Content-Type: multipart/mixed; boundary="-------------0612210138671" This is a multi-part message in MIME format. ---------------0612210138671 Content-Type: text/plain; name="06-368.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="06-368.keywords" Dirac operators, Weyl-Dirac operators, zero modes, zero resonances, the limiting absorption principle ---------------0612210138671 Content-Type: application/x-tex; name="saitoume.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="saitoume.tex" \documentclass[a4paper, 11pt,reqno]{amsart} \usepackage{amssymb} \usepackage{amsmath} \newtheorem{thm}{Theorem}[section] \newtheorem{prop}{Proposition}[section] \newtheorem{lem}{Lemma}[section] \newtheorem{rem}{Remark}[section] \newtheorem{cor}{Corollary}[section] \makeatletter \renewcommand{\theequation}{\thesection.\arabic{equation}} \@addtoreset{equation}{section} \makeatother \begin{document} %%%%%%%%\baselineskip=1.1\baselineskip \title{ The zero modes and the zero resonances\\ \hspace{-13pt}of Dirac operators } \author{Yoshimi Sait\={o} and Tomio Umeda\\} \address{ Department of Mathematics \\ University of Alabama at Birmingham, Birmingham, AL 35294, USA} \email{saito@math.uab.edu} \address{Department of Mathematics, University of Hyogo, Himeji 671-2201, Japan } \email{umeda@sci.u-hyogo.ac.jp} \date{} \maketitle \section{Introduction} This paper is concerned with the Dirac operator % \begin{equation} \label{eqn:1-1} H= \alpha \cdot D + Q(x), \quad D=\frac{1}{\, i \,} \nabla_x, \,\,\, x \in {\mathbb R}^3, \end{equation} % where $\alpha= (\alpha_1, \, \alpha_2, \, \alpha_3)$ is the triple of $4 \times 4$ Dirac matrices % \begin{equation*}\label{eqn:1-2} \alpha_j = \begin{pmatrix} \mathbf 0 &\sigma_j \\ \sigma_j & \mathbf 0 \end{pmatrix} \qquad (j = 1, \, 2, \, 3) \end{equation*} with the $2\times 2$ zero matrix $\mathbf 0$ and the triple of $2 \times 2$ Pauli matrices % \begin{equation*}\label{eqn:1-3} \sigma_1 = \begin{pmatrix} 0&1 \\ 1& 0 \end{pmatrix}, \,\,\, \sigma_2 = \begin{pmatrix} 0& -i \\ i&0 \end{pmatrix}, \,\,\, \sigma_3 = \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}, \end{equation*} % and $Q(x)$ is a $4\times 4$ matrix-valued function decaying at infinity. We would like to emphasize that one can regard the operator (\ref{eqn:1-1}) as a generalization of the operator % \begin{equation} \label{eqn:1-4} \alpha\cdot \big(D - A(x) \big) + q(x) I_4 , \end{equation} % where $(q, A)$ is an electromagnetic potential and $I_4$ is a $4\times 4$ identity matrix, by taking $Q(x)$ to be $- \alpha\cdot A(x) + q(x) I_4$. In the case where $q(x)\equiv 0$, the operator (\ref{eqn:1-4}) becomes of the form % \begin{equation*} \label{eqn:1-4-1} \alpha\cdot \big(D - A(x) \big) = \begin{pmatrix} \mathbf 0 &\sigma \cdot (D -A(x)) \\ \sigma \cdot (D -A(x)) & \mathbf 0 \end{pmatrix}. \end{equation*} % The component $\sigma \cdot (D -A(x))$ is called the Weyl-Dirac operator. See Balinsky and Evans \cite{BalinEvan2}. The zero modes of the Dirac operator $\alpha \cdot (D -A(x))$, the Weyl-Dirac operator $\sigma \cdot (D -A(x))$ and the Pauli operator $\{ \sigma \cdot (D -A(x)) \}^2 + q(x) I_2$ have attracted a considerable attention in recent years, because they have deep and fruitful implications from the view point of mathematics as well as physics. See Adam, Muratori and Nash \cite{AdamMuratoriNash1}, \cite{AdamMuratoriNash2}, \cite{AdamMuratoriNash3}, Balinsky and Evans \cite{BalinEvan1}, \cite{BalinEvan2}, \cite{BalinEvan3} and Elton \cite{Elton} for the three dimensional case, and Rozenblum and Shirokov \cite{RozenblumShirokov} for the two dimensional case. It should be remarked that the zero modes play significant roles in the study of Coulomb systems with magnetic fields and in the study of the Lieb-Thirring inequalities for the Pauli operators. See Fr\"ohlich, Lieb and Loss \cite{FrohlichLiebLoss}, Loss and Yau \cite{LossYau}, and Erd\"os and Solovej \cite{ErdosSolovej1}, \cite{ErdosSolovej2}, \cite{ErdosSolovej3}. We also should like to note that the operator (\ref{eqn:1-1}) also generalizes the Dirac operator of the form % \begin{equation} \label{eqn:1-5} \alpha\cdot D + m(x) \beta + q(x) I_4, \end{equation} % where $m(x)$ is considered to be a variable mass, and $\beta$ is the $4\times 4$ matrix defined by % \begin{equation*} \label{eqn:1-6} \beta = \begin{pmatrix} I_2 & \mathbf 0 \\ \mathbf 0 & - I_2 \end{pmatrix} \end{equation*} % Spectral properties of the operator (\ref{eqn:1-5}) have been extensively studied recent years. See Kalf and Yamada \cite{KalfYamada}, Kalf, Okaji and Yamada \cite{KalfOkajiYamada}, Schmidt and Yamada \cite{SchmidtYamada}, Pladdy \cite{Pladdy} and Yamada \cite{Yamada}. Finally, we would like to emphasize the significant role of the zero modes and the zero resonances in the analysis of the asymptotic behavior, around the origin of the complex plane, of the resolvent of the operator $H$ given by (\ref{eqn:1-1}). One can easily recognize the significance as the work \cite{JensenKato} by Jensen and Kato on the Schr\"odinger operator suggests. \vspace{10pt} \noindent \textbf{Notation.} \noindent The upper and lower half planes ${\mathbb C}_{\pm}$ are defined by % \begin{equation*} {\mathbb C}_{+} :=\{ \, z \in \mathbb C \; | \; \mbox{Im }z >0 \, \}, \qquad {\mathbb C}_{-} :=\{ \, z \in \mathbb C \; | \; \mbox{Im }z <0 \, \} \end{equation*} % respectively. By $S({\mathbb R}^3)$, we mean the Schwartz class of rapidly decreasing functions on ${\mathbb R}^3$, and we set $\mathcal S = [S({\mathbb R}^3)]^4$. By $L^2 =L^2({\mathbb R}^3)$, we mean the Hilbert space of square-integrable functions on ${\mathbb R}^3$, and we introduce a Hilbert space ${\mathcal L}^2$ by ${\mathcal L}^2 = [L^2({\mathbb R}^3)]^4$, where the inner product is given by % \begin{equation*} (f, g)_{{\mathcal L}^2} = \sum_{j=1}^4 (f_j, g_j)_{L^2} \end{equation*} % for $f = {}^t(f_1, f_2, f_3, f_4)$ and $g = {}^t(g_1, g_2, g_3, g_4)$. By $L^{2, s}({\mathbb R}^3)$, we mean the weighted $L^2$ space defined by % \begin{equation*} L^{2, s}({\mathbb R}^3) :=\{ \, u \; | \; \langle x \rangle^s u \in L^2({\mathbb R}^3) \, \} \end{equation*} % with the inner product % \begin{equation*} (u, \, v)_{L^{2, s}} := \int_{{\mathbb R}^3} \langle x \rangle^{2s} u(x) \, \overline{v(x)} \, dx, \end{equation*} % where % \begin{equation*} \label{eqn:bracket-x} \langle x \rangle = \sqrt{1 + |x|^2 \,}. \end{equation*} % We introduce the Hilbert space ${\mathcal L}^{2,s} = [L^{2,s}({\mathbb R}^3)]^4$ with the inner product % \begin{equation*} (f, g)_{{\mathcal L}^{2,s}} = \sum_{j=1}^4 (f_j, g_j)_{L^{2,s}}. \end{equation*} % % By $H^{\mu, s}({\mathbb R}^3)$, we mean the weighted Sobolev space defined by % \begin{equation*} H^{\mu, s}({\mathbb R}^3) :=\{ \, u \in {S^{\prime}({\mathbb R}^3)} \; | \; \langle x \rangle^s \langle D \rangle^{\mu} \, u \in L^2({\mathbb R}^3) \, \} \end{equation*} % with the inner product % \begin{equation*} (u, \, v)_{H^{\mu, s}} := \big( \langle x \rangle^{s} \langle D \rangle^{\mu} u , \langle x \rangle^{s}\langle D \rangle^{\mu} \, v \big)_{L^2}, \end{equation*} % where % \begin{equation} \label{eqn:bracket-D} \langle D \rangle = \sqrt{1 - \Delta \,}. \end{equation} % In a similar fashion, we introduce the Hilbert space ${\mathcal H}^{\mu,s} = [H^{\mu,s}({\mathbb R}^3)]^4$. Note that $H^{\mu, 0}({\mathbb R}^3)$ coincides with the Sobolev space of order $\mu$: $H^{\mu}({\mathbb R}^3)$, and by ${\mathcal H}^{\mu}$ we mean the Hilbert space $[H^{\mu}({\mathbb R}^3)]^4$. Also note that ${\mathcal H}^{0,0} ={\mathcal L}^2$ and ${\mathcal H}^{0,s} ={\mathcal L}^{2,s}$. By ${B}(\mu, s \, ; \,\nu, t )$, we mean the set of all bounded linear operators from $H^{\mu, s}({\mathbb R}^3)$ into $H^{\nu, t}({\mathbb R}^3)$, and by $\mathcal{B}(\mu, s \, ; \,\nu, t)$, the set of all bounded linear operators from ${\mathcal H}^{\mu, s}$ into ${\mathcal H}^{\nu, t}$. For an operator $W \in {B}(\mu, s \, ; \,\nu, t )$, we define a copy of $W \in \mathcal{B}(\mu, s \, ; \,\nu, t)$ by % \begin{gather} \label{eqn:mapsto} \begin{split} {\mathcal H}^{\mu, s} \ni f= \, &{}^t (f_1, \, f_2, \, f_3, \, f_4) \\ {}&\mapsto Wf ={}^t (Wf_1, \, Wf_2, \, Wf_3, \, Wf_4) \in {\mathcal H}^{\nu, t}. \end{split} \end{gather} % \vspace{15pt} \noindent \textbf{Assumption (A).} \noindent Each element $q_{jk}(x)$, $j, \, k =1, \, \cdots, \, 4$, of $Q(x)$ is a measurable function satisfying % \begin{equation} \label{eqn:2-1} | q_{jk}(x) | \le C \langle x \rangle^{-\rho} \quad ( \rho >1 ), \end{equation} % where $C(=C_q)$ is a positive constant. Moreover, $Q(x)$ is a Hermitian matrix for each $x \in {\mathbb R^3}$. \vspace{15pt} Note that under Assumption(A) the Dirac operator (\ref{eqn:1-1}) is a self-adjoint operator in $\mathcal L^2$ with $\mbox{Dom}(H) = {\mathcal H}^1$. The self-adjoint realization will be denoted by $H$ again. With an abuse of notation, we shall write $Hf$ \textit{in the distributional sense} for $f \in {\mathcal S}^{\prime}$ whenenver it makes sense. \vspace{15pt} \noindent D{\scriptsize EFINITION}. \ By a zero mode, we mean a function $f \in \mbox{Dom}(H)$ which satisfies % \begin{equation*} \label{eqn:1-7} Hf=0. \end{equation*} By a zero resonance, we mean a function $f \in {\mathcal L}^{2, -s}\setminus \mathcal L^2$, for some $s$ with $1/2 < s <3/2$, which satisfies % \begin{equation*} \label{eqn:1-8} Hf=0 \;\;\text{ in the distributional sense.} \end{equation*} % \vspace{15pt} It is evident that a zero mode of $H$ is an eigenfunction of $H$ corresponding to the eigenvalue $0$, i.e., a zero mode is an element of $\mbox{Ker}(H)$, the kernel of the self-adjoint operator $H$. We also remark that our definition of the zero resonance above is more general than that of Jensen and Kato \cite{JensenKato} with respect to the smoothness, though we shall restrict ourselves to $\rho > 5/2$ and $1/2 < s < \min \{ 3/2, \, \rho- 3/2 \}$ in dealing with zero resonances in section 2. Balinsky and Evans \cite{BalinEvan2} is particulary interesting from our view point in the sense that they dealt with the Weyl-Dirac operator $\sigma \cdot (D -A(x))$ and showed that the set of magnetic fields which give rise to zero modes is rather \lq\lq sparse.\rq\rq In this paper, we investigate the zero modes and the zero resonances of the operator (\ref{eqn:1-1}) under Assumption(A). Our goal is to establish a pointwise estimate of the zero modes as well as the continuity of the zero modes, and also to show that the zero resonasnces do not exist. \vspace{15pt} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Main results} \begin{thm} \label{thm:th-2-2} Suppose Assumption {\rm(A)} is verified. Let $f$ be a zero mode of the operator {\rm(\ref{eqn:1-1})}. Then {\rm (i)} the inequalities % \begin{equation} \label{eqn:2-2} |f(x)| \le C \begin{cases} \langle x \rangle^{-\rho +1} & (1 < \rho <3 ) \\ {} & \\ \langle x \rangle^{-2}\log ( 1 + \langle x \rangle) & ( \rho = 3) \\ {} & \\ \langle x \rangle^{-2}& ( \rho >3). \end{cases} \end{equation} % hold for all $x \in {\mathbb R}^3$, where the constant $C(=C_f)$ depends only on the zero mode $f$; \vspace{4pt} {\rm (ii)} the zero mode $f$ is a continuous function on ${\mathbb R}^3$. \end{thm} \vspace{15pt} \begin{rem} It is natural that zero modes only decay polynomially at infinity. In Loss and Yau \cite{LossYau}, they considered the Weyl-Dirac operator $\sigma \cdot (D-A(x))$, and constructed two examples of pairs of a vector potential $A$ and a zero mode $\psi$. One of their examples shows that $A(x) = O(|x|^{-2})$ and $\psi(x) = O(|x|^{-2})$ at infinity. Also, see examples in Adam, Muratori and Nash \cite{AdamMuratoriNash1}. It is not clear for us if the decay rates in Theorem \ref{thm:th-2-2} are optimal. \end{rem} \begin{rem} In Bugliaro,Fefferman and Graf \cite{BugliaroFeffermanGraf} and Erd\"os and Solovej \cite{ErdosSolovej2}, they established estimates of the density of zero modes of the Weyl-Dirac operator $\sigma \cdot (D-A(x))$. It is apparant that their estimates immediately imply estimates of each zero mode. These estimates of each zero mode are, however, quite unclear in terms of the decay rate at infinity. \end{rem} \vspace{10pt} Theorem \ref{thm:th-2-1} below means that zero resonances do not exist. In order to show this fact, we need a larger $\rho$ than Theorem \ref{thm:th-2-2}. \vspace{15pt} \begin{thm} \label{thm:th-2-1} Suppose Assumption {\rm(A)} is verified with $\rho > 5/2$. If $f$ belongs to ${\mathcal L}^{2, \, -s}$ for some $s$ with $1/2 < s < \min\{3/2, \, \rho -3/2 \}$ and satisfies $Hf=0$ in the distributional sense, then $f \in {\mathcal H}^1$. \end{thm} \vspace{15pt} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{A singular integral operator} \label{sec:SIO} One of the ingredients of the proofs of the main theorems is a singular integral operator acting on four component vector functions. The singular integral operator we deal with in this section is defined by % \begin{equation} \label{eqn:sio-1} A f(x) = \displaystyle{ \int_{{\mathbb R}^3} i \, \frac{\alpha \cdot (x - y)}{4\pi |x-y|^3} f(y) \, dy. } \end{equation} % for % \begin{equation*} \label{eqn:sio-2} f = {}^t(f_1, \, f_2, \, f_3, \, f_4) \in {\mathcal L}^2, \end{equation*} % where $\alpha \cdot (x - y)$ means the sum of the matrix operation $\alpha_j$ for the four-vector $(x_j - y_j) f$: % \begin{equation*} \label{eqn:sio-2+} \alpha \cdot (x - y) f = \sum_{j=1}^3 \alpha_j (x_j - y_j) f. \end{equation*} % We shall need a few estimates of $A$ on ${\mathcal L}^2$ and on its subspaces. \begin{lem} \label{lem:siolem-1} For each $f \in {\mathcal L}^2$, $Af(x)$ is defined for