Content-Type: multipart/mixed; boundary="-------------9905251508959" This is a multi-part message in MIME format. ---------------9905251508959 Content-Type: text/plain; name="99-193.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="99-193.keywords" Schrodinger Operators, Fundamental Solutions ---------------9905251508959 Content-Type: application/x-tex; name="s10.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="s10.tex" \documentstyle{amsppt} \loadeufm \magnification=\magstep 1 \baselineskip=18pt \NoBlackBoxes \parskip=5pt \define\r{{\Bbb{R}^n}} \define\loc{{\text{loc}}} \define\e{{\varepsilon}} \centerline {\bf On Fundamental Solutions of Generalized Schr\"odinger Operators} \medskip\medskip \centerline{\bf Zhongwei Shen \footnote{Supported in part by the AMS Centennial Research Fellowship and the NSF grant DMS-9732894}} \centerline{ Department of Mathematics} \centerline{University of Kentucky} \centerline{Lexington, KY 40506 } \centerline{U. S. A.} \centerline{E-mail: shenz\@ms.uky.edu} \noindent{\bf Abstract.}\ \ We consider the generalized Schr\"odinger operator $-\Delta +\mu$ where $\mu$ is a nonnegative Radon measure in $\Bbb{R}^n$, $n\ge 3$. Assuming that $\mu$ satisfies certain scale-invariant Kato condition and doubling condition, we establish the following bounds for the fundamental solution of $-\Delta +\mu$ in $\Bbb{R}^n$: $$ \frac{c\, e^{-\varepsilon_2 d(x,y,\mu)}} {|x-y|^{n-2}} \le \Gamma_\mu (x,y) \le \frac{C\, e^{-\varepsilon_1 d(x,y,\mu)}} {|x-y|^{n-2}} $$ where $d(x,y,\mu)$ is the distance function for the modified Agmon metric $m(x,\mu)dx^2$ associated with $\mu$. We also study the boundedness of the corresponding Riesz transforms $\nabla (-\Delta +\mu)^{-1/2}$ on $L^p(\Bbb{R}^n,dx)$. \noindent{\bf Keywords.} Schr\"odinger Operators, Fundamental Solutions, Riesz Transforms. \noindent{\bf 1991 Mathematics Subject Classification.} 42B20, 35J10. \bigskip \centerline{\bf Introduction} Consider the generalized Schr\"odinger operator $$ -\Delta +\mu \ \ \ \ \ \ \ \ \ \ \ \ \ \text{in } \ \Bbb{R}^n,\ n\ge 3 \tag 0.1 $$ where $\mu$ is a nonnegative Radon measure on $\Bbb{R}^n$. The main purpose of this paper is to establish optimal upper and lower bounds for the fundamental solution of $-\Delta +\mu$ under suitable conditions on the measure $\mu$. We will also study the boundedness of the operators $(-\Delta +\mu)^{i\gamma}$ $(\gamma\in \Bbb{R}$), and $\nabla(-\Delta +\mu) ^{-1/2}$ on $L^p(\Bbb{R}^n,dx)$. Throughout this paper we assume that $\mu$ satisfies the following conditions: there exist positive constants $C_0$, $C_1$ and $\delta$ such that $$ \mu(B(x,r))\le C_0\left(\frac{r}{R}\right)^{n-2+\delta} \mu(B(x,R)), \tag 0.2 $$ $$ \mu(B(x,2r))\le C_1\left\{ \mu (B(x,r))+r^{n-2}\right\} \tag 0.3 $$ for all $x\in \Bbb{R}^n$ and $00: \ \ \frac{\mu(B(x,r))}{r^{n-2}}\le C_1\right\} \tag 0.5 $$ where $C_1$ is the constant in (0.3) (see \cite{11}). With the modified Agmon metric $$ ds^2=m(x,\mu)\{ dx_1^2+\cdots +dx_n^2\}, \tag 0.6 $$ we define the distance function $$ d(x,y,\mu)=\inf_\gamma \int_0^1 m(\gamma(t),\mu) |\gamma^\prime(t)|dt \tag 0.7 $$ where $\gamma: [0,1]\to\r$ is absolute continuous and $\gamma(0)=x$, $\gamma(1)=y$ (see \cite{8, 12}). The following is one of the main results of the paper. \proclaim{\bf Theorem 0.8} Let $\mu$ be a nonnegative Radon measure in $\Bbb{R}^n, \ n\ge 3.$ Assume that $\mu$ satisfies the conditions (0.2)-(0.3). Let $\Gamma_\mu (x,y)$ denote the fundamental solution of $-\Delta +\mu$ in $\Bbb{R}^n$. Then we have $$ \frac{c\, e^{-\varepsilon_2 d(x,y,\mu)}}{|x-y|^{n-2}} \le \Gamma_\mu (x,y) \le \frac{C\, e^{-\varepsilon_1 d(x,y,\mu)}}{|x-y|^{n-2}} \tag 0.9 $$ where $C$, $c$, $\varepsilon_1$ and $\varepsilon_2$ are positive constants depending only on $n$ and constants $C_0$, $C_1$ and $\delta$ in (0.2)-(0.3). \endproclaim A few remarks are in order. \remark{\bf Remark 0.10} If $d\mu= V(x)dx$ and $V\ge 0$ is in the reverse H\"older class $(RH)_{n/2}$, i.e., $$ \left\{ \frac{1}{|B(x,r)|} \int_{B(x,r)} V(y)^{n/2}dy\right\}^{2/n} \le C\left\{ \frac{1}{|B(x,r)|} \int_{B(x,r)} V(y) dy\right\}, \tag 0.11 $$ then $\mu$ satisfies the conditions (0.2)-(0.3) for some $\delta>0$ \cite{11}. However, in general, measures which satisfy (0.2)-(0.3) need not to be absolute continuous with respect to the Lebesgue measure on $\Bbb{R}^n$. Indeed, if $ d\mu=d\sigma(x_1,x_2)dx_3\cdots dx_4$ where $\sigma$ is a doubling measure on $\Bbb{R}^2$, then $\mu$ satisfies (0.2)-(0.3) for some $\delta >0$. Also, if $\sigma$ is the surface measure on a Lipschitz graph $S=\left\{ (x^\prime, \varphi(x^\prime))\in \Bbb{R}^{n}: x^\prime\in \Bbb{R}^{n-1}\right\}$ and $\mu(E)\equiv\sigma (E\cap S)$, then $\mu$ satisfies (0.2)-(0.3) with $\delta=1$. \endremark \remark{\bf Remark 0.12} In the case $n=2$, it is possible to establish an essentially optimal upper bound of $\Gamma_\mu (x,y)$ under the assumptions (0.2)-(0.3). This was pointed out by M.~Christ in \cite{4}, who derived upper bounds for certain distribution kernels associated with the $\overline{\partial}$ equation in a weighted $L^2$ space. Our study of $-\Delta +\mu$ with conditions (0.2)-(0.3) is partially motivated by \cite{4}. In the case $n\ge 3$ and $d\mu=V(x)dx$, certain power-decay estimate of $\Gamma_\mu(x,y)$ was obtained by the author in \cite{11} under the condition $V\in (RH)_{n/2}$. It played an important role in the study of the $L^p$ boundedness of various operators associated with $-\Delta+V(x)$. Also see \cite{13} for the case of magnetic Schr\"odinger operator $H(\bold{a},V)=-(\nabla -i\bold{a})^2 +V$, and \cite{10} for some results on the heat kernel of $H(\bold{a},V)$. \endremark \remark{\bf Remark 0.13} It is easy to see that (0.2) implies (0.4). To see they are in fact equivalent, let $$ \phi(r)=\frac{\mu (B(x,r))}{r^{n-2}}. $$ By Fubini's Theorem, (0.4) implies $$ \int_0^R \phi(r)\, \frac{dr}{r}\le C\, \phi(R). $$ It follows that $$ \aligned \phi(R) &\ge c\int_0^R \phi(r)\, \frac{dr}{r}\ge c_1 \int_0^R\left\{ \int_0^r \phi(t)\frac{dt}{t}\right\} \frac{dr}{r}\\ &=c_1\int_0^R \phi(t)\ln \left(\frac{R}{t}\right)\frac{dt}{t}\\ &\ge c_1 \int_{r/2}^r \phi(t)\ln \left(\frac{R}{t}\right) \frac{dt}{t}\\ &\ge c_2\, \phi(r/2)\ln\left(\frac{R}{r}\right) \endaligned $$ where $01$ , it is possible to bound the first derivatives of $\Gamma_\mu(x,y)$ pointwise. \proclaim{\bf Theorem 0.17} Let $\mu$ be a nonnegative Radon measure in $\Bbb{R}^n$, $n\ge 3$. Assume that $\mu$ satisfies the conditions (0.2)-(0.3) for some $\delta>1$. Then there exist constants $C>0$, $\varepsilon>0$ such that $$ |\nabla_x \Gamma_\mu(x,y)| \le \frac{C\, e^{-\varepsilon d(x,y,\mu)}}{|x-y|^{n-1}}. \tag 0.18 $$ \endproclaim With the estimates on the fundamental solution at our disposal, in the second part of this paper, we study the boundedness of the operator $(-\Delta +\mu)^{i\gamma}$ ($\gamma\in \Bbb{R}$) and the Riesz transforms $\nabla (-\Delta +\mu)^{-1/2}$ on $L^p(\Bbb{R}^n,dx)$. Our basic argument will be similar to that in \cite {11}, where we assumed that $d\mu =V(x)dx$ and $V\in (RH)_{n/2}$. \proclaim{\bf Theorem 0.19} Let $\mu$ be a nonnegative Radon measure in $\Bbb{R}^n$, $n\ge 3$. Suppose that $\mu$ satisfies the conditions (0.2)-(0.3) for some $\delta>0$. Then, for $\gamma\in \Bbb{R}$, $(-\Delta+\mu)^{i\gamma}$ is a Calder\'on-Zygmund operator. \endproclaim \proclaim{\bf Theorem 0.20} Let $\mu$ be a nonnegative Radon measure in $\Bbb{R}^n$, $n\ge 3$. Suppose that $\mu$ satisfies (0.2)-(0.3) for some $\delta\in (0,1)$. Then $$ \| \nabla (-\Delta +\mu)^{-1/2}f\|_p \le C\, \|f\|_p \tag 0.21 $$ for $11$, then $\nabla(-\Delta +\mu)^{-1/2}$ and $\nabla(-\Delta +\mu)^{-1}\nabla$ are Calder\'on-Zygmund operators. \endproclaim In (0.21) and thereafter, $\|\cdot\|_p$ denotes the norm in $L^p(\Bbb{R}^n,dx)$. \remark{\bf Remark 0.22} We recall that an operator $T$ taking $C_c^\infty(\Bbb{R}^n)$ into $L^1_{\loc}(\Bbb{R}^n, dx)$ is called a Calder\'on-Zygmund operator if (a) $T$ extends to a bounded operator on $L^2(\Bbb{R}^n,dx)$, (b) there exists a kernel $K$ such that, for every $f\in L^\infty_c(\Bbb{R}^n, dx)$, $$ Tf(x)=\int_{\Bbb{R}^n} K(x,y) f(y)dy \ \ \ \ \text{a.e. on } \ \ \{\text{supp} f\}^c, \tag 0.23 $$ and, (c) the kernel $K$ satisfies $$ \left\{ \aligned |K(x,y)| &\le \frac{C}{|x-y|^n},\\ |K(x+h,y)-K(x,y)| & \le \frac{C\, |h|^\delta}{|x-y|^{n+\delta}},\\ |K(x,y+h)-K(x,y)| & \le \frac{C\, |h|^\delta}{|x-y|^{n+\delta}} \endaligned \right. \tag 0.24 $$ for $x$, $y$, $h \in \Bbb{R}^n$ with $|h|\le |x-y|/2$, and some $\delta >0$. It is well known that Calder\'on-Zygmund operators are bounded on $L^p(\Bbb{R}^n,dx)$ for $1 dx +\int_{\Bbb{R}^n} <\phi,\psi> d\mu \tag 1.5 $$ with domain $W^{1,2}(\Bbb{R}^n)\cap L^2(\Bbb{R}^n, d\mu)$. Clearly, $q [\, , \, ]$ is a semibound, symmetric closed form. Hence there exists a unique self-adjoint operator, which we call $-\Delta +\mu$, such that $$ q[\phi,\psi] =<(-\Delta +\mu)\phi, \psi>_{L^2(\Bbb{R}^n, dx)} $$ for any $\phi\in \text{Domain} (-\Delta +\mu)$ and $\psi \in W^{1,2}(\Bbb{R}^n)\cap L^2(\Bbb{R}^n, d\mu)$. Moreover, $$ \aligned & \text{Domain} (-\Delta +\mu)\\ & =\left\{ \phi\in W^{1,2}(\Bbb{R}^n)\cap L^2(\Bbb{R}^n, d\mu): \ \ (-\Delta +\mu)\phi \in L^2(\Bbb{R}^n, dx)\right\}. \endaligned \tag 1.6 $$ \proclaim{\bf Proposition 1.7} Suppose $\mu$ satisfies (0.2). Then $C^\infty_0 (\Bbb{R}^n)$ is form core for $q[\, , \, ]$, i.e., $C_0^\infty (\Bbb{R}^n)$ is dense in $W^{1,2}(\Bbb{R}^n)\cap L^2(\Bbb{R}^n, d\mu)$. \endproclaim \demo{Proof} Given $f\in W^{1,2}(\Bbb{R}^n)\cap L^2(\Bbb{R}^n, d\mu)$, we need to approximate it by functions in $C_0^\infty(\r)$. By multiplying a smooth cut-off function to $f$, we may assume that $f$ has compact support. Now suppose that supp$(f) \subset B$. Then $f\in W_0^{1,2}(B)$ and there exists a sequence $\{ f_j\} \subset C_0^\infty (B)$ such that $f_j\to f $ in $W_0^{1,2}(B)$. Since $W_0^{1,2}(B)\subset L^2(B,d\mu)$, we also have $f_j\to f $ in $L^2(B,d\mu)$. Thus $f_j\to f$ in $W^{1,2}(\Bbb{R}^n)\cap L^2(\Bbb{R}^n, d\mu)$. \enddemo Finally in this section we give the proof of the inequality (0.16). We will need some properties of the auxiliary function $m(x,\mu)$ defined by (0.5). \proclaim{\bf Proposition 1.8} Suppose $\mu$ satisfies (0.2)-(0.3). Then (a) \ \ $00$ such that, for $x$, $y\in \Bbb{R}^n$, $$ m(x,\mu)\le C m(y,\mu) \left\{ 1+ |x-y|m(y,\mu)\right\}^{k_0}. $$ \endproclaim \demo{Proof} Note that, by (0.2), $$ \lim_{r\to 0} \frac{\mu(B(x,r))}{r^{n-2}}=0 \ \ \text {and }\ \ \lim_{r\to \infty} \frac{\mu(B(x,r))}{r^{n-2}}=\infty. $$ Part (a) follows by definition. To see part (b), let $r=1/m(x,\mu)$. By definition, $$ \mu(B(x,r))=\lim_{t\to r^-} \mu(B(x,t)) \le C_1 r^{n-2}. $$ Also, by definition, $\mu (B(x,2r))> C_1(2r)^{n-2}$. It follows from (0.3) that $$ C_1 (2r)^{n-2}\le C_1 \left\{ \mu(B(x,r))+r^{n-2}\right\}. $$ This gives $\mu(B(x,r))\ge r^{n-2}$. To prove part (c), we again let $r=1/m(x,\mu)$. Suppose that $|x-y|0$. \enddemo \remark{\bf Remark 1.9} It follows easily from parts (c) and (d) of Proposition 1.8 that, for $x$, $y\in \Bbb{R}^n$, $$ 1+|x-y|m(x,\mu) \le C\left\{ 1+|x-y|m(y,\mu)\right\}^{k_0+1} \tag 1.10 $$ and $$ m(y,\mu )\ge \frac{c\, m(x,\mu)} {\left\{ 1+|x-y| m(x,\mu)\right\}^{\frac{k_0}{k_0+1}}}. \tag 1.11 $$ Also, by (0.3) and part (b) of Proposition 1.8, there exists $k_1>0$ such that $$ \frac{\mu(B(x,R))}{R^{n-2}} \le C \left\{ 1+R\, m(x,\mu)\right\}^{k_1} \tag 1.12 $$ for all $x\in \Bbb{R}^n$ and $R>0$. \endremark We are now ready to prove (0.16). \proclaim{\bf Theorem 1.13} Suppose that $\mu$ satisfies (0.2)-(0.3). Assume that $u\in W^{1,2}_{\loc} (\Bbb{R}^n)$ and $\nabla u\in L^2(\Bbb{R}^n, dx)$. (a) If $u\in L^2(\Bbb{R}^n, d\mu)$, then $m(\cdot, \mu)u\in L^2(\r, dx)$ and $$ \int_\r |u(x)|^2 m(x,\mu)^2 dx \le C\left\{ \int_\r |\nabla u|^2 dx + \int_\r |u|^2 d\mu\right\}. $$ (b) If $m(\cdot,\mu) u\in L^2 (\r, dx )$, then $u\in L^2 (\r, d\mu)$ and $$ \int_\r |u|^2 d\mu \le C \left\{ \int_\r |\nabla u|^2dx + \int_\r |u(x)|^2 m(x,\mu)^2 dx \right\}. $$ \endproclaim \demo{Proof} We will only prove part (a). Part (b) may be shown in the same manner. Let $B=B(y,R)$ where $R=1/m(y,\mu)$. It follows from Poincar\'e inequality (0.15) that $$ \mu(B)\int_B |u(x)|^2dx \le C R^2 \mu(3B) \int_B |\nabla u(x)|^2dx +2|B|\int_B |u(x)|^2 d\mu(x). \tag 1.14 $$ Note that, by part (b) of