\magnification=\magstep1 \input amstex \documentstyle{amsppt} \baselineskip=20pt \parskip=6pt \hsize=6.5truein \vsize=9truein \NoBlackBoxes \define\bb{\bold{B}} \define\pp{\Bbb{P}} \define\hh{\Bbb{H}} \centerline{\bf On Moments of Negative Eigenvalues for the Pauli Operator} \bigskip \centerline{Zhongwei Shen*\footnote{ Research supported in part by the AMS Centennial Research Fellowship, the MSRI at Berkeley, California, and the NSF grant DMS-9596266.}} \centerline{Department of Mathematics} \centerline{University of Kentucky} \centerline{Lexington, KY 40506} \centerline{E-mail: shenz\@ms.uky.edu} \centerline{Tel: (606) 257-3231} \centerline{Fax: (606) 257-4078} \bigskip \bigskip \centerline{\bf Dedicated to the Memory of Professor Ruilin Long} \bigskip \bigskip \noindent{\bf Abstract.}\ \ This paper concerns the three-dimensional Pauli operator $$\Bbb{P}=\big( \bold{\sigma} \cdot (\bold{p}-\bold{A}(x))\big)^2 +V(x)$$ with a non-homogeneous magnetic field $\bold{B}=\text{curl\,}\bold{A}$. The following Lieb-Thirring type inequality for the moment of negative eigenvalues is established: $$\sum_{\lambda_j<0} |\lambda_j| \le C_1 \int_{\Bbb{R}^3}|V(x)|_-^{5/2}dx + C_2\int_{\Bbb{R}^3} [b_p(x)]^{3/2} |V(x)|_-dx$$ where $p>3/2$ and $b_p(x)$ is the $L^p$ average of $|\bb|$ over certain cube centered at $x$ with a side length scaling like $|\bb|^{-1/2}$. We also show that, if $\bb$ has a constant direction, $$\sum_{\lambda_j<0} |\lambda_j|^\gamma \le C_{1,\gamma} \int_{\Bbb{R}^3}|V(x)|_-^{\gamma+3/2}dx + C_{2,\gamma} \int_{\Bbb{R}^3} b_p(x) |V(x)|_-^{\gamma +1/2}dx$$ where $\gamma>1/2$ and $p>1$. \vfill\eject \centerline{\bf 1. Introduction} Consider the Pauli operator $$\Bbb{P}=\Bbb{P}_0 +V(x) =\big(\bold{\sigma}\cdot(\bold{p}-\bold{A}(x))\big)^2 +V(x), \tag 1.1$$ acting on $L^2(\Bbb{R}^3,\Bbb{C}^2)$. Here $\bold{\sigma}=(\sigma_1,\sigma_2,\sigma_3)$ denotes the vector of Pauli matrices, $\bold{p}=-i\nabla$, and $\bold{A}(x)=(A_1(x),A_2(x),A_3(x))\in L^2_{loc}(\Bbb{R}^3,\Bbb{R}^3)$ is a vector potential. We shall use $\bold{B}=\text{curl\,}\bold{A}$ to denote the magnetic field generated by the potential $\bold{A}$. The goal of this paper is to establish the Lieb-Thirring type estimates for the moments of negative eigenvalues of $\Bbb{P}$: $$\Cal{M}_\gamma=\sum_{\lambda_j<0}|\lambda_j|^\gamma,\ \ \ \ \ \ \gamma >0. \tag 1.2$$ We remark that in quantum mechanics $\Bbb{P}$ is used to describe the motion of a charged spin-1/2 particle in an electromagnetic field. Lieb-Thirring type estimates are an important tool in the study of stability of matter and in semi-classical analysis \cite{13,10,11,12,2,4,5}. In the case of Schr\"odinger operator $-\Delta +V(x)$ in $\Bbb{R}^n$, the classical Lieb-Thirring inequality \cite{13} states that $$\Cal{M}_\gamma \le L_{\gamma,n}\int_{\Bbb{R}^n} |V(x)|_-^{\gamma +n/2}dx \tag 1.3$$ where $\gamma >0$, $n\ge 2$, and $|V|_-=\max (-V,0)$ denotes the negative part of $V$ (in the case $\gamma=0$ and $n\ge 3$, (1.3) is the well-known Rosenblum-Lieb-Cwickel estimate \cite{15}). For the Pauli operator $\Bbb{P}$ in $\Bbb{R}^3$ with a constant field, it was shown in \cite{11} that $$\Cal{M}_\gamma \le C_{1,\gamma}\int_{\Bbb{R}^3} |V(x)|_-^{\gamma+3/2}dx + C_{2,\gamma}\, |\bold{B}|\int_{\Bbb{R}^3} |V(x)|_-^{\gamma+1/2}dx \tag 1.4$$ for $\gamma >1/2$. Subsequently, L.~Erd\"os \cite{3} initiated the study of Lieb-Thirring estimates for Pauli operators with non-homogeneous fields. He observed that the direct extension of (1.4) to the case of the non-constant fields: $$\Cal{M}_\gamma \le C_{1,\gamma}\int_{\Bbb{R}^3} |V(x)|_-^{\gamma+3/2}dx + C_{2,\gamma} \int_{\Bbb{R}^3}|\bold{B}(x)|\, |V(x)|_-^{\gamma+1/2}dx \tag 1.5$$ as well as its consequence (by H\"older's inequality) $$\Cal{M}_\gamma \le \widetilde{C}_{1,\gamma}\int_{\Bbb{R}^3} |V(x)|_-^{\gamma+3/2}dx + \widetilde{C}_{2,\gamma} \int_{\Bbb{R}^3}|\bold{B}(x)|^{3/2} |V(x)|_-^{\gamma}dx \tag 1.6$$ is false without substantial regularity conditions on $\bb$. Moreover he conjectured that it would be necessary to replace $|\bold{B}|$ by some screened version of $|\bold{B}|$. Motivated by Erd\"os' observation, A.~Sobolev \cite{19,20} obtained estimates in the forms of (1.5) and (1.6), but with $|\bold{B}|$ replaced by a so-called effective (scalar) magnetic field $b(x)$: $$\sum_{\lambda_j<0} |\lambda_j| \le C_1\int_{\Bbb{R}^3} |V(x)|_-^{5/2}dx +C_2\int_{\Bbb{R}^3} b(x)^{3/2} |V(x)|_-dx.\tag 1.7$$ Roughly speaking, $b(x)$ is a slow varying function which dominates $|\bold{B}(x)|$ pointwise. We remark that the two-dimensional Pauli operator as well as the three-dimensional case with a constant direction field was also investigated in \cite{3,19,20}. Recently, L.~Bugliaro, C.~Fefferman, J.~Fr\"ohlich, G.~Graf, and J.~Stubbe \cite{2} established (1.7) with a $b(x)$ whose energy is comparable to that of $|\bb|$ : \ $\| b\|_{L^2}\approx \|\bb\|_{L^2}$. This in particular implies the following estimate of E.~Lieb, M.~Loss, and J.~Solovej \cite{10}: $$\Cal{M}_1 \le C_1\int_{\Bbb{R}^3}|V(x)|_-^{5/2}dx + C_2\left(\int_{\Bbb{R}^3}|\bold{B}(x)|^2dx\right)^{3/4} \left(\int_{\Bbb{R}^3}|V(x)|_-^4dx\right)^{1/4}. \tag 1.8$$ In this paper we further extend the results in \cite {3,19,20,2}. The main novelty of our work is that the effective field $b(x)$ we found is much simple and more natural than the previous ones in \cite{20,2}. Indeed, $b(x)=b_p(x)$ is defined to be the $L^p$ average of $|\bb|$ over a suitable cube centered at $x$ with a side length scaling like $|\bold{B}|^{-1/2}$. We believe that this choice of the effective field is optimal. To state our main results, we first define the basic length scale: $$\ell_p(x) =\sup\left\{ \ell>0:\ \ell^2\left(\frac{1}{\ell^3}\int_{Q(x,\ell)} |\bb(y)|^pdy\right)^{1/p}\le 1\right\}\tag 1.9$$ where $Q(x,\ell)$ denotes the cube centered at $x$ with side length $\ell$. It is easy to see that if $|\bb|\in L^p_{loc}(\Bbb{R}^3)$ for some $p>3/2$, then $0<\ell_p(x)<\infty$ unless $|\bb|\equiv 0$. Our effective field is now given by $$b_p(x)\equiv\frac{1}{\{ \ell_p(x)\}^2} =\left\{\frac{1}{\ell^3_p(x)}\int_{Q(x,\ell_p(x))} |\bb(y)|^pdy\right\}^{1/p}, \tag 1.10$$ i.e., $b_p(x)$ is the $L^p$ average of $|\bb|$ over the cube centered at $x$ with side length $\ell_p(x)$. In particular, we have $\|b_p\|_{L^q}\le C\|\bold{B}\|_{L^q}$ for any $q\ge p$. \proclaim{\bf Proposition 1.1} Let $p>3/2$. Then, for any $q\ge p$, there exists a constant $C_{p,q}>0$ such that $$\int_{\Bbb{R}^3} |b_p(x)|^q dx \le C_{p,q}\int_{\Bbb{R}^3} |\bb(x)|^q dx. \tag 1.11$$ \endproclaim The following are the main results of the paper. \proclaim{\bf Theorem 1.1} Let $p>3/2$ and $\gamma\ge 1$. Then there exist constants $C_1(\gamma,p) >0$ and $C_2 (\gamma, p)>0$ such that $$\sum_{\lambda_j<0} |\lambda_j|^\gamma \le C_1(\gamma, p)\int_{\Bbb{R}^3}|V(x)|_-^{\gamma +3/2}dx + C_2(\gamma, p)\int_{\Bbb{R}^3} \left[ b_p(x)\right]^{3/2} |V(x)|_-^\gamma dx.$$ \endproclaim We have a stronger estimate for magnetic fields with constant directions. \proclaim{\bf Theorem 1.2} Let $p>1$ and $\gamma>1/2$. Suppose that $\bb$ has a constant direction. Then there exist constants $C_3(\gamma,p) >0$ and $C_4 (\gamma, p)>0$ such that $$\sum_{\lambda_j<0} |\lambda_j|^\gamma \le C_3(\gamma, p)\int_{\Bbb{R}^3}|V(x)|_-^{\gamma +3/2}dx + C_4(\gamma, p)\int_{\Bbb{R}^3} b_p(x) |V(x)|_-^{\gamma+1/2} dx.$$ \endproclaim A few remarks are in order. \remark{\bf Remark 1.1} The definition of $b_p(x)$ is motivated by the auxiliary function $m(x,|\bb|)$, which is instrumental in the study of eigenvalue problems for the magnetic Schr\"odinger operator $-(\bold{p}-\bold{A}(x))^2 +V(x)$ with certain degenerate potentials \cite{16,17}. We should point out that, although the definition (1.9) is not rotation invariant, the basic length scales $\ell_p(x)$, hence $b_p(x)$, are equivalent in different coordinates systems. One can certainly use balls centered at $x$ instead of cubes in (1.9). We also remark that, if the components of $\bold{B}$ are polynomials, then $$b_p(x)\approx \sum_{|\alpha|\le k} |\nabla^\alpha \bold{B}(x)|^\frac{2}{|\alpha|+2}\tag 1.12$$ where $k$ is the degree of $\bb$. \endremark \remark{\bf Remark 1.2} Suppose that $\bold{B}=\bold{B}_1+\bold{B}_2$ where $|\bold{B}_1|\in L^\infty(\Bbb{R}^3)$ and $|\bold{B}_2|\in L^p (\Bbb{R}^3)$ for some $p>3/2$. Let $\ell =\ell_p(x)$, then $$\ell^2\left(\frac{1}{\ell^3}\int_{Q(x,\ell)}|\bold{B}(y)|^pdy\right)^{1/p}=1.