a.e. $x \in {\mathbb R}^3$. Moreover, $A$ is a bounded operator from ${\mathcal L}^2$ to ${\mathcal L}^6$, i.e., there exists a constant $C$ such that % \begin{equation*} \label{eqn:sio-3} \Vert A f \Vert_{{\mathcal L}^6} \le C \Vert f \Vert_{{\mathcal L}^2} \end{equation*} % for all $f \in {\mathcal L}^2$. \end{lem} \begin{proof} Since each $\alpha_j$, $j=1$, $2$, $3$, is a unitary matrix, we have % \begin{gather*} \begin{split} % \label{eqn:sio-4} | \alpha \cdot (x - y) \, f(y)| &\le \sum_{j=1}^3 |x_j - y_j| \, |f(y)| \\ {}& \le 3|x-y| \, |f(y)|. \end{split} \end{gather*} % (Note that $|x-y|$ and $|f(y)|$ are the Euclidean norms of ${\mathbb R}^3$ and ${\mathbb R}^4$ respectively.) % Therefore we get \begin{gather} \begin{split} \label{eqn:sio-5} |A f(x) | &\le \frac{3}{\,4\pi \,} \int_{{\mathbb R}^3} \frac{1}{|x-y|^2} |f(y)|\, dy \\ {} & \le \frac{\,3\pi \,}{2} I_1(\, |f| \, ) (x), \end{split} \end{gather} % where $I_1$ is the Riesz potential; see Stein \cite[p.117]{Stein}. We shall appeal two well-known facts (Stein \cite[p.119]{Stein}) that $I_1(\, u \, ) (x)$ is finite for a.e. $x \in {\mathbb R}^3$ if $u \in L^2({\mathbb R}^3)$, and that $I_1$ is a bounded operator from $L^2({\mathbb R}^3)$ to $L^6({\mathbb R}^3)$. These facts, together with (\ref{eqn:sio-5}), yield the conclusions of the lemma. \end{proof} \vspace{10pt} \begin{lem} \label{lem:siolem-2} Let $s > 3/2$. Then \vspace{3pt} {\rm(i)} $A\in{\mathcal B}( 0, s \, ; \, 0, 0)${\rm;} \vspace{3pt} {\rm(ii)} $A\in{\mathcal B}( 0, 0 \, ; \, 0, -s)$. \end{lem} % \begin{proof} It is shown in Umeda \cite[Lemma 5.1]{Umeda} that the Riesz potential $I_1$ is a bounded operator from $L^{2, s}({\mathbb R}^3)$ to $L^{2}({\mathbb R}^3)$ and that $I_1$ is a bounded operator from $L^{2}({\mathbb R}^3)$ to $L^{2, -s}({\mathbb R}^3)$. (Note that $I_1$ is written as $G_0$ in \cite{Umeda}.) The assertions (i) and (ii) of the lemma respectively follows from these two facts and (\ref{eqn:sio-5}). \end{proof} \vspace{10pt} We introduce a class of functions which is necessary to establish an ${\mathcal L}^{\infty}$ estimate of the operator $A$. For $q\ge 1$, we define % \begin{equation*} \label{eqn:sio-6} L^q_{ul}(\mathbb R^3) = \big\{\, u \in L^q_{loc}(\mathbb R^3) \; \big| \; \Vert u \Vert_{L^q_{ul}} := \sup_{x \in {\mathbb R}^3} \Vert u \Vert_{L^q{( B(x; \, 1) )}} < \infty \, \big\}, \end{equation*} % where $B(x; 1) = \{ \, y \in \mathbb R^3 \; | \;\, |x-y| \le 1 \, \}$, and define % \begin{equation*} \label{eqn:sio-7} {\mathcal L}^q_{ul} = \big[ L^q_{ul}(\mathbb R^3) \big]^4, \quad \Vert f \Vert_{{\mathcal L}^q_{ul}} = \sum_{k=1}^4 \Vert f_k \Vert_{L^q_{ul}}. \end{equation*} % \vspace{10pt} \begin{lem} \label{lem:siolem-3} Let $q >3$. Then there exists a constant $C_q$ such that % \begin{equation*} \label{eqn:sio-8} \Vert A f \Vert_{{\mathcal L}^{\infty}} \le C_q \big( \Vert f \Vert_{{\mathcal L}^2} +\Vert f \Vert_{{\mathcal L}^q_{ul}} \big). \end{equation*} for all $f \in {\mathcal L}^2 \cap {\mathcal L}^q_{ul}$. \end{lem} % \begin{proof} By virtue of (\ref{eqn:sio-5}), we only have to prove that there exists a constant $C_q^{\prime}$ such that % \begin{equation} \label{eqn:sio-9} \Vert I_1(u) \Vert_{{L}^{\infty}} \le C_q^{\prime} \big( \Vert u \Vert_{{L}^2} +\Vert u \Vert_{{L}^q_{ul}} \big) \mbox{ for } u \in L^2(\mathbb R^3) \cap L^q_{ul}(\mathbb R^3). \end{equation} % Since each $u \in L^2(\mathbb R^3) \cap L^q_{ul}(\mathbb R^3)$ can be docomposed as % \begin{align*} u &=v_+ -v_- + i (w_+ -w_-) \\%%%%%%\label{eqn:sio-10} \\ {}& v_{\pm} \ge 0, \;\; w_{\pm}\ge 0 \\%%%%%\label{eqn:sio-11} \\ {}& v_{\pm},\;\; w_{\pm} \in L^2(\mathbb R^3) \cap L^q_{ul}(\mathbb R^3) \\%%%\label{eqn:sio-12} \end{align*} % we shall prove (\ref{eqn:sio-9}) for $u\ge0$. Let $u \in L^2(\mathbb R^3) \cap L^q_{ul}(\mathbb R^3)$ be given, and let satisfy $u\ge0$. Then one can find a sequence $\{ \, \varphi_n \, \} \subset C_0^{\infty}(\mathbb R^3)$ such that % \begin{equation} \label{eqn:sio-13} 0 \le \varphi_n \le u, \quad \varphi_n \to u \;\; \mbox{ in } L^2(\mathbb R^3). \end{equation} % (First cut $u$ as ${\chi}_{B(0;\, n)}(x) u$ by multipying a characteristic function ${\chi}_{B(0;\, n)}$ of the ball with center at the origin and radius $n$, then use the mollifier.) For each $n$, we decompose as % \begin{gather} \begin{split} \label{eqn:sio-14} I_1(\varphi_n)(x) &= \int_{\mathbb R^3} \frac{1}{\, 2\pi^2 |x-y|^2 \,} \, \varphi_n (y) \, dy \\ \noalign{\vskip 4pt} &= h_0 * \varphi_n (x) + h_1 * \varphi_n (x), \end{split} \end{gather} % where % \begin{gather*} \begin{split} %%%%%%%\label{eqn:sio-16} h_0(x) &= \chi_{B(0;\, 1)}(x) \frac{1}{\, 2\pi^2 |x|^2 \,} \\ h_1(x) &= (1 -\chi_{B(0;\, 1)}(x) ) \frac{1}{\, 2\pi^2 |x|^2 \,}. \end{split} \end{gather*} % (One should note that the integral on the right hand side of (\ref{eqn:sio-14}) converges because of the fact that $\varphi_n \in C_0^{\infty}(\mathbb R^3)$.) If we apply the H\"older inequality to $h_0 * \varphi_n$, then we get % \begin{gather} \begin{split}\label{eqn:sio-17} |h_0 * \varphi_n(x) | &\le \frac{1}{\, 2\pi^2 \,} \Big\{ \int_{|x-y|\le 1} \frac{1}{\, |x-y|^{2p} \,} \, dy \Big\}^{1/p} \, \Vert \varphi_n \Vert_{L^q(B(x;1))} \\ \noalign{\vskip 4pt} {}& \quad \le C_{q}^{\prime\prime} \Vert u \Vert_{L^q(B(x;1))} \\ \noalign{\vskip 4pt} {}& \qquad ( p^{-1}= 1 - q^{-1} ) \end{split} \end{gather} % where we have used the fact that $2p < 3 \;\; (\because q >3$ by assumption) and (\ref{eqn:sio-13}). Similarly, if we apply the Schwarz inequality to $h_1 * \varphi_n$, we obtain % % \begin{gather} \begin{split}\label{eqn:sio-18} |h_1 * \varphi_n(x) | &\le \frac{1}{\, 2\pi^2 \,} \Big\{ \int_{|x-y|\ge 1} \frac{1}{\, |x-y|^{4} \,} \, dy \Big\}^{1/2} \, \Vert \varphi_n \Vert_{L^2({\mathbb R})} \\ \noalign{\vskip 4pt} {}& \quad \le C \Vert u \Vert_{L^2({\mathbb R}^3)} . \end{split} \end{gather} % If follows from (\ref{eqn:sio-14}), (\ref{eqn:sio-17}) and (\ref{eqn:sio-18}) that % \begin{gather} \begin{split}\label{eqn:sio-19} |I_1(\varphi_n)(x)| & \le C_{q}^{\prime} \big( \Vert u \Vert_{L^q(B(x;1))} + \Vert u \Vert_{L^2({\mathbb R}^3)} \, \big) \\ & \le C_q^{\prime} \big( \Vert u \Vert_{L^q_{ul}} + \Vert u \Vert_{L^2} \big) \end{split} \end{gather} % As was mentioned in the proof of Lemma \ref{lem:siolem-2}, the Riesz potential $I_1$ is a bounded operator from $L^2({\mathbb R}^3)$ to $L^{2, -s}({\mathbb R}^3)$ with $s > 3/2$. This fact, together with (\ref{eqn:sio-13}), implies that there exists a subsequence $\{ \varphi_{n^{\prime}} \}$ such that $I_1(\varphi_{n^{\prime}}) (x) \to I_1(u)(x)$ for a.e. $x \in {\mathbb R}^3$. Thus taking the limit of (\ref{eqn:sio-19}), along with the subsequence, gives (\ref{eqn:sio-9}). \end{proof} \vspace{15pt} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Estimates of the