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 domainW^{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 wherer=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_jwhere $$I_j(x)=\int\Sb y\in B\\ 2^{-j}r\le |y-x|<2^{-j+1}r \endSb \frac{d\mu (y)}{|y-x|^{n-1}}.$$ 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, for1\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 $$\left\{ \int_{B(y_0,r/2)} |K_1 (y,x_0)|^q dy\right\}^{1/q} \le \frac{C\, r^{\frac{n}{q}-n}} {\{ 1+r\, m(x_0, \mu)\} } \tag 7.12$$ for1\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 whereq=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 LetR=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 where1\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 anyp>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, forJ_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