\input amstex \magnification \magstep1 \documentstyle{amsppt} \NoPageNumbers \hsize=6truein \vsize 9truein \hoffset .15truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \centerline{\bf{THE EIGENVALUE SUM FOR A ONE-DIMENSIONAL POTENTIAL}}\smallskip \centerline{by} \smallskip \bigskip \centerline{\bf{C. Fefferman\footnote"*"{partially supported by NSF grant \#DMS--9104455.A01} $\quad$ and $\quad$ L. Seco}} \vglue 4pc \head Table of Contents\endhead \medskip \line{\sl \hfil Pages} \smallskip \line{Introduction\hfil 1} \smallskip \line{Review of Earlier Results\hfil 6}\smallskip \line{Truncated Eigenvalue Sums\hfil 19}\smallskip \line{The First WKB Eigenvalue Sum Theorem \hfil 45}\smallskip \line{Low Eigenvalues in a Potential with a Coulomb Singularity\hfil 49} \smallskip \line{The Second WKB Eigenvalue Sum Theorem\hfil 57}\smallskip \line{Eigenvalue Sums for Degenerate Potentials\hfil 71} \smallskip \line{The Third WKB Eigenvalue Sum Theorem\hfil 77} \smallskip \line{References\hfil 85} \smallskip \vfill\eject \pageno 1 \input amstex \documentstyle{amsppt} %\load eufm %\load msam %\load msbm %\load eufb %\UseAMSsymbols \magnification\magstep1 \NoBlackBoxes \hsize 6 truein \vsize 9truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \def\Icirc{\hbox{$\int$}\kern-6pt\raise 1pt\hbox{$\circ$}} \def\chm{\chi_{{}_-}} \def\chp{\chi_{{}_+}} %\input intro.tex %\input review.tex %\input trun.tex %\input trun2.tex %\input 1wkb.tex %\input loweigen.tex %\input 2wkb.tex %\input eigensums.tex %\input 3wkb.tex %\input ref.tex %%intro.tex %% \head Introduction\endhead \medskip This paper is part of a series [FS2$\ldots$7] proving the asymptotic formula announced in [FS1] for the ground-state energy of an atom. Our goal here is to give precise estimates for the sum of the negative eigenvalues of an ordinary differential operator $$H=-\frac{d^2}{dx^2}+V(x)\quad\text{on\ an\ interval}\ I_{\text{BVP}}\ .\tag"(1)"$$ \noindent We denote this sum by $\text{sneg}(H)$. The potentials of interest to us are large and slowly varying. A basic example is $$V(x)=\lambda^2V_1(x)\quad\text{on}\quad I_{\text{BVP}}=[-1,1]\ ,\tag"(2)"$$ \noindent with $V_1$ a fixed smooth function and $\lambda$ a large parameter. We suppose $$V_1(0)<0\ ,\ V_1^\prime(0)=0\ ,\ V_1^{\prime\prime}>c>0\ \text{on}\ [-1,1]\ ,\ \{V_1<0\}\subset\subset (-1,1)\ .\tag"(3)"$$ \noindent For such potentials, a standard approximate formula for the eigenvalue sum is $$\text{sneg}(H)\approx -\frac {2}{3\pi}\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+ ^{3/2}\ dx\ ,\tag"(4)"$$ \noindent where $t_+^s=t^s$ if $t>0$, $t_+^s=0$ if $t\le 0$. \noindent Unfortunately, (4) is too crude for our purposes. To give a sharper approximation, we set $$\phi(0)=\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{1/2}\ dx\ ,\quad \phi^\prime(0)=\frac 12 \int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{-1/2} \ dx\ ,\quad\text{and}\tag"(5)"$$ $$\tilde \chi(t)=\operatornamewithlimits{\min}_{k\in \Bbb Z}\ \bigl\{ |t-(k+\frac12 )|^2-\frac 1{12}\bigr\}\quad\text{for}\ t\in \Bbb R\ . \tag"(6)"$$ \noindent Our basic result for potentials of the form (2), (3) is as follows. \vglue 1pc \proclaim{Theorem 1} The sum of the negative eigenvalues of $H$ is given by $$\split {\roman{sneg}}(H)=-\frac{2}{3\pi}\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{3/2 }dx+\frac{1}{24\pi}\int_{I_{\text{BVP}}}V^{\prime\prime}(x) \bigl(-V(x)\bigr)_+^{-1/2}\ dx\\ +\frac \pi 2 \bigl(\phi^\prime(0)\bigr)^{-1} \tilde\chi\bigl(\frac 1\pi \phi(0)-\frac 12\bigr) +\ {\roman{Error}}\ ,\endsplit\tag"(7)"$$ \noindent with $|{\roman{Error}}|0$ and on $V_1$, but not on $\lambda$.\endproclaim \medskip On the right in (7), the first term has size $\sim \lambda^3$, while the next two terms are $\sim \lambda$. To motivate Theorem 1, we recall the textbook derivation of (4). At the heart of the argument is the WKB approximation (see Erd\'elyi [E]). According to WKB the eigenvalues $E_k$ of $H$ satisfy $$\phi(E_k)\approx \pi(k+1/2)\quad\text{for\ integers}\ k\ ,\ \text{with} \tag"(8)"$$ $$\phi(E)=\int\limits_{I_{\text{BVP}}}\bigl(E-V(x)\bigr)_+^{1/2}\ dx\ .\tag"(9)"$$ \noindent Defining $E(t)$ as the solution of $\phi(E)=\pi(t+1/2)$, we rewrite (8) as $E_k\approx E(k)$, which gives $$\text{sneg}(H)\approx \sum\limits_{k\in [a,b]\cap \Bbb Z}E(k)\ ,\ \text{with}\ a=\frac 1\pi \phi(\min V)-\frac 12,\ b=\frac 1\pi \phi(0)-\frac 12\ . \tag"(10)"$$ \noindent It is reasonable to guess that $$\sum\limits_{k\in [a,b]\cap \Bbb Z}E(k)\approx \int_a^bE(t)dt\ . \tag"(11)"$$ \noindent Elementary manipulations show that $\int_a^bE(t)dt=-\frac{2}{3\pi} \int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{-3/2}\ dx$. Hence, the usual approximation (4) for $\text{sneg}(H)$ follows from (8) and (11). Our previous papers [FS2,3] prove refinements of the basic approximations (8) and (11). Specifically, the eigenvalues $E_k$ satisfy $$\phi(E_k)+\frac{1}{48}\psi(E_k)\approx \pi(k+1/2)\ ,\ \text{where} \tag"(12)"$$ $$\psi(E)=\operatornamewithlimits{\lim}_{\delta\to 0+} \Bigl\{\int\limits_{E-V(x)>\delta}V^{\prime\prime}(x)\bigl(E-V(x)\bigr)_+^{-3/2}\ dx-q(E)\delta^{-1/2} \Bigr\}\tag"(13)"$$ \noindent and $q(E)$ is defined to make the limit finite in (13). \noindent Equation (8) holds modulo errors $O(\lambda^{-1})$, while (12) holds modulo errors $O(\lambda^{\varepsilon-2})$. (See [FS2].) Regarding (11), we studied $\sum\limits_{k\in \Bbb Z\cap [a,b]}f(k)$ for general, slowly varying functions $f$. We showed in [FS3] that $$\sum\limits_{k\in \Bbb Z\cap [a,b]}f(k)\approx \int_a^bf(t)dt- f(b)\chm(b)-f(a)\chp(a)+\frac 12f^\prime(b)\tilde\chi(b)-\frac 12 f^\prime(a)\tilde\chi(a)\tag"(14)"$$ \noindent with $$\chm(t)=(t-k-1/2)\ \text{for}\ k=\ \text{(greatest\ integer}\ \le t\ \text{);} \tag"(15)"$$ $$\chp(t)=(k-t-1/2)\ \text{for}\ k=\ \text{(least\ integer}\ \ge t\ \text{);\ and}\ \tilde\chi(t)\ \text{given\ by\ (6)}\ .\tag"(16)"$$ \noindent Here we take $f(t)=E(t)$. Equation (14) then holds modulo errors $O(1)$, while (11) merely holds modulo $O(\lambda)$. Using (12) and (14) in place of the crude approximations (8), (11) in the derivation of (4), we arrive at the sharp formula (7) for $\text{sneg}(H)$. This concludes our introductory remarks on Theorem 1. \medskip To understand atoms, we need to deal with potentials having a Coulomb singularity $$V(x)\approx\frac{\ell(\ell+1)}{x^2}-\frac Z x+E^0\ \text{near}\ x=0\ .\tag"(17)"$$ \noindent In our application [FS4,5], we have $Z>>1$, $E^0\sim Z^{4/3}$ and $\ell=O(Z^{1/3})$. When \ell<cS(x)\bigl(B(x)\bigr)^{-2};\ \text{while\ for}\\ |x-x_0|\ge c_1B(x_0)\ \text{we\ have}\ |V^\prime(x)|>cS(x)\bigl(B(x) \bigr)^{-1}\ .\endsplit\tag"(19)"$$The number that plays the r\^ ole of \lambda in Theorem 1 is$$ \Lambda=\Bigl(\int_{V(x)<0}\frac{dx}{\bigl(S(x)\bigr)^{1/2}\bigl(B(x) \bigr)^2}\Bigr)^{-1}\ ,\tag"(20)" \noindent as explained in [FS2]. Now we can state our first main result, modulo technicalities. \vglue 1pc \proclaim{First WKB Eigenvalue Sum Theorem} Let V satisfy (18), (19) and various technical conditions; and suppose \Lambda is large. Then {\roman{sneg}}(H) is given by (7), with |{\roman{Error}}|1 and N>K\varepsilon^{-10}. We define N^\prime=[\varepsilon N/500] and N^{\prime\prime}=\frac 32 \varepsilon N^\prime-K-33. Our goal is to understand the eigenvalues 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}\enddemo \noindent {\it Assumptions on V(x), S(x), B(x) in I\/} \roster \item"(Hyp0)" If x,y \in I and |x-y|cB(x_{\text{left}}), \text{dist}(x_{\text{rt}},\partial I)>cB(x_{\text{rt}}). \item"(Hyp3)" For x_{\text{left}}\le x\le x_{\text{left}}+c_1 B(x_{\text{left}}) we have -V^\prime(x)>cS(x_{\text{left}})B^{-1} (x_{\text{left}}), and for x_{\text{rt}}-c_1B(x_{\text{rt}})\le x\le x_{\text{rt}} we have +V^\prime(x)>cS(x_{\text{rt}})B^{-1}(x_{\text{rt}}). \item"(Hyp4)" For x_{\text{left}}+c_1B(x_{\text{left}})\le x\le x_{\text{rt}}-c_1B(x_{\text{rt}}) we have cS(x)E \quad for all\hfill\break x \in I_{\text{BVP}}\backslash [x_{\text{left}}(E), x_{\text{rt}}(E)]. \item"{(Hyp6)}" If x \in I_{\text{BVP}} satisfies xx_{\text{rt}}+\frac 12 \lambda_{\text{rt}} ^KB_{\text{rt}}, then V(x)\ge E_\infty+\frac{100}{|x-x_{\text{rt}}|^2}. \endroster \smallskip \noindent\demo{Technical Assumptions}\enddemo \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\smallskip \medskip \noindent \demo{WKB Condition}\enddemo \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 \smallskip \noindent\demo{Definitions and Basic Properties of Phases}\enddemo Assume hypotheses (Hyp0)\ldots(Hyp10). For |E-E_0|0 be given, with \varepsilon N\ge 100. Let V(x) be a potential defined on a (possibly unbounded) interval I_{\text{BVP}}. Let S, B be positive numbers, and let x_0 \in I_{\text{BVP}} be given. Define \lambda=S^{1/2}B. Let E_\infty be a given energy, with E_\infty>V(x_0). We make the following assumptions. \roster \item"{(H0^*)}" I=\{|x-x_0|\frac 12 \lambda^KB, we have V(x)\ge E_\infty +\frac{1000}{|x-x_0|^2}. \item"{(H6^*)}" \lambda is bounded below by a positive constant depending only on c, c^\prime, c^{\prime \prime}, C_\alpha in (H0^*)\ldots(H4^*), and on \varepsilon, K, N.