\magnification \magstep1 \input amstex \nopagenumbers \documentstyle{amsppt} \hsize=6truein \vsize 9truein \hoffset .15truein \baselineskip 20pt \hoffset .15truein \catcode`\@=11 \def\logo@{} \catcode`\@=12 \pageno=0 \centerline{\bf{THE DENSITY IN A THREE-DIMENSIONAL RADIAL POTENTIAL}}\smallskip \centerline{by} \smallskip \bigskip\bigskip\bigskip \centerline{\bf{C. Fefferman\footnote"*"{partially supported by NSF grant \#DMS--9104455.A02} $\quad$ and $\quad$ L. Seco\footnote"$^\dag$"{partialy supported by NSERC grant \#OGP0121848, and by a Connaught Fellowship}}} \centerline{\bf Table of Contents} \medskip \line{\sl \hfil Pages} \smallskip \line{Introduction\hfil 1--6} \smallskip \line{Review of Earlier Results\hfil 7--16}\smallskip \line{Separation of Variables\hfil 17}\smallskip \line{Elementary Properties of the Thomas-Fermi Potential\hfil 18--38}\smallskip \line{The Density in an Approximate Thomas-Fermi Potential\hfil 39--84}\smallskip \line{The WKB Density Theorems for Approximate T-F Potentials\hfil 85--86} \smallskip \line{References\hfil 87} \vfill\eject \def\Icirc{\hbox{$\int$}\kern-6pt\raise 1pt\hbox{$\circ$}} \def\chm{\chi_{{}_-}} \def\chp{\chi_{{}_+}} \def\rnt{\rho_{{}_{NT}}} \head Introduction\endhead \medskip Let $H=-\Delta+V$ be a Schr\"odinger operator on $\Bbb R^n$. The {\sl density\/} associated to $H$ is defined as $$ \rho(x)=\sum\limits_{E_k\le 0}|\varphi_k(x)|^2\ ,\tag"(1)" $$ \noindent where $E_k$, $\varphi_k$ are the eigenvalues and normalized eigenfunctions of $H$. A standard approximation to $\rho$ is the {\sl semiclassical density\/}, given by $$ \rho_{sc}(x)=c_n\bigl(-V(x)\bigr)_+^{n/2}\quad (x\in \Bbb R^n)\ ,\tag"(2)" $$ \noindent where $c_n>0$ depends only on the dimension $n$, and $t_+^s\equiv \cases t^s &\text{if $t>0$}\\ 0 &\text{otherwise}\endcases$. \bigskip \noindent The purpose of this paper is to estimate $\rho-\rho_{sc}$ for a particular potential $V_{TF}$ on $\Bbb R^3$, that arises in the study of atoms. Specifically, $V_{TF}(x)$ is the Thomas-Fermi potential for atomic number $Z$; we will describe $V_{TF}$ below. In [FS1], we announced the proof of an asymptotic formula for the ground-state energy of a non-relativistic atom. A crucial step in the proof of that formula is the estimate $$ \int_{\Bbb R^3}\int_{\Bbb R^3}\bigl(\rho(x)-\rho_{sc}(x)\bigr) \bigl(\rho(y)-\rho_{sc}(y)\bigr)\frac{dx\, dy}{|x-y|} \le CZ^{5/3-a}\ (a>0)\ , \tag"(3)" $$ \noindent where $\rho$ and $\rho_{sc}$ arise from $V_{TF}$ for large $Z$. The main result of this paper is that (3) holds, modulo an assumption which will be proven in our later papers [FS6,7]. The complete proof of the theorem in [FS1] is given by this paper, together with [FS2$\ldots$7]. The potential $V_{TF}$ for an atom is spherically symmetric. Hence, it is natural to prove (3) by separation of variables. Recall that the density $\rho$ associated to a radial potential $V(|x|)$ on $\Bbb R^3$ is given by $$ \rho(x)=\frac{1}{4\pi|x|^2}\sum\limits_{\ell\ge 0}(2\ell+1)\rho_\ell(|x|)\ , \tag"(4)" $$ \noindent where $\rho_\ell(r)$ is the density associated to $-\frac{d^2} {dr^2}+V_\ell(r)$ on $(0,\infty)$, with $$ V_\ell(r)=\frac{\ell(\ell+1)}{r^2}+V(r)\ .\tag"(5)" $$ \noindent The results of our earlier paper [FS4] allow us to compute $\rho_\ell(r)$ modulo errors whose total contribution to (4) will not affect (3). \medskip \noindent In fact, for suitable one-dimensional potentials $W$ on $(0,\infty)$, we proved in [FS4] that the density is well-approximated by $$ \tilde\rho_{sc}(x)=\frac1\pi \bigl(-W(x)\bigr)_+^{1/2}- \frac{\bigl(-W(x)\bigr)_+^{-1/2}}{\Cal N}\chm(\phi) \ ,\quad\text{where}\tag"(6)" $$ $$ \Cal N=\int_0^\infty\bigl(-W(x)\bigr)_+^{-1/2}dx\ ,\quad \phi=\frac 1\pi\int_0^\infty\bigl(-W(x)\bigr)_+^{1/2}dx-\frac 12\ ,\quad \text{and}\tag"(7)" $$ $$ \chm(t)=t-k-\frac 12\ ,\quad \text{with}\ k=\ (\text{greatest\ integer}\ \le t)\ . \tag"(8)" $$ \noindent On the right in (6), the first term is the usual semi-classical approximation, while the second term is a small correction. \smallskip \noindent Taking $W=V_\ell$ and putting (6) into (4), we see that $$\multline \rho(x)\approx \frac{1}{4\pi|x|^2}\ \sum\limits_{\ell\ge 0}\ (2\ell+1)\cdot \frac 1\pi\bigl(-V_\ell(|x|)\bigr)_+^{1/2}\\ -\frac{1}{4\pi|x|^2}\ \sum\limits_{\ell\ge 0}\ \frac{(2\ell+1)}{\Cal N_\ell} \chm(\phi_\ell)\cdot \bigl(-V_\ell(|x|)\bigr)_+^{-1/2}\ ,\endmultline\tag"(9)" $$ \noindent with $$ \Cal N_\ell=\int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{-1/2}dr\ ,\quad \phi_\ell=\frac 1\pi\int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{1/2}dr-\frac 12 \ .\tag"(10)" $$ \noindent The first term on the right in (9) is a Riemann sum, which closely approximates the integral $$ \int_0^\infty\frac{(2\lambda+1)}{4\pi|x|^2}\cdot \frac 1\pi\ \Bigl(-\frac{\lambda(\lambda+1)}{|x|^2}-V(x)\Bigr)_+^{1/2}\ d\lambda\ . $$ \noindent This integral is equal to the semiclassical density (2). Thus, (9) yields $$ \rho(x)-\rho_{sc}(x)\approx \rnt(x)\ ,\quad\text{with}\tag"(11)" $$ $$ \rnt(x)=-\frac{1}{4\pi|x|^2}\ \sum\limits_{\ell\ge 0}\ \frac{(2\ell+1)}{\Cal N_\ell}\chm(\phi_\ell)\cdot \bigl(-V_\ell(|x|)\bigr)_+ ^{-1/2}\ .\tag"(12)" $$ \noindent So the basic estimate (3) amounts to saying that $$ \int_{\Bbb R^3}\int_{\Bbb R^3}\rnt(x)\rnt(y)\ \frac{dxdy} {|x-y|}\le C\, Z^{\frac 53-a}\ (a>0)\ .\tag"(13)" $$ \noindent when $V$ is the Thomas-Fermi potential for atomic number $Z$. Let us see what is needed to prove (13). A glance at the definition (8) shows that $|\chm(t)|\le \frac 12$ for any $t$. Hence we have the trivial estimate $$ |\rnt(x)|\le \rho^+(x)\equiv \frac{1}{4\pi|x|^2}\ \sum\limits_{\ell\ge 0}\ \frac{(2\ell+1)}{\Cal N_\ell}\bigl(-V_\ell(|x|)\bigr)_+^{-1/2}\ . $$ \noindent One computes easily that $\int_{\Bbb R^3}\int_{\Bbb R^3} \rho^+(x)\rho^+(y)\frac{dxdy}{|x-y|}$ has the order of magnitude $Z^{5/3}$. Thus, (13) means that significant cancellation occurs in the sum (12). Such cancellation occurs if the $\phi_\ell$ are equidistributed modulo $1$, since $\chm(t)$ has period $1$ and mean zero. In this paper, we make assumptions on the equidistribution of the $\phi_\ell$ modulo $1$. These assumptions will be proven in [FS6,7]. To state our assumptions on the $\phi_\ell$, we introduce $\ell_{\text{max}}$, the largest angular momentum for which $V_\ell(r)$ is negative somewhere. (The order of magnitude of $\ell_{\text{max}}$ is $Z^{1/3}$.) Our assumptions on the $\phi_\ell$ are as follows. \smallskip \roster \item"{(A)}" There are at most $CZ^{-\frac 13-2a}$ integers \ $\ell\le \ell_{\text{max}}$ for which\hfill\break $|\phi_\ell-(\text{nearest\ integer})|\le \ell^{-6/43}$\ . \smallskip \item"{(B)}" For $Z^{(10^{-9})}\le \ell_1<\ell_2\le \ell_{\text{max}}$, with $\ell_2-\ell_1>\ell_{\text{max}}^{1-10a}$\ , we have\hfill\break $\Big|\sum\limits_{\ell_1\le\ell\le \ell_2}\frac{(2\ell+1)}{\Cal N_\ell} \chm(\phi_\ell)\Big|\le C\, Z^{-\frac 23 a}\sum\limits_{\ell_1\le \ell\le \ell_2}\frac{(2\ell+1)}{\Cal N_\ell}$. \endroster \smallskip \noindent Here, $0\le a<1/43$. The indices here are rather arbitrary, but at least (A) and (B) express the equidistribution of the $\phi_\ell$ modulo $1$. The main result of this paper is that (A) and (B) imply the estimate $$ \int_{\Bbb R^3}\int_{\Bbb R^3}\bigl(\rho(x)-\rho_{sc}(x)\bigr) \bigl(\rho(y)-\rho_{sc}(y)\bigr)\frac{dxdy}{|x-y|}\le C\, Z^{\frac 53-\ \frac 23 a}\ . $$ \noindent In [FS6,7], we show that (A) and (B) hold for some positive $a$. Although we forgo a serious discussion of (A) and (B) here, we should point out that they are intimately connected with the scarcity of periodic, zero energy orbits for the Thomas-Fermi potential $V_{TF}$. To illustrate this point, consider the harmonic oscillator $V_1(x) =A|x|^2-B$ or the hydrogenic potential $V_2(x)=E-\frac{Z}{|x|}$ on $\Bbb R^3$. These potentials scale like the Thomas-Fermi potential if we take $A\sim Z^2$, $B\sim Z^{4/3}$, $E\sim Z^{4/3}$. Of course, $V_1$ and $V_2$ are classical examples of potentials with many periodic orbits. One checks easily that (3), (13) and (B) all fail for $V_1$ and $V_2$. Next, we give a layman's explanation of the Thomas-Fermi potential. As a simplified model of an atom, we imagine a nucleus of charge $Z$ fixed at the origin, and an electron cloud with particle density $\rho(x)$ on $\Bbb R^3$. Each electron in the cloud feels the attraction of the nucleus and the repulsion of the electron cloud. Therefore, each electron behaves as if it were governed by the Hamiltonian $H=-\Delta+V$, with $$ V(x)=-\frac{Z}{|x|}+\int_{\Bbb R^3}\frac{\rho(y)dy}{|x-y|}\ .\tag"(14)" $$ \noindent On the other hand, the density of electrons governed by the Hamiltonian $H$ is given by (1). In the semiclassical approximation, therefore, $$ \rho(x)=(\text{const)}\bigl(-V(x)\bigr)^{3/2}\ .\tag"(15)" $$ \noindent The solution of equations (14), (15) is given by the {\sl Thomas-Fermi density\/} $\rho_{{}_{TF}}(x)$ and the {\sl Thomas-Fermi potential\/} $V_{TF}(x)$. These functions are radially symmetric, and (14), (15) reduce to an ordinary differential equation. In fact, one finds that $$ V_{TF}(x)=cZ^{4/3}V_1(cZ^{1/3}|x|)\quad\text{for\ a\ constant}\ c\ ,\ \text{with}\tag"(16)" $$ $$ V_1(t)=t^{-1}y(t)\ ,\tag"(17)" $$ \noindent and $y(t)$ defined as the solution of the {\sl Thomas-Fermi equation} $$ \frac{d^2}{dt^2}\ y(t)=t^{-1/2}\bigl(y(t)\bigr)^{3/2}\ \text{on}\ (0,\infty)\ ;\ \ y(0)=1\ ,\ \ y(\infty)=0\ .\tag"(18)" $$ \noindent From (16)$\ldots$(18), one can read off the basic properties of $V_{TF}$. For instance, the order of magnitude of $V_{TF}$ is given by $$ -V_{TF}(x)\sim \min\big\{Z|x|^{-1},|x|^{-4}\}\quad(x\in \Bbb R^3)\ . \tag"(19)" $$ \noindent A detailed discussion of Thomas-Fermi theory is given by Lieb [L]. We close this introduction by mentioning several open problems on the density (1). The first natural problem is to estimate $\rho-\rho_{sc}$ for the Thomas-Fermi potential of a molecule. The most we can hope for here is probably $$ \int_{\Bbb R^3}\int_{\Bbb R^3} \bigl(\rho(x)-\rho_{sc}(x)\bigr) \bigl(\rho(y)-\rho_{sc}(y)\bigr)\frac{dxdy}{|x-y|}=o(Z^{5/3}) \tag"(20)" $$ \noindent in place of (3). The proof of (20) ought to involve wave-equation methods since it brings in the aperiodicity of the Hamiltonian flow. The next problem concerns formula (11), whose precise meaning is as follows. $$ \rho(x)-\rho_{sc}(x)=\rnt(x)+g(x)\ ,\quad\text{with}\tag"(21)" $$ $$ \int\limits_{\Bbb R^3\times \Bbb R^3}\int\frac{g(x)g(y)}{|x-y|}dxdy\le CZ^{5/3-\gamma}\quad(\gamma>0)\ .\tag"(22)" $$ \noindent This is enough to accomplish the purpose of (11), namely the reduction of (3) to (13). However, we believe that (22) can be sharpened considerably, by taking $\gamma$ larger. If $\gamma$ is large enough, then (21) and (22) identify $\rnt$ as the leading correction to the semiclassical density for a given potential $V$. We would like to understand the function $\rnt$. This brings in analytic number theory. For instance, the computation of $\int_{\Bbb R^3}\rnt(x)dx=\sum\limits_{\ell\le \ell_{\text{max}}} (2\ell+1)\chm(\phi_\ell)$ seems very close to that of $\sum\limits_{\ell\le \ell_{\text{max}}}\chm(\phi_\ell)$, which in turn is clearly equivalent to counting the lattice points in the plane domain $\{(\lambda,\xi)\colon 0\le \xi<\Phi(\lambda)\}$, with $$ \Phi(\lambda)=\frac 1\pi\int_0^\infty\bigl(-\frac{\lambda(\lambda+1)} {r^2}-V_{TF}(r)\bigr)_+^{1/2}dr-\frac 12\ . $$ \noindent Our current understanding of lattice-point problems appears to be inadequate to answer basic questions about the size and fluctuations of $\rnt$. Finally, we point out that if $\rnt$ is indeed the leading correction to the semiclassical density, then we can write a corrected equation in place of (15). Combining that new equation with (14) and solving by perturbation theory, we hope to find the leading correction to the Thomas-Fermi density. This would suggest that the density of electrons in a non-relativistic atom has a number-theoretic character. We are grateful to Maureen Schupsky for expertly texing our paper. \vfill\eject \head Review of Earlier results\endhead \medskip In this section, we recall the main results from our previous paper [FS4] on one-dimensional potentials. \medskip \noindent{\bf{A. The WKB Density Theorem in One Dimension}} \noindent{\it{Set-Up\/}}: We are given positive numbers $\varepsilon$, $K$, $N$, $\hat c$; two intervals $I\subset I_{\text{BVP}}$ (possibly unbounded); a point $x_0\in I$; a potential $V(x)$ defined on $I_{\text{BVP}}$; and two positive functions $S(x)$, $B(x)$ defined on $I$. Our assumptions are as follows. \subhead Assumptions Concerning $V(x)$, $S(x)$, $B(x)$ on $I$\endsubhead \smallskip \roster \item"(Z0)" If $x,y\in I$ and $|x-y|cB(x_{\text{left}})$ and $\text{dist}\,(x_{\text{rt}},\partial I)>cB(x_{\text{rt}})$. \item"(Z3)" We have $V(x_0)<-cS(x_0)$, $V^\prime(x_0)=0$; and for $|x-x_0|\le c_1B(x_0)$ we have $V^{\prime\prime}(x)\ge cS(x_0)B^{-2} (x_0)$. \item"(Z4)" For $x_{\text{left}}\le x\le x_0-c_1B(x_0)$ we have $-V^\prime(x)>cS(x)B^{-1}(x)$; and for $x_0+c_1B(x_0)\le x\le x_{\text{rt}}$ we have $+V^\prime(x)>cS(x)B^{-1}(x)$. \endroster \noindent Define $\lambda(x)=S^{1/2}(x)B(x)$ for $x\in I$, and set $$ \Lambda=\Bigl(\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx} {\lambda(x)B(x)}\Bigr)^{-1}\ . $$ \subhead Assumptions Concerning $V(x)$ on all of $I_{\text{BVP}}$\endsubhead \smallskip \roster \item"(Z5)" We have $V(x)>0$ for all $x\in I_{\text{BVP}}\backslash [x_{\text{left}},x_{\text{rt}}]$. \item"(Z6)" For all $x\in I_{\text{BVP}}$ with $xx_{\text{rt}}+\Lambda^KB(x_{\text{rt}})$, we have $V(x)\ge \frac{1000} {|x-x_{\text{rt}}|^2}$. \endroster \smallskip \subhead Polynomial Growth Assumptions on $S(x)$, $B(x)$, $I$\endsubhead \smallskip \roster \item"(Z7)" We have $\max_{x\in I}B(x)<\Lambda^K\min_{x\in I}B(x)$; $\max_{x\in I}S(x)<\Lambda^K\min_{x\in I}S(x)$; and $|I|<\Lambda^K\cdot \min_{x\in I}B(x)$. \endroster \smallskip \subhead Smallness of the Constant $\hat c$\endsubhead \smallskip \roster \item"(Z8)" The constant $\hat c$ is bounded above by a certain small, positive number determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $C_\alpha$.