_2+\lambda^{-2}\{y_2,x\}+\ {\roman{Error}}$, with
$|{\roman{Error}}|\le C\lambda^{-N^{\prime}}$ for
$|x|<\lambda^{-\varepsilon}$.
Then $|(\frac{d}{dx})^\alpha(y_1-y_2)|<\overline C\lambda^{-N^{\prime\prime}}
$ for $|x|<\lambda^{-\varepsilon}$, $0\le \alpha\le 3$, with $\overline C$
depending only on $c$, $C$, $C_\alpha$, $N$, $\varepsilon$.\endproclaim


\vglue 1pc
\demo{Proof}  Let $P_i(x)$ be the $N^{\text{th}}$ order Taylor expansion
of $y_i(x)$ about $0$.  Taylor's theorem shows that $P_i(x)$ satisfy
the hypotheses assumed for $y_i(x)$, (with different $c$, $C$, $C_\alpha$
depending on the original $c$, $C$, $C_\alpha$) and that
$|\bigl(\frac{d}{dx}\bigr)^\alpha[P_1-P_2]|\le \overline C\lambda^{-N^{\prime\prime}}$ for $\alpha\le 3$ and
$|x|<\lambda^{-\varepsilon}$ implies $|\bigl(\frac{d}{dx}\bigr)^\alpha
[y_1-y_2]|\le \overline C\lambda^{-N^{\prime\prime}}$ for a different
$\overline C$.  Hence, it's enough to prove the lemma assuming the $y_i(x)$
are $N^{\text{th}}$ degree polynomials.\enddemo

By induction on $m$ $(0\le m\le N^{\prime\prime}+10)$ we will show
that $|y_1-y_2|\le \overline C\lambda^{-m}$ for
$|x|<\lambda^{-\varepsilon}$, with $\overline C$ depending only on
$c$, $C$, $C_\alpha$, N, $\varepsilon$.  This certainly holds for $m=0$.  Assume
it holds for a given $m$.  Then $|\bigl(\frac{d}{dx}\bigr)^\alpha y_1-
\bigl(\frac{d}{dx}\bigr)^\alpha y_2|\le \overline C\lambda^{-m+\varepsilon
\alpha}$ for $|x|<\lambda^{-\varepsilon}$ and $0\le \alpha\le 3$, since
the $y_i(x)$ are polynomials.  The Schwartzian $\{y_i,x\}$ is a function
$\Cal F\bigl(\frac{d}{dx}y_i, \frac{d^2}{dx^2}y_i, \frac{d^3}{dx^3}
y_i\bigr)$ with $\Cal F(t_1,t_2,t_3)$ smooth on $\{|t_1|,|t_2|, |t_3|
\le C\ \text{and}\ |t_1|\ge c>0\}$.  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 $x_0<x<\lambda^{-\varepsilon}$,
then $(\frac{dy_1}{dx})$, $(\frac {dy_3}{dx})$, $y_1(x)$ and $y_3(x)$ are
all positive, and $y_1(x)\sim (x-x_0)$, $y_3(x)\sim (x-x_0)$.  We apply
the elementary inequality $|a^{1/2}-b^{1/2}|\le \frac{|a-b|}
{a^{1/2}+b^{1/2}}$ with $a=(\frac{dy_1}{dx})^2y_1(x)$ and $b=(\frac{dy_3}
{dx})^2y_3(x)$ to derive from (12) that
$$
\Big|y_1^{1/2}\frac{dy_1}{dx}-y_3^{1/2}\frac{dy_3}{dx}\big|\le
\frac{\overline C\lambda^{-m+3\varepsilon-2}}{(x-x_0)^{1/2}}\quad\text{for}
\quad x_0<x<\lambda^{-\varepsilon}\ .\tag 13
$$

\noindent Since also $y_1(x_0)=y_3(x_0)=0$, the fundamental theorem of
calculus and (13) yield
$$
\big|y_1^{3/2}-y_3^{3/2}\big|\le \overline C\lambda^{-m+3\varepsilon-2}
\quad\text{for}\quad x_0<x<\lambda^{-\varepsilon}\ .\tag 14
$$

\noindent For $\frac 12 \lambda^{-\varepsilon}<x<\lambda^{-\varepsilon}$,
we have $y_1(x), y_3(x)\sim (x-x_0)\sim \lambda^{-\varepsilon}$
since $|x_0|<\overline C\lambda^{-2}$.  Hence (14) implies
$$
|y_1-y_3|\le \overline C\lambda^{-m+10\varepsilon-2}\quad
\text{for}\quad \frac 12 \lambda^{-\varepsilon}<x<\lambda^{-\varepsilon}\ .
$$

\noindent Since $y_1(x)$ and $y_3(x)$ are polynomials, it follows that
$|y_1-y_3|\le \overline C\lambda^{-m+10\varepsilon-2}\hfill\break
\quad\text{for}
\quad|x|<\lambda^{-\varepsilon}$.
Since $y_3(x)=y_2(x)-y_2(x_0)$ with $|y_2(x_0)|<\overline C
\lambda^{-m+3\varepsilon-2}$, we conclude that $|y_1-y_2|\le
\overline C\lambda^{-m+10\varepsilon-2}$ for $|x|<\lambda^{-\varepsilon}$.
To complete the induction step, all we need is $|y_1-y_2|\le \overline C
\lambda^{-m-1}$ for $|x|<\lambda^{-\varepsilon}$.  

Hence we know by induction that $|y_1-y_2|<\overline C\lambda^{-m}$ for
$|x|<\lambda^{-\varepsilon}$, $m=[N^{\prime\prime}+10]$.  Since $y_i(x)$ are
polynomials, it follows that $\big|\bigl(\frac{d}{dx}\bigr)^\alpha
[y_1-y_2]\big|\le \overline C\lambda^{-N^{\prime\prime}}$ for $|x|
<\lambda^{-\varepsilon}$, $0\le \alpha\le 3$, which is the conclusion
of the lemma.$\qquad\blacksquare$
\smallskip

After a rescaling, Lemma 3 applies to $x\mapsto Y(x,E)$ and
$x\mapsto (\lambda_\nu/\lambda_{\text{left}})^{2/3}Y_\nu(x,E)$ for fixed $E$
with $|E-E_\nu^\prime|<\hat c^\prime S_{\min}(E_\nu^\prime)$.  We
conclude that
$$\multline
\Big|\bigl(\frac{\partial}{\partial x}\bigr)^\alpha(\lambda_{\text{left}}
^{2/3}Y(x,E)-\lambda_\nu^{2/3}Y_\nu(x,E))\Big|\le C_{\#}
\lambda_{\text{left}}^{\frac 23 -N^{\prime\prime}}B_{\text{left}}^{-\alpha}
\quad (0\le \alpha\le 3)\\
\text{for}\quad |E-E_\nu^\prime|<\hat c^\prime S_{\min}(E_\nu^\prime),
|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\varepsilon}B_{\text{left}}
\ .\endmultline\tag 15
$$

\noindent Using (15), we can compare the right-hand side of (1) with its
analogue in which $\lambda_\nu$, $Y_\nu$ are replaced by
$\lambda_{\text{left}}$,$Y$.
We note that $\big|\frac{d}{d\xi}A^2(\xi)\big|\le C_{\#}$, hence
$$
|A^2(\lambda_{\text{left}}^{2/3}Y)-A^2(\lambda_\nu^{2/3}Y_\nu)|\le C_{\#}
\lambda_{\text{left}}^{2/3-N^{\prime\prime}}\ \text{in\ supp}\ g_\nu\ ,\
\text{by\ (15)}\ .\tag 16
$$

\noindent Since $\lambda_\nu^{2/3}\frac{\partial Y_\nu}{\partial x}\sim
\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}$ and
$|\lambda_{\text{left}}^{2/3}\frac{\partial Y}{\partial x}-\lambda_\nu
^{2/3}\frac{\partial Y_\nu}{\partial x}|\le C_{\#}\lambda_{\text{left}}
^{\frac 23-N^{\prime\prime}}B_{\text{left}}^{-1}$ in $\text{supp}\ g_\nu$
by (15), it follows that
$$
\Big|\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y}{\partial x}\bigr)^{-1}
-\lambda_\nu^{-2/3}\bigl(\frac{\partial Y_\nu}{\partial x}\bigr)^{-1}\Big|
\le C_{\#}\lambda_{\text{left}}^{-2/3-N^{\prime\prime}}B_{\text{left}}
\quad\text{in\ supp}\ g_\nu\ .\tag 17
$$
\noindent Hence in $\text{supp}\ g_\nu$ we have
$$\multline
\Big|\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y}{\partial x}
\Bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3}Y)-\lambda_\nu^{-2/3}
\bigl(\frac{\partial Y_\nu}{\partial x}\bigr)^{-1}A^2(\lambda_\nu^{2/3}
Y_\nu)\Big|\\
\le |\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y}{\partial x}\bigr)^{-1}
|\, |A^2(\lambda_{\text{left}}^{2/3}Y)-A^2(\lambda_\nu^{2/3}Y_\nu)|\\
+\Big|\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y}{\partial x}\bigr)
^{-1}
-\lambda_\nu^{-2/3}\bigl(\frac{\partial Y_\nu}{\partial x}\bigr)^{-1}\Big|
A^2(\lambda_\nu^{2/3}Y_\nu)\\
\le C_{\#}\lambda_{\text{left}}^{-2/3}B_{\text{left}}\cdot \lambda_{\text{left}}
^{\frac 23-N^{\prime\prime}}+C_{\#}\lambda_{\text{left}}^{-2/3-N^{\prime\prime}}
B_{\text{left}}\\
\text{(by\ (16),\ (17)\ and\ boundedness\ of\ the\ Airey\ function)}\
\le C_{\#}\lambda_{\text{left}}^{-N^{\prime\prime}}B_{\text{left}}\ .
\endmultline\tag 18
$$

\noindent Integrating (18) over the support of $g_\nu$, we get
$$\multline
\int_{I_{\text{BVP}}}\Big|\int_{-\infty}^0g_\nu(x,E)\lambda_\nu^{-2/3}
\bigl(\frac{\partial Y_\nu}{\partial x}\bigr)^{-1}A^2(\lambda_\nu^{2/3}
Y_\nu)\,dE\\
-\int_{-\infty}^0g_\nu(x,E)\lambda_{\text{left}}^{-2/3}
\bigl(\frac{\partial Y}{\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}
Y)\, dE\Big|\, dx\\
\le C_\ast S_{\min}(E_\nu^\prime)\lambda_{\text{left}}^{-N^{\prime\prime}}
B_{\text{left}}^2\ .\endmultline\tag 19
$$

\noindent Also, suppose $|E_\nu^\prime|\le 2\hat c^\prime S_{\min}
(E_\nu^\prime)$.  Then using (18) and the estimate
$$
\phi^\prime(0)=\frac 12 \int_{I_{\text{BVP}}}(-V(x))_+^{-1/2}\, dx\ge c_{\#}
S_{\text{left}}^{-1/2}B_{\text{left}}\ ,
\tag"(19 \text{bis})"
$$
\noindent we get
$$\multline
\int_{I_{\text{BVP}}}\big|g_\nu(x,0)\lambda_\nu^{-2/3}
\bigl(\frac{\partial Y_\nu(x,0)}{\partial x}\bigr)^{-1}A^2(\lambda_\nu^{2/3}
Y_\nu(x,0))[\phi^\prime(0)]^{-1}\\
-g_\nu(x,0)\lambda_{\text{left}}^{-2/3}\bigl(\frac{\partial Y(x,0)}
{\partial x}\bigr)^{-1}A^2(\lambda_{\text{left}}^{2/3}
Y(x,0))[\phi^\prime(0)]^{-1}\Big|\, dx\\
\le C_{\#}\lambda_{\text{left}}^{-N^{\prime\prime}}B_{\text{left}}^2
[\phi^\prime(0)]^{-1}\le C_\ast\lambda_{\text{left}}^{-N^{\prime\prime}}
B_{\text{left}}^2
(S_{\text{left}}^{-1/2}B_{\text{left}})^{-1}=C_\ast\lambda_{\text{left}}
^{1-N^{\prime\prime}}\\
\text{for}\quad |E_\nu^\prime|\le 2\hat c^\prime S_{\min}(E_\nu^\prime)\ .
\endmultline\tag 20
$$

\noindent For $|E_\nu^\prime|>2\hat c^\prime S_{\min}(E_\nu^\prime)$ 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 $|E-E_0|<\hat c(\delta E)$,
$|x-x_{\text{left}}(E)|<c_{\#}B_{\text{left}}$.  Changing variables
from $(x^\prime,E)$ to $(x^\prime,y)$ in (29), we get
$$
\Cal F(x)=\int_{-\infty}^{x}\int_{-\infty}^{Y(x^\prime,0)}
\lambda_{\text{left}}^{-2/3}h(y,x^\prime)\bigl[A^2(\lambda_{\text{left}}
^{2/3}y)-\frac 12(\lambda_{\text{left}}^{2/3}y)_+^{-1/2}\bigr]\,
dy\, dx^\prime\ .
$$

\noindent Changing the order of integration, we have
$$\multline
\Cal F(x)=\int_{-\infty}^{Y(x,0)}\Bigl[\int_{y<Y(x^\prime,0)<Y(x,0)}
\lambda_{\text{left}}^{-2/3}h(y,x^\prime)dx^\prime\Bigl]\\
\cdot \bigl[A^2(\lambda_{\text{left}}^{2/3}y)-\frac 12(\lambda_{\text{left}}
^{2/3}y)_+^{-1/2}\bigr]\, dy\ .\endmultline\tag 31
$$

\noindent To find the domain of integration in (31), we used the fact
that $x^\prime\mapsto Y(x^\prime,0)$ is strictly increasing for 
$|x^\prime-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}$.  (Recall that
$\lambda_{\text{left}}^2(\partial_xY)^2Y$ is a small perturbation of
$E-V(x)$ with $-V^\prime(x)>0$.  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 $y<Y(x,0)$, the inner integral in (31) may be written as
$$\multline
H(y,x)=\lambda_{\text{left}}^{-2/3}\int_{y}^{Y(x,0)}h(y,X(t))\frac{dX(t)}
{dt}\, dt\\
\text{with}\quad  X(t)\quad \text{defined\ as\ the\ solution\ of}\ Y(x,0)=t\ .
\endmultline\tag 36
$$

\noindent For $X(t)$ we have the estimates
$$
\Big|\bigl(\frac{d}{dt}\bigr)^\gamma X(t)\Big|\le
C_{\#}^\gamma B_{\text{left}}\quad (\gamma\ge 1)\ ,\tag 37
$$
\noindent since $Y(x,0)$ is a smooth function of $\frac{x-x_{\text{left}}
(0)}{B_{\text{left}}}$, and that smooth function has first derivative bounded
below.  The derivative $\partial_y^\beta\partial_t^\sigma h(y,X(t))$ is a sum
of terms of the form
$$\multline
\partial_{x^\prime}^\alpha\partial_y^\beta h(y,x^\prime)\bigm|_{x^\prime=
X(t)}\cdot \prod\limits_{\nu=1}^{\alpha}\Bigl[\bigl(\frac{d}{dt}\bigr)
^{\sigma_\nu}X(t)\Bigr]\quad\text{with}\quad \sigma_\nu\ge 1\ \text{and}\
\sigma_1+\ldots+\sigma_\alpha=\sigma\ .\endmultline\tag 38
$$

\noindent The term (38) is dominated by $C_\ast^{\beta\sigma}
S_{\text{left}}B_{\text{left}}(\delta x)^{-\alpha}\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-\beta} (B_{\text{left}}\bigr)^\alpha$. 
Thus,\hfill\break  $|\partial_y^\beta\partial_t^\sigma h(y,X(t))|\le
C_\ast^{\beta\sigma}S_{\text{left}}B_{\text{left}}\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-\sigma-\beta}$.  Together with (37), this
gives
$$\multline
|\partial_y^\beta\partial_t^\sigma\{h(y,X(t))\frac{dX(t)}{dt}\}|\le
C_\ast^{\beta\sigma}S_{\text{left}}B_{\text{left}}^2\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-\sigma-\beta}\\
=C_\ast^{\beta\sigma}\lambda_{\text{left}}^2\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-\sigma-\beta}\ .\endmultline\tag 39
$$

\noindent Equation (36) shows that
$$\multline
\lambda_{\text{left}}^{2/3}\partial_y^\gamma H(y,x)=\int_{y}^{Y(x,0)}  
\partial_y^\gamma\{h(y,X(t))\frac{dX}{dt}\}\, dt\\
+\sum\limits_{\gamma_1+\gamma_2=\gamma-1}\ \text{coeff}(\gamma_1,\gamma_2)
\partial_y^{\gamma_1}\partial_t^{\gamma_2}\{h(y,X(t))\frac{dX}{dt}\}
\bigm|_{t=y}\ .\endmultline
$$

\noindent From (39) we get therefore
$$
|\partial_y^\gamma H(y,x)|\le C_\ast^\gamma\lambda_{\text{left}}^{4/3}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-\gamma}
\Bigl[|y-Y(x,0)|+\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)\Bigr]\ .\tag 40
$$

\noindent We check the range of $y$ in which $H(y,x)$ is supported.
Since $g(x^\prime,E)$ is supported in $\{|x^\prime-x_{\text{left}}
(E)|\le\delta x\}$, and since $h(y,x^\prime)=g(x^\prime,E(y,x^\prime))\cdot
\{\text{STUFF}\}$, the support of $h(y,x^\prime)$ is contained in
$\{|x^\prime-x_{\text{left}}(E)|\le\delta x\}$ with $E=E(y,x^\prime)$,
i.e. $Y(x^\prime,E)=y$.  For such $(y,x^\prime)$ we have
$|y|\le |Y(x^\prime,E)-Y(x_{\text{left}}(E),E)|+|Y(x_{\text{left}}(E),E)|
\le \frac{C_{\#}}{B_{\text{left}}}|x^\prime-x_{\text{left}}(E)|+C_{\#}
\lambda_{\text{left}}^{-2}\le C_{\#}\frac{\delta x}{B_{\text{left}}}$,
so $h(y,x^\prime)$ is supported in $\{|y|\le C_{\#}\frac{\delta x}
{B_{\text{left}}}\}$.  So (40) implies 
$$\multline
|\partial_y^\gamma H(y,x)|\le C_\ast^\gamma\Bigl[\lambda_{\text{left}}^{4/3}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)\Bigr]\cdot\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-\gamma}\\
\text{if}\quad |Y(x,0)|<C_{\#}\frac{\delta x}{B_{\text{left}}}\endmultline\tag 41
$$
$$
|\partial_y^\gamma H(y,x)|\le C_\ast^\gamma\lambda_{\text{left}}^{4/3}
\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-\gamma}\quad\text{if}\quad |x-x_{\text{left}}
(0)|<c_{\#}B_{\text{left}}\ ;\tag 42
$$
\noindent and we know that
$$
\text{supp}\ H(y,x)\subset \{|y|\le C_{\#}\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)\}\quad\text{and}\quad H(Y(x,0),x)=0\ \text{by\ (36)}\ .
\tag 43
$$

\noindent We are interested in $x$ in the interval $\{|x-x_{\text{left}}(0)|
<c_{\#}B_{\text{left}}\}$.  Equations (31) and (36) give
$$\multline
\Cal F(x)=\int_{-\infty}^{Y(x,0)}H(y,x)\cdot \bigl[A^2(\lambda_{\text{left}}
^{2/3}y)-\frac 12 (\lambda_{\text{left}}^{2/3}y)_+^{-1/2}\bigr]\, dy\\
=\lambda_{\text{left}}^{-2/3}\tilde \Cal F(\lambda_{\text{left}}^{2/3}
Y(x,0))\quad\text{with}\endmultline\tag 44
$$
$$
\tilde\Cal F(\tilde x)=\int_{-\infty}^{\tilde x}\Theta(\tilde y,\tilde x)
\bigl[A^2(\tilde y)-\frac 12(\tilde y)_+^{-1/2}\bigr]\, d\tilde y\quad\text{and}
\tag 45
$$
$$
\Theta(\tilde y,\tilde x)=H(\lambda_{\text{left}}^{-2/3}\tilde y,x)\quad
\text{with}\ x\ \text{obtained\ by\ solving}\ \lambda_{\text{left}}^{2/3}
Y(x,0)=\tilde x\ .\tag 46
$$

Note that
$$
\Theta(\tilde y,\tilde x)\quad\text{is\ supported\ in}\quad
\{|\tilde y|<C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x}
{B_{\text{left}}}\}\quad \text{and\ satisfies}\tag"(46 \text{bis})"
$$
$$\multline
|\partial_{\tilde y}^\gamma\Theta(\tilde y,\tilde x)|\le C_\ast^\gamma
\bigl[\lambda_{\text{left}}^{4/3}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)
\Bigr]\bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\bigr)
^{-\gamma}\\
\text{if}\quad |\tilde x|\le C_{\#}(\lambda_{\text{left}}^{2/3}
\frac{\delta x}{B_{\text{left}}}\bigr)\ .\endmultline\tag 47
$$
$$
|\partial_{\tilde y}^\gamma\Theta(\tilde y,\tilde x)|\le C_\ast^\gamma
\lambda_{\text{left}}^{4/3}\cdot \bigl(\lambda_{\text{left}}^{2/3}
\frac{\delta x}{B_{\text{left}}}\bigr)^{-\gamma}\quad
\text{for}\quad|\tilde x|\le c_{\#}\lambda_{\text{left}}^{2/3}\ .
\tag 48
$$
$$
\Theta(\tilde y,\tilde x)=0\quad\text{when}\quad \tilde y=\tilde x\ .
\tag 49
$$

We shall study integrals of the form
$$
F(x)=\int_{-\infty}^x \theta(x,y)[A^2(y)-\frac 12 y_+^{-1/2}]\, dy\ ,\quad
\text{where\ we\ assume}\tag 50
$$
$$
|\partial_y^\gamma\theta(x,y)|\le C_\gamma R^{-\gamma}\ ,\ \text{supp}\
\theta(x,y)\subset \{|y|\le R\}\ , \theta(x,x)\equiv 0\ , R>10\ .
\tag 51
$$

We look separately at the cases $x>R$, $10<x\le R$, $|x|\le 10$, and
$x<-10$. Property (A3) from the section ``Review of Earlier Results" gives
$$\multline
\Big|\int_{-\infty}^{\infty}\theta(x,y)[A^2(y)-\frac 12 y_+^{-1/2}]\, dy\Big|
\le C^\prime R^{-5/2}\ ,\\
\text{with}\quad C^\prime\quad \text{depending\ only\ on}\quad
C_\gamma\quad\text{in\ (51)}\ .\endmultline\tag 52
$$

\noindent Since $\text{supp}\ \theta(x,y)\subset \{|y|\le R\}$,
estimate (52) shows that
$$
|F(x)|\le C^\prime R^{-5/2}\quad\text{for}\quad x>R\ ,\quad \text{and\ that}
\tag 53
$$
$$
F(x)=-\int_x^\infty \theta(x,y)[A^2(y)-\frac 12 y_+^{-1/2}]\, dy+O(R^{-5/2})
\quad\text{for}\quad 10<x\le R\ .\tag 54
$$

\noindent Recall that $A(y)=\text{Re}\, \Bigl[\frac{e^{\pm i\frac \pi 4}\,
e^{\frac 23 iy^{3/2}}}{y^{1/4}}\bigl(1\pm\frac{5i}{48}y^{-3/2}+O(y^{-3}))\Bigr]
\quad\text{for}\quad y>10$.  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
$$\multline
F(x)=\sum\limits_{k=\pm 1}c_k\int_x^\infty \theta(x,y)\frac{
e^{\frac 43 iky^{3/2}}}{y^{1/2}}\, dy
+\sum\limits_{k=\pm 1}\tilde c_k\int_x^\infty\theta(x,y)\frac{e^{\frac 43
iky^{3/2}}}{y^2}dy+O(R^{-1}x^{-3/2})\\
\text{for}\quad 10<x\le R\ .\endmultline\tag 55
$$
\noindent Let $q=\frac 12$ or $2$.

\noindent We use the elementary identity
$$\multline
\frac{\partial}{\partial y}\bigl[e^{\frac 43 iky^{3/2}}y^{-p}\partial_y^m
\theta(x,y)\big]=2ike^{\frac 43 iky^{3/2}}y^{\frac 12-p}\partial_y^m\theta
(x,y)\\
+e^{\frac 43iky^{3/2}}y^{-p}\partial_y^{m+1}\theta(x,y)-pe^{\frac 43 iky^{3/2}}
y^{-p-1}\partial_y^m\theta(x,y)\endmultline
$$
\noindent and induction on $M$ to show that
$$\multline
\frac{e^{\frac 43 iky^{3/2}}}{y^{q}}\theta(x,y)+\partial_y\Big[
\sum\Sb m\ge 0\\ p\ge 1\endSb a_{mp}^k e^{\frac 43 iky^{3/2}}y^{-p}
\partial_y^m\theta(x,y)\Bigr]\\
=\sum\Sb m\ge 0\\ p\ge M\endSb b_{mp}^k e^{\frac 43 iky^{3/2}} y^{-p}
\partial_y^m\theta(x,y)\quad (k=\pm 1)\endmultline\tag 56
$$
\noindent for a finite collection of coefficients $a_{mp}^k$ and $b_{mp}^k$
depending on $M$ and $q$.


We estimate the right-hand side, by recalling that $|\partial_y^m\theta(x,y)|
\le \hfill\break\cases CR^{-1} &\text{for $m\ge 1$}\\
CR^{-1}|x-y| &\text{for $m=0$} \endcases$.  Integrating (56), we
therefore find that
$$
\multline
\Big|\int_x^\infty \frac{e^{\frac 43 iky^{3/2}}}{y^{q}}\theta(x,y)dy-
\sum \Sb m\ge 0 \\ p\ge 1\endSb a_{mp}^k e^{\frac 43 ikx^{3/2}}x^{-p}
\partial_y^m\theta(x,y)\bigm|_{y=x}\Big|\\
\le \int_x^\infty \frac{C_M^\prime}{y^M}\bigl(R^{-1}+R^{-1}|x-y|\bigr)\, dy
\le C^\prime R^{-1}x^{-3/2}\quad \text{for}\quad 
M>100\ .\endmultline
$$

\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
$$
\Big|\int_x^\infty \frac{e^{\frac 43 iky^{3/2}}}{y^{q}}
\theta(x,y)dy\Big|\le C^\prime R^{-1}x^{-1}\quad\text{for}\quad
10<x\le R\ .\
$$

\noindent From (55) we get
$$
|F(x)|\le C^\prime R^{-1}x^{-1}\quad\text{for}\quad 10<x\le R\ .\tag 57
$$

Next we study $F(x)$ for $|x|<10$, where we write
$$
F(x)=\int_{-\infty}^{\min(x,0)}\theta(x,y)A^2(y)dy+\int_0^{\max(x,0)}
\theta(x,y)\bigl[A^2(y)-\frac 12 y^{-1/2}\bigr]\, dy\ .\tag 58
$$

\noindent Since $|\theta(x,y)|\le C^\prime R^{-1}|x-y|\le C^\prime 
R^{-1}(10+|y|)$ for $|x|\le 10$, while
$A^2(y)\le C_M(1+|y|)^{-M}$ for $y<0$, $|A^2(y)-\frac 12 y^{-1/2}|\le Cy^{-1/2}$
for $0<y\le 10$, (58) gives $|F(x)|\le
\int_{-\infty}^{\min(x,0)}C_M^\prime R^{-1}(10+|y|)\cdot
(1+|y|)^{-M}dy+\int_0^{\max(x,0)}C^\prime y^{-1/2}
R^{-1}(10+|y|)dy\le C^\prime R^{-1}$ for $|x|\le 10$.  Thus,
$$
|F(x)|\le C^\prime R^{-1}\quad\text{for}\quad |x|\le 10\ .\tag 59
$$

\noindent Similarly, for $-\infty<y<x\le -10$ we have $|\theta(x,y)|\le
C^\prime R^{-1}|x-y|$ and $A^2(y)\le C_m(1+|y|)^{-M}$, so
$$\multline
|F(x)|=\Big|\int_{-\infty}^{x}\theta(x,y)A^2(y)dy\Big|\le C_M^\prime
\int_{-\infty}^{x}R^{-1}|x-y|(1+|y|)^{-M}dy\\
\le C_m^\prime R^{-1}|x|^{-100}\\
\text{for\ large}\quad M\quad \text{and}\quad -\infty<x\le 10\ .\endmultline
$$
\noindent Thus,
$$
|F(x)|\le C^\prime R^{-1}|x|^{-100}\quad\text{for}\quad
-\infty<x\le -10\ .\tag 60
$$

\noindent Combining (53), (57), (59), (60) and throwing away information,
we get
$$
|F(x)|\le C^\prime R^{-1}(1+|x|)^{-1}\quad\text{for}\quad |x|\le R\tag 61
$$
$$
|F(x)|\le C^\prime R^{-5/2}\quad\text{for}\quad |x|\ge R\tag"(61 \text{bis})"
$$

\noindent We use (61) to control $\tilde \Cal F(\tilde x)$ in (45).

