\magnification \magstep1 \input amstex \nopagenumbers \documentstyle{amsppt} \hsize=6truein \vsize 9truein \hoffset .15truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \centerline{\bf{THE EIGENVALUE SUM FOR 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--2} \smallskip \line{Review of Earlier Results\hfil 3--19}\medskip \baselineskip 12pt \line{The Eigenvalue Sum for an Approximate TF Potential\hfil 20--66} \line{with an Exact Coulomb Singularity\hfil}\smallskip \baselineskip 20pt \line{Perturbation of Eigenvalue Sums\hfil 67--76}\smallskip \line{The WKB Eigenvalue Sum Theorem for Approximate TF Potentials\hfil 77} \smallskip \line{Estimates for Number-Theoretic Sums\hfil 78--94} \smallskip \line{The Main Theorems for Approximate TF Potentials\hfil 95--96} \line{References\hfil 97} \vfill\eject \pageno 1 \input amstex \documentstyle{amsppt} %\load eufm %\load msam %\load msbm %\load eufb \UseAMSsymbols \magnification\magstep1 \NoBlackBoxes \hsize 6truein \vsize 9truein \baselineskip 20pt \hoffset .15truein \catcode\@=11 \def\logo@{} \catcode\@=12 \def\leftheadline{{\rm \folio}\hfil {\sl Fefferman and Seco}} \def\rightheadline{{\sl The Eigenvalue Sum for a Three-Dimensional Radial Potential} \hfil{\rm \folio}} \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 $\Bbb R^3$, and let $E_k$, $\varphi_k(x)$ be the eigenvalues and (normalized) eigenfunctions of $H$. We will study the eigenvalue sum and density, defined by $$\text{sneg}(H)=\sum\limits_{E_k\le 0}E_k\ . \tag"(1)"$$ $$\rho(x)=\sum\limits_{E_k\le 0}|\varphi_k(x)|^2\quad (x\in\Bbb R^3)\ .\tag"(2)"$$ \noindent The standard semiclassical approximations" to these quantities are $$\text{sneg}(H)\approx -\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)_+ ^{5/2}dx\ ,\quad\text{and}\tag"(3)"$$ $$\rho(x)\approx\frac{1}{6\pi^2}\bigl(-V(x)\bigr)_+^{3/2}\ .\tag"(4)"$$ \noindent (See [L].) Here, $t_+^s=t^s$ if $t>0$, $t_+^s=0$ if $t\le 0$. In [FS1], we announced the proof of a precise asymptotic formula for the ground-state energy of a non-relativistic atom. To give the proof, one must understand and refine (3) and (4) for a particular radial potential $V_{TF}^Z$, the Thomas-Fermi potential for an atomic number $Z$. (See [L].) In [FS3], we reduced the asymptotic formula of [FS1] to the task of proving that $$\text{sneg}(H)=-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)_+^{5/2} dx+\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)_+^{1/2} \Delta Vdx+O(Z^{5/3-a})\tag"(5)"$$ \noindent and $$\int_{\Bbb R^3\times\Bbb R^3}\int\bigl[\rho(x)-\frac{1}{6\pi^2} \bigl(-V(x)\bigr)_+^{3/2}\bigr]\bigl[\rho(y)-\frac{1}{6\pi^2} \bigl(-V(y)\bigr)_+^{3/2}\bigr]\frac{dxdy}{|x-y|} =O(Z^{5/3-a})\tag"(6)"$$ \noindent for $V=V_{TF}^Z$, with $a>0$ independent of $Z$. \noindent Simpler and sharper reductions of [FS1] to (5) and (6) have been given by Bach [B], and by Graf-Solvej [GS]. The purpose of this paper is to prove (5) and (6), with $a=\frac{1}{150}$, for a class of radial potentials that includes $V_{TF}^Z$. This completes the proof of the results announced in [FS1]. Our proof of (5) and (6) is based on separation of variables. In [FS2, 4, 5] we made a careful study of ordinary differential operators. In [FS7], we used our ODE results to prove (6) for radial potentials $V$ that satisfy a non-resonance condition." The non-resonance condition is related to the scarcity of periodic orbits of a classical particle in the potential $V$. Here, we again use our results on ODE to show that (5) holds also, provided $V$ satisfies another non-resonance condition, similar to that of [FS7]. Then we will show that the non-resonance conditions hold for a class of radial potentials including $V_{TF}^Z$. Our proof of the non-resonance condition uses elementary number theory, together with an inequality for the Thomas-Fermi potential proved in [FS6]. We believe that (5) and (6) can be proven for the Thomas-Fermi potential for a molecule, with errors $o(Z^{5/3})$. Moreover, we believe that the leading number-theoretic corrections to the density and eigenvalue sum for an atom can be computed rigorously. See Cordoba-Fefferman-Seco [CFS], as well as the introduction to [FS7]. We thank Maureen Schupsky for the skill and effort she has devoted to Texing our long, highly technical papers. \vfill\eject \head Review of Earlier Results\endhead \medskip \subhead{\bf A. Eigenvalue Sums for Ordinary Differential Operators} \endsubhead\medskip The results of this section are taken from [FS5]. \noindent {\bf I.} The First WKB Eigenvalue Sum Theorem \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 $H=-\frac{d^2}{dx^2}+V(x)$ on $I_{\text{BVP}}$, with Dirichlet or Neumann boundary conditions; and let $\text{sneg}(H)$ be the sum of the negative eigenvalues of $H$. Our basic result on $\text{sneg}(H)$ is as follows. \proclaim{First WKB Eigenvalue Sum Theorem} \align {\roman{sneg}}(H)&=-\frac{2}{3\pi}\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{3/2}\ dx+\frac{1}{24\pi}\int_{I_{\text{BVP}}}V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{-1/2}\ dx\\ &\quad+\frac \pi 2\bigl(\phi^\prime(0)\bigr)^{-1}\tilde\chi\bigl( \frac 1\pi\phi(0)-\frac 12)+\ {\roman{Error}}\ , {\roman{with}}\\ &\qquad |{\roman{Error}}|\le \Lambda^{5\varepsilon-2}|V(x_0)|\ .\endalign\endproclaim \medskip \noindent Here, $\phi(0)=\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{1/2}dx$, and $\phi^\prime(0)=\frac 12\int_{I_{\text{BVP}}}\bigl(-V(x)\bigr)_+^{-1/2}dx$, and $\tilde\chi(t)=\operatornamewithlimits{\min}_{k\in\Bbb Z}|t-k-\frac 12|^2 -\frac{1}{12}$. \medskip \noindent{\bf II.} The Second WKB Eigenvalue Sum Theorem \medskip Suppose we are given a smooth potential $V(x)$ defined on $(0,\infty)$, an interval $I\subset (0,\infty)$ containing $\{V(x)<0\}$, and two positive functions $S(x)$, $B(x)$ defined on $I$. We will say that $V(x)$ has an {\sl exact Coulomb singularity\/} with parameters $(\ell,E_0,Z,x_\ast)$ if the following conditions are satisfied: (CS1) $V(x)=V_c(x)\equiv \frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$ for $0Z^{(10^{-9})}$. Set $H=-\frac{d^2}{dx^2}+V(x)$ on $(0,\infty)$ with Dirichlet boundary conditions. Then the sum of the negative eigenvalues of $H$ is given by the equation \align &{\roman{sneg}}(H)-{\roman{sneg}}(H_c)=\\ &\quad -\frac{2}{3\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}dx +\frac{1}{24\pi}\int_0^\infty V^{\prime\prime}(x)\cdot\bigl(-V(x)\bigr)_+^{-1/2} dx\\ &\quad\ \ +\frac \pi 2\bigl(\phi^\prime(0)\bigr)^{-1}\tilde\chi\bigl(\frac 1\pi \phi(0)-\frac 12 \bigr) +\frac {2}{3\pi}\int_0^\infty\bigl(-V_c(x)\big)_+^{3/2}dx\\ &\quad \ \ -\frac{1}{24\pi} \int_0^\infty V_c^{\prime\prime}(x)\cdot \bigl(-V_c(x)\bigr)_+^{-1/2} dx-\frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}- \sqrt{\ell(\ell+1)}-\frac 12\bigr)\\ &\quad \ \ +\ {\roman{Error}}\ ,\endalign \noindent with $H_c=-\frac{d^2}{dx^2}+V_c(x)$ on $(0,\infty)$ subject to Dirichlet boundary conditions, $V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$, and $|{\roman{Error}}|\le \Lambda^{5\varepsilon-2}\frac{Z}{x_\ast}$. Recall that $\phi(E)=\int_0^\infty\bigl(E-V(x)\bigr)_+^{1/2}dx$, $\phi^\prime (E)=\frac 12 \int_0^\infty\bigl(E-V(x)\bigr)_+^{-1/2}dx$, $\tilde\chi(t)=\min_{k\in \Bbb Z}|t-k-\frac 12|^2-\frac{1}{12}$. \endproclaim \medskip \noindent{\bf III.} The Third WKB eigenvalue Sum Theorem \medskip Let $\varepsilon,K,N>0$ be given, with $\varepsilon N\ge 100$. Let $V(x)$ be a potential defined on a (possibly unbounded) interval $I_{\text{BVP}}$. Let $S$, $B$ be positive numbers, and let $x_0 \in I_{\text{BVP}}$ be given. Define $\lambda=S^{1/2}B$. Let $E_\infty$ be a given energy, with $E_\infty>V(x_0)$. We make the following assumptions. \roster \item"{(H0$^*$)}" $I=\{|x-x_0|\frac 12 \lambda^KB$, we have $V(x)\ge E_\infty +\frac{1000}{|x-x_0|^2}$. \item"{(H6$^*$)}" $\lambda$ is bounded below by a positive constant depending only on $c$, $c^\prime$, $c^{\prime \prime}$, $C_\alpha$ in (H0$^*$)$\ldots$(H4$^*$), and on $\varepsilon$, $K$, $N$.\endroster \vglue 1pc \proclaim{Third WKB Eigenvalue Sum Theorem} Assume hypotheses (H0$^\ast$)$\ldots$(H6$^\ast$) with $E_\infty=0$, and assume also $-\lambda^{-3\varepsilon}ScB(x)$. \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 $cS(x)-\frac 43\, \frac{Z}{x_\ast}$ for $x>x_\ast$. Set $V_c (x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$. Then for $H=-\frac{d^2}{dx^2}+ V(x)$, $H_c=-\frac{d^2}{dx^2}+V_c(x)$ on $(0,\infty)$ with Dirichlet boundary conditions, we have $$\text{sneg}(H)-\text{sneg}(H_c)=-\frac{2}{3\pi}\int_0^\infty \big\{\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\} dx+\ {\roman{Error}}\ ,$$ \noindent with $|{\roman{Error}}|cS(x)B^{-1} (x)$. \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 \proclaim{Theorem 2} Suppose $V(x)$, $S(x)$, $B(x)$ satisfy hypotheses (Z$\hat 0$)$\ldots$(Z$\hat 9$), and suppose also that $V(x)$ has an exact Coulomb singularity (CS1)$\ldots$(CS6) for a given $(\ell,E_0,Z,x_\ast)$. Assume $V(x)>-\frac{2Z}{x_\ast}$ for $x>x_\ast$. Assume $V(2x_0) <-\frac{8Z}{x_\ast}$, and assume $$\int_0^\infty\bigl\{\bigl(-V(x)\bigr)_+^{1/2}-\bigl(V(\frac 12x_1)- V(x)\bigr)_+^{1/2}\bigr\}dx<\underline C\ ,\tag"(1)"$$ \noindent where $x_0$, $x_1$ are as in (Z$\hat 0$)$\ldots$(Z$\hat 9$). Set $V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx$. Then for $H=-\frac{d^2}{dx^2}+V(x)$, $H_c=-\frac{d^2}{dx^2}+V_c(x)$ on $(0,\infty)$ with Dirichlet boundary conditions, we have $$\text{sneg}(H)-\text{sneg}(H_c)=-\frac{2}{3\pi}\int_0^\infty\bigl\{\bigl(-V (x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\}dx+\ {\roman{Error}}\ ,$$ \noindent with $|{\roman{Error}}|< C\frac{Z}{x_\ast}$. Here $C$ depends only on $\underline C$ in (1), and on the constants in (Z$\hat 0$)$\ldots$(Z$\hat 9$) and in (CS1)$\ldots$(CS6).\endproclaim \medskip \noindent{\bf $\gamma$.} {\sl Set-up.} 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 $$16 x_0cS(x)B^{-1} (x). \item"(Z3^\dag)" \Lambda=\bigl(\int_I\frac{dx}{\lambda(x)B(x)}\bigr)^{-1} is greater than a certain large, positive number determined by c, C, C_\alpha in (Z0^\dag)\ldots(Z2^\dag). \item"(Z4^\dag)" |V(x)|\le \underline C/(x_0x) for x\in (0,x_0]. \item"(Z5^\dag)" V(x) is increasing and negative in [\frac{x_1}{8}, x_{\text{big}}], and satisfies there |V(x)|<\underline{C} x_1^{-2}. Also, x_{\text{big}}<\underline Cx_1. \item"(Z6^\dag)" V(x)\ge -10^{-9}x^{-2} for x\in [x_{\text{big}},\infty). \item"(Z7^\dag)" For E\in [V(\frac{x_\ast}{10}),0], we have\hfill\break \int_{x_0}^{x_{\text{crit}}}\bigl(E-V(x)\bigr)^{-1/2}dx\le \delta \cdot \int_{x_0}^{\frac 1{20} x_\ast}\bigl(E-V(x)\bigr)^{-1/2}dx. \endroster \vglue 1pc \proclaim{Theorem 3} Suppose V(x), S(x), B(x) satisfy hypotheses (Z0^\dagger)\ldots(Z7^\dagger). Suppose also that V(x) has an exact Coulomb singularity (CS1)\ldots(CS6) for given (\ell,E_0,Z,x_\ast). [We take the same x_\ast in (Z0^\dagger)\ldots (Z7^\dagger) as in (CS1)\ldots(CS6)]. We make the following additional assumptions:$$ V\bigl(\frac{x_\ast}{10}\bigr)<-\frac{3Z}{x_\ast}\tag"(2)"  V(x)>-\frac 43\ \frac {Z}{x_\ast}\quad\text{for}\quad x>x_\ast\tag"(3)"  V(x)>-Cx_{\text{big}}^2x^{-4}\quad \text{for}\ x>x_{\text{big}}\ . \tag"(4)"  \int_0^\infty\bigl\{\bigl(-V(x)\bigr)_+^{1/2}-\bigl(V(\frac 14x_1) -V(x)\bigr)_+^{1/2}\bigr\}dx<\underline C\ ,\tag"(5)" $$\noindent with x_1 as in (Z0^\dagger)\ldots(Z7^\dagger). Set V_c(x)=\frac{\ell(\ell+1)}{x^2}+E_0-\frac Zx. Then for H=-\frac{d^2}{dx^2}+V(x), H_c=-\frac{d^2}{dx^2}+V_c(x) on (0,\infty) with Dirichlet boundary conditions, we have$$ {\roman{sneg}}(H)-{\roman{sneg}}(H_c)=-\frac{2}{3\pi}\int_0^\infty \bigl\{\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x)\bigr)_+^{3/2}\bigr\}dx +\ {\roman{Error}}\ , $$\noindent with |{\roman{Error}}|<\frac{CZ}{x_\ast}. Here, C depends only on \underline C in (5), and on the constants in (Z0^\dagger)\ldots(Z7^\dagger), (CS1)\ldots(CS6), and (4).\endproclaim \medskip \subhead B. Approximate Thomas-Fermi Potentials\endsubhead \medskip The following results are taken from the section on the Density in an Approximate Thomas-Fermi Potential in [FS7]. Let V_Z^{TF}(r) be the Thomas-Fermi potential arising from a nucleus of charge +Z fixed at the origin. Thus, -\Delta_xV_Z^{TF}(|x|)= (\text{const})|V_Z^{TF}(|x|)|^{3/2} on \Bbb R^3\backslash\{0\}, and V_Z^{TF}(r)=-\frac Zr+O(Z^{4/3}) as r\to 0+. Recall that the size of V_Z^{TF}(r) and its derivatives is controlled by the weight functions$$\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)" $$\noindent with c_0 a small enough constant, determined by the C_\alpha in (1). In this section, we use c, C, C^\prime etc.\ to denote constants determined by the C_\alpha in (1), and by the constants \varepsilon, N, a to be introduced later. We assume that Z is large enough, depending on the C_\alpha in (1), and on \varepsilon, N, a. Our goal is to understand the eigenvalue sum arising from the Hamiltonian H=-\Delta_x+V(|x|) for an approximate T-F potential V. By separation of variables, we are led to consider the one-dimensional eigenvalue sums, arising from the potentials$$ 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 First WKB Eigenvalue Sum Theorem, 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 First WKB Eigenvalue Sum Theorem. 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 |\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 the hypotheses of the {\sl Third WKB Eigenvalue Sum Theorem\/}, with \lambda\sim S^{1/2}(\check r)\check r\sim\Omega. The constants in the hypotheses (H0^\ast)\ldots(H6^\ast) depend only on \varepsilon, N and the C_\alpha in (1). \endproclaim \vglue 1pc \demo{Remark} The version of Lemma 6 stated in [FS7] trivially implies the version stated here.\enddemo \medskip We shall also need to understand the function$$ W(r)=-r^2V(r)\tag"(16)" $$\noindent when V is an approximate T-F potential. Set$$ \Cal S(r)=r^2S(r)=\min\{Zr, r^{-2}\}\quad\text{for}\ r> 0\ . \tag"(17)" $$\noindent Then we have the following result. \vglue 1pc \proclaim{Lemma 7} The function W(r) satisfies$$ \big|\bigl(\frac{d}{dr}\bigr)^\alpha W(r)\big|\le C_\alpha\Cal S(r)r^{-\alpha} \quad\text{for}\ r>0\ ,\ \alpha\ge 0\ .\tag"(18)" $$\noindent Moreover, W(r) has a single critical point r=\check r, at which we have$$ cZ^{-1/3}<\check rc\Cal S(\check r)(\check r)^{-2}\ .\tag"(20)" $$\noindent Given any c_1>0 we can find c_2>0 depending on c_1 such that$$ |W^\prime(r)|>c_2\Cal S(r)r^{-1}\quad\text{for}\ |r-\check r|>c_1\check r\ . \tag"(21)" $$\endproclaim \vglue 1pc \demo{Sketch of Proof} Estimates (18) follow at once from (i) above. When V=V_{TF}, the estimates (19), (20), (21) follow by rescaling from the case Z=1, which in turn follows from the section on Elementary Properties of the TF Potential in [FS7]. Hence, (19), (20) and (21) hold for any V that satisfies (2). Details are left to the reader.\quad\blacksquare\enddemo \medskip We close this section by noting that the Thomas-Fermi potential satisfies$$\big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{E_0-\frac Zr-V_{TF}^Z(r)\bigr\}| \le C_\alpha Z^{3/2}r^{1/2-\alpha} \text{for}\ \alpha\ge 0\ \text{and}\ 00} \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 \medskip \subhead D. Approximating Sums by Integrals\endsubhead \medskip The following lemma is taken from [FS4]. For real numbers$t, define: \align \chp(t)&=k-t-\frac 12\quad\text{for}\ k\ \text{the\ smallest\ integer}\ \ge t\ ;\\ \chm(t)&=t-k-\frac 12\quad\text{for}\ k\ \text{the\ largest\ integer}\ \le t\ ;\\ \tilde\chi(t)&=\operatornamewithlimits{\min}_{k\in\Bbb Z} |t-k-\frac 12|^2-\frac{1}{12}\ .\endalign \vglue 1pc \proclaim{Lemma on Riemann Sums} Letf(t)$,$\sigma(t)$,$\tau(t)$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$|t_1-t_2|0}\bigl[-r^2V_{TF}^Z(r) \bigr]^{1/2}$. For$00$independent of$t$and$Z$.\endproclaim \medskip \subhead F. Elementary Integrals\endsubhead\medskip The following elementary identities come from [FS5]. \vglue 1pc \proclaim{Lemma 1} Let$V_c^\ell(r)=\frac{\ell(\ell+1)}{r^2}+E_0-\frac Zr$, and define $$n_\ell=\int_0^\infty\bigl(-V_c^\ell(r)\bigr)_+^{-1/2}dr\ ,\ \phi_\ell=\frac 1\pi\int_0^\infty\bigl(-V_c^\ell(r)\bigr)_+^{1/2} dr-\frac 12\ .$$ \noindent Then$\frac{\pi}{n_\ell}\tilde\chi(\phi_\ell)= \frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}} -\sqrt{\ell(\ell+1)}-\frac 12\bigr)$.