\endroster \smallskip \subhead The WKB Hypothesis\endsubhead \smallskip \roster \item"(Z9)" $\Lambda$ is bounded below by a certain large, positive number determined by $\varepsilon$, $K$, $N$, $c$, $C$, $c_1$, $\hat c$, $C_\alpha$. \endroster Let $E_k$ and $u_k(x)$ be the eigenvalues and (normalized) eigenfunctions of $-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{BVP}}$, with Dirichlet or Neumann boundary conditions. Then define the density $\rho(x)$ and its refined semiclassical approximation $\tilde\rho_{sc}(x)$ on $I_{\text{BVP}}$ by setting: $$\align \rho(x)&=\sum\limits_{E_k\le 0}|u_k(x)|^2\quad (x\in I_{\text{BVP}})\ ;\\ \tilde\rho_{sc}(x)&=\frac 1\pi \bigl(-V(x)\bigr)_+^{1/2}- \frac{\bigl(-V(x)\bigr)_+^{-1/2}}{\int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{-1/2} dy}\chm\Bigl(\frac 1\pi \int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{1/2}dy-\frac 12\Bigr)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad \qquad\qquad\qquad(x\in I_{\text{BVP}})\ . \endalign $$ \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 $$ H(x)=\int_{I_{\text{BVP}}\cap (-\infty,x]}\ (\rho(\overline x)-\tilde\rho_{sc} (\overline x))d\overline x \ . $$ \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 in this paper, and surely not optimal. Any exponent strictly less than $-1$ would serve our purpose.\enddemo \vglue 1pc \noindent{\bf{B. The Density for Degenerate One-Dimensional Potentials I}} 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 \noindent{\bf{Hypotheses}} \roster \item"(Z0$^\ast$)" $I=\{x\colon |x-x_0|0$. \item"(Z5$^\ast$)" For $x\in I_{\text{BVP}}$ with $|x-x_0|>\frac 12\lambda^KB$, we have $V(x)\ge \frac{1000}{|x-x_0|^2}$. \item"(Z6$^\ast$)" $\lambda$ is bounded below by a certain large, positive number determined by $\varepsilon$, $K$, $N$, $c$, $c^\prime$, $C_{\alpha}$. \endroster \noindent Let $E_k$, $u_k(x)$ be the eigenvalues and (normalized) eigenfunctions for $-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{BVP}}$, with Dirichlet or Neumann boundary conditions. As in the previous section, define the density $\rho(x)$ and its refined semiclassical approximation $\tilde\rho_{sc}(x)$ on $I_{\text{BVP}}$, by the formulas $$\align \rho(x)&=\sum\limits_{E_k\le 0}|u_k(x)|^2\\ \tilde\rho_{sc}(x)&=\frac 1\pi\bigl(-V(x)\bigr)_+^{1/2}- \frac{\bigl(V(x)\bigr)_+^{-1/2}} {\int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{-1/2}dy}\ \chm\bigl(\frac 1\pi \int_{I_{\text{BVP}}}\bigl(-V(y)\bigr)_+^{1/2}dy-\frac 12\bigr)\endalign $$ \noindent Then set $$ H(x)=\int_{I_{\text{BVP}}\cap (-\infty,x]}\bigl(\rho(\overline x)- \tilde\rho_{sc}(\overline x)\bigr)\, d\overline x \ . $$ \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 $I$. \vglue 1pc \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 \vglue 1pc \demo{Remark} Again, the precise exponents in Case I (B), (C) are irrelevant to us, and are surely not optimal.\enddemo \vglue 1pc \noindent{\bf{C. The Density for Degenerate One-dimensional Potentials II}} In this section we give (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 $1$. \noindent{\bf{Assumptions}}\smallskip \roster \item"(Z$\overline 0$)" For $x,y \in I$ with $|x-y|cB(x)$. \item"(Z$\overline 1$)" For $x\in I$ and $\alpha\ge 0$ we have $\big|\bigl(\frac{d}{dx}\bigr)^\alpha V(x)\big|\le C_\alpha S(x)B^{-\alpha}(x)$. \item"(Z$\overline 2$)" For $E_{\text{crit}}\le E\le 0$, the set $\{x\in I_{\text{BVP}}\mid V(x)\le E\}$ is a non-empty interval $(x_{\text{left}}(E), x_{\text{rt}}(E))$ contained in $I$, with $\text{dist}(x_{\text{left}}(E),\partial I)>cB(x_{\text{left}}(E))$ and $\text{dist}\,(x_{\text{rt}}(E),\partial I)>cB(x_{\text{rt}} (E))$. \item"(Z$\overline 3$)" For $E_{\text{crit}}\le E\le 0$, we have $-V^\prime(x)\ge cS(x)B^{-1}(x)$ for \hfill\break $x\in [x_{\text{left}}(E), x_{\text{left}}(E)+c_1B(x_{\text{left}}(E))]$ and $+V^\prime(x)\ge cS(x)B^{-1}(x)$ for $x\in [x_{\text{rt}}(E)- c_1B(x_{\text{rt}}(E)), x_{\text{rt}}(E)]$. \item"(Z$\overline 4$)" For $E_{\text{crit}}\le E\le 0$, we have $cS(x)cS(x)B^{-1} (x)$. \item"(Z$\hat 3$)" $\Lambda=\bigl(\int_I\frac{dx}{\lambda(x)B(x)}\bigr)^{-1}$ is greater than a certain large, positive number determined by $c$, $C$, $C_\alpha$ in (Z$\hat 0$)$\ldots$(Z$\hat 2$). \item"(Z$\hat 4$)" For $x\in (0,x_{\text{small}}]$ we have $V(x)\ge \underline cx_0^{-2}$ \item"(Z$\hat 5$)" For $x\in [x_{\text{small}},x_0]$ we have $|V(x)| \le \underline Cx_0^{-2}$ \item"(Z$\hat 6$)" We have $x_{\text{big}}<\underline Cx_1$ and $V(x)$ is increasing in $[x_1,x_{\text{big}}]$. \item"(Z$\hat 7$)" For $x\in \bigl[\frac{x_1}{8},x_{\text{big}}]$, we have $|V(x)|\le \underline Cx_1^{-2}$. \item"(Z$\hat 8$)" For $x\in [x_{\text{big}},\infty)$, we have $V(x)\ge 0$. \item"(Z$\hat 9$)" For $E\in [V(x_\ast),0]$ we have\hfill\break $\int_{x_0}^{x_{\text{crit}}}\bigl(E-V(x)\bigr)^{-1/2}dx\le \delta\cdot\int_{x_0}^{\frac 12 x_\ast}\bigl(E-V(x)\bigr)^{-1/2}dx$. \endroster When $\delta<<1$ hypothesis (Z$\hat 9$) shows that in the semiclassical approximation, most of the $L^2$-norm of an eigenfunction $u_k(x)$ with $E_k\in [V(x_\ast),0]$ is concentrated outside of $(0,x_{\text{crit}}]$. \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 \vglue 1pc \noindent{\bf{E. The Density for Degenerate One-Dimensional Potentials IV}} 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 $$ 16x_0cS(x)B^{-1} (x)$. \item"(Z3$^\dag$)" $\Lambda=\bigl(\int_I\frac{dx}{\lambda(x)B(x)}\bigr)^{-1}$ is greater than a certain large, positive number determined by $c$, $C$, $C_\alpha$ in (Z0$^\dag$)$\ldots$(Z2$^\dag$). \item"(Z4$^\dag$)" $|V(x)|\le \underline C/(x_0x)$ for $x\in (0,x_0]$. \item"(Z5$^\dag$)" $V(x)$ is increasing and negative in $[\frac{x_1}{8}, x_{\text{big}}]$, and satisfies there $|V(x)|<\underline{C} x_1^{-2}$. Also, $x_{\text{big}}<\underline Cx_1$. \item"(Z6$^\dag$)" $V(x)\ge -10^{-9}x^{-2}$ for $x\in [x_{\text{big}},\infty)$. \item"(Z7$^\dag$)" For $E\in [V(x_\ast),0]$, we have\hfill\break $\int_{x_0}^{x_{\text{crit}}}\bigl(E-V(x)\bigr)^{-1/2}dx\le \delta \cdot \int_{x_0}^{\frac 12 x_\ast}\bigl(E-V(x)\bigr)^{-1/2}dx$. \endroster \noindent As usual, (Z7$^\dag$) says that certain eigenfunctions live mostly outside of $[x_0,x_{\text{crit}}]$ in the semiclassical approximation. \vglue 1pc \proclaim{Fourth Degenerate Density Lemma} Assume (1) and (Z0$^\dag$)$\ldots$(Z7$^\dag$). Set\hfill\break $E_{\text{crit}}=V(x_\ast)$. Then $$ \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 $$ \noindent and $\int_0^\infty\rho(x)dx\le C_\ast+C_\ast\int_0^\infty \bigl(-V(x)\bigr)_+^{1/2}dx$ with $C_\ast$ depending only on $c$, $C$, $C_\alpha$, $\underline C$ in (Z0$^\dag$)$\ldots$(Z7$^\dag$).\endproclaim \vglue 1pc \noindent{\bf{F. Approximating Sums by Integrals}} Separation of variables leads to a sum over all angular momenta $\ell$. The following result lets us approximate such sums by integrals. For real numbers $t$, define: $\chp(t)=k-t-\frac 12$ for $k$ the smallest integer $\ge t$; $\chm(t)=t-k-\frac 12$ for $k$ the largest integer $\le t$; $\tilde\chi(t)=|t-k-\frac 12|^2-\frac{1}{12}$ for an integer $k$ that minimizes $|t-k-\frac 12|$. \vglue 1pc \proclaim{Lemma on Riemann Sums} Let $f(t)$, $\sigma(t)$, $\tau(t)$ be defined on a non--empty interval $[a,b]$. Suppose $\sigma(t)>0$, $\tau(t)\ge 1$ in $[a,b]$; and assume that whenever $t_1,t_2\in [a,b]$ with $|t_1-t_2|0\ ,\ w\in S^2\ . $$ \noindent Since $\sum\limits_m|Y_{\ell m}(w)|^2=\frac{2\ell+1} {4\pi}$ on $S^2$, it follows that $$\text{sneg}(H)= \sum\limits_{\ell\ge 0}(2\ell+1)\ \text{sneg}(H_\ell)\ ,\quad\text{and} $$ $$ 4\pi r^2\rho(rw)=\sum\limits_{\ell\ge 0}(2\ell+1)\rho_\ell(r)\quad \text{for} \quad r>0\ ,\ w\in S^2\ . $$ \noindent We will control $\text{sneg}(H_\ell)$ and $\rho_\ell(r)$ by using our ODE results, and then perform the sum over $\ell$. \vfill\eject \head Elementary Properties of the Thomas-Fermi Potential\endhead \medskip The potentials of interest to us arise by rescaling the potential $$ V_\Omega(x)=\frac\Omega{x^2}-\frac{y(x)}{x}\, \ x\in (0,\infty)\ ,\ \Omega\in [0,\infty)\ ,\ \tag"(1)" $$ \noindent where $y(x)$ is the solution of the Thomas-Fermi differential equation $$ \frac{d^2}{dx^2}y(x)=y^{3/2}(x)\cdot x^{-1/2}\ ,\ y(0)=1\ ,\ y(\infty)=0\ .\tag"(2)" $$ \noindent The basic properties of (2) are well-known. In particular, Hille [Hi] contains the following results. \vglue 1pc \proclaim{Lemma 1} $y(x)$ is smooth and strictly between $0$ and $1$ for $x\in (0,\infty)$, while $y^\prime(x)$ is bounded and negative on $(0,\infty)$. For $x$ small, $y(x)$ has a convergent series expansion of the form $$ y(x)=1-w x+\sum\limits_{k\ge 3}w_kx^{k/2}\ ,\text {with}\ w>0\ \text{and}\ w_k\ \text{real}\ . \tag"(3)" $$ \noindent For $x$ large, $y(x)$ has a convergent series expansion $$ y(x)=144\ x^{-3}\bigl(1+\sum\limits_{k\ge 1}b_kx^{-k\gamma})\quad \text{with}\ \gamma=\frac 12 (\sqrt{73}-7)\ .\tag"(4)" $$\endproclaim \bigskip We seldom need (3) and (4) in full strength. Instead, we will just use the following \vglue 1pc \proclaim{Corollary} Given $\eta>0$, there exist constants $00}\bigl(xy(x)\bigr)>0$. Thus, $V_\Omega\ge 0$ for $\Omega\ge \overline\Omega$. \vglue 1pc \proclaim{Lemma 2} Let $\Omega \in (0,\overline\Omega)$. Then $V_\Omega$ has exactly two zeros $x_1(\Omega)$ and $x_2(\Omega)$, and two critical points $x_0(\Omega)$, $x_m(\Omega)$, and these points may be taken to satisfy $$00$, there exist positive constants $c_i(\varepsilon)$ $(0\le i\le 5)$ with the following properties: For $\Omega\in (\varepsilon,\overline\Omega)$, we have $$ V_\Omega^{\prime\prime}\bigl(x_0(\Omega)\bigr)>c_0(\varepsilon) \tag"(8)" $$ $$ \Big|\bigl(\frac{d}{dx}\bigr)^kV_\Omega(x)\Big|c_4(\varepsilon)\ \text{for}\ x\in \bigl[x_0(\Omega) +c_5(\varepsilon)\ ,\frac{x_2(\Omega)+x_m(\Omega)}{2}\bigr] \tag"(11)" $$ $$\multline \text{For}\ |x-x_0(\Omega)|\le c_5(\varepsilon)\ ,\text{we\ have}\\ \frac{9}{20}V_\Omega^{\prime\prime}\bigl(x_0(\Omega)\bigr)\cdot \bigl(x-x_0(\Omega)\bigr)^2\le V_\Omega(x)-V_\Omega\bigl(x_0(\Omega)\bigr)\\ \le \frac{11}{20} V_\Omega^{\prime\prime}\bigl(x_0(\Omega)\bigr)\cdot \bigl(x-x_0(\Omega)\bigr)^2\ .\endmultline\tag"(12)" $$\endproclaim \medskip \noindent Immediately from Lemmas 1, 2, 3, we see that $V_\Omega^\prime (x)<0$ for $x0$. Therefore, $V_\Omega$ is a positive, smooth function on $I=(x_2(\Omega),\infty)$, and $V_\Omega$ tends to zero at the endpoints of $I$. Consequently, $V_\Omega\mid_I$ assumes a maximum at some point $x_c(\Omega)\in I$. In particular, $V_\Omega^{\prime\prime}(x_c(\Omega))\le 0$, $V_\Omega^\prime(x_c(\Omega))=0$, $x_c(\Omega)>x_2(\Omega)$. These last two conditions and Lemma 2 show that $x_c(\Omega)=x_m(\Omega)$. Thus, $V_\Omega^{\prime\prime}(x_m(\Omega))\le 0$, as asserted. Unfortunately, we will need to use a long list of additional elementary properties of $V_\Omega(x)$. These properties are surely well-known to anyone interested in atoms, but we don't know where to find them in the literature. Hence, we will prove them here, as consequences of Lemmas 1, 2, 3. Specifically, we will need the following result. \vglue 1pc \proclaim{Main Lemma on the Thomas-Fermi Potential} Let $S(x)=\min\{x^{-1} ,x^{-4}\}$. There exist universal constants $\overline c$, $c_1$, $c_2$, $\hat c$, $\hat C$, $C_\alpha>0$ for which the following properties hold. \noindent {\bf I.}\ Suppose $0<\Omega\le (1-\overline c)\overline\Omega$. Then we have the following. \smallskip {\bf A.} The Size and Sign of $V_\Omega(x)$ $$ \hat c\Omega x^{-2}\hat c\Omega\tag"(24)" $$ $$ \hat c\Omega^{-1/2}\hat c\Omega^{-1/2}\tag"(25)" $$ $$\split x_m(\Omega)>(1+2c_1)x_2(\Omega)\quad \text{and}\quad x_1(\Omega)<(1-c_1)x_0(\Omega)\\ \text{and}\quad x_0(\Omega)<(1-2c_1)x_2(\Omega)\ .\endsplit\tag"(26)" $$ \bigskip \noindent {\bf II.} Suppose $(1-\overline c)\overline\Omega\le \Omega< \overline\Omega$. Then we have the following. $$\split \Big|\bigl(\frac{d}{dx}\bigr)^\alpha V_\Omega(x)\Big|\le C_\alpha S(x)x^{-\alpha}\quad \text{and}\quad \hat cS(x)x^{-2}\hat c\Omega x^{-2}\quad\text{for}\quad x\not\in \bigl[ (1-c_2)x_0(\Omega), (1+c_2)x_0(\Omega)\bigr]\ . \tag"(28)" $$ $$ \hat c(\overline\Omega-\Omega)<-V_\Omega\bigl(x_0(\Omega)\bigr)< \hat C(\overline\Omega-\Omega)\ . \tag"(29)" $$\endproclaim \medskip The rest of this section is devoted to proving the above Main Lemma. We look separately at the three regions $0<\Omega<\varepsilon$, $\varepsilon\le \Omega\le \overline\Omega-\varepsilon$, $\overline\Omega -\varepsilon\le \Omega<\overline\Omega$, for a small, positive $\varepsilon$ to be picked later. We begin by proving the following preliminary result. \vglue 1pc \proclaim{Lemma 4} The functions $x_1(\Omega)$, $x_2(\Omega)$, $x_0(\Omega)$ are smooth on $(0,\overline\Omega)$, and $x_m(\Omega)$ is continuous on $(0,\overline\Omega)$.