\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2004(2004), No. 102, pp. 1--19.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2004/102\hfil Low regulairty for Dirac-Klein-Gordon
equations]
{Low regularity solutions for Dirac-Klein-Gordon equations in one
space dimension}

\author[Yung-fu Fang\hfil EJDE-2004/102\hfilneg]
{Yung-fu Fang}

\address{Department of Mathematics, National Cheng Kung University,
Tainan 701 Taiwan}
\email{fang@math.ncku.edu.tw}


\date{}
\thanks{Submitted August 13, 2004. Published August 27, 2004.}
\subjclass[2000]{35L70} 
\keywords{Dirac-Klein-Gordon; null form; low regularity; fixed point argument}


\begin{abstract}
 We establish the existence of local and global solutions for
 Dirac-Klein-Gordon equations in one space dimension.
 This is done using a null form estimate and a fixed point argument.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}


\section{Introduction and Main Results}

  In this paper, we study the Cauchy problem for the
Dirac-Klein-Gordon system. The unknown quantities are a spinor field
$\psi :\mathbb{R}\times\mathbb{R}^{1}\mapsto\mathbb{C}^{4}$ and a
scalar field $\phi:\mathbb{R}\times\mathbb{R}^{1}\mapsto\mathbb{R}$.
The evolution equations for these fields are
\begin{equation}
\begin{gathered}
{\mathcal{D}}\psi=\phi\psi, \quad(t,x)\in\mathbb{R}\times\mathbb{R}^1; \\
\square\phi=\overline{\psi}\psi; \\
\psi(0,x):=\psi_0(x),\quad\phi(0,x)=\phi_0(x),\quad\phi_{,t}(0,x)=
\phi_1(x), 
\end{gathered} \label{e0.1}
\end{equation}
where $\mathcal{D}$ is the Dirac operator,
$\mathcal{D}:=-i\gamma^\mu\partial_\mu$, $\mu = 0,1$, and $\gamma^\mu$
are the Dirac matrices, the wave operator $\square = -\partial_{tt}
+\partial_{xx}$,and $\overline \psi = \psi^\dagger \gamma^0$, and
$\dagger$ is the complex conjugate transpose.

  The purpose of this work is to demonstrate a variant
null form estimate, by employing the solution representations in
Fourier transform of the DKG equations. We will take advantage of the
null form structure depicted in the nonlinear term $\overline \psi
\psi$, which has been observed by \cite{KM} and \cite{Bo}. We interpret
the null form in a way that is different from that given in Bournaveas'
paper \cite{Bo}. Equipping with this estimate, we can lower the
regularity of the spinor field.

   For the DKG system, there are many conserved quantities which
are not positive definite, such as the energy. However the known
positive conserved quantity is the law of conservation of charge,
\begin{equation}
  \int |\psi(t)|^2\,dx   = \text{constant} \label{e0.2}
\end{equation}
which is applicable to lead to the global existence result, once the
local existence result is established, see \cite{Bo} and \cite{F2}.

  In 1973, Chadam showed that the Cauchy problem for the
DKG equations has a global unique solution for $\psi_0 \in H^1$,
$  \phi_0 \in H^1$, $  \phi_1 \in L^2$, see \cite{C}.
  In 1993, Zheng proved that there exists a global weak solution to the
Cauchy problem of a modified DKG equations, based on the technique
of compensated compactness, with $  \psi_0 \in L^2$, $  \phi_0
\in H^1$, $\phi_1 \in L^2$, see \cite{Z}.
  In 2000, Bournaveas derived a new proof of a global existence for the
DKG equations, via a null form estimate, if $\psi_0 \in L^2$,
$\phi_0 \in H^1$, $\phi_1 \in L^2$, see \cite{B}.
In 2002, Fang gave a direct proof for \eqref{e0.1}, based on a variant
null form estimate, which is straight forward, and the result
is parallel to Bournaveas', see \cite{F2}

  The outline of this paper is as follows. First we derive some
solutions representations in Fourier transform, depending on various
purposes. Next we prove some a priori estimates of solutions for Dirac
equation and for wave equation. Then we show a local and global results
for \eqref{e0.1}, employing the null form estimate together with other
estimates derived previously, and a fixed point argument. Finally we
show the null form estimate.