\endproclaim \noindent (See the paragraph just before the statement of the second WKB eigenvalue sum theorem in [FS5].) \vglue 1pc \proclaim{Lemma 2} If$Q$,$A$,$P$,$Q^2-4AP>0$, then $$\frac 1\pi\int_0^\infty\bigl(-(\frac{P}{t^2}-\frac Qt+A)\bigr)_+^{1/2}dt=\frac {Q}{2\sqrt A}-\sqrt P\ .$$\endproclaim \noindent (See equation (10) in the section of [FS5] on the second WKB eigenvalue sum theorem). \vfill\eject \head The Eigenvalue Sum for an Approximate TF Potential with an Exact Coulomb Singularity\endhead \medskip Let$V(r)$be a potential on$(0,\infty)$, which approximates the T-F potential$V_Z^{TF}(r)$in the following sense. $$\Big|\bigl(\frac{d}{dr}\bigr)^\alpha V(r)\Big|\le C_\alpha r^{-\alpha}\min\bigl\{\frac Zr, r^{-4}\bigr\},\quad \text{all}\ r\in (0,\infty), \alpha\ge 0\ .\tag"(1)"$$ $$\Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-V_Z^{TF} (r)\bigr\}\Big|\le c_0r^{-\alpha}\min\bigl\{\frac Zr, r^{-4}\bigr\}\quad \text{for}\ 0\le \alpha\le 2\ \text{and\ all}\ r \ .\tag"(2)"$$ \noindent Here$c_0$is a small constant determined by the$C_\alpha$in (1). $$\text{For}\ 00}\bigl(-r^2V(r)\bigr), and suppose the maximum is attained at \check r. Thus \Omega\sim Z^{1/3}, \check r\sim Z^{-1/3}. Similarly, define \Omega_c to be the positive root of \Omega_c(\Omega_{c}+1) =\operatornamewithlimits{\max}_{r>0}\bigl(-r^2V_c(r)\bigr), with V_c(r)=E_0-\frac Zr, and suppose the maximum is attained at \check r_c. Thus, \Omega_{c}\sim Z^{1/3} and \check r_c\sim Z^{-1/3}. \noindent For \ell\ge \Omega we have V_\ell(r)\ge 0 everywhere, so \text{sneg}\, (H_\ell)=0. Similarly, for \ell\ge \Omega_c we have \text{sneg}\, (H_\ell^c)=0. Note that for 0\le \ell\le Z^{+1/5} and x_\ast\sim Z^{\varepsilon- 3/5}, the potential V_\ell(r) satisfies conditions (CS1)\ldots(CS6) with any I\subset (0,\infty) containing \{V_\ell<0\}, with S(r)=\min\bigl\{\frac Zr, r^{-4}\bigr\} and B(r)=r, and with parameters (\ell, E_0, Z,x_\ast). (Conditions (CS1)\ldots(CS6) appear in the section on the second eigenvalue sum theorem.) \noindent In fact, (CS1) follows from (3); (CS2) follows from the definition of S(r), B(r); (CS3) holds since 0\le \ell\le Z^{1/5}\le \frac{1}{16}(Z\cdot cZ^{\varepsilon-3/5})^{1/2}; (CS4) is contained in (3); (CS5) is immediate from x_\ast\sim Z^{\varepsilon -3/5}; and (CS6) is contained in our assumption that Z exceeds a large constant determined by \varepsilon, c, C in (3) and C_\alpha in (1). Our plan is to use our theorems on the eigenvalue sum of an ODE to compute \text{sneg}\, (H_\ell)-\text{sneg}\, (H_\ell^c), and then to substitute the results into (4), (5). The following quantities, familiar from the discussion of the three-dimensional density, will play a role here as well:$$ \phi_\ell=\frac 1\pi \int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{1/2} dr-\frac 12=\frac 1\pi \int_0^\infty\Bigl(-\frac{\ell(\ell+1)}{r^2}- V(r)\Bigr)_+^{1/2}dr-\frac 12\tag"(6)"  n_\ell=\int_0^\infty\bigl(-V_\ell(r)\bigr)_+^{-1/2}dr=\int_0^\infty \Bigl(-\frac{\ell(\ell+1)}{r^2}-V(r)\Bigr)_+^{-1/2}dr\ .\tag"(7)" $$\noindent Let us begin computing\ \text{sneg}\, (H_\ell)-\text{sneg}(H_\ell^c). \proclaim{Lemma 1} Set x_0=\overline CZ^{-1}, x_{\text{crit}}=Z^{-9/10} x_\ast=Z^{\varepsilon-3/5}, x_1=1/\overline C, x_{\text{big}}= \overline C, \delta=\overline CZ^{-3/20} with \overline C a large constant depending on \varepsilon, c, C, C_\alpha in (1)\ldots(3). Then for \ell=0, the potential V_\ell(r) satisfies the hypotheses of Theorem 3 in the section on Eigenvalue Sums in Degenerate Potentials. \endproclaim \demo{Proof} By Lemma 1 of the section on Approximate TF Potentials, V(r) satisfies (Z0^\dag)\ldots(Z7^\dag) with x_\ast replaced by x_\ast^{\text{old}}=\ \text{(const.)}\ Z^{-8/10}. (See section A.IV\gamma for conditions (Z0^\dag)\ldots(Z7^\dag).) Now, x_\ast enters (Z0^\dag)\ldots(Z7^\dag) only in (Z7^\dag). Moreover, x_\ast^{\text{old}}cr^{-1}\min\bigl\{\frac Zr, r^{-4}\bigr\} and |V^\prime-(V_Z^{TF})^\prime |\le c_0r^{-1}\min\bigl\{\frac Zr, r^{-4}\bigr\}.) Hence (Z7^\dag) for x_\ast^{\text{old}} implies (Z7^\dag) for x_\ast. Thus (Z0^\dag)\ldots(Z7^\dag) hold for our present x_0, x_{\text{crit}}, x_\ast, x_1, x_{\text{big}}, \delta. With I=(0,\infty), conditions (CS1)\ldots(CS6) hold also, since \ell=0-\frac{4}{3}\ \frac{Z}{x_\ast}\quad\text{for}\ x>x_\ast\tag"(9)"$$ $$\int_0^\infty\bigl\{\bigl(-V(x)\bigr)_+^{1/2}-\bigl(V(\frac 14 x_1)-V(x)\bigr)_+ ^{1/2}\bigr\}dx\le \underline C\ .\tag"(10)"$$ $$V(x)>-Cx_{\text{big}}^2x^{-4}\quad\text{for}\ x>x_{\text{big}}\ .\tag"(10a)"$$ \noindent To verify these conditions, recall that $$|V_{TF}^Z(r)+\frac Zr-E_0|<\overline{\overline c}Z^{4/3}(Z^{1/3}r)^{1/2} \quad\text{for}\ r-\frac Zr for all r. \noindent Therefore for x_\ast=Z^{\varepsilon-3/5}, we have from (2) that \big|V(\frac{x_\ast}{10})+\frac{Z}{(x_\ast/10)}\big|<2c_0\frac{Z} {(x_\ast/10)}, so (8) is obvious. Also from (2), V(r)\ge -(1+c_0) \frac Zr for all r, so (9) is obvious. Since x_1=1/\overline C, (1), (2) show that V(\frac 14x_1)<0 and |V(\frac 14 x_1)|\le \underline C. Hence$$ \big|\bigl(-V(x)\bigr)_+^{1/2}-\bigl(V(\frac 14x_1)-V(x)\bigr)_+^{1/2} \big|\le \cases C\bigl(-V(x)\bigr)^{-1/2} &\text{if$-V(x)>\underline C$}\\ C\bigl(-V(x)\bigr)^{+1/2} &\text{if$-V(x)\le \underline C$}\endcases\ . $$\noindent Since -V(x)\sim\min\bigl\{\frac Zr,r^{-4}\bigr\} by (1), (2), it follows that$$ \big|\bigl(-V(r)\bigr)_+^{1/2}-\bigl(V(\frac 12x_1)-V(r)\bigr)_+^{1/2}\big| \le \cases C(\min\bigl\{\frac Zr, r^{-4}\bigr\}\bigr)^{-1/2}\le C^\prime &\text{if$0-\frac{2Z}{x_\ast}\quad\text{for}\ x>x_\ast\tag"(13)"  V_\ell(2x_0)<-\frac{8Z}{x_\ast}\tag"(14)" \multline \int_0^\infty\bigl\{\bigl(-V_\ell(x)\bigr)_+^{1/2}-\bigl(V_\ell (\frac 12 x_1)-V_\ell(x)\bigr)_+^{1/2}\bigr\}dx\le \underline C\ ,\\ \text{with}\ \underline C\ \text{allowed\ to\ depend\ on}\ \ell\ .\endmultline \tag"(15)" $$\noindent Recall that V(x)\ge -(1+c_0)\frac Zx for all x, and that \frac{\ell(\ell+1)}{x_0^2}\le 2(\overline C)^{-1}\frac{Z}{x_0} by definition of x_0. Hence for x>x_\ast we have V_\ell(x)>V(x) >-\frac{2Z}{x}>-\frac{2Z}{x_\ast}, which proves (13). Also$$ V_\ell(2x_0)\le 2(\overline C)^{-1}\frac{Z}{x_0}+V(2x_0)\le \bigl\{2(\overline C)^{-1}+c_0\bigr\}\frac{Z}{x_0}+V_Z^{TF}(2x_0)\ . $$\noindent Since 2x_0<>x_0. This proves (14). We control V_\ell(r) by equations (3)\ldots(15) in the section on Approximate TF Potentials. From those equations, we recall that$$ \{V_\ell(r)<0\}=(x_{\text{left}}(\ell), x_{\text{rt}}(\ell))\ \text{with}\ x_{\text{left}}(\ell)\sim \frac{\ell^2}{Z}\ , x_{\text{rt}} (\ell)\sim \frac 1\ell\ , $$\noindent and that |V_\ell(r)|\le C\min\bigl\{\frac Zr, r^{-4}\} in [x_{\text{left}}(\ell), x_{\text{rt}}(\ell)]. Since x_1=\frac{1}{\overline C\ell} and 1\le \ell\le Z^{10^{-9}}, we have \frac 12 x_1\in [x_{\text{left}}(\ell), x_{\text{rt}}(\ell)], so 0\le -V_\ell(\frac 12 x_1)\le C(\overline C\ell)^{4}. This implies$$ 0\le \bigl\{\bigl(-V_\ell(r)\bigr)_+^{1/2}-\bigl(V_\ell(\frac 12 x_1) -V_\ell(r)\bigr)_+^{1/2}\bigr\}\le \cases C(\overline C\ell)^4\bigl(-V_\ell(r) \bigr)^{-1/2} &\text{if $-V_\ell(r)>(\overline C\ell)^4$}\\ C(\overline C\ell)^{4/2} &\text{if $0<-V_\ell(r)\le (\overline C\ell)^4$}\\ 0 &\text{if $-V_\ell(r)<0$\ .}\endcases $$\noindent Hence$$ 0\le \bigl\{\bigl(-V_\ell(r)\bigr)_+^{1/2}-\bigl(V_\ell(\frac 12 x_1)-V_\ell (r)\bigr)_+^{1/2}\bigr\}\le (\text{Const.})\ell^2 \chi_{{}_{r\in [x_{\text{left}}(\ell), x_{\text{rt}}(\ell)]}}\ , $$\noindent which makes (15) obvious. The proof of Lemma 2 is complete. \qquad\blacksquare \medskip >From Lemma 2 above, and from Theorem 2 in the section on Eigenvalue Sums for Degenerate Potentials, we conclude that$$\multline \text{sneg}(H_\ell)-\text{sneg}(H_\ell^c)=-\frac{2}{3\pi}\int_0^\infty \bigl\{\bigl(-V_\ell(r)\bigr)_+^{3/2}-\bigl(-V_\ell^c(r)\bigr)_+^{3/2} \bigr\}dr+\ \text{Err}_\ell\endmultline\tag"(16)" $$\noindent for 1\le \ell\le Z^{10^{-9}}, with$$ |\text{Err}_\ell|\le C(\ell)\frac{Z}{x_\ast}=C(\ell)Z^{-\varepsilon+8/5}\ . \tag"(17)" $$\vglue 1pc \proclaim{Lemma 3} Pick I, x_{\text{crit}}, E_{\text{crit}}, \delta as in Lemma 3 in the section on Approximate TF Potentials, and let \overline C be the large constant mentioned in that lemma. Set x_\ast=Z^{\varepsilon-\frac 35}. Then for \overline C\le \ell\le Z^{10^{-9}}, V_\ell(r) satisfies the hypotheses of Theorem 1 in the section on Eigenvalue Sums for Degenerate Potentials.\endproclaim \vglue 1pc \demo{Proof} Hypotheses (Z\overline 0)\ldots(Z\overline 8) hold for V_\ell(r), by virtue of Lemma 3 in the section on Approximate TF Potentials.(For those hypotheses, see section A.IV\alpha.) Also, our I contains \{V_\ell<0\}= (x_{\text{left}} (\ell), x_{\text{rt}}(\ell)), and \ell\le Z^{1/5}, so (CS1)\ldots(CS6) hold as well. The only remaining hypotheses in Theorem 1 on Eigenvalue Sums for Degenerate Potentials are the following.$$ E_{\text{crit}}<-\frac{3Z}{x_\ast}\tag"(18)"  V_\ell(x)>-\frac 43\frac{Z}{x_\ast}\quad\text{for}\ x>x_\ast\ .\tag"(19)" $$We are using E_{\text{crit}}=-Z^{18/10}, x_\ast=Z^{\varepsilon-3/5}, so (18) is obvious. Since V_\ell(x)>V(x)>-(1+c_0)\frac Zx for all x>0, (19) is also obvious. The proof of the Lemma. is complete. \qquad\blacksquare\enddemo >From Lemma 3 above, and from Theorem 1 on Eigenvalue Sums for Degenerate Potentials, we conclude that$$\multline \text{sneg}(H_\ell)-\text{sneg}(H_\ell^c)=-\frac{2}{3\pi}\int_0^\infty \bigl\{\bigl(-V_\ell(r)\bigr)_+^{3/2}-\bigl(-V_\ell^c(r)\bigr)_+^{3/2}\bigr\} dr+\ \text{Err}_\ell\\ \text{for}\ \overline C\le \ell\le Z^{10^{-9}}\ ,\ \text{with}\endmultline \tag"(20)"  |{\roman{Err}}_{\ell}|\le C\frac{Z}{x_\ast}=CZ^{\frac 85-\varepsilon}\ .\tag"(21)" $$Next, suppose Z^{10^{-9}}<\ell\le Z^{+1/5}. Set x_\ast=Z^{\varepsilon-3/5}, and take I, K as in Lemma 4 from the section on Approximate TF Potentials. That lemma shows that (Z0)\ldots(Z9) hold for V_\ell(r), S(r)=\min\{\frac Zr, r^{-4}\}, B(r)=r, with \Lambda\sim\ell. Also \{V_\ell<0\}\subset I by (Z0)\ldots(Z9), and \ell\le Z^{+1/5}; hence, (CS1)\ldots(CS6) hold for V_\ell(r). Thus, the hypotheses of the Second WKB Eigenvalue Sum Theorem are satisfied. Applying that Theorem, we see that$$\multline \text{sneg}(H_\ell)-\text{sneg}(H_\ell^c)=\\ \ -\frac{2}{3\pi}\int_0^\infty\bigl\{\bigl(-V_\ell(x)\bigr)_+^{3/2} -\bigl(-V_\ell^c(x)\bigr)_+^{3/2}\bigr\}dr\\ \ +\frac{1}{24\pi}\int_0^\infty\bigl\{V_\ell^{\prime\prime}(x)\cdot \bigl(-V_\ell(x)\bigr)_+^{-1/2}-V_\ell^{c^{\prime\prime}}(x) \bigl(-V_\ell^c(x)\bigr)_+^{-1/2}\bigr\}dx\\ +\ \frac{\pi}{n_\ell}\tilde\chi(\phi_\ell)-\frac{2E_0^{3/2}} {Z}\tilde \chi\bigl(\frac{Z}{2E_0^{1/2}}-\sqrt{\ell(\ell+1)}-1/2\bigr) +\ \text{Err}_\ell\endmultline\tag"(22)" $$\noindent for Z^{10^{-9}}<\ell\le Z^{+1/5} with$$ |\text{Err}_\ell|\le \Lambda^{5\varepsilon-2}\frac{Z}{x_\ast} \le\ell^{5\varepsilon-2}Z^{8/5}\ .\tag"(23)" $$\noindent (See the definitions (6), (7) to check that (22) agrees with the conclusion of the Eigenvalue Sum Theorem.) Next, suppose Z^{1/5}\le \ell\le (1-\overline c)\Omega. Then Lemma 4 in the section on Approximate TF Potentials shows that V_\ell(r) satisfies (Z0)\ldots(Z9) with \Lambda\sim\ell. Therefore, the hypotheses of the First WKB Eigenvalue Sum Theorem are satisfied. Applying that theorem, we see that$$\multline \text{sneg}(H_\ell)=-\frac{2}{3\pi}\int_0^\infty\bigl(-V_\ell(x)\bigr)_+^{3/2} dx+\frac{1}{24\pi}\int_0^\infty V_\ell^{\prime\prime}(x)\cdot \bigl(-V_\ell(x)\bigr)_+^{-1/2}dx\\ +\frac{\pi}{n_\ell}\tilde\chi(\phi_\ell)+\ \text{Err}_\ell\endmultline\tag"(24)" $$\noindent for Z^{+1/5}\le \ell\le (1-\overline c)\Omega, with$$ |\text{Err}_\ell|\le \Lambda^{5\varepsilon-2}|\operatornamewithlimits{\min}\limits _{r>0} V_\ell(r)|\sim C\ell^{5\varepsilon-2}\cdot \bigl(\frac{Z^2}{\ell^2}\bigr) \ .\tag"(25)" $$\noindent To see that \operatornamewithlimits{\min}\limits_{r>0}V_\ell(r) \sim -\frac{Z^2}{\ell^2}, we refer to equations (3)\ldots(7) and (12), (13) in the section on Approximate TF Potentials. Next, suppose (1-\overline c)\Omega\le\ell\le\Omega-\Omega^{1-10\varepsilon}. Then Lemma 5 in the section on Approximate TF Potentials shows that (Z0)\ldots(Z9) hold for V_\ell(r), with \tilde S= \frac{\Omega(\Omega-\ell)}{\check r^2}, \tilde B and I picked suitably, and \Lambda\sim\Omega-\ell. Hence V_\ell(r), \tilde S, \tilde B, I satisfy the hypotheses of the First WKB Eigenvalue Sum Theorem. Applying that Theorem, we see that$$\multline \text{sneg}(H_\ell)=-\frac{2}{3\pi}\int_0^\infty\bigl(-V_\ell(x)\bigr)_+^{3/2} dx+\frac{1}{24\pi}\int_0^\infty V_\ell^{\prime\prime}(x)\cdot \bigl(-V_\ell(x)\bigr)_+^{-1/2}dx\\ +\frac{\pi}{n_\ell}\tilde\chi(\phi_\ell)+\ \text{Err}_\ell\endmultline\tag"(26)" $$\noindent for (1-\overline c)\Omega\le\ell\le \Omega-\Omega^{1-10\varepsilon}, with$$ |\text{Err}_\ell|\le \Lambda^{5\varepsilon-2}|\min V_\ell(r)|\sim (\Omega-\ell)^{5\varepsilon-2}\tilde S=\frac \Omega{\check r^2}(\Omega-\ell) ^{5\varepsilon-1}\ , $$\noindent i.e.$$ |\text{Err}_\ell|\le Z(\Omega-\ell)^{5\varepsilon-1}\ .\tag"(27)" $$Next we recall from the section on Approximate TF Potentials the following estimates. (See (14), (15) in that section.) For (1-\overline c)\Omega\le\ell<\Omega, there are an x_0(\ell)\sim Z^{-1/3} and an interval I=\{|x-x_0(\ell)|\frac{c\ell^2}{x^2}\quad\text{outside}\ I\ .\tag"(31)"$$ \noindent Here $0\frac{c\ell^2} {x^2}>\frac{c^\prime\ell^2}{|x-x_0|^2}>\frac{1000}{|x-x_0|^2}$ since $\ell\ge \Omega-\Omega^{1-10\varepsilon}\sim Z^{1/3}$. This implies (H5$^\ast$) at once, and (H4$^\ast$) also, since $\min\{E_\infty, V(x_0)+c^{\prime\prime}\lambda^{-2\varepsilon}S\}\le E_\infty=0$. (H6$^\ast$) follows from the fact that we take $Z$ large enough, since $\lambda=S^{1/2}B\sim (Z^{4/3})^{1/2}Z^{-1/3}=Z^{1/3}$. It remains to check that $-\lambda^{-3\varepsilon}Scx_{\text{left}}\ \text{and}\ \text{dist}(x_{\text{rt}}, \partial I)>cx_{\text{rt}}$. In fact, $\{V_\ell^c<0\}=\{x^2V_\ell^c(x)<0\}=\{\ell(\ell+1)-Zx+E_0x^2<0\}$, which has the form $(x_{\text{left}}, x_{\text{rt}})$ provided the discriminant $Z^2-4E_0\ell(\ell+1)>0$. Now $\ell<(1-\overline{\overline c})\Omega_c$ with $\Omega_c(\Omega_c+1)=\operatornamewithlimits{\max}\limits_{x>0} \bigl(-x^2(E_0-\frac Zx)\bigr)=\operatornamewithlimits{\max}\limits_{x>0} \bigl(Zx-E_0x^2)=\frac{Z^2}{4E_0}$. Hence $\ell(\ell+1) <(1-\overline{\overline c})\Omega_c(\Omega_c+1)=(1-\overline{\overline c})\frac{Z^2}{4E_0}$, so that $Z^2-4E_0\ell(\ell+1)>\overline{\overline c}Z^2>0$, as needed. \smallskip \noindent Note that $x_{\text{left}}=\frac{Z-\sqrt{Z^2-4E_0 \ell(\ell+1)}}{2E_0}$, $x_{\text{rt}}=\frac{Z+\sqrt{Z^2-4E_0\ell(\ell+1)}}{2E_0}$, so that $x_{\text{left}}>0$ and $x_{\text{rt}}<\frac{2Z}{2E_0}$. Hence $\{V_\ell^c<0\}=(x_{\text{left}},x_{\text{rt}})\subset (0,\frac{Z}{E_0})$. For $0\frac{\ell(\ell+1)}{x^2}-\frac Zx>0$, so $$\{V_\ell^c<0\}=(x_{\text{left}}, x_{\text{rt}})\subset \bigl[ \frac{\ell(\ell+1)}{Z},\frac{Z}{E_0}\big]\ .\tag"(33\text{bis})"$$ \noindent This implies $(x_{\text{left}}, x_{\text{rt}})\subset I$ and $\text{dist}(x_{\text{left}}, \partial I)>cx_{\text{left}}$ and $\text{dist}(x_{\text{rt}}, \partial I)>cx_{\text{rt}}$, by definition of $I$. This completes the proof of (Z2) and (Z5). \smallskip \noindent (Z3) is proven as follows. We have $x_0^2V_\ell^c(x_0)=\ell(\ell+1) -Zx_0+E_0x_0^2=\ell(\ell+1)-2\ell(\ell+1)+E_0\cdot\frac{4\ell^2(\ell+1)^2} {Z^2}=-\ell(\ell+1)\bigl[1-\frac{4E_0}{Z^2}\ell(\ell+1)\bigr]$. Also $0<\ell\le (1-\overline{\overline c})\Omega_c$, so $\ell(\ell+1)\le (1-\overline{\overline c}) \Omega_c(\Omega_c+1)=(1-\overline{\overline c})\operatornamewithlimits{\max} \limits_{x>0} \bigl(-x^2(E_0-\frac Zx)\bigr)$ by definition of $\Omega_c$. The max.\ of $-x^2(E_0-\frac Zx)=Zx-E_0x^2$ is $\frac{Z^2}{4E_0}$, attained at $x=\frac{Z}{2E_0}$, so that $\ell(\ell+1)\le (1-\overline{\overline c}) \cdot\frac{Z^2}{4E_0}$. Therefore, $-x_0^2V_\ell^c(x_0)=\ell(\ell+1) \cdot\bigl[1-\frac{4E_0}{Z^2}\ell(\ell+1)\bigr]\ge \overline{\overline c} \ell(\ell+1)$, so $-V_\ell^c(x_0)\ge \overline{\overline c} \frac{\ell(\ell+1)}{x_0^2}\ge c^\prime \frac{Z}{x_0}$ because $x_0=2\ell(\ell+1)Z^{-1}$. Thus $V_\ell^c(x_0)\le -c^\prime S(x_0)$ for $S(x)=\frac Zx$. The derivatives of $V_\ell^c$ are given by $$(V_\ell^c)^\prime(x)=-\frac{2\ell(\ell+1)}{x^3}+\frac{Z}{x^2} =\frac{Zx-2\ell(\ell+1)}{x^3}=\frac Z{x^2}\cdot\frac {[x-x_0]}{x}\tag"(34)"$$ \noindent and $$(V_\ell^c)^{\prime\prime}=+\frac{6\ell(\ell+1)}{x^4}-\frac{2Z}{x^3} =\frac{6\ell(\ell+1)-2Zx}{x^4}\ .