$$ It follows that $$\ell^2 \|\bold{B}_1\|_{L^\infty} +\ell^{2-3/p}\|\bold{B}_2\|_{L^p}\ge 1.$$ A simple computation yields that $$b_p(x) =\frac{1}{\ell^2} \le C\big\{ \|\bold{B}_1\|_{L^\infty} +\|\bold{B}_2\|_{L^p}^\frac{2p}{2p-3}\big\}. \tag 1.13$$ This, together with Theorem 1.1 ($p=2$), gives the estimate in \cite{3, Theorem 2.4, p.635}. Same argument also shows that Theorem 2.1 in \cite{3, p.632} follows from Theorem 1.2. \endremark \remark{\bf Remark 1.3} To compare our results with that in \cite{2}, we note that the effective field in \cite {2} is defined by $\widetilde{b}(x)=1/[r(x)]^2$ where $$\frac{1}{r(x)} =\int_{\Bbb{R}^3} \varphi\left(\frac{y-x}{r(x)}\right) |\bb(y)|^2dy\tag 1.14$$ and $\varphi(z)=(1+\frac{1}{2}z^2)^{-2}$. Let $r=r(x)$. It is easy to see that $$\int_{Q(x,r)}|\bb(y)|^2dy\le \frac{C}{r}.$$ Thus, if $\delta$ is small, we have $$(\delta r)^2\left\{\frac{1}{(\delta r)^3} \int_{Q(x,\delta r)} |\bb(y)|^2dy\right\}^{1/2} \le C\delta^{1/2}<1.$$ By definition, $\ell_2(x) \ge \delta r$. Hence, $$b_2(x)=\frac{1}{\{ \ell_2(x)\}^2} \le \frac{1}{(\delta r)^2} =\frac{\widetilde{b}(x)}{\delta ^2}.$$ Thus Theorem 1.1 extends the results in \cite {2}, with an effective field much easier to compute. One may deduce the estimates in \cite{20} from Theorems 1.1 and 1.2 with $p=\infty$ by a similar argument. \endremark \remark{\bf Remark 1.4} In Theorems 1.1 and 1.2, we have implicitly assumed that $\Bbb{P}_0+V$ is a self-adjoint operator. In fact, if the right hand side of the estimate in Theorem 1.1 (or 1.2) is finite, then $\Bbb{P}_0+V$ has a unique self-adjoint realization on $L^2(\Bbb{R}^3,\Bbb{C}^2)$ associated with its quadratic form. Indeed, under the assumption $\bold{A}\in L^2_{loc}(\Bbb{R}^3, \Bbb{R}^3)$, we may define $\Bbb{P}_0$ as the self-adjoint operator associated with the closed quadratic form $\| \Bbb{D}\psi\|_{L^2}^2$ where $\Bbb{D}=\sigma\cdot (\bold{p}-\bold{A})$, i.e., $\Bbb{P}_0 =\Bbb{D}^*\Bbb{D}$. Let $V_j=V\chi_{\{ |V|\le j\} }$. Since $V_j$ is bounded, the operator $\Bbb{P}_0-\frac{1}{\varepsilon}|V_j|$ is self-adjoint with a form core Domain$(\Bbb{D})$ for any $\varepsilon\in (0,1)$. We now apply the variation principle and Theorem 1.1 to $\Bbb{P}_0-\frac{1}{\varepsilon}|V_j|$. We obtain \aligned \int_{\Bbb{R}^3} |\sigma\cdot (\bold{p} & -\bold{A})\psi|^2dx -\frac{1}{\varepsilon}\int_{\Bbb{R}^3} |V_j|\, |\psi|^2dx\\ &\ge \lambda_1(\varepsilon) \int_{\Bbb{R}^3} |\psi|^2dx\ge -C(\delta, \varepsilon, V, \bold{B}) \int_{\Bbb{R}^3}|\psi|^2dx \endaligned for any $\psi\in\text{Domain} (\Bbb{D})$. Thus, $$\int_{\Bbb{R}^3} |V_j|\, |\psi|^2dx \le \varepsilon\int_{\Bbb{R}^3} |\sigma\cdot (\bold{p} -\bold{A})\psi|^2dx +\varepsilon\, C(\delta, \varepsilon, V, \bold{B}) \int_{\Bbb{R}^3}|\psi|^2dx.$$ Let $j\to \infty$. By Fatou's Lemma, we have $$\int_{\Bbb{R}^3} |V|\, |\psi|^2dx \le \varepsilon\int_{\Bbb{R}^3} |\sigma\cdot (\bold{p} -\bold{A})\psi|^2dx +\varepsilon\, C(\delta, \varepsilon, V, \bold{B}) \int_{\Bbb{R}^3}|\psi|^2dx$$ for $\psi\in \text{Domain} (\Bbb{D})$. It then follows from the KLMN Theorem \cite{14, p.167} that $\Bbb{P}_0+V$ can be extended to the unique self-adjoint operator associated with its quadratic form. Furthermore, Domain$(\Bbb{D})$ is a form core for $\Bbb{P}_0+V$. \endremark To prove Theorem 1.1, we will compare the Pauli operator $\pp_0$ with the magnetic Schr\"odinger operator $\hh_0 =(\bold{p}-\bold{A})^2$ in an appropriate scale, as in \cite{2,8,19,20}. This is possible since $\pp_0=\hh_0-\sigma\cdot\bb$. To this end, our first step is to localize the operators to cubes $Q$ over which the $L^p$ average of $|\bb|$ is small compare to $\ell(Q)^{-2}$ (the localization error in the kinetic energy), by a Calder\'on--Zygmund decomposition. More precisely, we divide $\Bbb{R}^3$ into a grid of disjoint cubes $\{ Q_j\}$ where each $Q_j$ is a maximal dyadic cube such that $$\left\{ \int_{12Q_j}|\bb(x)|^pdx\right\}^{1/p} \le \frac{\varepsilon}{[\ell (Q_j)]^{2-\frac{3}{p}}}. \tag 1.15$$ Here $\ell(Q_j)=\ell_j$ is the side length of $Q_j$, and $12Q_j$ denotes the cube which has the same center as $Q_j$ and side length $12 \ell(Q_j)$. It can be proved that $\ell_j\approx \ell_k$ if $4Q_j\cap 4Q_k \neq \emptyset$. Using this property, we construct a partition of unity for $\Bbb{R}^3$: $\sum_j \phi_j^2(x)\equiv 1$, with $\phi_j\in C_0^\infty (2Q_j,\Bbb{R})$. Next we apply the Birman-Schwinger principle which reduces the problem to the estimate of singular values of $|V|^{1/2}(\pp_0+\lambda)^{-1/2}$ for $\lambda>0$. We then use the resolvent identity to compare $\phi_j (\pp_0 +\lambda)^{-1}$ with $(\pp_0+\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1}\phi_j$ where $\Phi(x)=\sum_j \frac{1}{\ell_j^2}\, \phi_j^2(x) \approx b_p(x)$. We will show that, if $\varepsilon$ in (1.15) is sufficiently small, then $$\hh_0\le C\left\{ \pp_0+\Phi\right\} \tag 1.16$$ (Theorem 3.1). With (1.16) at our disposal, we are able to estimate the contribution from $(\pp_0+\Phi+\frac{1}{\ell_j^2} +\lambda)^{-1}\phi_j$. This leads to the first term in Theorem 1.1. Finally, to deal with the error term $\phi_j (\pp_0 +\lambda)^{-1}- (\pp_0+\Phi+\frac{1}{\ell_j^2}+ \lambda)^{-1}\phi_j$ which can be written as $$(\pp_0+\Phi+\frac{1}{\ell_j^2}+ \lambda)^{-1} \big\{ \big(\Phi+\frac{1}{\ell_j^2}\big) \phi_j+[\pp_0,\phi_j]\big\} (\pp_0 +\lambda)^{-1},$$ we use the resolvent identity again to compare $\psi_j (\pp_0+\Phi +\frac{1}{\ell_j^2})^{-1}$ with $(\hh_0+\frac{1}{\ell_j^2})^{-1}\psi_j$ where $\psi_j$ is a bump function such that $\psi_j\phi_j\equiv \phi_j$. The desired result follows from certain regularity estimates for the operator $\psi_j(\pp_0+\Phi +\frac{1}{\ell_j^2})^{-1}$ (see Lemma 4.2). A similar approach is used in the case of constant direction fields. Assuming $\bb=(0,0,B(x_1,x_2))$ without the loss of generality, we construct a partition of unity for $\Bbb{R}^2$ associated with $B$. The inequality $$\| (-\partial_{x_3}^2 +\lambda)^{1/2} (\pp_0 +\lambda)^{-1/2}\|_{L^2\to L^2} \le 1,\ \ \ \ \ \ \ \ \ \ \lambda>0 \tag 1.17$$ is used to exploit the fact that $\pp_0$ commutes with $\partial_{x_3}$. The paper is organized as follows. In Section 2 we collect some basic facts that will be used concerning the norm in the Neumann-Schatten classes. Section 2 also contains the proof of Proposition 1.1. The partition of unity will be constructed in Section 3. Section 4 is devoted to the proof of Theorem 1.1. Finally we study the case of constant direction fields in Section 5 where Theorem 1.2 is proved. Throughout this paper we will use $\|\psi\|_{L^p}$ to denote the $L^p$ norm of the function $\psi$. $\| T\|_{L^p\to L^q}$ will denote the operator norm, while $\| T\|_p$ is reserved for the norm in the Neumann-Schatten classes (see (2.1)). Finally we will use $C$ to denote constants, which are not necessarily the same at each occurrence, which may depend on $p$. \bigskip\medskip \centerline{\bf 2. Some Preliminaries} Most materials in this section on the Neumann-Schatten classes can be found in \cite {18}. We include them here for the reader's convenience. The proof of Proposition 1.1 will be given at the end of the section. Let $T$ be a compact operator on $L^2(\Bbb{R}^3,\Bbb{C}^2)$. We will use $n(s,T)$ to denote the number of singular values $\{ s_n(T)\}$ (counting multiplicity) of $T$ greater than $s$ where $s>0$. For $p\ge 1$, let $$\| T\|_p^p\equiv\sum_n |s_n(T)|^p=p\int_0^\infty s^{p-1}n(s,T)ds. \tag 2.1$$ The functional $\|T\|_p$ defines a norm on the Neumann-Schatten class $\Cal{S}_p$ which consists of compact operators $T$ with $\|T\|_p<\infty$. The following facts will be useful to us. \align n(s^2, T^*T)&=n(s,T), \tag 2.2\\ \|T\|_p^2&=\|T^* T\|_{p/2},\tag 2.3\\ n(s,T)&\le \frac{\| T\|_p^p}{s^p}.