resolvents} \label{sec:resolvent} Another ingredient of the proofs of the main theorems is the limiting absorption principle (LAP) for the free Dirac operator % \begin{equation} \label{eqn:reslvnDrc-00} H_0= \alpha \cdot D. \end{equation} % We note that $H_0$ with $\mbox{Dom}(H_0) ={\mathcal H}^1$ is a self-adjoint operator in ${\mathcal L}^2$. The self-adjoint realization will be denoted by $H_0$ again. It is well-known that the spectrum $\sigma(H_0)$ equals the whole real line $\mathbb R$. With an abuse of notation again, we shall write $H_0f$ for $f \in {\mathcal S}^{\prime}$. Our idea of proving the LAP for the free Dirac operator $H_0$ is based on a decomposition of the resolvent % \begin{equation} \label{eqn:reslvnDrc-0} R_0(z) = (H_0 +z ) \varGamma_0(z^2) \;\; \mbox{ on } \; {\mathcal L}^2, \quad \mbox{Im}\, z \not=0, \end{equation} % where % \begin{equation} \label{eqn:reslvnDrc-1} R_0(z) = (H_0 - z )^{-1}, \end{equation} % and $\varGamma_0(z)$ in (\ref{eqn:reslvnDrc-0}) denotes the copy of the resolvent $\varGamma_0(z)$ of the negative Laplacian % \begin{equation*} \label{eqn:laplace-0} -\Delta = -\big( \frac{{\partial}^2}{\partial x_1^2} + \frac{{\partial}^2}{\partial x_2^2} + \frac{{\partial}^2}{\partial x_3^2} \big). \end{equation*} % See (\ref{eqn:mapsto}) for the definition of the copy of an operator. In other words, we shall not distinguish between $\varGamma_0 (z)$ in $L^2({\mathbb R}^3)$ and $\varGamma_0 (z)$ in ${\mathcal L}^2$. We believe this will not cause any confusion. A formal computation shows that % \begin{equation} \label{eqn:diracsquare} H_0^2 = -\Delta, \end{equation} % from which one can deduce (\ref{eqn:reslvnDrc-0}). The decomposition (\ref{eqn:reslvnDrc-0}) was first exploited in Balslev and Hellfer \cite{BalslevHellfer}. Similar decomposition was also adopted in Pladdy, Sait\={o} and Umeda \cite{PSU1}, \cite{PSU2}. \vspace{10pt} \subsection{The resolvent of the negative Laplacian} It is well known that the resolvent of $-\Delta$ is represented as an integral operator: % \begin{equation} \label{eqn:resolventLapc} \varGamma_0 (z) u(x) = \int_{{\mathbb R}^3} \frac{e^{i \sqrt{z} |x-y|} }{\, 4\pi |x-y| \,} u(y) \, dy, \qquad u \in L^2({\mathbb R}^3 ) \end{equation} % for $z \in \mathbb C \setminus [0, \,+\infty)$, where $\mbox{Im} \sqrt{z} >0$. \begin{lem} \label{lem:hs} Let $s$, $s^{\prime} > 1/2$ and $s+ s^{\prime} >2$. Then % \begin{equation} \label{eqn:hs-2} \iint_{ \! {\mathbb R}^3 \times {\mathbb R}^3 } \, \langle x \rangle^{-2s^{\prime}} \frac{1}{ \, |x - y|^2 \,} \langle y \rangle^{-2s} \, dxdy < +\infty. \end{equation} % \end{lem} \begin{proof} We may assume, with no loss of generality, that $s < 3$. Then, by a well known estimate (e.g. Umeda \cite[Lemma 11.1]{Umeda}), we have % \begin{equation} \label{eqn:hs-3} \int_{ {\mathbb R}^3} \frac{1}{ \, |x - y|^2 \,} \langle y \rangle^{-2s} \, dy \le C_s \langle x \rangle^{-2s +1}. \end{equation} % Since $-2s^{\prime} -2s +1 < -3$ by assumption of the lemma, we see that (\ref{eqn:hs-3}) implies (\ref{eqn:hs-2}). \end{proof} \vspace{10pt} It follows from (\ref{eqn:resolventLapc}) and Lemma \ref{lem:hs} that the operator % \begin{equation} \label{eqn:khs-0} K(z) :=\langle x \rangle^{-s^{\prime}} \! \varGamma_0(z) \langle x \rangle^{-s}, \end{equation} % which is represented as % \begin{equation*} \label{eqn:khs-1} K(z) u (x) = \int_{{\mathbb R}^3} \langle x \rangle^{-s^{\prime}} \frac{e^{i \sqrt{z} |x-y|} }{\, 4\pi |x-y| \,} \langle y \rangle^{-s} u(y) \, dy, \end{equation*} % belongs to the Hilbert-Schmidt class on ${L}^2({\mathbb R}^3)$ for $z \in \mathbb C \setminus [0, \,+\infty)$: % \begin{equation*} \label{eqn:khs-1+} \Vert K(z) \Vert^2_{\mbox{\tiny{HS}}} = \iint_{ \! {\mathbb R}^3 \times {\mathbb R}^3 } \, \langle x \rangle^{-2s^{\prime}} \Big| \, \frac{e^{i \sqrt{z} |x-y|} }{\, 4\pi |x-y| \,} \, \Big|^2 \langle y \rangle^{-2s} \, dxdy < \infty, \end{equation*} % where $\Vert \cdot \Vert_{\mbox{\tiny{HS}}}$ denotes the Hilbert-Schmidt norm. Note that % \begin{gather} \label{eqn:hs-4} \begin{split} \Vert K(z_1) - K(z_2) \Vert^2_{\mbox{\tiny{HS}}} \qquad\qquad\qquad\quad \qquad\qquad \qquad\qquad\qquad\quad \\ \noalign{\vskip 4pt} = \iint_{ \! {\mathbb R}^3 \times {\mathbb R}^3 } \, \langle x \rangle^{-2s^{\prime}} \Big| \, \frac{e^{i \sqrt{z_1} |x-y|} }{\, 4\pi |x-y| \,} - \frac{e^{i \sqrt{z_2} |x-y|} }{\, 4\pi |x-y| \,} \, \Big|^2 \langle y \rangle^{-2s} \, dxdy \end{split} \end{gather} % for all $z_1$, $z_2 \in \mathbb C \setminus [0, \,+\infty)$. It follows from (\ref{eqn:hs-4}) that $K(z)$ is continous, with respect to the Hilbert-Schmidt norm topology, on $\mathbb C \setminus [0, \,+\infty)$. Furthermore, we can deduce from (\ref{eqn:hs-2}), (\ref{eqn:hs-4}) and the Lebesgue dominated convergence theorem that $K(z)$ can be continuously extended, with respect to the Hilbert-Schmidt norm topology, as follows: % \begin{equation} \label{eqn:khs-2} \widetilde K(z) = \begin{cases} K(z) & \mbox{if } \; z \in \mathbb C \setminus [0, \, +\infty), \\ {} & \\ K^+(\lambda) & \mbox{if } \; z= \lambda + i0, \; \lambda \ge 0, \\ {} & \\ K^-(\lambda) & \mbox{if } \; z= \lambda - i0, \; \lambda \ge 0, \end{cases} \end{equation} % where $K^+(\lambda)$ and $K^-(\lambda)$ for $\lambda >0$ are defined by % \begin{equation} \label{eqn:khs-3} K^{\pm} (\lambda) u(x) = \int_{{\mathbb R}^3} \langle x \rangle^{-s^{\prime}} \frac{e^{\pm i \sqrt{\lambda} |x-y|} }{\, 4\pi |x-y| \,} \langle y \rangle^{-s} u(y) \, dy, \end{equation} % and % \begin{equation} \label{eqn:khs-4} K^+(0)u(x)= K^-(0)u(x) =\int_{{\mathbb R}^3} \langle x \rangle^{-s^{\prime}} \frac{1}{\, 4\pi |x-y| \,} \langle y \rangle^{-s} u(y) \, dy. \end{equation} % For a later purpose, it is convenient to introduce a subset of the Riemann surface of $\sqrt{z \,}$ as follows: % \begin{gather} \label{eqn:riemann-1} \begin{split} {}&\;\; \Pi_{(0, \, +\infty)} \\ &:= \big( \mathbb C \setminus (0, \, +\infty) \big) \cup \{ \, z = \lambda + i0 \; | \; \lambda > 0 \, \} \cup \{ \, z = \lambda - i0 \; | \; \lambda > 0 \, \}. \end{split} \end{gather} % Thus, we can say that $\widetilde K(z)$ defined by (\ref{eqn:khs-2}) -- (\ref{eqn:khs-4}) is continuous on $\Pi_{(0, \, +\infty)}$ with respect to the Hilbert-Schmidt norm topology. In view of (\ref{eqn:khs-0}), we see that $\varGamma_0(z)$, $z \in \mathbb C \setminus [0, \,+\infty)$, is a Hilbert-Schmidt operator from $L^{2, s}({\mathbb R}^3)$ to $L^{2, -s^{\prime}}({\mathbb R}^3)$. Hence, in particular, $\varGamma_0(z) \in {B}(0, s \, ; \,0 , -s^{\prime})$, and % \begin{equation*} \label{eqn:hs-5-0} \Vert \varGamma_0(z)\Vert_{{B}(0, s \, ; \,0 , -s^{\prime})} \le \Vert K(z) \Vert_{\mbox{\tiny{HS}}} . \end{equation*} % Since we have the inequality % \begin{gather} \begin{split} \label{eqn:hs-5} \Vert \varGamma_0(z_1) - \varGamma_0(z_2) \Vert_{{B}(0, s \, ; \,0 , -s^{\prime})} \le \Vert K(z_1) - K(z_2) \Vert_{\mbox{\tiny{HS} } } \\ \noalign{\vskip 4pt} (z_1, \; z_2 \in \mathbb C \setminus [0, \, &+\infty)) \end{split} \end{gather} % we conclude from (\ref{eqn:khs-2}) and (\ref{eqn:hs-5}) that $\varGamma_0(z)\in {B}(0, s \, ; \,0 , - s^{\prime})$ can be continuously extended as follows: % \begin{equation} \label{eqn:laplace-1} \widetilde\varGamma_0(z) = \begin{cases} \varGamma_0(z) & \mbox{if } \; z \in \mathbb C \setminus [0, \, +\infty), \\ {} & \\ \varGamma_0^+(\lambda) & \mbox{if } \; z= \lambda + i0, \; \lambda \ge 0, \\ {} & \\ \varGamma_0^-(\lambda) & \mbox{if } \; z= \lambda - i0, \; \lambda \ge 0, \end{cases} \end{equation} % where % \begin{equation*} \label{eqn:laplace-2} \varGamma_0^{\pm}(\lambda) := \langle x \rangle^{s^{\prime}} K^{\pm}(\lambda) \langle x \rangle^{s} = \lim_{\epsilon \downarrow 0} \varGamma_0(\lambda \pm i \epsilon) \; \mbox{ in } \; {B}(0, s \, ; \,0 , - s^{\prime}). \end{equation*} % We must remark that % \begin{equation} \label{eqn:resolventLapc-1} \varGamma_0^+(0)u(x)= \varGamma_0^-(0)u(x) =\int_{{\mathbb R}^3} \frac{1}{\, 4\pi |x-y| \,} u(y) \, dy, \end{equation} % and that % \begin{equation*} \label{eqn:resolventLapc-2} \varGamma_0^{\pm} (\lambda) u(x) = \int_{{\mathbb R}^3} \frac{e^{\pm i \sqrt{\lambda} |x-y|} }{\, 4\pi |x-y| \,} u(y) \, dy. \end{equation*} % Thus % \begin{equation*} \label{eqn:resolventLapc-3} \widetilde\varGamma_0(\lambda + i0) \not= \widetilde\varGamma_0(\lambda -i0), \quad \lambda >0. \end{equation*} % Note that the equality (\ref{eqn:resolventLapc-1}) allows us to use the notation % \begin{equation} \label{eqn:resolventLapc-3-1} \widetilde\varGamma_0(0) \big( =\varGamma_0^+(0)= \varGamma_0^-(0) \big). \end{equation} % With the notation introduced in (\ref{eqn:riemann-1}), we can say that $\widetilde\varGamma_0(z)$ is a\linebreak ${B}(0, s \, ; \, 0, - s^{\prime})$-valued continuous function on $\Pi_{(0, \, +\infty)}$. Lemmas \ref{lem:lap-1} and \ref{lem:lap-1-1} below are variants of Lemma 2.1 of Jensen and Kato \cite{JensenKato}, although we shall give their proofs. \begin{lem} \label{lem:lap-1} Let $s$, $s^{\prime}$ satisfy the same assumption as in Lemma \ref{lem:hs}, and let $\mu \in \mathbb R$. Then $\widetilde\varGamma_0(z)$ is a ${B}(\mu, s \, ; \,\mu, - s^{\prime})$-valued continuous function on $\Pi_{(0, \, +\infty)}$. \end{lem} \begin{proof} As was mentioned before, $\widetilde\varGamma_0(z)$ defined by (\ref{eqn:laplace-1}) is a ${B}(0, s \, ; \,0 ,-s^{\prime})$-valued continuous function on $\Pi_{(0, \, +\infty)}$. Then the lemma directly follows from the inequalities % \begin{gather} \label{eqn:laplace-3} \begin{split} \Vert \varGamma_0(z) \Vert_{{B}(\mu, s \, ; \,\mu , -s^{\prime})} \le \Vert \varGamma_0(z) \Vert_{{B}(0, s \, ; \,0 , -s^{\prime})}, \quad z \in \mathbb C \setminus [0, \,+\infty) \end{split} \end{gather} % and % \begin{gather} \label{eqn:laplace-3+} \begin{split} \Vert \varGamma_0(z_1) - \varGamma_0(z_2) \Vert_{{B}(\mu, s \, ; \,\mu , -s^{\prime})} \le \Vert \varGamma_0(z_1) - \varGamma_0(z_2) \Vert_{{B}(0, s \, ; \,0 , -s^{\prime})} \\ \noalign{\vskip 4pt} ( z_1, \, z_2 \in \mathbb C \setminus [0, \,+\infty)). \qquad\qquad\qquad \end{split} \end{gather} % In order to show (\ref{eqn:laplace-3}), we shall use the fact that % \begin{equation} \label{eqn:laplace-4} \langle D \rangle^{\mu} \varGamma_0(z) u = \varGamma_0(z) \langle D \rangle^{\mu} u \end{equation} % for $u\in S({\mathbb R}^3)$ and $z\in \mathbb C \setminus [0, \,+\infty)$. We then have % \begin{gather} \label{eqn:laplace-5-0} \begin{split} \Vert \varGamma_0(z) u \Vert_{H^{\mu, -s{\prime}}} = \Vert \langle D \rangle^{\mu} \varGamma_0(z) u \Vert_{L^{2, -s^{\prime}}} = \Vert \varGamma_0(z) \langle D \rangle^{\mu}u \Vert_{L^{2, -s^{\prime}}} \qquad \\ \noalign{\vskip 4pt} \le \Vert \varGamma_0(z) \Vert_{B(0, s \, ; \,0, -s^{\prime})} \, \Vert \langle D \rangle^{\mu}u \Vert_{L^{2, s}} \qquad\qquad \qquad \qquad \\ \noalign{\vskip 4pt} = \Vert \varGamma_0(z) \Vert_{B(0, s \, ; \,0, -s^{\prime})} \, \Vert u \Vert_{H^{\mu, s}}, \qquad\qquad \qquad \qquad \quad \end{split} \end{gather} % which implies (\ref{eqn:laplace-3}). In a similar fashion, we can prove (\ref{eqn:laplace-3+}). \end{proof} \begin{rem} \label{rem:independence-1} We should remark that ${H}^{\mu, \, s}({\mathbb R}^3)$ in Lemma \ref{lem:lap-1} is a subset of ${L}^2({\mathbb R}^3)$ for $\mu \ge 0$, but not neccesarily for $\mu <0$. Thus $\widetilde\varGamma_0(z)$ may depend on $\mu$ and $s$. Nonetheless, we have the unique representation of $\widetilde\varGamma_0(z)$ on $S({\mathbb R}^3)$, a dense subset of ${H}^{\mu, \, s}({\mathbb R}^3)${\rm :} % \begin{equation} \label{eqn:kernelrep-1} \widetilde\varGamma_0(z) u(x) = \int_{{\mathbb R}^3} \frac{e^{i \sqrt{z} |x-y|} }{\, 4\pi |x-y| \,} u(y) \, dy, \qquad u \in S({\mathbb R}^3), \end{equation} % for every $z \in \Pi_{(0, \, +\infty)}$, where $\mbox{\rm Im}\sqrt{z} \ge 0$. This representation, together with the fact that $S({\mathbb R}^3)$ is dense in ${H}^{\mu, \, s}({\mathbb R}^3)$ for any pair of $\mu$ and $s$, ensures that the extension of $\widetilde\varGamma_0(z){\big|}_{S({\mathbb R}^3)}$ to ${H}^{\mu, \, s}({\mathbb R}^3)$ is independent of $\mu$ and $s$ in a certain sense. Howver, we shall not discuss about the uniqueness of the extension any longer. In the discussions below, we shall mostly deal with the extension of $\widetilde\varGamma_0(z){\big|}_{S({\mathbb R}^3)}$ to ${H}^{-1, \, s}({\mathbb R}^3)$. \end{rem} \begin{lem} \label{lem:lap-1-1} Let $s$, $s^{\prime}$ satisfy the same assumption as in Lemma \ref{lem:hs}, and let $\mu \in \mathbb R$. Then $\widetilde\varGamma_0(z)$ is a ${B}(\mu -2 , s \, ; \,\mu, - s^{\prime})$-valued continuous function on $\Pi_{(0, \, +\infty)}$. \end{lem} \begin{proof} We first note that % \begin{equation} \label{eqn:laplace-5} \langle D \rangle^{2} \varGamma_0(z) u = u + (z +1) \varGamma_0(z) u \end{equation} % for $u\in S({\mathbb R}^3)$ and $z\in \mathbb C \setminus [0, \,+\infty)$; cf. Jensen and Kato \cite[Lemma 2.1]{JensenKato}. (See (\ref{eqn:bracket-D}) for the definition of $\langle D \rangle$.) We then combine (\ref{eqn:laplace-5}) with Lemma \ref{lem:lap-1}, and obtain the conclusion if we appeal to the fact that $S({\mathbb R}^3)$ is dense in ${H}^{\mu -2, s}({\mathbb R}^3)$. \end{proof} \vspace{10pt} In the rest of the paper, we shall need a variant of Lemma \ref{lem:lap-1-1}, namely a version for four-component vector-valued functions, with $\mu = 1$ in the form described in Lemma \ref{lem:lap-1-2}. Thus $\widetilde\varGamma_0(z)$ in Lemma \ref{lem:lap-1-2} below denotes a copy of $\widetilde\varGamma_0(z)$; see (\ref{eqn:mapsto}). \begin{lem} \label{lem:lap-1-2} Let $s$, $s^{\prime}$ satisfy