\tag"(35)"$$ \noindent From (34), we have $(V_\ell^c)^\prime(x_0)=0$. If $|x-x_0| cS(x_0)B^{-2}(x_0)$ for $|x-x_0|1-\frac{1}{1+c_1}=c>0$. This is the remaining assertion of (Z4). \smallskip \noindent (Z5) has already been proven, together with (Z2). Let us compute $\Lambda=\Bigl(\int_{x_{\text{left}}}^{x_{\text{rt}}}\frac{dx} {S^{1/2}(x)x^2}\Bigr)^{-1}$. Since $|(V_\ell^c)^\prime(x)|\le \frac{CZ}{x^2}$ in $[x_{\text{left}},x_0]$ by (Z1), while $V_\ell^c(x_{\text{left}})=0$ by definition, and $(V_\ell^c)(x_0) <-\frac{cZ}{x_0}$ by (Z3), it follows that $x_{\text{left}}\le (1-c) x_0$ for a small, positive constant $c$. Hence $x_{\text{rt}}/ x_{\text{left}}>x_0/x_{\text{left}}>1+c^\prime$ for a small, positive constant $c^\prime$. So $\Lambda^{-1}=\int_{x_{\text{left}}}^{x_{\text{rt}}} \frac{dx}{(\frac Zx)^{1/2}x^2}=Z^{-1/2}\int_{x_{\text{left}}}^{x_{\text{rt}}} \frac{dx}{x^{3/2}}\sim Z^{-1/2}x_{\text{left}}^{-1/2}$. From (33 bis) and $x_{\text{left}}>1$. Hence $\Lambda^{-1}\sim Z^{-1/2}(\ell^2Z^{-1})^{-1/2} =\ell^{-1}$, so $\Lambda\sim\ell$ as claimed in the statement of Lemma 5. \smallskip \noindent (Z6) is proven as follows. The assertion of (Z6) about $x\in I_{\text{BVP}}$ with $xx_{\text{rt}}+\Lambda^KB(x_{\text{rt}})$, we must show that $V_\ell^c(x)\ge \frac{1000}{|x-x_{\text{rt}}|^2}$. \smallskip By the formula $x_{\text{rt}}=\frac{Z+\sqrt{Z^2-4E_0\ell(\ell+1)}}{2E_0}$, we have $x_{\text{rt}}\sim \frac{Z}{E_0}$. If $x>10^9\frac{Z}{E_0}$, then $V_\ell^c(x)\ge E_0-\frac Zx>\frac 12 E_0>1000$ since $E_0\sim Z^{4/3}$. For $x>x_{\text{rt}}+\Lambda^Kx_{\text{rt}}$ we have $x-x_{\text{rt}}> (c\ell)^K\bigl(\frac{cZ}{E_0}\bigr)\ge (cZ^{1/5})^K\cdot cZ^{-1/3} >Z$ since $K=10^{+9}$. Hence for $x>x_{\text{rt}}+\Lambda^Kx_{\text{rt}}$, we have $V_\ell^c(x)>1000>\frac{1000}{(x-x_{\text{rt}})^2}$, which is the assertion of (Z6). \smallskip \noindent (Z7) amounts to saying that $(x_{\text{rt}}/x_{\text{left}})\le \Lambda^K$. \noindent We saw that $x_{\text{left}}\sim \ell(\ell+1)Z^{-1}> Z^{-1}$ for $\ell\ge Z^{\frac 15}$, and that $x_{\text{rt}}\sim \frac{Z}{E_0}\sim Z^{-1/3}$. So (Z7) follows if we have $CZ^{2/3}\le \Lambda^K$. Since $\Lambda\sim\ell\ge Z^{1/5}$ while $K=10^{+9}$, this is obvious. So (Z7) holds. \smallskip \noindent (Z8) holds, simply because we pick $\hat c$ small enough, and \smallskip \noindent (Z9) holds because $\Lambda\sim\ell\ge Z^{1/5}$ and we take $Z$ large enough. The proof of Lemma 5 is complete. $\qquad\blacksquare$\enddemo \medskip Lemma 5 and the first WKB Eigenvalue Sum Theorem show that $$\multline \text{sneg}(H_\ell^c)=-\frac{2}{3\pi}\int_0^\infty \bigl(-V_\ell^c(x)\bigr)_+^{3/2} dx+\frac{1}{24\pi}\int_0^\infty (V_\ell^c)^{\prime\prime}(x)\cdot\bigl( -V_\ell^c(x)\bigr)_+^{-1/2}dx\\ +\frac{\pi}{n_\ell^c}\tilde\chi(\phi_\ell^c)+\ \text{Err}_\ell^c\quad \text{for}\ Z^{1/5}\le \ell\le (1-\overline{\overline c})\Omega_c\ ,\text{with} \endmultline\tag"(36)"$$ $$n_\ell^c=\int_0^\infty\bigl(-V_\ell^c(x)\bigr)_+^{-1/2}dx\tag"(37)"$$ $$\phi_\ell^c=\frac{1}{\pi}\int_0^\infty\bigl(-V_\ell^c(x)\bigr)_+^{1/2} dx-\frac 12\ ,\tag"(38)"$$ $$|\text{Err}_\ell^c|\le \Lambda^{5\varepsilon-2}|\operatornamewithlimits{\min} \limits_{x>0}V_\ell^c(x)|\sim \ell^{5\varepsilon-2}\cdot\frac{Z^2}{\ell^2}\ . \tag"(39)"$$ In the discussion of elementary integrals in the Review of Earlier Results, we saw that $$\frac{\pi}{n_\ell^c}\tilde\chi(\phi_\ell^c)=\frac{2E_0^{3/2}}{Z} \tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}-\sqrt{\ell(\ell+1)}-\frac 12\bigr)\quad \text{for}\ 0<\ell<\Omega_c\ .\tag"(40)"$$ \noindent Substituting this into (36), we obtain $$\multline \text{sneg}(H_\ell^c)=-\frac{2}{3\pi}\int_0^\infty\bigl(-V_\ell^c(x)\bigr)_+ ^{3/2}dx+\frac{1}{24\pi}\int_0^\infty V_\ell^{c^{\prime\prime}}(x) \bigl(-V_\ell^c(x)\bigr)_+^{-1/2}dx\\ +\frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}} -\sqrt{\ell(\ell+1)}-\frac 12\bigr)+\ \text{Err}_\ell^c\\ \text{for}\ Z^{1/5}<\ell\le (1-\overline{\overline c})\Omega_c ,\ \text{with} \endmultline\tag"(41)"$$ $$|\text{Err}_\ell^c|\le C\ell^{5\varepsilon-4}Z^2\ .\tag"(42)"$$ \noindent To compute $\text{sneg}(H_\ell^c)$ for $(1-\overline{\overline c}) \Omega_c\le \ell<\Omega_c$, we use the following elementary observation. \vglue 1pc \proclaim{Lemma 6} Suppose $(1-\overline{\overline c})\Omega_c\le\ell<\Omega_c$; and set $x_0=2\ell(\ell+1)Z^{-1}$, and $I=\{|x-x_0|<\frac{1}{10}x_0\}$. Then $$\big|\bigl(\frac{d}{dx}\bigr)^\alpha V_\ell^c(x)\big|\le C_\alpha Z^{4/3}x_0^{-\alpha}\quad\text{in}\ I\tag"(43)"$$ $$\bigl(\frac{d}{dx}\bigr)^2V_\ell^c(x)\ge cZ^{4/3}x_0^{-2}\quad\text{in}\ I\tag"(44)"$$ $$\bigl(\frac{d}{dx}V_\ell^c\bigr)(x_0)=0\quad\text{and}\quad -V_\ell^c(x_0)\sim \frac{\Omega_c}{x_0^2}(\Omega_c-\ell)\sim Z(\Omega_c-\ell)\tag"(45)"$$ $$V_\ell^c(x)\ge \frac{c\ell^2}{x^2}\quad\text{outside}\ I\ .\tag"(46)"$$\endproclaim \vglue 1pc \demo{Proof} Note that $\ell\sim Z^{1/3}$, $\Omega_c\sim Z^{1/3}$, $x_0\sim Z^{-1/3}$. We have $$\gather \bigl(\frac{d}{dx}\bigr)^\alpha\bigl\{\frac{\ell(\ell+1)}{x^2}\bigr\} =c_\alpha\frac{\ell(\ell+1)}{x^{2+\alpha}}\sim Z^{4/3}x_0^{-\alpha} \ \text{in}\ I\ ,\\ \bigl(\frac{d}{dx}\bigr)^\alpha\{E_0\}\sim Z^{4/3}x_0^{-\alpha} \ \ \text{if}\ \alpha=0\ ,\ 0\ \text{otherwise}\\ \bigl(\frac{d}{dx}\bigr)^\alpha\bigl\{\frac Z x\bigr\}=c_\alpha^\prime \frac{Z}{x^{1+\alpha}}\sim Z^{4/3}x_0^{-\alpha}\ \text{in}\ I\ .\endgather$$ \noindent This proves (43). To prove (44), recall that $(V_\ell^c)^{\prime\prime}(x)=\frac{6\ell(\ell+1)-2Zx}{x^4}\break\ge \frac{6\ell(\ell+1)-2Z\cdot\frac{11}{10}\cdot 2\ell(\ell+1)Z^{-1}} {x^4}$ for $x\in I$. Thus $x\in I$ implies $(V_\ell^c)^{\prime\prime}(x)\ge\frac{\ell(\ell+1)}{x^4}\sim \frac{\ell(\ell+1)}{x_0^2}\cdot \frac{1}{x_0^2}\sim Z^{4/3} x_0^{-2}$, proving (44). Next, we check (45). We have $$(V_\ell^c)^\prime(x_0)=-\frac{2\ell(\ell+1)}{x_0^3}+\frac{Z}{x_0^2}=0\ ,\ \text{and}\tag"(47)"$$ $$\multline -x_0^2V_\ell^c(x_0)=-\ell(\ell+1)+Zx_0-E_0x_0^2=-\ell(\ell+1) +2\ell(\ell+1)-E_0\cdot\frac{4\ell^2(\ell+1)^2}{Z^2}\\ =\ell(\ell+1)\bigl[1-\frac{4E_0}{Z^2}\ell(\ell+1)\bigr]\ .\endmultline \tag"(48)"$$ \noindent Since $\Omega_c(\Omega_c+1)=\operatornamewithlimits{\max}\limits_{x>0} \bigl(-x^2(E_0-\frac Zx)\bigr)=\frac{Z^2}{4E_0}$, (48) may be rewritten as $$\multline -x_0^2V_\ell^c(x_0)=\ell(\ell+1)\bigl[1-\frac{\ell(\ell+1)}{\Omega_c(\Omega_c +1)}\bigr]=\frac{\ell(\ell+1)}{\Omega_c(\Omega_c+1)} [\Omega_c(\Omega_c+1)-\ell(\ell+1)]\\ \sim[\Omega_c(\Omega_c+1)-\ell(\ell+1)]\sim \Omega_c(\Omega_c-\ell)\ \text{for}\ (1-\overline{\overline c})\Omega_c\le\ell<\Omega_c\ .\endmultline$$ \noindent Thus $-V_\ell^c(x_0)\sim \frac{\Omega_c}{x_0^2} (\Omega_c-\ell)\sim Z(\Omega_c-\ell)$. This and (47) prove (45). It remains to check (46). We argue as follows. \align x^2V_\ell^c(x)&=[\Omega_c(\Omega_c+1)+E_0x^2-Zx]+[\ell(\ell+1) -\Omega_c(\Omega_c+1)]\\ &=[\frac{Z^2}{4E_0}+E_0x^2-Zx]-[\Omega_c(\Omega_c+1)-\ell(\ell+1)]\\ &=E_0(x-\frac{Z}{2E_0})^2-[\Omega_c(\Omega_c+1)-\ell(\ell+1)]\\ &=E_0(x-2(\Omega_c)(\Omega_c+1)Z^{-1})^2-[\Omega_c(\Omega_c+1) -\ell(\ell+1)]\ .\tag"(49)"\endalign \noindent For $x\not\in I$ and $(1-\overline{\overline c})\Omega_c\le \ell<\Omega_c$, we have \align |x-2\Omega_c(\Omega_c+1)Z^{-1}|&\ge |x-2\ell(\ell+1)Z^{-1}|-2Z^{-1} |\Omega_c(\Omega_c+1)-\ell(\ell+1)|\\ &\ge \frac {1}{10}\cdot 2\ell(\ell+1)Z^{-1}-6Z^{-1}\Omega_c(\Omega_c-\ell)\\ &\ge \frac{1}{10}\Omega_c(\Omega_c+1)Z^{-1}-6\overline{\overline c}Z^{-1} \Omega_c^2\\ &\ge \frac{1}{20}Z^{-1}\Omega_c^2\quad\text{since}\ \overline{\overline c}\ \text{is\ small}\ .\tag"(50)"\endalign \noindent Also for $(1-\overline{\overline c})\Omega_c\le \ell<\Omega_c$ we have $$[\Omega_c(\Omega_c+1)-\ell(\ell+1)]\le 3\Omega_c(\Omega_c-\ell)\le 3\overline{\overline c}\, \Omega_c^2\ .\tag"(51)"$$ \noindent Putting (50) and (51) into (49), we see that $$x^2V_\ell^c(x)\ge E_0\cdot (\frac{1}{20}Z^{-1}\Omega_c^2)^2- 3\overline{\overline c}\, \Omega_c^2\ .$$ \noindent The first term on the right-hand side is $\sim Z^{4/3}\cdot (Z^{-1} Z^{2/3})^2\sim Z^{2/3}$, while the second term is $\sim \overline{\overline c} Z^{2/3}$. \noindent Since $\overline{\overline c}$ is taken small, we have $$x^2V_\ell^c(x)\ge cZ^{2/3}\sim \ell^2\quad\text{for}\ x\not\in I\ \text{and}\ (1-\overline{\overline c})\Omega_c\le\ell<\Omega_c\ .$$ \noindent This is equivalent to (46). The proof of Lemma 6 is complete. $\qquad\blacksquare$\enddemo\medskip Our first application of Lemma 6 is as follows. \vglue 1pc \proclaim{Lemma 7} Suppose $(1-\overline{\overline c})\Omega_c\le\ell <\Omega_c-c\Omega_c^{7/43}$. Set $x_0=2\ell(\ell+1)Z^{-1}$, take $\tilde S=\frac{\Omega_c(\Omega_c-\ell)}{\check r_c^2}$, $\tilde B= \frac{\check r_c(\Omega_c-\ell)^{1/2}}{\Omega_c^{1/2}}$, and define $I=[x_0-h, x_0+h]$, with $h=\min \bigl(\frac{1}{10}x_0, \underline C\tilde B)$ for a large constant $\underline C$. Let $\varepsilon$, $N$ be given. Set $K=100^{90}$, and let $\hat c$ be a small enough constant. Then the potential $V_\ell^c (r)$, the weights $\tilde S$, $\tilde B$ and the interval $I$ satisfy the hypotheses (Z0)$\ldots$(Z9) of the First WKB Eigenvalue Sum Theorem, with $\Lambda\sim(\Omega_c-\ell)$.\endproclaim\medskip \demo{Sketch of Proof} Repeat the proof of Lemma 5 in the section on the Density in an Approximate TF Potential in [FS7], replacing equations (14) and (15) in that section by Lemma 6 above.$\qquad\blacksquare$\enddemo >From Lemma 7 and the First WKB Eigenvalue Sum Theorem, we see that $$\multline \text{sneg}(H_\ell^c)=-\frac{2}{3\pi}\int_0^\infty\bigl(-V_\ell^c(x)\bigr)_+ ^{3/2}dx+\frac{1}{24\pi}\int_0^\infty(V_\ell^c)^{\prime\prime} (x)\cdot \bigl(-V_\ell^c(x)\bigr)_+^{-1/2}dx\\ +\frac{\pi}{n_\ell^c}\tilde\chi(\phi_\ell^c)+\ \text{Err}_\ell^c \endmultline$$ \noindent for $(1-\overline{\overline c})\Omega_c\le\ell\le \Omega_c-\Omega_c^{1-10\, \varepsilon}$, with $$|\text{Err}_\ell^c|\le \Lambda^{5\varepsilon-2}\cdot|\operatornamewithlimits {\min}\limits_{x>0}V_\ell^c(x)|\sim (\Omega_c-\ell)^{5\varepsilon-2} \tilde S=\frac{\Omega_c}{\check r_c^2}(\Omega_c-\ell)^{5\varepsilon-1}\ ,$$ \noindent i.e.\ $|\text{Err}_\ell^c|\le CZ(\Omega_c-\ell)^{5\varepsilon-1}$. \noindent Recalling the formula (40) for $\frac{\pi}{n_\ell^c} \tilde\chi(\phi_\ell^c)$, we obtain $$\multline \text{sneg}(H_\ell^c)=-\frac{2}{3\pi}\int_0^\infty\bigl(-V_\ell^c (x)\bigr)_+^{3/2}dx+\frac{1}{24\pi}\int_0^\infty(V_\ell^c)^{\prime\prime} (x)\cdot \bigl(-V_\ell^c(x)\bigr)_+^{-1/2}dx\\ +\frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}} -\sqrt{\ell(\ell+1)}-\frac 12\big)+\ \text{Err}_\ell^c\endmultline\tag"(52)"$$ \noindent for $(1-\overline{\overline c})\Omega_c\le\ell\le \Omega_c-\Omega_c^{1-10\, \varepsilon}$, with $$|\text{Err}_\ell^c|\le CZ(\Omega_c-\ell)^{5\varepsilon-1}\ .\tag"(53)"$$ The second application of Lemma 6 is as follows. \vglue 1pc \proclaim{Lemma 8} Suppose $\Omega_c-\Omega_c^{1-10\varepsilon}\le \ell<\Omega_c$, and let $\varepsilon$, $K$, $N$ be given. Suppose $Z$ is large enough, depending on $\varepsilon$, $K$, $N$ and the constants in (1), (3). Pick $x_0=2\ell(\ell+1)Z^{-1}$, $S=Z^{4/3}$, $B=x_0$, $I=\{|x-x_0|<\frac{1}{10} x_0\}$. Then $V_\ell^c(r)$ satisfies the hypotheses (H0$^\ast$)$\ldots$ (H6$^\ast$) of the Third WKB Eigenvalue Sum Theorem. Moreover, $\lambda\sim Z^{+1/3}$ and -\lambda^{-3\varepsilon}SZ^{10^{-9}}}(2\ell+1) \int_0^\infty\bigl\{V_\ell^{\prime\prime}(x)\cdot\bigl(-V_\ell(x)\bigr)_+^{-1/2} -V_\ell^{c^{\prime\prime}}(x)\cdot\bigl(-V_\ell^c(x)\bigr)_+^{-1/2}\bigr\}dx\\ &\ \ +\pi\sum\limits_{\Omega>\ell>Z^{10^{-9}}}\frac{(2\ell+1)}{n_\ell} \tilde\chi(\phi_\ell)\\ &\ \ -\sum\limits_{Z^{10^{-9}}<\ell<\Omega_c}(2\ell+1) \cdot\frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}} -\sqrt{\ell(\ell+1)}-\frac 12\bigr)\\ &\ \ +[\text{Error}^B+\text{Error}^C-\text{Error}^D]\ .\tag"(64)"\endalign $$\noindent Here we have used the fact that \bigl(-V_\ell(x)\bigr)_+^{3/2} \equiv 0 and \bigl(-V_\ell(x)\bigr)_+^{-1/2}\equiv 0 for \ell\ge \Omega; and that$$ \bigl(-V_\ell^c(x)\bigr)_+^{3/2}\equiv 0\quad\text{and}\quad \bigl(-V_\ell^c(x)\bigr)_+^{-1/2}\equiv 0\quad\text{for}\quad \ell\ge \Omega_c\ . $$\noindent From (59 bis), (61), (63), we have$$ \big|[\text{Error}^B+\text{Error}^C-\text{Error}^D]\big|\le C\, Z^{\frac 85+2\cdot 10^{-9}}\ .\tag"(65)" $$\noindent It is convenient to write (64) in a different form. \smallskip \noindent Fix a function \varphi(x) equal to 1 for x\ge 2\cdot Z^{-3/5}, equal to zero for x\le Z^{-3/5}, and satisfying the estimates$$ \big|\bigl(\frac{d}{dx}\bigr)^\alpha\varphi(x)\big|\le C_\alpha x^{-\alpha} \quad\text{for}\quad x\in (0,\infty)\ , \alpha\ge 0 \ .\tag"(66)" $$\noindent In view of (3), the expressions in curly brackets in (64) are supported in \{x>2Z^{-3/5}\}, where \varphi\equiv 1. Hence we may replace dx by \varphi(x)dx in the integrals in (64). Therefore, (64), (65) may be rewritten as follows.$$ \text{sneg}(H)-\text{sneg}(H_c)=X-X_c+Y-Y_c+W-W_c+\ \text{Error}^E\ , \tag"(67)" $$\noindent with$$ X=-\frac{2}{3\pi}\sum\limits_{\ell\ge 0}(2\ell+1)\int_0^\infty \bigl(-V_\ell(x)\bigr)_+^{3/2}\varphi(x)dx\tag"(68)"  X_c=-\frac{2}{3\pi}\sum\limits_{\ell\ge 0}(2\ell+1) \int_0^\infty\bigl(-V_\ell^c(x)\bigr)_+^{3/2}\varphi(x)dx\tag"(69)"  Y=\frac{1}{24\pi}\sum\limits_{\ell>Z^{10^{-9}}}(2\ell+1) \int_0^\infty V_\ell^{\prime\prime}(x)\cdot\bigl(-V_\ell(x)\bigr)_+^{-1/2} \varphi(x)dx\tag"(70)"  Y_c=\frac{1}{24\pi}\sum\limits_{\ell>Z^{10^{-9}}}(2\ell+1) \int_0^\infty V_\ell^{c^{\prime\prime}}(x)\cdot\bigl(-V_\ell^c(x)\bigr)_+ ^{-1/2}\varphi(x)dx\tag"(71)"  W=\pi\sum\limits_{Z^{10^{-9}}<\ell<\Omega}\frac{(2\ell+1)}{n_\ell} \tilde\chi(\phi_\ell)\tag"(72)"  W_c=\sum\limits_{Z^{10^{-9}}<\ell<\Omega_c}(2\ell+1) \cdot\frac{2E_0^{3/2}}{Z}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}} -\sqrt{\ell(\ell+1)}-\frac 12\bigr)\tag"(73)"  |\text{Error}^E|Z^{10^{-9}}}(2\ell+1)\int_{-\infty}^\infty \Bigl(-\big[\frac{W(x)}{x^2}\bigr]^{\prime\prime}+\frac{6\ell(\ell+1)} {x^4}\Bigr)\cdot\Bigl(\frac{W(x)-\ell(\ell+1)}{x^2}\Bigr)_+^{-1/2}\theta (x)dx\ .\tag"(76)" $$\noindent Here we assume \theta(x) is supported in |x-x_0|\le B and satisfies$$ \big|\bigl(\frac{d}{dx}\bigr)^\alpha\theta\big|\le C_\alpha B^{-\alpha}\ . \tag"(76\text{bis}$)" $$\noindent Regarding W(x), we assume$$ \big|\bigl(\frac{d}{dx}\bigr)^\alpha W\big|\le C_\alpha\Cal SB^{-\alpha} \quad\text{for}\ |x-x_0|\le B\ \qquad\text{and\ either}\tag"(77)"  |W^\prime(x)|>c\Cal SB^{-1}\quad\text{for}\quad |x-x_0|\le B\ ,\ \text{or\ else}\tag"(78)"  W^\prime(x_0)=0\quad\text{and}\quad -W^{\prime\prime}(x)>c\Cal SB^{-2}\quad \text{for}\ |x-x_0|\le B\ .\tag"(79)" $$We assume \Cal S\ge 1. Later, we will write X and X_c as sums of X(\theta_\nu,W), and similarly write Y, Y_c as sums of Y(\theta_\nu, W) for suitable \theta_\nu, W. To understand (75), (76), we first study the functions$$ F(\xi,\theta,W)=\int_{-\infty}^\infty\bigl(W(x)-\xi\bigr)_+^{3/2} \theta(x)dx\qquad\text{and}\tag"(80)"  G(\xi,\theta,W)=\int_{-\infty}^\infty\bigl(W(x)-\xi)_+^{-1/2} \theta(x)dx\tag"(81)" $$\noindent under assumptions (76 bis), (77), and either (78) or (79). Our basic result on F, G is as follows. \vglue 1pc \proclaim{Lemma 9} If (78) holds, then$$ \big|\bigl(\frac{d}{d\xi}\bigr)^mF(\xi,\theta,W)\big|\le C_m^\prime (\Cal S^{3/2}B)\Cal S^{-m}\quad\text{and}\tag"(82)"  \big|\bigl(\frac{d}{d\xi}\bigr)^mG(\xi,\theta,W)\big|\le C_m^\prime(\Cal S^{-1/2}B)\Cal S^{-m}\tag"(83)" $$\noindent for |\xi|c in \{|x-x_0|\le 1\}. (Otherwise, -W^\prime(x)>c, and we change variable from x to -x without changing F, G.) Change variable from x to y=W(x). The image of \{|x-x_0|\le 1\} is an interval [y_{\min},y_{\max}] with |y_{\min}|, |y_{\max}|c. In terms of y, we have \theta(x)dx=\tilde\theta(y)dy with \tilde\theta\in C_0^\infty (y_{\min}, y_{\max}) and the C^\infty seminorms of \tilde\theta bounded a-priori in terms of the constants in (76 bis), (77), (78). In terms of y, the integrals defining F, G become$$ F(\xi,\theta,W)=\int_{-\infty}^\infty(y-\xi)_+^{3/2}\tilde\theta(y) dy=\int_0^\infty t^{3/2}\tilde\theta(\xi+t)dt\tag"(84)" $$\noindent and$$ G(\xi,\theta,W)=\int_{-\infty}^\infty(y-\xi)_+^{-1/2}\tilde\theta(y) dy=\int_0^\infty t^{-1/2}\tilde\theta(\xi+t)dt\ .