\noindent Take
$$\align
\theta_{\text{in}}(\tilde x,\tilde y)&=\frac{\Theta(\tilde y,\tilde x)}
{\lambda_{\text{left}}^{4/3}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)}\
\chi_{{}_{|\tilde x|<C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x}
{B_{\text{left}}}}}\\
\theta_{\text{out}}(\tilde x,\tilde y)&=\frac{\Theta(\tilde y,\tilde x)}
{\lambda_{\text{left}}^{4/3}}\ \chi_{{}_{|\tilde x|<c_{\#}\lambda_{\text{
left}}^{2/3}}}\\
R&=C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}>10\ .
\endalign
$$


\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
$$
|\tilde \Cal F(\tilde x)|\le C_\ast^\prime[\lambda_{\text{left}}^{4/3}]
\bigl[\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\Bigr]^{-5/2}
\quad\text{for}\quad C_{\#}\lambda_{\text{left}}^{2/3}\frac{\delta x}
{B_{\text{left}}}<|\tilde x|<c_{\#}\lambda_{\text{left}}^{2/3}\ .
$$

\noindent Substituting these estimates in (44) gives
$$
|\Cal F(x)|\le C_\ast(1+\lambda_{\text{left}}^{2/3}
|Y(x,0)|)^{-1}\quad\text{if}\quad |x-x_{\text{left}}(0)|<C_{\#}
(\delta x)\tag 62
$$
$$
|\Cal F(x)|\le C_\ast\lambda_{\text{left}}^{-1}\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-5/2}\quad\text{if}\quad C_{\#}
(\delta x)<|x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}\ .\tag 63
$$

\noindent Since $1+\lambda_{\text{left}}^{2/3}|Y(x,0)|\sim 1+
\lambda_{\text{left}}^{2/3}|x-x_{\text{left}}(0)|B_{\text{left}}^{-1}$, (62)
and (63) imply
$$
\int_{|x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}}|\Cal F(x)|^2\, dx\le
C_\ast\lambda_{\text{left}}^{-2/3}B_{\text{left}}+C_\ast\lambda_{\text{left}}
^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}B_{\text{left}}\ .
\tag 64
$$

\noindent We proved (64) assuming $|E_0|<2\hat c(\delta E)$. Since
$\{|x-x_{\text{left}}(E_0)|<c_{\#}^\prime B_{\text{left}}\}\subset
\{|x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}\}$ when $|E_0|<2\hat c
(\delta E)$, estimate (64) gives
$$\multline
\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}\ |\Cal F(x)|^2dx\le
C_\ast\Bigl(\lambda_{\text{left}}^{-2/3}+\lambda_{\text{left}}^{-2}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\Bigr)B_{\text{left}}
\\
\text{if}\quad |E_0|<2\hat c(\delta E)\ .\endmultline\tag 65
$$

\noindent On the other hand, suppose $|E_0|\ge 2\hat c(\delta E)$.  Then as in
(31) we set $y=Y(x^\prime,E)$ and $h(y,x^\prime)=g(x^\prime,E)\bigl(
\frac{\partial Y}{\partial x^\prime}(x^\prime,E)\bigr)^{-1}\bigl(\frac
{\partial Y}{\partial E}(x^\prime,E)\bigr)^{-1}$.  This time, since
$g(x^\prime,E)$ is supported in $\{|E-E_0|<\hat c(\delta E)\}$, we have
$$\align
\Cal F(x)&=\int_{-\infty}^x\int_{-\infty}^\infty h(y,x^\prime)
\lambda_{\text{left}}^{-2/3}[A^2(\lambda_{\text{left}}^{2/3}y)-\frac 12
(\lambda_{\text{left}}^{2/3}y)_+^{-1/2}]\, dydx^\prime\\
&=\int_{-\infty}^\infty \bigl[\lambda_{\text{left}}^{-4/3}\int_{-\infty}^x
h(y,x^\prime)dx^\prime\bigr]\cdot \bigl[A^2(\lambda_{\text{left}}^{2/3}
y)-\frac 12(\lambda_{\text{left}}^{2/3}y)_+^{-1/2}\bigr]
\lambda_{\text{left}}^{2/3}\, dy\ . \tag 66\endalign
$$

\noindent The proof of estimate (35) is still valid (with trivial changes).
Also, $\text{supp}\
h(y,x^\prime)\hfill\break
\subset \{|x^\prime-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}},
|y|<C_{\#}\frac{\delta x}{B_{\text{left}}}\}$, since $g(x^\prime,E)$ appears
in $h$ as a factor and $y=Y(x^\prime,E)$.  Hence for $|x-x_{\text{left}}
(E_0)|<c_{\#}B_{\text{left}}$, we have
$$\align
\big|\partial_y^\gamma\bigl[\lambda_{\text{left}}^{-4/3}\int_{-\infty}^x
h(y,x^\prime)dx^\prime\bigr]\big|&\le C_\ast B_{\text{left}}\cdot
\lambda_{\text{left}}^{-4/3}S_{\text{left}}B_{\text{left}}\bigl(\frac{
\delta x}{B_{\text{left}}}\bigr)^{-\gamma}\\
&=C_\ast\lambda_{\text{left}}^{2/3}\bigl(\frac{\delta x}{B_{\text{left}}}
\bigr)^{-\gamma}\ , \endalign
$$
\noindent i.e.
$$\multline
|\partial_{\tilde y}^\gamma H(\tilde y,x)|\le C_\ast\lambda_{\text{left}}
^{2/3}\bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}
\bigr)^{-\gamma}\quad\text{for}\quad |x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}\ ,\\
\text{with}\quad H(\tilde y,x)=\lambda_{\text{left}}^{-4/3}
\int_{-\infty}^xh(\lambda_{\text{left}}^{-2/3}\tilde y,x^\prime)dx^\prime\ ,
\endmultline
\tag 67
$$

\noindent so that
$$
\Cal F(x)=\int_{-\infty}^\infty H(\tilde y,x)[A^2(\tilde y)-\frac 12 \tilde
y_+^{-1/2}]d\tilde y\quad\text{by\ (66)}\ ;\quad\text{and}\tag 68
$$
$$
\text{supp}\ H(\tilde y,x)\subset \{|\tilde y|<C_{\#}
\lambda_{\text{left}}^{+2/3}\frac{\delta x}{B_{\text{left}}}\}\ .\tag 69
$$

As above, we may apply (A3) from the section ``Review of Earlier Results"
to deduce from (67), (68), (69) that
$|\Cal F(x)|\le\hfill\break
C_\ast\lambda_{\text{left}}^{2/3}\cdot \bigl(\lambda_
{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\bigr)^{-5/2}$
for $|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}$, i.e.
$|\Cal F(x)|\le C_\ast\lambda_{\text{left}}^{-1}\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-5/2}$ for $|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}$, and thus
$$\multline
\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}\ |\Cal F(x)|^2dx\le
C_\ast\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}
\bigr)^{-5}B_{\text{left}}\\
\text{if}\quad |E_0|\ge 2\hat c(\delta E)\ .\endmultline\tag 70
$$

Next we suppose $|E_0|<2\hat c(\delta E)$ and estimate $\Cal G(x)$
in (30).  By definition,
$$\align
\Cal G(x)&=\int_{-\infty}^x g(x^\prime,0)\lambda_{\text{left}}^{-4/3}
\bigl(\frac{\partial Y(x^\prime,0)}{\partial x^\prime}\bigr)^{-2}
\bigl[A^2(\lambda_{\text{left}}^{2/3}Y(x^\prime,0))-\frac 12
(\lambda_{\text{left}}^{2/3}Y(x^\prime,0))_+^{-1/2}\bigr]\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\cdot\lambda_{\text{left}}^{2/3}\bigl(\frac{\partial Y}{\partial x^\prime}
(x^\prime,0)\bigr)\, dx^\prime\\
&=\int_{-\infty}^{\lambda_{\text{left}}^{2/3}Y(x,0)}\tilde g(\xi)
\bigl[A^2(\xi)-\frac 12 \xi_+^{-1/2}\bigr]\, d\xi\  \text{for}\quad
|x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}\\
&\text{with}\quad \tilde g(\xi)=\lambda_{\text{left}}^{-4/3}g(x^\prime,0)
\bigl(\frac{\partial Y(x^\prime,0)}{\partial x^\prime}\bigr)^{-2}\quad
\text{for}\quad \xi=\lambda_{\text{left}}^{2/3}Y(x^\prime,0)\ .\tag 71\endalign
$$

We estimate the derivatives of $\tilde g(\xi)$.  Since
$|\partial_{x^\prime}^\alpha g(x^\prime,0)|\le
C_\ast^\alpha(\delta x)^{-\alpha}$ and\hfill\break
$|\partial_{x^\prime}^\alpha\bigl(\frac{\partial Y(x^\prime,0)}{\partial 
x^\prime}\bigr)^{-2}|\le C_{\#}B_{\text{left}}^2\cdot 
B_{\text{left}}^{-\alpha}$, we have
$$
\Big|\partial_{x^\prime}^\alpha \{g(x^\prime,0)\bigl(\frac{\partial Y}
{\partial x^\prime}(x^\prime,0)\bigr)^{-2}\}\Big|\le C_\ast^\alpha
B_{\text{left}}^{2}(\delta x)^{-\alpha}\ .\tag 72
$$

\noindent On the other hand, $\big|\partial_{x^\prime}^\alpha
\bigl[\lambda_{\text{left}}^{2/3}Y(x^\prime,0)\bigr]\big|\le C_{\#}^\alpha
\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-\alpha}$ and
$\partial_{x^\prime}[\lambda_{\text{left}}^{2/3}Y(x^\prime,0)]>c_{\#}
\lambda_{\text{left}}^{2/3}B_{\text{left}}^{-1}$, 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
$$\multline
G(x)=\int_{-\infty}^x\theta(\xi)[A^2(\xi)-\frac 12\xi_+^{-1/2}]\, d\xi\ ,\quad
\text{with}\quad \big|\bigl(\frac{d}{d\xi}\bigr)^m\theta\big|\le C_mR^{-m}\ ,\\
\text{supp}\ \theta\subset \{|\xi|<R\}\ , R>10\ .\endmultline\tag 76
$$

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
$$
\Cal G(x)=\lambda_{\text{left}}^{-4/3}B_{\text{left}}^2G(\lambda_{\text{left}}
^{2/3}Y(x,0))\quad\text{for}\quad |x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}\ .\tag 77
$$

\noindent We study $G(x)$ separately in the cases $x>R$, $10<x\le R$,
$|x|\le 10$, $x<-10$.  For $x>R$ 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 
$10<x<R$ we get from (A3) that
$$
G(x)=O(R^{-5/2})-\int_x^\infty\theta(\xi)[A^2(\xi)-\frac 12 \xi_+^{-1/2}]\,
d\xi\ .\tag 78
$$

\noindent Since $[A^2(\xi)-\frac 12 \xi_+^{-1/2}]=\text{Re}\, \bigl[
\frac{e^{\pm i\frac \pi 2}\, e^{\frac 43 i\xi^{3/2}}}{2\xi^{1/2}}\bigr]
+\ \text{Re}\ \bigl[\frac{\text{(const)}e^{\frac 43 i\xi^{3/2}}}{\xi^2}\bigr]
+O(\xi^{-7/2})$ as we saw before, it follows from (78) that
$$\multline
G(x)=O(x^{-5/2})+\sum\limits_{k=\pm 1}c_k\int_x^\infty \theta(\xi)
\frac{e^{\frac 43 ik\xi^{3/2}}}{\xi^{1/2}}\, d\xi\\
+\sum\limits_{k=\pm 1}\tilde c_k\int_x^\infty\theta(\xi)
\frac{e^{\frac 43 ik\xi^{3/2}d\xi}}{\xi^2}\, d\xi\  .\endmultline
\tag"(78\text{bis})"
$$

\noindent The proof of (56) gives here
$$\multline
\frac{e^{\frac 43 ik\xi^{3/2}}}{\xi^{q}}\theta(\xi)+\frac{d}{d\xi}
\Bigl[\sum\Sb m\ge 0\\ p\ge 1\endSb a_{kmp}e^{\frac 43 ik\xi^{3/2}}
\Bigl[\bigl(\frac{d}{d\xi}\bigr)^m\theta(\xi)\Bigr]\xi^{-p}\Big]\\
=\sum\Sb p\ge M\\ m\ge 0\endSb b_{kmp}e^{\frac 43ik\xi^{3/2}}\xi^{-p}
\bigl(\frac{d}{d\xi}\bigr)^m\theta(\xi)\quad (k=\pm 1)\ ,\\
\text{with}\quad q=\frac 12\ \text{or}\ 2\ . \endmultline\tag 79
$$

\noindent The right-hand side is $O(\xi^{-M})$, so integrating (79)
over $[x,\infty)$ yields
$$\multline
\int_x^\infty \frac{e^{\frac 43 ik\xi^{3/2}}}{\xi^{q}}\theta(\xi)d\xi-
\sum\Sb m\ge 0\\ p\ge 1\endSb a_{kmp}e^{\frac 43 ikx^{3/2}}x^{-p}
\theta^{(m)}(x)\\
=O(x^{1-M})\quad (10<x<R)\ .\endmultline
$$

\noindent In particular,
$$
\Big|\int_x^\infty \frac{e^{\frac 43 ik\xi^{3/2}}}{\xi^{q}}
\theta(\xi)d\xi\Big|\le C\, x^{-1}\quad\text{for}\quad 10<x<R\ ,
$$
\noindent so (78 bis) implies
$$
|G(x)|\le Cx^{-1}\quad\text{for}\quad 10<x<R\ .
\tag 80
$$

\noindent For $|x|\le 10$ we have
$$\align
|G(x)|&\le \Big|\int_{\infty}^{\min(0,x)}\theta(\xi)A^2(\xi)d\xi\Big|+
\Big|\int_0^{\max(0,x)}\theta(\xi)[A^2(\xi)-\frac 12\xi^{-1/2}]\, d\xi\Big|\\
&\le C\int_{-\infty}^0 A^2(\xi)d\xi +C\int_0^{10}|A^2(\xi)-\frac 12
\xi^{-1/2}|d\xi\le C^\prime\ .\endalign
$$

\noindent For $x<-10$ and $-\infty<\xi<x$ we have $|A^2(\xi)|\le
\frac{C_M}{|\xi|^M}$, so $|G(x)|\le \int_{-\infty}^x |\theta(\xi)|
\frac{C_M}{|\xi|^M}d\xi\hfill\break\le \frac{C_M}{|x|^{M-1}}$. 
Combining our results
for the various cases, we get $|G(x)|\le \frac{C}{1+|x|}+CR^{-5/2}$.
Using this in (77) gives
$$\multline
|\Cal G(x)|\le C_\ast\frac{\lambda_{\text{left}}^{-4/3}B_{\text{left}}^2}
{1+\lambda_{\text{left}}^{2/3}|Y(x,0)|}+C_\ast\lambda_{\text{left}}^{-4/3}
B_{\text{left}}^2\bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}
{B_{\text{left}}}\bigr)^{-5/2}\\
\text{for}\quad |x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}\ .\endmultline
\tag 81
$$

\noindent Also $\phi^\prime(0)=\frac 12 \int_{V<0}(-V)^{-1/2}\ge c_{\#}
S_{\text{left}}^{-1/2}B_{\text{left}}$, so
$$\multline
|\Cal G(x)[\phi^\prime(0)]^{-1}|\le C_\ast\lambda_{\text{left}}^{-4/3}
B_{\text{left}}^2\cdot C_{\#}S_{\text{left}}^{1/2}B_{\text{left}}^{-1}
\cdot \Bigl[\frac{1}{1+\lambda_{\text{left}}^{2/3}|Y(x,0)|}+
\bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}\bigr)^{-5/2}
\Bigr]\\
=C_\ast\lambda_{\text{left}}^{-1/3}\cdot \Bigl[\frac{1}{1+\lambda_{\text{left}}
^{2/3}|Y(x,0)|}+\bigl(\lambda_{\text{left}}^{2/3}\frac{\delta x}
{B_{\text{left}}}\bigr)^{-5/2}\Bigr]\\
\text{for}\quad |x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}\ .\endmultline
$$

\noindent Hence
$$\multline
\int_{|x-x_{\text{left}}(0)|<c_{\#}B_{\text{left}}}|\Cal G(x)[\phi^\prime
(0)]^{-1}|^2\, dx\\
\le C_\ast\lambda_{\text{left}}^{-2/3}B_{\text{left}}
\bigl[\lambda_{\text{left}}^{-2/3}+(\lambda_{\text{left}}^{2/3}
\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\bigr]\ ,\endmultline\tag 82
$$

\noindent
again using $1+\lambda_{\text{left}}^{2/3}|Y(x,0)|\sim 1+\lambda_{\text{left}}
^{2/3}|x-x_{\text{left}}(0)|B_{\text{left}}^{-1}$.  Since
$\lambda_{\text{left}}^{2/3}\frac{\delta x}{B_{\text{left}}}>10$, we can
throw away information from (82) to get
$$\multline
\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}|\Cal G(x)[\phi^\prime
(0)]^{-1}|^2 dx\le C_\ast\lambda_{\text{left}}^{-2/3}B_{\text{left}}\\
\text{for}\quad |E_0|<2\hat c(\delta E)\ .\endmultline \tag 83
$$

\noindent Again we have used the inclusion $\{|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}\}\subset \{|x-x_{\text{left}}(0)|<c_{\#}^\prime 
B_{\text{left}}\}$ for $|E_0|<2\hat c(\delta E)$.

The information we just discarded is useless, since (83) will be combined
with (65). Note that
$$
\text{If}\quad |E_0|\ge 2\hat c(\delta E)\ ,\quad\text{then}\quad
g(x^\prime,0)\equiv 0\quad \text{and\ so}\quad \Cal G(x)=0\ .\tag"(83\text{bis})"
$$

We are ready to combine our results on $\Cal F(x)$, $\Cal G(x)$ with
estimates (25)$\ldots$(28) on $\rho(x,g)$.  From (25), (29), (30) we get
$$\multline
\rho(x,g)=\frac{1}{2\pi}\int_{-\infty}^0g(x,E)\bigl(\lambda_{\text{left}}^2
\bigl(\frac{\partial Y}{\partial x}(x,E)\bigr)^2Y(x,E)\bigr)_+^{-1/2}\, dE\\
-\frac 12 g(x,0)\bigl(\lambda_{\text{left}}^2\bigl(\frac{\partial Y(x,0)}
{\partial x}\bigr)^2Y(x,0)\bigr)_+^{-1/2}[\phi^\prime(0)]^{-1}\chm
(\frac 1\pi \phi(0)-\frac 12 )\\
+\frac {d}{dx}H(x)\endmultline\tag 84
$$

\noindent with
$$
H(x)=\frac 1\pi \Cal F(x)-\chm\bigl(\frac 1\pi\phi(0)-\frac 12\bigr)
\Cal G(x)[\phi^\prime(0)]^{-1}+\int_{\text{inf}\ I_{\text{BVP}}}^x\Cal E_3
(x^\prime)dx^\prime \ .\tag 85
$$

\noindent Note that $\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
\big|\int_{\text{inf}\ I_{\text{BVP}}}^x\Cal E_3(x^\prime)dx^\prime\big|^2dx
\le B_{\text{left}}\bigl(\int_{I_{\text{BVP}}}|\Cal E_3(x^\prime)|dx^\prime
\bigr)^2$.  Hence (65), (70), (83), (83 bis) and (85) show the following.
$$
\multline
\text{If}\quad |E_0|< 2\hat c(\delta E)\ ,\quad\text{then}\\
\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}|H(x)|^2dx\le
C_\ast\bigl(\lambda_{\text{left}}^{-2/3}+\lambda_{\text{left}}^{-2}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\bigr)B_{\text{left}}\\
+C_\ast
\bigl(\int_{I_{\text{BVP}}}|\Cal E_3(x^\prime)|dx^\prime\bigr)^2B_{\text{left}}
\ .\endmultline \tag 86
$$
$$\multline
\text{If}\quad |E_0|\ge2\hat c(\delta E)\ ,\quad\text{then}\\
\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}|H(x)|^2dx\le C_\ast
\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}
B_{\text{left}}\\
+C_\ast
\bigl(\int_{I_{\text{BVP}}}|\Cal E_3(x^\prime)|dx^\prime\bigr)^2B_{\text{left}}
\ .\endmultline \tag 87
$$

Combining (86), (87) with (26), (27), (28), we get the following
estimates for $H(x)$.
$$\multline
\text{If}\quad |E_0|>2\hat c(\delta E)\ \text{then}\\
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}}|H(x)|^2dx\le  C_\ast\lambda_{\text{left}}^{-2}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}+C_\ast
\Lambda_{\min}^{8\varepsilon-4}\lambda_{\text{left}}^2\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)\\
+C_\ast\Lambda_{\min}^{20-2N^{\prime\prime}}(\phi(\text{max}\ \Cal J))^2
+C_\ast\Lambda_{\min}^{4-2N^{\prime\prime}}\ .\endmultline\tag 88
$$
\newpage

$$\multline
\text{If}\quad |E_0|\le 2\hat c(\delta E)\ \quad\text{and}\quad
\min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|>\overline C_{\#}
\Lambda_{\min}^{-1}\ , \quad\text{then}\\
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}}|H(x)|^2dx\le C_\ast\lambda_{\text{left}}^{-2/3}+
C_\ast\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)
^{-5}\\
+C_\ast\Lambda_{\min}^{8\varepsilon-4}\lambda_{\text{left}}^2\bigl(\frac
{\delta x}{B_{\text{left}}}\bigr)+C_\ast\Lambda_{\min}^{-2}
\lambda_{\text{left}}^2\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^2\\
+C_\ast\Lambda_{\min}^{20-2N^{\prime\prime}}(\phi(\text{max}\ \Cal J))^2
+C_\ast\Lambda_{\min}^{4-2N^{\prime\prime}}\ .\endmultline \tag 89
$$
$$\multline
\text{If}\quad |E_0|\le 2\hat c(\delta E)\quad\text{and}\quad
\min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}
\Lambda_{\min}^{-1}\ , \quad\text{then}\\
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
|H(x)|^2dx\le C_\ast\lambda_{\text{left}}^{-2/3}+C_\ast\lambda_{\text{left}}
^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\\
+C_\ast\Lambda_{\min}^{-2}\lambda_{\text{left}}^2\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)
+C_\ast\Lambda_{\min}^{20-2N^{\prime\prime}}(\phi(\text{max}\ \Cal J))^2
+C_\ast\Lambda_{\min}^{4-2N^{\prime\prime}}\ .\endmultline \tag 90
$$

If in (84) we could replace $\lambda_{\text{left}}^2(\frac{\partial Y}
{\partial x})^2Y(x,E)$ by $E-V(x)$, then we would have
$\rho(x,g)=\rho_{sc}(x,g)+\frac{dH}{dx}$, with estimates (88)$\ldots$(90)
for $H$.  On the other hand, equation (7) shows that $\lambda_{\text{left}}^2
(\frac{\partial Y}{\partial x})^2Y(x,E)$ is close to $E-V(x)$, so we
proceed as follows.

\medskip
\proclaim{Lemma 4}  Let $X(t,\tau)$ be smooth on $\{|t|,|\tau|\le 1\}$,
with $\frac{\partial X}{\partial t}(t,0)>c>0$ 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 $F(s,\tau)=\break
\int_{\{0<\xi<X(s,\tau)\}}\varphi(T(\xi,\tau))
\xi^{-1/2}\bigl(\frac{\partial T(\xi,\tau)}{\partial\xi}\bigr)d\xi$.  Hence

\newpage

$$\multline
|F(s,\tau)-F(s,0)|\\
\le \int_{\min\{X(s,\tau),X(s,0)\}<\xi<\max\{X
(s,\tau),X(s,0)\}}|\varphi(T(\xi,\tau))|\ |\xi|^{-1/2}\ |\frac{\partial T
(\xi,\tau)}{\partial\xi}|\, d\xi\\
+\int_{0<\xi<X(s,0)}|\varphi(T(\xi,\tau))-\varphi(T(\xi,0))|\xi^{-1/2}
|\frac{\partial T(\xi,\tau)}{\partial\xi}|\, d\xi\\
+\int_{0<\xi<X(s,0)}|\varphi(T(\xi,0))|\xi^{-1/2}|\frac{\partial T(\xi,\tau)}
{\partial\xi}-\frac{\partial T(\xi,0)}{\partial\xi}|\, d\xi
\equiv A_1+A_2+A_3\ .\endmultline
$$

\noindent In $A_1$, the region of integration is contained in $\{|\xi-X(s,0)|
<C|\tau|\}$, and the integrand is dominated by $C|\xi|^{-1/2}$.  Hence
$A_1\le\frac{C|\tau|}{(|X(s,0)|+|\tau|)^{1/2}}$.


To handle $A_2$, note that $\xi\mapsto \varphi(T(\xi,\tau))$ and $\xi\mapsto
\varphi(T(\xi,0))$ are supported in intervals of length $\le C\delta$,
and  that $|\varphi^\prime|<C\delta^{-1}$, while $|T(\xi,\tau)
-T(\xi,0)|\le C\tau$. Hence
the integrand in $A_2$ is at most $C\delta^{-1}|\tau||\xi|^{-1/2}$, and the 
support of integrand is contained in a union of two intervals of length
$\le C\delta$.  Therefore, $A_2\le\hfill\break 
C\delta^{-1}|\tau|\max_{|J|<C\delta}
\int_J|\xi|^{-1/2}d\xi$, i.e. $A_2\le C\delta^{-1/2}|\tau|$.

In $A_3$, the integrand is dominated by $C|\tau|\xi^{-1/2}$, and the
region of integration is contained in $(0,C)$.  Thus, $A_3\le C|\tau|$.
Combining our estimates for $A_1$, $A_2$, $A_3$, we get $|F(s,\tau)-F(s,0)|
\le C|\tau|(|X(s,0)|+|\tau|)^{-1/2}+C\delta^{-1/2}|\tau|$ for $|s|\le 1$.
Since $\frac{\partial}{\partial s}X(s,0)>c>0$ 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
$$\multline
\frac{1}{B_{\text{left}}}\int_{|x^\prime-x_{\text{left}}(E)|<B_{\text{left}}}
\Big|\int_{-\infty}^{x^\prime}g(x,E)(\lambda_{\text{left}}^2(\partial_x
Y)^2Y)_+^{-1/2}dx\\
-\int_{-\infty}^{x^\prime}g(x,E)(\lambda_{\text{left}}^2(\partial_xY)^2Y+\{Y,x\})_+^{-1/2}dx\Big|^2dx^\prime\\
\le C_\ast B_{\text{left}}^{+4}\cdot \lambda_{\text{left}}^{-2}\cdot
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1}\lambda_{\text{left}}^{-4}
\quad\text{for}\quad |E-E_0|<2\hat c(\delta E)\ .\endmultline\tag 93
$$

\noindent In (93) we may shrink the region of the integration from
$\{|x^\prime-x_{\text{left}}(E)|<B_{\text{left}}\}$ to $\{|x^\prime-
x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}\}$.  Since $g(x,E)$
is supported in $\{|E-E_0|\le \hat c(\delta E)\}$, it follows from (93)
and the triangle inequality that
$$\multline
\frac{1}{2\pi}\int_{-\infty}^{x^\prime}
\int_{-\infty}^0 g(x,E)(\lambda_{\text{left}}^2(\partial_x
Y)Y)_+^{-1/2}dEdx\\
=\frac{1}{2\pi}\int_{-\infty}^{x^\prime}\int_{-\infty}^0g(x,E)\bigl(
\lambda_{\text{left}}^2(\partial_xY)^2Y+\{Y,x\}\bigr)_+^{-1/2}dEdx+\ \text{
Error}\,(x^\prime)\endmultline\tag 94
$$

\noindent with
$$\multline
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<
c_{\#}B_{\text{left}}}|\, \text{Error}(x)|^2dx\le
C_\ast B_{\text{left}}^4\lambda_{\text{left}}^{-6}\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^{-1}(\delta E)^2\\
\le C_\ast S_{\text{left}}^2B_{\text{left}}^4\lambda_{\text{left}}^{-6}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1}=C_\ast\lambda_{\text{left}}
^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1}\ .\endmultline \tag 95
$$

\noindent Also if $|E_0|\le 2\hat c(\delta E)$, then (93) gives
$$\multline
\frac 12 \int_{-\infty}^{x^\prime}g(x,0)\bigl(\lambda_{\text{left}}^2
\bigl(\frac{\partial}{\partial x}Y(x,0)\bigr)^2Y(x,0)\bigr)_+^{-1/2}
dx[\phi^\prime(0)]^{-1}\\
=\frac 12 \int_{-\infty}^{x^\prime}g(x,0)\bigl(\lambda_{\text{left}}^2
\bigl(\frac{\partial}{\partial x}Y(x,0)\bigr)^2Y(x,0)+
\{Y(x,0),x\}\bigr)_+^{-1/2}[\phi^\prime(0)]^{-1}dx+\\
\text{Error}^\prime(x^\prime)\ ,
\endmultline\tag 96
$$

\noindent with
$$\multline
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}}|\,\text{Error}^\prime(x)|^2dx\le C_\ast B_{\text{left}}^4
\lambda_{\text{left}}^{-6}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1}
[\phi^\prime(0)]^{-2}\\
\le C_\ast B_{\text{left}}^4\lambda_{\text{left}}^{-6}\bigl(\frac{
\delta x}{B_{\text{left}}}\bigr)^{-1}\cdot S_{\text{left}}B_{\text{left}}^{-2}
\quad\text{(by\ (19 bis))}\\
=C_\ast S_{\text{left}}B_{\text{left}}^2\lambda_{\text{left}}^{-6}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1}=C_\ast\lambda_{\text{left}}
^{-4}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-1}\ .\endmultline
\tag 97
$$

\noindent If $|E_0|>2\hat c(\delta E)$, 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,
$$\multline
\rho(x,g)=\frac{1}{2\pi}\int_{-\infty}^0 g(x,E)\bigl(\lambda_{\text{left}}^2
\bigl(\frac{\partial Y(x,E)}{\partial x}\bigr)^2Y(x,E)+\{Y(x,E),x\}\bigr)_+
^{-1/2}dE\\
-\frac 12 g(x,0)\bigl(\lambda_{\text{left}}^2\bigl(\frac{\partial Y(x,0)}
{\partial x}\bigr)^2Y(x,0)+\{Y(x,0),x\}\bigr)_+^{-1/2}[\phi^\prime(0)]^{-1}
\chm(\frac 1\pi \phi(0)-\frac 12)\\
+\frac{d}{dx}\Cal H(x)\ , \quad\text{with}\quad \Cal H(x)\quad
\text{satisfying\ the\ following}\ .\endmultline \tag 98
$$
$$\multline
\text{If}\quad |E_0|>2\hat c(\delta E)\ , \text{then}\
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}
B_{\text{left}}}|\Cal H(x)|^2dx
\le C_\ast\lambda_{\text{left}}^{-2}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\\
+C_\ast\Lambda_{\min}
^{8\varepsilon-4}\lambda_{\text{left}}^2\bigl(\frac{\delta x}{B_{\text{left}}}
\bigr)\\
+C_\ast\Lambda_{\min}^{20-2N^{\prime\prime}}(\phi(\text{max}\ \Cal J))^2+C_\ast
\Lambda_{\min}^{4-2N^{\prime\prime}}\ .\endmultline\tag 99
$$
$$\multline
\text{If}\quad |E_0|\le 2\hat c(\delta E)\quad\text{and}\quad
\min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|>\overline C_{\#}\Lambda_{\min}^{-1}\ ,
\ \text{then}\\
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
|\Cal H(x)|^2dx\le C_\ast\lambda_{\text{left}}^{-2/3}+C_\ast\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\\
+C_\ast\Lambda_{\min}^{8\varepsilon-4}\lambda_{\text{left}}^{2}
\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)\\
+C_\ast\Lambda_{\min}^{-2}\lambda_{\text{left}}^2\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)^2+C_\ast\Lambda_{\min}^{20-2N^{\prime\prime}}
(\phi(\text{max}\ \Cal J))^2+C_\ast\Lambda_{\min}^{4-2N^{\prime\prime}}\ .
\endmultline\tag 100
$$
$$\multline
\text{If}\quad |E_0|\le 2\hat c(\delta E)\quad\text{and}\quad 
\min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline C_{\#}
\Lambda_{\min}^{-1}\ , \quad\text{then}\\
\frac{1}{B_{\text{left}}}\int_{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}}
|\Cal H(x)|^2dx\le C_\ast\lambda_{\text{left}}^{-2/3}+C_\ast
\lambda_{\text{left}}^{-2}\bigl(\frac{\delta x}{B_{\text{left}}}\bigr)^{-5}\\
+C_\ast\Lambda_{\min}^{-2}\lambda_{\text{left}}^2\bigl(\frac{\delta x}
{B_{\text{left}}}\bigr)
+C_\ast\Lambda_{\min}^{20-2N^{\prime\prime}}(\phi(\text{max}\ \Cal J))^2\\
+C_\ast\Lambda_{\min}^{4-2N^{\prime\prime}}\ .\endmultline\tag 101
$$

\noindent The terms other than $\frac{d\Cal H}{dx}$ on the right in (98)
are very close to the semiclassical density $\rho_{sc}(x,g)$, by
virtue of (7).  In fact (7) shows that
$$\multline
\lambda_{\text{left}}^2\bigl(\frac{\partial Y}{\partial x}\bigr)^2Y+
\{Y,x\}=E-V(x)+f(x,E)\quad\text{with}\quad |f(x,E)|\le
C_{\#}\lambda_{\text{left}}^{-N^{\prime}}S_{\text{left}}\\
\text{in\ supp}\ g(x,E)\ .\endmultline
$$

\noindent Since also $|E-V(x)|\sim \frac{S_{\text{left}}}{B_{\text{left}}}
|x-x_{\text{left}}(E)|$ in $\text{supp}\ g(x,E)$, and since
(8), (9) yield 
$|E-V(x)+f(x,E)|\sim \frac{S_{\text{left}}}{B_{\text{left}}}|
x-x_{\text{perturbed}}(E)|$ in $\text{supp}\ g$ for some
$x_{\text{perturbed}}(E)$ 
we have for fixed $E$ (satisfying $|E-E_0|<2\hat c\delta E$) the following.
$$\multline
\int_{x\in I_{\text{BVP}}}|g(x,E)|\, |\bigl(E-V(x)+f(x,E))_+^{-1/2}-(E-V(x)
\bigr)_+^{-1/2}|\, dx\\
\le \int_{(\delta x)>|x-x_{\text{left}}(E)|>\lambda_{\text{left}}
^{-\frac 23 N^\prime}B_{\text{left}}}C_\ast\lambda_{\text{left}}^{-\frac 13
N^\prime}\bigl(\frac{S_{\text{left}}}{B_{\text{left}}}|x-x_{\text{left}}
(E)|\bigr)^{-1/2}\, dx\\
+\int_{|x-x_{\text{left}}(E)|<\lambda_{\text{left}}^{-\frac 23 N^\prime}
B_{\text{left}}}\Bigl\{C_\ast\bigl(\frac{S_{\text{left}}}{B_{\text{left}}}
|x-x_{\text{left}}(E)|\bigr)^{-1/2}\\
+C_\ast\bigl(\frac{S_{\text{left}}}
{B_{\text{left}}}|x-x_{\text{perturbed}}(E)|\bigr)^{-1/2}\Bigr\}dx\\
\le C_\ast\lambda_{\text{left}}^{-\frac 13 N^\prime}S_{\text{left}}^{-1/2}
B_{\text{left}}\ .\endmultline\tag 102
$$

\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 $I_{\text{left}}=\{|x-x_{\text{left}}(E_0)|<c_{\#}B_{\text{left}}\}$.