\endroster Let H=-\frac{d^2}{dx^2}+V(x) on L^2(I_{\text{BVP}}) with Dirichlet or Neumann conditions at the endpoints. For V(x_0)\delta \endSb V^{\prime\prime}(x)(E-V(x))^{-3/2}\,dx -q(E)\delta^{-1/2}\Bigr]\\ &=\lim_{\delta_{\text{left}},\delta_{\text{rt}}\to 0+} \Bigl[\int_{x_{\text{left}}(E)+\delta_{\text{left}}} ^{x_{\text{rt}}(E)-\delta_{\text{rt}}} V^{\prime\prime} (x)(E-V(x))^{-3/2}\,dx-q_{\text{left}}(E)\delta_{\text{ left}}^{-1/2}\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\,\,\,\,\, -q_{\text{rt}}(E)\delta_{\text{rt}}^{-1/2} \Bigr]\endalign \noindent withq(E)$,$q_{\text{left}}(E)$,$q_{\text{ rt}}(E)$uniquely specified by demanding the finiteness of the limits. \vglue 1pc \proclaim{Lemma B1} The phases$\phi(E)$,$\psi(E)$satisfy the estimates $$\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta \phi(E)\Big|\le C_{\#}^\beta\lambda S^{-\beta}$$ $$\Big|\Bigl(\frac{d}{dE}\Bigr)^\beta\psi(E)\Big|\le C_{\#}^\beta\lambda^{-1}S^{-\beta}$$ $$\frac{d}{dE}\phi(E)\ge c_{\#}\lambda S^{-1}$$ \noindent for$V(x_0)C_{\#}\lambda^{-2+4\varepsilon}$, then$k_{\text{max}}=\overline n$. In any case,$|k_{\text{max}}-\overline n|\le 1$. \item"{(b)}" If$0\le kcB(x_{\text{left}})$and$\text{dist}\,(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$. \item"(Z3)" We have$V(x_0)<-cS(x_0)$,$V^\prime(x_0)=0$; and for$|x-x_0|\le c_1B(x_0)$we have$V^{\prime\prime}(x)\ge cS(x_0)B^{-2} (x_0)$. \item"(Z4)" For$x_{\text{left}}\le x\le x_0-c_1B(x_0)$we have$-V^\prime(x)>cS(x)B^{-1}(x)$; and for$x_0+c_1B(x_0)\le x\le x_{\text{rt}}$we have$+V^\prime(x)>cS(x)B^{-1}(x)$. \endroster \noindent Define$\lambda(x)=S^{1/2}(x)B(x)$for$x\in I$, and set $$\Lambda=\Bigl(\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx} {\lambda(x)B(x)}\Bigr)^{-1}\ .$$ \subhead Assumptions Concerning$V(x)$on all of$I_{\text{BVP}}$\endsubhead \smallskip \roster \item"(Z5)" We have$V(x)>0$for all$x\in I_{\text{BVP}}\backslash [x_{\text{left}},x_{\text{rt}}]$. \item"(Z6)" For all$x\in I_{\text{BVP}}$with$xx_{\text{rt}}+\Lambda^KB(x_{\text{rt}})$, we have$V(x)\ge \frac{1000} {|x-x_{\text{rt}}|^2}$. \endroster \smallskip \subhead Polynomial Growth Assumptions on$S(x)$,$B(x)$,$I$\endsubhead \smallskip \roster \item"(Z7)" We have$\max_{x\in I}B(x)<\Lambda^K\min_{x\in I}B(x)$;$\max_{x\in I}S(x)<\Lambda^K\min_{x\in I}S(x)$; and$|I|<\Lambda^K\cdot \min_{x\in I}B(x)$. \endroster \smallskip \subhead Smallness of the Constant$\hat c$\endsubhead \smallskip \roster \item"(Z8)" The constant$\hat c$is bounded above by a certain small, positive number determined by$\varepsilon$,$K$,$N$,$c$,$C$,$c_1$,$C_\alpha$.\endroster \smallskip \subhead The WKB Hypothesis\endsubhead \smallskip \roster \item"(Z9)"$\Lambda$is bounded below by a certain large, positive number determined by$\varepsilon$,$K$,$N$,$c$,$C$,$c_1$,$\hat c$,$C_\alpha$. \endroster Let$E_k$and$u_k(x)$be the eigenvalues and (normalized) eigenfunctions of$-\frac{d^2}{dx^2}+V(x)$on$I_{\text{BVP}}$, with Dirichlet or Neumann boundary conditions. In [FS3] we studied the density$\rho(x)=\sum\limits _{E_k\le 0}|u_k(x)|^2$under the assumptions (Z0)$\ldots$(Z9). Our result was called the WKB Density Theorem." This paper will study$\text{sneg}(H)=\sum\limits_{E_k\le 0}E_k$under the same assumptions. \vglue 1pc \demo{Remark}\enddemo We have kept hypothesis (Z8) from [FS3], even though the constant$\hat c$plays no role in this paper, simply to keep the same hypotheses as before. \bigskip \subhead D. The WKB Eigenvalue Theorems for Potentials with Weak Turning Points\endsubhead \noindent\demo{Set--up} \enddemo 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. \noindent\demo{Hypotheses}\enddemo \roster \item"{(H$\hat 0$)}"$I_{\text{center}}$is non--empty, and for$x,y \in I_{\text{center}}$with$|x-y|cB(x)$. \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$cS(x)0$defined on$I$. Set$\lambda(x)=S^{1/2}(x)B(x)$. We are given an energy$E_0$. We make the following assumptions. \demo{Hypotheses}\enddemo \roster \item"{(H$\overline 0$)}" For$x,y \in I$with$|x-y|cB(x)$. \item"{(H$\overline 1$)}" For$x \in I$we have$|(\frac{d}{dx})^\alpha V(x)|\le C_\alpha S(x)B^{-\alpha}(x)$. \item"{(H$\overline 2$)}"$\{x \in I_{\text{BVP}}\mid V(x)cB(x_{\text{left}})$,$\text{dist}(x_{\text{right}},\partial I)>cB(x_{\text{right}})$. \item"{(H$\overline 3$)}" In$[x_{\text{left}},x_{\text{left}}+c_1 B(x_{\text{left}})]$we have$-V^\prime(x)\ge c S(x_{\text{left}})/ B(x_{\text{left}})$, and in$[x_{\text{right}}-c_1B(x_{\text{right}}), x_{\text{right}}]$we have$+V^\prime(x)\ge cS(x_{\text{right}})/B (x_{\text{right}})$. \item"{(H$\overline 4$)}" In$[x_{\text{left}}+c_1B(x_{\text{left}}), x_{\text{right}}-c_1B(x_{\text{right}})]$we have$cS(x)cS(x)B^{-1} (x)$. \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 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 E1} 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 \proclaim{Lemma E2} 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$\hat C_{\#}$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 \medskip \noindent {\bf F.} From the section The Density for Degenerate One-Dimensional Potentials IV" in [FS3], we recall the following results. 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_0cS(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(\frac{x_\ast}{10}),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 1{20} x_\ast}\bigl(E-V(x)\bigr)^{-1/2}dx. \endroster \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 F1} 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 \proclaim{Lemma F2} Suppose \tilde E=V(\tilde x) with \tilde x\in [\frac{1}{10}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_{\#} 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 \smallskip \noindent {\bf G.} In the section The Density for Degenerate One-Dimensional Potentials II" in [FS3], we studied the following class of degenerate potentials. \noindent{\it Set-Up\/}. We are given a potential V(x) defined on a (possibly unbounded) interval I_{\text{BVP}}; positive functions S(x), B(x), defined on a subinterval I\subset I_{\text{BVP}}; a point x_{\text{crit}}\in I_{\text{BVP}}; an energy E_{\text{crit}}\le 0; and a number \delta strictly between 0 and 1. \subhead Assumptions\endsubhead \smallskip \roster \item"(Z\overline 0)" For x,y \in I with |x-y|cB(x). \item"(Z\overline 1)" For x\in I and \alpha\ge 0 we have \big|\bigl(\frac{d}{dx}\bigr)^\alpha V(x)\big|\le C_\alpha S(x)B^{-\alpha}(x). \item"(Z\overline 2)" For E_{\text{crit}}\le E\le 0, the set \{x\in I_{\text{BVP}}\mid V(x)\le E\} is a non-empty interval (x_{\text{left}}(E), x_{\text{rt}}(E)) contained in I, with \text{dist}(x_{\text{left}}(E),\partial I)>cB(x_{\text{left}}(E)) and \text{dist}\,(x_{\text{rt}}(E),\partial I)>cB(x_{\text{rt}} (E)). \item"(Z\overline 3)" For E_{\text{crit}}\le E\le 0, we have -V^\prime(x)\ge cS(x)B^{-1}(x) for \hfill\break x\in [x_{\text{left}}(E), x_{\text{left}}(E)+c_1B(x_{\text{left}}(E))] and +V^\prime(x)\ge cS(x)B^{-1}(x) for x\in [x_{\text{rt}}(E)- c_1B(x_{\text{rt}}(E)), x_{\text{rt}}(E)]. \item"(Z\overline 4)" For E_{\text{crit}}\le E\le 0, we have cS(x)From [FS3] we recall the following result. \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|\frac{1}{20}. Lemma 1(B) from the section The Density for a One-Dimensional Potential" in [FS3] shows that V(x)-\tilde E_m, S_m, B_m, I_m satisfy hypotheses (Y0)\ldots (Y11) for a suitable \varphi_m(E). (For hypotheses (Y0)\ldots(Y11), see [FS3]). Therefore, again applying Lemma 1 from the section Combining the Microlocalized Results" in [FS3], we get the conclusion of Lemma 2.\quad\blacksquare \vglue 1pc \proclaim{Lemma 3} For a suitable \tilde E with \tilde E-V(x_0)\sim \lambda^{-2\varepsilon}(x_0)S(x_0), and with\break \min_{k^\prime\in \Bbb Z}|\phi(\tilde E)-\pi(k^\prime+1/2)|\ge \frac{1}{20}, the hypotheses (H0^\ast)\ldots(H6^\ast) of the WKB Theorem on Low Eigenvalues are satisfied, with V(x)-\tilde E in place of V(x); with S(x_0) in place of S; with B(x_0) in place of B; and with 100K in place of K. The constants called c, c^\prime, c^{\prime\prime}, C_\alpha in (H0^\ast)\ldots(H6^\ast) all have the form C_{\#}. The number called \lambda in (H0^\ast)\ldots(H6^\ast) is equal to \lambda(x_0).\endproclaim \vglue 1pc \demo{Proof}\enddemo For E-V(x_0)\sim\lambda^{-2\varepsilon}(x_0)S(x_0) we have \frac{d\phi(E)}{dE}\sim \lambda(x_0)S^{-1}(x_0). Hence we can pick an \tilde E with \tilde E-V(x_0)\sim \lambda^{-2\varepsilon} (x_0)S(x_0) and \min_{k^\prime\in \Bbb Z}|\phi(\tilde E)-\pi(k^\prime+1/2)|\ge \frac{1}{20}. With S=S(x_0), B=B(x_0) and with V(x)-\tilde E in place of V(x), we check (H0^\ast)\ldots(H6^\ast). (H0^\ast) is trivial from x_0\in [x_{\text{left}}(0),x_{\text{rt}}(0)]\subset I, \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)). (H1^\ast) is trivial from (Z1); (H2^\ast), (H3^\ast) are trivial from (Z3). To prove (H4^\ast), we argue as follows. Since \tilde E-V(x_0)\sim \lambda^{-2\varepsilon}(x_0)S(x_0), we have V(x)=\tilde E at x=x_{\text{left}}(\tilde E) and at x=x_{\text{rt}}(\tilde E), with x_{\text{left}}(\tilde E)\tilde E for x\in I_{\text{BVP}} with x\tilde E for x\in I_{\text{BVP}} with x>x_{\text{rt}} (\tilde E). Hence, V(x)-\tilde E>0 for x\in I_{\text{BVP}}\backslash \{|x-x_0|E_+ outside [x_{\text{left}}(E_+),x_{\text{rt}}(E_+)]. For |x-x_0|>\frac 12 \lambda^K(x_0)B(x_0) we have x\not\in [x_{\text{left}}(E_+),x_{\text{rt}} (E_+)], so V(x)-\tilde E\ge E_+-\tilde E\ge \lambda^{-2\varepsilon}(x_0) S(x_0)\ge \frac{1000}{|x-x_0|^2} as needed. To see the last inequality, we write \frac{1000}{|x-x_0|^2}\le \frac{4000}{\lambda^{2K}(x_0)B^2(x_0)}=\frac{4000 S(x_0)}{\lambda^{2K+2}(x_0)} \le \lambda^{-2\varepsilon}(x_0)S(x_0). The proof of (H5^\ast) is complete. Finally, (H6^\ast) is immediate from \lambda(x_0)\ge c_{\#}\Lambda and the WKB hypothesis (Z9). The proof of Lemma 3 is complete. \qquad\blacksquare \medskip We want to use the above Lemmas, together with the WKB Eigenvalue Theorem and the WKB Theorem on Low Eigenvalues, to give a precise description of the negative eigenvalues of -\frac{d^2}{dx^2}+V(x). To carry this out, and also to be able to apply the lemma on Riemann sums, we need to understand how the phases \phi(E), \psi(E) vary with E. For V(x_0)< E\le 0, define$$ S_{\min}(E)=\inf\{S(x)\mid x\in I, V(x)0$, define $S_{\min}(E)=S_{\min}(0)$ and $\Gamma(E) =\Gamma(0)$. Note that $$\multline S_{\min}(E)\sim S(x_0)\ \text{and}\ \Gamma(E)\sim \lambda(x_0)\\ \text{when}\ V(x_0)0. For such E, we have S_{\min}(E)\sim S(x_\ast) for some x_\ast with [x_\ast-c_{\#}B(x_\ast), x_\ast+c_{\#} B(x_\ast)]\subset \{x\in I\mid V(x)From (5)\ldots(11) follow these quantitative results:$$\align &\text{Suppose}\ E_1\ge E_0+c_{\#}S_{\min}(E_0)\ ,\text{with}\ E_0,E_1\in \bigl(V(x_0),c_{\#}S_{\min}(0)\bigr]\ .