\endproclaim \vglue 1pc \demo{Proof} To handle $x_0(\Omega)$, $x_1(\Omega)$, $x_2(\Omega)$, we need only apply the implicit function theorem, since $V_\Omega(x)$ is smooth on $(0,\infty)\times (0,\infty)$ and we know that $V_\Omega\bigl(x_1(\Omega)\bigr)=0$, $V_\Omega^\prime\bigl(x_1(\Omega)\bigr)<0$; $V_\Omega\bigl(x_2(\Omega)\bigr) =0$, $V_\Omega^\prime\bigl(x_2(\Omega)\bigr)>0$; $V_\Omega^\prime \bigl(x_0(\Omega)\bigr)=0$, $V_{\Omega}^{\prime\prime}\bigl(x_0(\Omega)\bigr) >0$ by Lemmas 2 and 3. For $x_m(\Omega)$ we need a different argument, because Lemmas 2 and 3 do not assert that $V_\Omega^{\prime\prime} \bigl(x_m(\Omega)\bigr)$ is strictly negative. We proceed as follows. We know from (6) that $\big|\frac{d}{dx}\bigl(y(x)x^{-1}\bigr)\big|\le Cx^{-5}$ for $x\ge C$, where $C>0$ is a large, universal constant. Hence, $V_\Omega^\prime(x)\le -2\Omega x^{-3}+Cx^{-5}<0$ for $x>\ \text{max}\bigl\{C,\bigl(\frac{C}{2\Omega}\bigr)^{1/2}\bigr\}$. Since $V_\Omega^\prime\bigl(x_m(\Omega)\bigr)=0$, it follows that $$ x_m(\Omega)\le\ \text{max}\bigl\{C,\bigl(\frac{C}{2\Omega}\bigr)^{1/2}\bigr\} \quad\text{for}\ \Omega\in (0,\overline\Omega)\ .\tag"(30)" $$ Now we suppose $x_m(\Omega)$ discontinuous and derive a contradiction. If $x_m(\Omega)$ is discontinuous, then we can find $\Omega_\nu\to \hat \Omega$ with $\hat\Omega\in (0,\infty)$ and $x_m(\Omega_\nu)\nrightarrow x_m(\hat\Omega)$. The $x_m(\Omega_\nu)$ are positive, and (30) shows that they remain bounded. Hence, by passing to a subsequence, we may assume $x_m(\Omega_\nu) \to \hat x$ with $\hat x\not= x_m(\hat \Omega)$. Since $x_m(\Omega_\nu) >x_2(\Omega_\nu)$ and $x_2(\Omega_\nu)\to x_2(\hat\Omega)>0$ we know that $\hat x$ is strictly positive. Noting that $V_{\Omega_\nu}^\prime \bigl(x_m(\Omega_\nu)\bigr)=0$ and passing to the limit, we see that $V_{\hat\Omega}^\prime(\hat x)=0$. Therefore, $\hat x=x_m(\hat\Omega)$ or $\hat x=x_0(\hat\Omega)$, by Lemma 2. We know that $\hat x\not= x_m(\hat\Omega)$, so $\hat x=x_0(\hat\Omega)$. Hence, $V_{\hat\Omega}^{\prime\prime}(\hat x)>0$, by Lemma 3. On the other hand, we have $V_{\Omega_\nu}^{\prime\prime}\bigl(x_m(\Omega_\nu)\bigr) \le 0$. Passing to the limit, we find that $V_{\hat\Omega}^{\prime\prime} (\hat x)\le 0$. Thus, $V_{\hat\Omega}^{\prime\prime} (\hat x)>0$ and $V_{\hat\Omega}^{\prime\prime}(\hat x)\le 0$. This contradiction completes the proof.\qed\enddemo \vglue 1pc \proclaim{Lemma 5} Given $\varepsilon>0$ there exists $\delta>0$ such that for every $c_1\in (0,\delta)$ there exist $\hat c$, $\hat C$, $C_\alpha$, depending only on $\varepsilon$ and $c_1$, such that (13),$\ldots$,(26) hold for all $\Omega\in [\varepsilon,\overline\Omega-\varepsilon]$. \endproclaim \vglue 1pc \demo{Proof} Let $c_A(\varepsilon)$, $c_B(\varepsilon)$, etc.\ denote positive constants depending only on $\varepsilon$. Similarly, let $c_A(\varepsilon,c_1)$, $c_B(\varepsilon,c_1)$ etc.\ denote positive constants depending only on $\varepsilon$ and $c_1$.\medskip\enddemo Lemma 4 and (7) show that we can find $c_A(\varepsilon)$, $C_B(\varepsilon)$ so that the following properties hold. $$ c_A(\varepsilon)(1+2c_A(\varepsilon))x_2(\Omega)\quad\text{and}\\ x_0(\Omega)<(1-2c_A(\varepsilon))x_2(\Omega)\ \text{and}\\ x_1(\Omega)<(1-c_A(\varepsilon))x_0(\Omega)\quad\text{for}\ \Omega\in [\varepsilon,\overline\Omega-\varepsilon]\ .\endmultline\tag"(32)" $$ \noindent Next, note that $-V_\Omega^\prime(x)\bigl[S(x)x^{-1}\bigr]^{-1}$ is continuous on $F=\{(\Omega,x)\colon\Omega\in[\varepsilon,\overline \Omega-\varepsilon]$, $\frac 12 c_A(\varepsilon)\le x\le 2C_B(\varepsilon) \}$ and strictly positive for $x=x_1(\Omega)$. Therefore, for suitable constants $c_D(\varepsilon)$, $C_E(\varepsilon)$, $c_F(\varepsilon)$, we have $$\multline c_D(\varepsilon)S(x)x^{-1}<-V_\Omega^\prime(x)0$, $x_2(\Omega)>0$ are continuous, we can find $c_M(\varepsilon,c_1)<\varepsilon$ such that $\Omega\in [\varepsilon,\overline\Omega-\varepsilon]$, $|\tilde\Omega-\Omega|\le c_M(\varepsilon,c_1)$ imply $|x_i(\Omega)- x_i(\tilde\Omega)|\le c_1x_i(\Omega)$ $(i=1,2)$, so that $[x_1(\Omega)\cdot (1-c_1),x_2(\Omega)\cdot (1+c_1)]\supset [x_1(\tilde\Omega),x_2(\tilde\Omega)]$. Set $\tilde\Omega=\Omega-c_M(\varepsilon,c_1)$. Since $V_\Omega(x)= c_M(\varepsilon,c_1)x^{-2}+V_{\tilde\Omega}(x)$ and $V_{\tilde\Omega}(x)>0$ outside $[x_1(\tilde\Omega),x_2(\tilde\Omega)]$, it follows that $$ V_\Omega(x)\ge c_M(\varepsilon,c_1)x^{-2}\quad \text{for}\ x\le (1-c_1)x_1(\Omega)\ ,\ \Omega\in [\varepsilon,\overline \Omega-\varepsilon]\ ,\text{and}\tag"(37)" $$ $$ V_\Omega(x)\ge c_M(\varepsilon,c_1)x^{-2}\quad \text{for}\ x\ge (1+c_1)x_2(\Omega)\ ,\ \Omega\in [\varepsilon, \overline\Omega-\varepsilon] \ . \tag"(38)" $$ \noindent Also, for all $\Omega$, $x$, we have $$ V_\Omega(x)=\Omega x^{-2}-y(x)x^{-1}<\Omega x^{-2}\ .\tag"(39)" $$ \noindent Equations (37), (38), (39) show that properties (13) and (19) hold for $\hat c$ small enough and $\hat C$ large enough, depending on $c_1$ and $\varepsilon$. Next, note that $-V_\Omega(x)S^{-1}(x)$ is strictly positive and continuous on $\{(\Omega,x)\colon\Omega\in [\varepsilon,\overline\Omega-\varepsilon]$, $(1+c_1)x_1(\Omega)\le x\le (1-c_1)x_2(\Omega)\}$. Hence, $$\multline c_N(\varepsilon,c_1)S(x)<-V_\Omega(x) (1-c_1)x_0(\Omega)$, so that $V_{\tilde\Omega}^\prime(x)<0$ for $x\le (1-c_1)x_0(\Omega)$. Since $V_\Omega(x)=c_P(\varepsilon,c_1)x^{-2} +V_{\tilde\Omega}(x)$, it follows that $$ V_\Omega^\prime(x)<-2c_P(\varepsilon,c_1)x^{-3}\quad\text{for} \quad x\le (1-c_1)x_0(\Omega)\ ,\ \Omega\in [\varepsilon,\overline\Omega -\varepsilon]\ .\tag"(40)" $$ \noindent On the other hand, for $x\le (1-c_1)x_0(\Omega)\le x_0(\Omega)\le C_B(\varepsilon)$, we have $$ |V_\Omega^\prime(x)|=|-2\Omega x^{-3}-y(x)x^{-2}+y^\prime(x)x^{-1}| \le C_Q(\varepsilon)x^{-3}\ .\tag"(41)" $$ \noindent Estimates (40) and (41) imply property (20) for $\Omega\in [\varepsilon,\overline\Omega-\varepsilon]$, if $\hat c$ is small enough and $\hat C$ large enough, depending on $c_1$ and $\varepsilon$. Property (21) follows at once from (35), (36), if $\hat c$ is small enough and $\hat C$ large enough, depending on $c_1$ and $\varepsilon$. Next, note that $V_\Omega^\prime(x)\cdot[S(x)x^{-1}]^{-1}$ is strictly positive and continuous on $\{(\Omega,x)\colon \Omega\hfill\break \in [\varepsilon, \overline\Omega-\varepsilon]$, $(1+c_1)x_0(\Omega)\le x\le (1+c_1)x_2(\Omega)\}$, since $\bigl(x_0(\Omega),(1+c_1)x_2(\Omega)\bigr] \break \subset\bigl(x_0(\Omega),x_m(\Omega)\bigr)$ by (32), (36). Therefore, $$\multline c_R(\varepsilon,c_1)S(x)x^{-1}K_2\ .\tag"(50)" $$ Then take $c_1=10^{-2}$. We will check that (13)$\ldots$(26) hold for suitable universal constants $\hat c$, $\hat C$, $C_\alpha$. We begin by locating $x_0(\Omega)$, $x_1(\Omega)$, $x_2(\Omega)$, $x_m(\Omega)$, for $0<\Omega\le \varepsilon_0$. Thus, fix $\Omega\in (0,\varepsilon_0]$. Set $f(t)=\Omega V_\Omega(\Omega t)=[t^{-2}-t^{-1}-(y(\Omega t)-1)t^{-1}]$. Applying (42) and (49), we see that $f(1-10^{-49})>0>f(1+10^{-49})$. Hence there is a point $\tilde x_1(\Omega)$ satisfying $$ |\tilde x_1(\Omega)-\Omega|<10^{-49}\Omega\quad\text{and}\quad V_\Omega(\tilde x_1(\Omega))=0\ .\tag"(51)" $$ Next, set $$ f(t)=4\Omega^2V_\Omega^\prime(2\Omega t)=-t^{-3} +t^{-2}+[y(2\Omega t)-1]t^{-2}-t^{-1}\frac{d}{dt}\{y(2\Omega t)-1\}\ . $$ \noindent Applying (42) and (49) again, we see that $f(1-10^{-49})<00>f (1+10^{-49})$. Hence there is a point $\tilde x_m(\Omega)$ satisfying $$ |\tilde x_m(\Omega)-12\sqrt 2\,\Omega^{-1/2}|<10^{-49}\cdot 12\sqrt 2 \,\Omega^{-1/2}\quad\text{and}\quad V_\Omega^\prime \bigl(\tilde x_m(\Omega)\bigr)=0\ . \tag"(54)" $$ \noindent Since $\Omega\le \varepsilon_0\le \frac{1}{10}$, equations (51)$\ldots$(54) show that $0<\tilde x_1(\Omega)<\tilde x_0(\Omega)< \tilde x_2(\Omega)<\tilde x_m(\Omega)$. Since $\tilde x_1(\Omega)$, $\tilde x_2(\Omega)$ are zeros of $V_\Omega$, and $\tilde x_0(\Omega)$ and $\tilde x_m(\Omega)$ are critical points, Lemma 2 shows that $\tilde x_1(\Omega)=x_1(\Omega)$, $\tilde x_2(\Omega)=x_2(\Omega)$, $\tilde x_0(\Omega)=x_0(\Omega)$, and $\tilde x_m(\Omega)=x_m(\Omega)$. Now we can verify properties (13)$\ldots$(26). In view of (51) and (53), we have $[x_1(\tilde\Omega),x_2(\tilde\Omega)]\subset ((1-10^{-2})x_1(\Omega),(1+10^{-2})x_2(\Omega))$ for $\tilde\Omega=(1-10^{-5})\Omega$. Since $V_\Omega(x)=10^{-5}\Omega x^{-2}+V_{\tilde\Omega}(x)$ and $V_{\tilde\Omega}(x)>0$ outside $[x_1(\tilde\Omega),x_2(\tilde\Omega)]$, we obtain $$ V_\Omega(x)>10^{-5}\Omega x^{-2}\quad\text{for}\quad x\le (1-10^{-2})x_1(\Omega)\ ,\quad\text{and}\tag"(55)" $$ $$ V_\Omega(x)>10^{-5}\Omega x^{-2}\quad\text{and}\quad x\ge (1+10^{-2}) x_2(\Omega)\ .\tag"(56)" $$ \noindent Properties (13) and (19) with $c_1=10^{-2}$ are immediate from (39), (55), (56). Similarly, $|x_0(\tilde\Omega)-x_0(\Omega)|<10^{-2}x_0(\Omega)$ for $\tilde\Omega=(1-10^{-5})\Omega$, and we have $V_\Omega^\prime(x)=-2\cdot 10^{-5}\Omega x^{-3}+V_{\tilde\Omega}^\prime(x)$, and $V_{\tilde\Omega}^\prime(x)<0$ for $x-2\Omega x^{-3}\ ,\tag"(58)" $$ \noindent since $y(x)$, $-y^\prime(x)>0$. Estimates (57), (58) yield property (20) with $c_1=10^{-2}$. They yield also $$ 10^{-6}\Omega^{-2}\le -V_\Omega^\prime(x)\le 10^6\Omega^{-2} \quad\text{for}\ \frac 12\Omega\le x\le \ \frac 32\Omega\ ,\tag"(59)" $$ \noindent since $\frac 32\Omega<(1-10^{-2})x_0(\Omega)$ by (52). Since $V_\Omega\bigl(x_1(\Omega)\bigr)=0$ and $10^{-6}\Omega^{-2}\le S\bigl(x_1(\Omega)\bigr)\bigl(x_1(\Omega)\bigr)^{-1}\le 10^6\Omega^{-2}$ by (51) and $\Omega\le \varepsilon_0\le \frac{1}{10}$, properties (14) and (15) with $c_1=10^{-2}$ follow from (59) and (51). Next, we verify property (16). We distinguish three cases. \medskip \noindent{\sl Case 1\/}: $(1+10^{-2})x_1(\Omega)\le x\le k_1$. \noindent Then (51) implies $$ (1+10^{-3})\Omega\le x\le k_1\ . $$ \noindent We have $V_\Omega(x)=(+\Omega x^{-2}-x^{-1})-x^{-1}\bigl(y (x)-1\bigr)<\Omega x^{-2}-x^{-1}+10^{-50}x^{-1}$ (by (42)) $=-x^{-1}\bigl(1-10^{-50}-\Omega x^{-1})<-x^{-1}(1-10^{-50}-\Omega\cdot [(1+ 10^{-3})\Omega]^{-1}\bigr)<-10^{-5}x^{-1}$. Thus, $$ V_\Omega(x)<-10^{-5}S(x)\ .\tag"(60)" $$ \noindent On the other hand, for all $x$ and $\Omega>0$ we have $$ V_\Omega(x)=\Omega x^{-2}-y(x)x^{-1}\ge -y(x)x^{-1}\ge -C\,S(x)\tag"(61)" $$ \noindent for a universal constant $C$, by Lemma 1. \noindent Property (16) in Case 1 follows from (60) and (61). \bigskip \noindent{\sl Case 2\/}: $k_1\le x\le K_2$. \noindent Then estimates (44) and (48) yield $$ -K_3S(x)x^{-4}\bigl\{144(1-10^{-50})-\Omega\cdot x^2\bigr\} >x^{-4}\bigl\{144(1-10^{-50})-\Omega\cdot (1-10^{-3})^2\cdot 144\, \Omega^{-1}\bigr\}=144\, x^{-4}\bigl\{1-10^{-50}-(1-10^{-3})^2\bigr\} >10^{-4}x^{-4}=10^{-4}S(x)$. Together with (61), this proves (16) in Case 3. The proof of property (16), with $c_1=10^{-2}$, is complete. Next, we verify property (22). The proof is similar to that of (16). Again we distinguish three cases. \medskip \noindent{\sl Case 1\/}: $(1+10^{-2})x_0(\Omega)\le x\le k_1$. \noindent Then (52) shows that $$ (1+10^{-3})\cdot 2\Omega\le x\le k_1\ . $$ \noindent We have $V_\Omega^\prime(x)=-2\Omega x^{-3}+x^{-2}+x^{-2}\{y(x)-1\} -x^{-1}y^\prime(x)$. The last two terms on the right are dominated by $10^{-50}x^{-2}$ by (42). Hence, $V_\Omega^\prime(x)\ge -2\Omega x^{-3} +(1-10^{-49})x^{-2}=x^{-2}[1-10^{-49}-2\Omega x^{-1}] \ge x^{-2}[1-10^{-49}-(1+10^{-3})^{-1}]>10^{-4}x^{-2}=10^{-4}S(x)x^{-1}$. On the other hand, for all $x$, $\Omega>0$ we have $$ V_\Omega^\prime(x)=-2\Omega x^{-3}+y(x)x^{-2}-y^\prime(x)x^{-1}\le y(x)x^{-2} -y^\prime(x)x^{-1}-2\Omega x^{-3}+(4\cdot 144-2\cdot 10^{-40})x^{-5} =x^{-5}\bigl\{4\cdot 144-2\cdot 10^{-40}-2\Omega x^2\}\\ &\ge x^{-5}\bigl\{4\cdot 144-2\cdot 10^{-40}-2\Omega\cdot (1-10^{-3})^2 \cdot 2\cdot 144\, \Omega^{-1}\bigr\}\\ &=x^{-5}\bigl\{4\cdot 144-2\cdot 10^{-40}-4\cdot 144\cdot (1-10^{-3})^2\bigr\} >10^{-3}x^{-5}=10^{-3}S(x)x^{-1}\ .\endalign $$ \noindent This and (62) complete the proof of property (22) in Case 3. Thus, we have verified property (22), with $c_1=10^{-2}$, in all cases. Next, we verify (17) and (18). Equations (52), (53), (54) show that $[(1-10^{-2})x_2(\Omega),(1+10^{-2})x_2(\Omega)]\subset [(1+10^{-2})x_0(\Omega), (1+10^{-2})x_2(\Omega)]$. Hence, property (22) shows that $$\multline k_7S(x)x^{-1}6\Omega x^{-4}-(2+10^{-40})x^{-3}>\frac{6\Omega} {(1+2\cdot 10^{-2})^4(2\Omega)^4}-\frac{(2+10^{-40})}{(1-2\cdot10^{-2})^3 (2\Omega)^3}\\ >10^{-2}\Omega^{-3}>10^{-7}\cdot x^{-3}=10^{-7}S(x)\cdot x^{-2}\ . \endmultline\tag"(67)" $$ \noindent On the other hand, (66) and (42) yield also $V_\Omega^{\prime\prime}(x)\le 6\Omega x^{-4}-(2-10^{-40})x^{-3}\le 6\Omega x^{-4}\le 10^5\Omega^{-3}$ by (65). Another application of (65) gives $\Omega^{-3}<10^7S(x)x^{-2}$, so that $$ V_\Omega^{\prime\prime}(x)<10^{12}S(x)x^{-2}\ .