\noindent (A) If $|E_0|>2\hat c(\delta E)$, 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
$|\tilde x_0-x_0|<cB(x_0)$ and $E_{\ell o}-V(\tilde x_0)>c_3S(\tilde x_0)$.

\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 $E>0$;
$$
\hat c\tau S(x_0)<E-V(x)<\hat C\tau S(x_0)\ .
$$
\noindent Moreover, $\hat c$, $\hat c_1>0$ 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 $G=\{(x,E)\mid E_{\ell o}+\hat c_1S_{\min}
(E_{\ell o})\le E\le E_{hi}$, $|x-x_0|<\hat cB(x_0)$,
$\hat c\tau S(x_0)<E-V(x)<\hat C\tau S(x_0)\}$ is non-empty.  For, if
$G$ were empty, then by (A3) we would have $g(x,E)=0$ for $E\le 0$,
so that the results
of this section on $\rho(x,g)-\rho_{sc}(x,g)$ would hold trivially.

If $G\ne \emptyset$, then one of the following must hold:

(i) $\tau \ge c_\ast>0$.

(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 $c_{\ast}<A/B<C_{\ast}$.

\noindent We give a discussion that applies to cases (i) and (ii).  Case (iii)
may be reduced to case (ii) by just studying $V(-x)$ in place of $V(x)$.

Next we recall what we have learned about the eigenvalues and
eigenfunctions of $-\frac{d^2}{dx^2}+V(x)$. Applying for each
$E_0\in[E_{\ell o},E_{hi}]$ the WKB Eigenfunction Theorem and
the lemma on phases
in the section on the WKB Theorems, as well as the Corollary to the
Reformulated Eigenvalue Theorem, we obtain the following results.

The eigenvalues in $[E_{\ell o}, E_{hi}]$ may be written as
$E_{k_{\ell o}},\ldots,E_{k_{hi}}$ with
$$
|\phi(E_k)-\pi (k+1/2)|\le C_{\#}\Lambda_{\min}^{-1}\quad \text{for}\quad
k_{\ell o}\le k\le k_{hi}\tag 1
$$
$$
\{k\mid k_{\ell o}\le k\le k_{hi}\}=\Bbb Z \cap [a,b],\quad\text{where}
\tag 2
$$
$$
a=a_0+\omega_{\ell o},\ b=b_0+\omega_{hi}\ ; \quad |\omega_{\ell o}|,
|\omega_{hi}|\le C_{\#}\Lambda_{\min}^{-1}\tag 3
$$
$$
a_0=\frac 1\pi \phi(E_{\ell o})-\frac 12, \quad b_0=\frac 1\pi
\phi(E_{hi})-\frac 12 \ . \tag 4
$$

The (normalized) eigenfunctions $u_k(x)$ corresponding to the $E_k$
may be written as
$$\split
u_k(x)=b_k\ \text{Re}\Bigl[\frac{e^{i\eta(x,E_k)}}{(E_k-V(x))^{1/4}}
\ (1+w_k(x))\Bigr]+\varepsilon_k(x)\\
\text{for}\ (x,E_k)\in \text{supp}\ g\ ,\endsplit\tag 5
$$

\noindent with
$$
\eta(x,E)=\pm \frac \pi 4+\int\limits_{x_{\text{left}}(E)}^x(E-V(t))^{1/2}\ dt
\tag 6
$$
$$\split
|w_k(x)|\le C_{\ast}\Lambda_{\min}^{-1}\tau^{-3/2},\ |\text{Re}\ w_k(x)|
\le C_{\ast}\Lambda_{\min}^{-2}\tau^{-3}\\
\text{for}\ (x,E_k)\in \text{supp}\ g\endsplit\tag 7
$$
$$
|\partial_x^\alpha w_k(x)|\le C_{\ast}^\alpha(\tau B(x_0))^{-\alpha}\quad
\text{for}\ (x,E_k)\in \text{supp}\ g\tag 8
$$
$$
|b_k^2\phi^\prime(E_k)-1|\le C_{\#}\Lambda_{\min}^{4\varepsilon-2}\tag 9
$$
$$
\int\limits_{I_{\text{BVP}}}\ |\varepsilon_k(x)|^2\ dx\le \Lambda^{-N^{\prime
\prime}}_{\min}\ .\tag 10
$$

The phase $\phi(E)=\int_I(E-V(x))_+^{1/2}dx$ satisfies the estimates
$$\multline
\Big|\Bigl(\frac{d}{dE}\Bigr)^m\phi(E)\Big|\le C_{\#}^m(S_{\min}(E)\phi^\prime
(E))(S_{\min}(E))^{-m}\ (m\ge 1)\\
\text{for}\quad E\in [E_{\min},E_{\max}]\equiv \operatornamewithlimits{\bigcup}
_{E_0\in [E_{\ell o},E_{hi}]}\ \{E\mid |E-E_0|<
c_{\#}S_{\min}(E_0)\}\ .\endmultline\tag 11
$$

To see (1)$\ldots$(4), we invoke the Corollary to the Reformulated
Eigenvalue Theorem.  (The hypotheses of that Corollary hold, by
{\it ASSUMPTION\/} (A1) and $G\neq \emptyset$.  In particular, $G\neq
\emptyset$ implies $\hat c_1S_{\min}(E_{\ell o})+E_{\ell o}\le E_{hi}$,
which yields $\phi(E_{hi})-\phi(E_{\ell o})>100$.)

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 $S_{\min}(E)=\ \text{inf}_{x_{\text{left}}(E)<x<x_{\text{rt}}
(E)}S(x)$.  This is slightly more general than our previous 
definition of $S_{\min}(E)$, since we do not assume $E\le 0$.

Note that on $[a_{\min},b_{\max}]=$ image of $[E_{\min},E_{\max}]$ under
$E\mapsto \frac 1\pi \phi(E)-\frac 12=t$, we may solve for $E$ as a 
function of $t$, and the solution $E(t)$ satisfies
$$\multline
\Big|\Bigl(\frac{d}{dt}\Bigr)^mE(t)\Big|\le C_{\#}^mS_{\min}(E)\cdot
(S_{\min}(E)\phi^\prime(E))^{-m}\ (m\ge 1)\\
\text{for}\ E=E(t), t\in [a_{\min}, b_{\max}] .\endmultline\tag 12
$$
\noindent This follows from (11).  Also from (11) we get
$$
\phi^\prime(E)\ \text{has\ constant\ order\ of\ magnitude\ on}\ 
\{E\mid |E-E_0|<c_{\#}S_{\min}(E_0)\}\ .\tag 13
$$
\noindent It will be useful to work with energy intervals of the form
$J(E)=\{E^\prime \mid |E^\prime -E|<c_{\#}(\phi^\prime(E))^{-1}\}$.

Since $\phi^\prime(E)\ge c_{\#}S^{-1/2}(\tilde x)B(\tilde x)$ for
$x_{\text{left}}(E)<\tilde x<x_{\text{rt}}(E)$, $E\in [E_{\min}, E_{\max}]$,
we have
$$
\phi^\prime(E)\ge c_{\#}S^{-1/2}(x_0)B(x_0)\quad\text{for}\quad
E\in [E_{\min}, E_{\max}]\ , \tag 14
$$
$$
\phi^\prime(E)S_{\min}(E)\ge c_{\#}\Lambda_{\min}\quad\text{for}\quad
E\in [E_{\min},E_{\max}]\ .
\tag 15
$$
\noindent (To prove (14), we take $\tilde x$ to be the point
$\tilde x_0$ given in (A2).
To prove (15), take $\tilde x$ so that $S(\tilde x)\sim
S_{\min}(E)$, and then recall that $\lambda(\tilde x)\ge \Lambda(E)\ge
c_{\#}\Lambda_{\min}$.)

>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 
$$
1\le C_{\#}\int\limits_{E_{\min}}^{E_{\max}}\chi_{{}_{E_k\in J(E)}}
\frac{dE}{|J(E)|}\quad\text{since}\quad |E-E_k|<c_{\#}^\prime
[\phi^\prime(E_k)]^{-1}\tag"(21a)"
$$

\noindent implies $E_k\in J(E)$ and $|J(E)|\sim (\phi^\prime(E_k))^{-1}$.
Hence
$$\multline
\sum\limits_k|g(x,E_k)|\frac{[\phi^\prime(E_k)]^{-1}\Lambda_{\min}^{4\varepsilon
-2}\tau^{-3}}{(E_k-V(x))^{1/2}}\\
\le C_{\#}\int\limits_{E_{\min}}^{E_{\max}}\Bigl[\operatornamewithlimits{\max}_
{\tilde E\in J(E)}\frac{|g(x,\tilde E)|(\phi^\prime(\tilde E))^{-1}}
{(\tilde E-V(x))^{1/2}}\ \Lambda_{\min}^{4\varepsilon-2}\tau^{-3}\Bigr]
\frac{\sum\limits_k \chi_{E_k\in J(E)}}{|J(E)|}\ dE\\
\le C_{\#}\int\limits_{E_{\min}}^{E_{\max}}\Bigl[\operatornamewithlimits{\max}_
{\tilde E\in J(E)} \frac{|g(x,\tilde E)|}{(\tilde E-V(x))^{1/2}}\Lambda_{\min}
^{4\varepsilon-2}\tau^{-3}\Bigr]\ dE \ ,\endmultline\tag 22
$$

\noindent since $[\phi^\prime(\tilde E)]^{-1}\sim |J(E)|$ for $\tilde E
\in J(E)$, and since $J(E)$ contains at most $C_{\#}$ of the $E_k$.

\noindent Estimate (15) gives
$$
|J(E)|\sim (\phi^\prime(E))^{-1}\le C_{\#}
\Lambda_{\min}^{-1}S_{\min}(E)<< \tau S(x_0)\ .\tag"(22a)"
$$

\noindent On the other hand,
$\hat c\tau S(x_0)<\tilde E-V(x)<\hat C\tau S(x_0)$
whenever $g(x,\tilde E)\ne 0$. (Here we used (A2) and (A4).)
Hence the right--hand side of (22) is dominated by
$$
C_{\ast}\Lambda_{\min}^{4\varepsilon-2}\tau^{-3}
\int\limits_{\frac {\hat c}{2}\tau S(x_0)<E-V(x)<2\hat C\tau S(x_0)}
\frac{dE}{(E-V(x))^{1/2}}
\le C_\ast \Lambda_{\min}^{4\varepsilon-2}\tau^{-3}\cdot (\tau S(x_0))^{1/2}\ .
$$
\noindent Therefore (20) and (22) yield
$|\rho_0(x,g)-\rho_1(x,g)|\le C_\ast\Lambda_{\min}^{4\varepsilon-2}
\tau^{-5/2}S^{1/2}(x_0)$, so that
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_0(x,g)-\rho_1(x,g)|\ dx\le
C_\ast\Lambda_{\min}^{4\varepsilon-2}\tau^{-5/2}\lambda(x_0)\ .
\tag 23
$$

Next we look at the function $E\mapsto \frac{g(x,E)[\phi^\prime(E)]^{-1}}
{2(E-V(x))^{1/2}}\equiv F_x(E)$.  We know that
$$
|\partial_E^\beta g(x,E)|\le C_\ast^\beta(\tau S(x_0))^{-\beta}
\qquad\text{for\ all}\quad (x,E)\ .
$$

Also, $|\partial_E^\beta\phi^\prime(E)|\le C_{\#}^\beta\phi^\prime(E)(S_{\min}
(E))^{-\beta}$ for $E\in [E_{\min},E_{\max}]$, so $|\partial_E^\beta[\phi^\prime
(E)]^{-1}|\le C_{\#}^\beta[\phi^\prime(E)]^{-1}(S_{\min}(E))^{-\beta}$
for $E\in [E_{\min},E_{\max}]$; and $|\partial_E^\beta(E-V(x))^{-1/2}|\break
\le C_{\ast}^\beta(\tau S(x_0))^{-1/2}(\tau S(x_0))^{-\beta}$ for $E-V(x)
\sim \tau S(x_0)$.  Together, these estimates yield
$$
\big|\partial_E^\beta F_x(E)\big|\le \frac{C_\ast^\beta[\phi^\prime(E)]^{-1}}
{(\tau S(x_0))^{1/2}}\bigr(\min\{\tau S(x_0),S_{\min}(E)\}\bigr)^{-\beta}
\quad\text{in\ supp}\ g\ .\tag 24
$$

\noindent Also, since $|\{\frac 1\pi \phi(E_k)-\frac 12\}-k|\le C_{\#}
\Lambda_{\min}^{-1}$ by (1), we have $|E_k-E(k)|\le C_{\#}\Lambda_{\min}^{-1}
\cdot \max\bigl\{[\phi^\prime(E)]^{-1}\mid E\ \text{between}\ E_k\ \text{and}\
E(k)\}$.

\noindent By (13), this means
$$
|E_k-E(k)|\le C_{\#}\Lambda_{\min}^{-1}[\phi^\prime(E_k)]^{-1}\ .\tag 25
$$

\noindent Estimates (13), (22a), (24) and (25) imply
$$\split
\sum\limits_{k=k_{\ell o}}^{k_{hi}}|F_x(E_k)-F_x(E(k))|\\
\le C_\ast\Lambda_{\min}^{-1}\sum\limits_{k\in \Cal K_x}\frac{[\phi^\prime(E_k)]
^{-2}}{(\tau S(x_0))^{1/2}}\bigl\{\frac{1}{\tau S(x_0)}+\frac{1}
{S_{\min}(E_k)}\bigr\}\ ,\endsplit\tag 26
$$

\noindent where $\Cal K_x= \{k\ \text{between}\ k_{\ell o}, k_{hi}\mid
E_k-V(x)\sim \tau S(x_0)\}$.  Recall that $\phi^\prime(E_k)\ge c_{\#}
S^{-1/2}(x_0)B(x_0)$ and $S_{\min}(E_k)\phi^\prime(E_k)\ge c_{\#}\Lambda_{\min}$.
Hence $[\phi^\prime(E_k)]^{-1}\bigl\{\frac{1}{\tau S(x_0)}+\frac {1}
{S_{\min}(E_k)}\bigr\}\le C_{\#}\bigl\{\frac{1}{\tau\lambda(x_0)}
+\frac{1}{\Lambda_{\min}}\bigr\}\le \frac{C_{\#}}{\tau\Lambda_{\min}}$, so that
(26) implies
$$
\sum\limits_{k=k_{\ell o}}^{k_{hi}}|F_x(E_k)-F_x(E(k))|\le C_\ast
\tau^{-1}\Lambda_{\min}^{-2}\sum\limits_{k\in \Cal K_x}\frac{[\phi^\prime(E_k)]^{-1}}{(\tau S(x_0))^{1/2}}\ .\tag 27
$$

\noindent To estimate the right--hand side of (27), we use the energy intervals
$J(E)$.  In fact, (21a) and (22a) yield
$$\multline
\sum\limits_{k\in \Cal K_x}[\phi^\prime(E_k)]^{-1}\le C_{\#}
\sum\limits_{k\in \Cal K_x}\int\Sb E-V(x)\sim \tau S(x_0)\\
E\in [E_{\min},E_{\max}]\endSb [\phi^\prime(E_k)]^{-1}\chi_{{}_{E_k\in J(E)}}
\frac{dE}{|J(E)|}\\
\le C_{\#}^\prime \int\Sb E-V(x)\sim \tau S(x_0)\\ E\in [E_{\min},E_{\max}]\endSb
\bigl\{\text{Number\ of}\ E_k\in J(E)\bigr\}\cdot \Bigl\{
\frac{\operatornamewithlimits{\max}_{\tilde E\in J(E)} 
[\phi^\prime(\tilde E)]^{-1}}{|J(E)|}\Bigr\}\ dE\\
\le C_{\#}^{\prime\prime}\int\limits_{E-V(x)\sim \tau S(x_0)}\ dE\ \text{(since\
the\ factors\ in\ curly\ brackets\ are}\ \le C_{\#})\\
= C_\ast^{\prime\prime}\tau S(x_0)\ .\endmultline\tag"(27 bis)"
$$

\noindent Putting this into (27), we get
$$
\sum\limits_{k=k_{\ell o}}^{k_{hi}}|F_x(E_k)-F_x(E(k))|\le
C_\ast\tau^{-1/2}\Lambda_{\min}^{-2}(S(x_0))^{1/2}\ ,
$$

\noindent so that
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}\sum\limits_{k=k_{\ell o}}^{k_{hi}}
|F_x(E_k)-F_x(E(k))|\ dx\le C_\ast\tau^{-1/2}\Lambda_{\min}^{-2}
\lambda(x_0)\ .\tag 28
$$

\noindent The definitions of $\rho_1(x,g)$ and $F_x(E)$ show that
$\rho_1(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}}F_x(E_k)$.  So, defining
$$
\rho_2(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}}F_x(E(k))
=\sum\limits_{k\in \Bbb Z\cap [a,b]}\ F_x(E(k))\ ,\quad \text{(see\ (2))}
\tag 29
$$

\noindent we learn from (28) that
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_1(x,g)-\rho_2(x,g)|\ dx\le
C_\ast\tau^{-1/2}\Lambda_{\min}^{-2}\lambda(x_0)\ .\tag 30
$$

\noindent We prepare to compute $\rho_2(x,g)$ using the lemma on Riemann sums.
This requires estimates for $(\frac{d}{dt})^mF_x(E(t))$, which is a sum of
terms of the form
$$
\bigl[\partial_E^\beta F_x\mid_{E=E(t)}\bigr]\cdot\prod\limits_{\nu=1}^\beta
\bigl(\frac{d}{dt}\bigr)^{m_\nu}E(t)\ ,\quad\text{with}\ m_\nu\ge 1\
\text{and}\ m_1+\ldots+m_\beta=m\ .\tag 31
$$
\noindent  In particular, $0\le \beta\le m$.

\noindent By (24) and (12), the term (31) is dominated by
$$\multline
\frac{C_\ast[\phi^\prime(E)]^{-1}}{(\tau S(x_0))^{1/2}}\bigl(\min
\{\tau S(x_0),S_{\min}(E)\}\bigr)^{-\beta}\cdot (S_{\min}(E))^\beta
\cdot (S_{\min}(E)\phi^\prime(E))^{-m}\\
\text{with}\ E=E(t),\ \text{for}\ (x,E)\in \text{supp}\ g\ \text{and}\
t\in [a_{\min},b_{\max}]\ .\endmultline\tag 32
$$

\noindent The expression (32) may be rewritten in the form
$$\align
&\frac{C_\ast[\phi^\prime(E)]^{-1}}{(\tau S(x_0))^{1/2}}\Bigl(\min
\bigl\{\frac{\tau S(x_0)}{S_{\min}(E)},1\bigr\}\Bigr)^{-\beta}\cdot
(S_{\min}(E)\phi^\prime(E))^{-m}\\
&\le \frac{C_\ast[\phi^\prime(E)]^{-1}}{(\tau S(x_0))^{1/2}}\Bigl(\min
\bigl\{\frac{\tau S(x_0)}{S_{\min}(E)},1\bigr\}\Bigr)^{-m}\cdot
(S_{\min}(E)\phi^\prime(E))^{-m}\\
&=\frac{C_\ast[\phi^{\prime}(E)]^{-1}}{(\tau S(x_0))^{1/2}}\Bigl(\min
\bigl\{\tau S(x_0)\phi^\prime(E),S_{\min}(E)\phi^\prime(E)\bigr\}\Bigr)^{-m}\\
&\le \frac {C_\ast[\phi^\prime(E)]^{-1}}{(\tau S(x_0))^{1/2}}
(\tau\Lambda_{\min})^{-m}\ ,\
\text{again\ because}\ S(x_0)\phi^\prime(E)\ge c_{\#}
\lambda(x_0)\\
&\qquad\qquad\qquad \text{and}\ S_{\min}(E)\phi^\prime(E)
\ge c_{\#}\Lambda_{\min}\ .
\endalign
$$
\noindent Hence,
$$
\Bigl|\bigl(\frac{d}{dt}\bigr)^mF_x(E(t))\Big|\le C_{\ast}^m\frac
{[\phi^\prime(E(t))]^{-1}}{(\tau S(x_0))^{1/2}}(\tau\Lambda_{\min})^{-m}\quad
\text{for}\quad t\in [a_{\min},b_{\max}]\ .\tag 33
$$

\noindent (We have dropped the requirement $(x,E(t))\in \text{supp}\ g$,
since the left side of (33) vanishes unless that requirement is met.)

>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
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\Cal E(x)|\ dx\le
\frac{C_\ast^{\overline N}}{\tau^{3/2}}\Lambda_{\min}^{-\overline N/3}
\lambda(x_0)+\frac{C_\ast}{\tau^{3/2}}\Lambda_{\min}^{-1}\\
\le\frac{C_\ast}{\tau^{3/2}}\Lambda_{\min}^{-2}\lambda(x_0)\ .
$$

\noindent Combining this with (34) and setting
$$
\rho_3(x,g)=\int_a^bF_x(E(t))\ dt-F_x(E(b))\chm(b)\ ,\tag 36
$$
\noindent we obtain
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_2(x,g)-\rho_3(x,g)|\ dx\le
C_\ast\tau^{-3/2}\Lambda_{\min}^{-2}\lambda(x_0)\ .\tag 37
$$

Next we compare $\rho_3(x,g)$ with
$$
\rho_4(x,g)=\int_a^{b_0}F_x(E(t))\ dt-F_x(E(b_0))\chm(b)\ .\tag 38
$$

\noindent Since $|F_x(E)|\le C_\ast\frac{[\phi^\prime(E)]^{-1}}
{(\tau S(x_0))^{1/2}}\chi_{{}_{E-V(x)\sim \tau S(x_0)}}$ and
$|b-b_0|\le \frac{C_{\#}}{\Lambda_{\min}}$, and since $\phi^\prime(E(t))$
has constant order of magnitude for $t$ between $b_0$ and $b$, we have
$$\split
\Big|\int_{b_0}^bF_x(E(t))\ dt\Big|\le \frac{C_\ast}{\Lambda_{\min}}
\frac{[\phi^\prime(E(b_0))]^{-1}}{(\tau S(x_0))^{1/2}}\
\chi_{{}_{\cases E-V(x)\sim \tau S(x_0)\ \text{for\ some}&{}\\
E\ \text{between}\ E(b_0)\ \text{and}\ E(b)&{}\endcases}}\ .\endsplit\tag 39
$$

\noindent As $t$ varies between $b_0$ and $b$, $E(t)$ varies by at most
$C_\ast[\phi^\prime(E(b_0))]^{-1}|b-b_0|\le C_\ast[\phi^\prime(E(b_0))]^{-1}
\Lambda_{\min}^{-1}\le C_\ast\lambda^{-1}(x_0)S(x_0)\Lambda_{\min}^{-1}
\ \text{(by\ (14))}\ <<\tau S(x_0)$, so (39) implies
$$
\Big|\int_{b_0}^bF_x(E(t))\ dt\Big|\le \frac{C_\ast}{\Lambda_{\min}}
\frac{[\phi^\prime(E_{hi})]^{-1}}{(\tau S(x_0))^{1/2}}\chi_{{}_{E_{hi}-V(x)\sim
\tau S(x_0)}}\ .$$

\noindent (Recall that $E(b_0)=E_{hi}$.)  Hence,
$$\align
&\int\limits_{|x-x_0|<c_{\#}B(x_0)}\Big|\int\limits_{b_0}^{b}F_x(E(t))\ dt\Big|
\ dx\\
&\quad\le \frac{C_\ast}{\Lambda_{\min}}\frac{[\phi^\prime(E_{hi})]^{-1}}
{(\tau S(x_0))^{1/2}}\int\limits_{|x-x_0|<c_{\#}B(x_0)}\chi_{{}_{E_{hi}-V(x)\sim
\tau S(x_0)}}\ dx\\
&\quad \le \frac{C_\ast}{\Lambda_{\min}}\frac{[\phi^\prime(E_{hi})]^{-1}}
{(\tau S(x_0))^{1/2}}\cdot (\tau B(x_0))\le \frac{C_\ast}{\Lambda_{\min}}
\frac{[S^{1/2}(x_0)B^{-1}(x_0)]}{(\tau S(x_0))^{1/2}}(\tau B(x_0))\\
&\quad \le \frac{C_\ast\tau^{1/2}}{\Lambda_{\min}}\le C_\ast\Lambda_{\min}^{-2}
\lambda(x_0)\ .\tag 40\endalign
$$

\noindent Similarly,
$$\align
&|F_x(E(b))-F_x(E(b_0))|\le |b-b_0|\operatornamewithlimits{\max}_
{t\ \text{between}\ b, b_0}\Big|\bigl(\frac{d}{dt}\bigr)F_x(E(t))\Big|\\
&\le C_{\#}\Lambda_{\min}^{-1}\cdot \max_{t\ \text{between}\ b,b_0}
C_\ast\frac{[\phi^\prime(E(t))]^{-1}}{(\tau S(x_0))^{1/2}}
(\tau\Lambda_{\min})^{-1}\chi_{{}_{E(t)-V(x)\sim \tau S(x_0)}}\\
&\le C_\ast\tau^{-1}\Lambda_{\min}^{-2}\frac{[\phi^\prime(E(b_0))]^{-1}}
{(\tau S(x_0))^{1/2}}\chi_{{}_{E(b_0)-V(x)\sim \tau S(x_0)}}\ ,\endalign
$$

\noindent since $[\phi^\prime(E(t))]^{-1}$ has constant order of magnitude
and $E(t)$ varies by $<<\tau S(x_0)$ as $t$ varies from $b_0$ to $b$.
Integrating in $x$, we get
$$\align
&\int\limits_{|x-x_0|<c_{\#}B(x_0)}|F_x(E(b))-F_x(E(b_0))|\ dx\\
&\le C_\ast\frac{\Lambda_{\min}^{-2}[\phi^\prime(E_{hi})]^{-1}}
{\tau(\tau S(x_0))^{1/2}}\int\limits_{|x-x_0|<c_{\#}B(x_0)}
\chi_{{}_{E_{hi}-V(x)\sim \tau S(x_0)}}\ dx\\
&\le \frac{C_\ast\Lambda_{\min}^{-2}[\phi^\prime(E_{hi})]^{-1}}
{\tau(\tau S(x_0))^{1/2}}(\tau B(x_0))\le C_\ast\frac
{\Lambda_{\min}^{-2}[S^{1/2}(x_0)B^{-1}(x_0)]}
{\tau(\tau S(x_0))^{1/2}}\ (\tau B(x_0))\\
&= C_\ast\Lambda_{\min}^{-2}\tau^{-1/2}\le C_\ast\Lambda_{\min}^{-2}
\lambda(x_0)\tau^{-1/2}\ .\tag 41\endalign
$$

\noindent Estimates (40), (41) and the definitions of $\rho_3(x,g)$ and
$\rho_4(x,g)$ show that
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_3(x,g)-\rho_4(x,g)|\ dx\le
C_\ast\Lambda_{\min}^{-2}\lambda(x_0)\tau^{-1/2}\ .\tag 42
$$

Next, we compare $\rho_4(x,g)$ with
$$
\rho_5(x,g)=\int_a^{b_0}F_x(E(t))\ dt-F_x(0)\chm(\frac 1\pi \phi(0)-
\frac 12 )\ .\tag 43
$$

\noindent Recall that $E(b_0)=E_{hi}$, and either $E_{hi}=0$ or else
$g(x,E_{hi})=g(x,0)=0$ for all $x$. Therefore to compare $\rho_4$, $\rho_5$
we may assume $E_{hi}=0$,
since otherwise $\rho_4(x,g)=\rho_5(x,g)$ for all $x$.  Under the assumption
$E_{hi}=0$ we have $\frac 1\pi \phi(0)-\frac 12=b_0$, and
$$
\rho_4(x,g)-\rho_5(x,g)=F_x(E(b_0))\cdot [\chm(b)-\chm(b_0)]\ .\tag 44
$$

If $\min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|\ge \overline C_{\#}
\Lambda_{\min}^{-1}$, then $|\chm(b)-\chm(b_0)|\le C_{\#}\Lambda_{\min}^{-1}$,
since $b_0=\frac 1\pi \phi(0)-\frac 12$ isn't too near the jumps of
$\chm$, while $|b-b_0|\le \frac{C_{\#}}{\Lambda_{\min}}$.  In any case,
$|\chm(b)-\chm(b_0)|\le C_{\#}$.  