\\ &\text{Then}\ \Phi(E_1)\ge \Phi(E_0)+c_{\#}^\prime\Gamma(E_0)\ . \tag"(13)"\endalign \align &\text{Suppose}\ E_1\le E_0-c_{\#}S_{\min}(E_0),\ \text{with}\ E_0, E_1\in \bigl(V(x_0),c_{\#}S_{\min}(0)\bigr]\ .\\ &\text{Then}\ \Phi(E_1)\le \Phi(E_0)-c_{\#}^\prime\Gamma(E_0)\ . \tag"(14)"\endalign $$\noindent From (13) and (14) follows at once$$\align &\text{Suppose}\ E_0, E_1\in \bigl(V(x_0), c_{\#}S_{\min}(0)\bigr]\ \text{and}\ |\Phi(E_0)-\Phi(E_1)|k_{hi}-1\ge k$. Hence there is some$E_0\in \bigl(V(x_0),0\bigr)$for which$\Phi(E_0)=k$. We distinguish three cases: (A)$E_0-V(x_0)\ge c_{\#}S(x_0)$(B)$c_{\#}\lambda^{-2\varepsilon}S(x_0)\le E_0-V(x_0)\le c_{\#} S(x_0)$(C)$E_0-V(x_0)\le c_{\#}\lambda^{-2\varepsilon}(x_0)S(x_0)$. First suppose we are in Case A. Lemma 1 shows that the WKB Eigenvalue Theorem applies. Note that our present phase function$\Phi(E)$is slightly different from the phase function called$\Phi(E)$in the WKB Eigenvalue Theorem, which we now call$\Phi_{\text{WKB}} (E)$. In fact,$\Phi_{\text{WKB}}(E)$is defined for$E\in [E_0-c_{\#}S_{\min}(E_0)$,$E_0+c_{\#}S_{\min}(E_0)]\cap (-\infty,0]$, and satisfies there$|\Phi_{\text{WKB}}(E)-\pi\Phi(E)|\le C_{\#}\Lambda^{-2}$. This is a part of the conclusion of the WKB Eigenvalue Theorem. Let$E_+=\min\bigl(0,E_0+c_{\#}S_{\min}(E_0)\bigr)$,$E_-=E_0-c_{\#}S_{\min} (E_0)$. These are the endpoints of the interval on which$\Phi_{\text{WKB}}$is defined. We shall check that$k$lies in the image of$\frac 1\pi\Phi_{\text{WKB}}$, by examining$\Phi_{\text{WKB}}(E_+)$and$\Phi_{\text{WKB}}(E_-)$. Let's start with$\Phi_{\text{WKB}}(E_+)$. Either$E_+=0$or$E_+=E_0+c_{\#} S_{\min}(E_0)$. If$E_+=0$, then $$\split \frac 1\pi\Phi_{\text{WKB}}(E_+)\ge \Phi(0)-C_{\#}\Lambda^{-2}\ge [k_{hi}-\overline C_{\#}\Lambda^{4\varepsilon-2}]-C_{\#}\Lambda^{-2}>k_{hi}-1\ge k\ ,\\ \text{so}\ \frac 1\pi\Phi_{\text{WKB}}(E_+)>k\ .\endsplit\tag"(16\text{bis})"$$ \noindent If$E_+=E_0+c_{\#}S_{\min}(E_0)$, then (5), (6), (7) give$\frac 1\pi\Phi_{\text{WKB}}(E_+)\ge \Phi(E_+)-C_{\#}\Lambda^{-2}\ge \Phi(E_0) +c_{\#}^\prime\Gamma(E_0)-C_{\#}\Lambda^{-2}>\Phi(E_0)=k$, so again$\frac 1\pi\Phi_{\text{WKB}}(E_+)>k$. Hence in all cases$\frac 1\pi\Phi_{\text{WKB}}(E_+)>k$. On the other hand,$\frac 1\pi\Phi_{\text{WKB}}(E_-)\le \Phi(E_-)+C_{\#}\Lambda^{-2} \le \Phi(E_0)-c_{\#}^\prime\Gamma(E_0)+C_{\#}\Lambda^{-2}<\Phi(E_0) =k$, so$\frac 1\pi\Phi_{\text{WKB}}(E_-)(k_{hi}-1) +\frac 12\ge k+\frac 12\ ,\tag"(18)" $$\noindent (17) cannot hold with \tilde E=0. Hence \tilde E is an eigenvalue. Also (17) and (18) yield \Phi(\tilde E)<\Phi(0), so \tilde E<0. We have proven (17) for a negative eigenvalue \tilde E. Thus the conclusion of Lemma 4 holds in Case (A). Next suppose we are in Case (B). Define an integer m by$$ 2^{-2(m+1)}c_{\#}S(x_0)>1$by (20), and recall that$\Phi(E_0)=k$. Then (21) and (25) imply$\Phi_{\text{WKB}}(E_+)>\pi k>\Phi_{\text{WKB}}(E_-)$, so that$\pi k\in \Phi_{\text{WKB}} \bigl([E_-,E_+]\bigr)$. Thus we can find$\tilde E=\tilde E_m$or an eigenvalue, satisfying (22). From (21) and (22), we get $$|\Phi(\tilde E)-k|\le C_{\#}\bigl(2^{-2m}\lambda(x_0)\bigr)^{-2}\ .\tag"(26)"$$ \noindent In particular, since$\Phi(E_0)=k$, we see from (25), (26) that$\Phi(\tilde E)<\Phi(E_+)$, hence$\tilde Ek_{hi}$. Hence there is some$E_0\in \bigl(V(x_0),0\bigr)$for which$\Phi(E_0)=k_{hi}$. \medskip Now we can repeat the proof of Lemma 4, using$k_{hi}$in place of$k$, and making the following modifications. Instead of (16 bis), we note that$\frac 1\pi \Phi_{\text{WKB}}(E_+)\ge \Phi(0)-C_{\#}\Lambda^{-2}\ge \bigl[k_{hi} +\overline C_{\#}\Lambda^{4\varepsilon-2}\bigr]-C_{\#} \Lambda^{-2}>k_{hi}$. Instead of (18), we note that$\Phi(0)\ge k_{hi}+\overline C_{\#}\Lambda^{4\varepsilon-2}$. \smallskip With these minor changes the proof of Lemma 4 goes through.$\qquad\blacksquare$\vglue 1pc \proclaim{Lemma 5} If$E_0\le 0$is an eigenvalue of$-\frac{d^2}{dx^2}+V(x)$, then$|\Phi(E_0)-k|\le \overline C_{\#}\Lambda^{4\varepsilon-2}$for an integer$k$\$(0\le k\le k_{hi})$.\endproclaim \vglue 1pc \demo{Proof}\enddemo We distinguish three cases: (A)$E_0-V(x_0)\ge c_{\#}S(x_0)$(B)$c_{\#}\lambda^{-2\varepsilon}(x_0)S(x_0)\le E_0-V(x_0)\le c_{\#}S(x_0)$(C)$E_0-V(x_0)\le c_{\#}\lambda^{-2\varepsilon}(x_0)S(x_0)$. \medskip First suppose we are in Case (A). Lemma 1 and the WKB Eigenvalue Theorem show that$|\Phi(E)-(\text{integer})|\le C_{\#}\Lambda^{-2}$for any eigenvalue$E\in [E_0-c_{\#}^\prime S_{\min} (E_0),E_0+c_{\#}^\prime S_{\min}(E_0)]\cap (-\infty,0]$. In particular,$|\Phi(E_0)-(\text{integer}) |\le C_{\#}\Lambda^{-2}$. Next, suppose we are in Case (B). Define an integer$m$by$2^{-2(m+1)}c_{\#}S(x_0)E_0$is as in Lemma 2.) In particular, for$E_0$itself we have$|\Phi(E_0)-(\text{integer})|\le C_{\#}\bigl(2^{-2m} \lambda(x_0)\bigr)^{-2}\le C_{\#}\bigl(\lambda^{1-2\varepsilon}(x_0)\bigr) ^{-2}\le C_{\#}\Lambda^{4\varepsilon-2}$. Next, suppose we are in Case (C). We apply Lemma 3 and the WKB Theorem for Low Eigenvalues. This shows that$|\Phi(E)-k|\le C_{\#}\bigl(\lambda (x_0)\bigr)^{4\varepsilon-2}$for any eigenvalue$E$satisfying $$E\le \min\bigl\{\tilde E,V(x_0)+c_{\#}\lambda^{-2\varepsilon}(x_0) S(x_0)\bigr\}\ .\tag"(29)"$$ \noindent (Here$\tilde E>E_0$is as in Lemma 3.) In particular,$E=E_0$satisfies (29), provided we pick$c_{\#}$small enough in the statement of Case (C). Hence,$|\Phi(E_0)-(\text{integer})|\le C_{\#} \bigl(\lambda(x_0)\bigr)^{4\varepsilon-2}\le C_{\#}\Lambda^{4\varepsilon-2}$. We have proven in all cases that $$|\Phi(E_0)-k|\le C_{\#}\Lambda^{4\varepsilon-2}\quad\text{for\ an\ integer}\ k\ .\tag"(30)"$$ \noindent As$E$varies from$V(x_0)$to$0$,$\Phi(E_0)$varies from$-\frac 12+O\bigl(\lambda^{-1}(x_0)\bigr)$to$\Phi(0)$. Hence (30) implies $$-\frac 12-C_{\#}\lambda^{-1}(x_0)-C_{\#}\Lambda^{4\varepsilon-2} \le k\le \Phi(0)+C_{\#}\Lambda^{4\varepsilon-2}\ .\tag"(31)"$$ \noindent Since$k$is an integer, (31) implies $$0\le k\le \Phi(0)+\overline C_{\#}\Lambda^{4\varepsilon-2}\ ,\tag"(32)"$$ \noindent provided we take$\overline C_{\#}$larger then the$C_{\#}$in (31)\ . \noindent Then from (32) and the definition of$k_{hi}$, we see that$0\le k\le k_{hi}$. Thus,$|\Phi(E_0)-k|\le \overline C_{\#} \Lambda^{4\varepsilon-2}$and$0\le k\le k_{hi}$, provided we also take$\overline C_{\#}$larger than the$C_{\#}$in (30). The proof of Lemma 5 is complete.$\qquad\blacksquare$\vglue 1pc \proclaim{Lemma 6} Suppose$\overline E_0$,$\overline E_1\le 0$are eigenvalues of$-\frac{d^2}{dx^2}+V(x)$. If$|\Phi(\overline E_0) -\Phi(\overline E_1)|\le C_{\#}\Lambda^{4\varepsilon-2}$then$\overline E_0=\overline E_1$.\endproclaim \vglue 1pc \demo{Proof}\enddemo Suppose not. We may suppose$\overline E_1<\overline E_0\le 0$. Then (16) implies $$\overline E_0-C_{\#}S_{\min}(\overline E_0)\Gamma^{-1} (\overline E_0)\Lambda^{4\varepsilon-2}\le \overline E_1<\overline E_0\le 0 \ .\tag"(33)"$$ \noindent We distinguish three cases: (A)$\overline E_0-V(x_0)\ge c_{\#}S(x_0)$(B)$c_{\#}\lambda^{-2\varepsilon}S(x_0)\le \overline E_0-V(x_0) \le c_{\#}S(x_0)$(C)$\overline E_0-V(x)\le c_{\#}\lambda^{-2\varepsilon}(x_0)S(x_0)$. \smallskip Suppose first we are in Case (A). Take$E_0=\overline E_0$, and apply Lemma 1 and the WKB Eigenvalue Theorem. Thus, the eigenvalues in$\Cal E= \bigl[\overline E_0-c_{\#}^\prime S_ {\min}(\overline E_0), \overline E_0+ c_{\#}^\prime S_{\min}(\overline E_0)\bigr]\cap (-\infty,0]$are among the$E_k$described in the WKB Eigenvalue Theorem, and distinct$E_k$have$|\Phi(E_k)-\Phi(E_{k^\prime})| \ge 1-C_{\#}\Lambda^{-2}$. Since (33) shows that$\overline E_0$,$\overline E_1\in \Cal E$, it follows that$|\Phi(\overline E_0)-\Phi(\overline E_1)|\ge 1-C_{\#}\Lambda^{-2}$, contradicting the hypothesis of the Lemma. Next suppose we are in Case (B). Define an integer$m$by$2^{-2(m+1)} c_{\#}S(x_0)<\overline E_0-V(x_0)\le 2^{-2m}c_{\#}S(x_0)$. Since we are in Case (B), we have $$c_{\#}\lambda^{-2\varepsilon}(x_0)<2^{-2m}\le 1\ .