\tag"(85)" $$If |\xi|\le C, then \tilde\theta(\xi+t)\equiv 0 for t>C^\prime =y_{\max}+C. Hence the t-integrals in (84), (85) may be taken over (0,C^\prime) instead of (0,\infty), without changing the value of the integrals. Thus, for |\xi|\le C we have$$ F(\xi,\theta,W)=\int_0^{C^\prime}t^{3/2}\tilde\theta(\xi+t)dt\qquad \text{and}\tag"(86)"  G(\xi,\theta,W)=\int_0^{C^\prime}t^{-1/2}\tilde\theta(\xi+t)dt\ .\tag"(87)" $$\noindent Since the constant C^\prime and the C^\infty-seminorms of \tilde\theta are bounded a-priori in terms of the constants in (76 bis), (77), (78), the formulas (86) and (87) imply trivially the conclusion of Lemma 9. On the other hand, assume (76 bis), (77) and (79). Then there is a smooth function y(x) defined on \{|x-x_0|\le 1\} with W(x)=W(x_0)-(y(x))^2 and y^\prime(x)>c>0. The C^\infty seminorms of y(x) are bounded a-priori in terms of the constants in (76 bis), (77), (79); and similarly, the lower bound c for y^\prime(x) is bounded below a-priori. Again we change variable from x to y=y(x) in the definitions of F, G. We have \theta(x)dx=\tilde\theta(y)dy for a function \tilde\theta whose C^\infty seminorms are bounded a-priori, and supported in [y_{\min}, y_{\max}] with |y_{\min}|, |y_{\max}| bounded a-priori. Hence,$$ F(\xi,\theta,W)=\int_{-\infty}^\infty\bigl([W(x_0)-\xi]-y^2)_+^{3/2} \tilde\theta(y)dy\qquad\text{and}\tag"(88)"  G(\xi,\theta,W)=\int_{-\infty}^\infty\bigl([W(x_0)-\xi]-y^2\bigr)_+^{-1/2} \tilde\theta(y)dy\ .\tag"(89)" $$\noindent Since \tilde\theta is C^\infty, we can write \tilde\theta (y)=\theta_1(y^2)+y\theta_2(y^2), with the C^\infty-seminorms of \theta_1, \theta_2 bounded a-priori. We substitute this in (88), (89), and note that the \theta_2-term may be dropped, since it contributes the integral of an odd function. Thus$$\gather F(\xi,\theta,W) =\int_{-\infty}^\infty\bigl([W(x_0)-\xi]-y^2\bigr)_+^{3/2} \theta_1(y^2)dy\qquad\text{and}\\ G(\xi,\theta,W)=\int_{-\infty}^\infty\bigl([W(x_0)-\xi]-y^2\bigr)_+^{-1/2} \theta_1(y^2)dy\ .\endgather $$\noindent If \xic^\prime\Cal Sx_0^{-1}\quad\text{for}\quad |x-x_0|\le cx_0\ ,\tag"(92)"$$ \noindent or else $$W(x_0)>0\ , W^\prime (x_0)=0\quad\text{and}\quad -W^{\prime\prime} >c^\prime\Cal Sx_0^{-2}\quad\text{for}\ |x-x_0|\le cx_0\ . \tag"(93)"$$ We assume also$\Cal S\ge 1$. \noindent Define $$\Cal F(\xi)=\int_0^\infty\Bigl(\frac{W(x)-\xi}{x^2}\Bigr)_+^{3/2} \theta(x)dx\qquad\text{and}\tag"(94)"$$ $$\Cal G(\xi)=\int_0^\infty\Bigl(-\bigl[\frac{W(x)}{x^2}\bigr]^{\prime\prime} +\frac{6\xi}{x^4}\Bigr)\cdot\Bigl(\frac{W(x)-\xi}{x^2}\Bigr)_+^{-1/2} \theta(x)dx\ .\tag"(95)"$$ \noindent Set$\theta_1(x)=\frac{x_0^3\theta(x)}{x^3}$and$\theta_2(x) =\frac{x[\frac{W(x)}{x^2}]^{\prime\prime}\theta(x)}{x_0^{-3}\Cal S}$. Both$\theta_1$and$\theta_2$satisfy the hypotheses of Lemma 9, with$B=cx_0$. (To see this, for$\theta_2$, note that$x[\frac{W(x)}{x^2}]^{\prime\prime}=\ \text{(const)}\ x^{-3} W(x)+\ \text{(const)}x^{-2}W^\prime(x)+\ \text{(const)}\ x^{-1} W^{\prime\prime}(x)$, so that$\bigl(\frac{d}{dx}\bigr)^\alpha \bigl\{x\big[\frac{W(x)}{x^2}\bigr]^{\prime\prime}\bigr\}$is a sum of terms$x^{-a}\bigl(\frac{d}{dx}\bigr)^bW$with$a+b=\alpha+3$. Hence$\big|(\frac{d}{dx}\bigr)^\alpha\{x[\frac{W(x)}{x^2}\big]^{\prime\prime} \bigr\}\big|\le C_\alpha\Cal Sx_0^{-3-\alpha}$in$\text{supp}\ \theta$, from which one verifies the estimate (76 bis) for$\theta_2$.) \noindent Immediately from the definitions (80), (81), (94), (95) we have $$\Cal F(\xi)=x_0^{-3}F(\xi,\theta_1,W)\quad\text{and}\tag"(96)"$$ $$\Cal G(\xi)=-x_0^{-3}\Cal S\cdot G(\xi,\theta_2,W)+6\xi x_0^{-3}G(\xi, \theta_1,W)\ .\tag"(97)"$$ \noindent The terms on the right are controlled by using Lemma 9, with$B=cx_0$. We deduce from (96), (97) and Lemma 9 that $$\big|\bigl(\frac{d}{d\xi}\bigr)^m\Cal F(\xi)\big|\le C_m(\Cal S^{3/2} x_0^{-2})\Cal S^{-m}\quad\text{for}\ \xi \in \Cal J\ ,\quad\text{and} \tag"(98)"$$ $$\big|\bigl(\frac{d}{d\xi}\bigr)^m\Cal G(\xi)\big|\le C_m (\Cal S^{1/2}x_0^{-2})\Cal S^{-m}\quad\text{for}\ \xi\in \Cal J\ ,\quad \text{where}\tag"(99)"$$ $$\Cal J=\{|\xi|Z^{10^{-9}}}g(\ell)\ .\tag"(104)"$$ \noindent Our next task is to estimate the derivatives of$f$and$g$. The derivative$(\frac{d}{dt})^k\Cal F(t(t+1))$is a sum of terms $$\split \big(\frac{d}{d\xi}\bigr)^m\Cal F(\xi)\bigm|_{\xi=t(t+1)}\cdot \prod\limits_{\nu=1}^m\bigl(\frac{d}{dt}\bigr)^{k_\nu} \{t(t+1)\}\\ \text{with}\ k_\nu\ge 1\ \text{and}\ k_1+\ldots+k_m=k\ .\endsplit\tag"(105)"$$ \noindent We are interested in$t\in [0,t_{\max})$, with$t_{\max}= C\Cal S^{1/2}$if (92) holds, and$t_{\max}=$the positive root of$t_{\max}(t_{\max} +1)=W(x_0)$if (93) holds. For$t$in this interval we have$\xi=t(t+1) \in\Cal J$, and$|(\frac{d}{dt})^{k_\nu}\{t(t+1)\}|\le C\Cal S^{1-\frac{k_\nu}{2}}$, since$|t|\le C\Cal S^{1/2}$. Hence by (98), the term (105) is dominated by $$C(\Cal S^{3/2}x_0^{-2})\Cal S^{-m}\cdot \prod\limits_{\nu=1}^m \Cal S^{1-\frac{k_\nu}{2}}=C(\Cal S^{3/2}x_0^{-2})\Cal S^{-\frac k2}\ .$$ \noindent Therefore, $$\big|\bigl(\frac{d}{dt}\bigr)^k\Cal F(t(t+1))\big|\le C_k(\Cal S^{3/2} x_0^{-2})\Cal S^{-k/2}\quad\text{for}\ t\in [0,t_{\max})\ .\tag"(106)"$$ \noindent Similarly, using (99) instead of (98), we find that $$\big|\bigl(\frac{d}{dt}\bigr)^k\Cal G(t(t+1))\big|\le C_k(\Cal S^{1/2} x_0^{-2})\Cal S^{-k/2}\ \text{for}\ t\in [0,t_{\max})\ .\tag"(107)"$$ \noindent Also,$|(\frac{d}{dt})^k(2t+1)|\le C\Cal S^{1/2-k/2}$for$t\in [0,t_{\max})$, since$t_{\max}t_{\max})\\ =\int_0^\infty(2t+1)\int_0^\infty\Bigl(\frac{W(x)-t(t+1)}{x^2}\Bigr)_+ ^{3/2}\theta(x)dxdt=\int_0^\infty\int_0^\infty \bigl(\frac{W(x)-\xi}{x^2}\bigr)_+^{3/2}\theta(x)dxd\xi\\ =\frac 25\int_0^\infty\bigl(\frac{W(x)}{x^2}\bigr)_+^{5/2} x^2\theta(x)dx\ .\endmultline\tag"(114)" $$\noindent Next, note that f(b) and f^\prime(b) both tend to zero as b\to t_{\max}-. In fact, f(b)=f^\prime(b)=0 for b near t_{\max}=C\Cal S^{1/2}, if (92) holds. If instead (93) holds, then for \xi\to W(x_0)- we have$$ \Cal F(\xi)=\int_0^\infty\bigl(\frac{W(x)-\xi}{x^2}\bigr)_+^{3/2} \theta(x)dx\to 0\ ,\quad\text{and}  \Cal F^\prime(\xi)=\ \text{(const)}\ \int_0^\infty(W(x)-\xi)_+^{1/2} \frac{\theta(x)}{x^3}dx\to 0\ . $$\noindent To see these equations, note that (W(x)-\xi)_+ is dominated by C\Cal S and supported in an interval about x_0 of length O((W(x_0)-\xi)^{1/2}). Thus f(b),f^\prime(b)\to 0 as b\to t_{\max}- in either case (92) or (93), since f(b)=(2b+1)\Cal F(b(b+1)), and b(b+1)\to W(x_0)-. Hence$$ \operatornamewithlimits{\lim}_{b\to t_{\max}-}\bigl\{-f(b)\chm(b)+\frac 12 f^\prime(b)\tilde\chi(b)\bigr\}=0\ .\tag"(115)" $$Next, recall that \chp(x)=k-x-\frac 12 for the smallest integer k\ge x. Thus \chp(0)=-\frac 12, so$$ -f(0)\chp(0)=+\frac 12 f(0)=\frac 12 \int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+ ^{3/2}\theta(x)dx\ .\tag"(116)" $$Similarly, \tilde\chi(0)=\operatornamewithlimits{\min}\limits_{k\in \Bbb Z} \bigl\{|0-k-1/2|^2-\frac{1}{12}\bigr\}=\frac 14-\frac{1}{12}=\frac 16, and from the definition (102) we get$$ f^\prime(0)=2\int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{3/2}\theta(x) dx+\ \text{(const)}\ \int_0^\infty (W(x))_+^{1/2}\frac{\theta(x)}{x^3}dx\ . $$\noindent The second term on the right is dominated by \Cal S^{1/2} x_0^{-2}, so$$ -\frac 12 f^\prime(0)\tilde\chi(0)=-\frac 16\int_0^\infty\Bigl(\frac{W(x)} {x^2}\Bigr)_+^{3/2}\theta(x)dx+\ \text{Err}\ ,\ \text{with}\ |\text{Err}|\le C\Cal S^{1/2}x_0^{-2}\ .\tag"(117)" $$\noindent Putting (113)\ldots(117) into (112), we see that$$ X(\theta,W)=\frac 25\int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{5/2} x^2\theta(x)dx+\frac 13 \int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{3/2} \theta(x)dx+\ \text{Error}_X^\prime\ ,\tag"(118)" $$\noindent with |\text{Error}_X^\prime|\le C\Cal Sx_0^{-2}. (Recall \Cal S\ge 1, so \Cal S^{1/2}\le \Cal S). Similarly, we use (109), (111) and the lemma on Riemann sums to compute Y(\theta,W). If t_{\max}\le Z^{10^{-9}}, then evidently Y(\theta,W)=0 by (111). Suppose t_{\max}>Z^{10^{-9}}. For the interval [a,b] in the Lemma on Riemann sums, we take [Z^{10^{-9}} +\varepsilon, t_{\max}-\varepsilon] and let \varepsilon\to 0+. We obtain the following crude result, which is enough for our purposes.$$ Y(\theta,W)=\int_{Z^{10^{-9}}}^{t_{\max}}g(t)dt+\ \text{Error}_Y\ , \ \text{with}\tag"(119)"  |\text{Error}_Y|\le C(\Cal Sx_0^{-2})+C_N(\Cal Sx_0^{-2})\Cal S^{-N/2} t_{\max}\le C^\prime\Cal Sx_0^{-2}\ ,\tag"(120)" $$\noindent since t_{\max}\le C\Cal S^{1/2} as we noted before. \noindent For t>t_{\max} we have g(t)\equiv 0. Moreover, \big|\int_0^{Z^{10^{-9}}}g(t)dt\big|\le C(\Cal Sx_0^{-2})\cdot Z^{10^{-9}} by (109). Hence, (119) and (120) imply$$ Y(\theta,W)=\int_0^\infty g(t)dt+\ \text{Error}_Y^\prime\ \text{with}\ |\text{Error}_Y^\prime|\le CZ^{10^{-9}}\Cal Sx_0^{-2}\ .\tag"(121)" $$\noindent By definition (103), we have$$\multline \int_0^\infty g(t)dt=\int_0^\infty(2t+1)\int_0^\infty\Bigl(\bigl[- \frac{W(x)}{x^2}\bigr]^{\prime\prime}+\frac{6t(t+1)}{x^4}\Bigr)\cdot \Bigl(\frac{W(x)-t(t+1)}{x^2}\Bigr)_+^{-1/2}\theta(x)dxdt\\ =\int_0^\infty\int_0^\infty\Bigl(\big[\frac{-W(x)}{x^2}\bigr]^{\prime\prime} +\frac{6\xi}{x^4}\Bigr)\cdot\Bigl(\frac{W(x)-\xi}{x^2}\Bigr)_+^{-1/2} \theta(x)dxd\xi\\ =\int_0^\infty\bigl[-\frac{W(x)}{x^2}\bigr]^{\prime\prime} x\theta(x)\bigl\{\int_0^\infty(W(x)-\xi)_+^{-1/2}d\xi\bigr\}dx\\ +\int_0^\infty\frac{6\theta(x)}{x^3}\bigl\{\int_0^\infty \xi\cdot(W(x)-\xi)_+^{-1/2}d\xi\bigr\}dx\ .\endmultline\tag"(122)" $$\noindent We evaluate the integrals in curly brackets:$$ \int_0^\infty\bigl(W(x)-\xi\bigr)_+^{-1/2}d\xi=2\bigl(W(x)\bigr)_+^{1/2}\ , \ \text{and} \multline \int_0^\infty\xi\bigl(W(x)-\xi\bigr)_+^{-1/2}d\xi=\int_0^\infty \bigl[W(x)-(W(x)-\xi)\bigr]\cdot\bigl(W(x)-\xi)_+^{-1/2}d\xi\\ =W(x)\cdot\int_0^\infty\bigl(W(x)-\xi\bigr)_+^{-1/2}d\xi-\int_0^\infty \bigl(W(x)-\xi\bigr)_+^{1/2}d\xi\\ =W(x)\cdot 2\bigl(W(x)\bigr)_+^{1/2} -\frac 23\bigl(W(x)\bigr)_+^{3/2}=\frac 43\bigl(W(x)\bigr)_+^{3/2}\ . \endmultline $$\noindent Hence (122) becomes$$ \int_0^\infty g(t)dt=2\int_0^\infty\Bigl[\frac{-W(x)}{x^2}\Bigr]^{\prime \prime}\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{1/2}\theta(x)x^2dx +8\int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{3/2}\theta(x)dx\ . $$\noindent Putting this into (121), we conclude that$$\split Y(\theta,W)=2\int_0^\infty\Bigl[-\frac{W(x)}{x^2}\Bigr]^{\prime\prime} \Bigl(\frac{W(x)}{x^2}\Bigr)_+^{1/2}\theta(x)x^2dx\\ +8\int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{3/2}\theta(x)dx+\ \text{Error} _Y^\prime\endsplit\tag"(123)" $$\noindent with$$\split |\text{Error}_Y^\prime|\le CZ^{10^{-9}}\Cal Sx_0^{-2}\quad\text{if}\ t_{\max}>Z^{10^{-9}}\ ;\\ Y(\theta,W)=0\ \text{if}\ t_{\max}\le Z^{10^{-9}}\ .\endsplit\tag"(124)" $$\noindent We record our results (118), (123), (124) on X(\theta,W), Y(\theta,W) in the following statement. \vglue 1pc \proclaim{Lemma 10} Suppose \theta(x), W(x) are defined on (0,\infty), with \theta supported in \{|x-x_0| \le cx_0\} (0c^\prime\Cal Sx_0^{-1} for |x-x_0| \le cx_0, in which case we set t_{\max}=C\Cal S^{1/2} \noindent or \item"(B)" W(x_0)>0, W^\prime(x_0)=0 and -W^{\prime\prime}> c^\prime\Cal Sx_0^{-2} for |x-x_0|\le cx_0 in which case we set t_{\max} equal to the positive root of t(t+1)=W(x_0). \endroster \noindent Then$$ X(\theta,W)=\sum\limits_{\ell\ge 0}(2\ell+1)\int_0^\infty \Bigl(\frac{W(x)-\ell(\ell+1)}{x^2}\Bigr)_+^{3/2}\theta(x)dx $$\noindent is given by$$ X(\theta,W)=\frac 25\int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{5/2} \theta(x)x^2dx+\frac 13\int_0^\infty\Bigl(\frac{W(x)}{x^2}\Bigr)_+^{3/2} \theta(x)dx+\ {\roman{Error}}_X\ , $$\noindent with |{\roman{Error}}_X|\le C\Cal Sx_0^{-2}. \noindent Also,$$ Y(\theta,W)=\sum\limits_{\ell>Z^{10^{-9}}}(2\ell+1)\int_0^\infty \Bigl(\big[-\frac{W(x)}{x^2}\bigr]^{\prime\prime}+ \frac{6\ell(\ell+1)}{x^4}\Bigr)\cdot\Bigl(\frac{W(x)-\ell(\ell+1)}{x^2} \Bigr)_+^{-1/2}\theta(x)dx $$\noindent is given by$$ Y(\theta,W)=2\int_0^\infty\Bigl[-\frac{W(x)}{x^2}\Bigr]^{\prime\prime} \Bigl(\frac{W(x)}{x^2}\Bigr)_+^{1/2}\theta(x)x^2dx+8\int_0^\infty \Bigl(\frac{W(x)}{x^2}\Bigr)_+^{3/2}\theta(x)dx+\ {\roman{Error}}_Y $$\noindent with$$ |{\roman{Error}}_Y|\le CZ^{10^{-9}}\Cal Sx_0^{-2}\ \text{if}\ t_{\max}>Z^{10^{-9}}\ ; $$\noindent Y(\theta,W)=0 if t_{\max}\le Z^{10^{-9}}.\endproclaim \medskip \noindent We apply Lemma 10 to calculate the numbers X, Y, X_c, Y_c defined by equations (68)\ldots(71). Define W(x)=-x^2V(x). Then:$$ W(x)\sim \Cal S(x)\equiv \min\{Zx, x^{-2}\}\quad\text{for}\ x\in (0,\infty)\tag"(125)"  \big|\bigl(\frac{d}{dx}\bigr)^\alpha W(x)\big|\le C_\alpha \Cal S(x)x^{-\alpha}\tag"(126)" \split W(x)\ \text{has\ a\ single\ critical\ point\ at}\ x=\check r\sim Z^{-1/3}\ , \text{where}\\ -W^{\prime\prime}>c\Cal S(\check r)\check r^{-2}\ . \endsplit\tag"(127)" \split \text{Outside\ any\ neighborhood}\ \{|x-\check r|c_2\Cal S(x)x^{-1}\ ,\\ \text{where}\ c_2\ \text{depends\ on}\ c_1\ .\endsplit\tag"(128)" $$These properties are contained in Lemma 7 in the section on approximate T-F potentials. Next, we use a partition of unity to write the function \varphi(x) in (68)\ldots(71) as a sum$$ \varphi(x)=\sum\limits_\nu\theta_\nu(x)+\theta_{\text{far}}(x)\ , \text{with\ the\ following\ properties}\ .\tag"(129)"  \text{Each}\ \theta_\nu(x)\ \text{is\ supported\ in}\ \{|x-x_\nu|cZ^{-3/5}\}$.) By definitions (68), (70) and the definition of$X(\theta,W)$,$Y(\theta,W)$in Lemma 10, we have $$X=-\frac{2}{3\pi}X(\varphi,W)=-\frac{2}{3\pi}\sum\limits_\nu X(\theta_\nu,W)-\frac{2}{3\pi}X(\theta_{\text{far}}, W)\ ,\text{and} \tag"(136)"$$ $$Y=\frac{1}{24\pi} Y(\varphi,W)=\frac{1}{24\pi}\sum\limits_\nu Y(\theta_\nu,W)+\frac {1}{24\pi}Y(\theta_{\text{far}},W)\ . \tag"(137)"$$ \noindent For$\ell>Z^{10^{-9}}$, we have$\bigl(-V_\ell(x)\bigr)_+^{-1/2}$supported in$[x_{\text{left}}(\ell), x_{\text{rt}}(\ell)]$. This interval is disjoint from the support of$\theta_{\text{far}}$, by virtue of (134) and the fact that$x_{\text{rt}}(\ell)\sim\frac 1\ell$. (Here it is important to take the constant$C_1$in (134) large enough). Consequently,$\int_0^\infty V_\ell^{\prime\prime}(x)\bigl(-V_\ell(x)\bigr)_+ ^{-1/2}\theta_{\text{far}}(x)dx=0$for$\ell>Z^{10^{-9}}$, so that $$Y(\theta_{\text{far}}, W)=0\ .\tag"(138)"$$ \noindent To estimate$X(\theta_{\text{far}}, W)$, we use the same observations as in (138) to see that$\int_0^\infty\bigl(-V_\ell(x)\bigr)_+^{3/2} \theta_{\text{far}}(x)dx=0$for$\ell>Z^{10^{-9}}$. Therefore, $$\multline |X(\theta_{\text{far}},W)|=\Big|\sum\limits_{0\le\ell\le Z^{10^{-9}}} (2\ell+1)\int_0^\infty\bigl(-V_\ell(x)\bigr)_+^{3/2}\theta_{\text{far}} (x)dx\Big|\\ \le CZ^{2\cdot 10^{-9}}\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2} |\theta_{\text{far}}(x)|dx\le C^\prime\Bigl(\int_{\frac 12 C_1Z^{-10^{-9}}} ^\infty x^{-6}dx\Bigr)Z^{2\cdot 10^{-9}}\\ \le Z^{10^{-8}}\ .\endmultline\tag"(139)"$$ \noindent To control$X(\theta_\nu,W)$and$Y(\theta_\nu,W)$we invoke Lemma 10. For each$\nu$, the functions$\theta_\nu$,$W$satisfy the hypotheses of Lemma 10, with$\Cal S=\Cal S(x_\nu)$and$t_{\max}\sim \Cal S^{1/2}(x_\nu)$. In fact, the hypothesis on$\text{supp}\ \theta_\nu$and the bounds assumed for$|(\frac{d}{dx})^\alpha\theta_\nu|$and$|(\frac{d}{dx})^\alpha W|$are contained in (130), (131) and (126). We have$\Cal S=\Cal S(x_\nu)\ge 1$, by (125), (132). We must show that (A) or (B) holds in the statement of Lemma 10. If$|x_\nu-\check r|>2cx_\nu$with$c$as in (130), then we take$x_0=x_\nu$in the statement of Lemma 10, and alternative (A) holds by virtue of (128). We have$t_{\max}\equiv C\Cal S^{1/2}(x_\nu)$in that case. If instead$|x_\nu-\check r|\le 2cx_\nu$, then we take$x_0=\check r$in the statement of Lemma 10, and we use there$10c$in place of$c$. Alternative (B) holds in this case, by virtue of (126), (127) and the fact that$\Cal S\equiv \Cal S(x_\nu)\sim \Cal S(\check r)\sim Z^{2/3}$. Moreover,$t_{\max}$is defined in this case as the positive root of$t(t+1)=W(\check r)\sim \Cal S(\check r)\sim Z^{2/3}$, so$t_{\max}\sim \Cal S^{1/2}(x_\nu)$. Thus in either case, the hypotheses of Lemma 10 hold, with$\Cal S=\Cal S(x_\nu)$and$t_{\max}\sim \Cal S^{1/2}(x_\nu)$. Applying Lemma 10 and recalling that$V(x)=-\frac{W(x)}{x^2}$, we get: $$X(\theta_\nu,W)=\frac 25\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2} \theta_\nu(x)x^2dx+\frac 13\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2} \theta_\nu(x)dx+\ \text{Error}_X(\nu)\tag"(140)"$$ \noindent with $$|\text{Error}_X(\nu)|\le C\Cal S(x_\nu)\cdot x_\nu^{-2}\ ;\ \text{and} \tag"(141)"$$ $$\multline Y(\theta_\nu,W)=2\int_0^\infty V^{\prime\prime}(x)\cdot \bigl(-V(x) \bigr)_+^{1/2}\theta_\nu(x)x^2dx\\ +8\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}\theta_\nu(x)dx+\ \text{Error}_Y (\nu)\endmultline\tag"(142)"$$ \noindent with $$\multline |\text{Error}_Y(\nu)|\le CZ^{10^{-9}}\Cal S(x_\nu)x_\nu^{-2}\quad\text{if}\ t_{\max}>Z^{10^{-9}}\ ;\\ Y(\theta_\nu,W)=0\quad\text{if}\ t_{\max}\le Z^{10^{-9}}\ .