Also,
$$\align
&\int\limits_{|x-x_0|<c_{\#}B(x_0)}|F_x(E(b_0))|\ dx\\
&\quad \le
\int\limits_{|x-x_0|<c_{\#}B(x_0)}C_\ast
\frac{[\phi^\prime(E(b_0))]^{-1}}{(\tau S(x_0))^{1/2}}
\chi_{{}_{E(b_0)-V(x)\sim \tau S(x_0)}}\ dx\\
&\quad \le C_\ast\frac{[\phi^\prime(E_{hi})]^{-1}}{(\tau S(x_0))^{1/2}}
\cdot \tau B(x_0)\le C_\ast\frac{[S^{1/2}(x_0)B^{-1}(x_0)]}
{(\tau S(x_0))^{1/2}}\tau B(x_0)=C_\ast \tau^{1/2}\ .\endalign
$$

\noindent Putting these remarks into (44), we get
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_4(x,g)-\rho_5(x,g)|\ dx\le
\frac{C_\ast\tau^{1/2}}{\Lambda_{\min}}\le C_\ast\Lambda_{\min}^{-2}\lambda(x_0)\\
\text{if}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|\ge \overline C_{\#}
\Lambda_{\min}^{-1}\ ,\endsplit\tag 45
$$
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_4(x,g)-\rho_5(x,g)|\ dx\le C_\ast
\tau^{1/2}\\
\text{if}\ \min_{k\in \Bbb Z}\ |\phi(0)-\pi(k+\frac 12)|\le \overline C_{\#}
\Lambda_{\min}^{-1}\ .\endsplit\tag 46
$$

\noindent Recalling the definition of $F_x(E)$ and changing variable from
$t$ to $E=E(t)$ in the integral in (43), we have
$$\split
\rho_5(x,g)=\frac{1}{2\pi}\int\limits_{E(a)}^{E_{hi}}
\frac{g(x,E)\ dE}{(E-V(x))^{1/2}}-\\
\frac 12\frac{g(x,0)}{(-V(x))^{1/2}}[\phi^\prime(0)]^{-1}
\chm(\frac 1\pi \phi(0)-\frac 12)\ .\endsplit\tag 47
$$

\noindent Since $g(x,E)=0$ for $E\in (-\infty,0]\ \backslash\ [E(a),E_{hi}]$, this
agrees with our definition of $\rho_{sc}(x,g)$.  (Note that $[\phi^\prime(0)]$
contains a factor of $\frac 12$.)  Hence $\rho_5(x,g)=\rho_{sc}(x,g)$.
Estimates (23), (30), (37), (42), (45), (46) therefore show that
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_0(x,g)-\rho_{sc}
(x,g)|\ dx\le C_{\ast}\Lambda_{\min}^{4\varepsilon-2}\tau^{-5/2}
\lambda(x_0)\\
\text{if}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|\ge \frac
{\overline C_{\#}}{\Lambda_{\min}}\endsplit\tag 48
$$
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\rho_0(x,g)-\rho_{sc}(x,g)|\ dx
\le C_\ast\Lambda_{\min}^{4\varepsilon-2}\tau^{-5/2}
\lambda(x_0)+C_\ast\tau^{1/2}\\
\text{if}\ \min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|\le
\frac{\overline C_{\#}}{\Lambda_{\min}}\ .\endsplit\tag 49
$$

\noindent So $\rho_0(x,g)$ lies near $\rho_{sc}(x,g)$ in $L^1$--norm.
Next we show that the oscillating terms $\rho_+(x,g)$ and $\rho_-(x,g)$
are small in $H^{-1}$--norm.  We study $\rho_+(x,g)$; the discussion of
$\rho_-(x,g)$ is analogous.  With $g_k(x)=\frac{g(x,E_k)b_k^2(1+w_k(x))^2}
{4(E_k-V(x))^{1/2}}$, we have
$$
\rho_+(x,g)=\sum\limits_{k=k_{\ell o}}^{k_{hi}}g_k(x)e^{2i\eta(x,E_k)}\quad
\text{by\ (18)}\ ;\tag 50
$$
$$
\text{supp}\ g_k\subset \{x\mid |x-x_0|<\hat cB(x_0), E_k-V(x)\sim
\tau S(x_0)\}\quad \text{by}\  (A3)\ ;\tag 51
$$
$$\split
|(\frac{d}{dx})^\alpha g_k(x)|\le
C_\ast^\alpha\frac{[\phi^\prime(E_k)]^{-1}} {(\tau S(x_0))^{1/2}}(\tau
B(x_0))^{-\alpha}\quad\text{by\ (51), (9), (8),}\ (A5)\ ,\\
\text{and}\ |(\frac{d}{dx})^\alpha V(x)|\le C_{\#}^\alpha S(x_0)B^{-\alpha}
(x_0)\ \text{for}\ |x-x_0|<\hat cB(x_0)\ .\endsplit \tag 52
$$

\noindent Recall that
$$
\frac{\partial}{\partial x}\eta(x,E_k)=(E_k-V(x))^{1/2}\ \text{in}\ \text{supp}\
g_k(x)\ .\tag 53
$$
\noindent We will show that any sum (50) with $g_k$ and $\eta$ satisfying
(51)$\ldots$(53) is small in $H^{-1}$--norm.

To estimate the $H^{-1}$--norm, we compute an indefinite integral for
$\rho_+(x,g)$, using the following observation.

Suppose $h_k(x)$ is supported in $\text{supp}\ g_k\subset \{|x-x_0|<\hat c
B(x_0)$, $E_k-V(x)\sim \tau S(x_0)\}$ and satisfies $|(\frac{d}{dx})^\alpha
h_k(x)|\le C_\ast^\alpha A(\tau B(x_0))^{-\alpha}$.  Then
$$
\frac{d}{dx}\Bigl(\frac{h_k(x)e^{2i\eta(x,E_k)}}
{2i(E_k-V(x))^{1/2}}\Bigr)=h_k(x)e^{2i\eta(x,E_k)}+
\tilde h_k(x)e^{2i\eta(x,E_k)}\ ,
$$

\noindent with $\tilde h_k(x)=\frac{d}{dx}
\bigl(\frac{h_k(x)}{2i(E_k-V(x))^{1/2}}\bigr)$ satisfying
$$\split
|(\frac{d}{dx})^\alpha\tilde h_k(x)|\le \frac{C_\ast^\alpha
A(\tau B(x_0))^{-1-\alpha}} {(\tau S(x_0))^{1/2}}=C_\ast^\alpha\tilde A(\tau
B(x_0))^{-\alpha},\\ \tilde A=\frac{A}{\tau^{3/2}\lambda(x_0)}\ .\endsplit
$$

\noindent Note that $|\tilde A|\le \Lambda_{\min}^{-\varepsilon}
|A|$, since we assume $\tau>\Lambda_{\min}^{\varepsilon-2/3}\ge 
\Lambda_{\min}^\varepsilon (\lambda(x_0))^{-2/3}$.

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
$\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\sum\limits_k\ \text{err}_k(x)|\ dx
\le C_\ast^\prime \Lambda_{\min}^{-N}\tau^{1/2}\lambda(x_0)$. For the indefinite
integral $\text{Err}(x)=\int\limits_{x_0}^x \sum\limits_k\ \text{err}_k(x^\prime)\
dx^\prime$ this implies
$$
|\text{Err}(x)|\le C_\ast\Lambda_{\min}^{-N}\lambda(x_0)\quad\text{for}\quad
|x-x_0|<c_{\#}B(x_0)\ ,\tag 59
$$

\noindent and (58) becomes
$$
\rho_+(x,g)=\frac{d}{dx}\bigl\{\sum\limits_k\sigma_k(x)
e^{2i\eta(x,E_k)}\bigr\}-\frac{d}{dx}\ \text{Err}(x)\ .\tag 60
$$

Thus, $H^{-1}$--estimates for $\rho_+(x,g)$ are reduced to $L^2$--estimates
\hfill\break for
$\sum\limits_k\sigma_k(x)e^{2i\eta(x,E_k)}$.  To make the $L^2$--estimates, we
study a single term
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}\sigma_k(x)\overline{\sigma_m(x)}
e^{2i[\eta(x,E_k)-\eta(x,E_m)]}\ dx\equiv U_{km}\ .\tag 61
$$

For $E_k$, $E_m$ far apart, we will estimate $U_{km}$ using the following
standard
\vglue 1pc

\proclaim{Stationary Phase Lemma} Suppose $a\in C_0^\infty(|x-\hat x_0|<b)$,
$\phi\in C^\infty$ on a neighborhood of $\text{supp}\ a$, with $\phi$ real
and $|\phi^{(k)}|\le C_k\alpha b^{-k}$ $(k\ge 1)$ on $\text{supp}\ a$,
and $|a^{(k)}|\le C_k\beta b^{-k}$ $(k\ge 0)$.  Assume also $|\phi^\prime|\ge
c\alpha b^{-1}$ on $\text{supp}\ a$.  Then $|\int_{-\infty}^\infty ae^{i\phi}
dx|\le C^\prime_m\beta b\alpha^{-m}$, with $C_m^\prime$ depending only on $m$
and on the constants $C_m$, $c$.\endproclaim

\vglue 1pc
\demo{Proof}  Extend $\frac{a}{i\phi^\prime}$ to a $C^\infty$ function 
by setting it equal to zero outside $\text{supp}\ a$.  Then 
$\frac{d}{dx}\bigl\{\frac{\alpha a}{i\phi^\prime}\bigr\}$ 
satisfies estimates analogous to $a$, and
$\int_{-\infty}^\infty ae^{i\phi}dx=\int_{-\infty}^\infty \frac{a}{i\phi^\prime}
\frac{d}{dx}\{e^{i\phi}\}dx=-\frac{1}{\alpha}\int_{-\infty}^\infty
\bigl[\frac{d}{dx}\bigl\{\frac{\alpha a}{i\phi^\prime}\bigr\}\bigr]e^{i\phi}\ 
dx$. (The integration by parts is justified, because $\phi$ is smooth on
a neighborhood of $\text{supp}\ a$.)  Repeated applications of this
identity yield the conclusion of the lemma. $\qquad\blacksquare$\enddemo
\medskip

To apply the stationary phase lemma to $U_{km}$, we take $\hat x_0\in\ \text{supp}\ \sigma_k$, $\phi(x)=2\eta(x,E_k)-2\eta(x,E_m)$, $a(x)=\sigma_k(x)\overline{
\sigma_m(x)}$, $b=C_\ast\tau B(x_0)$.  Evidently, $a(x)$ satisfies
the hypotheses
of the lemma, with $\beta=[\phi^\prime(E_k)]^{-1}[\phi^\prime(E_m)]^{-1}
(\tau S(x_0))^{-2}$, so it remains to investigate $\phi$.  In a neighborhood
of $\text{supp}\ a(x)$ we have $E_k-V(x)$, $E_m-V(x)\sim \tau S(x_0)$. 
Therefore, in a neighborhood of $\text{supp}\ a$, we have:
$$
\phi^\prime(x)=2(E_k-V(x))^{1/2}-2(E_m-V(x))^{1/2}=\int\limits_{E_m}^{E_k}
(E-V(x))^{-1/2}\ dE\ ,\qquad\text{hence}\tag"(61 bis)"
$$
$$
|\phi^\prime(x)|\ge c_\ast(\tau S(x_0))^{-1/2}|E_k-E_m|\quad \text{in}\
\text{supp}\ a\ 
.\tag 62
$$

Also, $\partial_x^\alpha(E-V(x))^{-1/2}$ is a sum of terms
$(E-V(x))^{-\frac 12-s}
\prod\limits_{\nu=1}^s \partial_x^{\alpha_\nu}V(x)$, with $\alpha_\nu
\break \ge 1$
and $\alpha_1+\ldots+\alpha_s=\alpha$.  In particular, $0\le s\le \alpha$.
For $x\in\ \text{supp}\ a$ and $E$ between $E_k$ and $E_m$, this term is
dominated by $C_\ast^\alpha(\tau S(x_0))^{-\frac 12-s}\prod\limits_{\nu=1}^s
\big\{S(x_0)B^{-\alpha_\nu}(x_0)\big\}=\frac{C_\ast^\alpha}{(\tau S(x_0))^{1/2}}
\tau^{-s}B^{-\alpha}(x_0)\le C_\ast^\alpha(\tau S(x_0))^{-\frac 12}
(\tau B(x_0))^{-\alpha}$.  Putting this into (61 bis), we get
$$\split
|(\frac{d}{dx})^\alpha\phi^\prime(x)|\le C_\ast^\alpha
(\tau S(x_0))^{-1/2}|E_k-E_m|\cdot (\tau B(x_0))^{-\alpha}\\
\text{in\ a\
neighborhood\ of\ supp}\ a\ .\endsplit\tag 63  
$$

\noindent Estimates (62) and (63) show that $\phi$ satisfies the hypotheses
of the stationary phase lemma, with $\alpha=(\tau S(x_0))^{-1/2}|E_k-E_m|
\cdot (\tau B(x_0))=\tau^{1/2}\lambda(x_0)\frac{|E_k-E_m|}{S(x_0)}$.
The stationary phase lemma shows therefore that
$$
|U_{km}|\le C_\ast^{\overline N}\ \frac{[\phi^\prime(E_k)]^{-1}[\phi^\prime
(E_m)]^{-1}}{(\tau S(x_0))^2}\cdot (\tau B(x_0))
\cdot\bigl[\frac{\tau^{1/2}\lambda(x_0)|E_k-E_m|}{S(x_0)}\bigr]^{-\overline N}\ ,
$$

\noindent with $\overline N$ as large as we please.  Also, replacing the
integrand in the definition by its absolute value, we get the trivial estimate
$|U_{km}|\le C_\ast\frac{[\phi^\prime(E_k)]^{-1}[\phi^\prime(E_m)]^{-1}}
{(\tau S(x_0))^2}\tau B(x_0)$.  Hence,
$$\multline
|U_{km}|\le \frac{C_\ast^{\overline N}[\phi^\prime(E_k)]^{-1}[\phi^\prime
(E_m)]^{-1}B(x_0)}{\tau S^2(x_0)}\bigl[1+\tau^{1/2}
\lambda(x_0)\frac{|E_k-E_m|}{S(x_0)}\bigr]^{-\overline N}\\
\text{with}\ \overline N\ \text{as\ large\ as\ we\ please}\ .\endmultline\tag 64
$$

\noindent We use (64) to estimate $\sum\limits_{km}|U_{km}|$.  Again we bring
in the energy intervals $J(E)$.  The length of $J(E)$ is $\sim(\phi^\prime
(E))^{-1}\le C_{\#}\frac{S(x_0)}{\lambda(x_0)}$ by (14).  It follows
that\hfill\break  $1+\tau^{1/2}\lambda(x_0)\frac{|E_k-E_m|}{S(x_0)}\sim
1+\tau^{1/2}\lambda(x_0)\frac{|E-E^\prime|}{S(x_0)}$ if $E_k\in
J(E)$ and $E_m\in J(E^\prime)$.  Also,
$$\align
[\phi^\prime(E_k)]^{-1}&\sim\int\limits_{E\in[E_{\min},E_{\max}]}\chi_{{}_{E_k\in
J(E)}}\ dE\quad\text{and}\\
[\phi^\prime(E_m)]^{-1}&\sim\int\limits_{E^\prime\in [E_{\min},E_{\max}]}
\chi_{{}_{E_m\in J(E^\prime)}}\ dE^\prime\ .\endalign
$$

\noindent Hence (64) implies
$$\split
\sum\limits_{km}|U_{km}|\le C_\ast^{\overline N}\int\limits_{E,E^\prime\in
[E_{\min},E_{\max}]}\bigl\{\text{No.\ of}\ E_k\in J(E)\bigr\}\\
\bigl\{\text{No.\ of}\ E_m\in J(E^\prime)\bigr\}\frac{B(x_0)}
{\tau S^2(x_0)} \Bigl(1+\tau^{1/2}\lambda(x_0)\frac{|E-E^\prime|}
{S(x_0)}\Bigr)^{-\overline N}\ dEdE^\prime\ .\endsplit
$$

\noindent Since $J(E)$, $J(E^\prime)$ contain at most $C_{\#}$ of the $E_j$,
this simplifies to
$$\multline
\sum\limits_{k,m}|U_{km}|\le \int\limits_{E_{\min}}^{E_{\max}}\int\limits_{E_{\min}}^{E_{\max}}C_\ast^{\overline N}\frac{B(x_0)}{\tau S^2(x_0)}\Bigl(1+
\tau^{1/2}\lambda(x_0)\frac{|E-E^\prime|}{S(x_0)}\Bigr)^{-\overline N}\ dEdE^\prime\\
\le C_\ast[E_{\max}-E_{\min}]\Bigl[\frac{S(x_0)}{\tau^{1/2}\lambda(x_0)}\Bigr]
\cdot \frac{B(x_0)}{\tau S^2(x_0)}\quad\text{if\ we\ take}\ \overline N=10\ .
\endmultline
$$

\noindent Since $[E_{\max}-E_{\min}]\le C_{\#}S(x_0)$, this implies
$$
\sum\limits_{k,m}|U_{km}|\le \frac{C_\ast B(x_0)}{\tau^{3/2}\lambda(x_0)}\ .
$$

\noindent By definition of $U_{km}$, it follows that
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}\Big|\sum\limits_k\sigma_k(x)
e^{2i\eta(x,E_k)}\Big|^2\ dx\le \frac{C_\ast B(x_0)}{\tau^{3/2}\lambda(x_0)}\ .
\tag 65
$$

\noindent Estimates (59), (65) and equation (60) give
$$
\rho_+(x,g)=\frac{d}{dx}\Cal H_+(x,g)\quad\text{with}\tag 66
$$
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\Cal H_+(x,g)|^2\ dx\le C_\ast
\frac{B(x_0)}{\tau^{3/2}\lambda(x_0)}+C_\ast\Lambda_{\min}^{-2N}\lambda^2(x_0)
B(x_0)\ .\endsplit\tag 67
$$

A completely analogous discussion of $\rho_-(x,g)$ shows that
$$
\rho_-(x,g)=\frac{d}{dx}\Cal H_-(x,g)\quad\text{with}\tag 68
$$
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\Cal H_-(x,g)|^2\ dx\le C_\ast
\frac{B(x_0)}{\tau^{3/2}\lambda(x_0)}
+C_\ast\Lambda_{\min}^{-2N}\lambda^2(x_0)B(x_0)\ .\tag 69
$$

\noindent Set $\Cal H_0(x,g)=\int\limits_{x_0}^x[\rho_0(x^\prime,g)-\rho_{sc}
(x^\prime,g)]dx^\prime$, so that
$$
\rho_0(x,g)=\rho_{sc}(x,g)+\frac{d}{dx}\Cal H_0(x,g)\ ;\tag 70
$$
\noindent  and by (48), (49), we have
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\Cal H_0(x,g)|^2\ dx\le C_\ast\Lambda_{\min}^
{8\varepsilon-4}\tau^{-5}\lambda^2(x_0)B(x_0)\\
\text{if}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|\ge \frac{\overline C_{\#}}{\Lambda_{\min}}\ ,\endsplit\tag 71
$$
$$\split
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\Cal H_0(x,g)|^2\ dx\le
C_\ast\Lambda_{\min}^{8\varepsilon-4}\tau^{-5}\lambda^2(x_0)B(x_0)+C_\ast
\tau B(x_0)\\
\text{if}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \frac{\overline C_{\#}}
{\Lambda_{\min}}\ .\endsplit\tag 72
$$

Finally, set $\Cal H_{\text{extra}}(x,g)=\int_{x_0}^x[\rho(x^\prime,g)-\rho_0
(x^\prime,g)-\sum\limits_\pm \rho_\pm(x^\prime,g)]\ dx^\prime$, so that
$$
\rho(x,g)=\rho_0(x,g)+\rho_+(x,g)+\rho_-(x,g)+\frac{d}{dx}\Cal H_{\text{extra}}
(x,g)\ ;\tag 73
$$
\noindent and by (16) we have
$$
\int\limits_{|x-x_0|<c_{\#}B(x_0)}|\Cal H_{\text{extra}}(x,g)|^2\ dx\le
C_\ast\Lambda_{\min}^{-N^{\prime\prime}}(\phi(0)+1)^2B(x_0)\ .\tag 74
$$

Define $\Cal H(x,g)=\Cal H_+(x,g)+\Cal H_-(x,g)+\Cal H_0(x,g)+\Cal H_{\text{extra}}
(x,g)$, so that
$$
\rho(x,g)=\rho_{sc}(x,g)+\frac{d}{dx}\Cal H(x,g)\ ,\quad\text{by\ (66),\
(68),\ (70),\ (73)}\ .\tag 75
$$

\noindent Combining our estimates (67), (69), (71), (72), (74), we obtain the
following lemma, the main result of this section.

\vglue 1pc
\proclaim{Oscillatory Density Lemma}  Suppose the potential $V(x)$, the
weight functions $S(x)$, $B(x)$, and the cutoff function $g(x,E)$ satisfy
assumptions (A1)$\ldots$(A6). Set 
$\phi(0)=\int_{I}(-V(x))_+^{1/2}dx$.  Then the microlocalized
density $\rho(x,g)$ 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
$I_{x_0}=\{|x-x_0|<c_{\#}B(x_0)\}$. 

\noindent (a)$\quad$ If $\min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge\overline C_{\#}
\Lambda_{\min}^{-1}$, then
$$\split
\frac{1}{B(x_0)}\int\limits_{I_{x_0}}|H(x)|^2\ dx\le
C_\ast\tau^{-3/2}\lambda^{-1}(x_0)+C_\ast\Lambda_{\min}^{8\varepsilon-4}
\tau^{-5}\lambda^2(x_0)\\
+C_\ast\Lambda_{\min}^{-N^{\prime\prime}}(\phi(0)+1)^2\ .\endsplit
$$

\noindent (b)$\quad$ If $\min_{k\in \Bbb Z}|\phi(0)-\pi(k+\frac 12)|\le \overline C_{\#}\Lambda_{\min}^{-1}$, then
$$\split
\frac{1}{B(x_0)}\int\limits_{I_{x_0}}|H(x)|^2\ dx\le C_\ast\tau+C_\ast
\tau^{-3/2}\lambda^{-1}(x_0)+C_\ast\Lambda_{\min}^{8\varepsilon-4}
\tau^{-5}\lambda^2(x_0)\\
+C_\ast\Lambda_{\min}^{-N^{\prime\prime}}(\phi(0)+1)^2\ .\endsplit
$$

\noindent Here, the constants $c_{\#}$, $\overline C_{\#}$, depend only on
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $c_3$, $C_\alpha$ in the
assumptions (A1), (A2).
The constant $C_\ast$ depends on these quantities, and also on
$\hat c$, $\hat c_1$, $\hat C$, $\hat C_{\alpha\beta}$ in assumptions 
(A3), (A5).\endproclaim



\vfill\eject

\heading{\bf{THE MICROLOCALIZED DENSITY NEAR THE MINIMUM OF THE POTENTIAL}}
\endheading
\medskip

Assume the hypotheses of the WKB Theorem on low eigenvalues, with $E_\infty=0$.
Take a function $g(E)$ satisfying
$$\multline
\text{supp}\ g(E)\subset \{|E-V(x_0)|<\tau^2S\}\quad\text{with}\quad
\lambda^{2\varepsilon-\frac 12}<\tau<\lambda^{-2\varepsilon},\quad\text{and}\\
\Big|\bigl(\frac{d}{dE}\bigr)^\beta g(E)\Big|\le \hat C_\beta(\tau^2S)^{-\beta}\
.\endmultline
$$

\noindent Regard $g$ as a function of $(x,E)$ which happens to be
independent of $x$, and form the microlocalized density $\rho(x,g)$.
Let $\rho_{sc}(x,g)$ be the corresponding semiclassical approximation.  Our
goal is to give crude estimates for an indefinite integral of
$\rho(x,g)-\rho_{sc}(x,g)$.

We write $c_{\#}$, $C_{\#}$ etc. for constants that depend only
on the constants in the hypotheses of the WKB Theorem on low eigenvalues.
We write $C_\ast$, $c_\ast$ etc. for constants that depend also on the
$\hat C_\beta$ above.

The WKB Theorem on low eigenvalues implies that the eigenvalues
in $\{|E-V(x_0)|\le \tau^2S \}\cap (-\infty,0]$ may be written as
$E_0,E_1,\ldots,E_{k_{hi}}$, with
$$\align
&\big|\frac{1}{\pi}\phi(E_k)-\frac 12-k|\le C_{\#}\lambda^{-1}\quad
\text{for}\ 0\le k\le k_{hi}\tag 1\\
&\{k\in \Bbb Z\mid 0\le k\le k_{hi}\}=\Bbb Z\cap [-\frac 12,b],\ \text{with}\\
&b=\frac 1\pi \phi(E_{hi})-\frac 12 +\omega_{hi},\ \  |\omega_{hi}|\le
C_{\#}\lambda^{-1}, E_{hi}= \min(0,V(x_0)+\tau^2S)\ .
\endalign
$$
\noindent Details (for once) are left to the reader.

\noindent For convenience, set $b_0=\frac 1\pi\phi(E_{hi})-\frac 12$,
$E_{\max}=V(x_0)+2\tau^2S$.  The phase $\phi(E)$ satisfies the following,
by the results in the section on WKB for low eigenvalues.
$$\gather
\phi(V(x_0))=0\\
\Big|\bigl(\frac{d}{dE}\bigr)^\beta\phi(E)\Big|\le C_{\#}^\beta
\lambda S^{-\beta}\quad\text{for}\quad E\in [V(x_0),E_{\max}]\\
\frac{d\phi(E)}{dE}\ge c_{\#}\lambda S^{-1}\quad\text{for}\quad
E\in [V(x_0),E_{\max}]\ .\endgather
$$

\noindent For $t\in [-\frac 12,\frac 1\pi\phi(E_{\max})-\frac 12]=\Cal J$,
we can therefore solve the equation $\frac 1\pi\phi(E)-\frac 12=t$,
obtaining a solution $E=E(t)\in [V(x_0),E_{\max}]$, with
$$\multline
\Big|\bigl(\frac{d}{dt}\bigr)^mE(t)\Big|\le C_{\#}^mS\lambda^{-m}(m\ge 1)
\quad\text{for}\ t\in \Cal J\ ,\quad\text{and}\\
\frac{d}{dt}E(t)\ge c_{\#}S\lambda^{-1}\quad\text{for}\quad t\in \Cal J\ .
\endmultline
$$

\noindent Also, $E(-\frac 12)=V(x_0)$.  For $0\le k\le k_{hi}$ we have
$k\in \Cal J$, so $E(k)$ is well-defined and lies in $[V(x_0),E_{\max}]$.  Also,
$E_k\in [V(x_0),E_{\max}]$, so $E_k=E(t_k)$ with
$t_k=\frac 1\pi\phi(E_k)-\frac 12\in \Cal J$.  We have $|t_k-k|\le
C_{\#}\lambda^{-1}$ by (1), so $|E_k-E(k)|=|E(t_k)-E(k)|\le C_{\#}
\lambda^{-1}\max_{t\in \Cal J}\big|\frac{dE(t)}{dt}\big|\le C_{\#}
S\lambda^{-2}$.  Note that $b_0\le \frac 1\pi \phi(E_{\max})=\frac 1\pi\phi
(E_{\max})-\frac 1\pi \phi(V(x_0))\le |E_{\max}-V(x_0)|\cdot
\max_{E\in [V(x_0),E_{\max}]}\bigl(\frac{d\phi(E)}{dE}\bigr)\le 2\tau^2S\cdot
C_{\#}\lambda S^{-1}=C_{\#}^\prime\tau^2\lambda$.  Since $k_{hi}$ is
the greatest integer in $b$ and $|b-b_0|\le C_{\#}\lambda^{-1}$, it
follows that $k_{hi}+1\le C_{\#}^\prime\tau^2\lambda+C_{\#}\le C_{\#}
^{\prime\prime}\tau^2\lambda$ (since $\tau>\lambda^{\varepsilon-\frac 12})$.  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 $x\in I_{\text{BVP}}$, $x<x_0-\lambda^\varepsilon
\tau B$ we have
$$\align
|H_-(x)|&=\Big|\int\limits_{x^\prime \in I_{\text{BVP}}\cap (-\infty,x]}
\rho(x^\prime,g)\ dx^\prime\Big|=\Big|\sum\limits_{k=0}^{k_{hi}}
g(E_k)\int\limits_{x^\prime\in I_{\text{BVP}}\cap (-\infty,x]}
u_k^2(x^\prime)\ dx^\prime\Big|\\
&\le C_\ast\sum\limits_{k=0}^{k_{hi}}\ \int\limits_{x^\prime \in I_{\text{BVP}}
\backslash \{|x^\prime -x_0|<\lambda^\varepsilon\tau B\}}
|u_k(x^\prime)|^2\ dx^\prime \le C_\ast(k_{hi}+1)\lambda^{-N^\prime}\\
&\le C_\ast\lambda^{1-N^\prime}\ ,\ \text{by\ (11)\ and\ our\ estimate\ for}\
k_{hi}\ .\tag 12\endalign
$$

\noindent Similarly, for $x\in I_{\text{BVP}}$, $x>x_0+\lambda^\varepsilon\tau B$
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,
$$\multline
\Big|\int\limits_{x^\prime\in I_{\text{BVP}}\cap (-\infty,x]}\Bigl\{
\int_{-\infty}^0 g(E)(E-V(x^\prime))_+^{-1/2}\ dE\Bigr\}\ dx^\prime\Big|\le\\
\qquad\qquad \int\Sb V(x^\prime)<E<0\\ |E-V(x_0)|<\tau^2S\endSb
\int C_\ast(E-V(x^\prime))_+^{-1/2}\ dEdx^\prime\le \\
C_\ast\int\limits_{\{V(x^\prime)\le \min(0,V(x_0)+\tau^2S)\}}
\Bigl\{\int\limits_{V(x^\prime)}^{V(x_0)+\tau^2S}\ (E-V(x^\prime))^{-1/2}
\ dE\Bigr\}\ dx^\prime \le\\
C_\ast\int\limits_{\{V(x^\prime)\le\min(0,V(x_0)+\tau^2S)\}}
\Bigl\{\int_{V(x^\prime)}^{V(x^\prime)+\tau^2S}
(E-V(x^\prime))^{-1/2}dE\Bigr\}\ dx^\prime \le\\
C_\ast\tau S^{1/2}\int\limits_{\{V(x^\prime)\le \min(0,V(x_0)+\tau^2S)\}}
\ dx^\prime
\le C_\ast\tau S^{1/2}\cdot \tau B=C_\ast\tau^2\lambda\ ,\endmultline
$$

\noindent and
$$\multline
\Big|\int\limits_{x^\prime \in I_{\text{BVP}}\cap (-\infty,x]}g(0)\cdot \frac 12
(-V(x^\prime))_+^{-1/2}[\phi^\prime(0)]^{-1}\chm(\frac 1\pi \phi(0)-\frac 12)\ 
dx^\prime\Big|\\
\le |g(0)|\ |\chm(\frac 1\pi \phi(0)-\frac 12)|\le C_\ast\ .\endmultline
$$

\noindent Putting these estimates together, we get
$$
\Big|\int\limits_{x^\prime \in I_{\text{BVP}}\cap (-\infty,x]}\rho_{sc}
(x^\prime,g)\ dx^\prime\Big|\le C_\ast(\tau^2\lambda)\ .
$$

\noindent This and (16) give
$$
|H_-(x)|\le C_\ast(\tau^2\lambda)\quad\text{for}\quad |x-x_0|\le \lambda^
\varepsilon\tau B\ .\tag 17
$$

\noindent Estimates (12), (14), (15), (17) are the main results of this section.
We summarize them in the following lemma.

\vglue 1pc
\proclaim{Lemma}  Assume the hypotheses of the WKB Theorem on low
eigenvalues, with $E_\infty=0$.  Suppose $g(E)$ satisfies:
$$
\multline
\text{supp}\ g(E)\subset \{|E-V(x_0)|<\tau^2S\}\quad\text{with}\quad
\lambda^{2\varepsilon-1/2}<\tau<\lambda^{-2\varepsilon}\ ;\quad\text{and}\\
\Big|\bigl(\frac{d}{dE}\bigr)^\beta g(E)\Big|\ \le \hat C_\beta(\tau^2S)^{-\beta}
\ .\endmultline
$$

Let $\rho(x,g)$, $\rho_{sc}(x,g)$ be the microlocalized density for $g$, and
its semiclassical approximation. 