\tag 2.4 \endalign We will also need \align &n(s_1+s_2,T_1+T_2)\le n(s_1,T_1)+n(s_2+T_2), \tag 2.5\\ &\|T_1 T_2\|_p\le \|T_1\|_p \|T_2\|_{L^2\to L^2},\tag 2.6\\ & \|T_1 T_2\|_p\le \| T_1\|_r \|T_2\|_s \ \ \ \ \ \ \ \ \text {if } \ \ \ \frac1p =\frac1r +\frac 1s,\ \ r, \, s\ge 1\tag 2.7 \endalign where $\|T\|_{L^2\to L^2}$ denotes the operator norm of $T$. The proof of the following lemma may be found in \cite{18}. \proclaim{\bf Lemma 2.1} If $f,\ g\in L^p(\Bbb{R}^3)$ with $2\le p<\infty$. Then $$\| f(x)g(-i\nabla)\|_p\le C_p \| f\|_{L^p} \| g\|_{L^p}.$$ \endproclaim It follows from Lemma 2.1 that, if $p\ge 2$, $\alpha > \frac{3}{2p}$, and $\lambda >0$, $$\| V (-\Delta +\lambda )^{-\alpha}\|_p \le C_{p, \alpha} \lambda^{\frac{3}{2p}-\alpha}\| V\|_{L^p}. \tag 2.8$$ Also, by Lemma 2.1, if $\alpha >1/2$ and $0 < \lambda_1\le \lambda_2$, $$\| V(-\partial _{x_3}^2+\lambda_1)^{-1/2} (-\Delta+\lambda_2)^{-\alpha}\|_2 \le C_\alpha \lambda_1^{-\frac14}\lambda_2^{-\alpha +\frac12}\|V\|_{L^2}. \tag 2.9$$ Thus, by the diamagnetic inequality \cite{9}, $$|\big( \hh_0+\lambda\big)^{-\alpha}\psi| \le (-\Delta +\lambda)^{-\alpha}|\psi|, \tag 2.10$$ where $\hh_0 =(\bold{p}-\bold{A}(x))^2$, and Theorem 2.13 in \cite{18, p.36}, we have \align &\|V( \hh_0+\lambda)^{-\alpha}\|_2 \le C_\alpha \lambda^{\frac34-\alpha} \|V\|_{L^2} \ \ \ \ \text{ for } \alpha >3/4, \tag 2.11\\ & \|V( \hh_0+\lambda)^{-\alpha}\|_4 \le C_\alpha \lambda^{\frac38-\alpha} \|V\|_{L^4} \ \ \ \ \text{ for } \alpha >3/8, \tag 2.12 \endalign and for $\alpha>1/2$ and $0<\lambda_1\le \lambda_2$, $$\| V(-\partial _{x_3}^2+\lambda_1)^{-1/2} (\hh_0+\lambda_2)^{-\alpha}\|_2 \le C_\alpha \lambda_1^{-\frac14}\lambda_2^{-\alpha +\frac12}\|V\|_{L^2}. \tag 2.13$$ The following two lemmas concern the operator norm of $V(-\Delta +\lambda)^{-\alpha}$ on $L^2(\Bbb{R}^3, \Bbb{C}^2)$. \proclaim{\bf Lemma 2.2} Let $\alpha >0$ and $\lambda >0$. Then \align & \| V (-\Delta +\lambda)^{-\alpha}\|_{L^2\to L^2} \le C_\alpha \| V\|_{L^\frac{3}{2\alpha}} \ \ \ \ \ \ \ \ \ \text{ if } \ \ \alpha <3/4,\tag 2.14\\ & \| V (-\Delta +\lambda)^{-\alpha}\|_{L^2\to L^2} \le C_\alpha \lambda^{\frac{3}{4}-\alpha} \| V\|_{L^2}\ \ \ \ \text { if } \ \ \alpha >3/4.\tag 2.15 \endalign \endproclaim \demo{\bf Proof} Note that, if $\alpha <3/4$, $$\|V(-\Delta+\lambda)^{-\alpha}\psi\|_{L^2} \le \|V\|_{L^{p_1}}\| (-\Delta +\lambda )^{-\alpha}\psi \|_{L^{p_2}} \le C_\alpha \|V\|_{L^{p_1}}\|\psi\|_{L^2}$$ where $\frac{1}{p_1}+\frac{1}{p_2}=\frac12$, $\frac{1}{p_2}=\frac12 -\frac{2\alpha}{3}$, and we have used H\"older's inequality and the well-known fractional integral estimate \cite{21}. (2.14) is proved since $p_1=3/(2\alpha)$. Now suppose $\alpha >3/4$. Using the Fourier transform, we have \aligned \|(-\Delta& +\lambda)^{-\alpha}\psi\|_{L^\infty} \le C\int_{\Bbb{R}^3} (|\xi|^2 +\lambda)^{-\alpha} |\hat{\psi}(\xi)|d\xi\\ &\le C \|\psi\|_{L^2} \left\{\int_{\Bbb{R}^3} (|\xi|^2 +\lambda)^{-2\alpha} d\xi\right\}^{1/2} \le C_\alpha \lambda^{\frac34 -\alpha}\|\psi\|_{L^2}. \endaligned \tag 2.16 It follows that $$\|V(-\Delta+\lambda)^{-\alpha}\psi \|_{L^2} \le \|V\|_{L^2} \| (-\Delta +\lambda)^{-\alpha}\psi \|_{L^\infty} \le C_\alpha \lambda^{\frac34 -\alpha} \|V\|_{L^2} \|\psi\|_{L^2}.$$ \enddemo \proclaim{\bf Lemma 2.3} Let $0<\alpha <1/2$ and $\lambda >0$. Suppose that $V$ depends only on the first two variables. Then $$\| V (-\Delta +\lambda )^{-\alpha}\|_{L^2\to L^2} \le C_\alpha \| V\|_{L^{1/\alpha}(\Bbb{R}^2)}.$$ \endproclaim \demo{\bf Proof} By the fractional integral estimates \cite{21}, if $\psi\in C^\infty (\Bbb{R}^3,\Bbb{C}^2)$ and decays at $\infty$, $$\|\psi\|_{L^q(\Bbb{R}^2)} \le C_\alpha \| (-\Delta_2 +\lambda )^\alpha\psi\|_{L^2(\Bbb{R}^2)}$$ where $\frac1q=\frac12 -\alpha$ and $\Delta_2 =\partial_{x_1}^2 +\partial_{x_2}^2$ denotes the Laplacian in $\Bbb{R}^2$. It then follows from H\"older's inequality that \align \|V\psi\|_{L^2(\Bbb{R}^2)} &\le \|V\|_{L^p(\Bbb{R}^2)} \|\psi\|_{L^q(\Bbb{R}^2)}\\ &\le C_\alpha \|V\|_{L^p(\Bbb{R}^2)} \| (-\Delta_2 +\lambda )^\alpha\psi\|_{L^2(\Bbb{R}^2)} \endalign where $\frac1p +\frac1q =\frac12$. Hence \align \int_{\Bbb{R}^3} |V|^2 |\psi|^2dx &\le C_\alpha^2 \|V\|_{L^p(\Bbb{R}^2)}^2 \int_{\Bbb{R}^3} |(-\Delta_2 +\lambda )^\alpha\psi|^2dx\\ &\le C_\alpha^2 \|V\|_{L^p(\Bbb{R}^2)}^2 \int_{\Bbb{R}^3} |(-\Delta +\lambda )^\alpha\psi|^2dx \endalign where the second inequality can be verified through the Fourier transform. Thus $$\| V\psi \|_{L^2} \le C_\alpha \|V\|_{L^p(\Bbb{R}^2)} \|(-\Delta +\lambda )^\alpha\psi\|_{L^2}.$$ The lemma now follows easily since $\frac1p =\frac12 -\frac1q=\alpha$. \enddemo We end this section with the \demo{\bf Proof of Proposition 1.1} It follows from (1.10) that $b_p(x)\le \{ M(|\bb|^p)(x)\}^{1/p}$ where $M$ is the Hardy-Littlewood maximal operator. This gives the desired estimate in the case $q>p$ since $M$ is bounded on $L^s$ for $s>1$ \cite{21}. For the case $q=p$, we appeal to an argument in \cite{2}. Let $\widetilde{b}_p(x)=1/[r_p(x)]^2$ where $r_p(x)$ is defined by $$\frac{1}{[r_p(x)]^{2p-3}} =\int_{\Bbb{R}^3} \varphi \left(\frac{x-y}{r_p(x)}\right)|\bb(y)|^pdy \tag 2.17$$ and $\varphi(z)=[1+\frac12 |z|^2]^{-2}$. Since $b_p(x)\le C\, \widetilde{b}_p(x)$ (see Remark 1.4 for $p=2$), it suffices to show that $$\int_{\Bbb{R}^3} | \widetilde{b}_p(x)|^pdx \le C\int_{\Bbb{R}^3} |\bb(x)|^pdx. \tag 2.18$$ Using (2.17) and Fubini's Theorem, one may reduce (2.18) to $$\int_{\Bbb{R}^3} \varphi \left(\frac{x-y}{r_p(x)}\right) \frac{dx}{[r_p(x)]^3} \le C<\infty. \tag 2.19$$ The proof of (2.19) relies on the following estimate $$r_p(y)g_-\left(\frac{|x-y|}{r_p(y)}\right) \le r_p(x) \le r_p(y)g_+\left(\frac{|x-y|}{r_p(y)}\right) \tag 2.20$$ where $g_+(s)$ and $g_-(s)$ are solutions of the equation $$s^2=2(t^{p+\frac12}-1)(1-t^{\frac32-p}) \tag 2.21$$ on $(1,\infty)$ and $(0,1)$ respectively. Since $p>3/2$, it is easy to see that $g_+(s)$, $g_-(s)$ are well defined. We omit the proof of (2.20), which is similar to that of Lemma 2 in \cite{2, p.569}. It is not hard to see that $g_+(s)\approx s^{\frac{4}{2p+1}}$ if $s\ge 1/2$, and $g_-(s)\ge c_0>0$ if $s<1/2$. This, together with (2.20), implies that \alignedat2 &r_p(x)\le C\, [r_p(y)]^{1-\frac{4}{2p+1}} |x-y|^{\frac{4}{2p+1}} \ \ & \text{ if }\ \ |x-y|\ge \frac12 r_p(y),\\ &r_p(x)\ge c \, r_p(y) & \text{ if }\ \ |x-y|<\frac12 r_p(y). \endalignedat It then follows that \aligned &\int_{\Bbb{R}^3} \varphi \left(\frac{x-y}{r_p(x)}\right) \frac{dx}{[r_p(x)]^3} \le \int_{\Bbb{R}^3} \frac{r_p(x)\, dx} {\{ [r_p(x)]^2 +\frac12 |x-y|^2\}^2}\\ &\le 4 \int_{|x-y|\ge \frac12 r_p(y)} \frac{r_p(x)\, dx} {|x-y|^4} + \int_{|x-y|< \frac12 r_p(y)} \frac{dx}{[r_p(x)]^3}\\ &\le C \, \big[r_p(y)\big]^{\frac{2p-3}{2p+1}} \int_{|x-y|\ge \frac12 r_p(y)} \frac{ dx} {|x-y|^{3+\frac{2p-3}{2p+1}}} +\frac{C}{[r_p(y)]^3} \int_{|x-y|< \frac12 r_p(y)} dx\\ &\le C. \endaligned The proof is finished. \enddemo \bigskip \centerline{\bf 3. A Partition of Unity} Throughout this section we fix $p>3/2$. We will assume that $|\bb|\in L^p_{loc}(\Bbb{R}^3)$ and $|\bb|\not\equiv 0$. Let $\varepsilon\in (0,1)$ be a small constant to be determined later. Let $\Cal{A}$ be the set of all dyadic cubes in $\Bbb{R}^3$ such that $$\left(\int_{12 Q}|\bb(x)|^pdx\right)^{1/p}\le \frac{\varepsilon}{[\ell(Q)]^{2-\frac{3}{p}}} \tag 3.1$$ where $\ell(Q)$ denotes the side length of $Q$. We say that $Q$ is a maximal element of $\Cal{A}$ if $Q\in\Cal{A}$ and $Q$ is not properly contained in any other cube in $\Cal{A}$. Let $\Cal{B}$ denote the set of all maximal elements of $\Cal{A}$. Clearly, by definition, the interiors of the cubes in $\Cal{B}$ are disjoint. \proclaim{\bf Lemma 3.1} Let $\Cal{B}=\left\{ Q_j,\ j=1,2,\dots\right\}$. Then, $$\Bbb{R}^3=\bigcup_j Q_j.