the same assumption as in Lemma \ref{lem:hs}. Then $\widetilde\varGamma_0(z)$ is a ${\mathcal B}(-1 , s \, ; \, 1, - s^{\prime})$-valued continuous function on $\Pi_{(0, \, +\infty)}$. \end{lem} \vspace{10pt} \subsection{The resolvent of the free Dirac operator $H_0$} In view of (\ref{eqn:reslvnDrc-0}), it is convenient for us to introduce the following operator valued-functions $\varOmega_0^{+}(z)$ defined on $\overline{{\mathbb C}}_{+}$ and $\varOmega_0^{-}(z)$ on $\overline{{\mathbb C}}_{-}$ as follows: % \begin{equation} \label{eqn:laplace-6-0} \varOmega_0^{\pm}(z) =\widetilde\varGamma_0(z^2), \quad z \in \overline{{\mathbb C}}_{\pm}, \end{equation} % in other words, % \begin{equation} \label{eqn:laplace-6} \varOmega_0^{\pm}(z) = \begin{cases} \varGamma_0(z^2) & \mbox{if } \; z \in {\mathbb C}_{\pm}, \\ {} & \\ \varGamma_0^{\pm}(\lambda^2) & \mbox{if } \; z= \lambda \ge 0, \\ {} & \\ \varGamma_0^{\mp}(\lambda^2) & \mbox{if } \; z= \lambda \le 0. \end{cases} \end{equation} % It follows from Lemma \ref{lem:lap-1-2} that $\varOmega_0^+(z)$ ( resp. $\varOmega_0^-(z)$) is a $\mathcal{B}(-1 , s \, ; \,1, - s^{\prime})$-valued continuous function on $\overline{\mathbb C}_+$ (resp. $\overline{\mathbb C}_-$). Also, it follows from (\ref{eqn:resolventLapc-3-1}) that % \begin{equation} \label{eqn:addtional1} \varOmega_0^+(0) = \varOmega_0^-(0) =\widetilde\varGamma_0(0). \end{equation} % In order to get expressions of the extended resolvents of the free Dirac operator $H_0$ in terms of $\varOmega_0^{\pm}(z)$ inroduced in (\ref{eqn:laplace-6}), we shall exploit the decomposition (\ref{eqn:reslvnDrc-0}) and a boundedness estimate of $H_0$ in some weighted Sobolev spaces which is given as follow. \begin{lem} \label{lem:dirac-bdd} Let $\mu$ and $s^{\prime}$ be in $\mathbb R$. Then % \begin{equation*} \label{eqn:dirac-bdd-2} H_0 \in \mathcal B (\mu, -s^{\prime} ; \mu-1, -s^{\prime}). \end{equation*} % \end{lem} \begin{proof} To prove the lemma, it is sufficient to show that each $D_j$, ($j=1$, $2$, $3$) is a bounded operator from ${\mathcal H}^{\mu, -s^{\prime}}$ to ${\mathcal H}^{\mu-1, -s^{\prime}}$. This fact is a direct consequence of Umeda \cite[Lemma 2.1]{Umeda0}. \end{proof} \vspace{5pt} \begin{lem} \label{lem:lap-2} Let $s$, $s^{\prime} > 1/2$, and $s + s^{\prime}>2$. Then $R_0(z) \in {\mathcal B}(-1, s \, ; \, 0, - s^{\prime})$ is continuous in $z \in {\mathbb C}_{\pm}$. Moreover, as ${\mathcal B}(-1, s \, ; \, 0, - s^{\prime})$-valued functions, they can possess continuous extenstions $R_0^{\pm}(z)$ to $\overline{\mathbb C}_{\pm}$ respectively, and % \begin{equation} \label{eqn:dirac-1} R_0^{\pm}(z) = (H_0 + z) \varOmega_0^{\pm}(z), \quad z \in \overline{\mathbb C}_{\pm}. \end{equation} \end{lem} % \begin{proof} We shall give the proof only for $z \in \overline{\mathbb C}_{+}$. The proof for $z \in \overline{\mathbb C}_{-}$ is similar. As was mentioned before Lemma \ref{lem:dirac-bdd}, $\varOmega_0^{+}(z)$ is a $\mathcal{B}(-1 , s \, ; \,1, - s^{\prime})$-valued continuous function on $\overline{\mathbb C}_+$. Combining this fact with (\ref{eqn:reslvnDrc-0}), (\ref{eqn:reslvnDrc-1}), the definition (\ref{eqn:laplace-6-0}) (or (\ref{eqn:laplace-6})) of $\varOmega_0^{+}(z)$, Lemma \ref{lem:lap-1-2} and Lemma \ref{lem:dirac-bdd} with $\mu=1$, we see that $R_0(z)= (H_0 + z) \varOmega_0^{+}(z) \in {\mathcal B}(-1, s \, ; \, 0, - s^{\prime})$ for any $z \in {\mathbb C}_{+}$. Now it is evident that the second assertion of the lemma follows from Lemma \ref{lem:lap-1-2} and Lemma \ref{lem:dirac-bdd} with $\mu=1$. \end{proof} Combining (\ref{eqn:dirac-1}) with (\ref{eqn:addtional1}), we obtain a corollary to Lemma \ref{lem:lap-2}. \begin{cor} \label{cor:origin0} Under the same assumption and the same notation as in Lemma \ref{lem:lap-2}, % \begin{equation*} \label{eqn:additional2} R_0^+(0)=R_0^-(0) = H_0\widetilde\varGamma(0). \end{equation*} % \end{cor} \vspace{4pt} \begin{rem} In \cite{IftMant}, Iftimovici and M\u{a}ntoiu showed that the limiting absorption principle for the the free Dirac operator $H_0= \alpha \cdot D + m \beta$, $m>0$, in ${\mathcal B}(0, 1 \, ; \, 0, - 1)$ holds on the whole real line. With the result exhibited in Lemma \ref{lem:lap-2}, together with the result in \cite{IftMant}, the limiting absorption principle for the the free Dirac operator $H_0= \alpha \cdot D + m \beta$ has been established for all $m \ge 0$. \end{rem} \begin{lem} \label{lem:ker3-2} For $f \in { \mathcal S }$ and $z \in {\mathbb C}_{\pm}$ % \begin{eqnarray} \label{eqn:3-2} {}&R_0(z)f (x) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ {}&= \displaystyle{ \int_{{\mathbb R}^3} \Big( i\, \frac{\alpha \cdot (x-y)}{|x-y|^2} \pm z \, \frac{\alpha \cdot (x-y)}{|x-y|} + zI_4 \Big) \frac{e^{\pm i z |x-y|}}{4\pi |x-y|} } f(y) \, dy. \nonumber \end{eqnarray} \end{lem} % \begin{proof} We first recall (\ref{eqn:kernelrep-1}), which we can write as % \begin{equation} \label{eqn:kernelrep-1-1} \varGamma_0(z^2) f(x) = \int_{{\mathbb R}^3} \frac{e^{ \pm i z |y|} }{\, 4\pi |y| \,} f(x-y) \, dy, \qquad f \in {\mathcal S}, \;\; z \in {\mathbb C}_{\pm}. \end{equation} % We then combine (\ref{eqn:laplace-6-0}) and (\ref{eqn:dirac-1}), and make differentiation under the integral sign in (\ref{eqn:kernelrep-1-1}), which gives \begin{gather} \label{eqn:3-2-0} \begin{split} {}&R_0(z)f (x) = \int_{{\mathbb R}^3} \frac{e^{\pm i z |y|}}{4\pi |y|} (\alpha \cdot D_x +z I_4) f(x-y) \, dy , \quad z \in {\mathbb C}_{\pm}. \end{split} \end{gather} % Noting the fact that % \begin{equation*} \label{eqn:differentiation} D_x f(x-y)= - D_y\big( f(x-y) \big), \end{equation*} % and making integration by parts on the right hand of (\ref{eqn:3-2-0}) implies that % \begin{gather} \label{eqn:3-2-1} \begin{split} R_0(z)f (x) &= \int_{{\mathbb R}^3} \Big( i\, \frac{\alpha \cdot y}{|y|^2} \pm z \, \frac{\alpha \cdot y}{|y|} + zI_4 \Big) \frac{e^{\pm i z |y|}}{4\pi |y|} f(x-y) \, dy \\ \noalign{\vskip 4pt} {}& \qquad\quad (z \in {\mathbb C}_{\pm}). \end{split} \end{gather} % A change of variables in (\ref{eqn:3-2-1}) yields (\ref{eqn:3-2}). (See also Thaller \cite[p.39]{Thaller}.) \end{proof} \begin{lem} \label{lem:riesz3-3} For $f \in { \mathcal S }$ % \begin{equation*} \label{eqn:3-3} R_0^{+}(0) f = R_0^{-}(0) f = A f, \end{equation*} where $A$ is the singular integral operator defined by {\rm(\ref{eqn:sio-1})}. \end{lem} % \begin{proof} In view of Corollary \ref{cor:origin0}, we only need to give the proof for $R_0^+(0)$. Let $f \in { \mathcal S }$, and let $\{ z_n \} \subset {\mathbb C}_{+}$ be a sequence such that $z_n \to 0$ as $n \to \infty$. It follows from Lemma \ref{lem:lap-2} that $R_0(z_n)f \to R^+_0(0)f$ in ${\mathcal L}^{2, -s^{\prime}}$ as $n \to \infty$. This fact implies that there exists a subsequece $\{ z_{n^{\prime}} \} \subset\{ z_n \}$ suth that % \begin{equation} \label{eqn:convergence-1} R_0(z_{n^{\prime}})f(x) \to R^+_0(0)f(x) \;\; \mbox{ a.e.