\tag"(68)" $$ \noindent Property (21) is immediate from (67) and (68). Let us verify property (23). Lemma 1 shows that $\bigl|\bigl(\frac{d} {dx}\bigr)^\alpha \bigl(\frac{y(x)}{x}\bigr)\big|\le C_\alpha S(x)x^{-\alpha}$ for all $\alpha\ge 0$, $x\in (0,\infty)$, with universal constants $C_\alpha$. Hence, it is enough to check that $\big|\bigl(\frac{d} {dx}\bigr)^\alpha(\Omega x^{-2})\big|\le C_\alpha^\prime S(x)x^{-\alpha}$. This amounts to saying that $\Omega x^{-2}\le KS(x) =K\min\{x^{-1},x^{-4}\bigr\}$. Thus, we must show that $$ \Omega\le Kx\quad\text{and}\quad \Omega x^{2}\le K\ , \quad \text{with}\ K\ \text{a\ universal\ constant}\ .\tag"(69)" $$ \noindent However, property (23) pertains only to $x\in [(1-10^{-2})x_1 (\Omega),(1+10^{-2})(x_2(\Omega))]$. Such $x$ satisfy (69), by virtue of (51) and (53). This proves property (23). It remains only to verify properties (24), (25), (26). These are immediate consequences of (51)$\ldots$(54). The proof of Lemma 6 is complete.$\qquad\qed$ \medskip Lemmas 5 and 6 produce different choices of $\hat c$, $\hat C$, $c_1$, $C_\alpha$. To reconcile them, we use the following trivial result. \bigskip\bigskip \proclaim{Lemma 7} Suppose we are given positive constants $c_1$, $\hat c$, $\hat C$, $C_\alpha$ $(c_1<1/2)$ and an interval $J\subset (0, \overline\Omega)$. Assume that (13),$\ldots$,(26) hold for $\Omega\in J$, with the given constants. \roster \item"{(A)}" If $c_1^\prime$, $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$ are positive constants, with $c_1^\prime=c_1$, $\hat c^\prime\le \hat c$, $\hat C^\prime\ge \hat C$, $C_\alpha^\prime \ge C_\alpha$, then for all $\Omega\in J$, (13)$\ldots$(26) hold with the new constants $c_1^\prime$, $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$. \item"{(B)}" Given any positive $c_1^\prime$ less than $c_1$, there exist positive constants $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$ such that for all $\Omega\in J$, (13)$\ldots$(26) hold with the new constants $c_1^\prime$, $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$. \endroster \endproclaim \bigskip\bigskip \demo{Sketch of proof}: To verify (A), we just check that each of the properties (13)$\ldots$(26) for the given $c_1$, $\hat c$, $\hat C$, $C_\alpha$ implies the corresponding property for the new constants $c_1^\prime$, $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$.\enddemo The proof of (B) is summarized in the following table.\smallskip %TABLE $$ \vbox{\offinterlineskip \def\strut{\vrule height 11 pt depth 5pt width 0pt} \def\vr{\vrule height 12 pt depth 5pt}\def\vrq{\vr\quad} \+ To prove the property listed below \hfill\quad\vr &\quad we use the following properties for\hfill\cr \+ for the new constants $c_1^\prime$, $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$\hfill\vr &\quad the old constants $c_1$, $\hat c$, $\hat C$, $C_\alpha$.\hfill\cr \hrule \+\hfill (13) \hfill\quad \vr &\quad\quad\quad\quad \hfill (13), (14), (24)\hfill&\cr \+\hfill (14)\hfill\quad \vr &\quad\quad\quad\hfill (14)\hfill&\cr \+\hfill (15)\hfill\quad \vr &\quad\quad\quad\hfill (15)\hfill&\cr \+\hfill (16)\hfill\quad \vr &\quad\quad\quad\quad \hfill (15), (16), (17)\hfill&\cr \+\hfill (17)\hfill\quad \vr &\quad\quad\quad\hfill (17)\hfill&\cr \+\hfill (18)\hfill\quad \vr &\quad\quad\quad\hfill (18)\hfill&\cr \+\hfill (19)\hfill\quad \vr &\quad\quad\quad\quad \hfill (18), (19), (25)\hfill&\cr \+\hfill (20)\hfill\quad \vr &\quad\quad\quad\quad \hfill (20), (21), (24)\hfill&\cr \+\hfill (21)\hfill\quad \vr &\quad\quad\quad\hfill (21)\hfill&\cr \+\hfill (22)\hfill\quad \vr &\quad\quad\quad\hfill (21), (22)\hfill&\cr \+\hfill (23)\hfill\quad \vr &\quad\quad\quad\hfill (23)\hfill&\cr \+\hfill (24)\hfill\quad \vr &\quad\quad\quad\hfill (24)\hfill&\cr \+\hfill (25)\hfill\quad \vr &\quad\quad\quad\hfill (25)\hfill&\cr \+\hfill (26)\hfill\quad \vr &\quad\quad\quad\hfill (26)\hfill&\cr} $$ \bigskip \noindent To illustrate, we work out the proofs of (16) and (13) for the new constants, which are the most complicated arguments in the above table. The proof of (16) is divided into cases. \medskip \noindent{\sl Case 1\/}: Suppose $(1+c_1^\prime)x_1(\Omega)\le x< (1+c_1)x_1(\Omega)$. Then (15) for the old constants yields $$ \hat cc_1^\prime S\bigl(x_1(\Omega)\bigr)<-V_\Omega(x)<\hat Cc_1S\bigl( x_1(\Omega)\bigr)\ .\tag"(70)" $$ \noindent Also, $S(x)$ is montone decreasing and satisfies $S(Ax) \ge A^{-4}S(x)$ for $A\ge 1$. Hence, $$ (1+c_1)^{-4}S\bigl(x_1(\Omega)\bigr)\le S(x)\le S\bigl(x_1(\Omega)\bigr)\ , \quad\text{since\ we\ are\ in\ Case\ 1}\ .\tag"(71)" $$ \noindent From (70) and (71) we get $(\hat cc_1^\prime)S(x)<-V_\Omega(x) <\bigl(\hat Cc_1(1+c_1)^4\bigr)S(x)$, which proves (16) in Case 1, provided we take $\hat c^\prime$ small enough and $\hat C^\prime$ large enough. \bigskip \noindent{\sl Case 2\/}: Suppose $(1+c_1)x_1(\Omega)\le x\le (1-c_1)x_2(\Omega)$. Then (16) for the old constants implies (16) for the new constants, provided we take $\hat c^\prime$ small enough and $\hat C^\prime$ large enough. \bigskip \noindent{\sl Case 3\/}: Suppose $(1-c_1)x_2(\Omega)0$, there exist positive constants $c_1$, $\hat c$, $\hat C$, $C_\alpha$ $(c_1<10^{-2})$ such that (13),$\ldots$,(26) hold for all $\Omega\in (0,\overline\Omega-\varepsilon]$. \endproclaim \vglue 1pc \demo{Proof} We may assume $\varepsilon<\varepsilon_0$, with $\varepsilon_0$ as in Lemma 6. Lemmas 6 and 7(B) show that for all sufficiently small $c_1$ there exist $\hat c^\prime$, $\hat C^\prime$, $C_\alpha^\prime$ such that (13)$\ldots$(26) hold when $\Omega\in (0,\varepsilon_0]$. Lemma 5 shows that for all sufficiently small $c_1$ there are $\hat c^{\prime\prime}$, $\hat C^{\prime\prime}$, $C_\alpha^{\prime\prime}$ such that (13)$\ldots$(26) hold when $\Omega\in [\varepsilon,\overline\Omega- \varepsilon]$. Taking $c_1$ sufficiently small, setting $\hat c=\min\{ \hat c^\prime,\hat c^{\prime\prime}\}$, $\hat C=\max\{\hat C^\prime, \hat C^{\prime\prime}\}$, $C_\alpha=\max\{C_\alpha^\prime,C_\alpha^{\prime \prime}\}$, and applying Lemma 7(A), we see that (13)$\ldots$(26) hold for all $\Omega\in (0,\overline\Omega-\varepsilon]$.$\qquad\qed$\enddemo \medskip We turn to the study of $V_\Omega(x)$ for $\Omega$ near $\overline\Omega$. \vglue 1pc \proclaim{Lemma 9} There exist positive constants $\overline c$, $c_2$, $\hat c$, $\hat C$, $C_\alpha$ such that properties (27)$\ldots$(29) hold for $(1-\overline c)\overline\Omega\le \Omega<\overline\Omega$. \endproclaim \vglue1pc \demo{Proof} Pick $\Omega_1<\overline\Omega$. For $\Omega\in [\Omega_1, \overline\Omega)$ we have $V_\Omega\ge V_{\Omega_1}$, so that $\{x\colon V_\Omega(x)<0\}\hfill\break \subset\{x\colon V_{\Omega_1}(x)<0\}$, i.e.\ $$ x_1(\Omega_1)\le x_1(\Omega)\bigl(x_1(\Omega_1)\bigr)^2\Cal E(\Omega)\ . \tag"(74)" $$ \noindent On the other hand, set $\hat\Omega=\Omega+\bigl(x_2(\Omega_1) \bigr)^2\Cal E(\Omega)$. Then for $x\le x_2(\Omega_1)$ we have $$ V_{\hat\Omega}(x)=V_\Omega(x)+(\hat\Omega-\Omega)x^{-2}\ge -\Cal E(\Omega)+\Bigl(\frac{x_2(\Omega_1)}{x}\Bigr)^2\Cal E(\Omega)\ge 0\ ; $$ \noindent while for $x\ge x_2(\Omega_1)$ we have $V_{\hat\Omega}(x) >V_{\Omega_1}(x)\ge 0$. Thus, $V_{\hat\Omega}(x)\ge0$ for all $x$, which shows that $\hat\Omega\ge\overline\Omega$, i.e.\ $$ \overline\Omega-\Omega\le \bigl(x_2(\Omega_1)\bigr)^2\Cal E(\Omega)\ . \tag"(75)" $$ \noindent If $\overline c>0$ is small enough that $(1-\overline c) \overline\Omega>\Omega_1$, then (74), (75) imply property (29). Next, observe that $\big|\bigl(\frac{d}{dx}\bigr)^\alpha V_\Omega(x)\big| \le C_\alpha$ for $x\in \bigl[\frac 12 x_1(\Omega_1),2x_2(\Omega_1)\bigr]$, $\Omega\in [\Omega_1,\overline\Omega)$. In the same region we have also $cc>0$ for $\Omega\in [\Omega_1,\overline\Omega)$ by (8), it follows that $c^\prime0$ outside $[(1-c_2)x_0(\Omega), (1+c_2)x_0(\Omega)]$. Taking $\Omega_3\in (\Omega_2,\overline\Omega)$, we conclude that $$\multline V_\Omega(x)=(\Omega-\Omega_2)x^{-2}+V_{\Omega_2}(x)> (\Omega_3-\Omega_2)x^{-2}\\ \text{if}\ \Omega\in (\Omega_3,\overline\Omega)\ \text{and}\ x\not\in \bigl[(1-c_2)x_0(\Omega), (1+c_2)x_0(\Omega)\bigr]\ . \endmultline $$ \noindent This implies property (28), and completes the proof of lemma 9. $\qquad\qed$\enddemo The {\bf Main Lemma on the Thomas-Fermi Potential} follows trivially from Lemmas 8 and 9. \vfill\eject \head The Density in an Approximate Thomas-Fermi Potential\endhead \medskip Let $V_Z^{TF}(r)$ be the Thomas-Fermi potential arising from a nucleus of charge $+Z$ fixed at the origin. Thus, $-\Delta_xV_Z^{TF}(|x|)= (\text{const})|V_Z^{TF}(|x|)|^{3/2}$ on $\Bbb R^3\backslash\{0\}$, and $V_Z^{TF}(r)=-\frac Zr+O(1)$ as $r\to 0+$. Recall that the size of $V_Z^{TF}(r)$ and its derivatives is controlled by the weight functions $$\multline S(r)=\frac Zr\quad\text{for}\quad r\le Z^{-1/3}\ ,\ S(r)=r^{-4}\quad \text{for}\ r\ge Z^{-1/3}\\ B(r)=r\quad\text{for\ all}\ r\in (0,\infty)\ .\endmultline\tag"(0)" $$ \noindent Specifically, we have \smallskip (i) $ \big|\bigl(\frac{d}{dr}\bigr)^\alpha V_Z^{TF}(r)\big|\le C_\alpha S(r)r^{-\alpha}\quad(\alpha\ge 0)$, (ii) $V_Z^{TF}(r)<-cS(r)$, and (iii) $\frac {d}{dr}V_Z^{TF}(r)>cS(r)r^{-1}$. \smallskip It will be important to study also small perturbations of the Thomas-Fermi potential. Thus, we say that $V(r)$ is an {\sl approximate T-F potential\/} if it satisfies the estimates $$ \Big|\bigl(\frac{d}{dr}\bigr)^\alpha V(r)\Big|\le C_\alpha S(r)r^{-\alpha}\quad (\text{all}\ \alpha\ge 0)\ ,\quad\text{and}\tag"(1)" $$ $$ \Big|\bigl(\frac{d}{dr}\bigr)^\alpha \bigl\{V(r)-V_Z^{TF}(r)\bigr\}\Big| \le c_0S(r)r^{-\alpha}\quad (0\le \alpha \le 2)\ ,\tag"(2)" $$ \noindent with $c_0$ a small enough constant, determined by the $C_\alpha$ in (1). In this section, we use $c$, $C$, $C^\prime$ etc.\ to denote constants determined by the $C_\alpha$ in (1), and by the constants $\varepsilon$, $N$, $a$ to be introduced later. We assume that $Z$ is large enough, depending on the $C_\alpha$ in (1), and on $\varepsilon$, $N$, $a$. Our goal is to understand the density $\rho$ arising from the Hamiltonian $H=-\Delta_x+V(|x|)$ for an approximate T-F potential $V$. By separation of variables, we are led to consider the one-dimensional densities $\rho_\ell(r)$, arising from the potentials $$ V_\ell(r)=\frac{\ell(\ell+1)}{r^2}+V(r)\ . $$ \noindent When $V=V_Z^{TF}$, the behavior of the potentials $V_\ell(r)$ is very thoroughly understood. Let us recall how $V_\ell(r)$ looks. Let $\Omega$ be the positive root of the equation $\Omega(\Omega+1)= \max\limits_{r> 0}\bigl(-r^2V(r)\bigr)$, and suppose the maximum is attained at $r=\check r$. (The sizes of these quantities are $\Omega\sim Z^{1/3}$ and $\check r\sim Z^{-1/3}$.) To describe $V_\ell(r)$, we distinguish between the two cases $1\le \ell\le (1-\overline c)\Omega$ and $(1-\overline c)\Omega\le \ell<\Omega$ for a small, universal constant $\overline c$. For $1\le \ell\le (1-\overline c)\Omega$, there are numbers $x_{\text{left}}(\ell)0$. \medskip \noindent (7)$\quad$ In $[(1+c_1)x_{\text{rt}}(\ell),\infty)$ we have $V_\ell(r)\sim \frac{\ell(\ell+1)}{r^2}$. \medskip Regarding the derivative of $V_\ell(r)$: \medskip \noindent (8)$\quad$ In $(0,(1-c_1)x_0(\ell)]\ \text{we\ have}\ -V_\ell^\prime(r)\sim \frac{\ell(\ell+1)}{r^3}$. \medskip \noindent (9)$\quad$ In $[(1-c_1)x_0(\ell)\ ,\ (1+c_1)x_0(\ell)]$ we have $V_\ell^{\prime\prime}(r)\sim S(r)r^{-2}$ and\hfill \break $\quad V_\ell^\prime(x_0(\ell))=0$. \medskip \noindent (10)$\quad$ In $[(1+c_1)x_0(\ell)\ ,\ (1+c_1)x_{\text{rt}}(\ell)]$ we have $V_\ell^\prime(r)\sim S(r)r^{-1}$. \medskip Regarding the higher derivatives of $V_\ell$: \medskip \noindent (11)$\quad$ In $I_\ell=\bigl[(1-c_1)x_{\text{left}}(\ell),(1+c_1) x_{\text{rt}}(\ell)\bigr]$ we have $\Big|\bigl(\frac{d}{dr}\bigr)^\alpha V_\ell(r)\Big|\le C_\alpha S(r) r^{-\alpha}$. \medskip Regarding the points $x_{\text{left}}(\ell)$, $x_0(\ell)$, $x_{\text{rt}}(\ell)$: $$ x_{\text{left}}(\ell)\ , x_0(\ell)\ ,\ |x_{\text{left}}(\ell)-x_0(\ell)|\sim \frac{\ell^2}{Z}\tag"(12)" $$ $$ x_{\text{rt}}(\ell)\sim\ell^{-1}\ .\tag"(13)" $$ Moreover, $$ x_{\text{left}}(\ell)<(1-c_1)x_0(\ell)\ ,\qquad x_0(\ell)<(1-2c_1)x_{\text{rt}}(\ell)\ , \tag"(13a)" $$ $$ c_1<1/2\ .\tag"(13b)" $$ On the other hand, suppose $(1-\overline c)\Omega\le \ell<\Omega$. Then there is a point $x_0(\ell)\sim Z^{-1/3}$ with the following properties: \medskip \noindent (14)$\quad$ In\ $[(1-c_2)x_0(\ell), (1+c_2)x_0(\ell)]$ we have $\big|\bigl(\frac{d}{dr}\bigr)^\alpha V_\ell(r)\big|\le C_\alpha S\bigl(x_0(\ell)\bigr)\bigl(x_0(\ell)\bigr)^{-\alpha}$ and $V_\ell^{\prime\prime}(r)\sim S(r)r^{-2}$. At $r=x_0(\ell)$ we have $V_\ell^\prime=0$ and $-V_\ell\sim\frac{\Omega(\Omega+1)-\ell(\ell+1)} {r^2}$. \medskip \noindent (15)$\quad$ Outside $[(1-c_2)x_0(\ell), (1+c_2)x_0(\ell)]$ we have $V_\ell(r)\ge \frac{c\ell(\ell+1)}{r^2}$. \noindent Here, $00$ and $N>1$. Let $\hat c$ be a small enough constant, depending on $\varepsilon$, $N$ and on the $C_\alpha$ in (1). Then for $Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega$, the potential $V_\ell(r)$ satisfies hypotheses (Z0)$\ldots$(Z9) of the {\sl WKB Density Theorem\/} in one dimension, with the weight functions $S(r)$ as in (0), $B(r)\equiv r$. The number called $\Lambda$ in (Z0)$\ldots$(Z9) is of the order of magnitude $\ell$. The constants in (Z0)$\ldots$(Z9) depend only on $\varepsilon$, $N$, and the $C_\alpha$ in (1).