\tag"(34)"$$ \noindent Take$E_0=\overline E_0$, and apply Lemma 2 and the WKB Eigenvalue theorem. Thus, the eigenvalues in$\Cal E\equiv \bigl[\overline E_0- c_{\#}^\prime 2^{-2m}S(x_0),\overline E_0+c_{\#}^\prime2^{-2m}S(x_0)\bigr]\cap (-\infty,\tilde E_m]$are among the$E_k$described in the WKB Theorem, so that distinct$E_k$have$|\Phi(E_k)-\Phi(E_{k^\prime})|\ge 1-C_{\#}\cdot \bigl(2^{-2m}\lambda(x_0)\bigr)^{-2}$. (Here$\tilde E_m>\overline E_0$is as in Lemma 2). In particular, (33) and (3) show that$\overline E_0-C_{\#}S(x_0)\lambda^{-1}(x_0)\Lambda^{4\varepsilon-2} \le \overline E_1<\overline E_0<\tilde E_m\le 0$, which implies$\overline E_0,\overline E_1\in \Cal E$by virtue of (34). Therefore,$|\Phi(\overline E_0)-\Phi(\overline E_1)|\ge 1-C_{\#}\bigl(2^{-2m}\lambda (x_0)\bigr)^{-2}\ge 1-C_{\#}^\prime\lambda^{4\varepsilon-2}(x_0)>\frac 12$, contradicting the hypothesis of the Lemma. Finally, suppose we are in Case (C). We apply Lemma 3 and the WKB Theorem on Low Eigenvalues. Thus, all the eigenvalues in$\Cal E=\bigl[V(x_0),V(x_0)+c_{\#}\lambda^{-2\varepsilon}(x_0) S(x_0)\bigr]\cap (-\infty,\tilde E]$are among the$E_k$described in the WKB Theorem on Low Eigenvalues, and distinct$E_k$satisfy$|\Phi(E_k)-\Phi(E_{k^\prime})|\ge 1-C_{\#}\lambda^{4\varepsilon-2} (x_0)$. Here$\tilde E\le 0$is as in Lemma 3, so that$\tilde E-V(x_0)\sim \lambda^{-2\varepsilon}(x_0)S(x_0)$. Since we are in Case (C), we have$\overline E_0<\tilde E$and$\overline E_0< V(x_0)+c_{\#}\lambda^{-2\varepsilon}(x_0)S(x_0)$, provided we take the$c_{\#}$in the statement of Case (C) small enough. Thus$\overline E_0\in \Cal E$. Since$V(x_0)<\overline E_1<\overline E_0$we have also$\overline E_1\in \Cal E$. Hence$|\Phi(\overline E_0) -\Phi(\overline E_1)|\ge 1-C_{\#}\lambda^{4\varepsilon-2}(x_0)>\frac 12$, contradicting the hypothesis of the Lemma. In all three cases (A), (B), (C), we have reached a contradiction, starting with the assumption$\overline E_1<\overline E_0$. The proof of Lemma 6 is complete.$\qquad\blacksquare$\vglue 1pc \proclaim{Lemma 7} The non-positive eigenvalues of$-\frac{d^2}{dx^2}+V(x)$are$E_0,E_1,\ldots,E_{k_{\max}}$, with $$|\Phi(E_k)-k|\le \overline C_{\#}\Lambda^{4\varepsilon-2}\ ,\tag"(35)"$$ $$k_{\max}=\ {\roman{greatest\ integer}}\ \le \Phi(0)+w_{hi}\ ,\tag"(36)"$$ \noindent and $$|w_{hi}|\le \overline C_{\#}\Lambda^{4\varepsilon-2}\ . \tag"(37)"$$ \endproclaim \vglue 1pc \demo{Proof}\enddemo This follows from Lemma 4 and its corollary, and from Lemmas 5 and 6. We take$k_{\max}=k_{hi}$or$k_{hi}-1$, depending on whether there is a non-positive eigenvalue$E$satisfying$|\Phi(E)-k_{hi}| \le \overline C_{\#}\Lambda^{4\varepsilon-2}$. For$0\le k\le k_{\max}$, we can find exactly one non-positive eigenvalue$E_k$satisfying (35). (That follows from Lemmas 4 and 6.) Moreover, every non-positive eigenvalue is one of the$E_k$, by Lemma 5. If$k_{\max}=k_{hi}$, then we take$w_{hi}=\overline C_{\#}\Lambda^{4\varepsilon-2}$, so that (36) and (37) hold by definition of$k_{hi}$and$w_{hi}$. If instead$k_{\max}=k_{hi}-1, then $$\Phi(0)From (43) and (46) we have therefore$$ |E^\prime(b)|\le C_{\#}\Lambda^{-1}S_{\min}(E_\ast)\ ,\ \text{and} \tag"(63)"  |E^\prime(b)-E^\prime(b_0)|\le C_{\#}\Lambda^{4\varepsilon-4}S_{\min} (E_\ast)\ .\tag"(64)" $$\noindent Applying (62), (63), (64), we get$$\multline \Big|\frac 12\ E^\prime(b)\tilde\chi(b)-\frac 12\ E^\prime(b_0)\tilde\chi \bigl(\frac 1\pi\phi(0)-\frac 12\bigr)\Big|\le\\ \frac 12\ \big|E^\prime(b)\big|\ \big|\tilde\chi(b)-\tilde\chi\bigl(\frac 1\pi \phi(0)-\frac 12\bigr)\big|+\frac 12|E^\prime(b)-E^\prime(b_0)\big|\ \big |\tilde\chi \bigl(\frac 1\pi \phi(0)-\frac 12)\big|\\ \le C_{\#}\Lambda^{-1}S_{\min}(E_\ast)\cdot\Lambda^{-1}+C_{\#} \Lambda^{4\varepsilon-4}S_{\min}(E_\ast)\le C_{\#}^\prime\Lambda^{-2} S_{\min}(E_\ast)\ .\endmultline\tag"(65)" $$\noindent By definition of E(t), we have also E^\prime(t)=\Bigl(\frac{d} {dE}\Phi(E)\Bigm|_{E=E(t)}\Bigr)^{-1}. Since E(b_0)=0, this means$$\multline E^\prime(b_0)=\Bigl(\frac{d}{dE}\Phi(E)\Bigm|_{E=0}\Bigr)^{-1}=\Bigl( \frac 1\pi\phi^\prime(0)+\frac{1}{48\pi}\psi^\prime(0)\Bigr)^{-1}\\ =\pi(\phi^\prime(0)\bigr)^{-1}\Bigl(1+\frac{1}{48} \frac{\psi^\prime(0)}{\phi^\prime(0)}\Bigr)^{-1}\ .\endmultline\tag"(66)" $$\noindent From (6), (7), we get \big|\frac{\psi^\prime(0)} {\phi^\prime(0)}\big|\le C_{\#}\Lambda^{-2}, so \Big|\bigl(1+\frac{1}{48}\ \frac{\psi^\prime(0)}{\phi^\prime(0)}\bigr)^{-1}-1\Big|\le C_{\#}\Lambda^{-2}. Hence, (66) implies$$ |E^\prime(b_0)-\pi(\phi^\prime(0))^{-1}|\le C_{\#}\Lambda^{-2} |E^\prime(b_0)|\le C_{\#}\Lambda^{-2}\sigma(b)\tau^{-1}(b)\ ,\tag"(67)" $$\noindent by (40), (42), (43) and the fact that |b-b_0|\le C_{\#}\Lambda^{4\varepsilon-2}. Applying (43) and (46) to (67), we get |E^\prime(b_0)-\pi(\phi^\prime(0))^{-1}|\le C_{\#}\Lambda^{-3} S_{\min}(E_\ast). Combining this with (65), we conclude that$$\multline +\frac 12\ \frac{d}{dt}E(t)\Bigm|_{t=b}\cdot\tilde\chi(b)=+\frac \pi 2 \bigl(\phi^\prime(0)\bigr)^{-1}\tilde\chi\bigl(\frac 1\pi\phi(0)-\frac 12\bigr) +\ \text{Error}_4\ ,\\ \text{with}\ |\text{Error}_4|\le C_{\#}\Lambda^{-2}S_{\min}(E_\ast)\ . \endmultline\tag"(68)" $$Next, we examine the term -\frac 12E^\prime(a)\tilde\chi(a). Since E(a)=E_\ast, we have (as in (66)) that$$\align E^\prime(a)=\Bigl(\frac{d\Phi}{dE}(E)\Bigm|_{E=E_\ast}\Bigr)^{-1}&= \Bigl(\frac 1\pi \phi^\prime(E_\ast)+\frac{1}{48\pi}\psi^\prime(E_\ast)\Bigr)^{-1} \\ &=\pi\bigl(\phi^\prime(E_\ast)\bigr)^{-1}\cdot\Bigl(1+\frac{\psi^\prime(E_\ast)} {48\phi^\prime(E_\ast)}\Bigr)^{-1}\ .\tag"(69)"\endalign $$\noindent Again, (6), (7) imply \big|\frac{\psi^\prime(E_\ast)} {\phi^\prime(E_\ast)}\big|\le C_{\#}\Lambda^{-2}, so that (69) yields$$ |E^\prime(a)-\pi(\phi^\prime(E_\ast))^{-1}\big|\le C_{\#} \Lambda^{-2}|E^\prime(a)|\le C_{\#}^\prime\Lambda^{-3} S_{\min}(E_\ast)\ , \tag"(70)" $$\noindent since$$ |E^\prime(a)|\le C_{\#}\sigma(a)\tau^{-1}(a)\le C_{\#}\Lambda^{-1} S_{\min}(E_\ast)\quad \text{by\ (40),\ (43),\ (46)}\ .\tag"(71)" $$\noindent We turn to \tilde\chi(a)=\tilde\chi\bigl(\frac 1\pi\phi(E_\ast)+ \frac{1}{48\pi}\psi(E_\ast)-\frac 12\bigr), using again our assumption$$ \frac 1\pi\phi(E_\ast)=m+\xi\ ,\ \text{with}\ m\ \text{an\ integer\ and}\ |\xi|\le \Lambda^{-2}\ . $$\vfill\eject %%trun2.tex % \noindent We have therefore (by definition of \tilde\chi)$$\multline \tilde\chi(a)=\tilde\chi\bigl(m-\frac12+\frac 1{48\pi} \psi(E_\ast)+\xi\bigr)=\operatornamewithlimits{\text{inf}}_ {k\in \Bbb Z}\ \big|\bigl(m-\frac 12+\frac{1}{48\pi}\psi(E_\ast) +\xi\bigr)-(k+\frac 12)\big|^2-\frac 1{12}\\ =\bigl(\frac{1}{48\pi}\psi(E_\ast)+\xi\bigr)^2-\frac{1}{12}\ ,\ \text{with}\ |\psi(E_\ast)|\le C_{\#}\Lambda^{-1}\ .\endmultline $$\noindent Hence$$ |\tilde\chi(a)+\frac 1{12}|\le C_{\#}\Lambda^{-2}\ .\tag"(72)" $$\noindent Applying (70), (71), (72), we get$$\multline \Big|-\frac 12 E^\prime(a)\tilde\chi(a)-\bigl(-\frac 12\bigr) \bigl(\pi(\phi^\prime(E_\ast))^{-1}\bigr)\cdot\bigl(-\frac 1{12}\bigr)\Big|\\ \le \frac 12 \big|E^\prime(a)-\pi(\phi^\prime(E_\ast))^{-1}\big|\cdot \big|\tilde\chi(a)|+\frac 12\ |\pi(\phi^\prime(E_\ast))^{-1}|\cdot|\tilde\chi (a)+\frac 1{12}|\\ \le C_{\#}\Lambda^{-3}S_{\min}(E_\ast)+C_{\#}\Lambda^{-1}S_{\min}(E_\ast) \cdot\Lambda^{-2}\le C_{\#}\Lambda^{-3}S_{\min}(E_\ast)\ .\endmultline $$\noindent That is,$$\split -\frac 12\ \frac{d}{dt}E(t)\Bigm|_{t=a}\cdot\tilde\chi(a)=+\frac\pi{24} \bigl(\phi^\prime(E_\ast)\bigr)^{-1}+\ \text{Error}_5\\ \text{with}\ |\text{Error}_5|\le C_{\#}\Lambda^{-3}S_{\min}(E_\ast)\ . \endsplit\tag"(73)" $$We can now substitute (59), (60), (61), (68), (73) into (57), to obtain the following result.$$\split \sum\limits_{k\in [a,b]}E_k=\int_a^{b_0}E(t)dt+\frac{E_\ast}{48\pi} \psi(E_\ast)+\frac \pi 2\bigl(\phi^\prime(0)\bigr)^{-1}\tilde\chi \bigl(\frac 1\pi\phi(0)-\frac 12 \bigr)+\frac \pi{24}\bigl(\phi^\prime (E_\ast)\bigr)^{-1}\\ +\ \text{Error}_6\ ,\quad \text{with}\endsplit\tag"(74)"  |\text{Error}_6|\le C_{\#}\Lambda^{4\varepsilon-2}S_{\min}(E_\ast)\ . \tag"(75)" $$Next we calculate the integral in (74). Changing variable from t to E=E(t) in that integral, we obtain$$\multline \int_a^{b_0}E(t)dt=\int_{E_\ast}^0E\frac{d\Phi}{dE}dE=\int_{E_\ast}^0 E\cdot\bigl(\frac 1\pi \phi^\prime(E)+\frac{1}{48\pi}\psi^\prime(E)\bigr)dE\\ =\bigl(\frac 1\pi\phi(E)+\frac 1{48\pi}\psi(E)\bigr)\cdot E\bigr]_{E_\ast}^0 -\int_{E_\ast}^0\bigl(\frac 1\pi\phi(E)+\frac 1{48\pi}\psi(E)\bigr)dE\\ =-\frac{E_\ast}{\pi}\phi(E_\ast)-\frac{E_\ast}{48\pi}\psi(E_\ast) -\frac 1\pi\int_{E_\ast}^0\phi(E)dE-\frac 1{48\pi} \int_{E_\ast}^0\psi(E)dE\ .\endmultline\tag"(76)" $$\noindent Thus, we have to integrate the phase functions \phi and \psi. To integrate \phi(E), we simply write$$\multline \int_{E_\ast}^0\phi(E)dE=\int\limits\Sb E_\ast\delta}V^{\prime\prime} (x)\cdot\bigl(E-V(x)\bigr)^{-3/2}dx-q(E)\delta^{-1/2}\Bigr]\ , \tag"(78)" $$\noindent exists uniformly for E\in [E_\ast,0]. Also, q(E) is continuous on [E_\ast,0].\endproclaim \vglue 1pc \demo{Proof}\enddemo It is enough to establish uniform convergence and continuity of q(E) for E\in [E_\ast,0] in a small neighborhood of a given E_0\in [E_\ast,0]. Because E_\ast is assumed to be strictly greater than the minimum of the potential, we can make a partition of unity V^{\prime\prime} (x)=G_{\text{left}}(x)+G_{\text{center}}(x)+G_{\text{rt}}(x) in a neighborhood of [x_{\text{left}}(E_0),x_{\text{rt}}(E_0)], with the following properties:$$\align &G_{\text{center}}(x)\ \text{is}\ C^\infty\ \text{and\ is\ supported\ in}\ \{E_0-V(x)>\delta_0\}\\ &\text{for\ some\ small,\ positive}\ \delta_0\ . \tag"(79)"\endalign \align &G_{\text{left}}(x)\ \text{is}\ C^\infty\ \text{and\ is\ supported\ in\ a\ small\ neighborhood\ of}\\ &x=x_{\text{left}}(E_0)\ .