\endmultline \tag"(143)"$$ \noindent Here,$t_{\max}$depends on$\nu$and has the order of magnitude$t_{\max}\sim \Cal S^{1/2}(x_\nu)$. \noindent In fact, (142) and (143) hold for all$\nu$. To see this, suppose$t_{\max}\le Z^{10^{-9}}$. We have $$\gather V^{\prime\prime}=\Bigl(-\frac{W(x)} {x^2}\Bigr)^{\prime\prime}=O(\Cal S(x_\nu)x_\nu^{-4})\ \text{in}\ \text{supp}\ \theta_\nu\ ,\quad \text{and}\\ V=-\frac{W(x)}{x^2}=O(\Cal S(x_\nu)x_\nu^{-2})\ \text{in}\ \text{supp}\ \theta_\nu\ ,\ \text{by\ (126)}\ .\endgather$$ \noindent Hence$\int_0^\infty V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{1/2}\theta_\nu(x)x^2dx$and$\int_0^\infty \bigl(-V(x)\bigr)_+^{3/2}\theta_\nu(x)dx$are dominated by$C\Cal S^{3/2}(x_\nu)\cdot x_\nu^{-2}\sim t_{\max}\Cal S(x_\nu) x_\nu^{-2}\le Z^{10^{-9}}\Cal S(x_\nu)\cdot x_\nu^{-2}$. Since also$Y(\theta_\nu,W)=0$in this case, (142) and (143) are obvious. Thus for all$\nu$we have (140)$\ldots$(143). \noindent Putting (139), (140), (141) into (136) we get $$X=-\frac{4}{15\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2}\bigl(\sum\limits_\nu \theta_\nu(x)\bigr)x^2dx-\frac {2}{9\pi}\int_0^\infty\bigl(-V(x)\bigr)_+ ^{3/2}\bigl(\sum\limits_\nu\theta_\nu(x)\bigr)dx+\ \text{Error}_X \tag"(144)"$$ \noindent with $$\multline |\text{Error}_X|\le CZ^{10^{-8}}+C\sum\limits_\nu\Cal S(x_\nu)x_\nu^{-2}\\ \le CZ^{10^{-8}}+C\sum\limits_{cZ^{-3/5}\le x_\nu\le Z^{-1/3}}(Zx_\nu) \cdot x_\nu^{-2}+C\sum\limits_{x_\nu>Z^{-1/3}}(x_\nu^{-2})x_\nu^{-2}\\ \le C^\prime Z^{8/5}\ .\quad \text{(Here\ we\ used (132)\ and (133).)} \endmultline\tag"(145)"$$ \noindent Since also$\varphi(x)=\sum\limits_\nu\theta_\nu(x)+\theta_{\text{far}}(x)$, and $$\Big|\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2}(\theta_{\text{far}} (x)\bigr)x^2dx\Big|\le C\int_{\frac 12 C_1Z^{-10^{-9}}}^\infty (x^{-4})^{5/2} x^2dx\le C^\prime Z^{10^{-6}}$$ $$\Big|\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}(\theta_{\text{far}}(x) \bigr)dx\Big|\le C\int_{\frac 12C_1Z^{-10^{-9}}}^\infty(x^{-4})^{3/2}dx \le C^\prime Z^{10^{-6}}\ ,$$ \noindent equations (144), (145) may be rewritten in the form $$X=-\frac{4}{15\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2}\varphi(x)x^2dx -\frac{2}{9\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}\varphi(x)dx + \text{Error}_X\tag"(146)"$$ \noindent with $$|\text{Error}_X|\le C\, Z^{8/5}\ .\tag"(147)"$$ \noindent Thus we have computed$X$. Similarly, we compute$Y$by putting (138), (142), (143) into (137). Thus, we obtain $$\multline Y=\frac{1}{12\pi}\int_0^\infty V^{\prime\prime}(x) \bigl(-V(x)\bigr)_+^{1/2}\bigl(\sum\limits_\nu\theta_\nu(x)\bigr) x^2dx+\frac{1}{3\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2} \bigl(\sum\limits_\nu\theta_\nu(x)\bigr)dx\\ +\ \text{Error}_Y\ ,\quad\text{with}\endmultline\tag"(148)"$$ $$|\text{Error}_Y|\le C\, Z^{10^{-9}}\sum\limits_\nu\Cal S(x_\nu) x_\nu^{-2}\le C^\prime Z^{8/5+10^{-9}}\ .\tag"(149)"$$ \noindent (The last inequality follows from the intermediate steps in (145).) As before, we use$\varphi(x)=\sum\limits_\nu\theta_\nu(x)+\theta_{\text{far}} (x)$and $$\Big|\int_0^\infty V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{1/2} \theta_{\text{far}}(x)x^2dx\Big|\le C\int_{\frac 12 C_1Z^{-10^{-9}}}^\infty x^{-6}(x^{-4})^{1/2}x^2dx\le C^\prime Z^{10^{-6}}$$ $$\Big|\int_0^\infty\bigl(-V(x)\bigr)_+^{3/2}\theta_{\text{far}} (x)dx\Big|\le C\int_{\frac 12C_1Z^{-10^{-9}}}^\infty(x^{-4})^{3/2} dx\le C^\prime Z^{10^{-6}}$$ \noindent to rewrite (148), (149) in the form $$Y=\frac{1}{12\pi}\int_0^\infty V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{1/2}\varphi(x)x^2dx+\frac{1}{3\pi}\int_0^\infty \bigl(-V(x)\bigr)_+^{3/2}\varphi(x)+\ \text{Error}_Y\tag"(150)"$$ \noindent with $$|\text{Error}_Y|\le C\, Z^{\frac 85+10^{-9}}\ .\tag"(151)"$$ \noindent Thus, we have computed$X$and$Y$in (67)$\ldots$(74). Next, we make an analogous computation of$X_c$,$Y_c$, defined by (69), (71) in terms of the Coulomb potential$V_c(x)=E_0-\frac Zx$. As in (125)$\ldots$(128), we now define$W_c(x)=-x^2V_c(x)=Zx-E_0x^2$. The analogues of (125)$\ldots$(128) are as follows. Set$\Cal S_c(x)=Zx$. Then: $$\big|\bigl(\frac{d}{dx}\bigr)^\alpha W_c(x)\big|\le C_\alpha\Cal S_c(x)x^{-\alpha}\quad\text{for}\ 0c\Cal S_c(\check r_c)\check r_c^{-2}\ \text{and}\ W\sim \Cal S_c(\check r_c)\ .\endmultline\tag"(153)"$$ $$\multline \text{For}\ x\in (0,\frac{10\, Z}{E_0})\ \text{outside\ the\ interval}\ \{|x-\check r_c|c_2\Cal S_c(x)x^{-1}\ ,\ \text{where}\ c_2\ \text{depends\ on}\ c_1\ .\endmultline\tag"(154)"$$ $$W_c(x)<0\quad\text{for}\ x\ge \frac{2Z}{E_0}\sim Z^{-1/3}\ .\tag"(155)"$$ \noindent We again use the decomposition (129) of$\varphi$. As in (136), (137), we have $$X_c=-\frac{2}{3\pi}\sum\limits_\nu X(\theta_\nu,W_c)-\frac{2}{3\pi} X(\theta_{\text{far}}, W_c)\quad\text{and}\tag"(156)"$$ $$Y_c=\frac{1}{24\pi}\sum\limits_\nu Y(\theta_\nu,W_c)+\frac{1}{24\pi} Y(\theta_{\text{far}}, W_c)\ .\tag"(157)"$$ \noindent Since$W_c(x)<0$in$\text{supp}\ \theta_{\text{far}}$by (134), (155), it follows immediately from the definitions of$X(\theta,W)$and$Y(\theta,W)$that $$X(\theta_{\text{far}}, W_c)=Y(\theta_{\text{far}}, W_c)=0\ .\tag"(158)"$$ \noindent Similarly, $$X(\theta_\nu, W_c)=Y(\theta_\nu, W_c)=0\quad\text{if}\ x_\nu>10\frac{Z} {E_0}\ .\tag"(159)"$$ For$x_\nu\le 10\frac{Z}{E_0}$, we can read off$X(\theta_\nu,W_c)$and$Y(\theta_\nu,W_c)$from Lemma 10, as in (139)$\ldots$(143). In fact, the hypotheses of Lemma 10 hold here for$\theta_\nu$,$W_c$, with$\Cal S=\Cal S_c(x_\nu)$and$t_{\max}\sim \Cal S_c^{1/2}(x_\nu)$. To check the hypotheses of Lemma 10, we argue as in the paragraphs following (139), with (152)$\ldots$(154) playing the role of (125)$\ldots$(128). Note that$t_{\max}\sim (Zx_\nu)^{1/2}>Z^{10^{-9}}$for all the$x_\nu$. Hence Lemma 10 yields the following results analogous to (140)$\ldots$(143). If$x_\nu\le 10 \frac{Z}{E_0}$, then $$X(\theta_\nu, W_c)=\frac 25\int_0^\infty\bigl(-V_c(x)\bigr)_+^{5/2} \theta_\nu(x)x^2dx+\frac 13\int_0^\infty\bigl(-V_c(x)\bigr)_+^{3/2} \theta_\nu(x)dx+\ \text{Error}_X^c(\nu)\tag"(160)"$$ \noindent with $$|\text{Error}_X^c(\nu)|\le C\Cal S_c(x_\nu)\cdot x_\nu^{-2}\ , \quad\text{and} \tag"(161)"$$ $$Y(\theta_\nu, W_c)=2\int_0^\infty V_c^{\prime\prime}(x)\cdot \bigl(-V_c(x)\bigr)_+^{1/2}\theta_\nu(x)x^2dx+8 \int_0^\infty\bigl(-V_c (x)\bigr)_+^{3/2}\theta_\nu(x)dx+\ \text{Error}_Y^c(\nu)\tag"(162)"$$ \noindent with $$|\text{Error}_Y^c(\nu)|\le CZ^{10^{-9}}\Cal S_c(x_\nu)\cdot x_\nu^2\ . \tag"(163)"$$ \noindent If instead$x_\nu>\frac{10\, Z}{E_0}$, then$V_c>0$on$\text{supp}\ \theta_\nu$by (130), (155). Therefore (160), (162) hold with$\text{Error}_X^c(\nu)=\ \text{Error}_Y^c(\nu)=0$, by virtue of (159). Hence, as in the proof of (144), (145), we get $$\multline X_c=-\frac{4}{15\pi}\int_0^\infty\bigl(-V_c(x)\bigr)_+^{5/2}\bigl( \sum\limits_\nu\theta_\nu(x)\bigr)x^2dx-\frac{2}{9\pi}\int_0^\infty \bigl(-V_c(x)\bigr)_+^{3/2}\bigl(\sum\limits_\nu\theta_\nu(x)\bigr) dx\\ + \text{Error}_X^c\endmultline\tag"(164)"$$ \noindent and $$|\text{Error}_X^c|\le C\sum\limits_{x_\nu<\frac{10 Z}{E_0}}\Cal S_c (x_\nu)x_\nu^{-2}\le C^\prime Z^{\frac 85}\ .\tag"(165)"$$ \noindent Since$V_c(x)>0$in$\text{supp}\ \theta_{\text{far}}$, we have also $$\int_0^\infty\bigl(-V_c(x)\bigr)_+^{5/2}(\theta_{\text{far}}(x)\bigr) x^2dx=\int_0^\infty\bigl(-V_c(x)\bigr)_+^{3/2}(\theta_{\text{far}}(x)\bigr) dx=0\ ,$$ \noindent so that (164), (165) may be rewritten in the form $$X_c=-\frac{4}{15\pi}\int_0^\infty\bigl(-V_c(x)\bigr)_+^{5/2} \varphi(x)x^2dx-\frac{2}{9\pi}\int_0^\infty\bigl(-V_c(x)\bigr)_+^{3/2} \varphi(x)dx+\ \text{Error}_X^c\tag"(166)"$$ \noindent with $$|\text{Error}_X^c|\le CZ^{\frac 85}\ .\tag"(167)"$$ Thus, we have succeeded in computing$X_c$. Similarly for$Y_c$, we obtain as in (148) the equation $$\multline Y_c=\frac{1}{12\pi}\int_0^\infty V_c^{\prime\prime}\cdot\bigl(-V_c(x)\bigr)_+^{1/2}\bigl(\sum\limits_\nu\theta_\nu(x)\bigr)x^2dx+\frac{1}{3\pi} \int_0^\infty\bigl(-V_c(x)\bigr)_+^{3/2}\bigl(\sum\limits_\nu\theta_\nu (x)\bigr)dx\\ +\ \text{Error}_Y^c\ ,\quad\text{with}\endmultline$$ $$|\text{Error}_Y^c|\le C\, Z^{\frac 85+10^{-9}}\ .$$ \noindent Again using the fact that$V_c(x)>0$in$\text{supp}\ (\theta_{\text{far}})$, we can rewrite these equations in the form: $$\multline Y_c=\frac{1}{12\pi}\int_0^\infty V_c^{\prime\prime}(x)\cdot\bigl(-V_c(x)\bigr)_+ ^{1/2}\varphi(x)x^2dx+\frac{1}{3\pi}\int_0^\infty\bigl(-V_c(x)\bigr)_+^{3/2} \varphi(x)dx\\ +\ \text{Error}_Y^c\endmultline\tag"(168)"$$ \noindent with $$|\text{Error}_Y^c|\le C\, Z^{\frac 85+10^{-9}}\ .\tag"(169)"$$ \noindent We have computed$X$,$Y$,$X_c$,$Y_c$in (67). Next we compute$W_c$, which is defined in (73). The function$\tilde\chi$is Lipschitz continuous and periodic with period$1. Hence \align &\tilde\chi\Bigl(\frac{Z}{2E_0^{1/2}}-\sqrt{\ell(\ell+1)}-\frac 12\Bigr)= \tilde\chi\Bigl(\frac{Z}{2E_0^{1/2}}-[\ell+\frac 12+O(\ell^{-1})]-\frac 12\Bigr)\\ &=\tilde\chi\Bigl(\frac{Z}{2E_0^{1/2}}-\ell-1\Bigr)+O(\ell^{-1})=\tilde\chi \Bigl(\frac{Z}{2E_0^{1/2}}\Bigr)+O(\ell^{-1})\ .\endalign \noindent Putting this into (73), we get $$\multline W_c=\sum\limits_{Z^{10^{-9}}<\ell<\Omega_c}(2\ell+1)\cdot\frac{2E_0^{3/2}}{Z} \tilde\chi\Bigl(\frac{Z}{2E_0^{1/2}}\Bigr)\\ +\sum\limits_{Z^{10^{-9}}<\ell<\Omega_c}(2\ell+1)\cdot\frac{2E_0^{3/2}} {Z}\cdot O(\ell^{-1})\\ =[\Omega_c^2+O(\Omega_c)]\cdot\frac{2E_0^{3/2}}{Z}\tilde\chi \Bigl(\frac{Z}{2E_0^{1/2}}\Bigr)+O(\Omega_c)\cdot\frac{E_0^{3/2}}{Z}\\ =\Omega_c^2\cdot 2E_0^{3/2} Z^{-1}\tilde\chi\Bigl(\frac{Z}{2E_0^{1/2}}\Bigr) +O(\Omega_cE_0^{3/2}Z^{-1})\ .\endmultline\tag"(170)"$$ \noindent Recall that\Omega_c$is defined as the positive root of$\Omega_c(\Omega_c+1)=\operatornamewithlimits{\max}\limits_{x>0} (-x^2V_c(x))=\operatornamewithlimits{\max}\limits_{x>0}(Zx-E_0x^2)=\frac 14 \frac{Z^2}{E_0}$. (The max.\ is attained at$x=\frac 12 \frac{Z}{E_0}$). Thus,$\Omega_c^2=\frac 14\frac{Z^2}{E_0}-\Omega_c$. Substituting this into (170), we get $$W_c=\bigl(\frac 14Z^2E_0^{-1})\cdot 2E_0^{3/2}Z^{-1}\tilde\chi \bigl(\frac{Z}{2E_0^{1/2}}\bigr)+O(\Omega_cE_0^{3/2}Z^{-1})\ ,$$ \noindent i.e.\ $$W_c=\frac 12 ZE_0^{1/2}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}\bigr)+\ \text{Error} _{W}\ ,\quad\text{with}\tag"(171)"$$ $$|\text{Error}_W|\le C\, Z^{4/3}\ .\tag"(172)"$$ \noindent (Recall that$\Omega_c\sim Z^{+1/3}$,$E_0\sim Z^{+4/3}$.) We prepare to substitute our results on$X$,$X_c$,$Y$,$Y_c$,$W_cinto (67). First of all, (146), (147), (166), (167) together yield: \align X-X_c&=-\frac{4}{15\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{5/2} -\bigl(-V_c(x)\bigr)_+^{5/2}\big]\varphi(x)x^2dx\\ &\ -\frac{2}{9\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{3/2}-\bigl(-V_c(x) \bigr)_+^{3/2}\bigr]\varphi(x)dx+\ \text{Error}_X^\ast\tag"(173)"\endalign \noindent with|\text{Error}_X^\ast|\le C\, Z^{8/5}$. We can omit the factor$\varphi(x)$in the two integrals in (173), since$V(x)=V_c(x)$for$\varphi(x)\neq1$by virtue of (3) and the definition of$\varphi. Therefore, \align X-X_c&=-\frac{4}{15\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{5/2} -\bigl(-V_c(x)\bigr)_+^{5/2}\bigr]x^2dx\\ &\ -\frac{2}{9\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{3/2} -\bigl(-V_c(x)\bigr)_+^{3/2}\bigr]dx+\ \text{Error}_X^\ast\ , \tag"(174)"\endalign \noindent with $$|\text{Error}_x^\ast|\le C\, Z^{8/5}\ .\tag"(175)"$$ Similarly, combining (150), (151) with (168), (169) and dropping the factors\varphi(x)from the resulting integrals, we find that \align Y-Y_c&=\frac{1}{12\pi}\int_0^\infty\bigl[V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{1/2}-V_c^{\prime\prime}(x)\cdot\bigl(-V_c(x)\bigr)_+^{1/2} \bigr]x^2dx\\ &\ +\frac{1}{3\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{3/2} -\bigl(-V_c(x)\bigr)_+^{3/2}\bigr]dx+\ \text{Error}_Y^\ast\ , \tag"(176)"\endalign \noindent with $$|\text{Error}_Y^\ast|\le C\, Z^{\frac 85+10^{-9}}\ .\tag"(177)"$$ Combining (174)\ldots(177), we get \align X-X_c+Y-Y_c&=-\frac{4}{15\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{5/2} -\bigl(-V_c(x)\bigr)_+^{5/2}\bigr]x^2dx\\ &\ +\frac{1}{12\pi}\int_0^\infty\bigl[V^{\prime\prime}(x)\cdot \bigl(-V(x)\bigr)_+^{1/2}-V_c^{\prime\prime}(x)\cdot\bigl(-V_c(x)\bigr)_+^{1/2} \bigr]x^2dx\\ &\ +\frac{1}{9\pi}\int_0^\infty\bigl[\bigl(-V(x)\bigr)_+^{3/2} -\bigl(-V_c(x)\bigr)_+^{3/2}\bigr]dx+\ \text{Error}_F\tag"(178)"\endalign \noindent with $$|\text{Error}_F|\le C\, Z^{\frac 85+10^{-9}}\ .\tag"(179)"$$ We can simplify the right-hand side of (178) by an elementary integration by parts. For\tilde V(x)$smooth on$(0,\infty)$, the function$\bigl(-\tilde V(x)\bigr)_+^{3/2}$belongs to$C^1(0,\infty)$, so that $$\int_{\varepsilon_1}^R\bigl(-\tilde V(x)\bigr)_+^{3/2}dx=x\cdot \bigl(-\tilde V(x)\bigr)_+^{3/2}\Bigr]_{\varepsilon_1}^R -\int_{\varepsilon_1}^R\frac 32\bigl(-\tilde V(x)\bigr)_+^{1/2}\cdot \bigl(-\tilde V^\prime(x)\bigr)xdx\ .$$ \noindent We apply this to$\tilde V=V$and to$\tilde V= V_c$. In either case, we may let$R\to\infty$, obtaining $$\gather \int_{\varepsilon_1}^\infty\bigl(-V(x)\bigr)_+^{3/2}dx=-\varepsilon_1 \bigl(-V(\varepsilon_1)\bigr)_+^{3/2}+\frac 34\int_{\varepsilon_1}^\infty \frac 2x V^\prime(x)\cdot\bigl(-V(x)\bigr)_+^{1/2}x^2dx\quad\text{and}\\ \int_{\varepsilon_1}^\infty\bigl(-V_c(x)\bigr)_+^{3/2}dx=-\varepsilon_1 \bigl(-V_c(\varepsilon_1)\bigr)_+^{3/2}+\frac 34 \int_{\varepsilon_1}^\infty\frac 2xV_c^\prime(x)\cdot \bigl(-V_c(x)\bigr)_+^{1/2}x^2dx\ .\endgather$$ \noindent Subtracting and recalling that$V(x)=V_c(x)$for$x\le \varepsilon_1$(if we pick$\varepsilon_10\ . \tag"(181)" $$\noindent In particular, this holds for all E\in (0,E_0] when Z, E_0, P, Z^2-4E_0P>0. Integrating (181) in E over (0,E_0], we obtain$$ -\frac{2}{3\pi}\int_0^\infty\Bigl(\frac Zx-\frac{P}{x^2}-E_0\Bigr)_+^{3/2} dx+\frac{2}{3\pi}\int_0^\infty\Bigl(\frac Zx-\frac{P}{x^2}\Bigr)_+^{3/2} dx=ZE_0^{1/2}-P^{1/2}E_0\ . $$\noindent Hence,$$\split -\frac{2}{3\pi}\int_0^\infty\Bigl(\frac Zx-\frac {P}{x^2}-E_0\Bigr)_+^{3/2} dx=ZE_0^{1/2}-P^{1/2}E_0+G(Z,P)\ ,\\ \text{whenever}\ Z, E_0, P>0\quad\text{and}\quad P0\quad\text{and}\quad Z^2-4E_0P>0\ .\endsplit\tag"(183)" $$\noindent Integrating (183) in P over the interval (0,\frac{Z^2}{4E_0}) and noting that (\frac Zx-\frac{P}{x^2}-E_0)_+^{5/2}\equiv 0 when P=\frac{Z^2}{4E_0}, we find that$$ -\frac{4}{15\pi}\int_0^\infty\Bigl(\frac Zx-E_0\Bigr)_+^{5/2}x^2dx=ZE_0^{1/2} \cdot\bigl(\frac{Z^2}{4E_0}\bigr)-\frac 23 E_0\bigl(\frac{Z^2}{4E_0} \bigr)^{3/2}-\frac 14 Z^2\cdot 2\bigl(\frac{Z^2}{4E_0}\bigr)^{1/2}\ , $$\noindent i.e.\$$ -\frac{4}{15\pi}\int_0^\infty\bigl(-V_c(x)\bigr)_+^{5/2}x^2dx= -\frac{1}{12}\ \frac{Z^3}{E_0^{1/2}}\ . $$\noindent Putting this equation into (180), we obtain$$\multline X-X_c+Y-Y_c\\ =-\frac{4}{15\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2}x^2dx+ \frac{1}{12\pi}\int_0^\infty\bigl(V^{\prime\prime}(x)+\frac 2x V^\prime(x)\bigr)\cdot\bigl(-V(x)\bigr)_+^{1/2}x^2dx\\ +\frac{1}{12}Z^3E_0^{-1/2}+\ \text{Error}_F\ ,\endmultline $$\noindent with \text{Error}_F estimated by (179). Combining this with the definition (72) of W and equations (171), (172) for W_c, we conclude that$$\multline X-X_c+Y-Y_c+W-W_c\\ =-\frac{4}{15\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2}x^2dx+\frac{1} {12\pi}\int_0^\infty\bigl(V^{\prime\prime}(x)+\frac 2x V^\prime(x)\bigr) \cdot\bigl(-V(x)\bigr)_+^{1/2}x^2dx\\ +\pi\sum\limits_{Z^{10^{-9}}<\ell<\Omega}\frac{(2\ell+1)}{n_\ell} \tilde\chi(\phi_\ell)+\frac{1}{12} Z^3E_0^{-1/2}-\frac 12 ZE_0^{1/2} \tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}\bigr)\\ +\ \text{Error}_G\ , \endmultline\tag"(184)" $$\noindent with$$ |\text{Error}_G|\le C\, Z^{\frac 85+10^{-9}}\ .