\endproclaim \vglue 1pc \proclaim{Lemma 5} Suppose $(1-\overline c)\Omega\le \ell<\Omega-c\Omega ^{7/43}$. Set $\tilde S=\frac{\Omega(\Omega-\ell)}{\check r^2}$, $\tilde B=\frac{\check r(\Omega-\ell)^{1/2}}{\Omega^{1/2}}$, and define $I=[x_0(\ell)-h, x_0(\ell)+h]$, with $h=\min(c_2x_0(\ell), \underline C \tilde B)$ and $\underline C$ a large constant determined by the $C_\alpha$ in (1). Let $\varepsilon>0$, $N>1$ be given. Set $K=100^{90}$, and let $\hat c$ be a small enough constant, depending on $\varepsilon$, $N$ and the $C_\alpha$ in (1). Then the potential $V_\ell(r)$, the weight functions $\tilde S$, $\tilde B$, and the interval $I$ satisfy the hypotheses (Z0)$\ldots$(Z9) of the {\sl WKB Density Theorem\/} in one dimension. The number called $\Lambda$ in (Z0)$\ldots$(Z9) is of the order of magnitude $(\Omega-\ell)$. The constants in (Z0)$\ldots$(Z9) depend only on $\varepsilon$, $N$ and the $C_\alpha$ in (1).\endproclaim \vglue 1pc \demo{Remark} The reason for using $\tilde S$, $\tilde B$, $I$ as above is that (as we will see) $|\min V_{\ell}|\sim \tilde S$, $V_{\ell}^{\prime\prime}\sim\tilde S\tilde B^{-2}$ at $x_0(\ell)$, and $I$ is comparable to $\{V_\ell<0\}$. \enddemo \vglue 1pc \proclaim{Lemma 6} Suppose $\Omega-c\Omega^{7/43}\le \ell<\Omega$. Set $x_0=x_0(\ell)$, $S=S(x_0)$, $B=x_0$. Let $\varepsilon>0$ and $N>1$ be given. Take $K=100^{90}$. Then the potential $V_\ell(r)$ satisfies hypotheses (Z0$^\ast)\ldots$(Z6$^\ast$) of the {\sl First Degenerate Density Lemma\/}, with $\lambda\sim S^{1/2}(\check r)\check r\sim\Omega$. The constants in (Z0$^\ast$)$\ldots$(Z6$^\ast$) depend only on $\varepsilon$, $N$ and the $C_\alpha$ in (1). \endproclaim \medskip We will use Lemma 1 for $\ell=0$, Lemma 2 for $1\le \ell\le\ \text{(Large\ Const)}$, Lemma 3 for $\text{(Large\ Const)}\le \ell\le Z^{10^{-9}}$, Lemma 4 for $Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega$, Lemma 5 for $(1-\overline c)\Omega\le \ell\le \Omega-c\Omega^{7/43}$, and Lemma 6 for $\Omega-c\Omega^{7/43}\le \ell<\Omega$. Thus, all the potentials $V_\ell(r)$ $(0\le \ell<\Omega)$ are covered by our ODE density results. For $\ell\ge \Omega$, the potential $V_\ell(r)$ is non-negative everywhere, so that $\rho_\ell(r)\equiv0$. We give the proofs of Lemmas 1$\ldots$6. \vglue 1pc \demo{Proof of Lemma 1} \noindent (Z0$^\dag$) is obvious from the definitions of $S(r)$, $B(r)$.\smallskip \noindent (Z1$^\dag$) is immediate from (1).\smallskip \noindent (Z2$^\dag$) is immediate from (ii), (iii) and (2).\smallskip \noindent (Z3$^\dag$) holds, since one computes from the definitions that $\Lambda\sim \overline C^{1/2}$, when $\overline C>>1$ and $Z>\overline C ^{\text{power}}$. \smallskip \noindent (Z4$^\dag$) follows from (1), since $x_0=\overline CZ^{-1}$. \smallskip \noindent (Z5$^\dag$) is proven as follows. From (ii), (iii) and (2), we see that $V(r)$ is increasing and negative on $(0,\infty)$. From the definitions we have $x_{\text{big}}<\underline Cx_1$. From (1) we have $|V(\frac{x_1}{8})|\le CS(\frac{x_1}{8})\le \underline Cx_1^{-2}$. This proves (Z5$^\dag$).\smallskip \noindent (Z6$^\dag$) is immediate from (1).\smallskip \noindent (Z7$^\dag$) is proven as follows. Since $V^\prime (x)\sim S(x)x^{-1}$, we have $V(x_\ast)-V(x)\sim \int_x^{x_\ast}S(t)t^{-1}dt$ for $x\le \frac{x_\ast}{2}$. Since $x_\astFrom (11), (12) and the definition of $S(r)$, we have $|V_\ell(r)|\le \frac{C\ell^2}{r^2}$ also in $[(1-c_1)x_{\text{left}}(\ell),x_0]$. Hence in $[x_{\text{small}},x_0]$ we have $|V_\ell(r)|\le \frac{C\ell^2}{r^2}\le \frac{C^\prime\ell^2}{x_0^2}$, which is (Z$\hat 5$) with a constant depending on $\ell$.\smallskip \noindent (Z$\hat 6$) is immediate from (10), (12), (13) and the definitions of $x_1$, $x_{\text{big}}$.\smallskip \noindent (Z$\hat 7$) is proven as follows. For $r\in [\frac{x_1}{8}, (1+c_1)x_{\text{rt}}(\ell)]$ we have $|V(r)|\le CS(r)= Cr^{-4}\le \frac{C^\prime\ell^2}{x_1^2}$ by (11), (13) and the definition of $x_1$. This implies (Z$\hat 7$) with a constant depending on $\ell$.\smallskip \noindent (Z$\hat 8$) follows from (7), since $x_{\text{big}}= (1+c_1)x_{\text{rt}}(\ell)$.\smallskip \noindent (Z$\hat 9$) is proven as follows. From (12), (13) and the definition of $x_0$, we have $(1+c_1)x_0(\ell)cx$ for any $x\in [x_{\text{left}} (\ell), x_{\text{rt}}(\ell)]$, in particular for $x=x_{\text{left}} (E), x_{\text{rt}}(E)$. This completes the proof of (Z$\overline 2$).\enddemo \smallskip \noindent (Z$\overline 3$) is proven as follows. If $|r-x_0(\ell)|< c_1x_0(\ell)$, then $V_\ell(r)\sim -S(r)\sim -S(x_0(\ell))$ by (5) and (13a), and we saw that $S(x_0(\ell))\sim \frac{Z^2}{\ell^2}$. On the other hand, $V_\ell\bigl(x_{\text{left}}(E)\bigr)=E$, and $E\in [E_{\text{crit}},0]$ implies $|E|<<\frac{Z^2}{\ell^2}$ since $E_{\text{crit}}=-Z^{18/10}$, $\overline C\le \ell\le Z^{10^{-9}}$. Hence $|x_{\text{left}}(E)-x_0(\ell)|>c_1x_0(\ell)$, and therefore $[x_{\text{left}}(E), x_{\text{left}}(E)\cdot (1+\frac{c_1}{10})]$ does not meet $[(1-\frac{c_1}{10})x_0(\ell), (1+\frac{c_1}{10}) x_0(\ell)]$. From (8), (9), (10) we get $|V_\ell^\prime(r)|\ge cS(r)r^{-1}$ in $[(1-c_1)x_{\text{left}}(\ell), (1+c_1) x_{\text{rt}}(\ell)]\backslash [(1-\frac{c_1}{10})x_0(\ell), (1+\frac{c_1}{10})x_0(\ell)]$. In particular, $|V_\ell^\prime(r)|\ge cS(r)r^{-1}$ in $[x_{\text{left}}(E), x_{\text{left}}(E)\cdot (1+\frac{c_1}{10})]$. Since $V_\ell(x_0(\ell))\sim -S(x_0(\ell)) >|E|$ for $E\in [E_{\text{crit}},0]$. Hence $E-V_\ell(x)\sim S(x)$. If instead $x\in [x_{\text{left}}(E)\cdot (1+\frac{c_1}{10}), (1-c_1)x_0(\ell)]$, then by (8) we have $-V_\ell^\prime (r)\sim S(r)r^{-1}$ for $r\in [x_{\text{left}}(E),x]$, and therefore $E-V_\ell(x)=V_\ell(x_{\text{left}} (E))-V_\ell(x)\sim \int_{x_{\text{left}}(E)}^{x}S(r)r^{-1}dr \ge \int_{(1-\frac{c_1}{100})x}^x S(r)r^{-1}dr$ (since $x\ge x_{\text{left}}(E)\cdot (1+\frac{c_1}{10}))\sim S(x)$.\smallskip Similarly, $E-V_\ell(x)\sim S(x)$ if $x\in [(1+c_1)x_0(\ell), x_{\text{rt}} (E)\cdot (1-\frac{c_1}{10})]$. So we know that $E-V_\ell(x)\sim S(x)$ in three intervals that cover $[(1+\frac{c_1}{10})x_{\text{left}} (E), (1-\frac{c_1}{10})x_{\text{rt}}(E)]$. This proves (Z$\overline 4$), with $c_1$ replaced by $\frac{c_1}{10}$. \smallskip \noindent (Z$\overline 5$) follows from (8) and (13a), which show $V_\ell^\prime(r)<0$ in $(0,x_{\text{left}}(\ell)]$; and from (1) and $V_\ell(r)=V(r)+\frac{\ell(\ell+1)}{r^2}$, which show that $V_\ell$ is $C^\infty$.\smallskip \noindent (Z$\overline 6$) holds, because we saw in the proof of (Z$\overline 3$) that $x_{\text{left}}(E)\le x_0(\ell)$. Hence $x_{\text{left}} (E)>|E_{\text{crit}}|\ge |E|$ for $r\in [x_0(\ell),x_\ast]$, $x_\ast= \text{(small\ const.)}\ \cdot Z^{-8/10}$. So we cannot have $x_{\text{rt}} (E)\in [x_0(\ell),x_\ast]$. We saw in the proof of (Z$\overline 3$) that $x_{\text{rt}}(E)\ge x_0(\ell)$. Therefore, $x_{\text{rt}}(E)>x_\ast$. Consequently, $$ \int_{x_{\text{left}}(E)}^y\bigl(E-V_\ell(x)\bigr)_+^{-1/2} dx\sim Z^{-1/2}y^{3/2} \text{if}\quad (1+\frac{c_1}{10})x_{\text{left}}(E)c_1x$. Hence we know (Z2) and (Z5). \smallskip \noindent (Z3) is proven as follows. With $x_0=x_0(\ell)$, we have $V_\ell(x_0)<-cS(x_0)$ by (5), (13a); and from (9) we have $V_\ell^\prime(x_0)=0$ and $V_\ell^{\prime\prime}>cS(x_0)x_0^{-2}$ in $[(1-c_1)x_0,(1+c_1)x_0]$. Hence we know (Z3). \smallskip \noindent (Z4) follows from (8), (10), since $\frac{\ell(\ell+1)}{r^3} \sim S(r)r^{-1}$ in $[x_{\text{left}}(\ell),x_0]$. \smallskip To compute $\Lambda$, recall that $\lambda(x)=\bigl(\frac Zx\bigr)^{1/2}x= Z^{1/2}x^{1/2}$ for $x\le Z^{-1/3}$, and $\lambda(x)=(x^{-4})^{1/2}x =x^{-1}$ for $x\ge Z^{-1/3}$. Hence if $\tilde x\sim Z^{-1/3}$, then $\lambda(x)\sim Z^{1/2}x^{1/2}$ for $x\le \tilde x$, and $\lambda(x) \sim x^{-1}$ for $x\ge \tilde x$. By (12), (13) we can find $\tilde x\sim Z^{-1/3}$ with $(1+c)x_{\text{left}}(\ell)<\tilde x <(1-c)x_{\text{rt}}(\ell)$. Then we have $\Lambda^{-1}\sim \int_{x_{\text{left}}(\ell)}^{\tilde x} \frac{dx}{(Z^{1/2}x^{1/2})x}+\int_{\tilde x}^{x_{\text{rt}}(\ell)} \frac{dx}{(x^{-1})x}\sim Z^{-1/2}(x_{\text{left}}(\ell))^{-1/2}+x_{\text{rt}}(\ell) \sim \ell^{-1}$. So $\Lambda\sim \ell$, as asserted in Lemma 4. \smallskip \noindent (Z5) was proven above, together with (Z2).\enddemo \smallskip \noindent (Z6) is proven as follows. Since $\Lambda\sim \ell\ge Z^{10^{-9}}$, the assertion about $x\in I_{\text{BVP}}=(0,\infty)$ with $xx_{\text{rt}}(\ell)+\Lambda^KB(x_{\text{rt}}(\ell))$, we have $V_\ell(x)\ge \frac{c\ell^2}{x^2}$ by (7). Also, $x\sim x-x_{\text{rt}} (\ell)$ for $x>x_{\text{rt}}(\ell)+\Lambda^KB(x_{\text{rt}}(\ell))$. Therefore $V_\ell(x)>\frac{c^\prime\ell^2}{(x-x_{\text{rt}}(\ell))^2}$ for $x>x_{\text{rt}}(\ell)+\Lambda^kB(x_{\text{rt}}(\ell))$, which implies (Z6). \smallskip \noindent (Z7) is proven as follows. For $00$ in $I$. Hence, $\{x\in I\mid V_\ell(x)<0\}$ is a (possibly empty) subinterval. Since $x_0(\ell)\in I$ and $V_\ell\bigl(x_0(\ell)\bigr)<0$ by (14), the subinterval is non-empty. It remains to show that the endpoints of $\{x\in I\mid V_\ell(x)<0\}$ have distance at least $c\tilde B$ from the endpoints of $I$. That is equivalent to saying that $V_\ell(x)>0$ if $x\in I$ and $\text{dist}(x,\partial I)c\tilde S$ at the endpoints of $I$. We distinguish the two cases $h=c_2x_0(\ell)$ and $h=\underline C\tilde B\frac{c\ell(\ell+1)}{r^2}\sim \frac {Z^{2/3}}{(Z^{-1/3})^2}=Z^{4/3}$, while $\tilde S\le \frac{\Omega^2}{\check r^2}\sim\frac{(Z^{1/3})^2} {(Z^{-1/3})^2}=Z^{4/3}$. Hence, $V_\ell(r)>c\tilde S$ at the endoints of $I$. In the other case $h=\underline C\tilde Bc_0\tilde B$. This is immediate from $-V_\ell(x_0)\sim \tilde S$, which we just saw, from $|V_\ell^{\prime}|\le C\tilde S\tilde B^{-1}$ in $I$, which follows from (Z1) above; and from the fact that $V_\ell\ge 0$ at the endpoints of $I$, which follows from (Z2) above. The proof of (Z3) is complete.\smallskip (Z4) follows from the fact that $V_\ell^{\prime\prime}\ge c\tilde S\tilde B^{-2}$ in $I$ (as we noted above), and that $V_\ell^\prime=0$ at $x_0(\ell)$. \smallskip Next we compute $\Lambda$. Since $V_\ell(x_0)\sim -\tilde S$ and $|V_\ell^\prime|\le C\tilde S\tilde B^{-1}$ in $I$ by (Z1) above, we see from (Z2) that $x_{\text{rt}}(\ell)-x_{\text{left}}(\ell)\ge c\tilde B$. On the other hand, $x_{\text{rt}}(\ell)-x_{\text{left}}(\ell)\le\ \text{diam}\ I=2h\le 2 \underline C\tilde B$. Thus $x_{\text{rt}}(\ell)-x_{\text{left}}(\ell) \sim \tilde B$. Also, $\lambda(x)=\tilde S^{1/2}\tilde B=(\Omega-\ell)$. Therefore, $\Lambda^{-1}=\int_{x_{\text{left}}(\ell)}^{x_{\text{rt}}(\ell)} \frac{dx}{\lambda(x)\tilde B}\sim (\Omega-\ell)^{-1}$,\ i.e. $\Lambda$ has the order of magnitude $(\Omega-\ell)$.\medskip \noindent (Z5) is proven as follows. In view of (Z2), it is enough to show that $V_\ell>0$ outside of $I$. We already know from (15) that $V_\ell>0$ outside $J\equiv \bigl[x_0(\ell)(1-c_2), x_0(\ell)(1+c_2)\bigr]$. Inside this interval we have $V_\ell^{\prime\prime}>c\tilde S\tilde B^{-2}$ by (14) and (16). Since $V_\ell\sim -\tilde S$, $V_\ell^\prime=0$ at $x_0(\ell)$, we conclude that $$ V_\ell(x)\ge -C\tilde S+\frac 12(c\tilde S\tilde B^{-2}) \cdot (\underline C\tilde B)^2 \quad\text{for}\ x\in J, |x-x_0|\ge \underline C\tilde B\ . $$ \noindent The constants $C$, $c$ here don't depend on $\underline C$. Hence if we pick $\underline C$ large enough, we get $V_\ell(x)>0$ for $x\in J$, $|x-x_0|\ge \underline C\tilde B$. So if $x\in (0,\infty)$ satisfies $V_\ell(x)<0$, then $x\in J$ and $|x-x_0|<\underline C\tilde B$. That is, $|x-x_0(\ell)|\le \min(c_2x_0(\ell),\underline C\tilde B)$, i.e.\ $x\in I$. Thus, $V_\ell>0$, outside $I$, completing the proof of (Z5). \smallskip \noindent (Z6) is proven as follows. We have $\Lambda\sim (\Omega-\ell) \ge c\Omega^{\frac{7}{43}}\sim Z^{\frac{7}{3\cdot43}}$, while $K=100^{90}$. Hence $\Lambda^K\ge Z^{100}$, so $\Lambda^K\tilde B>(1+c_2)x_0(\ell)$. This shows that $x_{\text{left}}(\ell)-\Lambda^K\tilde B<0$, so the assertion of (Z6) concerning $x\in (0,\infty)$ with $x(1+c_2)x_0(\ell)$. Therefore, (15) shows that $V_\ell(x)>\frac{c\ell(\ell+1)}{x^2}$ for $x>x_{\text{rt}}(\ell) +\Lambda^K\tilde B$. We have $x-x_{\text{rt}}(\ell)\sim x$ for $x>x_{\text{rt}}(\ell)+\Lambda^K\tilde B>x_{\text{rt}}(\ell) +(1+c_2)x_0(\ell)$, since $x_{\text{rt}}(\ell)\in I$ and thus $x_{\text{rt}} (\ell)\le (1+c_2)x_0(\ell)$. Therefore $V_\ell(x)> \frac{c^\prime\ell(\ell+1)}{(x-x_{\text{rt}}(\ell))^2}$ for $x>x_{\text{rt}}(\ell)+\Lambda^K\tilde B$, completing the proof of (Z6). \smallskip \noindent (Z7) is trivial, since $\tilde S$ and $\tilde B$ are constant functions, and since $|I|=2h\le 2\underline C\tilde B$, while $\Lambda\sim(\Omega-\ell)>c\ Z^{\frac{7}{3\cdot 43}}>>\underline C$.\smallskip \noindent (Z8) is trivial, since we picked $\hat c$ to be small enough. \smallskip \noindent (Z9) is trivial, since $\Lambda\sim (\Omega-\ell)\ge c\Omega^{7/43} \sim Z^{\frac{7}{3\cdot 43}}$, and we take $Z$ to be large enough, depending on $\varepsilon$, $N$ and $C_\alpha$ in (1).\smallskip \noindent The proof of Lemma 5 is complete.