\noindent Define
$$
H(x)=\int\limits_{x^\prime \in I_{\text{BVP}}\cap (-\infty,x]}
[\rho(x^\prime,g)-\rho_{sc}(x^\prime,g)]\ dx^\prime \ .
$$

\noindent Then
$$
|H(x)|\le C_\ast\lambda^{1-N^\prime}\quad\text{if}\quad
x\in I_{\text{BVP}}\ ,\ x<x_0-\lambda^\varepsilon\tau B\ .\tag"(a)"
$$
$$
|H(x)|\le C_\ast(\tau^2\lambda)\quad\text{if}\quad
|x-x_0|\le \lambda^\varepsilon\tau B\ .\tag"(b)"
$$
$$\split
|H(x)|\le C_\ast(\tau^2\lambda)^{-1}\quad\text{if}\quad
x\in I_{\text{BVP}}\ ,\ x>x_0+\lambda^\varepsilon\tau B\ ,\\
\operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge
\overline C_{\#}\lambda^{-1}\endsplit\tag"(c)"
$$
$$\split
|H(x)|\le C_\ast\quad\text{if}\quad x\in I_{\text{BVP}}\ ,\
x>x_0+\lambda^\varepsilon\tau B\ ,\\
\operatornamewithlimits{\min}_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le
\overline C_{\#}\lambda^{-1}\ .\endsplit\tag"(d)"
$$

\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 $|x-y|<cB(x)$, then $c<\frac{B(y)}
{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$.

\item"(Y1)" If $x\in I$ and $\alpha\ge 0$ then $|(\frac{d}{dx})^\alpha V(x)|
\le C_{\alpha}S(x)B^{-\alpha}(x)$.

\item"(Y2)" The set $\{x\in I\mid V(x)<0\}$ is a non-empty interval
$(x_{\text{left}}(0),x_{\text{rt}}(0))$, with
$\text{dist}(x_{\text{left}}(0),\partial I)>cB(x_{\text{left}}(0))$,
$\text{dist}(x_{\text{rt}}(0),\partial I)>cB(x_{\text{rt}}(0))$.

\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 $x<x_{\text{left}}(0)-
\Lambda^KB(x_{\text{left}}(0))$, we have $V(x)\ge \frac{1000}
{|x-x_{\text{left}}(0)|^2}$; and for all $x\in I_{\text{BVP}}$ with
$x>x_{\text{rt}}(0)+\Lambda^KB(x_{\text{rt}}(0))$, 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 $-\infty<E\le V(x_0)+c_2S(x_0)$ we have $\varphi(E)=0$.

\item"(Y10)" The constant $\hat c$ is bounded above by a certain small,
positive number determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$,
$C_\alpha$, $c_2$ above.\endroster

\subhead The WKB Hypothesis\endsubhead
\smallskip

\roster
\item"(Y11)" $\Lambda$ is bounded below by a certain large, positive
number determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$
,$c_2$, $\hat c$, $\hat C_\beta$ above.
\endroster

Let $u_k(x)$ and $E_k$ denote the (real, normalized) eigenfunctions and
eigenvalues of $H$, with $E_k\le 0$.

We denote by $c_{\#}$, $C_{\#}$ etc. constants that depend only on 
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$, $c_2$ in
(Y0)$\ldots$(Y11); while $c_\ast$, $C_\ast$ etc. denote constants that
depend only on $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$,
$c_2$, $\hat c$, $\hat C_\beta$ in (Y0)$\ldots$(Y11).

Our goal in this section is to compare $\rho(x,\varphi)$ with its semiclassical
approximation.  To carry this out, we construct a suitable partition of
unity.  More precisely, we will construct functions $g_{\nu k}(x,E)$ with
the following properties:

(i) Each $\rho(x,g_{\nu k})$ can be compared with its semiclassical
approximation, either by the Oscillatory density lemma or by the Airey density
lemma.

(ii) $\sum\limits_{\nu k}g_{\nu k}(x,E)=\varphi(E)$, except on a set where
microlocalized densities are negligibly small because $E<V(x)$.

We then apply (i) and (ii) to show that
$$
\rho(x,\varphi)=\sum\limits_{\nu k}\rho(x,g_{\nu k})+\rho(x,
\varphi-\sum\limits_{\nu k}g_{\nu k})\approx \sum\limits_{\nu k}
\rho_{sc}(x,g_{\nu k})\approx \rho_{sc}(x,\varphi)
$$
\noindent as desired.  We begin with the construction of the $g_{\nu k}$,
which proceeds as follows.  Introduce an interval
$\tilde I=[x_{\text{left}}(0)-c_{\#}^1B(x_{\text{left}}(0)), x_{\text{rt}}
(0)+c_{\#}^1B(x_{\text{rt}}(0))]\subset I$, for which we have:
$$\multline
S(x)\le C_{\#}S(x_{\text{left}}(0))\quad\text{and}\quad -V^\prime(x)>
c_{\#}S(x)B^{-1}(x)\\
\text{for}\quad x\in \tilde I\cap (-\infty,
x_{\text{left}}(0)]\ ;\quad\text{and}\endmultline
$$
$$\multline
S(x)\le C_{\#}S(x_{\text{rt}}(0))\quad\text{and}\quad
+V^\prime(x)>c_{\#}S(x)B^{-1}(x)\\
\text{for}\quad x\in \tilde I\cap [x_{\text{rt}}(0),+\infty)\ .
\endmultline
$$
\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.
$$
\chi_{{}_{k}}(t)\ \text{is\ supported\ in}\ \{2^{-k}\le t\le
C_\ast2^{-k}\}\ \text{for}\ 0\le k<k_{\max}\ .\tag 6
$$
$$
\chi_{{}_{k_{\max}}}(t)\ \text{is\ supported\ in}\ \{|t|\le C_\ast
2^{-k_{\max}}\}\tag 7
$$
$$
\Big|\Bigl(\frac{d}{dt}\Bigr)^m\chi_{{}_{k}}(t)\Big|\le C_\ast^m
(2^{-k})^{-m}\ \text{for}\ 0\le k\le k_{\max}\ \text{and\ all}\ t, m\ .\tag 8
$$
$$
\sum\limits_{0\le k\le k_{\max}}\chi_{{}_{k}}(t)=1\quad\text{for}\quad
-\Lambda^{-2/7}\le t\le 1/\hat c\ .\tag 9
$$

Then define $h_{\nu
k}(x,E)=\theta_\nu(x)\chi_{{}_{k}}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)$, and set
$g_{\nu k}(x,E)=\varphi(E)\cdot h_{\nu k}(x,E)$.  Thus, we have defined
$g_{\nu k}$.  We want to check properties (i) and (ii) above.  Let us begin
with (ii).  Equation (9) shows that
$$
\sum\limits_{0\le k\le k_{\max}}h_{\nu k}(x,E)=\theta_\nu(x)\ \text{if}\
E\in [V(x)-\Lambda^{-2/7}S(x_\nu), V(x)+\frac{S(x_\nu)}{\hat c}]\ .\tag 10
$$
\noindent If $x\in \text{supp}\ \theta_\nu$, then $c_{\#}S(x_{\nu})<S(x)<
C_{\#}S(x_{\nu})$ and $|V(x)|\le C_{\#}S(x)$.  Hence the energy interval
in (10) contains $[V(x)-c_{\#}\Lambda^{-2/7}S(x),0]$ by virtue of (Y10).
So $\sum\limits_{0\le k\le k_{\max}}h_{\nu k}(x,E)=\theta_\nu(x)$ if
$V(x)-c_{\#}\Lambda^{-2/7}S(x)\le E\le 0$ and $x\in \text{supp}\ \theta_\nu$.
The assumption $x\in \text{supp}\ \theta_\nu$ is unnecessary, since
$h_{\nu k}(x,E)=\theta_\nu(x)=0$ for $x\not\in\ \text{supp}\ \theta_\nu$.
Summing on $\nu$ and using (1), (4), we obtain
$$
\sum\limits_{\nu k}h_{\nu k}(x,E)=1\quad\text{if}\quad x\in \check I\quad
\text{and}\quad V(x)-c_{\#}\Lambda^{-2/7}S(x)\le E\le 0,\ \text{and}\tag 11
$$
$$
\Big|\sum\limits_{\nu k}h_{\nu k}(x,E)\Big|\le C_{\ast}\quad\text{for\ all}\
x, E\ .\tag 12
$$
\noindent Hypotheses (Y3), (Y4) show that $V(x)$ is decreasing in
$[x_{\text{left}}(0),x_0]$ and increasing in $[x_0,x_{\text{rt}}(0)]$.  Hence
for $V(x_0)<E\le 0$, the set $\{x\in [x_{\text{left}}(0),x_{\text{rt}}(0)]
|V(x)<E\}$ 
is an interval $(x_{\text{left}}(E),x_{\text{rt}}(E))$, with
$x_{\text{left}}(0)\le x_{\text{left}}(E)<x_0<x_{\text{rt}}(E)\le x_{\text{rt}}
(0)$, $V(x_{\text{left}}(E))=V(x_{\text{rt}}(E))=E$.  Also
$V(x)>E$ 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
$\{x\in I_{\text{BVP}}\mid
V(x)<E\}=\hfill\break (x_{\text{left}}(E),x_{\text{rt}}(E))$ for $V(x_0)<E\le 0$,
and $V(x)>E$ for $x\in I_{\text{BVP}}\backslash
[x_{\text{left}}(E),x_{\text{rt}}(E)]$.

If $V(x_0)<E\le 0$ and $x\in [x_{\text{left}}(E)-B(x_{\text{left}}(E))
\Lambda^{-3/7},x_{\text{rt}}(E)+B(x_{\text{rt}}(E))\Lambda^{-3/7}]$, then
we will check that
$$
V(x)-c_{\#}\Lambda^{-2/7}S(x)\le E\ ,\quad\text{with}\quad
c_{\#}\quad\text{as\ in\ (11)}\ .\tag 13
$$
\noindent In fact, this is obvious for $x\in [x_{\text{left}}(E),x_{\text{rt}}
(E)]$, since then already $V(x)\le E$.

For $x\in [x_{\text{left}}(E)-\Lambda^{-3/7}B(x_{\text{left}}(E)),
x_{\text{left}}(E)]$, we have
$V(x)\le V(x_{\text{left}}(E))+C_{\#}S(x)B^{-1}(x)|x-x_{\text{left}}(E)|\le
V(x_{\text{left}}(E))+C_{\#}^\prime\Lambda^{-3/7}S(x)
=E+C_{\#}^\prime\Lambda^{-3/7}S(x)$, which implies (13) by the WKB hypothesis
(Y11).  Similarly, (13) holds for $x\in [x_{\text{rt}}(E),x_{\text{rt}}(E)+
\Lambda^{-3/7}B(x_{\text{rt}}(E))]$.  This completes the verification of (13).

Combining (11) and (13), we get
$$\multline
\sum\limits_{\nu k}h_{\nu k}(x,E)=1\quad\text{and}\quad
\sum\limits_{\nu k}g_{\nu k}(x,E)=\varphi(E)\ \text{for}\ V(x_0)<E\le 0\
\text{and}\\
x\in[x_{\text{left}}(E)-\Lambda^{-3/7}B(x_{\text{left}}(E)), x_{\text{rt}}
(E)+\Lambda^{-3/7}B(x_{\text{rt}}(E))]\ .\endmultline\tag 14
$$
\noindent (Note that the $x$-interval in (14) is contained in $\check I$.)

\noindent Equation  (14) is the precise form of property (ii) above.

Next, we prepare to check property (i), by proving the basic estimates for
the derivatives of $h_{\nu k}(x,E)$ and $g_{\nu k}(x,E)$.
The derivative $\partial_x^\alpha\partial_E^\beta\chi_{{}_{k}}\bigl(
\frac{E-V(x)}{S(x_\nu)}\bigr)$ is a sum of terms
$$\multline
\Bigl[\Bigl(\frac{d}{dt}\Bigr)^m\chi_{{}_{k}}(t)
\mid_{t=\frac{E-V(x)}{S(x_\nu)}}\Bigr]
\cdot \prod\limits_{\mu=1}^m \Bigl[\partial_x^{\alpha_\mu}\partial_E
^{\beta_\mu}\Bigl(\frac{E-V(x)}{S(x_\nu)}\Bigr)\Bigr]\ ,\\
\text{with}\quad \alpha_\mu+\beta_\mu\ge 1, \alpha_1+\ldots+\alpha_m
=\alpha, \beta_1+\ldots+\beta_m=\beta.\\
\text{In\ particular,}\ 0\le m\le \alpha+\beta\ .\endmultline\tag 15
$$
\noindent If $\beta_\mu\ge 2$ then
$\partial_x^{\alpha_\mu}\partial_E^{\beta_\mu}\bigl(\frac{E-V(x)}
{S(x_\nu)}\bigr)=0$.

\noindent If $\beta_\mu=1$ and $\alpha_\mu\ge 1$, then again
$\partial_x^{\alpha_\mu}\partial_E^{\beta_\mu}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)=0$.

\noindent If $\beta_\mu=1$ and $\alpha_\mu=0$, then
$\partial_x^{\alpha_\mu}\partial_E^{\beta_\mu}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)
=S^{-1}(x_\nu)$.

\noindent If $\beta_\mu=0$ and $\alpha_\mu\ge 1$ then
$|\partial_x^{\alpha_\mu}\partial_E^{\beta_\mu}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)
|\le C_{\#}^{\alpha_\mu}B^{-\alpha_\mu}(x_\nu)$ for
$x\in\ \text{supp}\ \theta_\nu$.  Hence $|\partial_x^{\alpha_\mu}
\partial_E^{\beta_\mu}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)|\le C_{\#}
^{\alpha_\mu\beta_\mu}B^{-\alpha_\mu}(x_\nu)S^{-\beta_\mu}(x_\nu)$ for
$\alpha_\mu+\beta_\mu\ge 1$, $x\in\ \text{supp}\ \theta_\nu$.  This and
(8) show that the term (15) is dominated by \hfill\break $C_{\#}^{\alpha\beta}
(2^{-k})^{-m}B^{-\alpha}(x_\nu)S^{-\beta}(x_\nu)$, which in turn is
dominated by \hfill\break
$C_{\#}^{\alpha\beta}(2^{-k}B(x_\nu))^{-\alpha}(2^{-k}
S(x_\nu))^{-\beta}$, since $0\le m\le \alpha+\beta$.  Hence,
$$
\Big|\partial_x^\alpha\partial_E^\beta\chk\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)\Big|
\le C_{\#}^{\alpha\beta}(2^{-k}B(x_\nu))^{-\alpha}(2^{-k}S(x_\nu))^{-\beta}
\quad\text{for}\ x\in\ \text{supp}\ \theta_\nu \ .\tag 16
$$
\noindent Together with (3), this yields
$$
\Big|\partial_x^\alpha\partial_E^\beta
h_{\nu k}\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)\Big| \le
C_\ast^{\alpha\beta}(2^{-k}B(x_\nu))^{-\alpha}(2^{-k}S(x_\nu))^{-\beta} \
,\quad\text{all}\quad (x,E)\ .\tag 17 $$

\noindent To prove analogous estimates for $g_{\nu k}(x,E)$, we need to
compare $S(x_\nu)$ with $S(x_0)$.  Hence make the following argument.
For $x\in [x_{\text{left}}(0),x_{\text{rt}}(0)]$, define
$I_x=\hfill\break [x_{\text{left}}(0),x_{\text{rt}}(0)]\cap [x-c_{\#}^2B(x),
x+c_{\#}^2B(x)]$.  This interval has length at least $c_{\#}B(x)$, and
hypotheses (Y3), (Y4) show that $V(\tilde x)$ varies by at least
$c_{\#}S(x)$ as $\tilde x$ varies in $I_x$.  Hence,
$c_{\#}S(x)\le \operatornamewithlimits{\max}_{\tilde x\in I_x}
|V(\tilde x)|\le |V(x_0)|$ (since $V(x_0)\le V(\tilde x)\le 0$
in $I_x$) $\le C_{\#}S(x_0)$.  Thus, $S(x)\le C_{\#}S(x_0)$ for
$x\in [x_{\text{left}}(0),x_{\text{rt}}(0)]$.  The defining properties
of $\tilde I$ show that for every $x\in \tilde I$ there is an $x^\prime
\in [x_{\text{left}}(0), x_{\text{rt}}(0)]$ with $S(x)\le C_{\#}S(x^\prime)$.
(In fact, set $x^\prime=\ \text{either}\ x, x_{\text{left}}(0),\ \text{or}\
x_{\text{rt}}(0)$.) We
conclude that $S(x)\le C_{\#}S(x_0)$ for all $x\in \tilde I$, hence
$$
S(x_\nu)\le C_{\#}S(x_0),\ \text{all}\ \nu\ .\tag 18
$$
\noindent From (17), (18), and hypothesis (Y8) we get
$$
|\partial_x^\alpha\partial_E^\beta g_{\nu k}(x,E)|\le C_\ast^{\alpha\beta}
(2^{-k}B(x_\nu))^{-\alpha}(2^{-k}S(x_\nu))^{-\beta}\ \text{for\ all}\
(x,E)\ .\tag 19
$$  
\noindent The basic results needed to establish property (i) for $g_{\nu k}$
are as follows.

\vglue 1pc
\proclaim{Lemma 1} Suppose $V(x_0)+\frac{1}{10}c_2S(x_0)\le E_0\le 0$, with
$c_2$ as in hypothesis (Y9).  Then the hypotheses (Hyp 0)$\ldots$(Hyp 10) of
the WKB Theorems are satisfied, 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) depend only on the constants $\varepsilon$, $K$, $N$, $c$,
$C$, $c_1$, $c_2$, $C_\alpha$ in our present hypotheses (Y0)$\ldots$(Y11).
The number called $\Lambda$ in (Hyp 0)$\ldots$(Hyp 10) is greater than
or equal to the number $\Lambda$ appearing in (Y0)$\ldots$(Y11).
\endproclaim

\proclaim{Lemma 2} For each $\nu$, $k$ with $0\le k<k_{\max}$, one of the
following alternatives holds.

(A) $g_{\nu k}(x,E)=0$ for all $x\in I_{\text{BVP}}$, $E\in (-\infty,0]$.

(B) For suitable $\tilde x_0$, $E_{\ell o}$, $E_{hi}$, the functions
$V(x)$, $S(x)$, $B(x)$, $g_{\nu k}(x,E)$ satisfy the hypotheses (A1)$\ldots$(A6)
of the Oscillatory Density Lemma, with $x_\nu$ in place of $x_0$, with $300 K$
in place of $K$, and with $\tau=2^{-k}$.  The constants called
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$ in (A1), and
the constant called $c_3$ in (A2) depend only on the constants $\varepsilon$,
$K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$ in our present hypotheses
(Y0)$\ldots$(Y11).  The constants called $\hat c$, $\hat c_1$, $\hat C$,
$\hat C_{\alpha\beta}$, in (A2)$\ldots$(A5) depend only on the constants
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$, $\hat c$,
$\hat C_\beta$ in (Y0)$\ldots$(Y11).  The number called $\Lambda_{\min}$
in (A1)$\ldots$(A6) is greater than or equal to the number $\Lambda$
appearing in (Y0)$\ldots$(Y11).\endproclaim

\vglue 1pc
\proclaim{Lemma 3} For each $\nu$, one of the following alternatives holds.

(A) $g_{\nu k_{\roman{max}}}(x,E)=0$ whenever $E\le 0$.

(B) $V(x)$, $S(x)$, $B(x)$, $g_{\nu k_{\roman{max}}}(x,E)$ satisfy the
hypotheses (X0)$\ldots$(X12) of the Airey density lemma, with
$E_0=\min(V(x_\nu),0)$, with suitable $\delta E$, with
$c_\ast\Lambda^{-2/7}\le \frac{\delta x}{B(x_\nu)}<C_\ast
\Lambda^{-2/7}$, and with $300 K$ in place of $K$.  The constants called
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$ in (X0)$\ldots$(X12)
depend only on the constants $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$,
$C_\alpha$ in our present hypotheses (Y0)$\ldots$(Y11).  The constants
called $\hat c$, $\hat C_{\alpha\beta}$ in (X0)$\ldots$(X12) depend only on the
constants $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$
$\hat c$, $\hat C_\beta$ in (Y0)$\ldots$(Y11).  The number $\Lambda_{\min}$
in (X0)$\ldots$(X12) is greater than or equal to the number $\Lambda$
in (Y0)$\ldots$(Y11).  Also, $|x_\nu-x_{\text{left}}(E_0)|\le 2\hat c
B(x_\nu)$.

(C) The same conclusions as (B) hold, with the r$\hat o$les of ``left" and
``right" interchanged.  (In particular, in place of the Airey density
lemma, we use its analogue with the r$\hat o$les of ``left" and ``rt"
interchanged.)\endproclaim
\medskip

The proofs of Lemmas 1, 2, 3 are as follows.

\vglue 1pc
\demo{Proof of Lemma 1} We have to check (Hyp 0)$\ldots$(Hyp 10).  First
of all, (Hyp 0) and (Hyp 1) are our present hypotheses (Y0), (Y1).  To
check (Hyp 2), let $E\in (V(x_0),0]$.  We know that
$V(x_{\text{left}}(E))=V(x_{\text{rt}}(E))=E$ with $x_{\text{left}}(0)\le
x_{\text{left}}(E)<x_{\text{rt}}(E)\le x_{\text{rt}}(0)$, that
$$\align
V(x)<E\quad &\text{for}\quad x\in (x_{\text{left}}(E),x_{\text{rt}}(E))\ ,\quad
\text{and\ that}\\
V(x)>E\quad &\text{for}\quad x\in I_{\text{BVP}}\backslash [x_{\text{left}}
(E),x_{\text{rt}}(E)]\ .\endalign
$$
\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
$x_{\text{left}}(E_0)<x_0$.  Similarly, $+V^\prime(x_{\text{rt}}(E_0))
>c_{\#}^\prime S(x_{\text{rt}}(E_0))B^{-1}(x_{\text{rt}}(E_0))$.  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 $|x-x_0|<c_{\#}
B(x_0)$.  Since $E_0\ge V(x_0)+\frac{1}{10}c_2S(x_0)$, it follows that
$E_0-V(x)\ge \frac{1}{20}c_2S(x_0)>\tilde c_{\#}S(x)$ if
$|x-x_0|<c_{\#}B(x_0)$.  So it is enough to look at
$x\in [x_{\text{left}}(E_0)+c_{\#}^{\prime\prime}B(x_{\text{left}}
(E_0))$, $x_{\text{rt}}(E_0)-c_{\#}^{\prime\prime}B(x_{\text{rt}}(E_0))]
\backslash \{|x-x_0|<c_{\#}B(x_0)\}$.  For such $x$, either
$$
x_{\text{left}}(E_0)<x-\tilde c_{\#}^\prime B(x)<x<x_0-c_{\#}B(x_0)\\
$$
\noindent or
$$
x_{\text{rt}}(E_0)>x+\tilde c_{\#}^\prime B(x)>x>x_0+c_{\#}B(x_0)\ .
$$

\noindent Here we discuss only the first case since the second is analogous.
For $x_{\text{left}}(E_0)<\hfill\break x^\prime<x$ we have 
$-V^\prime(x^\prime)\ge
\tilde c_{\#}^{\prime\prime}S(x^\prime)B^{-1}(x^\prime)>0$, 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 $x<x_{\text{left}}(E_0)-\frac
12(\lambda(x_{\text{left}}(E_0)))^{300K}B(x_{\text{left}}(E_0))$.   We
have $\lambda(x_{\text{left}}(E_0))\hfill\break \ge c_{\#}\Lambda$, and
$|x_{\text{left}}(E_0)-x_{\text{left}}(0)|\le |I|\le \Lambda^K
B(x_{\text{left}}(E_0))$ by (Y7), so
$$\split
x<[x_{\text{left}}(0)+\Lambda^KB(x_{\text{left}}(E_0))]-\frac 12c_{\#}
\Lambda^{300K}B(x_{\text{left}}(E_0))\\<x_{\text{left}}(0)-
\Lambda^{200 K}B(x_{\text{left}}(E_0))\endsplit
$$
\noindent (by the WKB hypothesis (Y11)) $<x_{\text{left}}(0)-
\Lambda^{200K}\cdot\Lambda^{-K}B(x_{\text{left}}(0))$ (by another
application of (Y7)).

\noindent Hence $x<x_{\text{left}}(0)-\Lambda^KB(x_{\text{left}}(0))$, so
(Y6) yields $V(x)\ge \frac{1000}{|x-x_{\text{left}}(0)|^2}
\ge \hfill\break
\frac{1000}{|x-x_{\text{left}}(E_0)|^2}$, since $x<x_{\text{left}}(0)\le
x_{\text{left}}(E_0)$.  Thus, $V(x)\ge \frac{1000}{|x-x_{\text{left}}(E_0)|^2}$
if $x\in I_{\text{BVP}}$ and $x<x_{\text{left}}(E_0)-\frac
12(\lambda(x_{\text{left}}(E_0)))^{300K}B(x_{\text{left}}(E_0))$.  Similarly,
$V(x)\ge \frac{1000}{|x-x_{\text{rt}}(E_0)|^2}$ if $x\in I_{\text{BVP}}$
and $x>x_{\text{rt}}(E_0)+\frac 12(\lambda(x_{\text{rt}}(E_0)))^{300K}
B(x_{\text{rt}}(E_0))$.  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
$$\multline
c_{\#}S(x_\nu)B^{-1}(x_\nu)< (\text{sgn})\cdot V^\prime(x)<C_{\#}
S(x_\nu)B^{-1}(x_\nu)\\
\text{for}\quad |x-x_\nu|<c_{\#}^\prime B(x_\nu)\ .\endmultline\tag"(20.1)"
$$

\noindent (This follows from (Y3), (Y4) and the defining properties
of $\tilde I$.  Recall that $x_\nu\in \Tilde I$.)

For a small, positive constant $b$ to be picked later, we define
$$
\tilde x_0=x_\nu-b(\text{sgn})B(x_\nu)\ .\tag"(20.2)"
$$

\noindent If $b$ satisfies
$$
b<c_{\#}^\prime\quad\text{with}\quad c_{\#}^\prime\quad\text{as\ in\ (20.1),\ and}
\tag"(20.3)"
$$
$$
\hat c<\frac 12 b\ ,\tag"(20.4)"
$$

\noindent then from (20.1) we get
$$\multline
c_{\#}^{\prime\prime}bS(x_\nu)<V(x)-V(\tilde x_0)<C_{\#}^{\prime\prime}
bS(x_\nu)\\
\text{for}\quad |x-x_\nu|<\hat cB(x_\nu)\ .\endmultline\tag"(20.5)"
$$

\noindent We set $E_{hi}=0$ and $E_{\ell o}=V(\tilde x_0)+\frac 12 c_{\#}
^{\prime\prime}bS(x_\nu)$, with $c_{\#}^{\prime\prime}$ as in (20.5).
Note that $E_{\ell o}<0$.  In fact, if $E_{\ell o}\ge 0$, then (20.5) would
show that $V(x)>0$ 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
$|x_\nu-\tilde x_0|<c_{\#}^{\text{HYP}}B(x_\nu)$ since $\tilde x_0=x_\nu$;
and
$$\multline
E_{\ell o}=V(x_0)+\frac 34 c_2S(x_0)\ge V(x_\nu)+\frac 14
c_2S(x_0)\\
\text{(since\ we\ are\ in\ case\ (i))}\ \ge V(x_\nu)+c_{\#}S(x_\nu)
\endmultline
$$

\noindent (by (18)) $=V(\tilde x_0)+c_{\#}S(\tilde x_0)$.  This
proves (A2) in case (i), with $c_3$ of the form $c_{\#}$.  In case (ii),
the definition (20.2) yields $|x_\nu-\tilde x_0|<c_{\#}^{\text{HYP}}
B(x_\nu)$, provided we take
$$
b<c_{\#}^{\text{HYP}}\ .\tag"(20.9)"
$$
\noindent Also from (20.2) we get
$$
S(x_\nu)\ge c_{\#}S(\tilde x_0)\ ,\tag"(20.10)"
$$
\noindent provided we take
$$
b<c_{\#}\quad\text{for\ a\ suitable\ small\ constant}\ c_{\#}\ .\tag"(20.11)"
$$

\noindent Hence, $E_{\ell o}=V(\tilde x_0)+\frac 12 c_{\#}^{\prime\prime}
bS(x_\nu)\ge V(\tilde x_0)+[\frac 12 c_{\#}^{\prime\prime}bc_{\#}]
S(\tilde x_0)$.  Thus, (A2) holds in both cases (i) and (ii), with $x_\nu$
in place of $x_0$, and with $c_{\#}^{\text{HYP}}$ in place of $c$.  The
quantity called $c_3$ in (A2) either has the form $c_{\#}$ or $c_{\#}b$.

Next we check (A3), with $x_\nu$ in place of $x_0$, and with $\tau=2^{-k}$.
Thus, suppose we are given $(x,E)\in I_{\text{BVP}}\times \Bbb R^1$ with
$g_{\nu k}(x,E)\neq 0$.  We must have $\theta_\nu(x)\neq 0$
and $\chk\bigl(\frac{E-V(x)}{S(x_\nu)}\bigr)\neq 0$.  Therefore,
$$
|x-x_\nu|<\hat cB(x_\nu)\quad\text{and}\quad 
2^{-k}S(x_\nu)<E-V(x)<C_\ast2^{-k}S(x_\nu)\ .\tag"(20.12)"
$$

\noindent To complete the proof of (A3), it remains only to check that
$E_{\ell o}+\hat c_1S_{\min}(E_{\ell o})\le E\le E_{hi}$ or else
$E>0$.  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
$V(x_0)<E_{\ell o}$, we have $S_{\min}(E_{\ell o})=\ \text{inf}_{V(x)<E_{\ell o}}
S(x)\le S(x_0)$.  Since $g_{\nu k}(x,E)\neq 0$, we have $\varphi(E)\neq 0$
and hence $E\ge V(x_0)+c_2S(x_0)=E_{\ell o}+\frac 14 c_2S(x_0)\ge E_{\ell o}
+\frac 14 c_2S_{\min}(E_{\ell o})$, proving (20.13) with $\hat c_1=\frac 18 c_2$.

\noindent To check (20.13) in case (ii), we use (20.5), (20.10), (20.12) to
write
$$\multline
E>V(x)>V(\tilde x_0)+c_{\#}^{\prime\prime}bS(x_\nu)=E_{\ell o}+\frac 12
c_{\#}^{\prime\prime}bS(x_\nu)\\
\ge E_{\ell o}+\frac 12 c_{\#}^{\prime\prime}c_{\#}bS(\tilde x_0)\ .\endmultline
\tag"(20.14)"
$$

\noindent Also $V(\tilde x_0)<E_{\ell o}<0$, so $S_{\min}(E_{\ell o})=
\operatornamewithlimits{\text{inf}}_{V(x)<E_{\ell o}}S(x)\le S(\tilde x_0)$.
Hence (20.14) implies
$E\ge E_{\ell o}+(\frac 12 c_{\#}^{\prime\prime}c_{\#}b)S_{\min}(E_{\ell o})$,
which proves (20.13) with $\hat c_1=\frac 14 c_{\#}^{\prime\prime}c_{\#}b$.
The proof of (A3) is complete.


\noindent To prove (A4), we recall that $\Lambda_{\min}=\ \text{inf}_{E\in
[E_{\ell o},E_{hi}]}\Lambda(E)\ge \Lambda$ and that $\tau=2^{-k}$
with $\frac 12 \Lambda^{-2/7}<\tau\le 1$.  Hence
$\Lambda_{\min}^{\varepsilon-2/3}\le \Lambda^{\varepsilon-2/3}\le \frac 12
\Lambda^{-2/7}<\tau\le 1$, which yields (A4).


\noindent Since $\tau=2^{-k}$, hypothesis (A5) is merely our estimate (19).