Proposition 1.8, $$ \mu(B)\ge R^{n-2}\ \ \ \text{and }\ \ \mu(3B)\le C\left\{ \mu(B)+R^{n-2}\right\} \le CR^{n-2}. $$ Hence, (1.14) gives $$ \frac{1}{R^{n+2}} \int_B |u(x)|^2dx \le \frac{C}{R^n} \int_B |\nabla u(x)|^2 dx +\frac{C}{R^n} \int_B |u(x)|^2 d\mu(x). \tag 1.15 $$ Recall that $B=B(y,R)$ and $R=1/m(y,\mu)$. We integrate both sides of (1.15) with respect to $y$ over $\r$. By Fubini's Theorem, we obtain $$ \aligned \int_\r |u(x)|^2 dx & \int_{|x-y|<\frac{1}{m(y,\mu)}} \{ m(y,\mu)\}^{n+2} dy\\ & \le C\int_\r |\nabla u(x)|^2dx \int_{|x-y|<\frac{1}{m(y,\mu)}} \{ m(y,\mu)\}^n dy\\ &\ \ \ \ +C\int_\r |u(x)|^2d\mu(x) \int_{|x-y|<\frac{1}{m(y,\mu)}} \{ m(y,\mu)\}^n dy. \endaligned $$ The inequality in part (a) now follows, since part (c) of Proposition 1.8 yields $$ \int_{|x-y|<\frac{1}{m(y,\mu)}} \{ m(y,\mu)\}^{n+2} dy \ge c\{ m(x,\mu)\}^{n+2} \int_{|x-y|<\frac{c}{m(x,\mu)}} dy \ge c\, \{ m(x,\mu )\}^2$$ and $$ \int_{|x-y|<\frac{1}{m(y,\mu)}} \{ m(y,\mu)\}^n dy \le C \{ m(x,\mu)\}^n \int_{|x-y|<\frac{C}{m(x,\mu)}} dy \le C. $$ \enddemo \remark{\bf Remark 1.16} Let $$ \Cal{H} =\left\{ u\in W^{1,2}_\loc (\r): \ \nabla u\in L^2 (\r, dx)\ \text {and }\ m(\cdot, \mu )|u|\in L^2(\r ,dx)\right\} \tag 1.17 $$ equipped with the norm $$ \| u\|_{\Cal{H}} = \int_\r |\nabla u (x)|^2 dx +\int_\r m(x,\mu)^2 |u(x)|^2 dx. \tag 1.18 $$ Then $\Cal{H}$ is a Hilbert space. Using (1.11), one may show that $C_0^\infty (\r)$ is dense in $\Cal{H}$. By Theorem 1.13, we have $$ \aligned \Cal{H} &=\left\{ u\in W^{1,2}_\loc (\r):\ \nabla u\in L^2 (\r , dx) \ \ \text{and }\ u\in L^2(\r, d\mu)\right\}\\ & =\text{Domain} ((-\Delta +\mu)^{1/2}). \endaligned \tag 1.19 $$ \endremark \bigskip \centerline{\bf 2. The Fundamental Solution of $-\Delta +\mu$} \medskip \definition{\bf Definition 2.1} Let $\Omega$ be an open set in $\r$. Let $u\in W^{1,2}_\loc (\Omega)$ and $f\in L^1_\loc (\Omega, dx)$. $u$ is called a weak solution of $(-\Delta +\mu)u=f$ in $\Omega$ if $$ \int_\Omega <\nabla u, \nabla\psi> dx +\int_\Omega d\mu =\int_\Omega dx \tag 2.2 $$ for any $\psi\in C_0^1(\Omega)$. \enddefinition Using Remark 1.16 and the Lax-Milgram Theorem, we may deduce the following. \proclaim{\bf Proposition 2.3} Let $f\in L_\loc^2 (\r, dx)$. Assume $m(\cdot,\mu)^{-1}f\in L^2(\r, dx)$. Then $(-\Delta +\mu)u=f$ in $\r$ has a unique weak solution $u_f$ in $\Cal{H}$. \endproclaim Next we will show that there exists a unique kernel function $\Gamma_\mu(x,y)$ such that the solution $u_f$ in Proposition 2.3 is by $$ u_f(x)=\int_\r \Gamma_\mu (x,y)f(y) dy \tag 2.4 $$ for any $f\in L_c^2(\r, dx)$. We shall call $\Gamma_\mu(x,y)$ the fundamental solution of $-\Delta +\mu$ in $\r$. The proof of (2.4) is fairly standard. We include it for the sake of completeness. We remark that the fundamental solution as well as the Harnack inequality for the operator $-\Delta +\mu$ with a positive measure was investigated in \cite{5} by a probabilistic method. \definition{\bf Definition 2.5} Let $\Omega$ be an open set in $\r$ and $u\in W^{1,2}_\loc (\Omega)$. $u$ is called a subsolution of $(-\Delta +\mu)u=0$ in $\Omega$ if $$ \int_\Omega <\nabla u, \nabla \psi>dx +\int_\Omega d\mu \le 0 \tag 2.6 $$ for any nonnegative function $\psi$ in $C_0^1(\Omega)$. \enddefinition \proclaim{\bf Lemma 2.7} Let $u\in W^{1,2}_\loc (\Omega)$ be a subsolution of $(-\Delta +\mu)u=0$ in $\Omega$. Then $u^+=\max (u,0)$ is subharmonic in $\Omega$. \endproclaim \demo{Proof} We need to show that $$ \int_\Omega <\nabla u^+, \nabla \phi> dx \le 0\ \ \ \text{ for any } \ \phi\in C_0^1(\Omega),\ \ \phi\ge 0. \tag 2.8 $$ Note that $\nabla u^+=\nabla u$ on $\{ x\in \Omega: \ u(x)>0\}$, and $\nabla u^+=0$ on $\{ x\in\Omega:\ u(x)\le 0\}$. (2.8) follows by taking $$ \psi=\psi_\varepsilon =\phi\cdot\frac{u^+}{u^++\varepsilon} $$ in (2.6) and letting $\varepsilon\to 0$. We omit details. \enddemo \proclaim{\bf Lemma 2.9} Let $u\in W^{1,2}_\loc (\Omega)$ be a weak solution of $(-\Delta +\mu)u=0$ in $\Omega$. Then $u^+$ and $|u|$ are subharmonic in $\Omega$. Hence, $$ |u(x)|\le \frac{1}{|B|} \int_B |u(y)|\, dy \ \ \ \text{ if }\ B=B(x,R)\subset \Omega. \tag 2.10 $$ \endproclaim \demo{Proof} Since $u$ and $-u$ are subsolutions, $u^+$ and $u^- =(-u)^+$ are subharmonic in $\Omega$ by Lemma 2.7. It follows that $|u| =u^++u^-$ is also subharmonic. \enddemo \proclaim{\bf Lemma 2.11} Let $u\in W^{1,2}_\loc (\r)$ and $f\in L_\loc^1 (\r, dx)$. Suppose that $f\ge 0$ and $u$ is a weak solution of $(-\Delta +\mu)u=f$ in $\r$. Also assume that $$ \lim_{R\to\infty} \sup_{|x|=R} \frac{1}{R^n} \int_{B(x,R/2)} |u(y)|\, dy =0. \tag 2.12 $$ Then $u\ge 0$ in $\r$. \endproclaim \demo{Proof} Since $f\ge 0$, $-u$ is a subsolution in $\r$. By Lemma 2.7, $u^-=(-u)^+$ is subharmonic in $\r$. It follows from the maximal principle that $$ \sup_{\Omega_R} u^- \le \sup_{\partial \Omega_R} u^- $$ where $\Omega_R=B(0,R)$. Note that, for $x\in \partial \Omega_R$, $$ u^-(x)\le \frac{1}{|B(x,R/2)} \int_{B(x,R/2)} u^-(y)dy \le \frac{1}{B(x,R/2)} \int_{B(x,R/2)} |u(y)|dy. $$ Hence, by (2.12), $\sup_{\partial \Omega_R}u^-\to 0$ as $R\to\infty$. This implies that $u^-\equiv 0$. So $u=u^+\ge 0$. \enddemo \proclaim{\bf Lemma 2.13} Let $f\in L_c^2 (\r, dx)$ and $f\ge 0$. Let $u=u_f$ be the solution of $(-\Delta +\mu)u=f$ in $\r$ given by Proposition 2.3. Then $$ 0\le u(x)\le \int_\r \Gamma_0 (x,y) f(y)dy \tag 2.14$$ where $\Gamma_0(x,y)=c_n|x-y|^{2-n}$ is the fundamental solution of $-\Delta$ in $\r$. \endproclaim \demo{Proof} First, note that, by (1.11), $$ \frac{1}{R^2} \int_{R/2\le |x|\le 2R} |u|^2 dx \le C_\mu\int_{R/2\le |x|\le 2R} m(x,\mu)^2 |u|^2dx \le C_\mu \, \| u\|_\Cal{H}^2. \tag 2.15 $$ Hence $u$ satisfies the condition (2.12). It follows that $u\ge 0$ in $\r$. To prove the remaining inequality in (2.14), let $$ v(x)=\int_\r \Gamma_0 (x,y) f(y) dy. $$ Then $v\in W^{2,2}_\loc (\r)$, $-\Delta v=f$ in $\r$ and $v\ge 0$. Furthermore, $u-v$ is a subsolution of $(-\Delta +\mu)u=0$ in $\r$. Hence, by Lemma 2.7, $(u-v)^+$ is subharmonic in $\r$. This implies that $$ \sup_{\Omega_R} (u-v)^+ \le \sup_{\partial \Omega_R} (u-v)^+ \le \frac{C}{R^n} \int_{R/2\le |x|\le 2R} (|u|+|v|)dx $$ where $\Omega_R=B(0,R)$. It follows