$\qquad\blacksquare$\enddemo \medskip \demo{Proof of Lemma 6} Take $I=\{x\mid |x-x_0|\frac 12\lambda^KB=\frac 12\lambda^Kx_0$ must also satisfy $x>(1+c_2)x_0$. Hence $x\sim x-x_0$, and $V_\ell(x)\ge \frac{c\ell(\ell+1)}{x^2}$ by (15). Therefore, $V_\ell(x)>\frac{c^\prime\ell(\ell+1)}{|x-x_0|^2}$ which implies (Z5$^\ast$).\smallskip \noindent Finally (Z6$^\ast$) follows from our assumption that $Z$ is large (depending on $\varepsilon$, $N$ and the $C_\alpha$ in (1)), since $\lambda\sim Z^{1/3}$.\smallskip The proof of Lemma 6 is complete.$\qquad\blacksquare$\enddemo \medskip Our plan is to use separation of variables, Lemmas 1$\ldots$ 6, and our density results for ODE's to control the density $\rho$ arising from $-\Delta+V$ on $\Bbb R^3$. The precision of our results depends on number-theoretic properties of the potential $V$. Specifically, define $$ n_\ell=\int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{-1/2}dr\ ,\tag"(20)" $$ \noindent and $$ \phi_\ell=\frac 1\pi \int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{1/2}dr-\frac 12\ ,\ \text{for}\ 0\le \ell<\Omega\ .\tag"(21)" $$ \noindent We will get sharper results if the $\phi_\ell$ are approximately equidistributed modulo $1$. Define sets $\Cal L$, $\tilde{\Cal L}$ by $$ \Cal L=\big\{1\le \ell<\Omega\mid |\phi_\ell-\ \text{nearest\ integer}| <\frac C\ell\bigr\} \tag"(22)" $$ $$ \tilde{\Cal L}=\bigl\{1\le \ell<\Omega\mid |\phi_\ell-\ \text{nearest\ integer}|\ <\frac{C}{\ell^{7/43}}\bigr\}\ .\tag"(23)" $$ \noindent We say that the potential $V$ has {\sl Number-Theoretic Type\/} $a\ge 0$ if the following conditions hold. $$\multline \Cal D(T)\equiv \frac{[\text{Number of}\ \ell\le T\ \text{belonging\ to}\ \Cal L]}{T}\le C\Omega^{-2a}\\ \text{for}\quad \Omega^{1-4a}\le T<\Omega\ .\endmultline\tag"(24)" $$ $$ \tilde{\Cal D}\equiv \frac{[\text{number\ of}\ \ell\in \tilde{\Cal L}]} {\Omega}\le C\Omega^{-2a}\ .\tag"(25)" $$ \noindent Finally, $$\multline \Big|\sum\limits_{\ell_1\le \ell\le \ell_2}\frac {(2\ell+1)\chm(\phi_\ell)}{n_\ell}\Big|\le C\Omega^{-2a} \sum\limits_{\ell_1\le \ell\le \ell_2}\frac{(2\ell+1)}{n_\ell}\\ \text{for}\ Z^{10^{-9}}\le \ell_1\le \ell_2<\Omega\ \text{with}\ \ell_2-\ell_1\ge c\ell_2^{1-4a}\ \text{and}\ \ell_2\ge c\Omega^{1-4a}\ . \endmultline\tag"(26)" $$ Here, $\chm(\cdot)$ is the elementary function appearing in our ODE density results. We expect (26) with $a>0$ if the $\phi_\ell$ are equidistributed mod $1$, since $\chm$ is periodic and has mean value zero. In this section, we simply assume (24), (25), (26) for an $a\ge 0$. In [FS6] [FS7], we will show that $a$ can be taken strictly positive for the Thomas-Fermi potential $V_Z^{TF}$. Note that (24), (25), (26) hold trivially when $a=0$, and that (24), (25), (26) become stronger as $a$ increases. We are ready to analyze the three-dimensional density $\rho$. Let $V$ be an approximate T-F potential having number-theoretic type $0\le a<1/43$. Separation of variables tells us that $$\multline 4\pi r^2\rho(r)=\sum\limits_{0\le \ell<\Omega}(2\ell+1)\rho_{\ell}(r)\ ,\\ \text{where}\ \rho_{\ell}(\cdot)\ \text{is\ the\ density\ arising\ from}\ -\frac{d^2}{dr^2}+V_{\ell}(r)\ .\endmultline \tag"(27)" $$ \noindent For each $\ell$, set $$ \rho_{sc}^\ell(r)=\frac 1\pi\bigl(-V_\ell(r)\bigr)_+^{1/2}=\frac 1\pi \bigl(-\frac{\ell(\ell+1)}{r^2}-V(r)\bigr)_+^{1/2}\tag"(28)" $$ $$ \rho_{nt}^\ell(r)=-\Bigl(V_\ell(r)\Bigr)_+^{-1/2}\frac{\chm(\phi_\ell)} {n_\ell}\ ,\tag"(29)" $$ \noindent and define $\rho_{\text{error}}^\ell(r)=\rho_\ell(r) -\{\rho_{sc}^\ell(r)+\rho_{nt}^\ell(r)\}$, so that $$ \rho_\ell(r)=\rho_{sc}^\ell(r)+\rho_{nt}^\ell(r)+\rho_{\text{error}}^\ell(r)\ . \tag"(30)" $$ \noindent From (27), (30) we get $$ 4\pi r^2\rho(r)=\rho_{\text{LOW}}(r)+\rho_{sc}(r)+\rho_{NT}(r)+\rho_{\text{ERROR}}(r)\ ,\text{with}\tag"(31)" $$ $$ \rho_{\text{LOW}}(r)=\sum\limits_{0\le \ellZ^{2\cdot10^{-9}}}}\bigl(-r^2V(r)\bigr) ^{\frac{\varepsilon}{2}+\frac{41}{86}}\bigl(-V(r)\bigr)^{1/2}dr\\ &=C\int_0^{2R_0}\chi_{{}_{-r^2V(r)>Z^{2\cdot10^{-9}}}}\frac{\bigl(-r^2V (r)\bigr)}{r}^{\frac{\varepsilon}{2}+\frac{42}{43}}dr\ .\tag"(39)"\endalign $$ Similarly, $$\align &\text{ERR}_2(R_0)\le C\int_0^{2R_0}\Bigl\{\sum\limits_{Z^{10^{-9}} \le \ell\le (1-\overline c)\Omega}\chi_{{}_{\ell\in\Cal L}} \bigl(-\frac{\ell(\ell+1)}{r^2}-V(r)\bigr)_+^{1/2}\Bigr\}dr\\ &\le C\int_0^{2R_0}\Bigl\{\bigl(-V(r)\bigr)_+^{1/2}\sum\Sb Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega\\ \ell(\ell+1) <-r^2V(r)\endSb\chi_{{}_{\ell\in\Cal L}}\Bigr\}dr\\ &\le C\int_0^{2R_0}\Bigl\{\bigl(-V(r)\bigr)_+^{1/2} \chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}}\\ &\quad\quad\quad\qquad\qquad\qquad\qquad \big[\text{Number\ of}\ \ell\le \bigl(-r^2V(r)\bigr)^{1/2} \ \text{belonging\ to}\ \Cal L\big]\Bigr\}dr\\ &=C\int_0^{2R_0}\chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}}\bigl(-V(r)\bigr)_+^{1/2} \cdot\bigl(-r^2V(r)\bigr)^{1/2}\Cal D\bigl((-r^2V(r))^{1/2}\bigr)dr\\ &\le C\int_0^{2R_0}\chi_{{}_{Z^{2\cdot 10^{-9}}<-r^2V(r)1}}\bigl(-r^2V(r)\bigr)_+ ^{\varepsilon+\frac{42}{43}}\frac{dr}{r}\\ +C\Omega^{-2a}\int_0^{2R_0}\chi_{{}_{-r^2V(r)>1}} \bigl(-r^2V(r)\bigr)_+\frac{dr}{r}+C\int_0^{2R_0} \chi_{{}_{-r^2V(r)(1-\overline c)\Omega$. Therefore $$ \int_0^{2R_0}\bigl(-V_\ell(r)\bigr)_+^{1/2}dr\le C\chi_{{}_{R_0>c\check r}} \cdot(\Omega-\ell)\ . $$ \noindent Substituting this in (42), we see that $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{error}}^\ell (r)dr\big|^2\Bigr)^{1/2}\\ \le Z^{-N}+C\chi_{{}_{R_0>c\check r}}(\Omega-\ell)^{\varepsilon-\frac{2}{43}} +C\chi_{{}_{R_0>c\check r}}\chi_{{}_{\ell\in\Cal{\tilde L}}}\ .\endmultline $$ \noindent Therefore, by definition of $\rho_{\text{ERROR},2}$, we have the estimate $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{ERROR},2}(r)dr\big|^2 \Bigr)^{1/2}\\ \le \sum\limits_{(1-\overline c)\Omega<\ell\le \Omega-c\Omega^{7/43}} (2\ell+1)\Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R \rho_{\text{error}}^\ell(r)dr\big|^2\Bigr)^{1/2}\\ \le Z^{10-N}+C\chi_{{}_{R_0>c\check r}}\ \Omega^{\frac{84}{43}+\varepsilon} +C\chi_{{}_{R_0>c\check r}}\Omega^2\tilde{\Cal D}\endmultline $$ \noindent (by definition of $\tilde{\Cal D}$). Hence, by (25), we have $$ \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{ERROR},2}(r) dr\big|^2\Bigr)^{1/2}\le Z^{10-N}+C\chi_{{}_{R_0>c\check r}}\ \Omega^{2-2a}\ .\tag"(43)" $$ \noindent (Recall that $0\le a< 1/43$). So we have estimated $\rho_{\text{ERROR},2}$. We turn to $\rho_{\text{ERROR},3}$. For $\Omega-c\Omega^{7/43}<\ell<\Omega$, Lemma 6 and the {\sl First Degenerate Density Lemma\/} imply the estimate $$ \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{error}}^\ell (r)dr\big|^2\Bigr)^{1/2}\le Z^{-N}+C\chi_{{}_{R_0>c\check r}}\ . $$ \noindent (Here we have weakened drastically the conclusion of the {\sl First Degenerate Density Lemma}.) Therefore, by definition of $\rho_{\text{ERROR},3}$, we have $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R \rho_{\text{ERROR},3}(r)dr\big|^2\Bigr)^{1/2}\\ \le \sum\limits_{\Omega-c\Omega^{7/43}<\ell<\Omega}(2\ell+1) \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{error}}^\ell(r) dr\big|^2\Bigr)^{1/2}\\ \le Z^{10-N}+C\chi_{{}_{R_0>c\check r}}\sum\limits_{\Omega-c\Omega ^{7/43}<\ell<\Omega}(2\ell+1)\\ =Z^{10-N}+C\chi_{{}_{R_0>c\check r}}\Omega^{50/43}c\check r}}\Omega^{2-2a}\ ,\endmultline\tag"(44)" $$ \noindent again because $a\le 1/43$. \noindent Combining (36), (41), (43), (44), we obtain an estimate for $\rho_{\text{ERROR}}$, namely $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{ERROR}} (r)dr\big|^2\bigr)^{1/2}\\ \le Z^{10-N}+C\chi_{{}_{R_0>c\check r}}\Omega^{2-2a}\\ +C\int_0^{2R_0}\chi_{{}_{-r^2V(r)>1}}\bigl(-r^2V(r)\bigr)_+^{\varepsilon +\frac{42}{43}}\frac{dr}{r}+C\Omega^{-2a}\int_0^{2R_0} \chi_{{}_{-r^2V(r)>1}}\bigl(-r^2V(r)\bigr)_+\frac{dr}{r}\\ +C\int_0^{2R_0}\chi_{{}_{-r^2V(r)0}\bigl(-r^2V(r)\bigr)= \Omega(\Omega+1)\sim\Omega^2$. Taking $\varepsilon>0$ so small that $\varepsilon+\frac{42}{43}<1-a$, we can dominate the first two integrals on the right in (45) by $C\int_0^{2R_0}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr} {r}$. Hence, $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R\rho_{\text{ERROR}} (r)dr\big|^2\Bigr)^{1/2}\\ \le Z^{10-N}+C\chi_{{}_{R_0>c\check r}}\Omega^{2-2a}+C\int_0^{2R_0} \bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}\\ +C\int_0^{2R_0}\chi_{{}_{-r^2V(r)0$ be numbers that satisfy $\big|\sum\limits_{\ell_1\le \ell\le \ell_2}v_\ell\big|\le C\ell_2^{-2a}\sum\limits_{\ell_1\le \ell\le \ell_2}v_{\ell}^{\#}$ for $\ell_1\le \tilde\ell$, and $|v_\ell|\le Cv_\ell^{\#}$ for all $\ell$. (Here, $\tilde\ell$ needn't be an integer). Then $$ \Big|\sum\limits_{0\le \ell\le \ell_2}u_\ell v_\ell\Big|\le C\sum\limits_{\tilde\ell<\ell\le \ell_2}u_\ell v_\ell^{\#}+C\ell_2 ^{-2a}\sum\limits_{0\le \ell\le \ell_2}u_\ell v_\ell^{\#}\ . $$\endproclaim \vglue 1pc \demo{Proof} Take $\tilde v_\ell=C(\ell_2^{-2a}+\chi_{{}_{\ell>\tilde\ell}} )v_{\ell}^{\#}$, and apply the preceding Lemma. $\qquad\blacksquare$\enddemo \medskip We use the above Corollary and inequality (26) to make the following estimate of $\rho_{NT}$. \vglue 1pc \proclaim{Lemma 8} For $r>0$, let $\ell_2(r)$ be the largest integer $\ell\ge0$ with $\ell(\ell+1)<\bigl(-r^2V(r)\bigr)$. (If $-r^2V(r)\le 0$, then set $\ell_2(r)\equiv 0$.) Then we have the pointwise inequality $$\multline |\rho_{NT}(r)| \le C\Bigl[\sum\limits_{Z^{10^{-9}}\le\ell\le c\Omega^{1-4a}} (2\ell+1)\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}\Bigr]\\ +C\Bigl[\sum\limits_{Z^{10^{-9}}\le \ell\le \ell_2(r)}(2\ell+1) \ell^{-2a}\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}\bigr]\\ +C\Bigl[\sum\Sb \ell_2(r)-c(\ell_2(r))^{1-4a}\le \ell\le \ell_2(r)\\ \ell\ge Z^{10^{-9}}\endSb (2\ell+1)\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}} {n_\ell}\Bigr]\\ \equiv C\rho_{NT,1}(r)+C\rho_{NT,2}(r)+C\rho_{NT,3}(r)\ .\endmultline $$\endproclaim \vglue 1pc \demo{Proof} Recall that $\rho_{NT}(r)=-\sum\limits_{Z^{10^{-9}}\le \ell<\Omega}(2\ell+1)\bigl(-V_\ell(r)\bigr)_+^{-1/2} \frac{\chm(\phi_\ell)}{n_\ell}$. The summands are zero for $\ell> \ell_2(r)$; and also $\ell_2(r)\le \Omega$ by definition. Hence, $$ \rho_{NT}(r)=-\sum\limits_{Z^{10^{-9}}\le\ell\le \ell_2(r)} \frac{(2\ell+1)\chm(\phi_\ell)}{n_\ell}\bigl(-V_\ell(r)\bigr)_+^{-1/2}\ . $$\enddemo \noindent If $\ell_2(r)\le c\Omega^{1-4a}$, then already $|\rho_{NT}(r)| \le C\rho_{NT,1}(r)$, so the conclusion of the Lemma is trivial. Hence we may assume $\ell_2(r)>c\Omega^{1-4a}$. Set $u_\ell=\bigl(-V_\ell(r)\bigr)_+^{-1/2}\chi_{{}_{\ell\ge Z^{10^{-9}}}} =\bigl(-\frac{\ell(\ell+1)}{r^2}-V(r)\bigr)^{-1/2} \chi_{{}_{\ell\ge Z^{10^{-9}}}}$, $v_\ell=\chm(\phi_\ell)\cdot\frac{(2\ell+1)}{n_\ell} \chi_{{}_{\ell\ge Z^{10^{-9}}}}$, $v_\ell^{\#}=\chi_{{}_{\ell\ge Z^{10^{-9}}}} \frac{(2\ell+1)}{n_\ell}$ for $0\le\ell\le \ell_2(r)$. Also, set $\tilde\ell=\ell_2(r)-c(\ell_2(r))^{1-4a}$. Thus, $u_\ell$ is an increasing sequence of non-negative numbers, and $|v_\ell|\le Cv_{\ell}^{\#}$. Moreover, for $\ell_1\le \tilde\ell$ we have $$ \Big|\sum\limits_{\ell_1\le \ell\le \ell_2(r)}v_\ell\Big|\le C(\ell_2(r))^{-2a}\sum\limits_{\ell_1\le \ell\le \ell_2(r)}v_{\ell}^{\#} \ ,\quad\text{by\ (26)}\ .\tag"(47)" $$ \noindent In fact, $\ell_2(r)> c\Omega^{1-4a}$, and $\ell_2(r) -\ell_1\ge c(\ell_2(r))^{1-4a}$, so (26) implies (47) provided $\ell_1\ge Z^{10^{-9}}$. If instead $\ell_1c\Omega^{1-4a}$ and $\ell_2(r)-\ell_1^{\min}> c(\ell_2(r))^{1-4a}$, we again deduce (47) from (26). Thus, (47) is proven. We have verified all the hypotheses in the Corollary to Lemma 7. The conclusion of that Corollary shows that $|\rho_{NT}(r)|\le C\rho_{NT,2}(r)+ C\rho_{NT,3}(r)$. Hence the conclusion of Lemma 8 holds also when $\ell_2(r)>c\Omega^{1-4a}$. The proof of Lemma 8 is complete.$\qquad\blacksquare$\enddemo \medskip Let us estimate $\rho_{NT,1}(r)$. For $Z^{10^{-9}}\le \ell\le (1-\overline c) \Omega$ we have $$ \int_0^R\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell} dr\le \chi_{{}_{R>x_{\text{left}}(\ell)}}\ ,\tag"(48)" $$ \noindent by definition of $n_\ell$, and by virtue of the fact that $\bigl(-V_\ell(r)\bigr)_+^{-1/2}$ is supported in $[x_{\text{left}} (\ell),\infty)$. In that range of $\ell$ we have also $x_{\text{rt}} (\ell)>(1+c)x_{\text{left}}(\ell)$ by (12), so $$\multline \chi_{{}_{R>x_{\text{left}}(\ell)}}\le C\int_0^{2R}\chi_{{}_{r\in (x_{\text{left}}(\ell),x_{\text{rt}}(\ell))}}\frac{dr}{r}\\ =C\int_0^{2R}\chi_{{}_{\frac{\ell(\ell+1)}{r^2}+V(r)<0}}\frac{dr}{r} =C\int_0^{2R}\chi_{{}_{-r^2V(r)>\ell(\ell+1)}}\frac{dr}{r}\ .\endmultline $$ \noindent Combining this with (48), we get $$\multline \int_0^R\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}dr\le C\int_0^{2R}\chi_{{}_{\ell(\ell+1)<-r^2V(r)}}\frac{dr}{r}\ ,\\ \text{for}\quad Z^{+10^{-9}}\le \ell\le (1-\overline c)\Omega\ . \endmultline\tag"(49)" $$ \noindent In particular, (49) holds for $Z^{10^{-9}}\le \ell\le c\Omega^{1-4a}$. Hence, by definition of $\rho_{NT,1}(r)$, we have $$\multline \int_0^R\rho_{NT,1}(r)dr\le \sum\limits_{Z^{10^{-9}}\le \ell\le c\Omega^{1-4a}} (2\ell+1)\cdot C\int_0^{2R}\chi_{{}_{\ell(\ell+1)<-r^2V(r)}}\frac{dr}{r}\\ =C\int_0^{2R}\Bigl\{\sum\Sb Z^{+10^{-9}}\le \ell\le c\Omega^{1-4a}\\ \ell(\ell+1)<-r^2V(r)\endSb (2\ell+1)\Bigr\} \frac{dr}{r} \le C\int_0^{2R}\min\bigl\{(-r^2V(r)\bigr)_+, c\Omega^{2-8a}\bigr\} \frac{dr}{r}\ .