\noindent Now we pick the constant $b$.  We merely take
$b=c_{\#}^{\prime\prime\prime}$
small enough to satisfy (20.3), (20.8), (20.9), and (20.11).  From (Y10)
we see that (20.4) holds also.  Thus, we have proven (A1)$\ldots$(A5)
with constants $c$, $C$, $c_1$, $c_2$, $c_3$, $\hat c_1$, $C_\alpha$ of the
form $C_{\#}$, and with constants $\hat c$, $\hat C$, $\hat C_{\alpha\beta}$
of the form $C_\ast$.  Since $\Lambda_{\min}\ge \Lambda$, hypothesis (A6)
follows from (Y11).  We have proven everything asserted in Lemma 2.

\vglue 1pc
\demo{Proof of Lemma 3}  We have to establish one of the alternatives
(A), (B), (C) for a given $\nu$.  We distinguish several cases.

First of all, if $V(x_\nu)\le V(x_0)+\frac 12c_2S(x_0)$, then alternative
(A) holds.  To see this, note that $\theta_\nu(x)\chi_{{}_{k_{\max}}}\bigl(
\frac{E-V(x)}{S(x_\nu)}\bigr)$ is supported in $\{|E-V(x)|\le
C_\ast\Lambda^{-2/7}S(x_\nu)$, $|x-x_\nu|<\hat cB(x_\nu)\}$,
which is contained in $\{|E-V(x)|\le C_\ast\Lambda^{-2/7}S(x_\nu), |V(x)-
V(x_\nu)|\le C_{\#}\hat cS(x_\nu)\}$.  This in turn is contained
in $\{|E-V(x_\nu)|\le \frac{1}{10}c_2S(x_0)\}$ by (18) and hypotheses
(Y10), (Y11).  Consequently, if $V(x_\nu)\le V(x_0)+\frac 12c_2S(x_0)$,
then $h_{\nu k}(x,E)$ is supported in $\{E\le V(x_0)+\frac 23
c_2S(x_0)\}$.
Since $\varphi(E)=0$ for $E\le V(x_0)+c_2S(x_0)$, it follows that
$g_{\nu k_{\max}}(x,E)\equiv 0$.  Hence we are in alternative (A).

For the rest of the proof of Lemma 3, we may therefore assume that
$$
V(x_\nu)>V(x_0)+\frac 12 c_2S(x_0)\ ,\tag 21
$$
\noindent which implies $|x_\nu-x_0|>c_{\#}B(x_0)$.  Hence either
$x_\nu<x_0-c_{\#}B(x_0)$ or $x_\nu>x_0+c_{\#}B(x_0)$.  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 $|E-E_0|<c_{\#}S(x_\nu)$.  Hence
$$\split
|E-E_0|<\overline c_{\#}S(x_\nu)\quad\text{implies}\quad
c_{\#}S(x_{\text{left}}(E_0))<S(x_{\text{left}}(E))<C_{\#}S(x_{\text{left}}
(E_0))\\
\text{for\ a\ small\ constant}\ \overline c_{\#}\ .\endsplit\tag 29
$$

\noindent We define $(\delta E)=\min\{\overline c_{\#}S(x_\nu),
10^{-9}c_2S(x_0)\}$.  Thus, (28) gives
$$
\Cal J=[E_0-\delta E, E_0+\delta E]\cap (-\infty,0]\subset [V(x_0)+\frac 14
c_2S(x_0),0]\ ,\tag 30
$$
\noindent and (29) gives
$$
c_{\#}S(x_{\text{left}}(E_0))<S(x_{\text{left}}(E))<C_{\#}S(x_{\text{left}}
(E_0))\quad\text{for\ all}\quad E\in \Cal J\ .\tag 31
$$

\noindent Equation (30) and Lemma 1 show that for any $E\in \Cal J$, the
hypotheses of the WKB Theorems hold, with $E_\infty=0$ and with $E$ in place
of $E_0$.  This immediately implies hypotheses (X0)$\ldots$(X4) and
(X6), (X7) of the Airey density lemma.  The constants called
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$ in 
(X0)$\ldots$(X4), (X6), (X7)
are determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$
in (Y0)$\ldots$(Y11), and we have $\Lambda_{\min}=
\operatornamewithlimits{\text{inf}}_{E\in \Cal J}\Lambda(E)\ge \Lambda$.  These conclusions are all contained
in those of Lemma 1.

To check (X5), notice that
$$
c_{\#}S(x_\nu)<\delta E<C_{\#}S(x_\nu)\ ,\tag 32
$$
\noindent by (18) and the definition of $\delta E$.  Hence (X5) follows
at once from (31), and the inequality $|x_{\text{left}}(E_0)-x_\nu|=
|\overline x_\nu-x_\nu|<2\hat cB(x_\nu)$, which is contained in (26), (27).

Hypothesis (X8) follows trivially from (Y7), since $\Lambda_{\min}\ge
\Lambda$.

Next we check (X9) with suitable quantities in place of $\hat c$, $\delta x$.
Recall that $g_{\nu k_{\max}}(x,E)$ is supported in $\{|x-x_\nu|<
\hat cB(x_\nu)$, $\frac{|E-V(x)|}{S(x_\nu)}<C_\ast\Lambda^{-2/7}\}$, which is
contained in $\{|E-V(x_\nu)|\le C_{\#}\hat cS(x_\nu)+C_\ast\Lambda^{-2/7}
S(x_\nu)\}$, which in turn is contained in $\{|E-V(x_\nu)|\le
C_{\#}^\prime\hat c(\delta E)\}$ by (32) and (Y11).


>From (26), (32) we get also $|V(x_\nu)-E_0|<C_{\#}^{\prime\prime}
\hat cS(x_\nu)<C_{\#}^{\prime\prime\prime}\hat c(\delta E)$.  Hence
the support of $g_{\nu k_{\max}}(x,E)$ is contained in the set
$\{|E-E_0|<\tilde C_{\#}\hat c(\delta E)\}$.  So we take $\hat c_+=
\tilde C_{\#}\hat c$ in place of $\hat c$, and we have
$$
\text{supp}\ g_{\nu k_{\max}}(x,E)\subset \{|E-E_0|<\hat c_+(\delta E)\}\ .
\tag 33
$$

Again, $\text{supp}\ g_{\nu k_{\max}}(x,E)$ is contained in
$\{|x-x_\nu|<\hat cB(x_\nu)$, $|E-V(x)|<C_\ast\Lambda^{-2/7}
S(x_\nu)\}$, and for $|x-x_\nu|<\hat cB(x_\nu)$ and 
$|E-E_0|<\hat c_+(\delta E)$ we have $c_{\#}\frac{|E-V(x)|}{S(x_\nu)}
<\frac{|x-x_{\text{left}}(E)|}{B(x_\nu)}<C_{\#}\frac{|E-V(x)|}{S(x_\nu)}$,
by (23).  Hence $\text{supp}\ g_{\nu k_{\max}}(x,E)$ is contained in
$\{|x-x_{\text{left}}(E)|<C_\ast\Lambda^{-2/7}B(x_\nu)\}$.  So we take
$(\delta x)=C_\ast\Lambda^{-2/7}B(x_\nu)$, and we have
$$
\text{supp}\ g_{\nu k_{\max}}(x,E)\subset\{|x-x_{\text{left}}(E)|<
(\delta x)\}\ .\tag 34
$$
\noindent In view of (33), (34), the hypothesis (X9) is reduced to checking
that
$$
\lambda^{2\varepsilon-2/3}(x_{\text{left}}(E_0))\cdot B(x_{\text{left}}(E_0))
<\delta x<\frac{1}{20}\lambda^{-2\varepsilon}(x_{\text{left}}(E_0))\cdot
B(x_{\text{left}}(E_0))\ ,
$$
\noindent i.e.
$$
\lambda^{2\varepsilon-2/3}(\overline x_\nu)<\frac{\delta x}
{B(\overline x_\nu)}<\frac{1}{20}\lambda^{-2\varepsilon}(\overline x_\nu)
\qquad \text{(see\ (27))}\ .
$$
\noindent By (26) we see that $c_{\#}B(x_\nu)<B(\overline x_\nu)<C_{\#}
B(x_\nu)$ and $c_{\#}\lambda(\overline x_\nu)<\lambda(x_\nu)<C_{\#}
\lambda(\overline x_\nu)$.  Hence the hypothesis (X9) will follow from the
definition of $\delta x$, if we can prove
$$
\lambda^{3\varepsilon-2/3}(x_\nu)<\Lambda^{-2/7}<\lambda^{-3\varepsilon}
(x_\nu)\ .\tag 35
$$

\noindent The first inequality in (35) is obvious, since
$\lambda(x_\nu)\ge c_{\#}\Lambda$.  To verify the second inequality in
(35), we argue as follows.  Hypothesis (Y7) shows that
$$\multline
\Lambda^{-1}=\int_{x_{\text{left}}(0)}^{x_{\text{rt}}(0)}\frac{dx}
{\lambda(x)B(x)}\le \Lambda^{10K}\lambda^{-1}(x_\nu)\cdot
\int_I\frac{dx}{B(x)}\\
\le \Lambda^{10K}\lambda^{-1}(x_\nu)\cdot\int_I
\frac{dx}{\Lambda^{-K}|I|}=\Lambda^{11K}\lambda^{-1}(x_\nu)\ .\endmultline
$$
\noindent Hence
$\Lambda^{-(11K+1)}\le \lambda^{-1}(x_\nu)$, so $\Lambda^{-\frac 27}\le
\lambda^{-\frac 27\cdot\frac{1}{11K+1}}(x_\nu)\le \lambda^{-\frac{1}
{50 K}}(x_\nu)\le \lambda^{-3\varepsilon}(x_\nu)$ 
since $\varepsilon<\frac{1}{1000 K}$. 
The second inequality in (35) is verified and the proof of (X9) is
complete.

In view of our definition of $\delta x$ and the fact that
$B(x_{\text{left}}(E_0))=B(\overline x_\nu)\sim B(x_\nu)$ and
$S(x_{\text{left}}(E_0))=S(\overline x_\nu)\sim S(x_\nu)$ as in the
proof of (X9), the hypothesis (X10) amounts to the estimates
$|\partial_x^\alpha\partial_E^\beta g_{\nu k_{\max}}(x,E)|\le\hfill\break
C_\ast^{\alpha\beta}[\Lambda^{-2/7}B(x_\nu)]^{-\alpha}
[\Lambda^{-2/7}S(x_\nu)]^{-\beta}$.  These estimates are equivalent
to (19), since $2^{-k_{\max}}\sim \Lambda^{-2/7}$.  Thus (X10) holds.           

To verify (X11), we recall that in place of $\hat c$ we are using here
$\hat c_+=\tilde C_{\#}\hat c$.  Since $\tilde C_{\#}$ and the quantities
called $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$ in
(X0)$\ldots$(X10) are determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$,
$c_2$, $C_\alpha$ in our present hypotheses (Y0)$\ldots$(Y11), we see that
(X11) follows at once from (Y10).

The last hypothesis (X12) follows from the WKB hypothesis (Y11), since
$\Lambda_{\min}\hfill\break \ge \Lambda$ and the quantities called $\varepsilon$,
$K$, $N$, $c$, $c_1$, $C$, $C_\alpha$, $\hat c$, $\hat C_{\alpha\beta}$
in (X0)$\ldots$(X12) are determined by $\varepsilon$, $K$, $N$, $c$, $C$,
$c_1$, $C_\alpha$, $c_2$, $\hat c$, $\hat C_\beta$ in (Y0)$\ldots$(Y11).

Thus, we have verified all the hypotheses (X0)$\ldots$(X12) of the Airey
density lemma.  We have already checked that $\Lambda_{\min}\ge \Lambda$,
and equations (26), (27) yield $|x_\nu-x_{\text{left}}(E_0)|<2\hat cB(x_\nu)$.
Therefore, alternative (B) holds here.  The proof of Lemma 3 is complete.
$\qquad\blacksquare$\enddemo
\smallskip

Now we can apply Lemmas 2 and 3, the Airey density lemma, and the
Oscillatory density lemma to control $\rho(x,g_{\nu k})$.  Let
$I_\nu=\{x\mid |x-x_\nu|<\tilde c_{\#}B(x_\nu)\}$, with $\tilde c_{\#}$
small.  If $0\le k<k_{\max}$, then Lemma 2 and the Oscillatory density lemma
allow us to compare $\rho(x,g_{\nu k})$ with $\rho_{sc}(x,g_{\nu k})$ on 
$I_\nu$.  If $k=k_{\max}$, then according to Lemma 3, we are in one of the
cases (A), (B), (C).  In case (A), both $\rho(x,g_{\nu k})$ and
$\rho_{sc}(x,g_{\nu k})$ are identically zero.  In case (B), we can apply
the Airey density lemma to compare $\rho(x,g_{\nu k})$ with $\rho_{sc}(x,g_{\nu
k})$ on $I_\nu$.  (Note that $I_\nu$ is contained in the interval called
$I_{\text{left}}$ in the Airey density lemma, since $|x_\nu-x_{\text{left}}
(E_0)|\le 2\hat cB(x_\nu)$ in case (B).)  Similarly, in case (C) we can
apply the analogue of the Airey density lemma, with ``left" and ``rt" reversed,
to compare $\rho(x,g_{\nu k})$ with $\rho_{sc}(x,g_{\nu k})$ on $I_\nu$.
Thus, in all cases we obtain a comparison of $\rho(x,g_{\nu k})$ with
$\rho_{sc}(x,g_{\nu k})$.  From this process, we obtain the following results.
$$
\rho(x,g_{\nu k})-\rho_{sc}(x,g_{\nu k})=\frac{d}{dx}H_{\nu
k}(x)\quad\text{on}\quad I_\nu \ . \tag 36
$$
\medskip

\noindent{\it CASE I\/}: If $\min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge
\overline C_{\#}\Lambda^{-1}$, then:
$$\multline
(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\le C_\ast(2^{-k})^{-3/4}\lambda^{-1/2}
(x_\nu)+C_\ast\Lambda^{4\varepsilon-2}(2^{-k})^{-5/2}\lambda(x_\nu)\\
+C_\ast\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)\\
\text{for}\quad 0\le k<k_{\max}\ \endmultline\tag 37
$$
$$\multline
(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\le C_\ast\lambda^{-1/3}(x_\nu)+
C_\ast\lambda^{-1}(x_\nu)(\Lambda^{-2/7})^{-5/2}+C_\ast\Lambda^{4\varepsilon-2}
\lambda(x_\nu)(\Lambda^{-2/7})^{1/2}\\
+C_\ast\Lambda^{-1}\lambda(x_\nu)\Lambda^{-2/7}+C_\ast
\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)\\
\text{for}\quad k=k_{\max}\ .\endmultline\tag 38
$$

\medskip

\noindent{\it CASE II\/}: If $\min_{k \in \Bbb Z}|\phi(0)-\pi(k+1/2)|
\le \overline C_{\#}\Lambda^{-1}$, then:
$$\multline
(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\le C_\ast(2^{-k})^{1/2}+C_\ast(2^{-k})
^{-3/4}\lambda^{-1/2}(x_\nu)\\
+C_\ast\Lambda^{4\varepsilon-2}(2^{-k})^{-5/2}
\lambda(x_\nu)+C_\ast\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)\\
\text{for}\quad 0\le k<k_{\max} ;\endmultline\tag 39
$$
$$\multline
(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\le C_\ast\lambda^{-1/3}(x_\nu)
+C_\ast\lambda^{-1}(x_\nu)\cdot(\Lambda^{-2/7})^{-5/2}\\
+C_\ast\Lambda^{-1}
\lambda(x_\nu)(\Lambda^{-2/7})^{1/2}+C_\ast\Lambda^{-\frac{N^{\prime\prime}}
{2}}(\phi(0)+1)\\
\text{for}\quad k=k_{\max}\ .\endmultline\tag 40
$$

\noindent We apply (37)$\ldots$(40) to bound $\sum\limits_{0\le k\le k_{\max}}
(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}$ for fixed $\nu$.  In {\it Case I\/}, we
have to sum the right-hand side of (37) over $k=0,1, \ldots,k_{\max}-1$,
and then add the result to the right-hand side of (38).  Recalling that
$2^{-k_{\max}}\sim \Lambda^{-2/7}$ and $\lambda(x_\nu)\ge c_{\#}\Lambda$,
we obtain the following:
$$\multline
\sum\limits_{0\le k<k_{\max}}[1^{\text{rst}}\text{term\ on\ RHS\ of\ (37)}]\\
\le C_\ast(\Lambda^{-2/7})^{-3/4}(\Lambda^{-3/2}\lambda(x_\nu))=C_\ast
\Lambda^{-9/7}\lambda(x_\nu)\ .\endmultline
$$
$$\multline
\sum\limits_{0\le k<k_{\max}}[2^{\text{nd}}\text{term\ on\ RHS\ of\ (37)}]\\
\le C_\ast\Lambda^{4\varepsilon-2}(\Lambda^{-2/7})^{-5/2}\lambda(x_\nu)
=C_\ast\Lambda^{4\varepsilon-9/7}\lambda(x_\nu)\ .\endmultline
$$
$$\multline
\sum\limits_{0\le k<k_{\max}}[3^{\text{rd}}\text{term\ on\ RHS\ of\ (37)}]\\
\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)\log_2\Lambda\le
C_\ast\Lambda^{-\frac{N^{\prime\prime}}{3}}(\phi(0)+1)\ .\endmultline
$$
$$
[1^{\text{rst}}\ \text{term\ on\ RHS\ of\ (38)}]\le C_\ast\Lambda^{-4/3}
\lambda(x_\nu)<C_\ast\Lambda^{-9/7}\lambda(x_\nu)\ .
$$
$$
[2^{\text{nd}}\ \text{term\ on\ RHS\ of\ (38)}]\le C_\ast(\Lambda^{-2}
\lambda(x_\nu))(\Lambda^{-2/7})^{-5/2}=C_\ast\Lambda^{-9/7}\lambda(x_\nu)\ .
$$
$$
[3^{\text{rd}}\ \text{term\ on\ RHS\ of\ (38)}]\le C_\ast\Lambda^{-9/7}
\lambda(x_\nu)
$$
$$
[4^{\text{th}}\ \text{term\ on\ RHS\ of\ (38)}]\le C_\ast\Lambda^{-9/7}
\lambda(x_\nu)
$$
$$
[5^{\text{th}}\ \text{term\ on\ RHS\ of\ (38)}]=
C_\ast\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)\le
C_\ast\Lambda^{-\frac{N^{\prime\prime}}{3}}(\phi(0)+1)\ .
$$

\noindent In view of these observations, (37) and (38) imply
$$\multline
\sum\limits_{0\le k\le k_{\max}}(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}\le
C_\ast\Lambda^{4\varepsilon-9/7}\lambda(x_\nu)+C_\ast\Lambda^{-\frac{N^{\prime
\prime}}{3}}(\phi(0)+1)\\
\text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge
\overline C_{\#}\Lambda^{-1}\ .\endmultline\tag 41
$$

\noindent On the right in (41), the exponent of $4\varepsilon-9/7$
is not important to us.
Any exponent strictly less than $-1$ would be enough for our purposes.

Similarly, in {\it CASE II\/} we have to sum the right-hand side of (39)
over $k=0,1,\ldots,k_{\max}-1$, and then add the result to the right-hand
side of (40).

\noindent Again recalling that $2^{-k_{\max}}\sim \Lambda^{-2/7}$ and
$\lambda(x_\nu)\ge c_{\#}\Lambda$, we obtain:
$$
\sum\limits_{0\le k<k_{\max}}[1^{\text{rst}}\ \text{term\ on\ RHS\
in\ (39)}]\le C_\ast\le C_\ast\Lambda^{-1}\lambda(x_\nu)
$$
$$\sum\limits_{0\le k<k_{\max}}[2^{\text{nd}}\ \text{term\ on\ RHS\ in\
(39)}]\le C_\ast(\Lambda^{-2/7})^{-3/4}(\Lambda^{-\frac 32} \lambda(x_\nu))
=C_\ast\Lambda^{-9/7}\lambda(x_\nu)
$$
$$\multline
\sum\limits_{0\le k<k_{\max}}[3^{\text{rd}}\ \text{term\ on\ RHS\ in\ (39)}]
\\
\le C_\ast\Lambda^{4\varepsilon-2}(\Lambda^{-2/7})^{-5/2}\lambda(x_\nu)=
C_\ast\Lambda^{-9/7+4\varepsilon}\lambda(x_\nu)\endmultline
$$
$$\multline
\sum\limits_{0\le k<k_{\max}}[4^{\text{th}}\ \text{term\ on\ RHS\ in\
(39)}]\\
\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)\log_2
\Lambda<C_\ast\Lambda^{-\frac{N^{\prime\prime}}{3}}(\phi(0)+1)\endmultline
$$
$$
[1^{\text{st}}\ \text{term\ on\ RHS\ in\ (40)}]\le
C_\ast\Lambda^{-4/3}\lambda(x_\nu)<C_\ast\Lambda^{-9/7}\lambda(x_\nu)\ .
$$
$$
[2^{\text{nd}}\ \text{term\ on\ RHS\ in\ (40)}]\le
C_\ast(\Lambda^{-2}\lambda(x_\nu))\cdot(\Lambda^{-2/7})^{-5/2}=
C_\ast\Lambda^{-9/7}\lambda(x_\nu)\ .
$$
$$
[3^{\text{rd}}\ \text{term\ on\ RHS\ in\ 40)}]=C_\ast
\Lambda^{-8/7}\lambda(x_\nu)<C_\ast\Lambda^{-1}\lambda(x_\nu)
$$
$$
[4^{\text{th}}\ \text{term\ on\ RHS\ in\ (40)}]=
C_\ast\Lambda^{-\frac{N^{\prime\prime}}{2}}(\phi(0)+1)<C_\ast
\Lambda^{-\frac{N^{\prime\prime}}{3}}(\phi(0)+1)\ .
$$

\noindent In view of these observations, (39) and (40) imply
$$\multline
\sum\limits_{0\le k\le k_{\max}}(Av_{I_\nu}|H_{\nu k}|^2)^{1/2}
\le C_\ast\Lambda^{-1}\lambda(x_\nu)+C_\ast\Lambda^{-\frac{N^{\prime\prime}}
{3}}(\phi(0)+1)\\
\text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le
\overline C_{\#}\Lambda^{-1}\ .\endmultline\tag 42
$$

Let us examine how $H_{\nu k}(x)$ behaves on $I_\nu=\{|x-x_\nu|<
\tilde c_{\#} B(x_\nu)\}$.  We divide $I_\nu$ into three subintervals
$I_\nu^-$, $I_\nu^0$, $I_\nu^+$ by setting
$$\align
I_\nu^0&=\{|x-x_\nu|\le \hat cB(x_\nu)\}\\
I_\nu^-&=\ \text{part\ of}\ I_\nu\ \text{to\ the\ left\ of}\ I_\nu^0\\
I_\nu^+&=\ \text{part\ of}\ I_\nu\ \text{to\ the\ rt.\ of}\ I_\nu^0\ .
\endalign
$$

Recall that $g_{\nu k}(x,E)$ is supported in $\{(x,E)\mid x\in I_\nu^0\}$.
Hence, both $\rho(x,g_{\nu k})$ and $\rho_{sc}(x,g_{\nu k})$ are supported
in $I_\nu^0$.  Equation (36) therefore shows that $H_{\nu k}(x)$
is constant on $I_\nu^-$ and on $I_\nu^+$.  Say
$$
H_{\nu k}(x)=c_{\nu k}^-\ \text{for}\ x\in I_\nu^-\quad\text{and}\quad
H_{\nu k}(x)=c_{\nu k}^+\ \text{for}\ x\in I_\nu^+\ .\tag 43
$$
\noindent Since $I_\nu^-, I_\nu^+\subset I_\nu$ with $|I_\nu^-|,
|I_\nu^+|>c_\ast|I_\nu|$, 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 $\nu$ into 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 that $N^{\prime\prime}
>1000K$, 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 $\text{dist}
(\check x,[x_{\text{left}}(0),x_{\text{rt}}(0)])<C_{\#}
\Lambda^{-\varepsilon}B(\check x)$.  Applying
this to $\check x_\nu$ and using (58), we get
$$
\text{dist}(x_\nu,[x_{\text{left}}(0),x_{\text{rt}}(0)])\le \hat cB(x_\nu)
+C_{\#}\Lambda^{-\varepsilon}B(\check x_\nu)\le \hat cB(x_\nu)
+C_{\#}^\prime\Lambda^{-\varepsilon}B(x_\nu)\ 
$$
\noindent which implies (57) by the WKB hypothesis (Y11).

Next, we use Lemma 1 with $E=0$, to conclude that:
$$\multline
-V^\prime(x)\ge c_{\#}S(x_{\text{left}}(0))B^{-1}(x_{\text{left}}
(0))\\
\text{for}\quad x\in [x_{\text{left}}(0), x_{\text{left}}(0)
+\overline c_{\#}B(x_{\text{left}}(0))]\endmultline
$$
\noindent and hence
$$\multline
-V(x)>c_{\#}S(x_{\text{left}}(0))B^{-1}(x_{\text{left}}(0))\cdot
(x-x_{\text{left}}(0))\\
\text{for}\quad x\in [x_{\text{left}}(0),x_{\text{left}}(0)+\overline c_{\#}
B(x_{\text{left}}(0))]\ .\endmultline\tag 59
$$

\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
$$\multline
|h_{\text{err}}(x,E)|\le C_\ast\chi_{{}_{x\in I_{\text{BVP}}\backslash
[x_{\text{left}}(E)-\Lambda^{-3/7}B(x_{\text{left}}(E)),x_{\text{rt}}(E)+
\Lambda^{-3/7}B(x_{\text{rt}}(E))]}}\ ,\\
\text{for}\quad V(x_0)<E\le 0\ .\endmultline\tag 66
$$

\noindent This and hypothesis (Y9) show that $g_{\text{err}}(x,E)=0$
whenever $V(x)\le E\le 0$.  Hence $\rho_{sc}(x,g_{\text{err}})\equiv 0$,
by definition of the semiclassical density.  On the other hand,
$$
\rho(x,g_{\text{err}})=\sum\limits_k\varphi(E_k)h_{\text{err}}(x,E_k)
|u_k(x)|^2\ ,
$$
\noindent so that
$$\multline
\int\limits_{I_{\text{BVP}}}|\rho(x,g_{\text{err}})|\, dx
\le C_\ast\sum\limits_k|\varphi(E_k)|\\
\cdot \int\limits_{x\in I_{\text{BVP}}
\backslash [x_{\text{left}}(E_k)-\Lambda^{-3/7}B(x_{\text{left}}(E_k)),
x_{\text{rt}}(E_k)+\Lambda^{-3/7}B(x_{\text{rt}}(E_k))]}
|u_k(x)|^2dx\endmultline\tag 67
$$
\noindent by virtue of (66).  Here $E_k$ are the eigenvalues $\le 0$,
and $u_k$ are the eigenfunctions of $H$.