from (2.15) and $v(x)=O(|x|^{2-n})$ as $|x|\to\infty$ that $(u-v)^+=0$ in $\r$. Hence $u\le v$ in $\r$. \enddemo The following theorem is an easy consequence of Lemma 2.13. We omit its proof. \proclaim{\bf Theorem 2.16} For each $x\in \r$, there exists a unique $\Gamma_\mu(x,\cdot) \in L_\loc^p (\r, dy)$ ($1dx =\int_B<\nabla u,\nabla \psi>dx + \int_B<\nabla W,\nabla \psi> dx\\ & =\int_B<\nabla u, \nabla \psi> dx + \int_B u(y)d\mu(y) \int_B <\nabla_x \Gamma_0(x,y),\nabla_x\psi(x)>dx\\ &= \int_B<\nabla u,\nabla \psi> dx + \int_b d\mu (y) =0 \endaligned $$ since $u$ is a weak solution of $(-\Delta +\mu)u=0$ in $B$. We conclude that $v$ is harmonic in $B$. The proof is thus complete. \enddemo \proclaim{\bf Lemma 2.24} Let $B$ be a ball in $\r$. Then $W^{1,2}(B)$ is compactly imbedded in $L^2(B, d\mu)$. \endproclaim \demo{Proof} Let $Q$ be a closed cube containing $B$. It suffices to show that $W^{1,2}(Q)$ is compactly imbedded in $L^2(Q, d\mu)$. Let $R$ be the side length of $Q$. We divide $Q$ into a finite number of closed subcubes $\{ Q_j\}$ of equal size. Let $r$ denote the side length of $Q_j$. Note that the inequality (1.4) still holds if we replace the ball $B$ by a closed cube (the proof is the same). We apply (1.4) on each subcube $Q_j$ to obtain $$ \int_{Q_j}|\psi(x)|^2 d\mu(x) \le C \left\{ \frac{\mu (3Q_j)}{r^{n-2}} \int_{Q_j} |\nabla \psi (x)|^2 dx + \frac{\mu (Q_j)}{r^n} \int_{Q_j} |\psi(x)|^2 dx \right\} $$ where $\psi \in C^1 (Q)$. Summing in $j$ and using (0.2), we get $$ \int_Q|\psi|^2 d\mu \le C\, \frac{\mu(3Q)}{R^{n-2}} \left\{ \left(\frac{r}{R}\right)^\delta \int_Q |\nabla \psi|^2 dx + \frac{1}{r^2} \left(\frac{r}{R}\right)^\delta \int_Q |\psi|^2 dx\right\}. \tag 2.25 $$ By choosing $r$ small, we see that, for any $\epsilon>0$, there exists $C_\varepsilon>0$ such that $$ \int_Q |u|^2 d\mu \le C\, \frac{\mu (3Q)}{R^{n-2}} \left\{ \varepsilon \int_Q |\nabla u|^2 dx +C_\varepsilon \int_Q |u|^2 dx \right\} \tag 2.26 $$ for all $u\in W^{1,2}(Q)$. The compactness of the imbedding follows easily from (2.26). \enddemo We are now in a position to give the \noindent{\bf Proof of Theorem 2.18.} Fix $x_0$, $y_0\in \r$ such that $x_0\neq y_0$. Let $r =|x_0-y_0|$ and $$ f=\chi_{B(t,\varepsilon_1)},\ \ \ \ g=\chi_{B(s,\varepsilon_2)} $$ where $t\in B(x_0,r/8)$, $s\in B(y_0,r/8)$ and $0<\e_1,\, \e_2 dx &=\int_\r <\nabla u_f,\nabla u_g> dx + \int_\r d\mu\\ &= \int_\r dx \endaligned $$ where $u_f$ and $u_g$ are weak solutions, given by Proposition 2.3, with right hand sides $f$ and $g$ respectively. It follows from Theorem 2.16 that $$ \int_{B(t,\varepsilon_1)} dx \int_{B(s,\varepsilon_2)} \Gamma_\mu (x,y) dy =\int_{B(s, \varepsilon_2)} dy \int_{B(t,\varepsilon_1)} \Gamma_\mu (y,x)dx. \tag 2.27 $$ Since $\int_{B(s,\varepsilon_2)} \Gamma_\mu (x,y) dy$ is a weak solution of $(-\Delta +\mu)u=0$ in $B(x_0,r/2)$, it is continuous in $B(x_0,r/4)$ by Lemma 2.20. Thus, dividing both sides of (2.27) by $|B(t,\varepsilon_1)|$ and letting $\e_1\to 0$, we obtain $$ \int_{B(s,\e_2)} \Gamma_\mu (t,y) dy =\lim_{\e_1\to 0} \int_{B(s,\e_2)} dy \frac{1}{|B(t,\e_1)|} \int_{B(t,\e_1)} \Gamma_\mu (y,x) dx. \tag 2.28 $$ Now consider $$ u_{t,\e_1} (y) =\frac{1}{|B(t,\e_1)|} \int_{B(t,\e_1)} \Gamma_\mu (y,x)dx. \tag 2.29 $$ $u_{t,\e_1} $ is a nonnegative weak solution of $(-\Delta +\mu)u=0$ in $B(y_0,r/2)$. Since $u_{t,\e_1}$ is subharmonic in $B(y_0,r/2)$ by Lemma 2.7, we have Caccioppoli's inequality $$ \int_{B(y_0,r/4)} |\nabla u_{t,\e_1} |^2 dy \le \frac{C}{r^2} \int_{B(y_0,r/2)} |u_{t,\e_1}|^2 dy. $$ This, together with the size estimate (2.17), implies that $\| u_{t,\e_1}\|_{W^{1,2}(B(y_0,r/2))} \le C_r $ where $C_r$ does not depend on $t$ and $\e_1$. Hence there exist a sequence $\{ \e_{1,j}\}$ and $u_t\in W^{1,2}(B(y_0,r/4))$ such that, as $j\to\infty$, $\e_{1,j}\to 0$, $u_{t,\e_{1,j}}\to u_t $ weakly in $W^{1,2} (B(y_0, r/4))$ and $u_{t,\e_{1,j}}\to u_t $ strongly in $L^2(B(y_0,r/4), dy)$. Using Lemma 2.24, we may verify easily that $u_t$ is also a weak solution of $(-\Delta +\mu)u=0$ in $B(y_0, r/4)$. By (2.28), we have $$ \int_{B(s,\e_2)} \Gamma_\mu (t,y) dy =\int_{B(s,\e_2)} u_t (y) dy. $$ Since $s\in B(y_0, r/4)$ and $0<\e_2 \\ &= <\nabla u, \nabla u> |f|^2 +2 <\nabla u, \nabla f> uf + |u|^2 |\nabla f|^2\\ &= <\nabla u, \nabla (u |f|^2)> +|u|^2 |\nabla f|^2. \endaligned $$ Thus $$ \aligned & \int_\r |\nabla \psi|^2 dx +\int_\r |\psi|^2 d\mu\\ &= \int_\r <\nabla u, \nabla (u|f|^2)> dx + \int_\r d\mu +\int_\r |u|^2|\nabla f| ^2 dx\\ &= \int_\r |u|^2 |\nabla f|^2 dx \endaligned $$ where we have used the assumption that $(-\Delta +\mu) u=0$ in $\r\setminus B$, and $ u|f|^2\in W^{1,2}_0(\r\setminus B)$. It follows that $$ \aligned &\int_\r m(x,\mu)^2 |u\phi|^2 e^{2\e g} dx \le C\int_\r |u|^2 |\nabla f|^2 dx\\ &\le C \int_\r |u|^2 |\nabla \phi|^2 e^{2\e g} dx +\e^2 C \int_\r |u\phi|^2 e^{2\e g} |\nabla g|^2 dx\\ &\le C \int_\r |u|^2 |\nabla \phi|^2 e^{2\e g} dx +\e^2 C \int_\r m(x,\mu)^2 |u\phi|^2 e^{2\e g} dx. \endaligned $$ This implies that, if $\e^2C\le 1/2$, then $$ \int_\r m(x,\mu)^2 |u\phi|^2 e^{2\e g} dx \le 2C \int_\r |u|^2 |\nabla \phi|^2 e^{2\e g} dx . $$ \enddemo To use Lemma 3.1, we need to regularize the distance function $d(x,y,\mu)$ defined by (0.7). \proclaim{\bf Lemma 3.3} For each $y\in \r$, there exists a nonnegative function $\varphi_\mu (\cdot, y)\in C^\infty(\r)$ such that, for every $x\in \r$, $$ |\varphi_\mu (x,y)-d(x,y,\mu)|\le C,\tag 3.4 $$ and $$ |\nabla_x \varphi_\mu (x,y)|\le C\, m(x,\mu) . \tag 3.5 $$ \endproclaim \demo{Proof} Since $m(x,\mu)$ is a slow-varying function (part (c) of Proposition 1.8), there exist a sequence $\{ x_j\}$ in $\r$ and $\phi_j\in C_0^\infty (\r)$ such that $$ \r= \cup_{j=1}^\infty B_j\ \ \ \text{ where } \ \ B_j=B(x_j,\frac{1}{m(x_j,\mu)}) , \tag a $$ $$ \phi_j \in C_0^\infty (B_j),\ \ \ 0\le \phi_j\le 1\ \ \text {and }\ \ \sum_j \phi_j \equiv 1 , \tag b $$ $$ |\nabla \phi_j (x)|\le C\, m(x,\mu) , \tag c $$ $$ \sum_{j=1}^\infty \chi_{B_j}\le C . \tag d $$ We define $$ \varphi_\mu (x,y)=\sum_j d(x_j, y,\mu) \phi_j (x). \tag 3.6 $$ We omit the proof of (3.4)-(3.5), which may be found in \cite {12, p.4483}. \enddemo For technique reasons, we