$$ \endproclaim \demo{\bf Proof} It suffices to show that any point in $\Bbb{R}^3$ belongs to some cube $Q_j$ in $\Cal{B}$. To this end, fix $x\in \Bbb{R}^3$, let $\{ Q_{\alpha_k}\}_{k=-\infty}^\infty$ be an increasing sequence of dyadic cubes such that $x\in Q_{\alpha_k}$ for all $k$, $\ell(Q_{\alpha_k})\to 0$ as $k\to -\infty$, and $\ell(Q_{\alpha_k})\to\infty$ as $k\to\infty$. Note that, as $\ell(Q)\to 0$, the l.h.s. of (3.1) goes to $0$ while the r.h.s. goes to $\infty$. This implies that $Q_{\alpha_k}\in\Cal{A}$ if $\ell(Q_{\alpha_k})$ is small. Also, as $\ell(Q_{\alpha_k})\to\infty$, the l.h.s. of (3.1) goes to $\|\bb\|_{L^p} >0$ and the r.h.s. goes to 0. Thus, $Q_{\alpha_k}\notin\Cal{A}$ if $\ell(Q_{\alpha_k})$ is large. Hence there must be a maximal element $Q_j$ in $\Cal{A}$ (which may not be in $\{ Q_{\alpha_k}\}$) such that $x\in Q_j$. \enddemo \proclaim{\bf Lemma 3.2} Let $Q_j$, $Q_k \in \Cal{B}$. Suppose that $4Q_j\cap 4Q_k\neq\emptyset$. Then $$\frac{1}{2}\ell(Q_k) \le \ell(Q_j) \le 2\ell(Q_k).$$ \endproclaim \demo{\bf Proof} We adapt an argument found in \cite {8}. Suppose $\ell(Q_j)>2\ell (Q_k)$. Since $\ell(Q_j)$ and $\ell(Q_k)$ are powers of $2$, we have $\ell(Q_j)\ge 4\ell(Q_k).$ Let $Q_k^+$ be the dyadic parent of $Q_k$, i.e., $Q_k$ can be obtained by bisecting $Q_k^+$. Since $Q_k$ is a maximal element in $\Cal{A}$, $Q_k^+\notin \Cal{A}$. It follows that $$\left( \int_{12Q_k^+}|\bb|^pdx\right)^{1/p} >\varepsilon \left[\frac{1}{\ell(Q_k^+)}\right]^{2-\frac{3}{p}} =\varepsilon \left[\frac{1}{ 2\ell(Q_k)}\right]^{2-\frac{3}{p}} \ge \varepsilon \left[\frac{2}{\ell(Q_j)}\right]^{2-\frac{3}{p}} \tag 3.2$$ where we have used $p>3/2$ in the last inequality. We claim that $$12Q_k^+\subset 12 Q_j. \tag 3.3$$ This, together with (3.2), would imply that $$\left(\int_{12Q_j}|\bb|^pdx\right)^{1/p} \ge \left(\int_{12Q_k^+}|\bb|^pdx\right)^{1/p} > \varepsilon \left[\frac{2}{\ell(Q_j)}\right]^{2-\frac{3}{p}}$$ which contradicts with the assumption that $Q_j\in \Cal{A}$. To see (3.3), let $x_j$ and $x_k$ be the centers of $Q_j$ and $Q_k$ respectively. Since $4Q_j\cap 4Q_k\neq \emptyset$, there exists $z$ such that $$|z-x_j|_*\le 2\ell(Q_j)\ \ \ \text{ and } \ \ \ |z-x_k|_*\le 2\ell(Q_k)$$ where $|\cdot|_*$ is the norm in $\Bbb{R}^3$ defined by $$|x|_*=|(a,b,c)|_*=\max \{ |a|,|b|,|c|\}.$$ Thus, $$|x_j-x_k|_*\le 2\ell(Q_j)+2\ell(Q_k).$$ Now suppose $y\in 12 Q_k^+$. Then $|y-x_k|_* \le |y-x_k^+|_*+ |x_k^+-x_k|_* < 13 \ell(Q_k)$ where $x_k^+\in Q_k$ is the center of $Q_k^+$. It follows that $$|y-x_j|_*\le |y-x_k|_*+|x_k-x_j|_* < 15\ell(Q_k) +2\ell(Q_j) < 6\ell(Q_j).$$ Hence, $y\in 12 Q_j$. (3.3) is then proved. \enddemo \remark{\bf Remark 3.1} It follows easily from Lemma 3.2 that $\{ 4Q_j: Q_j\in \Cal{B}\}$ has the finite intersection property: $\sum_j \chi_{4Q_j}\le C$ where $C$ is an absolute constant. \endremark We are now in a position to construct the partition of unity associated with the field strength $|\bb|$. \proclaim{\bf Lemma 3.3} There exists a sequence of functions $\{ \phi_j\}$ such that (i)\ \ $\phi_j\in C_0^\infty(2Q_j)$ and $0\le \phi_j\le 1$, (ii)\ \ $|\nabla^\alpha \phi_j(x)|\le c_\alpha/\ell_j^{|\alpha|}$ where $\ell_j=\ell(Q_j)$, (iii)\ \ $\sum_j\phi_j^2\equiv 1$ in $\Bbb{R}^3$. \endproclaim \demo{\bf Proof} Choose $\eta\in C_0^\infty(Q(0,2))$ such that $\eta\equiv1$ in $Q(0,1)$ and $0\le \eta \le 1$. Let $\Cal{B}=\{ Q_j\}_{j=1}^\infty$ and $Q_j=Q(x_j,\ell_j)$. We define $$\eta_j(x)=\eta\left(\frac{x-x_j}{\ell_j}\right).$$ and $\varphi (x)=\sum_j \eta_j^2(x)$. Clearly, by Lemma 3.1 and Remark 3.1, $1\le \varphi \le C$ where $C$ is an absolute constant, and $\varphi\in C^\infty (\Bbb{R}^3)$. Finally let $\phi_j(x)=\eta_j(x)/\sqrt{\varphi (x)}$. It is easy to check that $\phi_j$ satisfies (i)-(iii). We omit the details. \enddemo Recall that $\ell_j=\ell(Q_j)$ and $Q_j$ is a maximal cube. Define $$\Phi=\sum_j\frac{1}{\ell_j^2}\, \phi_j^2.\tag 3.4$$ We will show in Section 4 that $\Phi(x)\approx b_p(x)$ (see Lemma 4.5). \proclaim{\bf Theorem 3.1} There exist constants $C>0$ and $\varepsilon_0>0$ such that, if $0<\varepsilon <\varepsilon_0$, we have $$\hh_0\le C\big\{ \pp_0+\Phi\big\}.$$ \endproclaim \demo{\bf Proof} We first show that, if $\varepsilon$ in (3.1) is small, then $$\int_{\Bbb{R}^3} |\, |\bb|^{1/2}\phi_j\psi|^2dx \le C\int_{\Bbb{R}^3} |\sigma\cdot (\bold{p}-\bold{A})(\phi_j\psi)|^2dx\tag 3.5$$ for any $\psi\in C^\infty_0(\Bbb{R}^3,\Bbb{C}^2)$. To this end, we note that, by H\"older's inequality $$\left\{\int_{2Q_j}|\bb|^{3/2}dx\right\}^{2/3} \le |2Q_j|^{\frac23-\frac1p} \left\{\int_{2Q_j}|\bb|^p dx\right\}^{1/p} \le 4\varepsilon.\tag 3.6$$ It then follows from the Sobolev embedding and the inequality $|\nabla |\psi|\,|\le |(\bold{p}-\bold{A})\psi|$ that \aligned \bigg\{ \int_{\Bbb{R}^3} |\, &|\bb|^{1/2}\phi_j\psi|^2dx\bigg\}^{1/2} \le \left\{\int_{2Q_j}|\bb|^{3/2}dx\right\}^{1/3} \left\{\int_{\Bbb{R}^3}|\phi_j\psi|^6dx\right\}^{1/6}\\ &\le C\varepsilon^{1/2} \left\{\int_{\Bbb{R}^3}|\nabla|\phi_j\psi|\,|^2dx\right\}^{1/2}\\ &\le C\varepsilon^{1/2} \left\{\int_{\Bbb{R}^3}|(\bold{p}-\bold{A})(\phi_j\psi)|^2 dx\right\}^{1/2}\\ &\le C\varepsilon^{1/2} \left\{\int_{\Bbb{R}^3}|\sigma\cdot (\bold{p}-\bold{A})(\phi_j\psi)|^2 dx\right\}^{1/2} +C\varepsilon^{1/2} \left\{ \int_{\Bbb{R}^3} |\, |\bb|^{1/2}\phi_j\psi|^2dx\right\}^{1/2} \endaligned where we have used the fact that $\pp_0 =\hh_0 -\sigma\cdot \bb$. This gives (3.5) if $C\varepsilon^{1/2}<1/2$. Next we use Lemma 3.3 and (3.5) to obtain \aligned &\int_{\Bbb{R}^3} |(\bold{p}-\bold{A})\psi|^2dx =\sum_j\int_{\Bbb{R}^3} |\phi_j(\bold{p}-\bold{A})\psi|^2dx\\ & \le 2\sum_j \int_{\Bbb{R}^3} |(\bold{p}-\bold{A})(\phi_j\psi)|^2dx +2\sum_j \int_{\Bbb{R}^3}|\nabla\phi_j|^2|\psi|^2dx\\ &\le 2\sum_j \int_{\Bbb{R}^3} |\sigma\cdot (\bold{p}-\bold{A})(\phi_j\psi)|^2dx +2\sum_j\int_{\Bbb{R}^3} |\, |\bb|^{1/2}\phi_j\psi|^2dx + 2\sum_j \int_{\Bbb{R}^3}|\nabla\phi_j|^2|\psi|^2dx\\ &\le C\sum_j \int_{\Bbb{R}^3} |\sigma\cdot (\bold{p}-\bold{A})(\phi_j\psi)|^2dx +2\sum_j \int_{\Bbb{R}^3}|\nabla\phi_j|^2|\psi|^2dx\\ &\le C\int_{\Bbb{R}^3} |\sigma\cdot (\bold{p}-\bold{A})\psi|^2dx +C\sum_j \frac{1}{\ell_j^2} \int_{2Q_j}|\psi|^2dx. \endaligned Clearly, the theorem follows if we have $$\sum_j\frac{1}{\ell_j^2}\, \chi_{2Q_j}\le C\, \Phi. \tag 3.7$$ We claim that (3.7) is an easy consequence of Lemma 3.2. Indeed, if $x\in Q_k$, the l.h.s. of (3.7) is bounded by $$\sum_{2Q_j\cap Q_k\neq\emptyset} \frac{1}{\ell_j^2} \le \frac{C}{\ell_k^2}\cdot \# \{ Q_j: \ 2Q_j\cap Q_k\neq\emptyset\} \le\frac{C}{\ell_k^2}\le C\, \Phi(x).$$ The proof is now complete. \enddemo It follows easily from Theorem 3.1 that, if $0<\lambda_1\le\lambda_2$, $$\| (\hh_0+\lambda_1)^{1/2} (\pp_0+\Phi +\lambda_2)^{-1/2}\|_{L^2\to L^2} \le C. \tag 3.8$$ The following lemma will be used in the next section. \proclaim{\bf Lemma 3.4} There exists a constant $C>0$ such that, for $\lambda>0$, $$\| (\pp_0+\Phi+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\phi_j (\pp_0+\lambda)^{1/2}\|_{L^2\to L^2}\le C.