\tag"(80)"\endalign  G_{\text{rt}}(x)\ \text{is}\ C^\infty\ \text{and\ is\ supported\ in\ a\ small\ neighborhood\ of}\ x=x_{\text{rt}}(E_0)\ .\tag"(81)" $$\noindent We write \psi(E) in (78) as a sum of \psi_{\text{left}}(E), \psi_{\text{center}}(E), \psi_{\text{rt}}(E) by replacing V^{\prime\prime}(x) respectively by G_{\text{left}}(x), G_{\text{center}}(x), G_{\text{rt}}(x). In place of q(E) we have functions q_{\text{left}}(E), q_{\text{center}}(E), q_{\text{rt}}(E). Evidently, for E near E_0, the limit$$ \psi_{\text{center}}(E)=\lim_{\delta\to 0+}\Bigl[\int_{E-V(x)>\delta} G_{\text{center}}(x)\bigl(E-V(x)\bigr)^{-3/2}dx-q_{\text{center}} (E)\delta^{-1/2}\Bigr] $$\noindent exists uniformly, and q_{\text{center}}(E)\equiv 0. The discussions for \psi_{\text{left}}(E) and \psi_{\text{rt}}(E) are analogous, so we just study$$ \psi_{\text{left}}(E)=\lim_{\delta\to 0+}\Bigl[\int_{E-V(x)>\delta} G_{\text{left}}(x)\cdot\bigl(E-V(x)\bigr)^{-3/2}dx-q_{\text{left}} (E)\delta^{-1/2}\Bigr]\ .\tag"(82)" $$\noindent For (x,E) near (x_{\text{left}}(E_0),E_0), we make the smooth change of variable (x,E)\to (\xi,E), \xi=E-V(x). (Since -V^\prime(x)>0 at x=x_{\text{left}}(E_0), this is a smooth change of variables. Here we make crucial use of our assumption E_\ast>V(x_0)=\min V.) The change of variable is well-defined for x\in \text{supp}\ G_{\text{left}} and E near to E_0, and produces |\xi|\le A for such (x,E). Here A is some positive number independent of x, E. Therefore, using \xi in place of x as the independent variable in (82), we get$$ \psi_{\text{left}}(E)=\lim_{\delta\to 0+}\Bigl[\int_\delta^A \theta(\xi,E)\cdot\xi^{-3/2}d\xi-q_{\text{left}}(E)\delta^{-1/2}\Bigr] \tag"(83)" $$\noindent for a smooth function \theta(\xi,E). Writing$$ \int_\delta^A\theta(\xi,E)\cdot\xi^{-3/2}d\xi=\int_\delta^A \bigl[\frac{\theta(\xi,E)-\theta(0,E)}{\xi}\bigr]\xi^{-1/2} d\xi+\theta(0,E)\int_\delta^A\xi^{-3/2} d\xi\ , $$\noindent we see that the limit in (83) exists uniformly, and that q_{\text{left}}(E)=2\theta(0,E) is continuous. Since the quantity in brackets in (83) is equal to that in (82) the proof is complete. \qquad\blacksquare Lemma 8 gives$$\split \int_{E_\ast}^0\psi(E)dE=\lim_{\delta\to 0+}\Bigl[\int\limits\Sb E_\ast\delta\endSb \int V^{\prime\prime}(x)\cdot\bigl(E-V(x)\bigr)^{-3/2} dxdE\\ -\int_{E_\ast}^0q(E)dE\cdot\delta^{-1/2}\Bigr]\ .\endsplit\tag"(84)" $$\noindent In particular, the limit in (84) exists. We have$$\multline \int\limits\Sb E_\ast\delta\endSb\int V^{\prime\prime}(x)\cdot \bigl(E-V(x)\bigr)^{-3/2}dxdE=\int\limits\Sb E<0\\ E-V(x)>\delta\endSb \int V^{\prime\prime}(x)\cdot\bigl(E-V(x)\bigr)^{-3/2}dxdE\\ -\int\limits\Sb E\delta\endSb\int V^{\prime\prime}(x)\cdot \bigl(E-V(x)\bigr)^{-3/2}dxdE\\ =\int_{V(x)<-\delta}V^{\prime\prime}(x)\Bigl[\int_{V(x)+\delta} ^0\bigl(E-V(x)\bigr)^{-3/2}dE\Bigr]dx\\ -\int_{V(x)2. However, contained in the hypotheses (Z0)$\ldots$(Z9) is the fact that $V(x)<-c_{\#}S(x_0)$ for |x-x_0|2\ \text{as\ needed}\ .\endsplit $$\noindent So the Lemma on Truncated Eigenvalue Sums holds for$$ E_{\ast}=V(x_0)+\tau^2\ , \tau\quad \text{small\ but\ nonzero}\ .\tag"(1)" $$\noindent Since all the eigenvalues of -\frac{d^2}{dx^2}+V(x) are greater than V(x_0)+\delta_0 for some small, fixed \delta_0, it follows that X=\ \text{sum\ of eigenvalues\ of}\ H\ \text{belonging\ to}\ [E_\ast,0]=\ \text{sneg}(H) for \tau\ne 0 small enough. Hence the Lemma on truncated eigenvalue sums gives for small non-zero \tau that$$\align \text{sneg}(H)&=-\frac{2}{3\pi}\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{3/2} \ dx+\frac{1}{24\pi}\int_{I_{\text{BVP}}}V^{\prime\prime}(x)\cdot\bigl( -V(x)\bigr)_+^{-1/2}\ dx\\ &\ \ +\frac{2}{3\pi}\int_{I_{\text{BVP}}}\bigl(E_\ast-V(x)\bigr)_+^{3/2}\ dx-\frac{1}{24\pi}\int_{I_{\text{BVP}}}V^{\prime\prime}(x)\cdot \bigl(E_\ast-V(x)\bigr)_+^{-1/2}\ dx\\ &\ \ +\frac{\pi}{24}\bigl(\phi^\prime(E_\ast)\bigr)^{-1}-\frac{E_\ast} {\pi}\phi(E_\ast)+\frac \pi 2\bigl(\phi^\prime(0)\bigr)^{-1} \tilde\chi\bigl(\frac 1\pi\phi(0)-\frac 12\bigr)+\ \text{Error}(\tau)\ ,\\ \tag"(2)"\endalign $$\noindent with$$ |\text{Error}(\tau)|\le C_{\#}\Lambda^{4\varepsilon-2}\cdot \inf\limits\Sb x\in I\\ V(x)0, so $h$ has a smooth square root $f(x)$ defined in a neighborhood of $x_0$, with $f(x_0)=\bigl(\frac 12V^{\prime\prime} (x_0)\bigr)^{1/2}$. Setting $\xi(x)=f(x)\cdot (x-x_0)$, we obtain $$V(x)=V(x_0)+\bigl(\xi(x)\bigr)^2\tag"(9)"$$ \noindent from (8), and $$\xi(x_0)=0\ , \quad \xi^\prime(x_0)=f(x_0)=\bigl(\frac 12 V^{\prime\prime}(x_0)\bigr)^{1/2}>0\tag"(10)"$$ \noindent from the definition of $\xi(x)$. \noindent Here, $\xi(x)$ is smooth in a neighborhood of $x_0$. Our change of variable is $s=\xi(x)$, which is inverted by a function $x=y(s)$, satisfying $$y(0)=x_0\ ,\quad y^\prime(0)=\bigl(\frac 12 V^{\prime\prime}(x_0)\bigr)^{-1/2} >0\ .\tag"(11)"$$ \noindent The function $y(s)$ is smooth in a neighborhood of the origin. Changing variable from $x$ to $s=\xi(x)$ gives $E_\ast-V(x)=\tau^2-s^2$ by (1), (9), so that $$\int_{I_{\text{BVP}}}V^{\prime\prime}(x)\bigl(E_\ast-V(x)\bigr)_+^{-1/2} \ dx =\int_{-\infty}^{\infty}\ V^{\prime\prime}\bigl(y(s)\bigr)\cdot (\tau^2-s^2)_+^{-1/2}y^\prime(s)\ ds\tag"(12)"$$ \noindent and $$\phi^\prime(E_\ast)=\frac 12 \int_{I_{\text{BVP}}}\bigl(E_\ast-V(x)\bigr)_+^{-1/2}\ dx =\frac 12 \int_{-\infty}^\infty (\tau^2-s^2)_+^{-1/2} y^\prime(s)\ ds \ .\tag"(13)"$$ \noindent For a general function $\theta(s)$ smooth in a neighborhood of the origin, and for small, nonzero $\tau$ we note that $$\split \int_{-\infty}^\infty(\tau^2-s^2)_+^{-1/2}\theta(s)ds=\theta(0)\cdot \Bigl[\int_{-\infty}^\infty(\tau^2-s^2)_+^{-1/2}ds\Bigr]\\ +\Bigl[\int_{-\infty}^\infty(\tau^2-s^2)_+^{-1/2}\{\theta(s)-\theta(0)\}\ ds\Bigr]\endsplit$$ \noindent The first integral on the right is identically equal to $\pi$. The second integral on the right is dominated by $C\int_{-\infty}^\infty |s|(\tau^2-s^2)_+^{-1/2}ds\le C|\tau|\int_{-\infty}^\infty(\tau^2- s^2)_+^{-1/2}ds=C\pi|\tau|$. It follows that $\lim_{\tau\to 0} \int_{-\infty}^{\infty}\theta(s)\cdot(\tau^2-s^2)_+^{-1/2}ds=\pi\theta(0)$ whenever $\theta(s)$ is smooth near the origin. Taking $\theta(s) =V^{\prime\prime}\bigl(y(s)\bigr)\cdot y^\prime(s)$ and $\theta(s)=y^\prime(s)$, we see from (12), (13) that $\lim_{\tau\to 0}\ \int_{I_{\text{BVP}}}\ V^{\prime\prime}(x)\cdot \bigl(E_\ast-V(x)\bigr)_+^{-1/2}dx=\pi V^{\prime\prime}\bigl(y(0)\bigr)y^\prime (0)$ and $\lim_{\tau\to 0}\phi^\prime(E_\ast)=\frac \pi 2y^\prime(0)$. Substituting (11), we get $$\lim_{\tau\to 0}\ \int_{I_{\text{BVP}}}\ V^{\prime\prime}(x)\cdot \bigl(E_\ast-V(x)\bigr)_+^{-1/2}\ dx=\pi V^{\prime\prime} (x_0)\cdot \bigl(\frac 12 V^{\prime\prime}(x_0)\bigr)^{-1/2}=\pi\sqrt{2} \bigl(V^{\prime\prime}(x_0)\bigr)^{+1/2}\tag"(14)"$$ \noindent and $$\lim_{\tau\to 0}\ \phi^\prime(E_\ast)=\frac \pi 2\ \bigl(\frac 12 V^{\prime\prime}(x_0)\bigr)^{-1/2}=\frac{\pi}{\sqrt 2}\bigl(V^{\prime\prime} (x_0)\bigr)^{-1/2}\ .\tag"(15)"$$ \noindent Hence, (4) is reduced to checking that $$\frac \pi{24}\Bigl[\frac{\pi}{\sqrt 2}\bigl(V^{\prime\prime} (x_0)\bigr)^{-1/2}\Bigr]^{-1}-\frac{1}{24\pi}\Bigl[\pi\sqrt 2\bigl( V^{\prime\prime}(x_0)\bigr)^{+1/2}\Bigr]=0\ ,$$ \noindent which is correct. The proof of the First WKB Eigenvalue Theorem is complete.$\qquad\blacksquare$ \vfill\eject %% loweigen.tex \head Low Eigenvalues in a Potential with a Coulomb Singularity\endhead \medskip In this section and the next, we prove a variant of the first WKB Theorem on Eigenvalue Sums, which applies to potentials that look like $$V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx\tag"(1)"$$ \noindent near $x=0$. Here, we show that if $V(x)$ is given exactly by (1) near the origin, then the low eigenvalues of $-\frac{d^2}{dx^2}+V(x)$ on $(0,\infty)$ are very close to the familiar eigenvalues of $-\frac{d^2}{dx^2}+V_c(x)$. In the next section, we combine this information with the Lemma on Truncated Eigenvalue Sums, to compare the eigenvalue sum for $-\frac{d^2}{dx^2}+V(x)$ with that for $-\frac{d^2}{dx^2}+V_c(x)$. We begin with a slight variant of the Agmon Lemma from [FS2]. \vglue 1pc \proclaim{Lemma 1} Suppose $-\frac{d^2}{dx^2}u+Vu=Eu$ on $(0,\infty)$, with $\Vert u\Vert_{L^2(0,\infty)}=1$. Suppose also that $V$ is smooth on $(0,\infty)$, and that $$V(x)-E\ge \frac 12 |E_\ast|\quad\text{for}\ x>x_\ast\ .$$ \noindent Then $$\Vert u\Vert_{L^2(2x_\ast,\infty)}^2\le C_N\bigl(x_\ast^2|E_\ast|)^{-N} \bigl\{1+x_\ast^2\operatornamewithlimits{\sup}_{\frac 18 x_\ast0 we have Av_{[y/2,y]}|u|^2=\frac 2y \Vert u\Vert^2_{L^2(y/2,y)}\le \frac 2y. Putting y/4 for y gives also Av_{[y/8,y/4]}|u|^2\le \frac 8y. Hence we can find x_1\in (\frac y8,\frac y4) and x_2\in (\frac y2,y) with |u(x_1)|^2\le \frac 8y, |u(x_2)|^2\le \frac 2y. The mean-value theorem produces an x\in [x_1,x_2]\subset(\frac y8,y) with 2u^\prime(x)u(x)=\frac{u^2(x_2)-u^2(x_1)}{x_2-x_1}, hence$$ |u^\prime(x)u(x)|\le \frac 12\ \frac{\frac 8y+\frac 2y}{y/4}=\frac{20}{y^2}\ . $$\noindent Putting y=x_\ast, we obtain an x_0\in (\frac {x_\ast}{8},x_\ast) with$$ |u^\prime(x_0)u(x_0)|\le 20 x_\ast^{-2}\ .