have to approximate $\varphi_\mu (x,y)$ by a sequence of $C^\infty$ bounded functions. \proclaim{\bf Lemma 3.7} For each $y\in \r$, there exists a sequence of nonnegative $C^\infty$ bounded functions $\{ \varphi_{\mu,j} (\cdot, y)\}$ such that, for every $ x\in \r$, $$ \varphi_{\mu,j} (x,y)\le \varphi_\mu (x,y) \ \ \ \text {and }\ \ \varphi_{\mu,j} (x,y)\to \varphi_\mu (x,y) \ \ \text {as }j\to\infty, \tag 3.8 $$ and $$ |\nabla_x \varphi_{\mu, j} (x,y)|\le C \, m(x,\mu) . \tag 3.9 $$ \endproclaim \demo{Proof} Fix $F\in C^\infty( (0,\infty))$ such that $F(t)=t$ if $t\in (0,1/2)$, $F(t)=0$ if $t\ge 2$, and $0\le F(t)\le t$ for all $t\ge 0$. Let $$ \varphi_{\mu, j} (x,y)=j F(\frac{\varphi_\mu (x,y)}{j}),\ \ \ \ j\ge 1. \tag 3.10 $$ It is easy to check that $\varphi_{\mu, j}$ satisfies (3.8)-(3.9). \enddemo We are now ready to prove the upper bound of $\Gamma_\mu (x,y)$. \proclaim{\bf Theorem 3.11} Assume that $\mu$ satisfies (0.2)-(0.3). Then $$ \Gamma_\mu (x,y) \le \frac{C\, e^{-\e_1 d(x,y,\mu)}}{ |x-y|^{n-2}} \tag 3.12 $$ for some $C>0$, $\e_1>0$. \endproclaim \demo{Proof} Fix $x_0$, $y_0\in \r$ and $x_0\neq y_0$. Without the loss of generality, we may assume that $y_0=0$. Since $$ d(x,y,\mu)\le C \ \ \ \text {if }\ |x-y|<\frac{C}{ m(x,\mu)}, \tag 3.13 $$ in view of (2.17), we may also assume that $|x_0|\ge C/m(0,\mu)$ and $$ B(x_0,\frac{2}{m(x_0,\mu )})\cap B(0,\frac{2}{m(0,\mu)}) =\emptyset. \tag 3.14 $$ Let $r=1/m(0,\mu)$. Since $\Gamma_\mu (x,0)$ is a weak solution of $(-\Delta +\mu)u=0$ in $\r\setminus \{ 0\}$ by Theorem 2.18, we may apply Lemma 3.1 with $u(x)=\Gamma_\mu (x,0)$, $g(x)=\varphi_{\mu,j} (x,0)$, and a suitable function $\phi$ in $C_0^\infty (B(0,2M)\setminus B(0,r))$ where $M\ge 4r$. We obtain $$ \aligned &\int_{2r\le |x|\le M} m(x,\mu)^2 |\Gamma_\mu (x,0)|^2 e^{2\e \varphi_{\mu,j} (x,0)} dx\\ &\ \ \le \frac{C}{r^2} \int_{r\le |x|\le 2r} |\Gamma_\mu (x,0)|^2 e^{2\e \varphi_{\mu,j}(x,0)} dx\\ &\ \ \ \ \ \ \ \ \ +\frac{C}{M^2} \int_{M\le |x|\le 2M} |\Gamma_\mu (x,0)|^2 e^{2\e \varphi_{\mu,j}(x,0)}dx. \endaligned \tag 3.15 $$ Since $\varphi_{\mu,j}(x,0)$ is bounded and $|\Gamma_\mu (x,0)|\le C/|x|^{n-2}$, the second term on the right hand side of (3.15) goes to zero as $M\to\infty$. This implies that $$ \aligned \int_{|x|\ge 2r}& m(x,\mu)^2 |\Gamma_\mu (x,0)|^2 e^{2\e \varphi_{\mu,j}(x,0)}dx\\ &\le \frac{C}{r^2} \int_{r\le |x|\le 2r} |\Gamma_\mu (x,0)|^2 e^{2\e \varphi_{\mu,j}(x,0)} dx\\ &\le \frac{C}{r^{n-2}}. \endaligned $$ Now let $j\to\infty$. By Fatou's Lemma, we have $$ \int_{|x|\ge 2r} m(x,\mu)^2 |\Gamma_\mu (x,0)|^2 e^{2\e\varphi_\mu (x,0)} dx \le \frac{C}{r^{n-2}}. $$ Using (3.14) and (3.4), we obtain $$ \int_{B(x_0,R)} m(x,\mu)^2 |\Gamma_\mu (x,0)|^2 e^{2\e d(x,0,\mu)} dx \le \frac{C}{r^{n-2}} \tag 3.16 $$ where $R=1/m(x_0,\mu)$. It follows that $$ \left\{ \frac{1}{R^n} \int_{B(x_0,R)} |\Gamma_\mu (x,0)|^2dx\right\}^{1/2} \le \frac{C\, e^{-\e d(x_0,0,\mu)}}{(Rr)^{\frac{n-2}{2}}}. \tag 3.17 $$ By Lemma 2.9, this gives $$ \aligned \Gamma_\mu (x_0,0) & \le \frac{C\, e^{-\e d(x_0, 0,\mu)}}{(Rr)^{\frac{n-2}{2}}}\\ &=C \left\{ m(x_0,\mu)m(0,\mu)\right\}^{\frac{n-2}{2}} e^{-\e d(x_0,0,\mu)}. \endaligned \tag 3.18 $$ To finish the proof, we claim that, if $|x-y|m(x,\mu)\ge 2$, then $$ d(x,y,\mu)\ge c\, \left\{ 1+|x-y| m(x,\mu)\right\}^{\frac{1}{k_0+1}} \tag 3.19 $$ for some $k_0>0$. Assume (3.19) for a moment. We have $$ |x_0|m(0,\mu) +|x_0|m(x_0,\mu) \le C_\e e^{\frac{\e}{2}d (x_0, 0,\mu)}. \tag 3.20 $$ for any $\e>0$. In view of (3.18), this gives $$ \Gamma_\mu (x_0,0) \le \frac{C\, e^{-\frac{\e}{2} d(x_0, 0,\mu)}}{|x_0|^{n-2}}. $$ It remains to prove (3.19). We choose $\gamma:\ [0,1]\to\r $ such that $\gamma(0)=x$, $\gamma(1)=y$ and $$ 2 d(x,y,\mu) \ge \int_0^1 m(\gamma(t),\mu)|\gamma^\prime (t)| dx. $$ It follows from (1.11) that $$ \aligned & 2d(x,y,\mu)\ge c\int_0^1 \frac{m(x,\mu) |\gamma^\prime (t)| dt} {\left\{ 1+|\gamma(t)-x| m(x,\mu)\right\}^{k_0/(k_0+1)}}\\ &\ge c\cdot \bigg\{ \text{ the geodesic distance from } x \text{ to } y \text{ in the metric}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{m(x,\mu)dz^2}{\big\{ 1+ |z-x| m(x,\mu)\}^{k_0/(k_0+1)}} \bigg\}\\ &= c\, \int_0^1 \frac{m(x,\mu) |y-x|dt} { \left\{ 1+ t|y-x|m(x,\mu)\right\}^{k_0/(k_0+1)}}\\ &\ge c\, \{ 1+|y-x| m(x,\mu)\} ^{\frac{1}{k_0+1}}. \endaligned $$ (3.19) is then proved. \enddemo \remark{\bf Remark 3.21} It follows from (3.19) and (3.12) that $$ \Gamma_\mu (x,y) \le C\, e^{-\e \{ 1+|x-y| m(x,\mu)\}^{\frac{1}{k_0+1}}}\cdot \frac{1} {|x-y|^{n-2}}. \tag 3.22 $$ We also point out that, using arguments similar to that in the proof of (3.19), one may show that, for any $x$, $y\in\r$, $$ d(x,y,\mu) \le C\, \{ 1+|x-y|m(x,y)\}^{k_0+1}. \tag 3.22 $$ \endremark \medskip \centerline{\bf 4. The Lower Bound of $\Gamma_\mu (x,y)$} \medskip In this section we establish the lower bound of $\Gamma_\mu (x,y)$ in the estimate (0.9). Our main tool is the following Harnack inequality. \proclaim{\bf Lemma 4.1} (Harnack inequality) Let $B=B(x_0,r)$ and $r<1/m(x_0,\mu)$. Let $u\in W^{1,2}(2B)$ be a nonnegative weak solution of $(-\Delta +\mu )u=0$ in $2B$. Then $$ \sup_{x\in B} u(x) \le C\, \inf_{x\in B} u(x) \tag 4.2 $$ \endproclaim \demo{Proof} If $d\mu =Vdx$ and $V$ satisfies $$ \sup_{x\in 2B} \int_{|y-x|dx + \int_B dx =0. \tag 4.6 $$ Clearly, as $j\to\infty$, $$ \int_B <\nabla v_{t_j},\nabla\psi>dx \to \int_B<\nabla v,\nabla \psi> dx. $$ We claim that, as $j\to\infty$, $$ \int_B dx \to \int_B d\mu. \tag 4.7 $$ Assume (4.7) for a moment. It follows that $v-u\in W_0^{1,2}(B)$ is a weak solution of $(-\Delta +\mu)u=0$ in $B$. Note that, in the definition of the weak solution, (2.2) in fact holds for any $\psi\in W_0^{1,2}(\Omega)$. It is easy to see that this implies $v-u\equiv 0$ in $B$. Hence, $v_{t_j}\to u$ a.e. on $B$. It remains to prove (4.7). We may assume that $\psi$ is a real function. We write $$ \int_B V_{t_j} v_{t_j} \psi dx =\int_B V_{t_j} (v_{t_j}-v)\psi dx +\int_B V_{t_j} v \psi dx. $$ Note that, by Minkowski's inequality, $$ \aligned & \left|\int_B V_{t_j} (v_{t_j}-v) \psi dx\right| =\left|\int_{2B} [(v_{t_j}-v)\psi] * \varphi_{t_j} d\mu\right|\\ &\le \mu (2B) \left\{\int_{2B} |[( v_{t_j}-v)\psi]* \varphi_{t_j}|^2 d\mu\right\}^{1/2}\\ & \le C_B \| (v_{t_j}-v)\psi\|_{L^2(2B, d\mu)}\to 0 \endaligned $$ as $j\to \infty$. Similarly, by (1.4), $$ \aligned & \left| \int_B V_{t_j} v\psi dx -\int_B v\psi d\mu \right| \le \mu(B)^{1/2} \left\{ \int_{2B} |(v\psi)*\varphi_{t_j} -v\psi|^2 d\mu\right\}^{1/2}\\ &\le C_B \| (v\psi)* \varphi_{t_j} -v\psi\|_{W^{1,2}(2B)}\to 0 \endaligned $$ as $j\to \infty$. Claim (4.7) is then proved. The proof of Lemma 4.1 is now complete. \enddemo Our next lemma compares $\Gamma_\mu (x,y)$ with $\Gamma_0(x,y)$ when $|x-y|$ is small. \proclaim{\bf Lemma 4.8} Let $x$, $y\in \r$. Suppose that $|x-y|<1/m(y,\mu)$. Then $$ |\Gamma_\mu (x,y) -\Gamma_0(x,y)| \le \frac{C\, \{ |x-y| m(y,\mu)\}^\delta}{|x-y|^{n-2}}. $$ \endproclaim \demo{Proof} By the uniqueness of the fundamental solution of $-\Delta$ in $\r$, it is not hard to show that, for any $x$, $y\in \r$, $$ \Gamma_0(x,y) =\Gamma_\mu (x,y) + \int_\r \Gamma_0 (x,z)\Gamma_\mu (z,y) d\mu (z). \tag 4.9 $$ It follows from the upper bound of $\Gamma_\mu (x,y)$ in Theorem 3.11 that $$ \aligned |\Gamma_\mu (x,y)-\Gamma_0 (x,y)| &\le C \int_\r \frac{e^{-\e d(z,y,\mu)} d\mu (z)} {|z-x|^{n-2} |z-y|^{n-2}}\\ = C\{ I_1 +I_2 +I_3\} \endaligned \tag 4.10 $$ where $I_1$, $I_2$ and $I_3$ denote the integrals over $B(x,r/2)$, $B(y,r/2)$ and $$ \{ z\in \r: \ |z-x|\ge r/2,\ |z-y|\ge r/2\} $$ respectively, and $r=|x-y|$. Clearly, by (0.4) and (0.2), $$ \aligned I_1 &\le \frac{C}{r^{n-2}}\int_{|z-x||z-y|\ge r/2} \frac{d\mu (z)}{|z-y|^{2n-4}} +C \int_{|z-y|\ge R/2} \frac{e^{-\e d(z,y,\mu)} d\mu (z)}{|z-y|^{2n-4}}\\ &= C\{ I_{31 }+I_{32}\}. \endaligned $$ By (0.2), we get $$ \aligned I_{31} &\le C \int_{r/2}^{2R} \frac{\mu(B(y,t))}{t^{n-2}} \cdot\frac{dt}{t^{n-1}}\\ &\le C\int_{r/2}^{2R}\left( \frac{t}{R}\right)^\delta \cdot \frac{\mu (B(x,R))}{R^{n-2}} \frac{dt}{t^{n-1}}\\ &\le C \left(\frac{r}{R}\right)^\delta \cdot\frac{1}{r^{n-2}}. \endaligned \tag 4.13 $$ Finally we use (3.19) to obtain $$ \aligned I_{32} &= \sum_{j=1}^\infty \int_{2^j R\le |z-y|<2^{j+1}R} \frac{e^{-\e d(z,y,\mu)}d\mu (z)}{|z-y|^{2n-4}}\\ &\le C \sum_{j=1}^\infty \frac{e^{-\e c\, j}}{(2^jR)^{2n-4}} \cdot \mu (B(y, 2^{j+1}R))\\ &\le C \sum_{j=1}^\infty \frac{e^{-\e c\, j}}{(2^jR)^{2n-4}} \cdot C^j\cdot R^{n-2}\\ &\le \frac{C}{R^{n-2}} \sum_{j=1}^\infty \frac{e^{-\e c\, j}}{(2^j)^{2n-4}} \cdot C^j\\ &\le \frac{C}{R^{n-2}}. \endaligned $$ This, together with (4.13), gives $$ I_3\le C\left(\frac{r}{R}\right)^\delta\cdot \frac{1}{r^{n-2}}. \tag 4.14 $$ The desired estimate now follows from (4.10)-(4.12) and (4.14). \enddemo We are now in a position to prove the lower bound of $\Gamma_\mu (x,y)$. \proclaim{\bf Theorem 4.15} Assume that $\mu$ satisfies the conditions (0.2)-(0.3). Then $$ \Gamma_\mu (x,y)\ge \frac{c\, e^{-\e_2 d(x,y,\mu)}}{|x-y|^{n-2}} \tag 4.16 $$ for some $c>0$, $\e_2>0$. \endproclaim \demo{\bf Proof} Given $x$, $y\in\r$. Since $\Gamma_0 (x,y)=c_n/|x-y|^{n-2}$, by Lemma 4.8, $$ \Gamma_\mu (x,y)\ge \frac{c_n}{2|x-y|^{n-2}} $$ if $|x-y|m(x,\mu)\le c$ and $c$ is small. By Harnack inequality (Lemma 4.1), this implies that $$ \Gamma_\mu (x,y)\ge\frac{c_A} {|x-y|^{n-2}}\ \ \ \text{ if } \ |x-y|m(x,\mu)\le A. \tag 4.17 $$ We fix a large $A$ so that $$ z\notin B(w,\frac{2}{m(w,\mu)}) \ \ \ \text{ whenever }\ \ |z-w|\ge \frac{A}{m(z,\mu)}. \tag 4.18 $$ By (4.17), it suffices to show (4.16) for any $x$, $y$ such that $|x-y|m(x,\mu)>A$. To this end, we choose $\gamma:\ [0,1]\to\r$ such that $\gamma(0)=x$, $\gamma(1)=y$ and $$ \int_0^1 m(\gamma(t),\mu) |\gamma^\prime (t)| dt \le 2 d(x,y,\mu). $$ Let $$ t_0=\sup \left\{ t\in [0,1]:\ \ |x-\gamma(t)|\le \frac{A}{ m(x,\mu)}\right\}. $$ If $|y-\gamma (t_0) |\le 1/m(\gamma(t_0),\mu)$, (4.16) follows from (4.17). Suppose $|y-\gamma(t_0)|>1/m(\gamma(t_0),\mu)$. We define $$ t_1=\inf\left\{ t\in[t_0,1]:\ \ |\gamma(t)-\gamma(t_0)|\ge \frac{1}{m(\gamma(t_0),\mu)}\right\}. $$ Since $m(\cdot,\mu)$ is locally bounded, we obtain $t_01$ in the second inequality. Now suppose that $|x_0-y_0|\le A/m(x_0,\mu)$. We choose $r=R=|x_0-y_0|/2$. Then $$ \sup_{B(x_0,r)} |u| \le\frac{C}{r^{n-2}}\ \ \ \text{ and }\ \frac{\mu (B(x_0,r))}{r^{n-2}}\le C. $$ We obtain $$ |\nabla_x \Gamma_\mu (x_0,y_0)| \le \frac{C}{r^{n-1}} \le \frac{C}{|x_0-y_0|^{n-1}} \le \frac{C_A\, e^{-\e d(x_0,y_0,\mu)}}{|x_0-y_0|^{n-1}}. \tag 5.3 $$ Finally, assume that $|x_0-y_0|> A/m(x_0,\mu)$. We may also assume that $y_0=0$ and $$ B(x_0,\frac{2}{m(x_0,\mu)})\cap B(0,\frac{2}{m(0,\mu)})=\emptyset \tag 5.4 $$ by choosing $A$ large. Let $r=\frac{1}{2\, m(x_0,\mu)}0$. We begin with a uniform H\"older estimate for the weak solutions of $(-\Delta +\mu)u=0$. \proclaim{\bf Lemma 6.1} Let $B=B(x_0,R)$. Suppose that $u\in W^{1,2}(B)$ is a weak solution of $(-\Delta +\mu)u=0$ in $B$. Then $$ |u(x)-u(y)|\le C \left(\frac{|x-y|}{R}\right)^\delta \sup_B |u| \tag 6.2 $$ for any $x$, $y\in B(x_0,R/4)$. \endproclaim \demo{Proof} By Lemma 2.20 and (1.12), it suffices to show that $$ \left\{ 1+ R\ m(x_0,\mu)\right\}^{k_1} \sup_{B(x_0,R/2)} |u| \le C\sup_{B(x_0,R)} |u|. \tag 6.3 $$ To this end, let $\varphi\in C_0^\infty (B(x_0, 3R/4))$ such that $0\le \varphi\le 1$, $\varphi =1$ on $ B(x_0, 5R/8) $, and $|\nabla \varphi|\le C/R$, $|\nabla^2 \varphi|\le C/R^2.