\tag"(185)" $$Substituting (184), (185) into (67), and recalling (74), we obtain the formula:$$\multline \text{sneg}(H)-\ \text{sneg}(H_c)\\ =-\frac{4}{15\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2} x^2dx+\frac{1}{12\pi}\int_0^\infty\bigl(V^{\prime\prime}(x)+\frac 2x V^\prime(x)\bigr)\cdot\bigl(-V(x)\bigr)_+^{1/2}x^2dx\\ +\pi\sum\limits_{Z^{10^{-9}}<\ell<\Omega}\frac{(2\ell+1)}{n_\ell}\tilde\chi (\phi_\ell)\\ +\frac{1}{12} Z^3E_0^{-1/2}-\frac 12 ZE_0^{1/2}\tilde\chi \bigl(\frac{Z}{2E_0^{1/2}}\bigr)+\ \text{Error}_H\ ,\endmultline \tag"(186)" $$\noindent with$$ |\text{Error}_H|\le C\, Z^{\frac 85+2\cdot 10^{-9}}\ .\tag"(187)" $$Next we compute \text{sneg}(H_c). Recall that the eigenvalues of H_c=-\Delta+E_0-\frac{Z}{|x|} on \Bbb R^3 are given by E_0-\frac{Z^2} {4n^2} with multiplicity n^2 (n=1,2,3,\ldots). Therefore,$$ \text{sneg}(H_c)=\sum\limits_{1\le n\le \frac{Z}{2E_0^{1/2}}} n^2 \bigl(E_0-\frac{Z^2}{4n^2}\bigr)=\sum\limits_{1\le n\le \frac{Z}{2E_0^{1/2}}} \bigl(E_0n^2-\frac{Z^2}{4}\bigr)\ .\tag"(188)" $$\noindent We evaluate the right-hand side of (188) by using the Lemma on Riemann sums, with$$ f(t)=E_0t^2-\frac{Z^2}{4}\ ,\quad [a,b]=\Bigl[1,\frac{Z}{2E_0^{1/2}}\Bigr]\ , \quad \sigma(t)\equiv Z^2\ ,\quad \tau(t)\equiv Z^{1/3}\ . $$\noindent The hypotheses of the Lemma on Riemann Sums are easily verified, since E_0\sim Z^{4/3}. Thus we obtain from (188) and the Lemma on Riemann sums:$$\multline \text{sneg}(H_c)=\sum\limits_{n\in\Bbb Z\cap [a,b]}f(n)\\ =\int_a^bf(t)dt-f(b)\chm(b)-f(a)\chp(a)+\frac 12 f^\prime(b)\tilde\chi(b)-\frac 12 f^\prime(a)\tilde\chi(a)+\ \text{Error}\ ,\\ \text{with}\ |\text{Error}|\le C\frac{\sigma(a)}{\tau^2(a)} +C\frac{\sigma(b)}{\tau^2(b)}+C\int_a^b\sigma(t)\tau^{-100}(t)dt\sim Z^{4/3}\ .\endmultline\tag"(189)" $$\noindent We have$$\multline \int_a^bf(t)dt=\int_1^{Z/(2E_0^{1/2})}\bigl(E_0t^2-\frac{Z^2}{4}\bigr)dt\\ =\Bigl(\frac 13 E_0\bigl[\frac{Z}{2E_0^{1/2}}\bigr]^3-\frac{Z^2}{4} \cdot\bigl[\frac{Z}{2E_0^{1/2}}\bigr]\Bigr)-\Bigl(\frac 13 E_0-\frac{Z^2}{4} \Bigr)\\ =-\frac{1}{12}Z^3E_0^{-1/2}+\frac{Z^2}{4}+O(Z^{4/3})\ ,\endmultline \tag"(190)"  f(b)\chm(b)=0\ ,\quad\text{since}\quad f(b)=E_0\bigl(\frac{Z}{2E_0^{1/2}} \bigr)^2-\frac{Z^2}{4}=0\tag"(191)" \multline f(a)\chp(a)=\bigl(E_0-\frac{Z^2}{4}\bigr)\chp(1)=\bigl(E_0-\frac{Z^2}{4}\bigr) \cdot(-\frac 12)\\ \text{(by\ definition\ of}\ \chp(-))=+\frac{Z^2}{8}+O(Z^{4/3})\endmultline \tag"(192)" \align \frac 12 f^\prime(b)\tilde\chi(b)&=\frac 12\cdot (2E_0b) \tilde\chi(b)=\frac 12\cdot\bigl(2E_0\cdot\frac{Z}{2E_0^{1/2}}\bigr) \tilde\chi(\frac{Z}{2E_0^{1/2}}\bigr)\\ &=\frac 12 ZE_0^{1/2}\tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}\bigr)\tag"(193)" \endalign  \frac 12 f^\prime(a)\tilde\chi(a)=\frac 12\cdot(2E_0)\tilde\chi(1) =O(Z^{4/3})\ .\tag"(194)" $$\noindent Putting (190)\ldots(194) into (189), we find that$$\align \text{sneg}(H_c)&=\Bigl(-\frac{1}{12}Z^3E_0^{-1/2}+\frac{Z^2}{4}\Bigr)- \bigl(\frac{Z^2}{8}\bigr)+\Bigl(\frac 12 ZE_0^{1/2}\tilde\chi \bigl(\frac{Z}{2E_0^{1/2}}\bigr)\Bigr)+O(Z^{4/3})\\ &=-\frac{1}{12}Z^3E_0^{-1/2}+\frac{Z^2}{8}+\frac 12 ZE_0^{1/2} \tilde\chi\bigl(\frac{Z}{2E_0^{1/2}}\bigr)+O(Z^{4/3})\ .\endalign $$\noindent Adding this equation to (186) and recalling (187), we obtain our basic formula for \text{sneg}(H), namely$$\multline \text{sneg}(H)\\ =-\frac{4}{15\pi}\int_0^\infty\bigl(-V(x)\bigr)_+^{5/2}x^2dx +\frac{Z^2}{8}+\frac{1}{12\pi}\int_0^\infty\bigl(V^{\prime\prime} (x)+\frac 2x V^\prime(x)\bigr)\cdot\bigl(-V(x)\bigr)_+^{1/2} x^2dx\\ +\pi\sum\limits_{Z^{10^{-9}}<\ell<\Omega}\frac{(2\ell+1)}{n_\ell} \tilde\chi(\phi_\ell)+\ \text{Error}_I\ ,\endmultline\tag"(195)" $$\noindent with$$ |\text{Error}_I|\le C\, Z^{\frac 85+2\cdot 10^{-9}}\ .\tag"(196)" \noindent These results hold under the assumptions (1), (2), (3) above, with \phi_\ell, n_\ell given by (6) and (7). It will be convenient to modify slightly the sum in (195). To do so, we give a lower bound for n_\ell. Recall that \{V_\ell(r)<0\}= (x_{\text{left}}(\ell), x_{\text{rt}}(\ell)) with x_{\text{left}}(\ell) \sim\frac{\ell^2}{Z}, x_{\text{rt}}(\ell)\sim\frac 1\ell for Z^{10^{-9}}<\ell<\Omega. In (x_{\text{left}}(\ell), x_{\text{rt}}(\ell)) we have 0<-V_\ell(r)=-V(r)-\frac{\ell(\ell+1)}{r^2}<-V(r)0}\bigl(-r^2V(r)\bigr)\\ n_\ell&=\int_0^\infty\bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\bigr)_+^{-1/2}dr\\ \phi_\ell&=\frac 1\pi \int_0^\infty\bigl(-V(r)- \frac{\ell(\ell+1)}{r^2}\bigr)_+^{1/2}dr-\frac 12\ .\endalign \noindent The constant $C^\prime$ depends only on $c$, $C$, $C_\alpha$, $\varepsilon$ in (A), (B), (C).\endproclaim \vglue 1pc \demo{Proof} We saw that the sum in (195) could be restricted to a smaller range of $\ell$ as in the statement of the Main Lemma. Since $\Delta V(x) =\bigl(V^{\prime\prime}+\frac 2rV^\prime\bigr)\bigm|_{r=|x|}$ and $\text{dvol}\ =4\pi r^2dr$, our present conclusion is equivalent to (195), (196).$\qquad \blacksquare$\enddemo \vfill\eject \head Perturbation of Eigenvalue Sums\endhead \medskip In the previous section, we computed the sum of the negative eigenvalues of $-\Delta+V(|x|)$ under the restrictive assumption that $V(r)=E_0-\frac Zr$ exactly when $r$ is small. In this section, we use simple perturbation theory, together with our three-dimensional density results, to remove the restrictive assumption. We take $V(r)$ to satisfy the following conditions. $$\Big|\bigl(\frac{d}{dr}\bigr)^\alpha V(r)\Big|\le C_\alpha \min\bigl\{\frac Zr, r^{-4}\bigr\}\cdot r^{-\alpha}\quad\text{for}\ r\in (0,\infty), \alpha\ge 0\tag"(1)"$$ $$\multline \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-V_Z^{TF}(r)\bigr\}\Big|\le c_0\cdot\min\bigl\{\frac Zr, r^{-4}\bigr\}\cdot r^{-\alpha}\ \text{for}\ r\in (0,\infty)\ , 0\le\alpha\le 2\ ;\\ \text{with}\ c_0\ \text{a\ small,\ positive\ constant\ depending\ on}\ C_\alpha\ .\endmultline \tag"(2)"$$ $$\multline \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-\bigl[E_0-\frac Zr\bigr] \bigr\}\Big|\le C_\alpha Z^{\frac 32}r^{\frac 12}\cdot r^{-\alpha}\\ \text{for}\ r\in (0,2Z^{-\frac 35+2\varepsilon}), \alpha\ge 0\ .\ \text{Here}\ 0<\varepsilon<10^{-12}\ ,\text{and}\ cZ^{4/3}From (3) and the properties of \theta, we get$$\multline \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{\theta(r)\cdot \bigl[E_0-\frac Zr-V(r)\bigr]\bigr\}\Big|\le C_\alpha^\prime Z^{\frac 32} r^{\frac 12-\alpha} = C_\alpha^\prime \frac Zr\cdot r^{-\alpha}\cdot (Z^{1/2}r^{3/2})\\ \le C_\alpha^{\prime\prime}\frac Zr\cdot r^{-\alpha}\cdot (Z^{1/2} \cdot Z^{-\frac{9}{10}+3\varepsilon})\\ = C_\alpha^{\prime\prime}\frac Zr\cdot r^{-\alpha}\cdot Z^{-\frac 25+3\varepsilon}\ \text{in\ supp}\ \theta\ .\endmultline\tag"(4)" $$\noindent Therefore,$$ \Big|\bigl(\frac{d}{dr}\bigr)^\alpha V_1(r)\Big|\le C_\alpha\min \bigl\{\frac Zr, r^{-4}\bigr\}r^{-\alpha}+C_\alpha^{\prime\prime} Z^{-\frac 25+3\varepsilon}\min\bigl\{\frac Zr, r^{-4}\bigr\}r^{-\alpha} $$\noindent by (1); and for 0\le \alpha\le 2 we have$$\multline \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V_1(r)-V_Z^{TF}(r)\bigr\}\Big|\le \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-V_Z^{TF}(r)\bigr\}\Big|\\ +C^{\prime\prime}Z^{-\frac 25+3\varepsilon}\min\bigl\{\frac Zr, r^{-4}\bigr\} r^{-\alpha} \le 2c_0\min\bigl\{\frac Zr, r^{-4}\bigr\}r^{-\alpha}\ \text{by\ (2)}\ . \endmultline $$These estimates imply hypotheses (A) and (B) for the potential V_1. Thus, we may apply the Main Lemma of the previous section to V_1, obtaining$$\multline \text{sneg}(-\Delta+V_1)=-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V_1(x)\bigr) ^{5/2}dx+\frac{Z^2}{8}+\frac{1}{48\pi ^2}\int_{\Bbb R^3}(\Delta V_1(x)\bigr) \cdot\bigl(-V_1(x)\bigr)^{1/2}dx\\ +\pi \sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)+\ \text{Error}_1\ , \endmultline\tag"(5)" $$\noindent with$$ |\text{Error}_1|\le C^\prime Z^{\frac 85+2\cdot 10^{-9}}\tag"(6)"  \Omega=\ \text{pos.\ root\ of}\ \Omega(\Omega+1)=\operatornamewithlimits {\max}\limits_{r>0}\bigl(-r^2V_1(r)\bigr)\tag"(7)"  n_\ell=\int_0^\infty\Bigl(-V_1(r)-\frac{\ell(\ell+1)}{r^2}\Bigr)_+^{-1/2} dr\quad (Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega)\tag"(8)"  \phi_\ell= \frac 1\pi\int_0^\infty\Bigl(-V_1(r)-\frac{\ell(\ell+1)}{r^2}\Bigr) _+^{1/2}dr-\frac 12\quad (Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega)\ .\tag"(9)" $$\noindent Here we write V, V_1 both for functions of one variable, and for the corresponding radial functions on \Bbb R^3. In (7), (8), (9), we can replace V_1(r) by V(r) without changing the values of \Omega, n_\ell, \phi_\ell. In fact, (2) shows that -r^2V(r) is an increasing function on (0,2Z^{-\frac 35+2\varepsilon}). Similarly, -r^2V_1(r) is increasing on (0,2Z^{-\frac 35+3\varepsilon}), by virtue of hypotheses (A) and (B) of the Main Lemma, which we know to be valid for V_1. Hence, the maxima of -r^2V(r) and -r^2V_1(r) are both attained in (2Z^{-\frac 35+2\varepsilon},\infty), where V\equiv V_1. \smallskip \noindent Consequently, changing V_1(r) to V(r) in (7) leaves the value of \Omega unchanged. To see that (8), (9) are also unchanged, suppose \ell>Z^{\frac 8{25}+10\varepsilon}. Then for 0Z^{\frac{8}{25}+10\varepsilon} and r<2Z^{-\frac 35+2\varepsilon}. This implies$$\multline \int_0^\infty\Bigl(-V_1(r)-\frac{\ell(\ell+1)}{r^2}\Bigr)_+^{-1/2} dr=\int_{2Z^{-\frac 35+2\varepsilon}}^\infty\Bigl(-V_1(r) -\frac{\ell(\ell+1)}{r^2}\Bigr)_+^{-1/2}dr\\ =\int_{2Z^{-\frac 35+2\varepsilon}}^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)}{r^2} \Bigr)_+^{-1/2}dr=\int_0^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\Bigr)_+ ^{-1/2}dr\endmultline $$\noindent for \ell>Z^{\frac{8}{25}+10\varepsilon}, so (8) is unaffected when we replace V_1(r) by V(r). Similarly for (9). Thus,$$\multline \text{sneg}(-\Delta+V_1)=-\frac{1}{15\pi ^2}\int_{\Bbb R^3}\bigl(-V_1(x) \bigr)^{5/2}dx+\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}\bigl(\Delta V_1(x) \bigr)\cdot\bigl(-V_1(x)\bigr)^{1/2}dx\\ +\pi\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)+\ \text{Error}_1\ ,\quad\text{with} \endmultline\tag"(10)" \Omega=\ \text{pos.\ root\ of}\ \Omega(\Omega+1)=\operatornamewithlimits{\max} \limits_{r>0}\bigl(-r^2V(r)\bigr)\tag"(11)"  n_\ell=\int_0^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\Bigr)_+^{-1/2} dr\tag"(12)"  \phi_\ell=\frac 1\pi\int_0^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\Bigr) _+^{1/2}dr-\frac 12\tag"(13)"  |\text{Error}_1|\le C^\prime Z^{\frac 85+2\cdot 10^{-9}}\ .\tag"(14)" $$We want to compare \text{sneg}(-\Delta+V_1) with \text{sneg}(-\Delta+V). Let E_1,\ldots,E_N be the non-positive eigenvalues of -\Delta+V_1, and let \varphi_1(x)\ldots\varphi_N(x) be the corresponding normalized eigenfunctions. We introduce the N-particle wave function$$ \psi(x_1\ldots x_N)=\frac{1}{\sqrt{N!}}\sum\limits_\sigma(\text{sgn}\ \sigma) \varphi_{\sigma 1}(x_1)\ldots\varphi_{\sigma N}(x_N)\ ,\tag"(15)" $$\noindent where \sigma denotes a permutation of \{1\ldots N\}. Thus, \psi is antisymmetric and has norm 1. Moreover,$$\multline \text{sneg}(-\Delta+V_1)=E_1+\ldots+E_N=\langle \sum\limits_{k=1}^N \bigl(-\Delta_{x_k}+V_1(x_k)\bigr)\psi,\psi\rangle\\ =\langle \sum\limits_{k=1}^N \bigl(-\Delta_{x_k}+V(x_k)\bigr)\psi,\psi\rangle +\langle\sum\limits_{k=1}^N\bigl(V_1(x_k)-V(x_k)\bigr)\psi,\psi\rangle\\ \ge\ \text{sneg}(-\Delta+V)+\langle\sum\limits_{k=1}^N \bigl(V_1(x_k)-V(x_k)\bigr)\psi,\psi\rangle\ ,\endmultline\tag"(16)" $$\noindent as follows by expanding \psi(x_1\ldots x_N) in terms of the eigenfunctions of -\Delta+V(x). For the wave function (15), we have$$ \langle\sum\limits_{k=1}^N\bigl(V_1(x_k)-V(x_k)\bigr)\psi,\psi\rangle= \int_{\Bbb R^3}\rho_1(x)\bigl(V_1(x)-V(x)\bigr)dx\ ,\tag"(17)" $$\noindent where$$ \rho_1(x)=\sum\limits_{k=1}^N|\varphi_k(x)|^2=\ \text{density\ associated\ to}\ -\Delta+V_1\ .\tag"(18)" $$\noindent Putting (17) into (16), we get the inequality$$ \text{sneg}(-\Delta+V_1)\ge\ \text{sneg}(-\Delta+V)+\int_{\Bbb R^3} \rho_1(x)\bigl(V_1(x)-V(x)\bigr)dx\ .\tag"(19)" $$\noindent Similarly, reversing the roles of V_1 and V in the last paragraph, we find that$$ \text{sneg}(-\Delta+V)\ge\ \text{sneg}(-\Delta+V_1)+\int_{\Bbb R^3} \rho(x)\cdot\bigl(V(x)-V_1(x)\bigr)dx\ ,\tag"(20)" $$\noindent where \rho(x) is the density associated to -\Delta+V. \noindent In (19), (20) we want to replace \rho_1 and \rho by their semiclassical approximations. Hence we rewrite these inequalities in the form$$\multline \text{sneg}(-\Delta+V_1)\ge\ \text{sneg}(-\Delta+V) +\frac{1}{6\pi^2}\int_{\Bbb R^3}\bigl(-V_1(x)\bigr)^{3/2}\bigl(V_1(x)- V(x)\bigr)dx\\ +\int_{\Bbb R^3}\bigl[\rho_1(x)-\frac{1}{6\pi^2}\bigl(-V_1(x)\bigr)^{3/2} \bigr]\bigl(V_1(x)-V(x)\bigr)dx\endmultline\tag"(21)" \multline \text{sneg}(-\Delta+V)\ge\ \text{sneg}(-\Delta+V_1)+ \frac{1}{6\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2} \bigl(V(x)-V_1(x)\bigr)dx\\ +\int_{\Bbb R^3}\bigl[\rho(x)-\frac{1}{6\pi^2}\bigl(-V(x)\bigr)^{3/2}\bigr] \bigl(V(x)-V_1(x)\bigr)dx\ .\endmultline\tag"(22)" $$\noindent We can control the integrals at the far right in (21), (22) by using Theorem 1 in the section on the WKB Density Theorems for Approximate TF Potentials. Since$$ V_1(x)-V(x)=\theta(r)\cdot\bigl[E_0-\frac Zr-V(r)\bigr]\ (r=|x|)\ , $$\noindent estimate (3) and the properties of \theta(r) allow us to write$$ V_1(x)-V(x)=\sum\limits_{k=1}^\infty(Z^{3/2}\delta_k^{1/2})U_k(x)\ ,\tag"(23)" $$\noindent where: U_k(x) is a radial function supported in \delta_k <|x|<2\delta_k, |U_k(x)|\le C, |\nabla U_k(x)|\le C\delta_k^{-1}, \delta_{k+1}\le (1-c)\delta_k, \delta_1\sim Z^{-\frac 35+2\varepsilon}. \noindent From (23) and the Lebesgue dominated convergence theorem, we get$$\multline \int_{\Bbb R^3}\Bigl[\rho_1(x)-\frac{\bigl(-V_1(x)\bigr)^{3/2}}{6\pi^2}\Bigr] \cdot\bigl(V_1(x)-V(x)\bigr)dx\\ =\sum\limits_{k=1}^\infty Z^{3/2}\delta_k^{1/2}\int_{\Bbb R^3} \Bigl[\rho_1(x)-\frac{\bigl(-V_1(x)\bigr)^{3/2}}{6\pi^2}\Bigr]U_k (x)dx\ .\endmultline\tag"(24)" $$\noindent Theorem 1 in the section on the Density for Approximate TF Potentials shows that$$ \Big|\int_{\Bbb R^3}\bigl[\rho_1(x)-\frac{\bigl(-V_1(x)\bigr)^{3/2}}{6\pi^2} \bigr]U_k(x)dx\Big|\le C^\prime\bigl(Z\delta_k+Z^{\frac 13+2\cdot 10^{-9}} \bigr)\ . $$\noindent This and (24) imply$$\multline \Big|\int_{\Bbb R^3}\bigl[\rho_1(x)-\frac{1}{6\pi^2}\bigl(-V_1(x)\bigr)^{3/2} \bigr]\cdot\bigl(V_1(x)-V(x)\bigr)dx\Big|\\ \le C\sum\limits_{k=1}^\infty Z^{3/2}\delta_k^{1/2}\bigl(Z\delta_k +Z^{\frac 13+2\cdot 10^{-9}}\bigr)\\ =C\sum\limits_{k=1}^\infty Z^{5/2}\delta_k^{3/2}+C\sum\limits_{k=1}^\infty Z^{\frac{11}{6}+2\cdot 10^{-9}}\delta_k^{1/2}\\ \sim Z^{5/2}\delta_1^{3/2}+Z^{\frac{11}{6}+2\cdot 10^{-9}}\delta_1^{1/2}\\ \sim Z^{5/2}\bigl(Z^{-\frac 35+2\varepsilon}\bigr)^{\frac 32} +Z^{\frac{11}{6}+2\cdot 10^{-9}}\bigl(Z^{-\frac 35+2\varepsilon}\bigr)^{1/2} \sim Z^{\frac 85+3\varepsilon}\ .\endmultline\tag"(25)" $$\noindent Similarly,$$ \Big|\int_{\Bbb R^3}\bigl[\rho(x)-\frac{1}{6\pi^2}\bigl(-V(x)\bigr)^{3/2} \bigr]\cdot\bigl(V_1(x)-V(x)\bigr)dx\Big|\le CZ^{\frac 85+3\varepsilon}\ . \tag"(26)" $$\noindent Putting (25), (26) into (21), (22), we find that$$\multline \text{sneg}(-\Delta+V_1)\ge\ \text{sneg}(-\Delta+V)+\frac{1}{6\pi^2} \int_{\Bbb R^3}\bigl(-V_1(x)\bigr)^{3/2}\bigl(V_1(x)-V(x)\bigr)dx\\ -CZ^{\frac 85+3\varepsilon}\endmultline\tag"(27)" $$\noindent and$$\multline \text{sneg}(-\Delta+V)\ge\ \text{sneg}(-\Delta+V_1)+\frac{1}{6\pi^2} \int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2}\bigl(V(x)-V_1(x)\bigr)dx\\ -CZ^{\frac 85+3\varepsilon}\ .\endmultline\tag"(28)" $$\noindent In the integral in (27) we want to replace \bigl(-V_1(x)\bigr)^{3/2} by \bigl(-V(x)\bigr)^{3/2}. Recall that V_1(r)-V(r) is supported in \bigl[0,2Z^{\frac 35+2\varepsilon}\bigr] and is dominated by Z^{3/2}r^{1/2}; while -V(r)\sim Zr^{-1} in that interval. Hence we can write$$ V_1(r)=\bigl(1+f_1(r)\bigr)V(r)\ ,\qquad f_1(r)=\bigl(V(r)-V_1(r)\bigr) \bigl(-V(r)\bigr)^{-1}\tag"(29)" $$\noindent with f_1(r) supported in \bigl[0,2Z^{-\frac 35+2\varepsilon}\bigr], and with$$ |f_1(r)|\le CZ^{1/2}r^{3/2}\ .\tag"(30)" $$\noindent Therefore, for any exponent p we have$$\align \bigl(-V_1(r)\bigr)^p&=\bigl(1+pf_1(r)+O(|f_1(r)|^2)\bigr)\cdot\bigl(-V(r)\bigr)^p\\ &=\bigl(1+pf_1(r)\bigr)\cdot\bigl(-V(r)\bigr)^p+O(Zr^3\cdot Z^pr^{-p})\\ &=\bigl(-V(r)\bigr)^p+p\bigl(V(r)-V_1(r)\bigr)\cdot \bigl(-V(r)\bigr)^{p-1}+O(Z^{p+1}r^{3-p})\ .