$$ \endproclaim \demo{\bf Proof} By duality, it suffices to show that $$\|(\pp_0+\lambda)^{1/2}\phi_j (\pp_0+\Phi+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_{L^2\to L^2} \le C. \tag 3.9$$ To this end, we note that, for any $\psi\in C^\infty_0(\Bbb{R}^3,\Bbb{C}^2)$, \aligned &\int_{\Bbb{R}^3} |(\pp_0+\lambda)^{1/2}\phi_j (\pp_0+\Phi+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\psi|^2dx\\ & = \int_{\Bbb{R}^3}|\sigma\cdot (\bold{p}-\bold{A})\phi_j (\pp_0+\Phi+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\psi|^2dx +\lambda \int_{\Bbb{R}^3} |\phi_j(\pp_0+\Phi+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\psi|^2dx\\ & \le 4\int_{\Bbb{R}^3}|\psi|^2 +2\int_{\Bbb{R}^3} |\nabla \phi_j|^2 |(\pp_0+\Phi+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\psi|^2dx\\ & \le C\int_{\Bbb{R}^3} |\psi|^2dx \endaligned where we have used $|\nabla \phi_j|\le C/\ell_j$ and \align &\| \sigma \cdot (\bold{p}-\bold{A})(\pp_0+\Phi+\lambda)^{-1/2}\|_{L^2 \to L^2}\le 1,\tag 3.10\\ &\|(\pp_0+\Phi+\lambda)^{-1/2}\|_{L^2 \to L^2}\le \frac{1}{\sqrt{\lambda }}. \tag 3.11 \endalign The proof is finished. \enddemo \bigskip \centerline{\bf 4. The Proof of Theorem 1.1} For $\lambda >0$, we denote by $N(\lambda, \Bbb{P}_0+V)$ the number of eigenvalues (counting multiplicity) of $\Bbb{P}_0+V$ smaller than $-\lambda$. Since $V\ge -\lambda -|V+\lambda|_-$, we have $$N(2\lambda, \Bbb{P}_0+V)\le N(\lambda, \Bbb{P}_0-|V+\lambda|_-).$$ It then follows from the Birman-Schwinger principle and (2.2) that $$N(2\lambda, \Bbb{P}_0+V) \le n(1,Y(\Bbb{P}_0+\lambda)^{-1}Y ) = n(1,Y(\pp_0+\lambda)^{-1/2}) \tag 4.1$$ where $Y=|V+\lambda|^{1/2}_-$. As in \cite {19,20}, our approach will be based on the following resolvent identity: $$\varphi H_1^{-1} =H_2^{-1}\varphi + H_2^{-1} \big\{ (H_2-H_1)\varphi +[H_1,\varphi]\big\} H_1^{-1} \tag 4.2$$ where $\varphi\in C^\infty_0(\Bbb{R}^3,\Bbb{R})$, $H_1,\, H_2$ are two operators, and $[H_1,\varphi]=H_1\varphi-\varphi H_1$ denotes the commutator between $H_1$ and $\varphi$. Using (4.2), we may write \aligned Y(\pp_0+\lambda)^{-1/2}&=\sum_j Y\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} \phi_j(\pp_0+\lambda)^{1/2}\\ &\ \ \ \ \ \ +\sum_j Y\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} (\Phi +\frac{1}{\ell_j^2})\phi_j (\pp_0+\lambda)^{-1/2}\\ &\ \ \ \ \ \ +\sum_jY\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\pp_0,\phi_j](\pp_0+\lambda)^{-1/2}\\ &=I_1 +I_2 +I_3 \endaligned where $\Phi$ is the function defined by (3.4). Thus, by (2.5), $$n(1, Y(\pp_0+\lambda)^{-1/2}) \le n(1/3, I_1)+n(1/3,I_2)+n(1/3, I_3). \tag 4.3$$ We begin with the estimate of $n(1/3, I_1)$. \proclaim{\bf Lemma 4.1} There exists a constant $C>0$ such that, for any $\lambda>0$, $$n(1/3, I_1)\le \frac{C}{\sqrt{\lambda}}\int_{\Bbb{R}^3}|Y|^4 dx.$$ \endproclaim \demo{\bf Proof} First observe that, by (2.4), $$n(1/3, I_1)\le C\|\sum_j Y\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} \phi_j(\pp_0+\lambda)^{1/2}\|_4^4. \tag 4.4$$ Let $$I_{1j}=Y\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} \phi_j(\pp_0+\lambda)^{1/2}. \tag 4.5$$ It follows from (2.6) and Lemma 3.4 that \aligned \|I_{1j}\|_4 &\le \|Y\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1/2} \|_4 \|(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1/2} \phi_j(\pp_0+\lambda)^{1/2}\|_{L^2\to L^2}\\ &\le C\|Y\phi_j(\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_4\\ &\le C\|Y\phi_j (\hh_0+\lambda)^{-1/2}\|_4 \| (\hh_0+\lambda)^{1/2} (\pp_0+\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_{L^2\to L^2}\\ &\le C\|Y\phi_j (\hh_0+\lambda)^{-1/2}\|_4 \endaligned where we also used (3.8) in the last inequality. We now apply (2.12) with $\alpha =1/2$ to obtain $$\|I_{1j}\|_4 \le C\lambda^{-1/8}\|Y\phi_j\|_{L^4}. \tag 4.6$$ Next we note that, by (2.3), $$\|\sum_jI_{1j}\|_4^4 =\|\sum_j\sum_k\sum_m\sum_n I_{1j} I_{1k}^* I_{1m}I_{1n}^*\|_1 \le \sum_j\sum_k\sum_m\sum_n \|I_{1j} I_{1k}^* I_{1m}I_{1n}^*\|_1.$$ Since $\phi_j$ is supported in $2Q_j$ and $\phi_j (\pp_0+\lambda)^{1/2} (\pp_0+\lambda)^{1/2}\phi_k = \phi_j (\pp_0+\lambda)\phi_k=0$ unless $2Q_j\cap 2Q_k \neq \emptyset$, it is not very hard to see that $$I_{1j} I_{1k}^* I_{1m}I_{1n}^*=0$$ unless $$2Q_j\cap2Q_k\neq\emptyset,\ \ 2Q_k\cap 2Q_m\neq\emptyset,\ \ \text {and } \ \ 2Q_m\cap 2Q_n\neq \emptyset . \tag 4.7$$ On the other hand, by Lemma 3.2, if we fix any one of indices $j,k,m,n$, the number of remaining indices which satisfy (4.7) is bounded by an absolute constant. Thus, by (2.7), \aligned n(1/3,I_1) &\le C\|\sum_jI_{1j}\|_4^4 \le C\sum \|I_{1j}\|_4 \|I_{1k}\|_4\|I_{1m}\|_4\|I_{1n}\|_4\\ &\le C\sum \big\{ \|I_{1j}\|_4^4+ \|I_{1k}\|_4^4 +\|I_{1m}\|_4^4 +\|I_{1n}\|_4^4\big\}\\ &\le C\sum_j\|I_{1j}\|^4_4 \endaligned where the second and third sums are over all $(j,k,m,n)$ satisfying (4.7). The lemma now follows from (4.6). \enddemo Using $|Y|\le |V|^{1/2}\chi_{\{ x\in\Bbb{R}^3: V(x) <-\lambda\} }$, we may deduce easily from Lemma 4.1 that $$\int_0^\infty n(1/3, I_1)\, d\lambda \le C\int_{\Bbb{R}^3} |V(x)|^{5/2}dx. \tag 4.8$$ Let $\psi_j\in C^\infty_0(3Q_j, \Bbb{R})$ such that $0\le \psi_j\le 1$, $\phi_j\psi_j\equiv \phi_j$ and $|\nabla^\alpha \psi_j |\le C_\alpha/\ell_j^{|\alpha|}$. We will need the following lemma in the estimates of $n(1/3, I_2)$ and $n(1/3,I_3)$. \proclaim{\bf Lemma 4.2} There exist constants $C>0$ and $\varepsilon_0>0$ such that, if $0<\varepsilon<\varepsilon_0$, \align &\| \, [\pp_0,\psi_j] (\pp_0+\Phi +\frac{1}{\ell_j^2})^{-1} \|_{L^2\to L^2} \le C, \tag 4.9 \\ &\| \psi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\|_{L^2\to L^\infty} \le C \ell_j^{\frac12}, \tag 4.10 \\ & \| (\hh_0 +\frac{1}{\ell_j^2})^{-\frac14 +\delta} (\sigma\cdot \bb)\psi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} \|_{L^2\to L^2} \le C_\delta \, \ell_j^{\frac12 -2\delta} \tag 4.11 \endalign where $0<\delta <\min\big(1/4, (p-(3/2))/2\big)$. \endproclaim \demo{\bf Proof} To show (4.9), we note that $$[\pp_0,\psi_j]= \big[\Bbb{D}^*,[\Bbb{D},\psi_j]\big] +[\Bbb{D}^*,\psi_j]\Bbb{D} +[\Bbb{D},\psi_j]\Bbb{D}^* \tag 4.12$$ where $\pp_0=\Bbb{D}^*\Bbb{D}$ and $\Bbb{D}=\sigma\cdot (\bold{p}-\bold{A})$ ($\pp_0$ is defined as the operator associated with the quadratic form $\|\Bbb{D}\psi\|_{L^2}^2$. We do not need that $\Bbb{D}$ is self-adjoint). Since \align & |\, [\Bbb{D}^*,\psi_j]\psi | +|\, [\Bbb{D},\psi_j]\psi |\le 2|\nabla \psi_j|\, |\psi| \le \frac{C}{\ell_j}|\psi|, \tag 4.13 \\ & |\, \big[\Bbb{D}^*,[\Bbb{D},\psi_j]\big]\psi|\le |\nabla^2\psi_j|\, |\psi | \le \frac{C}{\ell_j^2}|\psi|, \tag 4.14 \endalign we have $$\|\, [\pp_0,\psi_j] (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi\|_{L^2} \le C\| \psi\|_{L^2}. \tag 4.15$$ where we also used (3.10)--(3.11), $\Bbb{D}^* =\Bbb{D}$ on the domain of $\Bbb{D}$, and $\|(\pp_0+\Phi +\lambda)^{-1}\|_{L^2\to L^2} \le \lambda^{-1}$. To prove (4.10), we use the resolvent identity (4.2) and $\pp_0 =\hh_0 -\sigma\cdot \bb$ to write \aligned &\psi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}= (\hh_0+\frac{1}{\ell_j^2})^{-1}\psi_j\\ &\ \ \ \ \ \ +(\hh_0+\frac{1}{\ell_j^2})^{-1} \big\{ (\sigma\cdot\bb -\Phi)\psi_j +[\pp_0,\psi_j]\big\} (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}. \endaligned \tag 4.16 Hence, by the diamagnetic inequality (2.10) and (2.16), we get \aligned &|\psi_j(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi|\\ &\le (-\Delta +\frac{1}{\ell_j^2})^{-1} |\psi_j\psi| +(-\Delta +\frac{1}{\ell_j^2})^{-1}|\bb| \psi_j |(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi|\\ & \ \ \ \ \ \ \ +(-\Delta +\frac{1}{\ell_j^2})^{-1} \big| (-\Phi\psi_j +[\pp_0, \psi_j]) (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi\big|\\ & \le C\ell_j^{\frac12}\|\psi\|_{L^2} +C\ell_j^{2\delta} \| (-\Delta +\frac{1}{\ell_j^2})^{\delta-\frac14} |\bb|\psi_j |(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi|\, \|_{L^2} \endaligned \tag 4.17 where $0<\delta <\min\big(1/4, (p-(3/2))/2\big)$ and we also used (4.15) in the second inequality. To deal with the second term in r.h.s. of (4.17), we apply (2.14) to obtain \aligned &\| (-\Delta +\frac{1}{\ell_j^2})^{\delta-\frac14} |\bb|\psi_j |(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi|\, \|_{L^2}\\ &\le \|(-\Delta +\frac{1}{\ell_j^2})^{\delta-\frac14} |\bb|^{-\delta +\frac14}\chi_{3Q_j}\|_{L^2\to L^2} \|\, |\bb|^{\delta +\frac34}\psi_j |(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi|\, \|_{L^2}\\ &\le C \| \, |\bb|^{-\delta +\frac14}\|_{L^{\frac{6}{1-4\delta}}(3Q_j)} \|\, |\bb|^{\delta +\frac34}\|_{L^2(3Q_j)} \|\psi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi \|_{L^\infty}\\ &\le C\ell_j^{2-2\delta-\frac{3}{p}} \left\{ \int_{3Q_j} |\bb|^pdx\right\}^{1/p} \| \psi_j(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi \|_{L^\infty}\\ &\le C\ell_j^{-2\delta}\varepsilon \|\psi_j(\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\psi\|_{L^\infty} \endaligned \tag 4.18 where we also used H\"older's inequality, $(3/2)+2\delta 3/2$, we may deduce that the function is in $L^q$ for any $q<\infty$. It then follows that it is in $L^\infty$ by the same argument. Finally (4.11) follows from (4.18) and (4.10). \enddemo We now are ready to estimate $n(1/3, I_2)$ and $n(1/3, I_3)$. \proclaim {\bf Lemma 4.3} There exists a constant $C>0$ such that $$\int_0^\infty n(1/3,I_2)\, d\lambda \le C\int_{\Bbb{R}^3}|V(x)|^{5/2}dx +C\int_{\Bbb{R}^3} \Phi(x)^{3/2} |V(x)|\, dx.