$ Since $$ (-\Delta +\mu)(u\varphi) =-2\nabla u\cdot \nabla\varphi -u\Delta \varphi\ \ \text{ in }\ \r, \tag 6.4 $$ we have $$ u(x)\varphi(x)= \int_\r \Gamma_\mu (x,y) \left\{ -2\nabla u\cdot \nabla \varphi - u\Delta \varphi \right\} dy. $$ It follows that, for any $x\in B(x_0, R/2)$, $$ \aligned |u(x)| &\le\frac{C}{R} \int_{5R/8\le |y-x_0|\le 3R/4} \Gamma_\mu (x,y) |\nabla u(y)|dy\\ &\ \ \ \ +\frac{C}{R^2} \int_{5R/8\le |y-x_0|\le 3R/4} \Gamma_\mu (x,y) |u(y)|dy\\ &\le \frac{C}{R^2} \left\{ \int_{5R/8\le |y-x_0|\le 3R/4} |\Gamma_\mu (x,y)|^2 dy\right\}^{1/2} \left\{ \int_{B(x_0,R)} |u(y)|^2 dy\right\}^{1/2} \endaligned $$ where we have used H\"older inequality and Caccioppoli's inequality. Using the upper bound of $\Gamma_\mu (x,y)$ in Theorem 0.8, we obtain $$ |u(x)| \le\frac{C}{R^{n/2}} \sup_{B(x_0,R)} |u| \left\{ \int_{5R/8\le |y-x_0|\le 3R/4} e^{-2\e d(x,y,\mu)} dy\right\}^{1/2} \tag 6.5 $$ for $x\in B(x_0,R/2)$. It follows from (3.19) and (1.10) that, for $x\in B(x_0, R/2)$ $$ |u(x)| \le \frac{C}{\{ 1+Rm(x_0,\mu)\}^{k}} \sup_{B(x_0,R)}|u| $$ for any $k>0$. (6.3) is then proved. \enddemo We are now in a position to give the \noindent{\bf Proof of Theorem 0.19} Since $-\Delta +\mu$ is a positive self-adjoint operator, by the spectral theorem, $(-\Delta +\mu)^{i\gamma}$ is bounded on $L^2(\r,dx)$. We need to show that the kernel associated with $(-\Delta +\mu)^{i\gamma}$ satisfies the estimate (0.24). By functional calculus, we may write, at least formally $$ (-\Delta +\mu)^{i\gamma} =-\frac{\sin (\pi \gamma i )}{\pi} \int_0^\infty \lambda^{i\gamma} (-\Delta +\mu +\lambda)^{-1} d\lambda. \tag 6.6 $$ Hence, $$ (-\Delta +\mu)^{i\gamma} f(x) =\int_\r K(x,y) f(y) dy \tag 6.7 $$ where $$ K(x,y)= -\frac{\sin (\pi\gamma i)}{\pi} \int_0^\infty \lambda^{i\gamma} \Gamma_{\mu +\lambda}(x,y) d\lambda. \tag 6.8 $$ We remark that (6.7)-(6.8) may be justified by a limiting argument, using $$ A^{i\gamma} f \equiv \lim_{\e \to 0^+} A^{i\gamma -\e} f. $$ Also note that the measure $\mu +\lambda$ satisfies (0.2)-(0.3) with constants independent of $\lambda\ge 0$. Since $$ \aligned & m(x,\mu +\lambda)\ge m(x,\lambda)\ge c\, \sqrt{\lambda},\\ & d(x,y,\mu +\lambda)\ge d(x,y,\lambda)\ge c\, \, \sqrt{\lambda} \, |x-y| \endaligned \tag 6.9 $$ for $\lambda\ge 0$, by Theorem 0.8, we have $$ 0\le \Gamma_{\mu +\lambda} (x,y)\le \frac{ C\, e^{-c\, \sqrt{\lambda}|x-y|}} {|x-y|^{n-2}}. \tag 6.10 $$ This, together with (6.8), gives $$ |K(x,y)|\le \frac{C}{|x-y|^{n-2}}. \tag 6.11 $$ Next, consider $u(x)=\Gamma_{\mu +\lambda}(x,y_0)$. $u$ is a weak solution of $(-\Delta +\mu +\lambda)u=0$ in $B(x_0,R)$ where $R=|x_0-y_0|/2$. It follows from Lemma 6.1 that, if $|h|2-\delta. \tag 7.7 $$ To prove (7.7), we write $$ \aligned Tf(x)&= \int_{|y-x|>r} K_1(y,x) f(y) dy + \int_{|y-x|\le r} \{ K_1(y,x)-K_1^0(y,x)\} f(y) dy\\ &\ \ \ \ \ \ \ \ \ +\int_{|y-x|\le r} K_1^0(y,x) f(y) dy \endaligned \tag 7.8 $$ where $r=1/m(x,\mu)$ and $K_1^0 (x,y)$ is the kernel for the operator $\nabla (-\Delta)^{-1/2}$. To estimate the first term in the right hand side of (7.8), we need the following lemma. \proclaim{\bf Lemma 7.9} Let $B=B(x_0,r)$ and $$ I(x)=\int_B \frac{d \mu(y)}{|y-x|^{n-1}}. $$ Suppose that $\mu$ satisfies (0.2) for some $\delta\in (0,1)$. Then, for $1\le q <(2-\delta)/(1-\delta)$, $$ \| I\|_{l^q (B,dx)} \le C\, \frac{\mu(3B)}{r^{n(1-\frac{1}{q})-1}}. $$ \endproclaim \demo{Proof} Write $I=\sum_{j=0}^\infty I_j$ where $$ I_j(x)=\int\Sb y\in B\\ 2^{-j}r\le |y-x|<2^{-j+1}r \endSb \frac{d\mu (y)}{|y-x|^{n-1}}. $$ Then $$ \aligned \|I_j\|_{L^\infty (B,dx)} &\le \frac{1}{(2^{-j}r)^{n-1}} \cdot \sup_{x\in B} \mu \{ y\in B: |y-x|\le 2^{-j+1} r\}\\ &\le \frac{C}{2^{-j}r} \cdot \frac{\mu (3B)}{r^{n-2}} \cdot (2^{-j})^\delta \endaligned $$ where we have used (0.3). Also, note that, $$ \aligned \| I_j\|_{L^1(B,dx)} &\le \int_{x\in B} dx \int\Sb y\in B\\ 2^{-j} r\le |y-x|<2^{-j+1} r \endSb \frac{d\mu (y)}{|y-x|^{n-1}}\\ &=\int_{y\in B} d\mu (y) \int\Sb x\in B\\ 2^{-j}r \le |y-x| <2^{-j+1}r \endSb \frac{dx}{|y-x|^{n-1}}\\ &\le C\, 2^{-j} r \, \mu (B) \endaligned $$ It follows that, for $1\le q\le \infty$, $$ \aligned \| I_j\|_{L^q(B,dx)} &\le \| I_j\|_{L^\infty (B)}^{1-\frac{1}{q}} \|I_j\|_{L^1(B)}^{\frac{1}{q}}\\ &\le C\, (2^{-j})^{\frac{2}{q}-1 +\delta(1-\frac{1}{q})} \cdot \frac{\mu (3B)}{r^{n(1-\frac{1}{q})-1}}. \endaligned $$ Since $(2/q)-1+\delta(1-(1/q))>0$ if $q<(2-\delta)/(1-\delta)$, we have $$ \|I\|_{L^q(B,dx)} \le \sum_{j=0}^\infty \| I_j\|_{L^q(B,dx)} \le C\, \frac{\mu (3B)}{r^{n(1-\frac{1}{q})-1}}. $$ \enddemo \proclaim{\bf Lemma 7.10} Suppose $\mu$ satisfies (0.2)-(0.3) for some $\delta\in (0,1)$. Then $$ \left|\int_{|y-x|>R} K_1(y,x)f(y) dy\right| \le C_p\left\{ M(|f|^p)(x)\right\}^{1/p} $$ for $p>2-\delta$, where $R=1/m(x,\mu)$ and $M(g)$ denotes the Hardy-Littlewood maximal function of $g$. \endproclaim \demo{Proof} Fix $x_0$, $y_0\in\r$ with $x_0\neq y_0$. Let $u(x)=\Gamma_{\mu+\lambda} (x,x_0)$ where $\lambda\ge 0$. Then $u$ is a weak solution of $(-\Delta +\mu+\lambda)u=0$ in $B(y_0, 2r)$ where $r=|x_0-y_0|/4$. It follows from the proof of Theorem 0.17 that $$ |\nabla u(x)|\le \frac{C}{r} \sup_{B(y_0,r)} |u|+C \int_{B(y_0,r)} \frac{d\mu (y)}{|y-x|^{n-1}} \sup_{B(y_0,r)} |u| \tag 7.11 $$ for any $x\in B(y_0, r/2)$. Hence, by Lemma 7.9, if $1\le q\le (2-\delta)/(1-\delta)$, $$ \aligned \left\{ \int_{B(y_0,r/2)} |\nabla u(x)|^q dx\right\}^{1/q} &\le \frac{C}{r} \sup_{B(y_0,r)} |u|\cdot r^{n/q} +C\cdot \frac{\mu(3B)}{r^{n(1-\frac{1}{q})-1}}\sup_{B(y_0,r)} |u|\\ &=C\, r^{\frac{n}{q}-1} \sup_{B(y_0,r)} |u| \left\{ 1+\frac{\mu(3B)}{r^{n-2}} \right\}\\ &\le C\, r^{\frac{n}{q}-1} \sup_{B(y_0,2r)} |u| \endaligned $$ where the last inequality follows from (1.12) and (6.3). This gives $$ \aligned \left\{ \int_{B(y_0, r/2)} |\nabla_x \Gamma_{\mu +\lambda} (y, x_0)|^q dy\right\}^{1/q} &\le C\, r^{\frac{n}{q}-1} \sup_{y\in B(y_0,2r)} |\Gamma_{\mu +\lambda} (y,x_0)|\\ &\le C\, r^{\frac{n}{q}-n +1} \sup_{y\in B(y_0,2r)} e^{-\e d(y,x_0,\mu +\lambda)}\\ &\le C\, r^{\frac{n}{q}-n+1} \cdot \frac{ e^{-c\sqrt{\lambda} r} } {\{ 1+r\, m(x_0,\mu)\} } \endaligned $$ where we have used (3.19) in the last inequality. Using (7.5) and Minkowski's inequality, we now obtain $$ \left\{ \int_{B(y_0,r/2)} |K_1 (y,x_0)|^q dy\right\}^{1/q} \le \frac{C\, r^{\frac{n}{q}-n}} {\{ 1+r\, m(x_0, \mu)\} } \tag 7.12 $$ for $1\le q \le (2-\delta)/(1-\delta)$. Finally, for $p>2-\delta$, we have $$ \aligned &\left| \int_{|y-x_0|\ge R} K_1(y,x_0)f(y) dy\right|\\ &\le \sum_{j=1}^\infty \int_{2^{j-1}R\le |y-x_0|< 2^j R} |K_1 (y,x_0)| \, |f(y)| dy\\ &\le \sum_{j=1}^\infty \left\{ \int_{2^{j-1}R\le |y-x_0|< 2^j R} |K_1(y,x_0)|^q dy\right\}^{1/q} \left\{ \int_{|y-x_0|\le 2^j R} |f(y)|^p dy\right\}^{1/p}\\ &\le C \sum_{j=1}^\infty \frac{ (2^jR)^{\frac{n}{q}-n}} { 1+2^j} \left\{ \int_{|y-x_0|\le 2^j R} |f(y)|^p dy\right\}^{1/p}\\ &\le C \left\{ M(|f|^p)(x_0)\right\}^{1/p} \endaligned $$ where $q=p^\prime$, $R=1/m(x_0,\mu)$, and we used (7.12) in the third inequality. \enddemo The next lemma handles the second term in (7.8). \proclaim{\bf Lemma 7.13} Suppose that $\mu$ satisfies (0.2)-(0.3) for some $\delta\in (0,1)$. Then $$ \left|\int_{B(x,R)} \left\{ K_1 (y,x)-K_1^0 (y,x)\right\} f(y) dy\right| \le C_p \left\{ M(|f|^p)(x) \right\}^{1/p} $$ for $p>2-\delta$ and $R=1/m(x,\mu)$. \endproclaim \demo{Proof} Since $$ \Gamma_\lambda (y,x) =\Gamma_{\mu +\lambda} (y,x) +\int_\r \Gamma_\lambda (y,z) \Gamma_{\mu +\lambda} (z,x) d\mu (z), \tag 7.14 $$ we have $$ \aligned & |\nabla_y \left\{ \Gamma_{\mu +\lambda} (y,x) -\Gamma_\lambda (y,x) \right\} |\\ &\le \int_\r |\nabla_y \Gamma_\lambda (y,z)|\, \Gamma_{\mu +\lambda} (z,x) d\mu (z)\\ &\le C\int_\r \frac{ e^{-c\sqrt{\lambda} |y-z|} e^{-c\sqrt{\lambda}|x-z|} e^{-\e d(z,x,\mu)}} {|y-z|^{n-1} |z-x|^{n-2}} \, d \mu(z). \endaligned $$ Let $R=1/m(x,\mu)$ and $r=|x-y|/2$. Using the same argument as in the proof of Lemma 4.8, we may show that, if $|x-y|\le 1/m(x,\mu)$, $$ \aligned & |\nabla_y \left\{ \Gamma_{\mu +\lambda} (y,x) -|\Gamma_\lambda (y,x)\right\}|\\ &\le C\, e^{-c\sqrt{\lambda}\, r} \left\{ \frac{1}{r^{n-2}} \int_{B(y,r)} \frac{ d\mu (z)} {|z-y|^{n-1}} + \frac{ (r\, m(x,\mu))^\delta}{r^{n-1}} \right\}. \endaligned \tag 7.15 $$ By (7.5), this implies that $$ |K_1(y,x)-K_1^0 (y,x)| \le C \left\{ \frac{1}{r^{n-1}} \int_{B(y,r)} \frac{ d\mu (z)} {|z-y|^{n}} + \frac{ (r\, m(x,\mu))^\delta}{r^n} \right\} $$ This, together with Lemma 7.9, gives $$ \aligned & \left\{ \int_{ 2^{-j}R<|y-x|\le 2^{-j +1}R} |K_1 (y,x) -K_1^0 (y,x)|^q dy\right\}^{1/q}\\ &\le C \left\{ \frac{1}{(2^{-j} R)^{n-1}} \cdot \frac{\mu (B(x, 2^{-j +3}R))}{ (2^{-j}R)^{n(1-\frac{1}{q})-1} } + (2^{-j}R)^{\frac{n}{q}-n} \cdot (2^{-j})^\delta \right\}\\ &\le C\, (2^{-j})^\delta \cdot (2^{-j}R)^{n(\frac{1}{q}-1)} \endaligned $$ where $1\le q <(2-\delta(/(1-\delta)$. It follows that, for $p>2-\delta$, $$ \aligned &\left| \int_{B(x,R)} \left\{ K_1(y,x)-K_1^0 (y,x) \right\} f(y) dy\right|\\ &\le \sum_{j=1}^\infty \left\{ \int_{2^{-j}R<|y-x|\le 2^{-j+1} R} |K_1(y,x)-K_1^0(y,x)|^{p^\prime} dy \right\}^{1/p^\prime} \left\{ \int_{B(x, 2^{-j+1}R)} |f(y)|^p dy \right\}^{1/p}\\ &\le C \left\{ M(|f|^p)(x)\right\}^{1/p} \sum_{j=1}^\infty (2^{-j})^\delta\\ &\le C \left\{ M(|f|^p)(x)\right\}^{1/p}. \endaligned $$ \enddemo We are now ready to give the \noindent{\bf Proof of Theorem 7.1} In view of (7.8), by Lemmas 7.10 and 7.13, we have $$ |Tf(x)|\le C \left\{ M(|f|^p)(x)\right\}^{1/p} +2 \sup_{\e >0} \left| \int_{|y-x|>\e} K_1^0 (y,x) f(y) dy \right| $$ for any $p>2-\delta$. (7.7) follows from the well-known estimates for the Hardy-Littlewood maximal function and Riesz transforms $\nabla (-\Delta )^{1/2}$ \cite{14}. \medskip \remark{\bf Remark 7.16} It is interesting to note that, for any $1\le p\le \infty$, we have $$ \| m(\cdot, \mu )(-\Delta +\mu )^{-1/2} f\|_p \le C\| f\|_p. \tag 7.17 $$ This follows easily from the fact that the integral kernel of $(-\Delta +\mu)^{-1/2}$ is bounded by $$ \frac{C\, e^{-\e d(x,y,\mu)}}{|x-y|^{n-1}} $$ and (3.22). \endremark The rest of this section deals with the remaining part of Theorem 0.20. \proclaim{\bf Theorem 7.18} Assume that $\mu$ satisfies (0.2)-(0.3) for some $\delta >1$. Then $ \nabla (-\Delta +\mu)^{-1/2} $ is a Calder\'on-Zygmund operator. \endproclaim \demo{Proof} By Theorem 7.1, $\nabla (-\Delta +\mu)^{-1/2}$ is bounded on $L^p (\r, dx)$ for any $11$. It follows from (0.2) that $$ \aligned J_1 &\le \frac{C_k e^{-c\sqrt{\lambda}\, r}}{ r^{n-2} \{ 1+r\, m(y_0,\mu)\}^k } \left\{ \left(\frac{|h|}{r}\right)^{\delta -1} \cdot\frac{\mu (B(x_0, r))}{r^{n-1}} +|h| \int_{2|h|}^r \frac{\mu (B(x_0, t))}{t^{n-2}}\cdot\frac{dt}{t^3}\right\}\\ &\le \frac{C_k e^{-c\sqrt{\lambda}\, r}}{ r^{n-2} \{ 1+r\, m(y_0,\mu)\}^k } \left\{ \left(\frac{|h|}{r}\right)^{\delta -1} \cdot\frac{\mu (B(x_0, r))}{r^{n-1}} +|h| \int_{2|h|}^r \left(\frac{t}{r}\right)^\delta\cdot \frac{dt}{t^3} \frac{\mu (B(x_0,r))}{r^{n-2}}\right\}\\ &\le \frac{C_k e^{-c\sqrt{\lambda}\, r}}{ r^{n-2} \{ 1+r\, m(y_0,\mu)\}^k } \cdot \left(\frac{|h|}{r}\right)^{\delta -1} \cdot\frac{\mu (B(x_0, r))}{r^{n-1}}. \endaligned $$ Since $$ \frac{\mu (B(x_0,r))}{r^{n-2}} \le C\, \frac{\mu (B(y_0, 4r))}{(4r)^{n-2}} \le C\, \left\{ 1+r\, m(y_0, \mu)\right\}^{k_1}, $$ we get $$ J_1 \le C\left(\frac{|h|}{r}\right)^{\delta-1} \frac{e^{-c\sqrt{\lambda}}}{r^{n-1}}. \tag 7.24 $$ Next, for $J_2$, we have $$ \aligned J_2 &\le \frac{C\, |h| e^{-c\sqrt{\lambda}\, r}}{r^n} \int_\r \frac{e^{-\e d(z,y_0,\mu)}}{|z-y_0|^{n-2}} d\mu (z)\\ &\le \frac{C\, |h| e^{-c\sqrt{\lambda}\, r}}{r^n} \endaligned $$ where the second inequality follows from (3.19) and (1.12). This, together with (7.23)-(7.24), yields that $$ \aligned |\nabla_x &\Gamma_{\mu +\lambda} (x_0+h, y_0) -\nabla_x \Gamma_{\mu +\lambda} (x_0,y_0)|\\ &\le C\left(\frac{|h|}{|x_0-y_0|}\right)^{\delta -1} \cdot \frac{e^{-c\sqrt{\lambda}|x_0-y_0|}}{|x_0-y_0|^{n-1}}. \endaligned \tag 7.25 $$ In view of (7.4)-(7.5), we obtain $$ |K_1(x_0 +h,y_0) -K_1 (x_0,y_0)| \le C \left (\frac{|h|}{|x_0-y_0|}\right)^{\delta -1} \cdot \frac{1}{|x_0-y_0|^n}. \tag 7.26 $$ Finally, we estimate $|K_1(x_0, y_0+h)-K_1(x_0,y_0)|$. Using Theorem 0.17, Caccioppli's inequality and Lemma 2.24, one may show that $\nabla_x \Gamma_{\mu +\lambda} (x_0,y)$ is a weak solution of $(-\Delta +\mu +\lambda) u=0$ in $\r\setminus \{ x_0\}$. It follows from Lemma 6.1 and Theorem 0.17 that $$ \aligned & |\nabla_x \Gamma_{\mu +\lambda} (x_0,y_0+h) -\nabla_x \Gamma_{\mu +\lambda} (x_0,y_0)|\\ &\le C\left (\frac{|h|}{r}\right)^{\delta_1} \sup_{B(y_0,r)} |\nabla_x \Gamma_{\mu +\lambda}(x_0,y)|\\ &\le C\, \left(\frac{|h|}{r}\right)^{\delta_1} \frac{e^{-c\sqrt{\lambda}\, r}}{r^{n-1}} \endaligned $$ where $\delta_1 \in (0,1)$. By (7.4)-(7.5), this implies that $$ |K_1 (x_0, y_0 +h) -K_1 (x_0,y_0)| \le C\left( \frac{|h|}{|x_0-y_0|} \right)^{\delta_1} \frac{1}{|x_0-y_0|^n}. \tag 7.27 $$ The proof is complete. \enddemo \proclaim{\bf Theorem 7.28} Assume that $\mu$ satisfies (0.2) and (0.3) for some $\delta>1$. Then $\nabla (-\Delta +\mu)^{-1}\nabla$ is a Calder\'on-Zygmund operator. \endproclaim \demo{Proof} By duality, Theorem 7.1 implies that $\nabla (-\Delta +\mu )^{-1} =\nabla (-\Delta +\mu)^{-1/2} \cdot (-\Delta +\mu)^{-1/2}\nabla $ is bounded on $L^p(\r, dx)$ for any $1