\tag"(31)"\endalign $$\noindent In particular,$$\align \big|\bigl(-V_1(r)\bigr)^{3/2}-\bigl(-V(r)\bigr)^{3/2}\big|&\le C|V(r)-V_1(r)|\cdot\bigl(-V(r)\bigr)^{1/2}+CZ^{\frac 52}r^{3/2}\\ &\le CZ^{3/2}r^{1/2}\cdot(Zr^{-1})^{1/2}+CZ^{5/2}r^{3/2}\\ &\le CZ^2\quad\text{for}\ r\in [0,2Z^{-\frac 35+2\varepsilon}]\ , \endalign $$\noindent so$$\multline \Big|\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2}\bigl(V_1(x)-V(x)\bigr)dx-\int_{\Bbb R^3} \bigl(-V_1(x)\bigr)^{3/2}\bigl(V_1(x)-V(x)\bigr)dx\Big|\\ \le \int_{|x|<2Z^{-\frac 35+2\varepsilon}}CZ^2|V_1(x)-V(x)|\ \text{dvol}(x)\\ \le \int_0^{2Z^{-\frac 35+2\varepsilon}}CZ^2\cdot (CZ^{3/2}r^{1/2}) \cdot r^2dr\sim Z^{\frac 72}\cdot (2Z^{-\frac 35+2\varepsilon})^{\frac 72}\ \sim Z^{\frac 75+7\varepsilon}\ .\endmultline $$\noindent Combining this with (27), we get$$\multline \text{sneg}(-\Delta+V_1)\ge\ \text{sneg}(-\Delta+V)+\frac{1}{6\pi^2} \int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2}\bigl(V_1(x)-V(x)\bigr)dx\\ -CZ^{\frac 85+3\varepsilon}\ ,\endmultline $$\noindent which together with (28) implies$$ \text{sneg}(-\Delta+V)=\ \text{sneg}(-\Delta+V_1)+\frac{1}{6\pi^2} \int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2}\bigl(V(x)-V_1(x)\bigr) dx+\ \text{Error}_2\tag"(32)" $$\noindent with |\text{Error}_2|\le CZ^{\frac 85+3\varepsilon}. \noindent Putting (10) into the right-hand side of (32), we obtain$$\multline \text{sneg}(-\Delta+V)=\Bigl\{-\frac{1}{15\pi^2}\int_{\Bbb R^3} \bigl(-V_1(x)\bigr)^{5/2}dx+\frac{1}{6\pi^2}\int_{\Bbb R^3} \bigl(-V(x)\bigr)^{3/2}\bigl(V(x)-V_1(x)\bigr)dx\Bigr\}\\ +\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}\bigl(\Delta V_1(x)\bigr) \cdot\bigl(-V_1(x)\bigr)^{1/2}dx\\ +\pi\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)+\ \text{Error}_3\ , \endmultline\tag"(33)" $$\noindent with$$ |\text{Error}_3|\le CZ^{\frac 85+2\cdot 10^{-9}}\ .\tag"(34)" $$>From (31) with p=\frac 52, we get$$\multline \Bigl\{-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V_1(x)\bigr)^{5/2}dx+\frac{1} {6\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2}\bigl(V(x)-V_1(x)\bigr)dx\Bigr\}\\ =-\frac{1}{15\pi^2}\int_{\Bbb R^3}\Bigl[\bigl(-V(x)\bigr)^{5/2} +\frac 52\bigl(V(x)-V_1(x)\bigr)\cdot\bigl(-V(x)\bigr)^{3/2} +g(x)\Bigr]dx\\ +\frac{1}{6\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2}\bigl(V(x)-V_1 (x)\bigr)dx\endmultline $$\noindent with$$ |g(x)|\le CZ^{7/2}|x|^{1/2}\chi_{{}_{|x|<2Z^{-\frac 35+2\varepsilon}}}\ . $$\noindent That is,$$\multline \Bigl\{-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V_1(x)\bigr)^{5/2} dx+\frac{1}{6\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{3/2} \bigl(V(x)-V_1(x)\bigr)dx\Bigr\}\\ =-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{5/2}dx-\frac{1}{15\pi^2} \int_{\Bbb R^3}g(x)dx\ ,\endmultline $$\noindent with the last term on the right dominated by$$ \int_0^{2Z^{-\frac 35+2\varepsilon}}Z^{\frac 72}r^{1/2} \cdot r^2dr\sim Z^{\frac 72}(2Z^{-\frac 35+2\varepsilon})^{\frac 72}\sim Z^{\frac 75+7\varepsilon}\ . $$\noindent So the expression in curly brackets is (33) is equal to -\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{5/2}dx+O(Z^{\frac 75 +7\varepsilon}). Consequently, (33) and (34) may be rewritten as$$\multline \text{sneg}(-\Delta+V)=-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr) ^{5/2}dx+\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}\bigl(\Delta V_1(x) \bigr)\bigl(-V_1(x)\bigr)^{1/2}dx\\ +\pi\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega}\frac{(2\ell+1)} {n_\ell}\tilde\chi(\phi_\ell)+\ \text{Error}_4\endmultline\tag"(35)" $$\noindent with$$ |\text{Error}_4|\le CZ^{\frac 85+2\cdot10^{-9}}\ . \tag"(36)" $$\noindent We want to replace V_1 by V in the second integral in (35). We have$$\multline \Big|\int_{\Bbb R^3}(\Delta V_1)\bigl(-V_1(x)\bigr)^{1/2}dx-\int_{\Bbb R^3} (\Delta V)\bigl(-V(x)\bigr)^{1/2}dx\Big|\\ \le \int_{\Bbb R^3}|\Delta V-\Delta V_1|\cdot\bigl(-V_1(x)\bigr)^{1/2}dx +\int_{\Bbb R^3}|\Delta V|\cdot|(-V_1)^{1/2}-(-V)^{1/2}|dx\ .\endmultline \tag"(37)" $$\noindent From (3) and the properties of \theta, we have$$ \big|\Delta(V-V_1)\big|=\Big|\bigl(\frac{d^2}{dr^2}+\frac 2r\frac{d}{dr}\bigr) \bigl\{\theta(r)\bigl[E_0-\frac Zr-V(r)\bigr]\bigr\}\Big|\le CZ^{3/2}r^{-3/2}\ ,\ r=|x|\ . $$\noindent Also,$$ \bigl(-V_1(x)\bigr)^{1/2}\sim Z^{1/2}r^{-1/2}\quad\text{in}\quad \text{supp}\,(V-V_1)\subset\bigl\{|x|<2Z^{-\frac 35+2\varepsilon}\bigr\}\ . $$\noindent Hence$$\multline \int_{\Bbb R^3}|\Delta V-\Delta V_1|\cdot\bigl(-V(x)\bigr)^{1/2} dx\le C\int_0^{2Z^{-\frac 35+2\varepsilon}}(Z^{3/2}r^{-3/2})\cdot (Z^{1/2}r^{-1/2})\cdot r^2dr\\ \sim Z^2\cdot 2Z^{-\frac 35+2\varepsilon}\sim Z^{\frac 75+2\varepsilon}\ . \endmultline\tag"(38)" $$\noindent On the other hand, (1) yields |\Delta V|\le CZr^{-3} in \text{supp}\,(V_1-V); and (31) shows that$$\align \big|\bigl(-V(x)\bigr)^{1/2}-\bigl(-V_1(x)\bigr)^{1/2}\big| &\le C|V(x)-V_1(x)|\cdot\bigl(-V(x)\bigr)^{-1/2}+CZ^{\frac 32} r^{\frac 52}\\ &\le C(Z^{3/2}r^{1/2})\cdot(Zr^{-1})^{-1/2}+CZ^{\frac 32} r^{5/2}\\ &=CZr+CZ^{\frac 32} r^{5/2}=CZr(1+Z^{\frac 12}r^{\frac 32})\\ &\le C^\prime Zr\qquad\text{in}\quad \text{supp}\,(V-V_1)\ .\endalign $$\noindent Therefore,$$\multline \int_{\Bbb R^3}|\Delta V(x)|\cdot\big|\bigl(-V(x)\bigr)^{1/2} -\bigl(-V_1(x)\bigr)^{1/2}\big|dx\\ \le C\int_0^{2Z^{-\frac 35+2\varepsilon}}(Zr^{-3})\cdot(Zr)\cdot r^2dr \sim Z^2\cdot (2Z^{-\frac 35+2\varepsilon})\sim Z^{\frac 75+2\varepsilon}\ . \endmultline $$\noindent Putting this and (38) into (37), we see that$$ \Big|\int_{\Bbb R^3}\bigl(+\Delta V_1)\cdot(-V_1)^{1/2}dx-\int_{\Bbb R^3} (+\Delta V)\cdot(-V)^{1/2}dx\Big|\le CZ^{\frac 75+2\varepsilon}\ . $$\noindent Hence, equations (35), (36) may be rewritten in the form$$\multline \text{sneg}(-\Delta+V)=-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{5/2} dx+\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}\bigl(\Delta V(x)\bigr) \cdot\bigl(-V(x)\bigr)^{1/2}dx\\ +\pi\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)+\ \text{Error}_5\ , \endmultline\tag"(39)" $$\noindent with$$ |\text{Error}_5|\le CZ^{\frac 85+2\cdot10^{-9}}\ .\tag"(40)" $$We record equations (11), (12), (13), (39), (40) in the Theorem of the next section. \vfill\eject \head The WKB Eigenvalue Sum Theorem for Approximate TF Potentials \endhead \medskip Suppose V(r) is defined on (0,\infty), and satisfies the following conditions.$$ \Big|\bigl(\frac{d}{dr}\bigr)^\alpha V(r)\Big|\le C_\alpha\min \big\{\frac Zr,r^{-4}\bigr\}\cdot r^{-\alpha}\quad \text{for\ all}\ r\in (0,\infty)\ ,\alpha\ge 0\ .\tag"(1)" \multline \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-V_Z^{TF}(r)\bigr\}\Big| \le c_0\min\bigl\{\frac Zr, r^{-4}\bigr\}\cdot r^{-\alpha}\\ \text{for}\ 0\le \alpha\le 2\ \text{and}\ r\in (0,\infty)\ ,\ \text{with}\ c_0>0\ \text{determined\ by\ the}\ C_\alpha\ \text{in\ (1)}\ .\endmultline\tag"(2)" \multline \Big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{E_0-\frac Zr-V(r)\bigr\}\Big|\le C_\alpha Z^{\frac 32}\, r^{\frac 12-\alpha}\\ \text{for}\ \alpha\ge 0\ \text{and}\ r\in (0,2Z^{-\frac 35+2\varepsilon})\ ,\ \text{with}\ cZ^{4/3}0}\bigl(-r^2V(r)\bigr), and define \align n_\ell&=\int_0^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\Bigr)_+^{-1/2}dr\ ,\\ \phi_\ell&=\frac 1\pi\int_0^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)} {r^2}\Bigr)_+^{1/2}dr-\frac 12\qquad\text{for}\ 1\le \ell<\Omega\ . \endalign Then the sum of the negative eigenvalues ofH$is given by $$\multline \text{sneg}(H)=-\frac{1}{15\pi^2}\int_{\Bbb R^3}(-V)^{5/2} dx+\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}(\Delta V)\cdot (-V)^{1/2}dx\\ +\pi\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)+ \text{Error}\ ,\endmultline$$ \noindent with $$|\text{Error}|\le C^\prime Z^{\frac 85+2\cdot10^{-9}}\ .$$ \noindent The constant$C^\prime$depends only on$C_\alpha$,$c$,$C$,$\varepsilon$in (1), (2), (3). \vglue 1pc \demo{Proof} The conclusion is contained in equations (11), (12), (13), (39), (40) in the preceding section.\quad$\blacksquare$\enddemo \vfill\eject \head Estimates for Number-Theoretic Sums\endhead \medskip Let$V(r)$,$\Omega$,$n_\ell$,$\phi_\ell$be as in the WKB Eigenvalue Sum Theorem for Approximate TF Potentials. Thus, we assume: $$\bigl|\bigl(\frac{d}{dr}\bigr)^\alpha V(r)\big|\le C_\alpha \min\bigl\{\frac Zr, r^{-4}\bigr\}\cdot r^{-\alpha} \quad\text{for}\ \alpha\ge 0\, \ r>0\ ;\tag"(1)"$$ $$\big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-V_Z^{TF}(r)\bigr\}\big| 0\ ,\tag"(2)"$$ \noindent with small$c_0>0$determined by the$C_\alpha$in (1); $$\big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{E_0-\frac Zr-V(r)\bigr\}\big| \le C_\alpha Z^{\frac 32}r^{\frac 12-\alpha}\quad\text{for}\ \alpha\ge 0\ ,\ 00}\bigl(-r^2V(r)\bigr)\ ;\tag"(4)"$$ $$n_\ell=\int_0^\infty\bigl(-V(r)-\frac{\ell(\ell+1)}{r^2}\bigr)_+^{-1/2} \quad\text{for}\ 1\le \ell<\Omega\ ;\tag"(5)"$$ $$\phi_\ell=\frac 1\pi\int_0^\infty\Bigl(-V(r)-\frac{\ell(\ell+1)}{r^2} \Bigr)_+^{1/2}dr-\frac 12\quad\text{for}\ 1\le \ell<\Omega\ .\tag"(6)"$$ To make full use of the WKB Density and Eigenvalue Sum Theorems for Approximate TF Potentials, we want to prove the following estimates for some$a>0$. $$\multline \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}\\ \text{for}\ Z^{10^{-9}}\le\ell_1<\ell_2<\Omega\ \text{with}\ \ell_2-\ell_1>\Omega^{1-10a}\ .\endmultline\tag"(7)"$$ $$\multline \text{There\ are\ at\ most}\ C\Omega^{1-6a}\ \text{integers}\ \ell<\Omega\ \text{for\ which}\\ |\phi_\ell-\ \text{nearest\ integer}\ |\le\ell^{-6/43}\ . \endmultline\tag"(8)"$$ $$\Big|\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)\Big|\le C\Omega^{-2a}Z^{5/3}\ . \tag"(9)"$$ These estimates are clearly related to the equidistribution of the$\phi_\ell$modulo$1$. To prove them, we need an important extra assumption on the potential$V(r)$. With $$\Theta(t)=\int_0^\infty\Bigl(-V(r)-\frac{t^2}{r^2}\Bigr)_{+}^{1/2}dr\quad \text{for}\ 00}\, [-r^2V(r)])^{1/2}\ ,\tag"(10)"$$ \noindent we assume $$\Big|\frac{d^2}{dt^2}\Theta(t)\Big|\ge cZ^{-1/3}\quad\text{for}\ Z^{10^{-9}}0}\, [-r^2V(r)])^{1/2}\ .\tag"(11)"$$ \noindent In [FS6] we proved (11) for$V(r)$close enough to the Thomas-Fermi potential$V_Z^{TF}(r)$in a suitable metric. Here, we simply assume (11). Note that (11) fails for$V(r)=E_0-\frac Zr$and for$V(r)=\lambda^2 (r^2-\mu)$(with, say,$\lambda\sim Z$,$\mu\sim Z^{-2/3}$), i.e.\ for the hydrogen atom and the harmonic oscillator. In these examples,$\frac{d^2}{dt^2}\Theta(t)=0$, the$\phi_\ell$are not equidistributed modulo$1$, the eigenvalues of$-\Delta+V$are highly degenerate, and the desired semiclassical approximations to the density and eigenvalue sum are inaccurate. The failure of (11) is closely related to the periodicity of zero-energy orbits of a classical particle in the potential$V$. All the number theory we need is contained in the following standard, elementary result. (Something more sophisticated would be needed to get the sharpest$a>0$in (7), (8), (9).) \proclaim{Lemma 1} Suppose$f(t)$,$\phi(t)$are defined on an interval$[0,S]$and satisfy there the estimates $$|f(t)|\le C\ ,\ |f^\prime(t)|\le CR^{-1}\ ,\ bR^{-1}\le \phi^{\prime\prime} (t)\le CR^{-1}\ , \text{with}\ R\ge S\ge 10\ .\tag"(12)"$$ \noindent Let$\chi(t)$be a periodic function of period$1$, having average zero and total variation on$[0,1]$given by $${\roman{Var}}_{[0,1]}\chi\le C\ .\tag"(13)"$$ \noindent Then we have the estimate $$\Big|\sum\limits_{0\le k\le S}f(k)\chi(\phi(k))\Big|\le \frac{C^\prime}{b}R^{2/3}\ell nR\ ,$$ \noindent with$C^\prime$depending only on$C$in (12), (13).\endproclaim \vglue 1pc \demo{Proof} Define a finite sequence$k_0, k_1,\ldots, k_{\nu_{\max}}$inductively, as follows. We put$k_0=0$, and pick an integer$N$comparable to$R^{1/3}$. Suppose we have already defined$k_0, k_1, \ldots, k_\nu$. By the pigeon-hole principle, we can find integers$p_\nu$,$q_\nu$with $$1\le q_\nu\le N\quad\text{and}\quad |\phi^\prime(k_\nu)-\frac{p_\nu}{q_\nu}| \le \frac{C}{q_\nu N}\ .\tag"(14)"$$ \noindent After dividing$p_\nu$,$q_\nu$by their greatest common divisor, we obtain (14) for$p_\nu$and$q_\nu$relatively prime. If$k_\nu+q_\nu \le S$, then define$k_{\nu+1}=k_\nu+q_\nu$. \smallskip \noindent If instead$k_\nu+q_\nu>S$, then define$\nu_{\max}=\nu$, so that the sequence$k_0,k_1,\ldots$terminates at the given$k_\nu$. \smallskip \noindent Note that the sequence must terminate eventually, since$k_{\nu+1}\ge k_\nu+1$for$0\le \nu<\nu_{\max}$. Using the integers$k_0\ldots k_{\nu_{\max}}$, we break up$\sum\limits_{0\le k\le S}f(k)\chi(\phi(k))$into a sum of $$X_\nu=\cases \sum\limits_{k_\nu\le kS. Hence the sum defining X_{\nu_{\max}} has at most N terms. The terms are bounded, so$$ |X_{\nu_{\max}}|\le CN\ .\tag"(16)" $$\noindent For the other X_\nu we will derive a much better estimate. Fix \nu<\nu_{\max}, and set \xi=\phi(k_\nu). Then for k=k_\nu+m with 0\le m0, the set F(p,q) must be a closed interval, say F(p,q)= [t_{\text{low}}, t_{\text{hi}}]. Since \phi^{\prime\prime} \ge bR^{-1} and t_{\text{low}}, t_{\text{hi}}\in F(p,q), we have \frac{2C}{qN}\ge \phi^\prime(t_{\text{hi}})-\phi^\prime(t_{\text{low}}) \ge bR^{-1}\cdot (t_{\text{hi}}-t_{\text{low}}), i.e.\ \text{length}\,(F(p,q))=t_{\text{hi}}-t_{\text{low}}\le \frac{2CR}{qNb}. On the other hand, for \nu\in \Cal E(p,q) the interval I_\nu has length k_{\nu+1}-k_\nu=q_\nu=q. So the I_\nu for \nu\in\Cal E(p,q) are pairwise disjoint intervals of length q, all contained in an interval F(p,q) of length at most \frac{2CR}{qNb}. Consequently, there can be at most \frac{2CR}{q^2Nb} distinct \nu in \Cal E(p,q). For a given q, (1\le q\le N), how many p can have \Cal E(p,q) non-empty? Since 0<\phi^{\prime\prime}\le CR^{-1} and [0,S] has length S\le R, it follows that \phi^\prime(t) varies by at most C, as t varies over [0,S]. If \nu\in\Cal E(p,q) and \tilde\nu\in\Cal E(\tilde p,q), then |\phi^\prime(k_\nu) -\frac pq\big|=\big|\phi^\prime(k_\nu)-\frac{p_\nu}{q_\nu}\big|\le \frac{C}{q_\nu N}=\frac{C}{qN}, and similarly |\phi^\prime (k_{\tilde \nu})-\frac{\tilde p}{q}\big|\le \frac{C}{qN}. We know that |\phi^\prime(k_\nu)-\phi^\prime (k_{\tilde\nu})|\le C, so we must have |p-\tilde p|\le Cq. So for a given q we find that \Cal E(p,q) can be non-empty for at most Cq distinct p. Combining this with our bound for the number of \nu in a given \Cal E(p,q), we see that \operatornamewithlimits{\bigcup}\limits_{p}\Cal E(p,q) contains at most \frac{C^\prime R}{qNb} distinct \nu. Since \{0,q,\ldots,\nu_{\max}-1\}=\operatornamewithlimits{\bigcup}\limits_{ 1\le q\le N}\bigl(\operatornamewithlimits{\bigcup}\limits_p\Cal E(p,q)\bigr), it follows that \nu_{\max}\le \sum\limits_{1\le q\le N}\frac{C^\prime R} {qNb}\le \frac{C^{\prime\prime}R\ell nN}{Nb}. Hence, (26) implies the estimate$$ \Big|\sum\limits_{0\le k\le S} f(k)\chi(\phi(k))\Big|\le CN+\frac{CR\ell nN} {Nb}\ . $$\noindent Since N\sim R^{1/3} and b\le C, the right-hand side is comparable to \frac{CR^{2/3}\ell nR}{b}, which proves the conclusion of the Lemma.\qquad\blacksquare\enddemo \vglue 1pc \demo{Trivial Remarks} In place of bR^{-1}\le \phi^{\prime\prime}(t) \le CR^{-1} in the above Lemma, we could just as easily assume bR^{-1}\le -\phi^{\prime\prime}(t)\le CR^{-1} for t\in [0,S]. Also, we can replace [0,S] by any other interval of length \le R.\enddemo \medskip To prove (7), (8), (9) we will use Lemma 1 with$$ \phi(t)=\frac 1\pi\int_0^\infty\Bigl(-V(r)-\frac{t(t+1)}{r^2}\Bigr)_+^{1/2} dr-\frac 12\ ,\ 00$, we have $$\big|\bigl(\frac{d}{dr}\bigr)\bigl(-r^2V(r)\bigr)\big|>c_2\Cal S(r)r^{-1} \text{for}\ |r-\check r|>c_1\check r\ ,\tag"(31)"$$ \noindent with $c_2>0$ depending on $c_1$. We introduce a partition of unity $\operatornamewithlimits{\sum}\limits_\nu\theta_\nu(r)=1$ on $(0,\infty)$, with $\theta_\nu$ supported in $\{|r-r_\nu|1+c^\prime$, $\big|\bigl(\frac{d}{dr}\bigr)^\alpha \theta_\nu\big|\le C_\alpha r_\nu^{-\alpha}$; and then write $$\Phi^\prime(\xi)=-\sum\limits_\nu\int_0^\infty\bigl(-r^2V(r)- \xi\bigr)_+^{-1/2}\theta_\nu(r)\frac{dr}{2\pi r}\equiv -\sum\limits_\nu Q_\nu(\xi)\ .\tag"(32)"$$ \noindent Lemma 9 from the section on the Eigenvalue Sum in an Approximate TF Potential with an Exact Coulomb singularity applies here, so that we get the following estimates for $Q_\nu(\xi)$. $$\multline \Big|\bigl(\frac{d}{d\xi}\bigr)^mQ_\nu(\xi)\Big|\le C_m\Cal S^{-\frac 12-m} (r_\nu)\ \text{for}\ 0<\xic\check r \ .