\endmultline\tag"(50)" $$ \noindent Thus, we have estimated $\rho_{NT,1}(r)$. Next, we estimate $$\align \rho_{NT,2}(r)&=\sum\limits_{Z^{10^{-9}}\le \ell<\Omega} (2\ell+1)\ell^{-2a}\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}\\ &=\Bigl[\sum\limits_{Z^{10^{-9}}\le \ell<(1-\overline c)\Omega}(2\ell+1) \ell^{-2a}\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}\Bigr]\\ &\ \ +\Bigl[\sum\limits_{(1-\overline c)\Omega\le \ell<\Omega} (2\ell+1)\ell^{-2a}\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}\Bigr]\\ &\equiv \rho_{NT,2}^{\text{LO}}(r)+\rho_{NT,2}^{\text{HI}}(r)\ .\tag"(51)" \endalign $$ \noindent We can estimate $\rho_{NT,2}^{\text{LO}}(r)$ by the same idea as in (50). In fact, (49) yields $$\align \int_0^R\rho_{NT,2}^{\text{LO}}(r)dr&\le \sum\limits_{Z^{10^{-9}}\le \ell<(1-\overline c)\Omega} (2\ell+1)\ell^{-2a}\cdot C\int_0^{2R}\chi_{{}_{\ell(\ell+1)}<-r^2V(r)} \frac{dr}{r}\\ &\le C\int_0^{2R}\Bigl\{\sum\Sb \ell(\ell+1)<-r^2V(r)\\ \ell\ge 0\endSb \ell^{1-2a}\Bigr\}\frac{dr}{r}\\ &\le C\int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\ \frac{dr}{r}\ .\tag"(52)" \endalign $$ To handle $\rho_{NT,2}^{\text{HI}}$, we simply note that $$ \int_0^R\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}dr \le \chi_{{}_{R>x_{\text{left}}(\ell)}}\le \chi_{{}_{R>c\check r}}\quad \text{for}\ \ (1-\overline c)\Omega\le \ell<\Omega\ .\tag"(53)" $$ \noindent That's obvious from the definition of $n_\ell$ and the fact that $\bigl(-V_\ell(r)\bigr)_+^{-1/2}$ is supported in $[x_{\text{left}} (\ell),\infty)$. (See the opening paragraphs of this section for the definition of $\check r$.) Summing (53), we find that $$\multline \int_0^R\rho_{NT,2}^{\text{HI}}(r)dr\le \sum\limits_{(1-\overline c) \Omega\le \ell<\Omega}(2\ell+1)\ell^{-2a}\int_0^R\frac {\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}dr\\ \le C\chi_{{}_{R>c\check r}}\sum\limits_{(1-\overline c)\Omega\le \ell<\Omega}(2\ell+1)\ell^{-2a}\le C\chi_{{}_{R>c\check r}}\Omega^{2-2a} \ .\endmultline\tag"(54)" $$ \noindent Combining (51), (52), (54), we obtain $$ \int_0^R\rho_{NT,2}(r)dr\le C\chi_{{}_{R>c\check r}}\Omega^{2-2a} +C\int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}\ .\tag"(55)" $$ \noindent Thus, we have estimated $\rho_{NT,2}$. Next, we estimate $\rho_{NT,3}$. Define $$ \rho_{\text{JUNK}}^\ell(r)=\frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell} \chi_{{}_{\ell_2(r)-c(\ell_2(r))^{1-4a}\le \ell\le \ell_2(r)}}\ ,\tag"(56)" $$ \noindent so that $$ \rho_{NT,3}=\sum\limits_{Z^{10^{-9}}\le \ell<\Omega}(2\ell+1) \rho_{\text{JUNK}}^\ell\ .\tag"(57)" $$ \noindent We investigate the support and integral of $\rho_{\text{JUNK}}^\ell$. \noindent First suppose $Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega$. Then Lemma 4 applies, so $V_\ell(r)$ satisfies (Z0)$\ldots$(Z9). In particular, $$ -V_\ell(r)\sim\frac{S(x_{\text{left}}(\ell))}{x_{\text{left}}(\ell)} \cdot (r-x_{\text{left}}(\ell))\quad\text{in}\ J_{\text{left}}\equiv \bigl[x_{\text{left}}(\ell), (1+c_1)x_{\text{left}}(\ell)\bigr] \tag"(58)" $$ $$ -V_\ell(r)\sim S(r)\quad\text{in}\quad J_{\text{mid}}\equiv \bigl[(1+c_1)x_{\text{left}}(\ell), (1-c_1)x_{\text{rt}}(\ell)]\tag"(59)" $$ $$ -V_\ell(r)\sim\frac{S(x_{\text{rt}}(\ell))}{x_{\text{rt}}(\ell)} \cdot (x_{\text{rt}}(\ell)-r)\quad\text{in}\ J_{\text{rt}} \equiv \bigl[(1-c_1)x_{\text{rt}}(\ell), x_{\text{rt}}(\ell)\bigr]\ , \tag"(60)" $$ \noindent and $V_\ell(r)>0$ outside $J_{\text{left}}\cup J_{\text{mid}} \cup J_{\text{rt}}$. \smallskip \noindent We know that $\bigl(-V_\ell(r)\bigr)_+^{-1/2}$, hence also $\rho_{\text{JUNK}}^\ell(r)$, is supported in $J_{\text{left}} \cup J_{\text{mid}}\cup J_{\text{rt}}$. In fact, $\rho_{\text{JUNK}}^\ell (r)\equiv 0$ in $J_{\text{mid}}$. To see this, fix $r\in J_{\text{mid}}$. By (59) we have $-V_\ell(r)>-cV(r)$, i.e.\ $0>(1-c)V(r)+\frac{\ell(\ell+1)} {r^2}$, i.e.\ $\ell(\ell+1)<(1-c)\cdot \bigl(-r^2V(r)\bigr)$. This implies $\ell\le (1-c^\prime)\ell_2(r)$, so that we cannot have $\ell_2(r)-c(\ell_2(r))^{1-4a}\le \ell\le \ell_2(r)$. (This argument uses the fact that $\ell\ge Z^{10^{-9}}>>1$ to avoid trivial problems arising for $\ell=0$, $-r^2V(r)<<1$. When $a=0$, we must take care to pick $c$ small enough in $\ell_2(r)-c(\ell_2(r))^{1-4a}\le \ell \le \ell_2(r)$ in order to contradict $\ell\le (1-c^\prime) \ell_2(r)$.) Hence, the characteristic function $\chi_{{}_{\ell_2(r)-c(\ell_2(r))^{1-4a}\le \ell\le \ell_2(r)}}$ is zero when $r\in J_{\text{mid}}$, so $\rho_{\text{JUNK}}^\ell\equiv 0$ in $J_{\text{mid}}$, as claimed. Next, we study $\rho_{\text{JUNK}}^\ell$ in $J_{\text{left}}$. We need the observatioin $$ (x_{\text{left}}(\ell))^2S(x_{\text{left}}(\ell))\ge c\ell^2\ . \tag"(61)" $$ \noindent This can be read from (12) and the definition of $S(\cdot)$, or we can invoke Lemma 4 to see that $$ c\ell\le \Lambda\le C\lambda(x_{\text{left}}(\ell))=CS^{1/2} (x_{\text{left}}(\ell))\cdot x_{\text{left}}(\ell)\ . $$ \noindent Hence we know (61). Now suppose $r\in J_{\text{left}}$ and $\ell_2(r)-c(\ell_2(r))^{1-4a}\le \ell\le \ell_2(r)$. Since $\ell\ge Z^{10^{-9}}>>1$, this implies $\ell_2(r)<\ell+C\ell^{1-4a}$. Hence, $\ell_2(r)+1\le \ell+C^\prime\ell^{1-4a}$. By defintion of $\ell_2(r)$, this shows that $-r^2V(r)<(\ell+C^\prime\ell^{1-4a})(\ell+C^\prime\ell^{1-4a}+1)$, which implies $-r^2V(r)<\ell(\ell+1)+C^{\prime\prime}\ell^{2-4a}$, i.e.\ $$\multline -V_\ell(r)=-V(r)-\frac{\ell(\ell+1)}{r^2}(1+c)x_{\text{left}}(\ell)$, so $\int_0^{2R}\chi_{{}_{V_\ell(r)<0}}\frac{dr}{r}=\int_0^{2R} \chi_{{}_{\frac{\ell(\ell+1)}{r^2}+V(r)<0}}\frac{dr}{r}\ge c\chi_{{}_{R\in [x_{\text{left}}(\ell),\infty)}}$. Hence (66) implies $$\multline \int_0^R\rho_{\text{JUNK}}^\ell(r)\le C\ell^{-2a}\int_0^{2R} \chi_{{}_{\frac{\ell(\ell+1)}{r^2}+V(r)<0}}\frac{dr}{r}\\ \text{for}\quad Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega\ . \endmultline\tag"(67)" $$ \noindent This is our basic estimate for $\rho_{\text{JUNK}}^\ell$ when $Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega$. \smallskip Next we derive an analogue of (67) for $(1-\overline c)\Omega\le \ell \le \Omega-c^\prime\Omega^{1-4a}$. Here, $c^\prime$ is a small constant to be fixed after a few paragraphs. Fix $\ell$ in this range, and apply Lemma 5. Thus we have $$ -V_\ell(r)\sim\frac{\tilde S}{\tilde B}(r-x_{\text{left}}(\ell)) \quad\text{in}\ J_{\text{left}}\equiv [x_{\text{left}}(\ell), x_{\text{left}}(\ell)+c_1\tilde B]\tag"(68)" $$ $$ -V_\ell(r)\sim\tilde S\quad\text{in}\ J_{\text{mid}}\equiv [x_{\text{left}}(\ell)+c_1\tilde B, x_{\text{rt}}(\ell)-c_1\tilde B]\tag"(69)" $$ $$ -V_\ell(r)\sim\frac{\tilde S}{\tilde B}(x_{\text{rt}}(\ell)-r)\quad\text{in} \ J_{\text{rt}}\equiv [x_{\text{rt}}(\ell)-c_1\tilde B, x_{\text{rt}}(\ell)] \tag"(70)" $$ \noindent and $V_\ell(r)>0$ outside $J_{\text{left}}\cup J_{\text{mid}} \cup J_{\text{rt}}$. \smallskip \noindent Here, $\tilde S=\frac{\Omega(\Omega-\ell)}{\check r^2}$; $|J_{\text{left}}|$, $|J_{\text{rt}}|, |J_{\text{mid}}|\sim \tilde B$; and $$ r\sim\check r\quad\text{for}\ r\in J_{\text{left}}\cup J_{\text{mid}} \cup J_{\text{rt}}\ .\tag"(71)" $$ \noindent Therefore, $$ n_\ell=\int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{-1/2}dr\sim\tilde S^{-1/2} \tilde B\ .\tag"(72)" $$ \noindent Note that $\bigl(-V_\ell(r)\bigr)_+^{-1/2}$, hence also $\rho_{\text{JUNK}}^\ell(r)$, is supported in $J_{\text{left}}\cup J_{\text{mid}}\cup J_{\text{rt}}$. Suppose $r\in J_{\text{left}} \cup J_{\text{mid}}\cup J_{\text{rt}}$ belongs to the support of $\rho_{\text{JUNK}}^\ell$. Then $\ell_2(r)-c(\ell_2(r))^{1-4a} \le \ell\le \ell_2(r)$ with a small constant $c$. Since $(1-\overline c)\Omega\le \ell<\Omega$, this implies $\ell_2(r)\le \ell +2c\Omega^{1-4a}$, hence $\ell_2(r)+1\le \ell+3c\Omega^{1-4a}$. By definition of $\ell_2(r)$, this yields $$ -r^2V(r)\le (\ell+3c\Omega^{1-4a})(\ell+3c\Omega^{1-4a}+1)\le \ell(\ell+1)+10\, c\Omega^{2-4a}\ . $$ \noindent Therefore $$ -V_\ell(r)=-V(r)-\frac{\ell(\ell+1)}{r^2}<\frac{10\, c\Omega^{2-4a}} {r^2}\quad\text{for}\ r\in\ \text{supp}\ \rho_{\text{JUNK}}^\ell\ . \tag"(73)" $$ \noindent If $r\in J_{\text{mid}}$, then (69) and (73) show that $\tilde S<\frac{\tilde c\Omega^{2-4a}}{r^2}$. We can make $\tilde c$ small by taking $c$ small. From (71) and the definition of $\tilde S$, we conclude that $$ (\Omega-\ell)\le \tilde c^\prime\Omega^{1-4a}\ .\tag"(74)" $$ \noindent On the other hand, we are studying $\ell$ in the range $(1-\overline c)\Omega\le \ell\le \Omega-c^\prime\Omega^{1-4a}$. This contradicts (74), provided we take $c$ small enough in (56) and pick $c^\prime=2\tilde c^\prime$. Therefore, no $r\in J_{\text{mid}}$ can belong to the support of $\rho_{\text{JUNK}}^\ell$. Now we know that $\text{supp}\ \rho_{\text{JUNK}}^\ell\subset J_{\text{left}}\cup J_{\text{rt}}$. Suppose $r\in (\text{supp}\ \rho_{\text{JUNK}}^\ell)\cap J_{\text{left}}$. >From (68) and (73) we get $$ \frac{\tilde S}{\tilde B}(r-x_{\text{left}}(\ell))\le C\frac{\Omega^{2-4a}}{r^2}\le C^\prime\frac{\Omega^{2-4a}}{\check r^2}\ \quad\text{(by\ (71))}. $$ \noindent By definition of $\tilde S$, this is equivalent to $\frac{(\Omega-\ell)}{\tilde B}(r-x_{\text{left}}(\ell))\le C^\prime\Omega^{1-4a}$, i.e.\ $$ r\le x_{\text{left}}(\ell)+\frac{C^\prime\Omega^{1-4a}}{\Omega-\ell} \tilde B\quad \text{for}\ r\in J_{\text{left}}\cap\ \text{supp}\ \rho_{\text{JUNK}}^\ell\ . \tag"(75)" $$ \noindent Putting (68), (72), (75) into the definition of $\rho_{\text{JUNK}} ^\ell$, we obtain the inequality $$\multline \rho_{\text{JUNK}}^\ell(r)\le C\chi_{{}_{0c\check r}} \Bigl(\frac{\Omega^{1-4a}}{\Omega-\ell}\Bigr)^{1/2}\ ,\\ \text{for}\ (1-\overline c)\Omega\le \ell\le \Omega-c^\prime\Omega^{1-4a}\ . \endmultline\tag"(76)" $$ \noindent This is our basic estimate for $\rho_{\text{JUNK}}^\ell$ when $(1-\overline c)\Omega<\ell\le \Omega-c^\prime\Omega^{1-4a}$. For $\Omega-c^\prime\Omega^{1-4a}<\ell<\Omega$, we use the obvious inequality $\int_0^R\rho_{\text{JUNK}}^\ell(r)dr\le \int_0^\infty \frac{\bigl(-V_\ell(r)\bigr)_+^{-1/2}}{n_\ell}dr=1$ (any $\ell$). Since $\text{supp}\ \rho_{\text{JUNK}}^\ell\subset\ \text{supp}(V_\ell(r))_+^{-1/2} \subset [c\check r, \infty)$ for the $\ell$ in question, it follows that $$ \int_0^R\rho_{\text{JUNK}}^\ell(r)dr\le \chi_{{}_{R>c\check r}} \quad\text{for}\ \Omega-c^\prime\Omega^{1-4a}<\ell<\Omega\ .\tag"(77)" $$ \noindent We can now control $\rho_{NT,3}(r)$ by using (57), (67), (76), (77). >From (57) we get $$ \rho_{NT,3}(r)=\rho_{\text{JUNK},1}(r)+\rho_{\text{JUNK},2}(r)+ \rho_{\text{JUNK},3}(r)\quad\text{with}\tag"(78)" $$ $$ \rho_{\text{JUNK},1}(r)=\sum\limits_{Z^{10^{-9}}\le \ell\le (1-\overline c) \Omega}(2\ell+1)\rho_{\text{JUNK}}^\ell(r)\tag"(79)" $$ $$ \rho_{\text{JUNK},2}(r)=\sum\limits_{(1-\overline c)\Omega<\ell\le \Omega-c^\prime\Omega^{1-4a}}(2\ell+1)\rho_{\text{JUNK}}^\ell(r) \tag"(80)" $$ $$ \rho_{\text{JUNK},3}(r)=\sum\limits_{\Omega-c^\prime\Omega^{1-4a} <\ell<\Omega}(2\ell+1)\rho_{\text{JUNK}}^\ell(r)\ .\tag"(81)" $$ Equations (67) and (79) yield $$\multline \int_0^R\rho_{\text{JUNK},1}(r)dr\le C\sum\limits_{Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega}(2\ell+1) \int_0^{2R}\ell^{-2a}\chi_{{}_{\frac{\ell(\ell+1)}{r^2}+V(r)<0}} \frac{dr}{r}\\ \le C^\prime\int_0^{2R}\Bigl\{\sum\limits_{Z^{10^{-9}}\le \ell\le (1-\overline c)\Omega}\ell^{1-2a}\chi_{{}_{\ell<(-r^2V(r))_+^{1/2}}}\Bigr\}\frac{dr}{r}\\ \le C^{\prime\prime}\int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r} \endmultline\tag"(82)" $$ \noindent Equations (76) and (80) yield $$\multline \int_0^R\rho_{\text{JUNK},2}(r)dr=\sum\limits_{(1-\overline c)\Omega <\ell\le \Omega-c^\prime\Omega^{1-4a}}(2\ell+1)\int_0^R \rho_{\text{JUNK}}^\ell(r)dr\\ \le C\chi_{{}_{R>c\check r}}\sum\limits_{(1-\overline c)\Omega<\ell\le \Omega-c^\prime\Omega^{1-4a}}(2\ell+1)\Bigl(\frac{\Omega^{1-4a}} {\Omega-\ell}\Bigr)^{1/2} \le C^\prime\chi_{{}_{R>c\check r}}\Omega^{2-2a}\ .\endmultline\tag"(83)" $$ \noindent Equations (77) and (81) yield $$\multline \int_0^R\rho_{\text{JUNK},3}(r)dr=\sum\limits_{\Omega-c^\prime \Omega^{1-4a}<\ell<\Omega}(2\ell+1)\int_0^R\rho_{\text{JUNK}}^\ell(r)dr\\ \le \sum\limits_{\Omega-c^\prime\Omega^{1-4a}<\ell<\Omega} (2\ell+1)\chi_{{}_{R>c\check r}}\le C\Omega^{2-4a}\chi_{{}_{R>c\check r}} \le C\chi_{{}_{R>c\check r}}\Omega^{2-2a}\ .\endmultline\tag"(84)" $$ \noindent Combining (78) with (82), (83), (84), we obtain $$ \int_0^R\rho_{NT,3}(r)dr\le C\chi_{{}_{R>c\check r}}\Omega^{2-2a} +C\int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}\ .\tag"(85)" $$ \noindent At last we can estimate $\rho_{NT}$. In fact, Lemma 8 and estimates (50), (55) and (85) together tell us that $$\multline \big|\int_0^R\rho_{NT}(r)dr\big|\le C\chi_{{}_{R>c\check r}} \Omega^{2-2a}+C\int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}\\ +C\int_0^{2R}\min\bigl\{(-r^2V(r))_+, c\Omega^{2-8a}\bigr\} \frac{dr}{r}\ .