$$ \endproclaim \demo{\bf Proof} We first note that, since $\| (\pp_0+\lambda)^{-1/2}\|_{L^2\to L^2} \le 1/\sqrt{\lambda}$, \align n(1/3,I_2) &=n\big (1/3,\sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Phi+\frac{1}{\ell_j^2}) \phi_j (\pp_0 +\lambda)^{-1/2}\big)\\ &\le n\big(\sqrt{\lambda}/3, \sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Phi+\frac{1}{\ell_j^2}) \phi_j\big)\\ &\le n\big(\sqrt{\lambda}/6, \sum_j Y\phi_j \big\{ (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} -(\pp_0 +\Phi +\frac{1}{\ell_j^2} )^{-1}\big\} (\Phi+\frac{1}{\ell_j^2}) \phi_j\big)\\ &\ \ \ \ \ \ + n\big(\sqrt{\lambda}/6, \sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\Phi+\frac{1}{\ell_j^2}) \phi_j\big)\\ & =K_1+K_2. \endalign Using $$(\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} -(\pp_0 +\Phi +\frac{1}{\ell_j^2} )^{-1} =-\lambda (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1}$$ and (2.4), we have \aligned K_1 &=n\big(\sqrt{\lambda}/6, \sum_j Y\phi_j \lambda (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Phi+\frac{1}{\ell_j^2}) \phi_j\big)\\ &\le C{\lambda^2} \| \sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Phi+\frac{1}{\ell_j^2}) \phi_j\|_4^4\\ &\le C\lambda^2 \sum_j \| Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Phi+\frac{1}{\ell_j^2}) \phi_j\|_4^4\\ &\le C\lambda^2 \sum_j \| Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} \phi_j\|_4^4\cdot\frac{1}{\ell_j^8} \endaligned where we have used the finite intersection property of the supports of $\phi_j$ as in the proof of Lemma 4.1. We now use (2.6) and the $(L^2, L^2)$ bound of $(\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-\alpha}$ ($\alpha =1, 1/2$) to obtain \align \| Y\phi_j (\pp_0 &+\Phi +\frac{1}{\ell_j^2})^{-1} (\pp_0 +\Phi +\frac{1}{\ell_j^2} +\lambda)^{-1} \phi_j\|_4\\ &\le \| Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\|_4 \cdot\frac{1}{\lambda +\frac{1}{\ell_j^2}}\\ &\le \| Y\phi_j (\pp_0 +\Phi+\frac{1}{\ell_j^2})^{-1/2}\|_4 \cdot\frac{\ell_j}{\lambda +\frac{1}{\ell_j^2}}\\ &\le \|Y\phi_j (\hh_0 +\frac{1}{\ell_j^2})^{-1/2}\|_4 \| (\hh_0 +\frac{1}{\ell_j^2})^{1/2} (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1/2}\|_{L^2\to L^2} \cdot\frac{\ell_j}{\lambda +\frac{1}{\ell_j^2}}\\ &\le C\|Y\phi_j (\hh_0 +\frac{1}{\ell_j^2})^{-1/2}\|_4 \cdot\frac{\ell_j}{\lambda +\frac{1}{\ell_j^2}}\\ &\le C\|Y\phi_j\|_{L^4} \cdot\frac{\ell_j^2}{\lambda^{5/8}} \endalign where we also used (3.8) and (2.12) in the last two inequality. It then follows that $$K_1\le \frac{C}{\sqrt{\lambda}} \sum_j \int_{\Bbb{R}^3} |Y\phi_j|^4dx \le \frac{C}{\sqrt{\lambda}}\int_{\{x\in \Bbb{R}^3:V(x)< -\lambda\} } |V|^2dx. \tag 4.20$$ It remains to estimate $K_2$. Since $Y\le |V|^{1/2}$, we have $$K_2 \le n\big(\sqrt{\lambda}/6, \sum_j |V|^{1/2}\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\Phi+\frac{1}{\ell_j^2})\phi_j\big).$$ It follows that \aligned \int_0^\infty K_2\, d\lambda &\le C\| \sum_j |V|^{1/2}\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} (\Phi+\frac{1}{\ell_j^2})\phi_j\|_2^2\\ &\le C\sum_j \|\, |V|^{1/2}\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} \phi_j\|_2^2\cdot\frac{1}{\ell_j^4}. \endaligned \tag 4.21 Using $\phi_j\psi_j\equiv \phi_j$, (4.16), and (4.9), we have \align &\|\, |V|^{1/2}\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1} \phi_j\|_2\\ & \le \|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-1} \phi_j\|_2 +\|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-1} (\sigma\cdot \bb)\psi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\phi_j\|_2\\ & +\|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-1} (-\Phi\psi_j +[\pp_0,\psi_j]) (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\phi_j\|_2\\ &\le C \|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-1}\|_2\\ &\ \ \ \ + \|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-\frac34 -\delta} \|_2 \|(\hh_0+\frac{1}{\ell_j^2})^{-\frac14+\delta} (\sigma\cdot \bb)\psi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2})^{-1}\phi_j\|_{L^2\to L^2}\\ &\le C \|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-1}\|_2 +C\ell_j^{\frac12-2\delta} \|\, |V|^{1/2}\phi_j (\hh_0+\frac{1}{\ell_j^2})^{-\frac34 -\delta} \|_2\\ &\le C\ell_j^{\frac12}\|\, |V|^{1/2}\phi_j\|_{L^2} \endalign where we also used (4.11) in the third and (2.11) in the last inequality. Inserting the estimate above into the r.h.s. of (4.21), we obtain $$\int_0^\infty K_2 \, d\lambda \le C\sum_j \frac{1}{\ell_j^3} \int_{\Bbb{R}^3} |V|\, \phi_j^2\, dx \le C\int_{\Bbb{R}^3} \Phi(x)^{3/2} |V(x)|dx.$$ This, together with (4.20), gives the desired estimate. \enddemo \proclaim{\bf Lemma 4.4} There exists a constant $C>0$ such that $$\int_0^\infty n(1/3, I_3)\, d\lambda \le C\int_{\Bbb{R}^3} |V(x)|^{5/2}dx +\int_{\Bbb{R}^3} \Phi(x)^{3/2} |V(x)|dx.$$ \endproclaim \demo{\bf Proof} It follows from (2.5) and (4.12) that \aligned &n(1/3, I_3) =n\big(1/3, \sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\pp_0,\phi_j](\pp_0+\lambda)^{-1/2}\big)\\ &\le n\big(1/6, \sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D}^*, [\Bbb{D},\phi_j]](\pp_0 +\lambda)^{-1/2}\big)\\ &\ \ \ \ \ + n\big(1/6, 2\sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} \left\{ [\Bbb{D}^*, \phi_j] \Bbb{D} +[\Bbb{D}, \phi_j] \Bbb{D}^*\right\} (\pp_0 +\lambda)^{-1/2}\big) \endaligned \tag 4.22 where $\Bbb{D}=\sigma\cdot (\bold{p}-\bold{A})$. Since $|\big[\Bbb{D}^*, [\Bbb{D},\phi_j]\big]\psi|\le |\nabla^2 \phi_j|\, |\psi|$, the first term on the r.h.s. of (4.22) can be treated exactly as $n(1/3, I_2)$. With $\| \Bbb{D} (\pp_0 +\lambda )^{-1/2}\|_{L^2\to L^2}\le 1$ and $\| \Bbb{D}^* (\pp_0 +\lambda )^{-1/2}\|_{L^2\to L^2}\le 1$, we may use the same argument as in the proof of Lemma 4.1 to bound the second term by \aligned C\| \sum_j Y\phi_j (\pp_0 +\Phi & +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D}, \phi_j]\|_4^4 +C\| \sum_j Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D}^*, \phi_j]\|_4^4\\ &\le C\sum_j \|Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1}\|_4^4 \cdot\frac{1}{\ell_j^4}\\ & \le C\sum_j \|Y\phi_j (\pp_0 +\Phi +\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_4^4\\ &\le C\sum_j \|Y\phi_j (\hh_0+\lambda)^{-1/2}\|_4^4\\ &\le\frac{C}{\sqrt{\lambda}} \int_{\{ x\in \Bbb{R}^3:V(x)< -\lambda\} } |V|^2dx. \endaligned The lemma then follows by integration. \enddemo We need one more lemma before we carry out the proof of Theorem 1.1. \proclaim{\bf Lemma 4.5} There exist two constants $C_1>0$, $C_2>0$ depending on $\varepsilon$ and $p$, such that, for every $x\in \Bbb{R}^3$, $$C_1 \Phi (x)\le b_p(x)\le C_2 \Phi(x).