Recall that $\lambda(x_{\text{left}}(E_k)),\lambda(x_{\text{rt}}(E_k))
\ge c_{\#}\Lambda(E_k)\ge c_{\#}\Lambda$, so $\Lambda^{-3/7}\hfill\break >
\lambda^{\varepsilon-2/3}(x_{\text{left}}(E)), \lambda^{\varepsilon-2/3}
(x_{\text{rt}}(E))$. 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 $|E-E_k|<c_{\#}S_{\min}(E_k)$ we have $|x_{\text{left}}(E)-x_{\text{left}}
(E_k)|<c_{\#}^\prime B(x_{\text{left}}(E_k))$ and $|x_{\text{rt}}(E)-x_{\text{rt}}(E_k)|<c_{\#}^\prime B(x_{\text{rt}}(E_k))$, so $S_{\min}(E)\sim S_{\min}
(E_k)$, provided we take $c_{\#}$ small.  Recall that $S_{\min}(E)\sim
S(\overline x)$ for some $\overline x\in [x_{\text{left}}(E)+
c_{\#}B(x_{\text{left}}(E)),x_{\text{rt}}(E)
\hfill\break -c_{\#}B(x_{\text{rt}}(E))]$ and
therefore 
$$\multline
\frac{d\phi}{dE}(E)=\frac 12 \int_I\bigl(E-V(x)\bigr)_+^{-1/2}\, dx
\ge c_{\#}S^{-1/2}(\overline x)B(\overline x)\quad \text{(by\ Lemma\ 1)}\\
\ge c_{\#}^\prime S_{\min}^{-1}(E)\cdot \lambda(\overline x)\ge c_{\#}
^{\prime\prime}S_{\min}^{-1}(E)\Lambda(E)\ge c_{\#}^{\prime\prime}
S_{\min}^{-1}(E)\Lambda\ .\endmultline
$$

\noindent Here $\frac{d\phi}{dE}(E)\ge c_{\#}^\prime S_{\min}^{-1}
(E_k)\Lambda$ for $|E-E_k|\le c_{\#}S_{\min}(E_k)$.  (We have used
this estimate repeatedly in other sections.)  Also, $\frac{d\phi}{dE}
\ge 0$ for all $E\in (V(x_0),0]$.  So if we are given $E$ satisfying
$E_k+c_{\#}S_{\min}(E_k)\le E\le 0$, then
$$
\phi(E)\ge \phi(E_k)+\int\limits_{E_k}^{E_k+c_{\#}S_{\min}(E_k)}
\phi^\prime(\tilde E)d\tilde E\ge \phi(E_k)+c_{\#}^\prime\Lambda
>\pi(k^\prime+1/2)+c_{\#}^{\prime\prime}\Lambda
$$

\noindent by (69).  In particular, we cannot have $|\phi(E)-\pi(k^\prime+1/2)|
\le C_{\#}\Lambda^{-1}$.   Similarly, if $V(x_0)<E\le E_k-
c_{\#}S_{\min}(E_k)$, then
$$
\phi(E)\le \phi(E_k)-\int\limits_{E_k-c_{\#}S_{\min}(E_k)}^{E_k}
\phi^\prime(\tilde E)d\tilde E\le \phi(E_k)-c_{\#}^\prime
\Lambda<\pi(k^\prime+1/2)-c_{\#}^{\prime\prime}\Lambda
$$

\noindent by (69), so again we cannot have $|\phi(E)-\pi(k^\prime+1/2)|\le
C_{\#}\Lambda^{-1}$.  These remarks and (70) show that $E_k$ is the only
eigenvalue $E\in (V(x_0),0]$ satisfying $|\pi(k^\prime+1/2)-\phi(E)|\le
C_{\#}\Lambda^{-1}$.  It follows that the number of eigenvalues $E_k\in
[V(x_0)+c_2S(x_0),0]$ is less than or equal to the number of integers
$k^\prime$ for which $|\phi(E)-\pi(k^\prime+1/2)|\le C_{\#}
\Lambda^{-1}$ for some $E\in (V(x_0),0]$. As $E$ ranges over $(V(x_0),0]$,
$\phi(E)$ ranges over $(0,\phi(0)]$, so the number of $k^\prime$
as above is dominated by $C_{\#}(\phi(0)+1)$.  Hence
$$
(\text{Number\ of}\ E_k\in [V(x_0)+c_2S(x_0),0])\le C_{\#}
(\phi(0)+1)\le C_{\#}\Lambda^{5K}\ \text{by\ (55)}\ .\tag"(70 \text{bis})"
$$

\noindent Substituting this in (68), we get
$$
\int\limits_{I_{\text{BVP}}}|\rho(x,g_{\text{err}})|\, dx\le \Lambda^{-\frac{
N^{\prime\prime}}{4}}\ .\tag 71
$$

\noindent Recalling that $\rho_{sc}(x,g_{\text{err}})\equiv 0$ and setting
$H_{\text{err}}(x)\hfill\break
=\int\limits_{I_{\text{BVP}}\cap (0,x]}(\rho(x^\prime,
g_{\text{err}})-\rho_{sc}(x^\prime,g_{\text{err}}))\, dx^\prime$,
we get from (71) the estimate $|H_{\text{err}}(x)|\le \Lambda^{-\frac{N^{\prime
\prime}}{4}}$ for all $x\in \Bbb R$.  In particular
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}|H_{\text{err}}(x)|^2\Bigr)^{1/2}
\le \Lambda^{-\frac{N^{\prime\prime}}{4}}\ , \text{for}\ \tilde x\
\text{with}\ \{|x-\tilde x|<\hat cB(\tilde x)\}\subset I\ .\endmultline
\tag 72
$$

\noindent We now know the following:
$$
\varphi(E)=\sum\limits_{\nu k}g_{\nu k}(x,E)+g_{\text{err}}(x,E)
\quad (\text{definition\ of}\ g_{\text{err}})\ .
$$
$$
\tilde H(x)=\int\limits_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime,
\sum\limits_{\nu k}g_{\nu k})-\rho_{sc}(x^\prime,\sum\limits_{\nu k}
g_{\nu k}))\, dx^\prime\quad (\text{definition\ of}\ \tilde H)
$$
$$
H_{\text{err}}(x)=\int\limits_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime,
g_{\text{err}})-\rho_{sc}(x^\prime,g_{\text{err}}))\, dx^\prime\quad
(\text{definition\ of}\ H_{\text{err}})\ .
$$

\noindent Therefore,
$$
H(x)=\int\limits_{I_{\text{BVP}}\cap (-\infty,x]}(\rho(x^\prime,\varphi)
-\rho_{sc}(x^\prime,\varphi))\, dx^\prime=\tilde H(x)+H_{\text{err}}(x)\ .
\tag 73
$$

\noindent Combining estimates (64), (65), (72), we obtain;
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\, |H(x)|^2\Bigr)^{1/2}
\le C_\ast\Lambda^{4\varepsilon-9/7}\int\limits_{I_{\text{BVP}}
\cap (-\infty,\tilde x+C_{\#}\hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}\,
dx\\
+2\Lambda^{-\frac{N^{\prime\prime}}{4}}\\
\text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\ge \overline C_{\#}
\Lambda^{-1}\ ; \endmultline\tag 74
$$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\, |H(x)|^2\Bigr)^{1/2}\le
C_\ast\Lambda^{-1}\int\limits_{I_{\text{BVP}}\cap (-\infty,\tilde x
+C_{\#}\hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}\, dx+2\Lambda^{
\frac{-N^{\prime\prime}}{4}}\\
\text{in\ case}\quad \min_{k\in \Bbb Z}|\phi(0)-\pi(k+1/2)|\le \overline
C_{\#} \Lambda^{-1}\ .\endmultline\tag 75
$$

\noindent Estimates (74), (75) have been proven for arbitrary $\tilde x$
for which $\{|x-\tilde x|<\hat cB(\tilde x)\}\hfill\break\subset I$.  The set
of
such $\tilde x$ is slightly smaller than $I$.  We also want estimates for
$H(x)$, with $x$ not necessarily belonging to $I$, or even to
$I_{\text{BVP}}$.  To derive these estimates, we use Lemma 1 and the WKB
Eigenfunction Theorem, to see that $E_k\in [V(x_0)+c_2S(x_0),0]$ implies\
$\int\limits_{I_{\text{BVP}}\backslash [x_{\text{left}}(0)-\hat cB
(x_{\text{left}}(0)),x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))]}|u_k(x)|^2\, 
dx\le \Lambda^{-N^{\prime\prime}}$. \hfill\break Therefore
$$\multline
\int\limits_{I_{\text{BVP}}\backslash [x_{\text{left}}(0)-\hat
cB(x_{\text{left}}(0)),x_{\text{rt}}(0)+\hat
cB(x_{\text{rt}}(0))]}|\rho(x,\varphi)|\, dx\\ \le
C_\ast\Lambda^{-N^{\prime\prime}}\cdot (\text{Number\ of}\ E_k\in
[V(x_0)+c_2S(x_0),0])\le \Lambda^{-\frac{N^{\prime\prime}} {4}}\ \text{by\ (70
bis)}\ .\endmultline\tag 76 $$

\noindent Also, for $x\in I_{\text{BVP}}\backslash [x_{\text{left}}
(0)-\hat cB(x_{\text{left}}(0)),x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))]$
we have $V(x)>0$, 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
$-\infty<x\hfill\break \le x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))$
we have $|H(x)|\le 10\Lambda^{-\frac{N^{\prime\prime}}{4}}$; for
$x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le \tilde x\hfill\break
\le x_{\text{rt}} (0)+\hat cB(x_{\text{rt}}(0))$ we have $\bigl(Av_{|x-\tilde
x|<\hat c B(\tilde x)}|H(x)|^2\bigr)^{1/2}\hfill\break 
\le C_\ast\Lambda^{4\varepsilon-9/7}
\int\limits_{I_{\text{BVP}}\cap (-\infty,\tilde x+C_{\#}\hat cB(\tilde x)]}
\bigl(-V(x)\bigr)_+^{1/2}\, dx+10\Lambda^{-\frac{N^{\prime\prime}}{4}}$;
and for\hfill\break  $x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\le x<+\infty$ we
have
$$
|H(x)|\le C_\ast\Lambda^{4\varepsilon-9/7}\int\limits_{I_{\text{BVP}}}
\bigl(-V(x^\prime)\bigr)_+^{1/2}\, dx^\prime+10\Lambda^{-\frac{N^{\prime\prime}}
{4}}\ .
$$
\medskip

\noindent{\it CASE II\/}:  Suppose $\min_{k\in \Bbb Z}\, |\phi(0)-\pi
(k+1/2)|\le \overline C_{\#}\Lambda^{-1}$.  Then for $-\infty<x\hfill\break\le
x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))$ we have $|H(x)|\le
10\Lambda^{-\frac{N^{\prime\prime}}{4}}$; for $x_{\text{left}}(0)-\hat
cB(x_{\text{left}}(0))\le \tilde x\le  x_{\text{rt}}(0)+\hat
cB(x_{\text{rt}} (0))$ we have $\bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}
|H(x)|^2\bigr)^{1/2}\hfill\break
\le C_\ast\Lambda^{-1}\int\limits_{I_{\text{BVP}}
\cap (-\infty,\tilde x+C_{\#}\hat cB(\tilde x)]}\bigl(-V(x)\bigr)_+^{1/2}
\, dx+10\Lambda^{-\frac{N^{\prime\prime}}{4}}$; and for\hfill\break
$x_{\text{rt}} (0)+\hat cB(x_{\text{rt}}(0))\le x<+\infty$ we have
$$
|H(x)|\le C_\ast\Lambda^{-1}\int\limits_{I_{\text{BVP}}}\bigl(-V(x^\prime)
\bigr)_+^{1/2}\, dx^\prime+10\Lambda^{-\frac{N^{\prime\prime}}{4}}\ .
$$

\noindent The constants $C_{\#}$, $\overline C_{\#}$ depend only on
$\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$, $C_\alpha$ in (Y0)$\ldots$(Y11)
while $C_\ast$ depends only $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$,
$C_\alpha$, $\hat c$, $\hat C_\beta$ in (Y0)$\ldots$(Y11).
\endproclaim

\vfill\eject

\heading{\bf{THE DENSITY FOR A ONE-DIMENSIONAL POTENTIAL}}\endheading
\medskip

Our goal here is to extend the results of the previous section on
$\rho(x,\varphi)-\rho_{sc}(x,\varphi)$ to cover the case $\varphi\equiv 1$.
Thus, we at last gain control over the full density.

We suppose we are given a potential $V(x)$ defined on a (possibly
unbounded) interval $I_{\text{BVP}}$, together with a subinterval
$I$ on which are defined positive functions $S(x)$, $B(x)$.  We are
given also positive numbers $K$, $\varepsilon$, $N$ with $K>100$,
$\varepsilon<\frac{1}{1000 K}$, $N>\frac{K}{\varepsilon^{50}}$;
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 $0\le k<k_{\text{last}}$, and we use our results on
the microlocalized density near the minimum of the potential to control
$H_{k_{\text{last}}}(x)$.  The $\varphi_k(E)$ may be taken to satisfy the
following conditions, for a small constant $c_{\#}^2>0$ 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 $1\le k<k_{\text{last}}$,
$$
\varphi_k(E)\quad\text{is\ supported\ in}\quad
\{2^{-2(k+3)}c_{\#}^2S(x_0)<E-V(x_0) <2^{-2(k-3)}c_{\#}^2S(x_0)\}\tag 3
$$
$$
\Bigl|\bigl(\frac{d}{dE}\bigr)^\beta\varphi_k(E)\Big|\le C_{\#}^\beta
(2^{-2k}S(x_0))^{-\beta}\quad\text{for\ all}\quad E\ , \beta\ .\tag 4
$$
\noindent Regarding $k=k_{\text{last}}$,
$$
\varphi_{k_{\text{last}}}(E)\quad\text{is\ supported\ in}\quad
\{|E-V(x_0)|<2^{-2(k_{\text{last}}-3)}c_{\#}^2S(x_0)\}\ ,\quad\text{and}
\tag 5
$$
$$
\Big|\bigl(\frac{d}{dE}\bigr)^\beta\varphi_{k_{\text{last}}}(E)\Big|\le
C_{\#}^\beta(2^{-2k_{\text{last}}}S(x_0))^{-\beta}\quad\text{for\ all}\quad
E\ , \beta\ .\tag 6
$$
$$
\sum\limits_{0\le k\le k_{\text{last}}}\varphi_k(E)=1\quad\text{for}\quad
V(x_0)\le E\le 0\ .\tag 7
$$

\noindent Since all the eigenvalues $E_k\le 0$ satisfy $V(x_0)\le E_k\le 0$,
equation (7) is enough to show that
$$
H(x)=\sum\limits_{0\le k\le k_{\text{last}}}H_k(x) \quad\text{for\ all}\ x\ , 
\tag 8
$$
\noindent with $H(x)$, $H_k(x)$ defined as above.

We take $k_{\text{last}}$ to satisfy $2^{-k_{\text{last}}-1}<(\lambda(x_0))^
{-\frac{18}{43}}\le 2^{-k_{\text{last}}}$.  We pick $c_{\#}^2$ 
small enough that $V(x_0)+10^3c_{\#}^2S(x_0)<0$,
$$
V(x_0)+c_{\#}S(x_0)B^{-2}(x_0)\cdot(x-x_0)^2\le V(x)\le V(x_0)+C_{\#}
S(x_0)B^{-2}(x_0)\cdot (x-x_0)^2\ ,\tag 9
$$
\noindent and
$$\multline
c_{\#}S(x_0)B^{-1}(x_0)<\frac{V^\prime(x)}{(x-x_0)}<C_{\#}S(x_0)B^{-1}(x_0)\\
\text{for}\quad V(x_0)<V(x)\le V(x_0)+10^3c_{\#}^2S(x_0)\ .
\endmultline\tag 10
$$

\noindent For $V(x_0)<E<V(x_0)+10^3c_{\#}^2S(x_0)$, we note that $\phi^\prime(E)
\sim\lambda(x_0)(S(x_0))^{-1}$.  Hence, as $E$ varies in $\bigl\{\frac{E-V(x_0)}
{S(x_0)}\in \bigl[\frac{10^3}{3}c_{\#}^22^{-2k},\frac{10^3}{2}c_{\#}^2
2^{-2k}\bigr]\bigr\}$, we find that $\phi(E)$ varies by $\sim \lambda(x_0)
\cdot 2^{-2k}\ge \lambda(x_0)2^{-2k_{\max}}\sim \lambda(x_0)\cdot
\lambda^{-\frac{36}{43}}(x_0)=\lambda^{7/43}(x_0)>c_{\#}\Lambda^{7/43}
\hfill\break >>1$ 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 $I_k=\{|x-x_0|<C_{\#}^1B_k\}$, with
$C_{\#}^1$  and $c_{\#}^2$ picked as in the following lemma.

\vglue 1pc
\proclaim{Lemma 1}  We can pick $c_{\#}^2$, $C_{\#}^1$ so that
$V(x_0)+10^3c_{\#}^2S(x_0)<0$ and (9) and (10) hold, and the following
conditions are satisfied.

\roster
\item"(A)" Hypotheses (Y0)$\ldots$(Y11) in the previous section
are satisfied, with $\varphi_0(E)$ in place of $\varphi(E)$, with
$c_{\#}^2$ in place of $c_2$, and with each $\hat C_\beta$ in
(Y8) of the form $C_{\#}^\beta$.

\item"(B)" Suppose $1\le k<k_{\text{last}}$.  Then hypotheses (Y0)$\ldots$(Y11)
are satisfied, with $\varphi_k(E-\tilde E_k)$ in place of $\varphi(E)$,
$V_k(x)$ in place of $V(x)$, $S_k$ in place of $S(x)$,
$B_k$ in place of $B(x)$, $I_k$ in place of $I$, $c_{\#}^2$ in place
of $c_2$.
\endroster

The constants called $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $c_2$,
$C_\alpha$, $\hat C_\beta$ in (Y0)$\ldots$(Y11) are all of the form $C_{\#}$. \
In particular, they do not depend on $k$.

\roster
\item"(C)" The hypotheses of the WKB Theorem on Low Eigenvalues
((H0$^\ast$)$\ldots$(H6$^\ast$)) are satisfied, with $V_{k_{\text{last}}}(x)$
in place of $V(x)$, $S(x_0)$ and $B(x_0)$ in place of $S$ and $B$,
and with $E_\infty=0$.  Also, the function $\varphi_{k_{\text{last}}}(E)$
satisfies the assumptions on $g(E)$ which we made in the section on the
microlocalized density near the minimum of the potential, with $C_{\#}
(\lambda(x_0))^{-18/43}$ in place of $\tau$, and with the constants
$\hat C_\beta$ of the form $C_{\#}$.
\endroster
\endproclaim

\vglue 1pc
\demo{Sketch of Proof}

{\it{Part (A)\/}} follows from (1) and (2), together with our assumptions\hfill
\break
(Y0)$\ldots$(Y7) and (Y10), (Y11$^\ast$).  In fact, (1) and (2) amount
to (Y8), (Y9) for $\varphi_0(E)$; and (Y11) follows from (Y11$^\ast$),
since each $\hat C_\beta$ has the form $C_{\#}$.

{\it{Part (B)\/}}.  We have to verify (Y0)$\ldots$(Y11).  (Y0)$\ldots$(Y5)
are trivial, and we leave them to the reader.

Here, the quantity called $\Lambda$ in (Y0)$\ldots$(Y11) for $V_k$,
$S_k$, $B_k$, etc. is 
$\Lambda_{(V_k)}=2^{-2k}\lambda(x_0)\ge 2^{-2k_{\max}}\lambda(x_0)\ge
c_{\#}(\lambda(x_0))^{7/43}>c_{\#}\Lambda^{7/43}$.  To check (Y6),
note that $V(x)=\tilde E_k$ at the point $x_{k,\text{left}}$ that plays
the r$\hat o$le 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 $x<x_{\text{left}}(0)$.  Hence for
$x<x_k^-$ we have $V(x)>V(x_k^-)\ge \tilde E_k+c_{\#}S_k$, so
$V_k(x)=V(x)-\tilde E_k\ge c_{\#}S_k=c_{\#}\lambda_k^2B_k^{-2}$.

\noindent If $x<x_{k,\text{left}}-C_{\#}B_k$, then $B_k^{-2}\ge
(x-x_{k,\text{left}})^{-2}$ and $x<x_k^-$, so $V_k(x)\ge
c_{\#}\lambda_k^2B_k^{-2}\ge c_{\#}\lambda_k^2(x-x_{k,{\text{left}}})^{-2}$.
Similarly, if $x_{k,\text{rt}}$ plays the r$\hat o$le of $x_{\text{rt}}(0)$
for $V_k(x)$, and if $x>x_{k,\text{rt}}+C_{\#}B_k$, 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$\hat o$le 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 $H_k(x)$ $(1\le k\le k_{\text{last}})$:
$$
|H_k(x)|\le C_\ast\lambda_k^{-\frac{N^{\prime\prime}}{4}}\quad\text{for}\quad
x<x_{\text{left}}(\tilde E_k)-\hat cB_k\tag 13
$$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB_k}\ |H_k(x)|^2\Bigr)^{1/2}\le
C_\ast\lambda_k^{4\varepsilon-2/7}\quad\text{for}\\
x_{\text{left}}(\tilde E_k)-\hat cB_k\le \tilde x\le x_{\text{rt}}(\tilde E_k)
+\hat cB_k\endmultline\tag 14
$$
$$
|H_k(x)|\le C_\ast\lambda_k^{4\varepsilon-2/7}\quad\text{for}\quad
x>x_{\text{rt}}(\tilde E_k)+\hat cB_k\ .\tag 15
$$

\noindent These estimates in turn trivially imply the following.
$$
|H_k(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\quad\text{for}
\quad x<x_{\text{left}}(\tilde E_1)-\hat cB(x_0)\quad (1\le
k<k_{\text{last}})\tag 16 $$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(x_0)}\ |H_k(x)|^2\Bigr)^{1/2}\le
C_\ast\lambda_k^{-\frac 27+4\varepsilon}\\
\text{for\ all}\quad \tilde x\in \Bbb R\quad(1\le k<k_{\text{last}})
\endmultline\tag 17
$$
$$
|H_k(x)|\le C_\ast\lambda_k^{-\frac 27+4\varepsilon}\quad\text{for}
\quad x>x_{\text{rt}}(\tilde E_1)
+\hat cB(x_0)\quad (1\le k<k_{\text{last}})\ .\tag"(17\text{bis})"
$$

Next, applying Lemma 1 (A) and the Lemma of the previous section, we learn
the following:
\smallskip

\noindent{\it{Case I\/}}:  Suppose $\min_{k\in \Bbb Z}\ |\phi(0)-\pi(k+1/2)|
\ge \overline C_{\#}\Lambda^{-1}$.  Then
$$
|H_0(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\quad\text{for}
\quad x<x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\ .\tag 18
$$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\ |H_0(x)|^2\Bigr)^{1/2}\\
\le C_\ast\Lambda^{4\varepsilon-\frac 97}\int_{I_{\text{BVP}}\cap
(-\infty,\tilde x+C_{\#}\hat cB(\tilde x)]}(-V(x))_+^{1/2}dx
+C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\\
\text{for}\quad x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le \tilde x\le
x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ .\endmultline\tag 19
$$
$$\multline
|H_0(x)|\le C_\ast\Lambda^{4\varepsilon-9/7}\int_{I_{\text{BVP}}}
(-V(x^\prime))_+^{1/2}dx^\prime
+C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\\ \quad\text{for}\quad
x>x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 20
$$

\medskip
\noindent{\it{Case II\/}}:  Suppose $\min_{k\in \Bbb Z}\ |\phi(0)
-\pi(k+1/2)|\le \overline C_{\#}\Lambda^{-1}$.  Then
$$
|H_0(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}{1000}}\quad\text{for}
\quad x<x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\ .\tag 21
$$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\ |H_0(x)|^2\Bigr)^{1/2}\\
\le C_\ast\Lambda^{-1}\int_{I_{\text{BVP}}\cap (-\infty,\tilde x+C_{\#}
\hat cB(\tilde x)]}(-V(x))_+^{1/2}dx+C_\ast\Lambda^{-\frac{N^{\prime\prime}}
{1000}}\\
\text{for}\quad x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le \tilde x\le
x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ .\endmultline\tag22
$$
$$\multline
|H_0(x)|\le
C_\ast\Lambda^{-1}\int_{I_{\text{BVP}}}(-V(x^\prime))_+^{1/2}dx^\prime +C_\ast
\Lambda^{-\frac{N^{\prime\prime}}{1000}}\\ \text{for}\quad
x>x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ . \endmultline\tag 23
$$

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
$$
|H_{k_{\text{last}}}(x)|\le C_\ast(\lambda(x_0))^{1-N^\prime}
\quad\text{if}\quad x<x_0-(\lambda(x_0))^{\varepsilon-\frac{18}{43}}
B(x_0)\tag 24
$$
$$\multline
|H_{k_{\text{last}}}(x)|\le C_\ast(\lambda(x_0))^{2\cdot(-\frac{18}{43})+1}
=C_\ast\lambda^{\frac{7}{43}}(x_0)\\
\text{if}\quad |x-x_0|<\lambda^{\varepsilon-\frac{18}{43}}(x_0)B(x_0)
\endmultline\tag 25
$$
$$
|H_{k_{\text{last}}}(x)|\le C_\ast\lambda^{-\frac{7}{43}}(x_0)\quad
\text{if}\quad x>x_0+(\lambda(x_0))^{\varepsilon-\frac{18}{43}}B(x_0)\ .
\tag 26
$$

\noindent From (24) we get at once
$$
|H_{k_{\text{last}}}(x)|\le C_\ast\Lambda^{-\frac{N^{\prime\prime}}
{1000}}\quad\text{if}\quad x<x_0-C_{\#}\hat cB(x_0)\ .\tag 27
$$

\noindent For any $\tilde x\in \Bbb R$, (24)$\ldots$(26) imply
$$\multline
\int_{|x-\tilde x|<\hat cB(x_0)}
|H_{k_{\text{last}}}(x)|^2dx\le \int_{|x-x_0|<\lambda^{\varepsilon-18/43}
(x_0)B(x_0)} |H_{k_{\text{last}}}(x)|^2dx\\
+\int_{\{|x-\tilde x|<\hat cB(x_0)\}\backslash
\{|x-x_0|<\lambda^{\varepsilon-18/43}(x_0)B(x_0)\}}\ |H_{k_{\text{last}}}
(x)|^2dx\\
\le C_\ast\Bigl(\lambda^{\frac{7}{43}}(x_0)\Bigr)^2\cdot \lambda^{\varepsilon-
\frac{18}{43}}(x_0)B(x_0)+C_\ast\Bigl(\lambda^{-\frac{7}{43}}(x_0)\Bigr)^2B(x_0)
\le C_\ast\lambda^{\varepsilon-\frac{4}{43}}(x_0)B(x_0)\ ,\endmultline
$$

\noindent so
$$
\Bigl(Av_{|x-\tilde x|<\hat cB(x_0)}\ |H_{k_{\text{last}}}(x)|^2
\Bigr)^{1/2}\le C_\ast\lambda^{\varepsilon-\frac{2}{43}}(x_0)\ .
\tag 28
$$

\noindent From (26) we get at once
$$
|H_{k_{\text{last}}}(x)|\le C_\ast\lambda^{-\frac{7}{43}}(x_0)\quad
\text{if}\quad x>x_0+\hat cB(x_0)\ .\tag 29
$$

\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 $\Big|\sum\limits_{
1\le k\le k_{\text{last}}}H_k(x)\Bigr|\le \sum\limits_{1\le k<k_{\text{last}}}
C_\ast\lambda_k^{4\varepsilon-2/7}+C_\ast\lambda^{-\frac{7}{43}}(x_0)$.
Recall that $\lambda_k\sim 2^{-2k}\lambda(x_0)$, which decreases
geometrically to $\lambda_{k_{\max}}\sim\hfill\break
(\lambda^{-\frac{18}{43}}(x_0))^2
\lambda(x_0)\sim \lambda^{+\frac{7}{43}}(x_0)$, so $\sum\limits_k
\lambda_k^{4\varepsilon-\frac 27}\le C_\ast(\lambda(x_0))^{\varepsilon-
\frac {2}{43}}$. Thus  
$$
\Big|\sum\limits_{1\le k\le k_{\text{last}}}H_k(x)\Big|
\le C_\ast\bigl(\lambda(x_0)\bigr)^{\varepsilon-\frac{2}{43}}\quad
\text{for}\quad x>x_{\text{rt}}(\tilde E_1)+\hat cB(x_0)\ .\tag 31
$$

\noindent From (17) and (28), we get for any $\tilde x\in \Bbb R$ that
$$
\Bigl(Av_{|x-\tilde x|<\hat cB(x_0)}\Big|\sum\limits_{1\le k\le
k_{\text{last}}}H_k(x)\Big|^2\Bigr)^{1/2}\\
\le \sum\limits_{1\le k<k_{\text{last}}}C_\ast\lambda_k^{4\varepsilon-\frac 27}
+C_\ast\lambda^{\varepsilon-\frac{2}{43}}(x_0)\ .
$$

\noindent As in the proof of (31), the right-hand side is dominated by
$(\lambda(x_0))^{\varepsilon-\frac{2}{43}}$, so
$$
\Bigl(Av_{|x-\tilde x|<\hat cB(x_0)}\Big|
\sum\limits_{1\le k\le k_{\text{last}}}H_k(x)\Big|^2\Bigr)^{1/2}\le
C_\ast\bigl(\lambda(x_0)\bigr)^{\varepsilon-\frac{2}{43}}\ .\tag 32
$$

\noindent From (30), (31), (32) we conclude that:
$$
\Big|\sum\limits_{1\le k\le k_{\max}}H_k(x)\Big|\le
C_\ast\Lambda^{-10^{-4}N^{\prime\prime}}\quad\text{if}\quad
x<x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\ .\tag 33
$$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\ \Big|
\sum\limits_{1\le k\le k_{\max}}H_k(x)\Big|^2\Bigr)^{1/2}\le C_\ast
\Lambda^{-10^{-4}N^{\prime\prime}}\\
\text{if}\quad x_{\text{left}}(0)-\hat cB(x_{\text{left}}(0))\le
\tilde x<x_{\text{left}}(\tilde E_1)-C_{\#}\hat cB(x_0)\endmultline\tag 34
$$
$$\multline
\Bigl(Av_{|x-\tilde x|<\hat cB(\tilde x)}\
\Big|\sum\limits_{1\le k\le k_{\max}}H_k(x)\Big|^2\Bigr)^{1/2}
\le C_\ast\bigl(\lambda(x_0)\bigr)^{\varepsilon-\frac{2}{43}}\\
\text{if}\quad x_{\text{left}}(\tilde E_1)-C_{\#}\hat cB(x_0)\le \tilde x<
x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\endmultline\tag 35
$$
$$
\Big|\sum\limits_{1\le k\le k_{\max}}H_k(x)\Big|\le C_\ast
\bigl(\lambda(x_0)\bigr)^{\varepsilon-\frac 2{43}}\quad\text{if}
\quad x>x_{\text{rt}}(0)+\hat cB(x_{\text{rt}}(0))\ .\tag 36
$$

\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 $|x-y|<cB(x)$, then $c<\frac{B(y)}
{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$.
\item"(Z1)" If $x\in I$ and $\alpha \ge 0$, then $\big|\bigl(\frac{d}
{dx}\bigr)^\alpha V(x)\big|\le C_\alpha S(x)B^{-\alpha}(x)$.

\item"(Z2)" The set $\{x\in I\mid V(x)<0\}$ is a non-empty interval
$(x_{\text{left}},x_{\text{rt}})$, with
$\text{dist}\,(x_{\text{left}},\partial I)>cB(x_{\text{left}})$
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 $x<x_{\text{left}}-\Lambda^K
B(x_{\text{left}})$, we have $V(x)\ge \frac{1000}{|x-x_{\text{left}}|^2}$;
and for all $x\in I_{\text{BVP}}$ with
$x>x_{\text{rt}}+\Lambda^KB(x_{\text{rt}})$, 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$^\ast$)" $I=\{x\colon |x-x_0|<cB\}\subset I_{\text{BVP}}$.

\item"(Z1$^\ast$)" $\big|\bigl(\frac{d}{dx}\bigr)^\alpha V(x)\Big|\le
C_\alpha SB^{-\alpha}$ on $I$.

\item"(Z2$^\ast$)" $V^{\prime\prime}(x)\ge c^\prime SB^{-2}$ on $I$.
\endroster

Set $\lambda=S^{1/2}B$, and make the following further assumptions.
\roster
\item"(Z3$^\ast$)" $V^\prime(x_0)=0$ and $-\lambda^{-\frac{36}{43}}S
\le V(x_0)<0$.

\item"(Z4$^\ast$)" For $x\in I_{\text{BVP}}\backslash I$ we have
$V(x)>0$.

\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) $|H(x)|\le C_\ast\lambda^{1-N^\prime}\quad\text{if}\ x \in I_{\text{BVP}}\ ,
x<x_0-\lambda^{\varepsilon-\frac{18}{43}}B$

(b) $|H(x)|\le C_\ast\lambda^{\frac{7}{43}}\quad\text{if}\quad
|x-x_0|<\lambda^{\varepsilon-\frac{18}{43}}B$.

(c) $|H(x)|\le C_\ast\lambda^{-\frac{7}{43}}\quad\text{if}\quad
x\in I_{\text{BVP}}\ , x>x_0+\lambda^{\varepsilon-\frac{18}{43}}B$
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 $|x-y|<cB(x)$, we have
$c<\frac{B(y)}{B(x)}<C$ and $c<\frac{S(y)}{S(x)}<C$, and
$|I|>cB(x)$.

\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
$cS(x)<E-V(x)<CS(x)$ for $x\in [x_{\text{left}}(E)+c_1B
(x_{\text{left}}(E)), x_{\text{rt}}(E)-c_1B(x_{\text{rt}}(E))]$

\item"(Z$\overline 5$)" $V(x)$ is decreasing and $C^\infty$ on
$I_{\text{BVP}}^{\text{interior}}\cap (-\infty,x_{\text{left}}(0)]$.

\item"(Z$\overline 6$)" For $E_{\text{crit}}\le E\le 0$, we
have $x_{\text{left}}(E)+cB(x_{\text{left}}(E))\le x_{\text{crit}}$.

\item"(Z$\overline 7$)" For $E_{\text{crit}}\le E\le 0$, we have
$$
\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{crit}}]}\bigl(E-V(t)\bigr)_+^{-1/2}dt\le \delta \int_{I_{\text{BVP}}}\bigl(E-V(t)\bigr)_+^{-1/2}dt
$$

\item"(Z$\overline 8$)" $\Lambda\equiv \Bigl(\int_{x_{\text{left}}(0)}
^{x_{\text{rt}}(0)}\frac{dx}{\lambda(x)B(x)}\Bigr)^{-1}$ is greater
than a certain large, positive number determined by $c$, $C$, $c_1$,
$C_\alpha$ above.
\endroster

Here, $\lambda(x)=S^{1/2}(x)B(x)$ as usual.

If $\delta<<1$, then assumption (Z$\overline 7$) says that in the
semiclassical approximation, eigenfunctions $u_k(x)$ with eigenvalues
$E_k\in [E_{\text{crit}},0]$ are almost entirely concentrated in
$(x_{\text{crit}},\infty)\cap I_{\text{BVP}}$.  We want to show that
the true density 
$$
\rho(x)=\sum\limits_{E_k\le 0}|u_k(x)|^2
$$
\noindent is then highly concentrated in the same interval.  Here, $E_k$
and $u_k(x)$ are the eigenvalues and (normalized) eigenfunctions of
$-\frac{d^2}{dx^2}+V(x)$ with Dirichlet boundary conditions on $I_{\text{BVP}}$.  The precise result is as follows.