} x \in {\mathbb R}^3 \;\;\; \mbox{ as } n^{\prime} \to \infty. \end{equation} % On the other hand, Lemma \ref{lem:ker3-2}, togher with the Lebesgue dominated convergence theorem, implies that % \begin{equation} \label{eqn:convergence-2} R_0(z_{n})f(x) \to \int_{{\mathbb R}^3} i\, \frac{\alpha \cdot (x-y)}{4\pi |x-y|^3} f(y) \, dy = Af(x) \;\;\; \mbox{ as } n \to \infty \end{equation} % for each $x \in {\mathbb R}^3$. The conclusion of the lemma now follows from (\ref{eqn:convergence-1}) and (\ref{eqn:convergence-2}). \end{proof} \begin{lem} \label{lem:riesz3-4} Let $s$, $s^{\prime} > 1/2$, and $s + s^{\prime}>2$. Then $A$ can be continuously extended to an operator in ${\mathcal B}(-1, s \, ; \, 0, - s^{\prime})$. \end{lem} % % \begin{proof} Since $\mathcal S$ is dense in ${\mathcal H}^{-1, s}$, Lemmas \ref{lem:lap-2} and \ref{lem:riesz3-3} directly imply the lemma. \end{proof} In the rest of the paper, we shall denote the extension in Lemma \ref{lem:riesz3-4} by $A$ again. Thus we have % \begin{equation*} R_0^+(0) =R_0^-(0) = A \; \mbox{ in } \; {\mathcal B}(-1, s \, ; \, 0, - s^{\prime}). \end{equation*} % \begin{lem} \label{lem:riesz3-5-0} Let $s > 1/2$. Then % \begin{equation} \label{eqn:adjoint-1} H_0 A g = g \end{equation} % for all $g \in {\mathcal L}^{2, s}$. \end{lem} % \begin{proof} Let $g \in {\mathcal L}^{2, s}$ be given. We then start with the fact that % \begin{equation} \label{eqn:adjoint-1+} (H_0 - z) R_0(z)g =g \quad ( \forall z \in {\mathbb C}_+ ). \end{equation} % Choose $s^{\prime} > 1/2$ so that $s + s^{\prime} >2$. We see from Lemmas \ref{lem:lap-2}, \ref{lem:riesz3-3} and \ref{lem:riesz3-4} that % \begin{equation} \label{eqn:pair-4} R_0(i/n) g \to R_0^+(0)g = Ag \;\; \mbox{ \rm in } \;\; {\mathcal L}^{2, -s^{\prime}} \quad {\mbox \rm as } \;\; n \to \infty . \end{equation} % Lemma \ref{lem:dirac-bdd}, with $\mu=0$, and (\ref{eqn:pair-4}) imply that % \begin{equation} \label{eqn:pair-5} (H_0 -\frac{i}{\, n \,}) R_0(\frac{i}{\, n \,}) g \to H_0 Ag \;\; \mbox{ \rm in } \;\; {\mathcal H}^{-1, -s^{\prime}} \quad {\mbox \rm as } \;\; n \to \infty. \end{equation} % Since, by (\ref{eqn:adjoint-1+}), % \begin{equation*} \label{eqn:pair-5+} (H_0 - \frac{i}{\, n \,} ) R_0(\frac{i}{\, n \,})g =g \quad \mbox{ for }\forall n, \end{equation*} % we find that (\ref{eqn:pair-5}) yields (\ref{eqn:adjoint-1}). \end{proof} \vspace{5pt} We shall need Lemma 2.4 of Jensen and Kato \cite{JensenKato}, which we shall rewrite in a suitable form to our setting (cf. Lemma \ref{lem:riez3-5-7} below), where the operators $-\Delta$ and $\widetilde\varGamma_0(0)$ act on four-component vector functions. The reader should note that $\widetilde\varGamma_0(0)$ is the same as $G_0$ in the Jensen-Kato paper. See (\ref{eqn:resolventLapc-1}) and (\ref{eqn:resolventLapc-3-1}). \begin{lem}[\textbf{Jensen-Kato}] \label{lem:riez3-5-7} Let $s >1/2$. Then \vspace{3pt} {\rm(i)} $(-\Delta) \widetilde\varGamma_0(0) g = g$ \ for all \ $g \in {\mathcal H}^{-1, s}.$ \vspace{2pt} {\rm(ii)} $ \widetilde\varGamma_0(0) (-\Delta)f = f \;$ if $\; f \in {\mathcal L}^{2, -3/2} \ and \ (-\Delta)f \in {\mathcal H}^{-1, s}$. \end{lem} % \vspace{4pt} \begin{lem} \label{lem:riez3-5-8} Let $s >1/2$. Then \ $\widetilde\varGamma_0(0) H_0 \, g = Ag$ \ for all $g \in {\mathcal L}^{2, s}.$ \end{lem} % % \begin{proof} Let $g \in {\mathcal L}^{2, s}$ be given. Noting that $H_0^2 = -\Delta$ (cf. (\ref{eqn:diracsquare})), we have % \begin{equation} \label{eqn:identity-10} (-\Delta) A g =H_0 (H_0 Ag) =H_0 \, g \end{equation} % where we have used Lemma \ref{lem:riesz3-5-0} in the second equality. Since $H_0 \, g \in {\mathcal H}^{-1, s}$ by Lemma \ref{lem:dirac-bdd}, it follows from (\ref{eqn:identity-10}) that $(-\Delta) A g \in {\mathcal H}^{-1, s}$. On the other hand, we find, by Lemma \ref{lem:riesz3-4}, that $Ag \in {\mathcal L}^{2, -3/2}$, because we can choose $s^{\prime}$ so that $1/2 < s^{\prime} \le 3/2$ \ and $s + s^{\prime} >2$. (Choose $s^{\prime}$ so that $\max (s, \, 2-s) < s^{\prime} \le 3/2$.) Now we can apply Lemma \ref{lem:riez3-5-7}(ii) with $f$ replaced by $Ag$, and obtain % \begin{equation} \label{eqn:identity-11} \widetilde\varGamma_0(0)(-\Delta) A g =Ag. \end{equation} % It follows from (\ref{eqn:identity-10}) that the left hand side of (\ref{eqn:identity-11}) equals $\widetilde\varGamma_0(0) H_0g$. This proves the conclusion of the lemma. \end{proof} % % \vspace{2pt} % \begin{lem} \label{lem:riesz3-5} If $f \in {\mathcal L}^{2, -3/2}$ and $H_0 f \in {\mathcal L}^{2, s}$ for some $\, s > 1/2$, then $AH_0 f = f$. \end{lem} % % \begin{proof} Put % \begin{equation*} \label{eqn:identity-12} g= H_0 f. \end{equation*} % By assumption of the lemma, we see that $g \in {\mathcal L}^{2, s}$ for some $\, s > 1/2$. It follows from Lemma \ref{lem:riez3-5-8} that $Ag = \widetilde\varGamma_0(0) H_0 \, g$, {\it i.e.}, % \begin{equation*} \label{eqn:identity-13} AH_0f = \widetilde\varGamma_0(0) H_0 H_0f = \widetilde\varGamma_0(0) (-\Delta)f. \end{equation*} % Since $(-\Delta)f=H_0g \in {\mathcal H}^{-1, s}$ by Lemma \ref{lem:dirac-bdd}, it follows from assertion (ii) of Lemma \ref{lem:riez3-5-7} that $\widetilde\varGamma_0(0) (-\Delta)f=f$. Thus $AH_0f =f$. \end{proof} \vspace{15pt} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Pointwise estimates for zero modes and their continuity} \begin{proof}[{\bf Proof of Theorem \ref{thm:th-2-2}}] We first prove assertion (i). Let $f$ be a zero mode of the operator (\ref{eqn:1-1}). Then we have % \begin{equation} \label{eqn:proof-1} Hf = \big( \alpha \cdot D + Q(x) \big) f =0, \quad f \in \mbox{Dom}(H) = {\mathcal H}^1 . \end{equation} % It follows from (\ref{eqn:proof-1}) and Assumption (A) that % \begin{equation} \label{eqn:proof-2} H_0f =(\alpha \cdot D) f = - Q(x) f \in {\mathcal L}^{2, \rho}. \end{equation} % (Recall (\ref{eqn:reslvnDrc-00}) for the definiton of $H_0$.) Since $\rho >1 >1/2$ by assumption of the theorem, we can apply Lemma \ref{lem:riesz3-5} to (\ref{eqn:proof-2}) and get % \begin{equation} \label{eqn:proof-3} f = AH_0 f = - A Q f. \end{equation} %%%%%where we have used Lemmas \ref{lem:riesz3-3} and %%%%%%\ref{lem:riesz3-4} in the third equality of %%%%%%%(\ref{eqn:proof-3}). It follows from (\ref{eqn:proof-3}) and Lemma \ref{lem:siolem-1} that $f \in {\mathcal L^2} \cap {\mathcal L}^6$. It follows from (\ref{eqn:proof-3}) again and Lemma \ref{lem:siolem-3} that $f \in {\mathcal L}^{\infty}$. This fact, together with (\ref{eqn:proof-3}) and Assumption(A), implies that % \begin{equation} \label{eqn:proof-4} | f(x) | \le C \Vert f \Vert_{{\mathcal L}^{\infty}} \int_{{\mathbb R}^3} \frac{1}{\, |x-y|^2 \, \langle y \rangle^{\rho} \,} \, dy. \end{equation} % Noting that $\rho >1$ by assumption, and applying Lemma 11.1 in Appendix of Umeda \cite{Umeda} to the integral in (\ref{eqn:proof-4}), we get the desired inequalities. We next prove assertion (ii) by utilizing (\ref{eqn:proof-3}): % \begin{equation*} \label{eqn:proof-4+1} f(x) = -\displaystyle{ \int_{{\mathbb R}^3} i \, \frac{\alpha \cdot (x - y)}{4\pi |x-y|^3} Q(y) f(y) \, dy. } \end{equation*} % Let $x_0$ be any point in ${\mathbb R}^3$, and let $\varepsilon >0$ be given. We choose $r>0$ so that % \begin{equation} \label{eqn:proof-4+2} \displaystyle{ \int_{|y| \le 2r} \frac{1}{\, |y|^2 \,} \, dy < \varepsilon}. \end{equation} % We then decompose $f(x)$ into two parts: % \begin{gather} \label{eqn:proof-4+3} \begin{split} f(x) &= - \Big( \int_{B(x, \,2r)} + \int_{E(x,\,2r)} \Big) \, i \, \frac{\alpha \cdot (x - y)}{4\pi |x-y|^3} Q(y) f(y) \, dy \\ \noalign{\vskip 4pt} &=: f_b(x) + f_e(x), \end{split} \end{gather} % where % \begin{equation*} \label{eqn:proof-4+4} B(x, \, 2r) = \{ \, y \; \big| \,|x - y| \le 2r \, \}, \quad E(x, \,2r) = \{ \, y \; \big| \,|x - y| > 2r \, \}. \end{equation*} % Since each $\alpha_j$ is a unitary matrix, it follows from (\ref{eqn:proof-4+2}) and (\ref{eqn:proof-4+3}) that % \begin{equation} \label{eqn:proof-4+5} |f_b(x)| < \frac{3}{\, 4\pi \,} C_q C_f \,\varepsilon \quad \mbox{ for }\; \forall x \in {\mathbb R}^3 , \end{equation} % where $C_q$ is a constant determined by (\ref{eqn:2-1}) in Assumption (A) and $C_f$ is a constant described in the inequality (\ref{eqn:2-2}), which we have just proved in the first half of the proof. It follows from the definition of $f_e(x)$ that % \begin{gather} \label{eqn:proof-4+6} \begin{split} {}&f_e(x) - f_e(x_0) \\ \noalign{\vskip 4pt} &= \int_{{\mathbb R}^3} \! \Big\{ 1_{E(x_0, \, 2r)}(y) \frac{\alpha \cdot (x_0 - y)}{4\pi |x_0-y|^3} - 1_{E(x, \, 2r)}(y) \frac{\alpha \cdot (x - y)}{4\pi |x-y|^3} \Big\} \\ \noalign{\vskip 4pt} &\qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \times Q(y) f(y) \, dy . \end{split} \end{gather} % To apply the Lebesgue dominated convergence theorem to the integral in (\ref{eqn:proof-4+6}), we need the following two facts that % \begin{equation} \label{eqn:proof-4+7} |x_0 - y | \ge r \quad \mbox{ if } \; |x_0 - x| 2r. \end{equation} % and that % \begin{equation} \label{eqn:proof-4+8} |x - y | \ge \frac{2}{\, 3 \,} |x_0 -y| \quad \mbox{ if } \; |x_0 - x| 2r \end{equation} % (use the inequality $|x-y| \ge |x_0 -y| - |x_0 -x|$). We can deduce from (\ref{eqn:proof-4+7}) and (\ref{eqn:proof-4+8}) that % \begin{gather} \label{eqn:proof-4+9} \begin{split} {}&\Big| 1_{E(x, \, 2r)}(y) \frac{\alpha \cdot (x - y)}{4\pi |x-y|^3} Q(y) f(y) \Big| \\ &\quad \le 1_{E(x_0, \, r)}(y) \frac{3}{\, 4\pi \,} \Big( \frac{2}{\, 3 \,} |x_0 -y| \Big)^{-2} \, \big| Q(y) f(y) \big| \end{split} \end{gather} % whenever $|x_0 - x| < r$. It is straightforward that the estimate (\ref{eqn:proof-4+9}) implies % \begin{gather} \label{eqn:proof-4+10} \begin{split} {}&\big| \mbox{the integrand in (\ref{eqn:proof-4+6})} \big| \\ \noalign{\vskip 4pt} &\le 1_{E(x_0, \, r)}(y) \frac{3}{\, 4\pi \,} \Big( 1+ (\frac{3}{\, 2 \,})^2 \Big) |x_0 -y|^{-2} \, \big| Q(y) f(y) \big| \end{split} \end{gather} % whenever $|x_0 - x| < r$. In view of (\ref{eqn:2-1}) in Assumption (A) and the inequality (\ref{eqn:2-2}), the function on the right hand side of (\ref{eqn:proof-4+10}) is integrable on ${\mathbb R}^3$. Thus, we can apply the Lebesgue dominated convergence theorem to the integral in (\ref{eqn:proof-4+6}), and conclude that % \begin{equation} \label{eqn:proof-4+11} \lim_{x \to x_0} \big( f_e(x) - f_e(x_0) \big)=0. \end{equation} % Combining (\ref{eqn:proof-4+11}) with both (\ref{eqn:proof-4+3}) and (\ref{eqn:proof-4+5}) yields % \begin{equation*} \label{eqn:proof-4+12} \limsup_{x \to x_0} \big| f(x) - f(x_0) \big| \le 2\times \frac{3}{\, 4\pi \,} C_q C_f \,\varepsilon . \end{equation*} % Since $\varepsilon$ was arbitrary, this completes the proof of assertion (ii). \end{proof} \vspace{15pt} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Nonexistence of zero resonances} \begin{lem} \label{lem:sobolev-1} If $f \in {\mathcal L}^{2}$ and $(\alpha \cdot D) f \in {\mathcal L}^{2}$, then $f \in{\mathcal H}^{1}$. \end{lem} % % \begin{proof} We take the Fourier transform of $(\alpha \cdot D)f$, and we have % \begin{equation} \label{eqn:proof-5} {\mathcal F}[ (\alpha \cdot D)f] = (\alpha \cdot \xi) \hat f, \end{equation} % where % \begin{equation*} \label{eqn:proof-5+} {\mathcal F}[f](\xi) = \hat f(\xi) = (2\pi)^{-2/3} \int_{{\mathbb R}^3} e^{-ix\cdot \xi} \, f(x) \, dx. \end{equation*} % Then by using assumption of the lemma and (\ref{eqn:proof-5}), we see that % \begin{gather} \label{eqn:proof-6} \begin{split} \infty > \Vert (\alpha \cdot D)f \Vert_{{\mathcal L}^2}^2 &= \int_{{\mathbb R}^3} | (\alpha \cdot \xi) \hat f (\xi) | ^2 \, d\xi \\ &= \int_{{\mathbb R}^3} < \!\! (\alpha \cdot \xi) \hat f (\xi) , (\alpha \cdot \xi) \hat f (\xi) \!\! >_{\!{}_{\mathbb C}} \, d\xi \\ &= \int_{{\mathbb R}^3} |\xi|^2 |\hat f (\xi)|^2 \, d\xi, \end{split} \end{gather} % where $< \cdot, \, \cdot>_{\!{}_{\mathbb C}}$ denotes the inner product of ${\mathbb C}^4$. In the third equality of (\ref{eqn:proof-6}), we have used the anti-commutation relation of the Dirac matrices: $\alpha_j \alpha_k + \alpha_k \alpha_j = 2\delta_{jk}$, $\delta_{jk}$ being $=1$ if $j=k$, and $=0$ if $j\not=k$. Since, by assumption of the lemma, $f \in {\mathcal L}^{2}$, the coclusion of the lemma follows from (\ref{eqn:proof-6}). \end{proof} \begin{proof}[{\bf Proof of Theorem \ref{thm:th-2-1}}] Let $f$ satisfy the assumption of the theorem: $f\in {\mathcal L}^{2,-s}$ for some $s$ with $1/2 < s < \min\{3/2, \, \rho -3/2\}$. In the same manner as in (\ref{eqn:proof-2}) and (\ref{eqn:proof-3}), we can show that % \begin{equation} \label{eqn:proof-7} H_0f=(\alpha \cdot D) f =- Qf \in {\mathcal L}^{2,\rho-s}, \end{equation} % and that % \begin{equation} \label{eqn:proof-8} f= -A Qf. \end{equation} % Note that $\rho -s >3/2>1/2$, which we have used to apply Lemma \ref{lem:riesz3-5} in showing (\ref{eqn:proof-8}). It follows from Lemma \ref{lem:siolem-2}(i) that $f \in {\mathcal L}^2$. This fact, together with (\ref{eqn:proof-7}) and Lemma \ref{lem:sobolev-1}, gives the conclusion of the theorem. \end{proof} \vspace{20pt} {\bf Acknowledgement:} T.U. would like to express his gratitude to Michael Loss for the hospitality during his visit to Georgia Institute of Technology, USA, in April, 2002. Discussions with Michael were one of the motivations of the present paper. Also, he would like to express his thanks to the Department of Mathematics, the University of Alabama at Birmingham, USA, for their hospitality. 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