$$ \endproclaim \demo{\bf Proof} Suppose $x\in Q_j$ for some $j$. Then $\Phi (x)\approx 1/\ell_j^2$. Let $\delta\in (0,1)$ and $\ell_j=\ell(Q_j)$. Since $Q(x,\delta \ell_j)\subset 12 Q_j$, we have \align (\delta \ell_j)^2 \bigg(\frac{1}{(\delta\ell_j)^3} \int_{Q(x,\delta \ell_j)}& |\bb(y)|^pdy\bigg)^{1/p} = (\delta \ell_j)^{2-\frac{3}{p}} \left(\int_{Q(x,\delta \ell_j)} |\bb(y)|^pdy\right)^{1/p}\\ &\le (\delta \ell_j)^{2-\frac{3}{p}} \left(\int_{12 Q_j} |\bb(y)|^pdy\right)^{1/p}\\ &\le (\delta \ell_j)^{2-\frac{3}{p}} \cdot \frac{\varepsilon}{\ell_j^{2-\frac{3}{p}}} =\delta^{2-\frac{3}{p}} \varepsilon <1. \endalign Hence, by the definition of $\ell_p(x)$ (see (1.9)), we have $\ell_p(x)\ge \ell_j$. It follows that $$b_p(x)=\frac{1}{\{ \ell_p(x)\}^2 } \le\frac{1}{\ell_j^2}\le C_2 \Phi (x).$$ To show $b_p(x)\ge C_1 \Phi(x)$, we note that, if $N$ is large, $12 Q_j^+\subset Q(x,N\ell_j)$ where $Q_j^+$ is the dyadic parent of $Q_j$. Hence \align (N\ell_j)^2 \bigg(\frac{1}{(N\ell_j)^3} \int_{Q(x,N\ell_j)}& |\bb(y)|^pdy\bigg)^{1/p} = (N\ell_j)^{2-\frac{3}{p}} \left( \int_{Q(x,N\ell_j)} |\bb(y)|^pdy\right)^{1/p}\\ &\ge (N\ell_j)^{2-\frac{3}{p}} \left( \int_{12 Q_j^+} |\bb(y)|^pdy\right)^{1/p}\\ &> (N\ell_j)^{2-\frac{3}{p}}\cdot \frac{\varepsilon} {(2\ell_j)^{2-\frac{3}{p}}} =\varepsilon \left(\frac{N}{2}\right)^{2-\frac{3}{p}} >1 \endalign if $N$ is large enough. Again, by definition, this implies that $\ell_p(x)\le N\ell_j$. Therefore $$b_p(x)=\frac{1}{\{\ell_p(x)\}^2 } \ge \frac{1}{N^2\ell_j^2} \ge C_1 \Phi (x).$$ \enddemo Finally we are in a position to give the \demo{\bf Proof of Theorem 1.1} It suffices to prove the theorem for $\gamma=1$ (see \cite{20}). We may also assume, without the loss of generality, that $V\le 0$ a.e.. By means of (4.1) and (4.3), we have \align \Cal{M}_1&=\sum_{\lambda_j<0} |\lambda_j| =2\int_0^\infty N(2\lambda, \pp_0+V)\, d\lambda\\ &\le 2\int_0^\infty n(1, Y(\pp_0+\lambda)^{-1/2})\, d\lambda\\ &\le 2\sum_{j=1}^3 \int_0^\infty n(1/3, I_j)\, d \lambda. \endalign The theorem now follows from (4.8) and Lemmas 4.3--4.5. \enddemo \bigskip \centerline{\bf 5. Fields with Constant Directions} In this section we study the special case where the magnetic field $\bb$ has a constant direction. The goal is to prove Theorem 1.2 stated in the Introduction. Without the loss of generality, we may assume that $\bold{A}=(A_1(x_1,x_2), A_2(x_1,x_2),0)$ and $\bb=(0,0,B)$ where $B=\frac{\partial A_2}{\partial x_1} -\frac{\partial A_1}{\partial x_2}$. With this assumption, we have the following identity: $$\int_{\Bbb{R}^3} |\sigma\cdot (\bold{p}-\bold{A})\psi|^2dx =\int_{\Bbb{R}^3} | [\sigma_1(p_1-A_1)+\sigma_2(p_2-A_2)]\psi|^2dx +\int_{\Bbb{R}^3} |\frac{\partial \psi}{\partial x_3}|^2dx.$$ This in particular implies that, for $\lambda>0$, $$\| (-\partial_{x_3}^2 +\lambda)^{1/2} (\pp+\lambda)^{-1/2}\|_{L^2\to L^2}\le 1. \tag 5.1$$ We shall use $x^\prime =(x_1,x_2)$, $y^\prime=(y_1,y_2)$ to denote points in $\Bbb{R}^2$. Our approach to the case of constant direction fields is similar to that in Section 4 for arbitrary fields. We begin by observing that, if $\bb=(0,0,B(x^\prime))$, the basic length scale $\ell_p(x)$ defined by (1.9) is reduced to $$\ell_p(x^\prime) =\sup \left\{ \ell>0:\ \ \ell^2 \left\{ \frac{1}{\ell^2}\int_{S(x^\prime,\ell)} |B(y^\prime)|^p dy^\prime\right\}^{1/p}\le 1 \right\} \tag 5.2$$ where $S(x^\prime,\ell)$ denotes the square in $\Bbb{R}^2$ centered at $x^\prime$ with side length $\ell$. We will assume that $B\not\equiv 0$ and $B\in L^p_{loc}(\Bbb{R}^2)$ for some $p>1$. With this we have $0<\ell_p(x^\prime)<\infty$ for any $x^\prime\in \Bbb{R}^2$. We now sketch the construction of the partition of unity for $\Bbb{R}^2$ associated with $B$, which is parallel to that in Section 3. First we write $$\Bbb{R}^2 =\bigcup_j S_j \tag 5.3$$ where $\{ S_j\}_{j=1}^\infty$ are maximal elements in the set of all dyadic squares $S$ in $\Bbb{R}^2$ such that $$\left\{ \int_{12 S} |B(x^\prime)|^pd x^\prime\right\}^{1/p} \le \frac{\varepsilon} {[\ell(S)]^{2-\frac{2}{p}}} \tag 5.4$$ where $\varepsilon\in (0,1)$ is a constant to be determined later. Next we use the same argument as in the proof of Lemma 3.2 to show that, $$\frac12\ell(S_k) \le \ell(S_j) \le 2\ell(S_k), \ \ \ \ \text{ if } \ \ 4S_j\cap 4S_k\neq\emptyset. \tag 5.5$$ It follows from (5.5) that there exists a sequence of functions $\{ \varphi_j\}$ such that \align &\varphi_j\in C_0^\infty (2S_j, \Bbb{R})\ \ \ \text{ and }\ \ 0\le \varphi_j\le 1, \tag 5.6\\ & |\nabla^\alpha \varphi_j|\le C_\alpha/\ell_j^{|\alpha|} \ \ \ \text{ where }\ \ \ell_j=\ell(S_j), \tag 5.7\\ & \sum_j\varphi_j^2\equiv 1\ \ \ \text{ in } \ \Bbb{R}^2. \tag 5.8 \endalign Let $$\Psi =\sum_j\frac{1}{\ell_j^2}\, \varphi_j^2. \tag 5.9$$ \proclaim{\bf Theorem 5.1} There exist constants $C>0$ and $\varepsilon_0>0$ such that, if $\varepsilon$ in (5.4) is less than $\varepsilon_0$, then $$\hh_0\le C\left\{ \pp_0 +\Psi\right\}.$$ \endproclaim \demo{\bf Proof} We will only show that, if $\varepsilon$ in (5.4) is small, then $$\int_{\Bbb{R}^3} |\, |B|^{1/2} \varphi_j\psi|^2dx \le C\int_{\Bbb{R}^3} |\sigma\cdot (\bold{p}-\bold{A})(\varphi_j\psi)|^2dx \tag 5.10$$ for any $\psi\in C^\infty_0(\Bbb{R}^3,\Bbb{C}^2)$. The rest of the proof is exactly the same as that of Theorem 3.1. To show (5.10), we need the following Sobolev inequality $$\left(\int_S |f|^qdx^\prime\right)^{1/q} \le C_q |S|^{1/q} \left( \int_S |\nabla_2 f|^2dx^\prime\right)^{1/2} \tag 5.11$$ for $f\in C_0^\infty (S,\Bbb{R})$ where $S$ is a square, $\nabla_2 =(\partial_{x_1},\partial_{x_2})$, and $10$ such that, for any $\lambda>0$, $$n(1/3, J_1) \le \frac{C}{\sqrt{\lambda}} \int_{\{ x\in \Bbb{R}^3: V(x)<-\lambda\} } |V(x)|^2dx.$$ \endproclaim The proof of Lemma 5.1, which uses (5.12)--(5.13) and (2.12), is similar to that of Lemma 4.1. We leave it to the reader. Let $\xi_j\in C_0^\infty (3S_j,\Bbb{R})$ such that $0\le \xi_j\le 1$, $\xi_j\varphi_j\equiv \varphi_j$, and $|\nabla^\alpha\xi_j|\le C_\alpha/\ell_j^{|\alpha|}$. The following lemma is needed to estimate $n(1/3,J_2)$ and $n(1/3,J_3)$. \proclaim{\bf Lemma 5.2} Let $0<\delta \le \frac12 (1-\frac1p)$. There exists a constant $C_\delta>0$ such that, for any $\lambda >0$, $$\| (\hh_0 +\frac{1}{\ell_j^2} +\lambda)^{-\frac12 +\delta} (\sigma\cdot \bb)\xi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} \|_{L^2\to L^2} \le C_\delta \, \ell_j^{1-2\delta}.$$ \endproclaim \demo{\bf Proof} Note that, by (5.12), \aligned \| (\hh_0 +&\frac{1}{\ell_j^2} +\lambda)^{-\frac12 +\delta} (\sigma\cdot \bb)\xi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} \|_{L^2\to L^2}\\ &\le \| (\hh_0 +\frac{1}{\ell_j^2} +\lambda)^{-\frac12 +\delta} (\sigma\cdot \bb)\xi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1/2} \|_{L^2\to L^2}\cdot \ell_j\\ &\le \| (\hh_0 +\frac{1}{\ell_j^2} +\lambda)^{-\frac12 +\delta} (\sigma\cdot \bb)\xi_j (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-1/2} \|_{L^2\to L^2}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \cdot \|(\hh_0+\frac{1}{\ell_j^2}+\lambda)^{1/2} (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1/2} \|_{L^2\to L^2}\cdot \ell_j\\ &\le C\ell_j \| (\hh_0 +\frac{1}{\ell_j^2} +\lambda)^{-\frac12 +\delta} (\sigma\cdot \bb)\xi_j (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-\frac12} \|_{L^2\to L^2}\\ &\le C\ell_j^{1+2\delta} \| (\hh_0 +\frac{1}{\ell_j^2} +\lambda)^{-\frac12 +\delta} (\sigma\cdot \bb)\xi_j (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-\frac12 +\delta} \|_{L^2\to L^2}. \endaligned Thus, by the diamagnetic inequality (2.10) and duality, it suffices to show that $$\|\, |B|^{1/2}\chi_{3S_j} (-\Delta +\frac{1}{\ell_j^2} +\lambda )^{-\frac12 +\delta} \|_{L^2\to L^2} \le C_\delta\, \ell_j^{-2\delta}. \tag 5.15$$ We claim that (5.15) follows from Lemma 2.3. Indeed, by Lemma 2.3, the l.h.s. of (5.15) is bounded by $$C_\delta \, \|\, |B|^{1/2}\|_{L^{\frac{2}{1-2\delta}}(3S_j)} \le C_\delta \ell_j^{-2\delta}$$ where we used $p\ge 1/(1-2\delta)$, H\"older's inequality, and (5.4) in the inequality. The proof is complete. \enddemo We are now ready to deal with $n(1/3,J_2)$ and $n(1/3,J_3)$. \proclaim{\bf Lemma 5.3} There exists a constant $C>0$ such that $$n(1/3, J_2) \le \frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)<-\lambda\} } \Psi (x) |V(x)|dx.