\tag"(2)" $$\noindent Instead letting y\to \infty, we obtain a sequence x_\nu\to \infty, with$$ u^\prime(x_\nu)u(x_\nu)\to 0\ . \tag"(3)" $$\noindent Now suppose \phi\in C^\infty(0,\infty). Integration by parts yields$$\multline 0=\int_{x_0}^{x_\nu}e^\phi u\bigl\{-\frac {d^2}{dx^2}u+(V-E)u\bigr\}dx= -e^\phi uu^\prime\bigr]_{x_0}^{x_\nu}\\ +\int_{x_0}^{x_\nu}e^\phi\{(u^\prime)^2+(V-E)u^2+\phi^\prime uu^\prime\}dx\\ \ge -e^\phi uu^\prime\bigr]_{x_0}^{x_\nu}+\int_{x_0}^{x_\nu}e^\phi \{\frac 12 (u^\prime)^2+(V-E-100(\phi^\prime)^2)u^2\}dx\endmultline\tag"(4)" $$\noindent since \phi^\prime uu^\prime\ge -\frac 12(u^\prime)^2-100(\phi^\prime)^2 u^2. \noindent We want to take \phi(x)=0 for x\frac 14 |E_\ast|>0. Hence (4) implies$$\multline |u(x_0)u^\prime(x_0)|+|e^{\phi(x_\nu)}u(x_\nu)u^\prime(x_\nu)|+ \int_{x_0}^{x_\ast}|V(x)-E|\cdot|u(x)|^2dx\\ \ge \int_{x_\ast}^{x_\nu}e^\phi\bigl\{\frac 12(u^\prime)^2 +\frac 14 |E_\ast|(u(x))^2\bigr\}dx\ .\endmultline $$\noindent Note that \phi(x_\nu)=\phi(y) since x_\nu\ge y, and that$$\split \int_{x_0}^{x_\ast}|V(x)-E|\ |u(x)|^2dx\le\operatornamewithlimits{\sup}_{x\in (x_0,x_\ast)}|V(x)-E|\cdot\Vert u\Vert^2\\ \le \operatornamewithlimits{\sup}_{\frac 18x_\ast2x_\ast$and let$\nu\to \infty$. We obtain from (2), (3), (5) that $$\multline \exp(-10^{-3}|E_\ast|^{1/2}x_\ast)\cdot\bigl\{20x_\ast^{-2}+ \operatornamewithlimits{\sup} _{\frac 18 x_\ast2x_\ast, so we can let y\to \infty to get:$$ \int_{2x_\ast}^\infty(u^\prime)^2dx\le 40x_\ast^{-2}\bigl\{1+ x^2_\ast\sup_{\frac 18x_\astx_\ast$, with a given $E_\ast<0$. Define $H_1=-\frac{d^2}{dx^2}+V_1(x)$ and $H_2=-\frac{d^2}{dx^2} +V_2(x)$ on $(0,\infty)$ with Dirichlet boundary conditions. Let $E_1x_\ast$ we have $V_1(x)-E_1\ge \frac 12 E_\ast-E_\ast= \frac 12|E_\ast|$, so Lemma 1 applies. In view of our estimate on $\operatornamewithlimits{\sup}_{\frac 18 x_\ast\frac 12$ by (6), if we take $N_1$ large enough. Also $$\split (H_2-E_1)w=\bigl(-\frac{d^2}{dx^2}+V_2-E_1)(\chi u)\\ = \chi\bigl(-\frac{d^2}{dx^2}+V_2-E_1)u-2\chi^\prime u^\prime-\chi^{\prime\prime}u\\ =-2\chi^\prime u^\prime-\chi^{\prime\prime}u\ , \endsplit\tag"(8)"$$ \noindent since $V_2=V_1$ in $\text{supp}\ \chi$, and $(H_1-E_1)u=0$. \noindent Since $\chi^\prime$, $\chi^{\prime\prime}$ are supported in $\{x>2x_\ast\}$ and $|\chi^\prime(x)|\le Cx_\ast^{-1}$, $|\chi^{\prime\prime} (x)|\le Cx_\ast^{-2}$, it follows from (6), (7), (8) that $$\multline \Vert(H_2-E_1)w\Vert^2\le Cx_\ast^{-2}\Vert u^\prime\Vert_{L^2(2x_\ast,\infty)} ^2+Cx_\ast^{-4}\Vert u\Vert_{L^2(2x_\ast,\infty)}^2\\ \le C_{N_1}x_\ast^{-4}(x_\ast^2|E_\ast|)^{K+1-N_1}= C_{N_1} |E_\ast|^2(x_\ast^2|E_\ast|)^{K-1-N_1}\ .\endmultline$$ \noindent Taking $N_1>K-1+2N$ and assuming $x_\ast^2|E_\ast|$ large enough, we conclude that $$\Vert(H_2-E_1)w\Vert\le \frac 12 |E_\ast|(x_\ast^2|E_\ast|)^{-N}\ .\tag"(9)"$$ \noindent Since $\Vert w\Vert>\frac 12$, it follows from spectral theory and (9) that $|E_1-E_2|\le |E_\ast|(x_\ast^2|E_\ast|)^{-N}$ for some $E_2$ in the spectrum of $H_2$. This is the conclusion of the lemma.$\qquad\blacksquare$ \vglue 1pc \proclaim{Corollary} Under the assumptions of Lemma 2, let $u$ be the normalized eigenfunction of $H_1$ corresponding to $E_1$. Then there is a function $w$ on $(0,\infty)$, satisfying $\Vert u-w\Vert\le (|E_\ast|x_\ast^2)^{-N}$ and $\Vert(H_2-E_1)w\Vert\le |E_\ast|\cdot(x_\ast^2|E_\ast|)^{-N}$. \endproclaim \vfill\eject \demo{Proof}\enddemo The second estimate is (9) and the first is immediate from (6), provided we take $N_1$ large enough and use the function $w$ constructed in the proof of Lemma 2.$\qquad\blacksquare$ \vglue 1pc \proclaim{Lemma on Low Eigenvalues in Coulomb Singularities} Suppose we are given $E_0$, $E_\ast$, $Z$, $\ell$, $x_\ast$ satisfying the following (a) $cZ^{4/3}Z^{-500}$ (d) $\ell$ is a non-negative integer, and $\ellZ^{\frac{1}{100}}$ and $x_\ast>Z^{-1}$ (f) $Z$ is greater than a large positive number determined by $c$ and $C$ in (a) and (d). Let $V(x)$ be a smooth potential in $(0,\infty)$, equal to $V_c(x)\equiv \frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$ for $x\le 10 x_\ast$. Assume also that $V(x)$, $V_c(x)>\frac 12 E_\ast$ for $x>x_\ast$. Let $H=-\frac{d^2} {dx^2}+V(x)$ on $(0,\infty)$, with Dirichlet boundary conditions. Then the spectrum of $H$ in $(-\infty,E_\ast)$ consists of eigenvalues $E_0-\frac{Z^2}{4n^2}+\ \text{Error}_n$ for all integers $n$ in the range $\ellx_\ast$ we have $V(x)\ge \frac 12 E_\ast>E_\ast$, and therefore the spectrum of $H$ in $(-\infty,E_\ast)$ consists at most of finitely many eigenvalues. Also, note that $V(x)\ge \frac 12 E_\ast\ge -\frac 12 Z^2$ for $x>x_\ast$, while $V(x)=V_c(x)\ge -\frac Zx$ for $x\le x_\ast$. Hence $V(x)\ge -\frac Zx-\frac 12 Z^2$ for all $x\in (0,\infty)$, which implies that the lowest eigenvalue of $H$ is greater than or equal to that of $-\frac{d^2}{dx^2}-\frac Zx-\frac 12 Z^2$, which is $-\frac 34 Z^2$. \vfill\eject If $E<0$ is an eigenvalue of either $H$ or $H_c=-\frac{d^2}{dx^2}+V_c(x)$, then $-\frac 34 Z^2\le E<0$, and $$\multline \operatornamewithlimits{\sup}_{\frac 18 x_\ast>CZ^{8/3}\ \text{for\ large}\ K\ . \endmultline$$ \noindent Hence $$\operatornamewithlimits{\sup}_{\frac 18 x_\ast\ell, so we have E_2=E_0-\frac{Z^2}{4n^2}, n>\ell. Hence |E-(E_0-\frac{Z^2}{4n^2})|cZ^2\cdot\bigl[\frac{Z}{2(E_0-E_\ast)^{1/2}}\bigr]^{-3} =c^{\prime}Z^{-1}(E_0-E_\ast)^{3/2}\\ \ge c^{\prime\prime}Z^{-1}(Z^{4/3})^{3/2}=c^{\prime\prime}Z>1\ .\endmultline$$ \noindent It follows by spectral theory that $$\Vert(H_c-[E_0-\frac{Z^2}{4n^2}])w\Vert=\Vert(H_c-[E_0-\frac{Z^2}{4n^2}]) \varphi\Vert\ge \Vert\varphi\Vert\ .$$ \noindent Hence (16) implies $\Vert\varphi\Vert\le 2Z^{-500}$, and similarly $\Vert\tilde\varphi\Vert \le 2Z^{-500}$. Thus, (17) and (18) show that $$\Vert w-a\psi_n\Vert\ ,\ \Vert\tilde w-\tilde a\psi_n\Vert\le 2Z^{-500}\ . \tag"(19)"$$ \noindent On the other hand, $\Vert\psi_n\Vert=1$ and $\Vert w\Vert$, $\Vert\tilde w\Vert\sim 1$ by (14) and (15), since $\Vert u\Vert=\Vert \tilde u\Vert=1$. Therefore, (19) implies that $|a|$, $|\tilde a|\sim 1$. Putting this back into (19), we conclude that $\Vert w-a^\prime\tilde w\Vert4\ell(\ell+1)E$ so that $\frac{\ell(\ell+1)}{x^2}+E-\frac Zx$ is negative somewhere in $(0,\infty)$. Then $$\int_0^\infty\Bigl(-\bigl(\frac{\ell(\ell+1)}{x^2}+E-\frac Zx\bigr)\Bigr)_+^{1/2} dx=\pi \Bigl(\frac{Z}{2E^{1/2}}-\sqrt{\ell(\ell+1)}\, \Bigr)\ .$$\endproclaim \vglue 1pc \demo{Proof}\enddemo For $00\ . \endsplit\tag"(10)" $$\noindent The conclusion of the Lemma is immediate from (10).\qquad \blacksquare In view of Lemma 1, we will have$$ \phi_c(E_\ast)=\int_0^\infty\bigl(E_\ast-V_c(x)\bigr)_+^{1/2}dx= \pi\bigl(\frac{Z}{2\sqrt{E_0-E_\ast}}-\sqrt{\ell(\ell+1)}\, \bigr)=\pi m_\ast \tag"(11)" $$\noindent for an integer m_\ast, provided we take$$ E_\ast=E_0-\frac{Z^2}{4[m_\ast+\sqrt{\ell(\ell+1)}]^2}\ ,\tag"(12)" $$\noindent and provided$$ Z^2>4\ell(\ell+1)(E_0-E_\ast)\ .\tag"(13)" $$\noindent We will pick$$ m_\ast=\ \text{largest\ integer}\ <\frac{1}{16}(Zx_\ast)^{1/2}\ ,\tag"(14)" $$\noindent and check that (13) holds, since$$\split 4\ell(\ell+1)(E_0-E_\ast)=4\ell(\ell+1)\cdot \frac{Z^2} {4[m_\ast+\sqrt{\ell(\ell+1)}]^2}\\ <4\ell(\ell+1)\cdot \frac{Z^2}{4[\sqrt{\ell(\ell+1)}]^2}=Z^2\ .\endsplit $$\vglue 1pc \proclaim{Lemma 2} Suppose V(x) is smooth on (0,\, \infty) and satisfies the hypotheses (Z0)\ldots(Z9) and (CS1)\ldots(CS6). Assume also that \ell>Z^{1/K}+10^{9}. Define E_\ast by (12), (14). Then all the hypotheses of the Lemma on Low Eigenvalues in Coulomb Singularities are satisfied.\endproclaim \vglue 1pc \demo{Proof}\enddemo The hypotheses (a)\ldots(f) of that Lemma are as follows. (a) says that cZ^{4/3} -\frac{Z^2}{4[m_\ast+\sqrt{\ell(\ell+1)}]^2}>-Z^2, which is the desired lower bound for E_\ast. To prove the desired upper bound for E_\ast, it's enough to check that$$ \frac{Z^2}{4[m_\ast+\sqrt{\ell(\ell+1)}]^2}>c\, Z^{\frac 43+\frac{1}{100}}\ , \tag"(14$\text{bis}$)" $$\noindent i.e.\ [m_\ast+\sqrt{\ell(\ell+1)}]^2Z^{-500}, i.e.\ \min_{n\in \Bbb Z}\big|\frac{Z^2}{ 4[m_\ast+\sqrt{\ell(\ell+1)}]^2}-\frac{Z^2}{4n^2}\big|>Z^{-500}, i.e.\$$ \min_{n\in \Bbb Z}\big|[m_\ast+\sqrt{\ell(\ell+1)}]^{-2}-n^{-2}\big| >4\ Z^{-502}\ .\tag"(15)" $$\noindent We have \sqrt{\ell(\ell+1)}=\ell\sqrt{1+\ell^{-1}}=\ell+\frac 12 +O(\ell^{-1}). Since \ell>Z^{1/K}+10^{9}, it follows that \big|\sqrt{\ell(\ell+1)}-(\ell+\frac 12)\big|<\frac{1}{100}. Thus, m_\ast+\sqrt{\ell(\ell+1)}=(m_\ast+\ell+\frac 12)+\xi with |\xi|<\frac{1}{100}. So in (15), the minimum is attained by n=m_\ast+\ell or by n=m_\ast+\ell+1. In either case we have \frac 14<|n-(m_\ast+\sqrt{\ell(\ell+1)})|<1, so |[m_\ast+\sqrt{\ell(\ell+1)}]^{-2}-n^{-2}|\sim n^{-3}\sim (m_\ast +\ell)^{-3}>c(Zx_\ast)^{-3/2} since we saw in the proof of (b) that m_\ast,\ell\le\frac{1}{16}(Zx_\ast)^{1/2}. Thus,$$ \min_{n\in \Bbb Z}\big|[m_\ast+\sqrt{\ell(\ell+1)}]^{-2}-n^{-2}\big| >c(Zx_\ast)^{-3/2}>c\bigl(Z^{\frac 23-\frac{1}{100}}\bigr)^{-3/2} >> 4Z^{-502}\ , $$\noindent completing the proof of (c). (d) says that \ell is an integer and 0\le \ell\le CZ^{1/3}, which is immediate from (CS3) and (CS5). (e) says that x_\ast>Z^{-1} and |E_\ast|x_\ast^2>Z^{\frac{1}{100}}. Now x_\ast>Z^{-1} is contained in (CS5). From (CS4), (12), (14 bis), we have |E_\ast|\sim \frac{Z^2}{4[m_\ast+\ell]^2}. Hence, since \ell\le\frac{1}{16}(Zx_\ast)^{1/2}\sim m_\ast, we get |E_\ast|x_\ast^2\sim \frac{Z^2x_\ast^2}{[m_\ast+\ell]^2}\sim \frac{Z^2x_\ast^2}{m_\ast^2}\sim \frac{Z^2x_\ast^2}{(Zx_\ast)} (by (14)) =Zx_\ast, so |E_\ast|x_\ast^2>cZ^{\frac{2}{100}} by (CS5). This completes the proof of (e). (f) asserts that Z is large, which is just (CS6). Thus we have proven (a)\ldots(f).