\endmultline\tag"(33)"$$ $$\multline \Big|\bigl(\frac{d}{d\xi}\bigr)^mQ_\nu(\xi)\Big|\le C_m\Cal S^{-\frac 12-m} (r_\nu)\ \text{for}\ 0<\xi<\operatornamewithlimits{\max}\limits_{r>0}\, [-r^2V(r)]\ ,\\ \text{provided}\ \text{dist}(\check r,\ \text{supp}\ \theta_\nu)\le c\check r\ .\endmultline\tag"(34)"$$ \noindent Immediately from the definition of $Q_\nu$ we have $$Q_\nu(\xi)=0\ \text{for}\ \xi>C\Cal S(r_\nu)\ .\tag"(35)"$$ \noindent Combining (33), (34), (35), we find that $$\Big|\bigl(\frac{d}{d\xi}\bigr)^mQ_\nu(\xi)\Big|\le C_m\Cal S^{-\frac 12-m} (r_\nu)\quad \text{for}\ 0<\xi<\operatornamewithlimits{\max}_{r>0}\,[-r^2V(r)]\ . \tag"(36)"$$ \noindent Putting (35) and (36) into (32), we obtain for $0<\xi<\operatornamewithlimits{\max}\limits_{r>0}\,[-r^2V(r)]$ that $$\multline \Big|\bigl(\frac{d}{d\xi}\bigr)^m\Phi^\prime(\xi)\Big|\le C_m\sum\limits_{C\Cal S(r_\nu)>\xi}\Cal S^{-\frac 12-m}(r_\nu)\\ \sim \sum\Sb CZr_\nu>\xi\\ r_\nu\le Z^{-1/3}\endSb (Zr_\nu)^{-\frac 12-m} +\sum\Sb Cr_\nu^{-2}>\xi\\ r_\nu>Z^{-1/3}\endSb (r_\nu^{-2})^{-\frac 12-m}\ , \endmultline$$ \noindent i.e.\ $$\big|\bigl(\frac{d}{d\xi}\bigr)^m\Phi^\prime(\xi)\big|\le C_m\xi^{-\frac 12-m}\ \text{for}\ 0<\xi<\operatornamewithlimits{\max}_{r>0}[-r^2V(r)]\ ,\ m\ge 0\ .\tag"(37)"$$ \noindent Upper and lower bounds for $\phi^{\prime\prime}(t)$ come from (37), (11) and the close relation among $\phi(t)$, $\Phi(\xi)$, $\Theta(t)$. In fact, $\phi(t)=\Phi(t(t+1))$, so $$\phi^{\prime\prime}(t)=(2t+1)^2\Phi^{\prime\prime}(t(t+1))+2\Phi^\prime (t(t+1))\ .$$ \noindent Hence, (37) yields $$|\phi^{\prime\prime}(t)|\le C(2t+1)^2[t(t+1)]^{-3/2}+C[t(t+1)]^{-1/2}\ \text{for}\ 00}\bigl(-r^2V(r)\bigr)]^{1/2} we get from (37) that$$ |\Theta^\prime(t)|=|2\pi t\Phi^\prime(t^2)|\le (2\pi t) \cdot C[t^2]^{-1/2}\le C^\prime\ .\tag"(39)" $$\noindent Also, \phi(t)=\frac 1\pi\Theta(\sqrt{t(t+1)})-\frac 12=\frac 1\pi \Theta(H(t))-\frac 12 with H(t)=\sqrt{t(t+1)}, so$$ \phi^{\prime\prime}(t)=\frac 1\pi(H^\prime(t))^2\Theta^{\prime\prime} (H(t))+\frac 1\pi H^{\prime\prime}\Theta^\prime(H(t))\ \text{for}\ 0100$we have$H(t)=\sqrt{t(t+1)}=t+\frac 12 +\operatornamewithlimits{\sum}\limits_{k\ge 1}c_kt^{-k}$(convergent power series), so$H^\prime(t)\sim 1$and$|H^{\prime\prime}(t)|\le Ct^{-3}$. Hence (40) implies $$|\phi^{\prime\prime}(t)|\ge c|\Theta^{\prime\prime}(H(t))|-Ct^{-3}|\Theta^\prime (H(t))|\quad\text{for}\ 100c\check r \endmultline\tag"(46)"$$ $$\multline \big|\bigl(\frac{d}{d\xi}\bigr)^mP_\nu(\xi)\big|\le C_m\Cal S^{-\frac 12-m} (r_\nu)\cdot r_\nu^2\ \text{for}\ 0<\xi<\operatornamewithlimits{\max}_{r>0} [-r^2V(r)]\\ \text{provided}\ \text{dist}(\check r,\ \text{supp}\ \theta_\nu)\le c\check r\ .\endmultline\tag"(47)"$$ \noindent Immediately from the definition we get $$P_\nu(\xi)=0\quad\text{for}\quad \xi>C\Cal S(r_\nu)\ .\tag"(48)"$$ \noindent Combining (46)$\ldots$(48), we get $$\big|\bigl(\frac{d}{d\xi}\bigr)^mP_\nu(\xi)\big|\le C_m\Cal S ^{-\frac 12-m}(r_\nu)\cdot r_\nu^2\chi_{{}_{C\Cal S(r_\nu)>\xi}}\ \text{for}\ 0<\xi<\operatornamewithlimits{\max}\limits_{r>0} \, [-r^2V(r)]\ .\tag"(49)"$$ \noindent Putting (49) into (45) yields $$\multline \big|\bigl(\frac{d}{d\xi}\bigr)^mn_1(\xi)\big|\le C_m \sum\limits_{C\Cal S(r_\nu)>\xi}r_\nu^2\Cal S^{-\frac 12-m} (r_\nu)\\ \sim \sum\Sb CZr_\nu>\xi\\ r_\nu\xi\\ r_\nu\ge Z^{-1/3}\endSb r_\nu^2(r_\nu^{-2})^{-\frac 12-m}\ . \endmultline\tag"(50)"$$ \noindent The first sum on the right has the order of magnitude $$(Z^{-\frac 13})^2\cdot (Z\cdot Z^{-\frac 13})^{-\frac 12-m} =Z^{-\frac 23}\cdot Z^{-\frac 13-\frac 23 m}=Z^{-1-\frac 23 m}$$ \noindent for$m=0,1$. \noindent The second sum on the right has the order of magnitude$(\xi^{-1/2})^2\cdot (\xi)^{-\frac 12-m}=\xi^{-\frac 32-m}\ $. The second term$\xi^{-\frac 32-m}$dominates the first term$Z^{-1-\frac 23m}$since$0<\xi<\operatornamewithlimits{\max} \limits_{r>0}\, [-r^2V(r)]\sim Z^{2/3}$. Therefore (50) implies $$n_1(\xi)\le C\xi^{-3/2}\ \text{for}\ 0<\xi<\operatornamewithlimits{\max} _{r>0}\, [-r^2V(r)]\ ,\tag"(51)"$$ \noindent and $$\big|\bigl(\frac{d}{d\xi}\bigr)n_1(\xi)\big|\le C\xi^{-5/2} \ \text{for}\ 0<\xi<\operatornamewithlimits{\max}_{r>0}\, [-r^2V(r)]\ .\tag"(52)"$$ Next, set$n_2(t)=\int_0^\infty\bigl(-V(r)-\frac{t(t+1)}{r^2}\bigr)_+^{-1/2} dr=n_1(t(t+1))$for$0(1+c^\prime)x_{\text{left}} (\ell)$. Therefore, $$\multline n_2(\ell)\ge \int_{x_{\text{left}}(\ell)}^{x_{\text{rt}}(\ell)} c\cdot\Bigl[\min\bigl\{\frac Zr, r^{-4}\bigr\}\Bigr]^{-1/2} dr\ge c\int_{(1+c^\prime)^{-1}x_{\text{rt}}(\ell)}^{x_{\text{rt}}(\ell)} r^2dr\\ \ge c^{\prime\prime}(x_{\text{rt}}(\ell))^3\ge c^{\prime\prime\prime} \ell^{-3}\ .\endmultline\tag"(55)"$$ On the other hand, if$(1-\overline c)\Omega\le \ell<\Omega$, then we know from (3)$\ldots$(15) that$V_\ell(r)=\frac{\ell(\ell+1)}{r^2}+V(r)$has a minimum at$r=x_0(\ell)\sim Z^{-1/3}, and that \align -&V_\ell(x_0(\ell))\sim\frac{\Omega(\Omega-\ell)}{[x_0(\ell)]^2} \sim Z(\Omega-\ell)\\ &V_\ell^\prime(x_0(\ell))=0\\ &V_\ell^{\prime\prime}\sim\frac{Z}{x_0(\ell)}\cdot [x_0(\ell)]^{-2} \sim Z^2\quad\text{for}\ |x-x_0(\ell)|CZ^{1/9} and L_1>cL_2. Then the phase function \phi(t) in (26 bis) satisfies cZ^{-1/3}\le |\phi^{\prime\prime}(t)|\le C(L_2)^{-1}\quad\text{in}\ [L_1, L_2]\ , $$\noindent by virtue of (38) and (42). Thus, \phi(t) on [L_1, L_2] satisfies the hypotheses of Lemma 1, with R=L_2, b=L_2Z^{-1/3}, S=L_2-L_1\le R. We take \chi(t) in Lemma 1 to be \chm(t), and note that it has period 1, average zero, and bounded variation on [0,1]. Thus, \chi satisfies the hypotheses of Lemma 1. For f(t) we take the function (43). Estimates (57), (58) show that (L_2)^{-4}f(t) satisfies the hypotheses of Lemma 1 on [L_1, L_2]. Therefore, Lemma 1 applies, and it tells us that$$\align \Big|&\sum\limits_{L_1\le\ell\le L_2}\frac{f(\ell)}{(L_2)^4}\chm (\phi(\ell))\Big|\le \frac CbR^{2/3}\ell nR\ ,\ \ \text{i.e.}\\ &\qquad \Big|\sum\limits_{L_1\le\ell\le L_2}\frac{(2\ell+1)}{n_\ell} \chm(\phi_\ell)\Big|\le \frac{C}{L_2Z^{-1/3}}(L_2^{2/3}\ell n L_2)\cdot L_2^4\ ,\ \ \text{i.e.}\\ &\qquad\qquad \Big|\sum\limits_{L_1\le\ell\le L_2}\frac{(2\ell+1)}{n_\ell} \chm(\phi_\ell)\Big|\le CZ^{1/3}L_2^{\frac{11}{3}}\ell nZ\\ &\qquad\qquad\qquad\qquad \text{for}\quad CZ^{1/9}\le L_1cL_2\ . \tag"(60)"\endalign $$\noindent Similarly, taking \tilde\chi in place of \chm above, we get$$\multline \Big|\sum\limits_{L_1\le\ell\le L_2}\frac{(2\ell+1)}{n_\ell} \tilde\chi(\phi_\ell)\Big|\le CZ^{1/3}L_2^{11/3}\ell nZ\\ \text{for}\quad CZ^{1/9}\le L_1cL_2\ . \endmultline\tag"(61)" $$\noindent Estimate (61) easily implies (9). In fact, we divide the integers \ell between Z^{\frac{8}{25}+10\varepsilon} and \Omega into disjoint intervals \{L_1^\nu\le \ell\le L_2^\nu\} with L_2^\nu/L_1^\nu between 2 and 3. \smallskip \noindent Applying (61) to each of these intervals and summing on \nu, we obtain$$ \Big|\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)\Big|\le CZ^{1/3} \Omega^{\frac{11}{3}}\ell nZ\le C^\prime Z^{\frac{14}{9}}\ell nZ\ . \tag"(62)" $$Thus, (9) holds provided Z^{\frac{14}{9}}\ell nZ\Omega^{1-10a}\ .\tag"(63)"$$ \noindent Then\ell_2>\Omega^{1-10a}\sim (Z^{1/3})^{4/5}>>Z^{1/9}$. Assume for a moment that$\ell_1\ge c\ell_2$. Then (60) applies, with$\ell_1$,$\ell_2$in place of$L_1$,$L_2$. Hence, $$\Big|\sum\limits_{\ell_1\le\ell\le \ell_2}\frac{(2\ell+1)}{n_\ell} \chm(\phi_\ell)\Big|\le CZ^{1/3}\ell_2^{\frac{11}{3}}\ell nZ\ .$$ \noindent On the other hand, (59) and$\ell_1\ge c\ell_2$show that $$\Omega^{-2a}\sum\limits_{\ell_1\le\ell\le \ell_2}\frac{(2\ell+1)}{n_\ell} \sim\Omega^{-2a}\cdot \ell_2^4\cdot (\ell_2-\ell_1)\ .$$ \noindent Consequently, (7) holds, provided$Z^{\frac 13}\ell_2^{\frac{11}{3}} \ell nZ\le C\Omega^{-2a}\ell_2^4(\ell_2-\ell_1)$, i.e.\$Z^{1/3}\ell nZ\le C\Omega^{-2a}\ell_2^{1/3}(\ell_2-\ell_1)$. This in turn follows from$Z^{1/3}\ell nZ\le C\Omega^{-2a}(\ell_2-\ell_1)^{4/3}$, which is a consequence of$Z^{\frac 13}\ell nZ\le C\Omega^{-2a} (\Omega^{1-10a})^{\frac 43}$, i.e.\$Z^{\frac 13}\ell nZ\le C \Omega^{\frac 43-\frac{46}{3}a}$, which is true for$a=\frac{1}{50}$. Therefore, (7) holds with$a=\frac{1}{50}$, provided$\ell_1\ge c\ell_2$. On the other hand, assume (63) with$\ell_1CZ^{1/9}$, then (60) yields $$\Big|\sum\limits_{\ell_1^\nu\le\ell\le \ell_2^\nu}\frac{(2\ell+1)} {n_\ell}\chm(\phi_\ell)\Big|\le CZ^{\frac 13}(\ell_2^\nu)^{\frac{11}{3}} \ell nZ\ .\tag"(64)"$$ \noindent If instead$\ell_1^\nu\le CZ^{1/9}$, then we use (59) to make the trivial estimate $$\Big|\sum\limits_{\ell_1^\nu\le\ell\le \ell_2^\nu}\frac{(2\ell+1)}{n_\ell} \chm(\phi_\ell)\Big|\le C\sum\limits_{\ell_1^\nu\le\ell\le \ell_2^\nu} \frac{(2\ell+1)}{n_\ell}\le C^\prime (\ell_2^\nu)^5\ .\tag"(65)"$$ \noindent Summing (64), (65) over$\nu$, we get $$\Big|\sum\limits_{\ell_1\le\ell\le\ell_2}\frac{(2\ell+1)}{n_\ell} \chm(\phi_\ell)\Big|\le CZ^{\frac 13}(\ell_2)^{\frac{11}{3}}\ell nZ+CZ^{5/9} \ .\tag"(66)"$$ \noindent On the other hand, (59) and$\ell_1\le c\ell_2$imply $$\sum\limits_{\ell_1\le\ell\le \ell_2}\frac{(2\ell+1)}{n_\ell} \sim\sum\limits_{\ell_1\le\ell\le \ell_2}\ell^4\sim \ell_2^5\ .$$ \noindent Hence, (7) holds, provided $$Z^{\frac 13}\ell_2^{\frac{11}{3}}\ell nZ+Z^{\frac 5 9}\le C\Omega^{-2a} \ell_2^5\ ,\quad\text{i.e.}\tag"(67)"$$ $$Z^{\frac 13}\ell nZ\le C\Omega^{-2a}\ell_2^{\frac 43}\quad\text{and} \quad Z^{\frac 59}\le C\Omega^{-2a}\ell_2^5\ .\tag"(68)"$$ Since$\ell_2>\ell_2-\ell_1>\Omega^{1-10a}$and$a=\frac 1{50}$, we have $$\Omega^{-2a}\ell_2^{4/3}>\Omega^{-2a}\Omega^{\frac 43-\frac{40a}{3}} =\Omega^{\frac 43-\frac{46}{3}a}>\Omega\,\ell n\,\Omega>cZ^{1/3}\ell nZ$$ \noindent and $$\Omega^{-2a}\ell_2^5>\Omega^{-2a}\Omega^{5-50a}=\Omega^{4-2a}>Z^{\frac 59} \ .$$ \noindent Thus, (68) holds, and therefore (7) is valid also for$\ell_1\le c\ell_2$. Hence, (7) holds with$a=\frac{1}{50}$in all cases. Next we prove (8). For$\Omega\ge 2L>CZ^{1/9}$, and$\phi(t)$given by (26 bis) we know that $$cZ^{-1/3}\le |\phi^{\prime\prime}(t)|\le\frac CL\quad\text{for}\ t\in [L,2L] ,\quad\text{by\ virtue\ of\ (38),\ (42)}\ .$$ \noindent Thus,$\phi(t)$satisfies the hypotheses of Lemma 1 with$R=S=L$,$b=Z^{-1/3}L$. For$\chi(t)$we take the function $$\chi(t)=\cases 1-\tilde cL^{-6/43} &\text{if |t-\ nearest\ integer|\ \le 10L^{-6/43}}\\ -\tilde cL^{-6/43} &\text{otherwise}\ .\endcases$$ \noindent This function has period$1$and bounded variation on$[0,1]$; and for a suitable$\tilde c\sim 1$,$\chi(t)$has average zero. Therefore,$\chi(t)$satisfies the hypotheses of Lemma 1. For$f(t)$in Lemma 1, we just take$f(t)\equiv 1$. Thus, all the hypotheses of Lemma 1 are satisfied. Applying the Lemma, we learn that $$\multline \Big|\bigl[\text{Number\ of}\ \ell\in [L,2L]\ \text{with}\ |\phi_\ell-\ \text{nearest\ integer}| \le 10L^{-6/43}\bigr]\\ -\tilde cL^{-6/43}\cdot L\Big|\le \frac{CL^{2/3}\ell nL}{Z^{-\frac 13}L} \le \frac{C^\prime Z^{1/3}\ell nZ}{L^{1/3}}\ .\endmultline$$ \noindent Therefore, the number of$\ell\in [L,2L]$with$|\phi_\ell- \text{nearest\ integer}|\le \ell^{-6/43}$is at most$\tilde cL^{\frac{37}{43}}+\frac {C^\prime Z^{1/3}\ell nZ}{L^{1/3}}$. We know this for$CZ^{1/9}\le 2L<\Omega$. \noindent Let us apply the above estimate for$L=L_m\equiv 2^{-m}\cdot (\Omega/2)$and$m=0,1,\ldots,m_{\max}$with$m_{\max}$taken so that$L_{m_{\max}}\sim Z^{\frac 14}$. Summing on$m$, we see that $$\multline (\text{Number\ of}\ \ell\in [Z^{\frac 14},\Omega)\ \text{with}\ |\phi_\ell-\ \text{nearest\ integer}|\le \ell^{-6/43})\\ \le \sum\limits_{m=0}^{m_{\max}}\bigl(\tilde cL_m^{\frac{37}{43}} +C^\prime Z^{1/3}\frac{\ell nZ}{L_m^{1/3}}\bigr)+C\sim \Omega^{\frac{37}{43}} +\frac{Z^{1/3}\ell nZ}{Z^{1/12}}\sim \Omega^{\frac{37}{43}}\ .\endmultline$$ \noindent Combining this with the trivial estimate $$(\text{Number\ of}\ \ell\le Z^{1/4}\ \text{with}\ |\phi_\ell- \text{nearest\ integer}|\ \le \ell^{-6/43}) \le Z^{1/4}<<\Omega^{\frac{37}{43}}\ ,$$ \noindent we obtain $$\multline (\text{Number\ of}\ \ell<\Omega\ \text{with}\ |\phi_\ell-\ \text{nearest\ integer}|\le\ell^{-6/43})< C\Omega^{\frac{37}{43}}\\ \le C\Omega^{1-6a}\quad\text{for}\ a\le \frac{1}{43}\ .\endmultline$$ \noindent Thus, (8) holds, provided$a\le \frac{1}{43}$. The following result summarizes our knowledge of (7), (8), (9). \vglue 1pc \proclaim{Lemma 2} Assume$V(r)$satisfies (1), (2), (3), (11). Set$a=1/50$. \roster \item"(A)" Given$\ell_1$,$\ell_2$integers, with$Z^{10^{-9}} \le \ell_1< \ell_2<\Omega$and$\ell_2-\ell_1>\Omega^{1-10a}$, we have $$\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}\ .$$ \item"(B)" There are at most$C\Omega^{1-6a}$integers$\ell<\Omega$for which$|\phi_\ell-\ \text{nearest\ integer}|\le \ell^{-6/43}$. \item"(C)" $$\Big|\sum\limits_{Z^{\frac{8}{25}+10\varepsilon}<\ell<\Omega} \frac{(2\ell+1)}{n_\ell}\tilde\chi(\phi_\ell)\Big|\le C\Omega^{-2a} Z^{5/3}\ .$$ \endroster\endproclaim \medskip This lets us make use of our previous results on the density and eigenvalue sum for$-\Delta+V$. We spell out the conclusions in the next section. \vfill\eject \head The Main Theorems for Approximate TF Potentials\endhead \medskip Recall that$V_Z^{TF}(x)$denotes the screened Thomas-Fermi potential on$\Bbb R^3$. Thus,$-\Delta V_Z^{TF}= (\text{const})(-V_Z^{TF})^{3/2}$on$\Bbb R^3\backslash \{0\}$, and$V_Z^{TF}(x)=-\frac{Z}{|x|}+O(Z^{4/3})$as$x\to 0$. Write$V_Z^{TF}(r)$for the corresponding function on$(0,\infty)$. Suppose$V(r)$is a real-valued function on$(0,\infty)$. Assume the following estimates. $$\big|\bigl(\frac{d}{dr}\bigr)^\alpha V(r)\big|\le C_\alpha\min \bigl\{\frac Zr, r^{-4}\bigr\}\cdot r^{-\alpha}\quad\text{for}\ r>0\ , \alpha\ge 0\ .\tag"(1)"$$ $$\multline \big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{V(r)-V_Z^{TF}(r)\bigr\}\big| \le c_0\min\bigl\{\frac Zr, r^{-4}\bigr\}\cdot r^{-\alpha}\ \quad\text{for}\ r>0\ ,\ 0\le \alpha\le 2\\ \text{with}\ c_0>0\ \text{determined\ by\ the}\ C_\alpha\ \text{in\ (1)}\ . \endmultline\tag"(2)"$$ $$\big|\bigl(\frac{d}{dr}\bigr)^\alpha\bigl\{E_0-\frac Zr-V(r)\bigr\}\big| \le C_\alpha Z^{3/2}r^{\frac 12-\alpha}\quad\text{for}\ \ \alpha\ge 0\ ,\ 00}\bigl(-r^2V(r)\bigr) \bigr]^{1/2}\ .\tag"(4)"$$ \noindent Let$H=-\Delta+V(|x|)$on$\Bbb R^3$, and let$E_1\ldots E_N$,$\varphi_1(x)\ldots \varphi_N(x)$be the non-positive eigenvalues of$H$and their corresponding normalized eigenfunctions. Define$\text{sneg}(H)= E_1+\ldots+E_N, \align \rho(x)&=\sum\limits_{k=1}^N|\varphi_k(x)|^2\quad\text{for}\ \ x\in \Bbb R^3\ ,\\ \rho_{sc}(x)&=\frac{1}{6\pi^2}\bigl(-V(x)\bigr)^{3/2}\quad\text{for}\ \ x\in \Bbb R^3\ .\endalign \noindent Then we have the following result. \vglue 1pc \proclaim{Main Theorem} If (1)\ldots$(4) hold, then $$\int_{\Bbb R^3\times\Bbb R^3}[\rho(x)-\rho_{sc}(x)]\cdot [\rho(y)-\rho_{sc}(y)]\frac{dxdy}{|x-y|}\le C^\prime Z^{\frac 53- \frac{1}{75}}$$ \noindent and $${\roman{sneg}}(H)=-\frac{1}{15\pi^2}\int_{\Bbb R^3}\bigl(-V(x)\bigr)^{5/2} dx+\frac{Z^2}{8}+\frac{1}{48\pi^2}\int_{\Bbb R^3}(\Delta V(x))\cdot \bigl(-V(x)\bigr)^{1/2}dx+\ {\roman{Error}}\ ,$$ \noindent with$|{\roman{Error}}|\le C^\prime Z^{\frac 53-\frac{1}{75}}$. \endproclaim \vglue 1pc \demo{Proof} The first conclusion follows from Lemma 2 in the previous section, and from Theorem 2 in the section on the Density for an Approximate TF Potential. The second conclusion follows from Lemma 2 in the previous section, and from the WKB Eigenvalue Sum Theorem for Approximate TF Potentials.$\quad\blacksquare$\enddemo \vglue 1pc \demo{Remark} By using [CFS] in place of Lemma 2, one obtains a sharper bound for the error in the formula for$\text{sneg}(-\Delta+V_{TF}^Z)\$.\enddemo \vfill\eject \head References\endhead \medskip \widestnumber\key{FS7\ \ } \ref\key B \by V.\ Bach \paper Error Bound for the Hartree-Fock Energy of Atoms and Molecules, \jour Comm. Math. Phys. \endref \ref\key CFS \by A.\ Cordoba, C. Fefferman, and L.\ Seco, \toappear \endref \ref\key FS1 \by C. Fefferman and L. Seco \paper The Ground-State Energy of a Large Atom \jour Bull. A.M.S. \vol 23 \moreref no 2 \yr 1990 \pages 525--530 \endref \ref\key FS2 \bysame %C. Fefferman and L. Seco \paper Eigenvalues and Eigenfunctions of Ordinary Differential Operators \jour Advances in Math \vol 95 \yr 1992 \pages 145--305 \endref \ref\key FS3 \bysame %C. Fefferman and L. 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