$$ \endproclaim \demo{\bf Proof} First we note that, by (2.4), (2.6), and (5.1), \aligned n(1/3, J_2) &=n\big(1/3, \sum_j Y\varphi_j (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Psi +\frac{1}{\ell_j^2})\varphi_j (\pp_0+\lambda)^{-1/2}\big)\\ &\le C \| \sum_j Y\varphi_j (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Psi +\frac{1}{\ell_j^2})\varphi_j (\pp_0+\lambda)^{-1/2} \|_2^2\\ &\le C \| \sum_j Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Psi +\frac{1}{\ell_j^2})\varphi_j \|_2^2\\ &\ \ \ \ \ \ \ \ \cdot \| (-\partial_{x_3}^2+\lambda)^{1/2} (\pp_0+\lambda)^{-1/2}\|^2_{L^2\to L^2}\\ &\le C \|\sum_j Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Psi +\frac{1}{\ell_j^2})\varphi_j \|_2^2\\ &\le C \sum_j \|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} (\Psi +\frac{1}{\ell_j^2})\varphi_j \|_2^2\\ &\le C \sum_j \|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1}\varphi_j \|_2^2\cdot\frac{1}{\ell_j^4} \endaligned where we used the finite intersection property of supp$\varphi_j$ in the fourth inequality. The fact that $\pp_0+\Psi$ commutes with $\partial_{x_3}$ is essential in the estimate above. Next we use the resolvent identity (4.2) to obtain \aligned &\|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1}\varphi_j \|_2\\ &= \|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} \xi_j(\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1}\varphi_j \|_2\\ &\le \|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\hh_0+\frac{1}{\ell_j^2} +\lambda)^{-1}\varphi_j \|_2\\ &\ \ \ \ \ +\|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\hh_0+\frac{1}{\ell_j^2} +\lambda)^{-1}(\sigma\cdot \bb)\xi_j (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1}\varphi_j \|_2\\ &\ \ \ \ \ +\|Y\varphi_j(-\partial_{x_3}^2+\lambda)^{-1/2} (\hh_0+\frac{1}{\ell_j^2} +\lambda)^{-1} \left\{ -\Psi\xi_j +[\pp_0,\xi_j]\right\} (\pp_0 +\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1}\varphi_j \|_2\\ &=K_1+K_2+K_3. \endaligned By (2.13), $$K_1\le C\lambda^{-1/4} (\frac{1}{\ell_j^2}+\lambda)^{-1/2} \|Y\varphi_j\|_{L^2} \le C\lambda^{-1/4}\ell_j \|Y\varphi_j\|_{L^2}. \tag 5.16$$ To estimate $K_2$, we use (2.6), Lemma 5.2, and (2.13) to obtain \aligned K_2&\le \|Y\varphi_j (-\partial_{x_3}^2+\lambda)^{-1/2} (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-\frac12 -\delta}\|_2\\ &\ \ \ \ \ \ \ \ \cdot \| (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-\frac12 +\delta} (\sigma\cdot\bb )\xi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1}\|_{L^2\to L^2}\\ &\le C\, \|Y\varphi_j (-\partial_{x_3}^2+\lambda)^{-1/2} (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-\frac12 -\delta}\|_2 \cdot \ell_j^{1-2\delta}\\ &\le C\, \lambda^{-1/4} (\frac{1}{\ell_j^2}+\lambda)^{-\delta} \|Y\varphi_j\|_{L^2}\cdot \ell_j^{1-2\delta}\\ &\le C\, \lambda^{-1/4}\ell_j \|Y\varphi_j\|_{L^2} \endaligned \tag 5.17 where $\delta=\frac12 (1-\frac1p)$. To bound $K_3$, we observe that $$\| \left\{ -\Psi\xi_j +[\pp_0,\xi_j]\right\} (\pp_0+\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} \|_{L^2\to L^2}\le C$$ using the same argument as in the proof of (4.9). It follows that \aligned K_3 &\le \| Y\varphi_j (-\partial_{x_3}^2 +\lambda)^{-1/2} (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-1}\|_2\\ &\ \ \ \ \ \ \ \ \ \ \cdot \| \left\{ -\Psi\xi_j +[\pp_0,\xi_j]\right\} (\pp_0+\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} \|_{L^2\to L^2}\\ &\le C\lambda^{-1/4}\ell_j \|Y\varphi_j\|_{L^2}. \endaligned \tag 5.18 Finally, we add (5.16), (5.17) and (5.18) to conclude that $$\|Y\varphi_j (-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0+\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} \varphi_j \|_2 \le K_1+K_2 +K_3 \le C\lambda^{-1/4}\ell_j \|Y\varphi_j\|_{L^2}.$$ It follows that \aligned n(1/3,J_2) &\le C\sum_j \|Y\varphi_j (-\partial_{x_3}^2+\lambda)^{-1/2} (\pp_0+\Psi +\frac{1}{\ell_j^2} +\lambda)^{-1} \varphi_j \|_2^2\cdot\frac{1}{\ell_j^4}\\ &\le\frac{C}{\sqrt{\lambda}} \sum_j \frac{1}{\ell_j^2} \int_{\Bbb{R}^3} |Y\varphi_j|^2dx\\ &\le \frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } \Psi(x) |V(x)|dx \endaligned since $|Y|\le |V|^{1/2}\chi_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} }$. The proof is finished. \enddemo \proclaim{\bf Lemma 5.4} There exists a constant $C>0$ such that $$n(1/3,J_3) \le\frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } |V(x)|^2dx + \frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } \Psi(x) |V(x)|dx.$$ \endproclaim \demo{\bf Proof} Using $[\pp_0,\varphi_j]=\big[\Bbb{D}^*,[\Bbb{D},\varphi_j]\big] +[\Bbb{D},\varphi_j]\Bbb{D}^* +[\Bbb{D}^*,\varphi_j]\Bbb{D}$, we have \aligned J_3&= \sum_jY\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\pp_0,\varphi_j] (\pp_0 +\lambda)^{-1/2}\\ &= \sum_jY\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} \big[\Bbb{D}^*,[\Bbb{D},\varphi_j]\big] (\pp_0 +\lambda)^{-1/2}\\ &\ \ \ \ +\sum_jY\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} \left\{ [\Bbb{D},\varphi_j] \Bbb{D}^* + [\Bbb{D}^*,\varphi_j] \Bbb{D} \right\} (\pp_0 +\lambda)^{-1/2}\\ &=J_{31} +J_{32}. \endaligned Thus, $$n(1/3,J_3)\le n(1/6, J_{31}) +n(1/6, J_{32}).$$ Since $| [\Bbb{D}^*,[\Bbb{D},\varphi_j]]\psi|\le |\nabla^2 \varphi_j|\, |\psi|$, we may use the same argument as in the proof of Lemma 5.3 to show that $$n(1/6, J_{31}) \le\frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } \Psi (x) |V(x)|dx.$$ On the other hand, using $\| \Bbb{D} (\pp_0+\lambda)^{-1/2} \|_{L^2\to L^2}\le 1$, $\| \Bbb{D}^* (\pp_0+\lambda)^{-1/2} \|_{L^2\to L^2}\le 1$, and (5.12), we may bound $n(1/6, J_{32})$ by \aligned &C\| \sum_jY\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} \left\{ [\Bbb{D}^*,\varphi_j] \Bbb{D} + [\Bbb{D},\varphi_j] \Bbb{D}^*\right\} (\pp_0 +\lambda)^{-1/2}\|_4^4\\ &\le C\| \sum_jY\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D}^*,\varphi_j]\|_4^4 + C\| \sum_jY\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D},\varphi_j]\|_4^4\\ &\le C \sum_j\|Y\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D}^*,\varphi_j]\|_4^4 + C \sum_j\|Y\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1} [\Bbb{D},\varphi_j]\|_4^4\\ &\le C \sum_j\|Y\varphi_j (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_4^4 \cdot\frac{1}{(\frac{1}{\ell_j^2}+\lambda)^2} \cdot \frac{1}{\ell_j^4}\\ &\le C \sum_j\|Y\varphi_j (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_4^4 \| (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{1/2} (\pp_0+\Psi +\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_{L^2\to L^2}^4\\ &\le C \cdot \sum_j\|Y\varphi_j (\hh_0+\frac{1}{\ell_j^2}+\lambda)^{-1/2}\|_4^4\\ &\le \frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } |V(x)|^2dx \endaligned where (2.12) is used in the last inequality. The proof is complete. Finally we are in a position to give the \demo{\bf Proof of Theorem 1.2} It follows from (5.14), Lemmas 5.1, 5.3, and 5.4 that \aligned N(2\lambda, \pp_0 +V) &\le n(1/3, J_1) +n(1/3, J_2) +n(1/3, J_3)\\ &\le \frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } |V(x)|^2dx +\frac{C}{\sqrt{\lambda}} \int_{\{ x\in\Bbb{R}^3: V(x)< -\lambda\} } \Psi (x)|V(x)|dx. \endaligned Thus, if $\gamma>1/2$, \aligned \Cal{M}_\gamma &=\sum_{\lambda_j<0} |\lambda|^\gamma =\gamma \int_0^\infty \lambda^{\gamma -1} N(\lambda,\pp_0+V)\, d\lambda\\ &= \gamma 2^\gamma \int_0^\infty \lambda^{\gamma -1} N(2\lambda,\pp_0+V)\, d\lambda\\ &\le C_\gamma \int_{\Bbb{R}^3} |V(x)|_-^{\gamma+3/2}dx +C_\gamma \int_0^\infty \Psi(x) |V(x)|_-^{\gamma+1/2}dx. \endaligned Noting that $\Psi(x)\approx b_p(x)$ for $p>1$ by an argument similar to that in the proof of Lemma 4.5, we are done. \enddemo \Refs \widestnumber\key{20} \ref\key 1 \by J.~Avron, I.~Herbst, and B.~Simon \paper Schr\"odinger Operators with Magnetic Fields. 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