\medskip The remaining hypotheses of the Lemma on Low Eigenvalues in Coulomb Singularities are V\in C^\infty(0,\infty), V(x)=V_c(x) for x\le 10 x_\ast, and V(x), V_c(x)>\frac 12 E_\ast for x>x_\ast. Since (CS1)\ldots(CS6) include the second of these assertions, and we are assuming the first, it remains only to show that$$ V(x), V_c(x)>\frac 12 E_\ast\quad \text{for}\ x\ge x_\ast\ .\tag"(16)" $$Now V_c(x) is decreasing for xx_0. We have x_0<\frac{4\ell^2}{Z} \le \frac 4Z\bigl[\frac{1}{16}(Zx_\ast)^{1/2}\bigr]^2=\frac{1}{64} x_\ast, so V_c(x) is increasing in [x_\ast,\infty). Hence to check V_c(x)>\frac 12 E_\ast for x\ge x_\ast it's enough to check that V_c(x_\ast)=\frac{\ell(\ell+1)}{x_\ast^2}+ E_0-\frac {Z}{x_\ast}>\frac 12 E_\ast. This follows from \frac 12 E_0-\frac{Z}{x_\ast}>\frac 12 E_\ast, i.e.\ \frac{Z}{x_\ast}<\frac 12 \bigl(E_0-E_\ast)=\frac{Z^2}{ 8[m_\ast+\sqrt{\ell(\ell+1)}]^2}, i.e.\$$ \bigl[m_\ast+\sqrt{\ell(\ell+1)}\, \bigr]^2<\bigl(\frac{Zx_\ast}{8}\bigr)\ . \tag"(17)" $$\noindent Since m_\ast<\frac{1}{16}(Zx_\ast)^{1/2} and \sqrt{\ell(\ell+1)} <\ell+1\le\frac{2}{16}(Zx_\ast)^{1/2}, we have [m_\ast+\sqrt{\ell(\ell+1)}]^2 <\bigl(\frac{3}{16}\bigr)^2Zx_\ast<\frac{Zx_\ast}{8}, proving (17). So V_c(x)>\frac 12 E_\ast for x\ge x_\ast. To handle V(x), we argue as follows. There is a local minimum of V_c(x) at x=x_00 for x>x_{\text{rt}}. We know that V(x_\ast)=V_c(x_\ast) >\frac 12 E_\ast. Now suppose x>x_\ast. Either x\le x_{\text{rt}}, in which case V(x)\ge V(x_\ast)>\frac 12 E_\ast; or else x>x_{\text{rt}}, in which case V(x)>0>\frac 12 E_\ast. Thus V(x)>\frac 12 E_\ast for all x\ge x_\ast. The proof of Lemma 2 is complete.\qquad\blacksquare By Lemma 2 and the Lemma on Low Eigenvalues in Coulomb Singularities, we have$$ |Y-Y_c|Z^{(10^{-9})}$ and with $K=10^9$ and $\hat c$ suitably picked. Moreover, with $\lambda_c(x)=S_c^{1/2}(x)B_c(x)$ and $\Lambda_c=\bigl(\int_{V_c<0}\frac{dx}{\lambda_c(x)B_c(x)}\bigr)^{-1}$, we have $\Lambda_c\sim \ell$.\endproclaim \demo{Proof}\enddemo We provide the elementary details for the reader's convenience. (Z0) says that $c0$, so $(x_{\text{left}},x_{\text{rt}})\subset (0,\infty)$. Note that $$x_{\text{rt}}=\frac{Z+\sqrt{Z^2-4E_0\ell(\ell+1)}}{2E_0}>\frac{Z} {2E_0}>cZ^{-1/3}\ .\tag"(22 \text{bis})"$$ \noindent For $x\le 10^{-6}\frac{\ell^2}{Z}$ we have $\frac{\ell(\ell+1)} {x^2}+E_0-\frac Zx>\frac Z{x^2}\bigl\{\frac{\ell(\ell+1)}{Z}-x\bigr\} >0$, so $(x_{\text{left}},x_{\text{rt}})=\bigl\{\frac{\ell(\ell+1)}{x^2} +E_0-\frac Zx<0\bigr\}\subset (10^{-6}\frac {\ell^2}{Z},\infty)$. For $x\ge 10^{+6}\frac{Z}{E_0}$ we have $\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx \ge \frac{E_0}{x}\bigl\{x-\frac{Z}{E_0}\bigr\}>0$, so $(x_{\text{left}}, x_{\text{rt}})\subset (10^{-6}\frac{\ell^2}{Z}, 10^{+6}\frac{Z}{E_0})$ as claimed. We have proven (Z2) and (Z5).\medskip (Z3) says that $V_c(x_0)<-c\frac{Z}{x_0}$, $V^\prime(x_0)=0$, $V^{\prime\prime} (x)>c\frac {Z}{x^3}$ for |x-x_0| \frac{6Z}{x^3}\ \bigl\{\frac{100}{101}\ \frac{\ell(\ell+1)}{Zx_0} -\frac 13 \bigr\}\\ =\frac{6Z}{x^3}\ \bigl\{\frac{100}{101}\cdot\frac 12-\frac 13\bigr\}> \frac{cZ}{x^3}\ . \endsplit $$\noindent This proves (Z3) with c_1=\frac{1}{100}. (Z4) says that$$\gather {\align V_c^{\prime}(x)<-\frac{cZ}{x^2}\quad &\text{for}\ x_{\text{left}}+\frac{cZ}{x^2}\quad &\text{for}\ \frac{101}{100} x_0\frac{101}{100} x_0 we have $$V_c^\prime(x)=\frac{Z}{x^2}\bigl\{\frac{-2\ell(\ell+1)}{Zx} +1\bigr\}>\frac{Z}{x^2}\bigl\{-\frac{100}{101}\cdot\frac{2\ell(\ell+1)} {Zx_0}+1\bigr\}=+\frac{1}{101}\frac{Z}{x^2}\ ,$$ \noindent which proves (Z4). Next we compute $\lambda_c(x)$ and $\Lambda_c=\bigl(\int_{x_{\text{left}}} ^{x_{\text{rt}}}\ \frac{dx}{\lambda_c(x)B_c(x)}\bigr)^{-1}$. We have $\lambda_c(x)=\bigl(\frac{Z}{x}\bigr)^{1/2}\cdot x=Z^{1/2}x^{1/2}$, so $$\Lambda_c^{-1}=\int_{x_{\text{left}}}^{x_{\text{rt}}}\ \frac{dx} {Z^{1/2}x^{3/2}}=\ \text{(const)}\ Z^{-\frac 12}\cdot(x_{\text{left}}^{-1/2} -x_{\text{rt}}^{-1/2})\ .$$ Now $x_{\text{left}}=\inf \{x\mid V_c(x)<0\}\le x_0$ since $V_c(x_0)<0$ by (Z3). Thus $x_{\text{left}}<\frac{2\ell(\ell+1)} {Z}$, while $x_{\text{rt}}>cZ^{-1/3}$. Since $\ell\le\frac{1}{16} (Zx_\ast)^{1/2}\frac{5000}{x^2}\quad\text{for}\ \Lambda_c^{(10^9)}x_{\text{rt}} \frac Z{E_0}$, hence for $x>CZ^{-1/3}$, hence for $x>C^\prime x_{\text{rt}}$, by virtue of (22 bis). This implies (23 bis) since we are taking here $\ell\ge Z^{(10^{-9})}$. So we have proven (Z6). (Z7) says that $(\max I_c)/(\min I_c)<\Lambda_c^{10^9}$, i.e.\ $10^{18}\frac{Z^2}{\ell^2E_0}Z$. This proves (Z7). (Z8) says that the constant $\hat c$, which hasn't yet been specified, is small enough. Just pick $\hat c$ small, and (Z8) is satisfied. (Z9) says that $\Lambda_c$ is large, i.e.\ that $\ell$ is large. Here we take $\ell>Z^{(10^{-9})}$ with $Z$ very large. Hence $\ell$ is large. So (Z9) is proven. Thus (Z0)$\ldots$(Z9) are all satisfied.$\qquad\blacksquare$ \medskip Now we know that both $V(x)$ and $V_c(x)$ satisfy (Z0)$\ldots$(Z9). To apply the Lemma on Truncated Eigenvalue Sums, we still must check the assumptions made there on $E_\ast$, namely $\frac 1\pi \phi(E_\ast)$, $\frac 1\pi \phi_c(E_\ast)$ are integers, $\phi(E_\ast)<\phi(0)-1$, $\phi_c(E_\ast)<\phi_c(0)-1$. Here $\phi(E)=\int_0^\infty\bigl(E-V(x)\bigr)_+^ {1/2}dx$, $\phi_c(E)=\int_0^\infty\bigl(E-V_c(x)\bigr)_+^{1/2}dx$. Equation (11) shows that $\frac 1\pi \phi_c(E_\ast)$ is an integer. Moreover, for $E\le \frac 12 E_\ast$ we have $\phi_c(E)=\phi(E)$. In fact, Lemma 2 shows that $V(x)$, $V_c(x)>\frac 12 E_\ast$ for $x\ge x_\ast$, hence the region of integration in the definitions of $\phi(E)$, $\phi_c(E)$ may be changed to $(0,x_\ast)$. In that region, $V(x)=V_c(x)$, so $\phi(E)=\phi_c(E)$ for $E\le \frac 12E_\ast$. In particular, $\frac 1\pi \phi(E)$ is an integer. To check that $\phi(E_\ast)< \phi(0)-1$ and $\phi_c(E_\ast)<\phi_c(0)-1$, we prove the stronger statements $\phi(E_\ast)<\phi(\frac12 E_\ast)-1$ and $\phi_c (E_\ast)<\phi_c(\frac 12 E_\ast)-1$. These two statements are the same, so we may just look at $\phi_c$, which is given by $\frac 1\pi \phi_c(E)=\frac{Z}{2(E_0-E)^{1/2}}-\sqrt{\ell(\ell+1)}$, by Lemma 1. Thus we must show that $$\frac{\pi Z}{2}(E_0-E_\ast)^{-1/2}<\frac {\pi Z}{2}(E_0-\frac 12 E_\ast)^{-1/2}-1 \tag"(24)"$$ \noindent in order to verify the hypotheses of the Lemma on Truncated Eigenvalue Sums. From (14 bis) we have $-E_\ast>cZ^{\frac 43+\frac{1}{100}}$, hence $-E_\ast>>E_0$. Therefore, $(E_0-E_\ast)>\frac 32 (E_0-\frac 12 E_\ast)$, so $(E_0-\frac 12 E_\ast)^{-1/2}-(E_0-E_\ast)^{-1/2} >(E_0-\frac 12 E_\ast)^{-1/2}\cdot \bigl(1-(\frac 32)^{-1/2}\bigr) >c|E_{\ast}|^{-1/2}$. Hence, (24) holds provided $|E_\ast|Z^{-1+\frac{1}{100}}$. This completes the proof of (24) and allows us to use the Lemma on Truncated Eigenvalue Sums. We get: \align X&=-\frac{2}{3\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}dx+\frac{1} {24\pi}\int_0^\infty V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{-1/2} dx\\ &\quad +\frac \pi 2\bigl(\phi^\prime(0)\bigr)^{-1}\tilde\chi(\frac 1\pi \phi(0)-\frac 12\bigr) +\bigl\{\frac{2}{3\pi}\int_0^\infty\bigl(E_\ast-V(x)\bigr)_+^{3/2} dx\\ &\quad -\frac{1}{24\pi}\int_0^\infty V^{\prime\prime}(x)\bigl(E_\ast-V(x)\bigr)_+ ^{-1/2}\ dx+ \frac{\pi}{24}\bigl(\phi^\prime(E_\ast)\bigr)^{-1}- \frac{E_\ast}\pi\phi(E_\ast)\bigr\}+\ \text{Error}\tag"(26)"\endalign \noindent with $$|\text{Error}|\le C_{\#}\Lambda^{4\varepsilon-2}\operatornamewithlimits{\inf} \Sb x\in I\\ V(x)E_\ast for x\ge x_\ast, and V(x)=V_c(x) for x\frac 12 E_\ast for x\ge x_\ast. Hence the inf's in (31) occur over subsets of (0,x_\ast), where V(x) =V_c(x) and S(x)=S_c(x). Also, from (Z0)\ldots(Z9) applied to V and V_c, we know that V(x)>Z^{-100}. Thus, (5), (18), (30) and (36) imply the following result, which strengthens the First WKB Eigenvalue Sum Theorem when \ell<<(Zx_\ast)^{1/2}. \vglue 1pc \proclaim{Second WKB Eigenvalue Sum Theorem} Suppose V(x) is a smooth potential defined on (0,\infty), I\subset (0,\infty) is an interval containing \{V(x)<0\}, and S(x), B(x) are positive functions defined on I. Also, let \varepsilon, K, N, \ell, E_0, Z, x_\ast be given. Assume the hypotheses (Z0)\ldots(Z9) of the WKB Density Theorem, and assume that V(x) has an exact Coulomb singularity with parameters (\ell,E_0,Z,x_\ast). (That is, assume (CS1)\ldots(CS6)). Finally, suppose \ell>Z^{(10^{-9})}. Set H=-\frac{d^2}{dx^2}+V(x) on (0,\infty) with Dirichlet boundary conditions. Then the sum of the negative eigenvalues of H is given by the equation$$\align &{\roman{sneg}}(H)-{\roman{sneg}}(H_c)=\\ &\quad -\frac{2}{3\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}dx +\frac{1}{24\pi}\int_0^\infty V^{\prime\prime}(x)\cdot\bigl(-V(x)\bigr)_+^{-1/2} dx\\ &\quad\ \ +\frac \pi 2\bigl(\phi^\prime(0)\bigr)^{-1}\tilde\chi\bigl(\frac 1\pi \phi(0)-\frac 12 \bigr) +\frac {2}{3\pi}\int_0^\infty\bigl(-V_c(x)\big)_+^{3/2}dx\\ &\quad \ \ -\frac{1}{24\pi} \int_0^\infty V_c^{\prime\prime}(x)\cdot \bigl(-V_c(x)\bigr)_+^{-1/2} dx-\frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}- \sqrt{\ell(\ell+1)}-\frac 12\bigr)\\ &\quad \ \ +\ {\roman{Error}}\ ,\endalign $$\noindent with H_c=-\frac{d^2}{dx^2}+V_c(x) on (0,\infty) subject to Dirichlet boundary conditions, V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx, and |{\roman{Error}}|\le \Lambda^{5\varepsilon-2}\frac{Z}{x_\ast}. Recall that \phi(E)=\int_0^\infty\bigl(E-V(x)\bigr)_+^{1/2}dx, \phi^\prime (E)=\frac 12 \int_0^\infty\bigl(E-V(x)\bigr)_+^{-1/2}dx, \tilde\chi(t)=\min_{k\in \Bbb Z}|t-k-\frac 12|^2-\frac{1}{12}. \endproclaim \vfill\eject %%eigensums.tex%% \head Eigenvalue Sums for Degenerate Potentials\endhead \medskip In this section, we derive crude versions of the Second WKB Eigenvalue Sum Theorem that hold under weakened hypotheses. The idea is as follows. Suppose H=-\frac{d^2}{dx^2}+V(x) on (0,\infty), (Dirichlet b.c.) with V(x) having an exact Coulomb singularity (i.e.