\endmultline\tag"(86)" $$ \noindent This is our basic result for $\rho_{NT}$. Let us review what has happened so far. Equation (31) expresses the density $\rho(r)$ in terms of $\rho_{\text{LOW}}$, $\rho_{sc}$, $\rho_{NT}$, $\rho_{\text{ERROR}}$, which are defined by (32)$\ldots$(35). We have estimated $\rho_{\text{ERROR}}$ and $\rho_{NT}$ by (46) and (86). It remains to understand $\rho_{sc}$ and estimate $\rho_{\text{LOW}}$. Next we study $\rho_{sc}$, which is of course the main term in $\rho$. We will prove the following elementary result, from which the behavior of $\rho_{sc}$ can be read off easily. \vglue 1pc \proclaim{Lemma 9} For all real numbers $W$, we have $$\multline \sum\limits_{\ell\ge Z^{10^{-9}}}(2\ell+1)\bigl(W-\ell(\ell+1)\bigr)_+^{1/2} =\frac 23W^{3/2}\chi_{{}_{W>Z^{2\cdot 10^{-9}}}}\\ +O\bigl(Z^{2\cdot 10^{-9}}W^{1/2}\chi_{{}_{W>Z^{2\cdot 10^{-9}}}}+ W^{3/4}\chi_{{}_{W>Z^{2\cdot 10^{-9}}}}\bigr)\ .\endmultline $$\endproclaim \vglue 1pc \demo{Proof} If $W\le 2Z^{2\cdot 10^{-9}}$, then the conclusion of the Lemma is trivial. Suppose $W>2Z^{2\cdot 10^{-9}}$.\enddemo \medskip Let $t_{\max}$ be the positive root of $t(t+1)=W$, and set $$ G(t)=\bigl(W-t(t+1)\bigr)^{1/2}\quad\text{for}\ Z^{10^{-9}}\le t\le t_{\max}\ . $$ \noindent We will bound the derivatives of $G(t)$. In fact, $\bigl(\frac{d}{dt}\bigr)^mG(t)$ is a sum of terms of the form $$ \multline \bigl(W-t(t+1)\bigr)^{\frac 12-A}\cdot\prod\limits_{\nu=1}^A \bigl(\frac{d}{dt}\bigr)^{m_\nu}\{t(t+1)\}\ ,\\ \text{with}\ m_\nu\ge 1\ \text{and}\ m_1+\ldots+m_A=m\ . \endmultline\tag"(87)" $$ \noindent For a nonzero term we must have $m_\nu\le 2$, so $1\le m_\nu \le 2$ and $A\le m_1+\ldots+m_A\le 2A$. Thus $A\le m\le 2A$, i.e.\ $\frac m2\le A\le m$. Since $\big|\bigl(\frac{d}{dt}\bigr)^{m_\nu} \{t(t+1)\}\big|\le Ct^{2-m_\nu}$ for $t\ge Z^{10^{-9}}$, the term (87) is dominated by $\bigl(W-t(t+1)\bigr)^{\frac 12-A}t^{2A-m}$. Therefore, $$\align &\Big|\bigl(\frac{d}{dt}\bigr)^mG(t)\Big|\le C_m\sum\limits_{\frac m2\le A\le m}\bigl(W-t(t+1)\bigr)^{\frac 12-A}\ t^{2A-m}\\ &= C_m\sum\limits_{\frac m2\le A\le m}\bigl(W-t(t+1)\bigr)^{1/2} t^{-m}\cdot\bigl(\frac{t^2}{W-t(t+1)}\bigr)^A\ .\tag"(88)"\endalign $$ \noindent We distinguish the cases $t\le \frac 12 t_{\max}$, $t>\frac 12 t_{\max}$. If $t\le \frac 12 t_{\max}$, then $\frac{t^2}{W-t(t+1)} \le C$, so the largest term on the right of (88) comes from $A=\frac m2$. Hence $$\multline \big|\bigl(\frac{d}{dt}\bigr)^mG(t)\big|\le C_m^\prime\bigl(W-t(t+1)\bigr) ^{\frac 12-\frac m2}\le C_m^{\prime\prime}W^{\frac 12-\frac m2}\\ \text{if}\quad Z^{10^{-9}}\le t\le \frac 12 t_{\max}\ .\endmultline\tag"(89)" $$ \noindent If $t_{\max}\ge t\ge \frac 12 t_{\max}$, then $$ \bigl[W-t(t+1)\bigr]\sim W^{1/2}(t_{\max}-t)\ .\tag"(90)" $$ \noindent Therefore, $\frac{t^2}{W-t(t+1)}\sim \frac{W}{W^{1/2}(t_{\max} -t)}>c$, so the largest term in (88) comes from $A=m$. Thus, (88) and (90) yield $$\align \big|\bigl(\frac{d}{dt}\bigr)^mG(t)\big|&\le C_m^\prime \bigl[W^{1/2}(t_{\max}-t)\bigr]^{\frac 12-m}[W^{1/2}]^m\\ &= C_m^\prime\bigl[W^{1/2}(t_{\max}-t)\bigr]^{1/2}\cdot (t_{\max} -t)^{-m}\ ,\tag"(91)"\endalign $$ \noindent for $\frac 12 t_{\max}\le t2Z^{2\cdot 10^{-9}}$.$\qquad\blacksquare$ To control $\rho_{sc}$ using Lemma 9, we simply recall the definitions (28), (33) to write $$\align \rho_{sc}(r)&=\sum\limits_{\ell\ge Z^{10^{-9}}}(2\ell+1)\cdot \frac1\pi \bigl(-\frac{\ell(\ell+1)}{r^2}-V(r)\bigr)_+^{1/2}\\ &=\frac{1}{\pi r}\sum\limits_{\ell\ge Z^{10^{-9}}}(2\ell+1) \bigl(-r^2V(r)-\ell(\ell+1)\bigr)_+^{1/2}\ .\endalign $$ \noindent Lemma 9, with $W=-r^2V(r)$, gives $$ \rho_{sc}(r)=\frac{2}{3\pi r}\bigl(-r^2V(r)\bigr)^{3/2}\chi_{{}_{-r^2 V(r)>Z^{2\cdot 10^{-9}}}}+\tilde\rho_{\text{EXTRA}}(r)\ ,\text{with} \tag"(102)" $$ $$ |\tilde\rho_{\text{EXTRA}}(r)|\le C\chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}} \Bigl\{\frac{\bigl(-r^2V(r)\bigr)^{3/4}}{r}+Z^{2\cdot 10^{-9}} \frac{\bigl(-r^2V(r)\bigr)^{1/2}}{r}\Bigr\}\ .\tag"(103)" $$ \noindent We rewrite (102), (103) slightly as $$ \rho_{sc}(r)=4\pi r^2\cdot\frac{\bigl(-V(r)\bigr)^{3/2}}{6\pi^2} \chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}}+\tilde\rho_{\text{EXTRA}}(r)\ ,\quad \text{or} $$ $$ \rho_{sc}(r)=4\pi r^2\cdot \frac{\bigl(-V(r)\bigr)_+^{3/2}}{6\pi^2}+ \rho_{\text{EXTRA}}(r)\ ,\quad\text{with}\tag"(104)" $$ $$\multline |\rho_{\text{EXTRA}}(r)|\le Cr^2\bigl(-V(r)\bigr)_+^{3/2} \chi_{{}_{-r^2V(r)\le Z^{2\cdot 10^{-9}}}}\\ +C\frac{\bigl(-r^2V(r)\bigr)_+^{3/4}}{r}\cdot\chi_{{}_{-r^2V(r) >Z^{2\cdot 10^{-9}}}}+CZ^{2\cdot 10^{-9}} \frac{\bigl(-r^2V(r)\bigr)_+^{1/2}}{r}\chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}} \endmultline\tag"(105)" $$ \noindent These are our basic results on $\rho_{sc}$. Next, we estimate $\rho_{\text{LOW}}$. Fix $\overline C$ as in Lemma 3. For $\overline C\le \ellV(r)\sim -S(r)$ by (i), (ii), (2), (106) and (107) imply $$ \int_0^{Z^{-9/10}}\rho_\ell(r)dr\le C+CZ^{-3/20}\int_0^\infty S^{1/2} (r)dr+C\int_{S(r)>cZ^{18/10}}S^{1/2}(r)dr $$ \noindent and $$ \int_0^\infty\rho_\ell(r)dr\le C+C\int_0^\infty S^{1/2}(r)dr\ . $$ \noindent Recalling the definition of $S(r)$, we get $$ \int_0^{Z^{-9/10}}\rho_\ell(r)dr\le CZ^{11/60}\qquad\text{for}\quad \overline C\le \ellZ^{-9/10}}}Z^{\frac 13+2\cdot10^{-9}}\ .\tag"(116)" $$ \noindent This is our basic estimate for $\rho_{\text{LOW}}$. \noindent At last we have learned enough to draw conclusions about the three-dimensional density $\rho$. From (31) and (104) we have $$ 4\pi r^2\bigl[\rho(r)-\frac{1}{6\pi^2}\bigl(-V(r)\bigr)_+^{3/2}\bigr] =\rho_{\text{LOW}}(r)+\rho_{\text{EXTRA}}(r)+\rho_{NT}(r)+\rho_{\text{ERROR}} (r)\ .\tag"(117)" $$ \noindent The integrals of $\rho_{\text{LOW}}(r)$, $\rho_{\text{EXTRA}}(r)$, $\rho_{NT}(r)$, on $[0,R]$ are estimated by (116), (105) and (86). Hence we obtain $$\multline \Big|\int_0^R\bigl[\rho_{\text{LOW}}(r)+ \rho_{\text{EXTRA}}(r) +\rho_{NT}(r)]dr\Big| \le \bigl[CZ^{\frac{11}{60}+2\cdot 10^{-9}}+C\chi_{{}_{R>Z^{-9/10}}} Z^{\frac 13+2\cdot10^{-9}}\bigr]\\ \ +\bigl[C\int_0^Rr^2\bigl(-V(r)\bigr)_+^{3/2} \chi_{{}_{-r^2V(r)\le Z^{2\cdot 10^{-8}}}}dr +C\int_0^R\bigl(-r^2V(r)\bigr)_+^{3/4}\chi_{{}_{-r^2V(r)> Z^{2\cdot 10^{-9}}}}\frac{dr}{r}\\ +CZ^{2\cdot 10^{-9}}\int_0^R\bigl(-r^2V(r)\bigr)_+^{1/2} \chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}}\frac{dr}{r}\bigr]\\ +\bigl[C\chi_{{}_{R>c\check r}}\Omega^{2-2a}+C\int_0^{2R} \bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r} +C\int_0^{2R}\min\bigl\{\bigl(-r^2V(r)\bigr)_+, c\Omega^{2-8a} \bigr\}\frac{dr}{r}\bigr]\ .\endmultline\tag"(118)" $$ \noindent (For convenience, we have harmlessly changed a $10^{-9}$ to $10^{-8}$ in (118). This merely weakens the estimate.) \noindent We will check that several of the terms on the right-hand side of (118) may be dropped, because they are dominated by the remaining terms on the right. In fact, $$\multline Z^{2\cdot 10^{-9}}\int_0^R\bigl(-r^2V(r)\bigr)_+^{1/2} \chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}}\frac{dr}{r}\\ \le \int_0^Rr^2\bigl(-V(r)\bigr)_+^{3/2} \chi_{{}_{-r^2V(r)\le Z^{2\cdot 10^{-8}}}}dr +\int_0^R\bigl(-r^2V(r)\bigr)_+^{3/4} \chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}} \frac{dr}{r}\endmultline\tag"(119)" $$ $$ \int_0^R\bigl(-r^2V(r)\bigr)_+^{3/4}\chi_{{}_{-r^2V(r)>Z^{2\cdot 10^{-9}}}} \frac{dr}{r}\le \int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r} \tag"(120)" $$ $$\multline \int_0^Rr^2\bigl(-V(r)\bigr)_+^{3/2}\chi_{{}_{-r^2V(r)\le Z^{2\cdot 10^{-8}}}} dr\\ \le C\int_0^Rr^2S^{3/2}(r)\chi_{{}_{r^2S(r)\le CZ^{2\cdot10^{-8}}}}dr \le CZ^{3\cdot 10^{-8}}\ \text{(by\ (0))}\\ \le CZ^{\frac{11}{60}+2\cdot 10^{-9}}\ .\endmultline\tag"(121)" $$ \noindent Hence on the right-hand side of (118) we may delete the left-hand side of (119), (120), (121) without changing the total order of magnitude. Thus, (118) becomes $$\multline \big|\int_0^R[\rho_{\text{LOW}}(r)+\rho_{\text{EXTRA}}(r)+\rho_{NT}(r)] dr\big|\\ \le CZ^{\frac 15}+C\chi_{{}_{R>Z^{-9/10}}}\cdot Z^{\frac 13+2\cdot 10^{-9}} +CZ^{\frac 23-\frac 23 a}\ \chi_{{}_{R>cZ^{-\frac 13}}}\\ +C\int_0^{2R}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}+C\int_0^{2R} \min\bigl\{\bigl(-r^2V(r)\bigr)_+, cZ^{\frac 23- \frac 83a}\bigr\}\frac{dr}{r}\ .\endmultline $$ \noindent (Here we dominated $Z^{\frac{11}{60}+2\cdot 10^{-9}}$ by $Z^{\frac{12} {60}}=Z^{1/5}$, and we recalled that $\Omega\sim Z^{1/3}$, $\check r\sim Z^{-1/3}$.) This implies trivially that $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R[\rho_{\text{LOW}}(r) +\rho_{\text{EXTRA}}(r)+\rho_{NT}(r)]dr\big|^2\Bigr)^{1/2}\\ \le CZ^{\frac 15}+C\chi_{{}_{R_0>\frac 12Z^{-9/10}}}\cdot Z^{\frac 13 +2\cdot 10^{-9}}+CZ^{\frac 23-\frac 23a}\chi_{{}_{R_0>cZ^{-1/3}}}\\ +C\int_0^{4R_0}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}+C\int_0^{4R_0} \min\bigl\{\bigl(-r^2V(r)\bigr)_+, cZ^{\frac 23-\frac 83a} \bigr\}\frac{dr}{r}\ .\endmultline\tag"(122)" $$ This estimate can be combined with estimate (46) for $\rho_{\text{ERROR}}$. In view of (117), we obtain the following result. $$\multline \Bigl(Av_{|R-R_0|<\frac{1}{10}R_0}\big|\int_0^R[\rho(r) -\frac{1}{6\pi^2}(-V(r))_+^{3/2}\bigr]4\pi r^2dr\big|^2\bigr)^{1/2}\\ \le \Bigl[CZ^{\frac 15}+CZ^{\frac 13+2\cdot 10^{-9}}\chi_{{}_{R_0>\frac 12 Z^{-9/10}}}+CZ^{\frac 23-\frac 23a}\chi_{{}_{R_0>cZ^{-1/3}}}\\ +C\int_0^{4R_0}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r} +C\int_0^{4R_0}\min\bigl\{(-r^2V(r))_+, cZ^{\frac 23-\frac 83a} \bigr\}\Bigr]\\ +\Bigl[Z^{10-N}+CZ^{\frac 23-\frac 23a}\chi_{{}_{R_0>cZ^{-1/3}}}\\ +C\int_0^{2R_0}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}+ C\int_0^{2R_0}\bigl(-r^2V(r)\bigr)_+\chi_{{}_{-r^2V(r)\frac 12 Z^{-9/10}}}+CZ^{\frac 23-\frac 23a}\chi_{{}_{R_0>cZ^{-1/3}}}\\ +C\int_0^{4R_0}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r} +C\int_0^{4R_0}\min\bigl\{\bigl(-r^2V(r)\bigr)_+, cZ^{\frac 23-\frac 83 a} \bigr\}\frac{dr}{r}\ .\endmultline\tag"(123)" $$ \noindent The right-hand side simplifies further. For $r\sim Z^{-1/3}$ we have $-r^2V(r)\sim r^2S(r)=Zr\sim Z^{2/3}$, so that $Z^{\frac 23-\frac 23 a}\chi_{{}_{R_0>cZ^{-1/3}}}\le C\int_0^{4R_0} \bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r}$. Thus, the term $CZ^{\frac 23-\frac 23 a}\chi_{{}_{R_0>cZ^{-1/3}}}$ may be deleted from (123). \noindent Also, $$\multline \int_0^{4R_0}\bigl(-r^2V(r)\bigr)_+^{1-a}\frac{dr}{r} \le \int_0^\infty\bigl(-r^2V(r)\bigr)_+^{1-a} \chi_{{}_{-r^2V(r)<1}}\frac{dr}{r}\\ +\int_0^{4R_0}\chi_{{}_{-r^2V(r)\ge 1}}\cdot \min\bigl\{ \bigl(-r^2V(r)\bigr)_+, C\Omega^{2(1-a)}\bigr\}\frac{dr}{r}\\ \le C\int_0^\infty\bigl(r^2S(r)\bigr)^{1-a}\chi_{{}_{r^2S(r)\frac 12 Z^{-9/10}}} +C\int_0^{4R_0}\min\{r^2S(r), Z^{\frac 23-\frac 23 a}\}\frac{dr}{r}\ . \endmultline\tag"(124)" $$ \noindent By definition (0), $$ \min\{r^2S(r), Z^{\frac 23-\frac 23 a}\} =\cases Zr&\text{if $00$. (See (24)$\ldots$(26)). With $$ G(R)=\int_0^R\bigl[\rho(r)-\frac{\bigl(-V(r)\bigr)^{3/2}}{6\pi ^2} \bigr]4\pi r^2dr\ , $$ \noindent we have $$\multline \Big|\int_0^\infty U(r)\rho(r)4\pi r^2dr-\int_0^\infty U(r) \cdot\frac{\bigl(-V(r)\bigr)^{3/2}}{6\pi ^2}4\pi r^2dr\Big|\\ =\Big|\int_0^\infty U(r)G^\prime(r)dr\big|=\Big|\int_0^\infty U^\prime(r) G(r)dr\Big|\\ \le C\int_0^\infty\Bigl(\frac{1}{R_0}\int_{|R-R_0|<\frac{1}{10} R_0}|U^\prime(R)|\ |G(R)|dR\Bigr)dR_0\\ \le C\int_0^\infty\Bigl(\frac{1}{R_0}\int_{|R-R_0| <\frac{1}{10}R_0}|U^\prime(R)|^2dR\Bigr)^{1/2} \Bigl(\frac{1}{R_0}\int_{|R-R_0|<\frac{1}{10}R_0}|G(R)|^2dR\Bigr)^{1/2} dR_0\\ \le C\int_0^\infty \delta^{-1}\chi_{{}_{c\delta0} \bigl(-r^2V(r)\bigr)$. Thus $\Omega\sim Z^{1/3}$. For integers $0\le \ell<\Omega$, we define $$ n_\ell=\int_0^\infty\bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\bigr)_+^{-1/2}\ dr \quad\text{and} $$ $$ \phi_\ell=\frac 1\pi \int_0^\infty\bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\bigr)_+^{1/2} dr-\frac 12\ . $$ \vglue 1pc \proclaim{Theorem 2} Suppose the numbers, $n_\ell$, $\phi_\ell$ satisfy the following conditions, with $0\le a<1/43$. \roster \item"(A)" There are at most $C\Omega^{1-6a}$ integers $\ell\le \Omega$ for which $|\phi_\ell-\ \text{(nearest\ integer)}|\le \ell^{-6/43}$. \item"(B)" For $Z^{10^{-9}}\le \ell_1<\ell_2<\Omega$ with $\ell_2-\ell_1> \Omega^{1-10a}$, we have $$ \Big|\sum\limits_{\ell_1\le \ell\le \ell_2}\frac{(2\ell+1)}{n_\ell} \chm (\phi_\ell)\Big|\le C\Omega^{-2a}\sum\limits_{\ell_1\le \ell\le \ell_2} \frac{(2\ell+1)}{n_\ell}\ . $$\endroster \noindent Finally, suppose $Z$ is greater than a certain large, positive constant determined by $C$, $a$ in (A), (B); and by the $C_\alpha$ in (1). Then $\int_{\Bbb R^3\times \Bbb R^3}\rho_{\text{error}}(x) \rho_{\text{error}}(y)\frac{dxdy}{|x-y|}\le C^\prime Z^{\frac 53-\frac 23 a}$. The constant $C^\prime$ depends only on $C$, $a$ and the $C_\alpha$ in (1).\endproclaim \vglue 1pc \demo{Proof} (A) and (B) imply (24)$\ldots$(26) in the previous section, so our Lemma is just (142) of that section.$\qquad\blacksquare$\enddemo \medskip Clearly, we shall have to do some work to establish (A) and (B) for a positive $a$. Note that the analogues of (A) and (B) are false for the harmonic oscillator and for the Hydrogen atom, as mentioned in the introduction. \vfill\eject \magnification \magstep1 \input amstex \documentstyle{amsppt} %\NoBlackBoxes \hsize 6 truein \vsize 9truein \baselineskip 20 pt \hoffset .15 truein \catcode`\@=11 \def\logo@{} \catcode`\@=12 \head References\endhead \medskip \widestnumber\key{FS7} \ref\key FS1 \by C. Fefferman and L. Seco \paper The Ground-State Energy of a Large Atom \jour Bull. 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