\vglue 1pc
\proclaim{Second Degenerate Density Lemma} Assume (Z$\overline 0$)$\ldots$
(Z$\overline 8$).  Then
$$\multline
\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{crit}}]}\rho(x)dx\le
C_{\#}+C_{\#}\delta\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{1/2}dx\\
+C_{\#}\int_{I_{\text{BVP}}}\bigl(E_{\text{crit}}-V(x)\bigr)_+^{1/2}dx
\endmultline
$$
\noindent  and $\int_{I_{\text{BVP}}}
\rho(x)dx\le C_{\#}+C_{\#}\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{1/2}dx$,
with $C_{\#}$ depending only on $c$, $C$, $c_1$, $C_\alpha$
in (Z$\overline 0$)$\ldots$(Z$\overline 8$).\endproclaim
\medskip

\demo{Proof}  Let $E_k$ be an eigenvalue in $[E_{\text{critical}},0]$, and
let $u_k(x)$ be the corresponding (normalized) eigenfunction.  The
hypotheses (H$\overline 0$)$\ldots$(H$\overline 6$) of Theorem 2 in
the discussion of WKB Theory with weak turning points are satisfied,
with $E_k$ in place
of $E_0$, and with constants depending only on the constants in our present
hypotheses (Z$\overline 0$)$\ldots$(Z$\overline 8$). (See ``Review of
Earlier Results.")  Theorem 2 shows that
$$\multline
|u_k(x)|^2\le C_{\#}\Bigl[\int_{I_{\text{BVP}}}\bigl(E_k-V(t)\bigr)_+^{-1/2}
dt\Bigr]^{-1}\cdot \bigl(E_k-V(x)\bigr)^{-1/2}\\
\text{for}\quad x_{\text{left}}(E_k)<x<x_{\text{rt}}(E_k)\endmultline
\tag 1
$$
\noindent and
$$\multline
|u_k(x)|^2\le C_{\#}\Bigl[\int_{I_{\text{BVP}}}\bigl(E_k-V(t)\bigr)_+^{-1/2}
dt\Bigr]^{-1}\cdot \Bigl[S(x_{\text{left}}(E_k))\cdot \lambda^{-2/3}
(x_{\text{left}}(E_k))\Bigr]^{-1/2}\\
\cdot\exp\Bigl(-c_{\#}\lambda^{2/3}(x_{\text{left}}(E_k))\cdot
\frac{(x_{\text{left}}(E_k)-x)}{B(x_{\text{left}}(E_k))}\Bigr)\\
\text{for}\quad x\in I_{\text{BVP}}\ , x\le x_{\text{left}}(E_k)\ .
\endmultline\tag 2
$$

We write $C_{\#}$, $c_{\#}$ etc. for constants determined by
$c$, $C$, $c_1$, $C_\alpha$ in (Z$\overline 0$)$\ldots$(Z$\overline 8$).
Integrating (2), we find that
$$\multline
\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{left}}(E_k)]}|u_k(x)|^2dx\\
\le C_{\#}\bigl[\int_{I_{\text{BVP}}}\bigl(E_k-V(t)\bigr)_+^{-1/2}dt\bigr]
^{-1}S^{-1/2}(x_{\text{left}}(E_k))\lambda^{+1/3}(x_{\text{left}}
(E_k))\\
\cdot\frac{B(x_{\text{left}}(E_k))}{\lambda^{2/3}
(x_{\text{left}}(E_k))}\ .\endmultline\tag 3
$$

\noindent With $c$ in (Z$\overline 6$) we have
$$\multline
S^{-1/2}(x_{\text{left}}(E_k))B(x_{\text{left}}(E_k))
\le C_{\#}\int_{x_{\text{left}}(E_k)}^{x_{\text{left}}(E_k)
+cB(x_{\text{left}}(E_k))}\bigl(E_k-V(t)\bigr)^{-1/2}dt\\
\le C_{\#}\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{crit}}]}
\bigl(E_k-V(t)\bigr)_+^{-1/2}dt\quad (\text{by\ (Z}\overline 6))\\
\le C_{\#}\delta\Bigl[\int_{I_{\text{BVP}}}\bigl(E_k-V(t)\bigr)_+^{-1/2}
dt\Bigr]\quad (\text{by\ (Z}\overline 7))\ ,\endmultline
$$

\noindent so estimate (3) implies
$$
\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{left}}(E_k)]}
|u_k(x)|^2dx\le C_{\#}\delta \lambda^{-1/3}(x_{\text{left}}(E_k))\le
C_{\#}\delta\ ,\tag 4
$$


\noindent because $\lambda(x_{\text{left}}(E_k))\ge c_{\#}\Lambda\ge
c_{\#}$ by (Z$\overline 8$).

We know from (Z$\overline 6$) that $x_{\text{left}}(E_k)<x_{\text{crit}}$,
and we know from (Z$\overline 7$) and $\delta<1$ that
$x_{\text{crit}}<x_{\text{rt}}(E_k)$.  Hence we may integrate (1)
to prove the estimate
$$\align
&{}\int_{x_{\text{left}}(E_k)}^{x_{\text{crit}}}|u_k(x)|^2dx\\
&\qquad\le C_{\#}
\Bigl[\int_{I_{\text{BVP}}}\bigl(E_k-V(t)\bigr)_+^{-1/2}dt\Bigr]^{-1}
\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{crit}}]}\bigl(E_k-V(t)\bigr)_+
^{-1/2}dt\\
&\qquad \le C_{\#}\delta\quad\text{by\ (Z}\overline 7)\ .\endalign
 $$

\noindent Combining this with (4), we get
$$
\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{crit}}]}|u_k(x)|^2dx\le
C_{\#}\delta\quad\text{if}\quad E_{\text{crit}}\le E_k\le 0\ .
\tag 5
$$

\noindent If $E_k<E_{\text{crit}}$, then at least we know that
$$\multline
\int_{I_{\text{BVP}}\cap(-\infty,x_{\text{crit}}]}|u_k(x)|^2dx\le 1\ ,\\
\text{since}\quad u_k\quad\text{is\ a\ normalized\ eigenfunction}\ .
\endmultline\tag 6
$$

\noindent From (5), (6) and the definition of the density $\rho(x)$, we get
$$\multline
\int_{I_{\text{BVP}}\cap(-\infty,x_{\text{crit}}]}\rho(x)dx\le C_{\#}
\delta \cdot\,(\text{Number\ of}\ E_k\in [E_{\text{crit}},0])\\
+C_{\#}\cdot(\,\text{Number\ of}\ E_k<E_{\text{crit}})\\
\le C_{\#}\delta\cdot\,(\text{Number\ of}\ E_k\le 0)+C_{\#}
\cdot(\,\text{Number\ of}\ E_k\le E_{\text{crit}})\endmultline\tag 7
$$

\noindent Again we invoke Theorem 2 on WKB Theory
with weak turning points.  This time, for $E_0$ we take either $0$
or $E_{\text{crit}}$.  Hypotheses (H$\overline 0$)$\ldots$(H$\overline 6$)
are easily verified for both values of $E_0$, and the constants in
(H$\overline 0$)$\ldots$(H$\overline 6$) depend only on the constants
in our present hypotheses (Z$\overline 0$)$\ldots$(Z$\overline 8$). 
Theorem 2 shows that
$$\multline
\Big|(\text{Number\ of}\ E_k\le 0)- \frac 1\pi\int_{I_{\text{BVP}}}
\bigl(-V(x)\bigr)_+^{1/2}dx\big|\le C_{\#}\ ,\quad\text{and}\\
\Big|(\text{Number\ of}\ E_k\le E_{\text{crit}})-\frac 1\pi
\int_{I_{\text{BVP}}}\bigl(E_{\text{crit}}-V(x)\bigr)_+^{1/2}dx\Big|\le C_{\#}\
. \endmultline\tag 8 $$

\noindent These estimates and (7) imply
$$\multline
\int_{I_{\text{BVP}}\cap(-\infty,x_{\text{crit}}]}\rho(x)dx\le
C_{\#}+C_{\#}\delta\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{1/2}
dx\\
+C_{\#}\int_{I_{\text{BVP}}}\bigl(E_{\text{crit}}-V(x)\bigr)_+^{1/2}dx
\ ,\endmultline
$$

\noindent which is the conclusion of the lemma regarding 
$\int_{I_{\text{BVP}}\cap (-\infty,x_{\text{crit}}]}\rho(x)dx$.  The conclusion
regarding $\int_{I_{\text{BVP}}}\rho(x)dx$ is immediate from(8).
$\qquad\blacksquare$\enddemo




\vfill\eject







\heading{\bf{THE DENSITY FOR DEGENERATE ONE-DIMENSIONAL POTENTIALS III}}
\endheading
\medskip

The results in this section are for application to three-dimensional
problems, with angular momentum $\ell$ in the range $[1,\ \text{Large\
Constant}\,]$.  Our setting is as follows.

We are given a potential $V(x)$, smooth on $(0,\infty)$.  We take
$B(x)=x$, and let $S(x)$ be a positive function on $I=[x_0,x_1]\subset
(0,\infty)$.  As usual, we set $\lambda(x)=S^{1/2}(x)B(x)$ on $I$.  In
addition to $x_0$, $x_1$, we are given other points $x_{\text{small}}$,
$x_{\text{big}}$, $x_{\text{crit}}$, $x_\ast\in
(0,\infty)$, with
$$
0<x_{\text{small}}<\frac 12 x_0\ , 2x_0<x_{\text{crit}}
<\frac 12 x_\ast\ , x_\ast<\frac {1}{16}x_1\ , 2x_1<x_{\text{big}}\ .\tag 1
$$
\noindent Set $H=-\frac{d^2}{dx^2}+V(x)$ on $(0,\infty)$ with Dirichlet boundary
conditions.  Let $E_k$, $u_k(x)$ be the eigenvalues and (normalized)
eigenfunctions of $H$.  Define the density $\rho(x)$ as usual by
$\rho(x)=\sum\limits_{E_k\le 0}|u_k(x)|^2$.

Our goal is to make a (very crude) estimate of $\int_0^{x_{\text{crit}}}
\rho(x)dx$.  In addition to (1), we make the following assumptions.
\medskip

\subhead Hypotheses\endsubhead
\smallskip

\roster
\item"(Z$\hat 0$)" If $x,y\in I$ and $|x-y|<\frac 12 B(x)$, then
$c<S(y)/S(x)<C$.

\item"(Z$\hat 1$)" If $x\in I$, then $\big|\bigl(\frac{d}{dx}\bigr)^\alpha
V(x)\big|\le C_\alpha S(x)B^{-\alpha}(x)$.

\item"(Z$\hat 2$)" If $x\in I$, then $V(x)<-cS(x)$ and $V^\prime(x)>cS(x)B^{-1}
(x)$.

\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 
$B(x)\equiv x$, $2x_0<x_1$.  (H$\hat 1$) follows from (Z$\hat 1$)
and (Z$\hat 2$).  (H$\hat 2$) follows from (Z$\hat 3$).  (H$\hat 3$)
is verified as follows.  By definition, $I_{\text{left}}$ and $I_{\text{rt}}$
are non-empty.  Since $0<x_{\text{small}}<x_0$, we have $|I_{\text{left}}|
\le x_0=B(x_0)=B(x_{\text{left}})$.  Since $0<x_{\text{big}}<\underline C
x_1$ by (Z$\hat 6$), we have $|I_{\text{rt}}|\le \underline Cx_1=
\underline CB(x_{\text{rt}})$.  By (Z$\hat 2$) we have
$S(x)\le C_{\#}|V(x)|$ for $x\in I$, hence for $x=x_0,x_1$.  Therefore,
$\lambda(x_{\text{left}})=S^{1/2}(x_0)B(x_0)=S^{1/2}(x_0)\cdot
x_0\le C_{\#}|V(x_0)|^{1/2}\cdot x_0\le C_\ast$ by (Z$\hat 5$); and
similarly, $\lambda(x_{\text{rt}})=S^{1/2}(x_1)B(x_1)=S^{1/2}(x_1)\cdot
x_1\le C_{\#}|V(x_1)|^{1/2}\cdot x_1\le C_\ast$ by (Z$\hat 7$).
This completes the verification of (H$\hat 3$).
(H$\hat 4$) follows from (Z$\hat 5$), since $|V(x)|\le \underline Cx_0^{-2}
\le C_\ast|I_{\text{left}}|^{-2}$ in $I_{\text{left}}$, and we can take
$\hat x_{\text{far\ left}}$ so that $(x-\hat x_{\text{far\ left}})\sim
|I_{\text{left}}|$ in $I_{\text{left}}$.  (H$\hat 5$) follows from
(Z$\hat 7$) and the equation $|I_{\text{rt}}|\sim x_1$ contained in (Z$\hat 6$).
(H$\hat 6$) follows from (Z$\hat 4$) since $0<x_{\text{small}}<\frac 12 x_0$.
(H$\hat 7$) follows from (Z$\hat 8$).$\qquad\blacksquare$\enddemo


\vglue 1pc
\proclaim{Lemma 2}  Suppose $2x_0<\tilde x<\frac 12 x_1$.  Set 
$\tilde E=V(\tilde x)$, and define:  $\tilde V(x)=V(x)-\tilde E$,
$\tilde E_0=0$, $\tilde I_{\text{center}}=[x_0,\tilde x-\hat C_{\#}
\lambda^{-2/3}(\tilde x)\cdot\tilde x]$ with $\hat C_{\#}$ to be picked
large enough, $\tilde I_{\text{left}}=[x_{\text{small}},x_0]$,
$\tilde I_{\text{far\ left}}\ = (0,x_{\text{small}}]$,
$\tilde I_{\text{rt}}=[\tilde x-\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot
\tilde x,\tilde x+\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot
\tilde x]$, $\tilde I_{\text{far\ rt}}\ =[\tilde x+\hat C_{\#}
\lambda^{-2/3}(\tilde x)\cdot\tilde x,\infty)$.  On $\tilde I_{\text{center}}$,
define $\tilde B(x)=\min(x,\tilde x-x)$, $\tilde S(x)=S(x)\cdot
\bigl(\frac{\tilde x-x}{\tilde x}\bigr)$.  Then $\tilde V(x)$,
$\tilde E_0$, $\tilde I_{\text{far\ left}}\ldots\tilde I_{\text{far\ rt}}$,
$\tilde S(x)$, $\tilde B(x)$ satisfy hypotheses 
(H$\hat 0$)$\ldots$(H$\hat 7$) in the section on WKB Theory with
weak turning points.  The
constants called $c$, $C$, $C_\alpha$ in (H$\hat 0$)$\ldots$(H$\hat 7$)
may be taken to have the form $C_{\#}$.  The constants called
$\underline c$, $\underline C$ in (H$\hat 0$)$\ldots$(H$\hat 7$)
may be taken to have the form $C_\ast$.\endproclaim

\vglue 1pc
\demo{Remark}  We have $\lambda(\tilde x)\ge c_{\#}\Lambda$, so (Z$\hat 3$)
shows that $\tilde I_{\text{rt}}\subset I$ and $\tilde I_{\text{center}}
\supset [x_0,\frac {9}{10}\tilde x]$.\enddemo

\vglue 1pc
\demo{Proof}  The above remark shows that $\tilde I_{\text{center}}$
contains $[x_0,\frac{9}{10}\cdot 2x_0]$, so $|\tilde I_{\text{center}}|\ge c_{\#}
\tilde B(x)$ for $x=x_0$, hence for all $x\in \tilde I_{\text{center}}$.
If $x,y\in \tilde I_{\text{center}}$ with $|x-y|<c_{\#}
\tilde B(x)$, then one checks that $\tilde B(y)\sim \tilde B(x)$,
$\bigl(\frac{\tilde x-x}{\tilde x}\bigr)\sim\bigl(\frac{\tilde x-y}
{\tilde x}\bigr)$, and $S(x)\sim S(y)$ since $|x-y|<c_{\#}B(x)$.  Hence
$\tilde B(y)\sim \tilde B(x)$ and $\tilde S(y)\sim \tilde S(x)$.  These 
remarks prove (H$\hat 0$).

To prove (H$\hat 1$) we argue as follows.  For $x\in \tilde I_{\text{center}}$,
we know that $x\le \tilde x$ and that $x,\tilde x\in I$.  From (Z$\hat 2$)
we get $V(x)<V(\tilde x)<0$, so $|V(x)-\tilde E|=|V(x)-V(\tilde x)|\le
|V(x)|\le C_{\#}S(x)\le C_{\#}^\prime\tilde S(x)$
provided $|x-\tilde x|>c_{\#}\tilde x$.  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 $\alpha=0$, $|x-\tilde x|<c_{\#}
\tilde x$.  However in this case, we have $|\tilde V(x)|=|V(x)-\tilde E|=
|V(x)-V(\tilde x)|\le C_{\#}S(\tilde x)\cdot \bigl(\frac{\tilde x-x}
{\tilde x}\bigr)$ by our estimates for $V^\prime(x)$, and the right-hand side
is $\le C_{\#}\tilde S(x)$. Thus we know that $\big|\bigl(\frac{d}{dx}\bigr)
^\alpha\tilde V(x)\big|\le C_\ast^{\alpha}\tilde S(x)\tilde B^{-\alpha}
(x)$, for all $\alpha\ge 0$, $x\in \tilde I_{\text{center}}$.

To complete the proof of (H$\hat 1$), we need to show that
$-\tilde V(x)>c_{\#}\tilde S(x)$ 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 $V(x_0)<V(\tilde x)<0$, so it's enough to check that $|V(x_0)|\le
C_\ast x_0^{-2}$.  This follows from another application of (Z$\hat 5$),
completing the proof of (H$\hat 4$).

To prove (H$\hat 5$), we need to show that $|V(x)-V(\tilde x)|\le C_\ast
(2\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot\tilde x)^{-2}$ for
$|x-\tilde x|\le \hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot\tilde x$.
The range of $x$ is contained in $I$, by the remark following the statement
of Lemma 2.  Hence $|V(x)-V(\tilde x)|\le C_{\#}S(\tilde x)B^{-1}(\tilde x)
\cdot |x-\tilde x|\le C_{\#}S(\tilde x)\cdot \hat C_{\#}
\lambda^{-2/3}(\tilde x)=C_{\#}[\lambda^2(\tilde x)\cdot \tilde x^{-2}]
\cdot \hat C_{\#}\lambda^{-2/3}(\tilde x)=C_{\#}\hat C_{\#}\lambda^{4/3}
(\tilde x)\cdot \tilde x^{-2}=C_{\#}\hat C_{\#}^3(2\hat C_{\#}\lambda^{-2/3}
(\tilde x)\cdot \tilde x)^{-2}$, as required.  The proof of (H$\hat 5$)
is complete.

To prove (H$\hat 6$), we need to show that $V(x)-V(\tilde x)\ge c_\ast
x_0^{-2}$ for $x\in (0,x_{\text{small}}]$.  From (Z$\hat 4$) get
$V(x)\ge c_\ast x_0^{-2}$, and (Z$\hat 2$) gives $V(\tilde x)<0$.  Hence
(H$\hat 6$) is trivially correct.  Finally, (H$\hat 7$) asserts that
$V(x)-V(\tilde x)\ge \frac{-10^{-9}}{(x-\max \tilde I_{\text{center}})^2}$ for
$x\ge \tilde x+\hat C_{\#}\lambda^{-2/3}(\tilde x)\cdot\tilde x$.  We will
prove the stronger inequality $V(x)-V(\tilde x)\ge 0$ for $x\ge \tilde x$.

In fact, in $[\tilde x,x_1]$ and in $[x_1,x_{\text{big}}]$, the function
$V$ is increasing, by (Z$\hat 2$), (Z$\hat 6$).  Hence, $\tilde x\le x
\le x_{\text{big}}$ implies $V(x)-V(\tilde x)\ge 0$.  On the other
hand, if $x\ge x_{\text{big}}$, then $V(x)\ge 0$ by (Z$\hat 8$), while
$V(\tilde x)<0$ by (Z$\hat 2$).  Hence again $V(x)-V(\tilde x)\ge 0$,
completing the proof.$\qquad\blacksquare$\enddemo
\smallskip

Now set $\tilde E_{\text{low}}=V(x_\ast)$, $\tilde E_{\text{hi}}=V(\frac 14
x_1)$. For $E_k\in [\tilde E_{\text{low}},\tilde E_{\text{hi}}]$ we can write
$$
E_k=V(\tilde x)=\tilde E\quad\text{for\ some}\quad
\tilde x\in[x_\ast,\frac 14x_1]\ ,\quad\text{by\ (Z}\,\hat 2)\ .
$$
\noindent Lemma 2 and Theorem 1 on WKB Theory with
weak turning points yield the following estimates for the normalized
eigenfunction $u_k$. (See ``Review of Earlier Results.")
$$\multline
|u_k(x)|^2\le C_\ast\Bigl(\int_{\tilde I_{\text{center}}}\bigl(E_k-V(y)\bigr)
^{-1/2}dy\Bigr)^{-1}\bigl(E_k-V(x)\bigr)^{-1/2}\\
\text{for}\quad x\in \tilde I_{\text{center}}\ ;\endmultline\tag 2
$$
$$\multline
|u_k(x)|^2\le C_\ast\Bigl(\int_{\tilde I_{\text{center}}}\bigl(E_k-V(y)\bigr)
^{-1/2}dy\Bigr)^{-1}\bigl(E_k-V(x_0)\bigr)^{-1/2}
\exp\Bigl(\frac{-c_\ast(x_0-x)}{x_0}\Bigr)\\
\text{for}\quad x\le x_0\ .\endmultline
$$

\noindent In particular,
$$\align
\int_0^{x_0}|u_k(x)|^2dx&\le C_\ast\Bigl(\int_{\tilde I_{\text{center}}}
\bigl(E_k-V(y)\bigr)^{-1/2}dy\Bigr)^{-1}\bigl(E_k-V(x_0)\bigr)^{-1/2}x_0\\
&\le C_\ast\Bigl(\int_{\tilde I_{\text{center}}}\bigl(E_k-V(y)\bigr)^{-1/2}
dy\Bigr)^{-1}S^{-1/2}(x_0)\cdot x_0\endalign
$$

\noindent because $E_k=V(\tilde x)\ge V(2x_0)\ge V(x_0)+c_{\#}S(x_0)
B^{-1}(x_0)\cdot (c_{\#}x_0)$ by (Z$\hat 2$).

Also, $E_k=V(\tilde x)\in [V(x_0),0]$, so $|E_k|\le C_{\#}
S(x_0)$; and $|V(x)|\le C_{\#}S(x_0)$ for $x\in [x_0,\frac 32 x_0]$.
For such $x$ we have also $E_k-V(x)>0$.  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 $V(x)<V(\frac 14 x_1)=\tilde E_{\text{hi}}
<0$, so
$$
\big|\bigl(-V(x)\bigr)^{1/2}-\bigl(\tilde E_{\text{hi}}-V(x)\bigr)^{1/2}\big|
\le C_{\#}|\tilde E_{\text{hi}}|\bigl(-V(x)\bigr)^{-1/2}\le C_{\#}
|\tilde E_{\text{hi}}|S^{-1/2}(x) \ .
$$

\noindent Hence
$$\multline
\int_{x_0}^{x_1/8}\big\{\bigl(-V(x)\bigr)^{1/2}-\bigl(\tilde E_{\text{hi}}
-V(x)\bigr)^{1/2}\bigr\}dx\le C_{\#}|\tilde E_{\text{hi}}|
\int_{x_0}^{x_1}\frac{dx}{S^{1/2}(x)}\\
\le C_{\#}|\tilde E_{\text{hi}}|x_1^2\int_{x_0}^{x_1}\frac{dx}
{S^{1/2}(x)x^2}=\frac{C_{\#}}{\Lambda}|\tilde E_{\text{hi}}|x_1^2
\quad (\,\text{by\ definition\ of}\ \Lambda)\\
\le |\tilde E_{\text{hi}}|x_1^2\quad (\,\text{by\ (Z}\,\hat 3))=|V(\frac{x_1}
{4})|x_1^2\le C_\ast\quad\text{by\ (Z}\,\hat 7).\endmultline
$$
\noindent Thus, all terms on the right in (9) are dominated by $C_\ast$,
so that (9) becomes
$$
\bigl(\,\text{Number\ of}\ E_k\in (\tilde E_{\text{hi}},0])\le C_\ast
\tag 10
$$

\noindent Substituting (7), (8), (10) into (5), we obtain
$$\multline
\int_0^{x_{\text{crit}}}\rho(x)dx\le C_\ast\delta\int_{x_0}^{x_1}
\bigl(-V(x)\bigr)^{1/2}dx\\
+C_\ast\int_{x_0}^{x_1}\bigl(\tilde E_{\text{low}}-V(x)\bigr)_+^{1/2}
dx+C_\ast\ .
\endmultline\tag 11
$$

Immediately from (8), (11) and the definition $\tilde E_{\text{low}}=
V(x_\ast)$, we get the following.

\vglue 1pc
\proclaim{Third Degenerate Density Lemma} Assume (1) and 
(Z$\hat 0$)$\ldots$(Z$\hat 9$).  Set $E_{\text{crit}}=V(x_\ast)$.  Then
$$
\int_0^{x_{\text{crit}}}\rho(x)dx\le C_\ast+C_\ast\delta\int_0^\infty
\bigl(-V(x)\bigr)_+^{1/2}dx\\
+C_\ast\int_0^\infty\bigl(E_{\text{crit}}-V(x)\bigr)_+^{1/2}dx
$$
\noindent and
$$
\int_0^\infty\rho(x)dx\le C_\ast+C_\ast\int_0^\infty\bigl(-V(x)\bigr)_+^{1/2}
dx
$$
\noindent with $C_\ast$ depending only on $c$, $C$, $C_\alpha$, $\underline c$,
$\underline C$, in (Z$\hat 0$)$\ldots$(Z$\hat9$).
\endproclaim

\vfill\eject



\heading{\bf{THE DENSITY FOR DEGENERATE ONE-DIMENSIONAL POTENTIALS IV}}
\endheading
\medskip

The results in this section will be used to handle angular momentum
$\ell=0$ in three-dimensional problems.  Our setting is as follows.

We are given a smooth potential $V(x)$ on $(0,\infty)$.  We take
$B(x)=x$, and let $S(x)$ be a positive function on $I=[x_0,x_1]\subset
(0,\infty)$.  Let $\lambda(x)=S^{1/2}(x)B(x)$ as usual.  We are given
$x_{\text{crit}}$, $x_\ast$, $x_{\text{big}}$, satisfying
$$
16x_0<x_{\text{crit}}\ , 16x_{\text{crit}}<x_\ast\ , 16x_\ast<x_1\ ,
16x_1<x_{\text{big}}\ .\tag 1
$$

\noindent Set $H=-\frac{d^2}{dx^2}+V(x)$ on $(0,\infty)$, with Dirichlet
boundary conditions.  Let $E_k$, $u_k(x)$ be the eigenvalues and (normalized)
eigenfunctions of $H$.  Define the density $\rho(x)$ as usual by
$$
\rho(x)=\sum\limits_{E_k\le 0}|u_k(x)|^2\ .
$$

\noindent Our goal is to make a crude estimate of $\int_0^{x_{\text{crit}}}
\rho(x)dx$.  In addition to (1), we make the following assumptions.


\subhead Hypotheses\endsubhead
\smallskip

\roster
\item"(Z0$^\dag$)" If $x,y\in I$ and $|x-y|<\frac 12 B(x)$, then
$c<S(y)/S(x)<C$.

\item"(Z1$^\dag$)" If $x\in I$ and $\alpha\ge 0$, then $\big|\bigl(\frac{d}
{dx}\bigr)^\alpha V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$.

\item"(Z2$^\dag$)" If $x\in I$, then $V(x)<-cS(x)$ and $V^\prime(x)>cS(x)B^{-1}
(x)$.

\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 $V(x_0)<V(\tilde x)<0$.  Hence $|\tilde V(x)|=|V(x)-
V(\tilde x)|\le |V(x)|+|V(\tilde x)|\le
\frac{2C_\ast}{x_0x}$, proving (H$\hat 4$).
(H$\hat 5$) is proven here just as it was in the proof of
Lemma 2 in the previous section. (H$\hat 6$) holds vacuously,
since $\tilde I_{\text{far\ left}}=\emptyset$.  (H$\hat 7$) asserts that
$V(x)-V(\tilde x)\ge -10^{-9}(x-[\tilde x-\hat C_{\#}\lambda^{-2/3}
(\tilde x)\cdot \tilde x])^{-2}$ for $x\ge \tilde x+\hat C_{\#}\lambda^{-2/3}
(\tilde x)\cdot \tilde x$.  We shall verify the following inequalities,
which together are stronger than (H$\hat 7$).
$$
V(x)-V(\tilde x)\ge 0\quad\text{for}\quad \tilde x\le x\le x_{\text{big}}
\tag 2
$$
$$
V(x)-V(\tilde x)\ge -10^{-9}x^{-2}\quad\text{for}\quad x\ge x_{\text{big}}\ .
\tag 3$$
\noindent From (Z2$^\dag$), (Z5$^\dag$), we have $V$ increasing on
$[\tilde x,x_1]$ and on $[x_1,x_{\text{big}}]$, so (2) holds.  From
(Z2$^\dag$), (Z6$^\dag$) we have $V(\tilde x)\le 0$ and $V(x)\ge -10^{-9}
x^{-2}$ for $x\ge x_{\text{big}}$, so (3) holds.  This completes the proof
of (H$\hat 7$).$\qquad\blacksquare$\enddemo

Using Lemmas 1 and 2 above, and Theorem 1 in the section on WKB with weak
turning points, we derive the following result, exactly as in the previous
section.

\vglue 1pc
\proclaim{Fourth Degenerate Density Lemma}  Assume (1) and
(Z0$^\dag$)$\ldots$(Z7$^\dag$).  Set\hfill\break
$E_{\text{crit}}=V(x_\ast)$.  Then
$$
\int_0^{x_{\text{crit}}}\rho(x)dx\le C_\ast+C_\ast\delta\int_0^\infty
\bigl(-V(x)\bigr)_+^{1/2}dx+C_\ast\int_0^\infty\bigl(E_{\text{crit}}
-V(x)\bigr)_+^{1/2}dx
$$
\noindent and $\int_0^\infty\rho(x)dx\le C_\ast+C_\ast\int_0^\infty
\bigl(-V(x)\bigr)_+^{1/2}dx$ with $C_\ast$ depending only on $c$, $C$, 
$C_\alpha$, $\underline C$ in (Z0$^\dag$)$\ldots$(Z7$^\dag$).\endproclaim




\vfill\eject

\head References \endhead
\medskip

\widestnumber\key{FS7}

\ref\key E 
\by A. Erd\'elyi
\book Asymptotic Expansions
\bookinfo Dover
\yr 1956
\endref

\ref\key FS1 
\by C. Fefferman and L. Seco
\paper The Ground-State Energy of a Large Atom
\jour Bull. A.M.S.
\yr 1990
\endref

\ref\key FS2
\bysame %C. Fefferman and L. Seco
\paper Eigenvalues and Eigenfunctions of Ordinary Differential Operators
\jour Advances in Math
\toappear
\endref

\ref\key FS3
\bysame %C. Fefferman and L. Seco
\paper The Eigenvalue Sum in a One-Dimensional Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS4
\bysame
\paper The Density in a Three-Dimensional Radial Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS5
\bysame
\paper The Eigenvalue Sum for a Three-Dimensional Radial Potential
\jour Advances in Math
\toappear
\endref

\ref\key FS6
\bysame
\paper On the Dirac and Schwinger Corrections to the Ground-State Energy
of an Atom
\jour Advances in Math
\toappear
\endref

\ref\key FS7
\bysame
\paper Aperiodicity of the Hamiltonian Flow in the Thomas-Fermi Potential
\jour Advances in Math
\toappear
\endref





\vfill\eject

\end