\ smooth on (0,\infty) and satisfying (CS1)\ldots(CS6) for suitable (\ell,E_0,Z,x_\ast)). We pick E_\ast\in \bigl(-\frac{8Z}{x_\ast}, -\frac{8Z}{3x_\ast}\bigr) so that$$ \min_{n\in \Bbb Z}\big|E_\ast-\bigl(E_0-\frac{Z^2}{4n^2}\bigr)\big|>Z^{-500}\ . $$\noindent If V(x)>\frac 12E_\ast for x>x_\ast, then the hypotheses of the Lemma on Low Eigenvalues in Coulomb Singularities apply. Hence, with H_c=-\frac{d^2}{dx^2}+V_c(x), V_c(x)=\frac{\ell(\ell+1)}{x^2} +E_0-\frac Zx, we have$$ \Big|\sum\limits_k(E_\ast-E_k)_+-\sum\limits_k(E_\ast-E_k^c)_+\Big| \le Z^{-100}\ ,\tag"(1)" $$\noindent where \{E_k\} are the negative eigenvalues of H and \{E_k^c\} are those of H_c. Now let$$\align \Cal N(E)&=\ \text{number\ of\ eigenvalues\ of}\ H\ \text{which\ are}\ E_\ast$for$x>x_\ast$, it follows that$\bigl(E_\ast-V(x)\bigr)_+^{3/2} =\bigl(E_\ast-V_c(x)\bigr)_+^{3/2}everywhere. Thus \align &-\frac 1\pi\int_{E_{\ast}}^0\int_0^\infty\bigl(E-V(x)\bigr)_+^{1/2} dx\, dE+\frac 1\pi\int_{E_\ast}^0\int_0^\infty\bigl(E-V_c(x)\bigr)_+^{1/2} dx\, dE\\ &= -\frac{2}{3\pi}\int_0^\infty\bigl\{\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\}dx\ .\tag"(8)"\endalign \noindent Note that the quantity in curly brackets vanishes nearx=0$, although$\bigl(-V(x)\bigr)^{3/2}$fails to be integrable when$V(x)\sim-\frac Zx$near$x=0$. Putting (8) into (6), (7), we get $$\text{sneg}(H)-\text{sneg}(H_c)=-\frac{2}{3\pi}\int_0^\infty \bigl\{\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\}dx+\ \text{Error} \tag"(9)"$$ \noindent with $$|{\roman{Error}}|< C\frac{Z}{x_\ast}\ .\tag"(10)"$$ So proving (9), (10) for potentials with exact Coulomb singularities is reduced to proving (2), which concerns the number of eigenvalues$-\frac 43\, \frac{Z}{x_\ast}$for$x>x_\ast$. Set$V_c (x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$. Then for$H=-\frac{d^2}{dx^2}+ V(x)$,$H_c=-\frac{d^2}{dx^2}+V_c(x)$on$(0,\infty)$with Dirichlet boundary conditions, we have $$\text{sneg}(H)-\text{sneg}(H_c)=-\frac{2}{3\pi}\int_0^\infty \big\{\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\} dx+\ {\roman{Error}}\ ,$$ \noindent with$|{\roman{Error}}|Z^{-500}$. Then$E_{\text{crit}}\le E_\ast\le 0$, and$V(x)>\frac 12E_\ast$for$x>x_\ast$. For$E\in [E_{\text{crit}},0]$, and hence for$E\in [E_\ast,0]$, Theorem D2 from the section on WKB Theory with weak turning points implies (2). Hence (9), (10) follow, and Theorem 1 is proven.$\qquad\blacksquare$\medskip For Hypotheses (Z$\overline 0$)$\ldots$(Z$\overline 8$), see Review of Previous Results", part G. \vglue 1pc \proclaim{Theorem 2} Suppose$V(x)$,$S(x)$,$B(x)$satisfy hypotheses (Z$\hat 0$)$\ldots$(Z$\hat 9$), and suppose also that$V(x)$has an exact Coulomb singularity (CS1)$\ldots$(CS6) for a given$(\ell,E_0,Z,x_\ast)$. Assume$V(x)>-\frac{2Z}{x_\ast}$for$x>x_\ast$. Assume$V(2x_0) <-\frac{8Z}{x_\ast}$, and assume $$\int_0^\infty\bigl\{\bigl(-V(x)\bigr)_+^{1/2}-\bigl(V(\frac 12x_1)- V(x)\bigr)_+^{1/2}\bigr\}dx<\underline C\ ,\tag"(11)"$$ \noindent where$x_0$,$x_1$are as in (Z$\hat 0$)$\ldots$(Z$\hat 9$). Set$V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$. Then for$H=-\frac{d^2}{dx^2}+V(x)$,$H_c=-\frac{d^2}{dx^2}+V_c(x)$on$(0,\infty)$with Dirichlet boundary conditions, we have $$\text{sneg}(H)-\text{sneg}(H_c)=-\frac{2}{3\pi}\int_0^\infty\bigl\{\bigl(-V (x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\}dx+\ {\roman{Error}}\ ,$$ \noindent with$|{\roman{Error}}|< C\frac{Z}{x_\ast}$. Here$C$depends only on$\underline C$in (11), and on the constants in (Z$\hat 0$)$\ldots$(Z$\hat 9$) and in (CS1)$\ldots$(CS6).\endproclaim \medskip For hypotheses (Z$\hat 0$)$\ldots$(Z$\hat 9$) see Review of Earlier Results," section E. \vglue 1pc \demo{Proof}\enddemo Take$E_\ast\in (-\frac{8Z}{x_\ast},-\frac{4Z}{x_\ast})$, such that$\operatornamewithlimits{\min}\limits_n|E_\ast-(E_0-\frac{Z^2} {4n^2})|>Z^{-500}$. Then$V(x)>\frac 12E_\ast$for$x>x_\ast$, and$V(2x_0)-\frac 43\ \frac {Z}{x_\ast}\quad\text{for}\quad x>x_\ast\tag"(13)"  V(x)>-Cx_{\text{big}}^2x^{-4}\quad \text{for}\ x>x_{\text{big}}\ . \tag"(14)"  \int_0^\infty\bigl\{\bigl(-V(x)\bigr)_+^{1/2}-\bigl(V(\frac 14x_1) -V(x)\bigr)_+^{1/2}\bigr\}dx<\underline C\ ,\tag"(15)" $$\noindent with x_1 as in (Z0^\dagger)\ldots(Z7^\dagger). Set V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx. Then for H=-\frac{d^2}{dx^2}+V(x), H_c=-\frac{d^2}{dx^2}+V_c(x) on (0,\infty) with Dirichlet boundary conditions, we have$$ {\roman{sneg}}(H)-{\roman{sneg}}(H_c)=-\frac{2}{3\pi}\int_0^\infty \bigl\{\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\}dx +\ {\roman{Error}}\ , $$\noindent with |{\roman{Error}}|<\frac{CZ}{x_\ast}. Here, C depends only on \underline C in (15), and on the constants in (Z0^\dagger)\ldots(Z7^\dagger), (CS1)\ldots(CS6), and (14).\endproclaim \medskip For hypotheses (Z0^\dagger)\ldots(Z7^\dagger), see Review of Earlier Results," part F. \vglue 1pc \demo{Proof} \enddemo Take E_\ast\in \bigl(-\frac{3Z}{x_\ast},-\frac 83\ \frac{Z} {x_\ast}\bigr), with \operatornamewithlimits{\min}\limits_n\big| E_\ast-\bigl(E_0-\frac{Z^2}{4n^2}\bigr)\big|>Z^{-500}. Then V(x)>\frac 12 E_\ast for x>x_\ast, by (13). As before, it is enough to prove (2), and by (15) we need only look at E=0 and at E\in [E_\ast,V(\frac 14x_1)]. By (12), [E_\ast,V(\frac 12x_1)]\subset [V(\frac 1{10}x_\ast), V(\frac 14x_1)]. For E\in [V(\frac{1}{10}x_\ast),V(\frac 14x_1)], Lemma F2 in the section Review of Earlier Results" applies. Therefore, Theorem D1 in the section on WKB Theory with Weak Turning Points gives$$ \Cal N(E)=O(1)+\frac 1\pi\int_{x_0}^{\tilde x-\hat C_{\#} \lambda^{-2/3}(\tilde x)\cdot\tilde x}\ \bigl(E-V(x)\bigr)_+^{1/2} dx\ ,\tag"(16)" $$\noindent where V(\tilde x)=E, \frac{1}{10}x_\ast\le \tilde x\le \frac 14 x_1. >From (Z4^\dagger) we have$$ \int_0^{x_0}\bigl(E-V(x)\bigr)_+^{1/2}dx\le \int_0^{x_0} \bigl(-V(x)\bigr)_+^{1/2}dx=O(1)\ .\tag"(17)" $$\noindent Lemma F2 shows that |E-V(x)|\le C(\lambda^{-2/3} (\tilde x)\cdot \tilde x)^{-2} for x\in [\tilde x-\hat C_{\#} \lambda^{-2/3}(\tilde x)\cdot\tilde x,\tilde x], because that's contained in condition (H\hat 5) in the conclusion of Lemma F2. This and (16), (17) imply$$ \Cal N(E)=O(1)+\frac 1\pi \int_0^{\tilde x}\bigl(E-V(x)\bigr)_+^{1/2} dx\ .\tag"(18)" $$\noindent Now (Z2^\dagger), (Z5^\dagger) show that V(x) is increasing in [\tilde x,x_{\text{big}}]\subset [x_0,x_1]\cup [x_1,x_{\text{big}}]. Since V(\tilde x)=E, we conclude that V(x)>E for x\in [\tilde x,x_{\text{big}}]. Hence, (18) is equivalent to$$ \Cal N(E)=O(1)+\frac{1}{\pi}\int_0^{x_{\text{big}}}\bigl(E-V(x)\bigr)_+^{1/2} dx\ .\tag"(19)" $$\noindent From (14) we get \bigl(E-V(x)\bigr)_+\le \bigl(-V(x)\bigr)_+ \le Cx_{\text{big}}^2x^{-4} for x>x_{\text{big}}, so \int_{x_{\text{big}}}^\infty\bigl(E-V(x)\bigr)_+^{1/2}dx=O(1), and (19) implies \Cal N(E)=O(1)+\frac 1\pi \int_0^\infty\bigl(E-V(x)\bigr)_+^{1/2} dx, which is (2). Thus we have proven (2) for E\in [V(\frac{1}{10}x_\ast),V(\frac 14x_1)]. To complete the proof of Theorem 3, we must check (2) for E=0. Lemma F1 and Theorem D1 from the section on WKB Theory with Weak Turning Points together yield$$ \Cal N(0)=O(1)+\frac 1\pi \int_{x_0}^{x_1}\bigl(-V(x)\bigr)_+^{1/2} dx\ .\tag"(20)" $$\noindent From (Z4^\dagger) we get \int_0^{x_0}\bigl(-V(x)\bigr)_+^{1/2} dx=O(1). \noindent From (Z5^\dagger) we get \int_{x_1}^{x_{\text{big}}} \bigl(-V(x)\bigr)_+^{1/2}dx=O(1). \noindent From (14) we get \int_{x_{\text{big}}}^\infty\bigl(-V(x)\bigr)_+^{1/2} dx=O(1). Therefore, (20) implies \Cal N(0)=O(1)+\frac 1\pi \int_0^\infty \bigl(-V(x)\bigr)_+^{1/2}dx, which is (2). The proof of Theorem 3 is complete.\qquad\blacksquare \vfill\eject %% 3wkb.tex%% \head The Third WKB Eigenvalue Sum Theorem\endhead \medskip Suppose we are in the setting of the WKB Theorem on Low Eigenvalues, with E_\infty=0. We assume the hypotheses (H0^\ast)\ldots(H6^\ast) of that Theorem, and we make the further assumption that$$ -\lambda^{-3\varepsilon}S\frac 12 \lambda^{-2\varepsilon} S$by (1), it follows from (6) that $$t_{\max}=\Phi(E_{\max})\ge c_{\#}\lambda S^{-1}\cdot \frac 12 \lambda^{-2\varepsilon}S+\Phi(0)=c_{\#}^\prime\lambda^{1-2\varepsilon} +b_0\ .$$ \noindent Thus, $$t_{\min} C_{\#}\lambda^{4\varepsilon-2}\ .\tag"(16)"$$ \noindent We leave to the reader the task of deducing (12)$\ldots$(16) from the WKB Theorem on Low Eigenvalues. Note that$a\le b$. If$|a_0-b_0|>C_{\#}\lambda^{4\varepsilon-2}$, then this follows from (9), (13), (14), (15). If$|a_0-b_0|\le C_{\#} \lambda^{4\varepsilon-2}$, then (10) gives$|b_0+\frac 12|\le C_{\#}^\prime \lambda^{-1}<<1$, so$\min_{k\in \Bbb Z}|b_0-k|>\frac 12-C_{\#}\lambda^{-1} >\frac 14$, and (16) gives$w_{hi}=0$. Then (9), (13), (14) show that$a\le b$. Hence$a\le b$in all cases. We use (11)$\ldots$(15) and the Lemma on Riemann Sums to control $$\text{sneg}(H)=\sum\limits_{k=0}^{k_{\max}}E_k\ .$$ \noindent Since$t_{\min}