  The main result in this work is as follows.

\begin{theorem}[Local Existence] \label{thm0.1}
Let $ 0<\epsilon\leq\frac14$
and $0<\delta\leq2\epsilon$. If the initial data of \eqref{e0.1}
$\psi_0 \in H^{-\frac14+\epsilon}$, $  \phi_0 \in H^{\frac12+\delta}$,
$\phi_1 \in H^{-\frac12+\delta}$, then there is a unique local solution
for \eqref{e0.1}.
\end{theorem}

\begin{theorem}[Global Existence] \label{thm0.2}
Let $\delta>0$. If the initial
data of \eqref{e0.1} $  \psi_0 \in L^2$, $  \phi_0 \in
H^{\frac12+\delta}$, $  \phi_1 \in H^{-\frac12+\delta}$, then there
is a unique global solution for \eqref{e0.1}.
\end{theorem}

\subsection*{Remarks}
{\bf 1.} The $DKG$ equations follow from the Lagrangian
\begin{equation}
\int_{\mathbb{R}^{1+1}} \big\{ |\nabla\phi|^2 - |\phi_t|^2 -
\overline\psi\mathcal{D}\psi - \phi\overline\psi\psi \big\} dx \,dt.
\label{e0.3}
\end{equation}

\noindent{\bf 2.} The Dirac-Klein-Gordon system must be
\begin{equation}
\mathcal{D} \psi = \phi\psi; \\
\square \phi + m^2\phi = \overline\psi\psi, \label{e0.4}
\end{equation}

\noindent{\bf 3.} $\widehat{\mathcal{D}}^2 = \widehat\square I$, where
$I$ is the $4\times4$ identity matrix.

\noindent{\bf 4.} $\overline\psi\psi = \psi^\dagger \gamma^0 \psi =
|\psi_1|^2 + |\psi_2|^2 - |\psi_3|^2 - |\psi_4|^2$, where $
\psi_j$ are the component functions of the vector function $\psi$,
which take values in $\mathbb{C}$.
\smallskip

  The case $\delta=0$ is critical in the following sense. Assuming
that the initial data $(\phi_0,\phi_1)$ are in $\displaystyle
H^{\frac12}\times H^{-1/2}$ does not imply that $\phi(t,\cdot)$ is
bounded. In fact, it is a BMO function. One of the motivations for
proving the existence of global solution with low regularity, is based
on an observation made by Grillakis, which is that the initial data of
\eqref{e0.1}: $  \psi_0 \in L^2$, $  \phi_0 \in H^{\frac12}$, $  \phi_1
\in H^{-\frac12}$, is a right space for the existence of an invariant
measure, see \cite{B} and \cite{Ku}, resulted from the DKG equations.

\section{Solution Representation}

  In what follows, we denote by $(t, x)$ the time-space variables
and by $(\tau, \xi)$ the dual variables with respect to the Fourier
transform. We will use $ \alpha=\frac14-\epsilon$ in this paper. We
will also often skip the constant in the inequalities. For convenience,
we also denote the multipliers by
\begin{gather*}
\widehat{E}(\tau,\xi)=|\tau|+|\xi|+1 \\ % \label{e1.1a}\\
\widehat{S}(\tau,\xi)=\big||\tau|-|\xi|\big|+1 \\ %\label{e1.1b}\\
\widehat{W}(\tau,\xi)=\tau^2-|\xi|^2 \\ %\label{e1.1c}\\
\widehat{D}(\tau,\xi)=\gamma^0\tau+\gamma^1\xi \\ %\label{e1.1d}\\
\widehat{M}(\xi)=|\xi|+1\,. %\label{e1.1e}
\end{gather*}
We use $  \widehat W$ and $  \widehat D$ as the symbols of the wave and
Dirac operators respectively. Consider the Dirac equation,
\begin{equation}
\begin{gathered}
\mathcal{D}\psi=G,\quad(t,x)\in\mathbb{R}^1\times\mathbb{R}^1, \\
\psi(0)=\psi_0.
\end{gathered} \label{e1.2}
\end{equation}

  First by taking the Fourier transform on \eqref{e1.2} over
the space variable and solving the resulting ODE, we can formally
write down the solution as follows.
\begin{equation}
\begin{aligned}
\widetilde\psi(t,\xi)
&= \frac{e^{it|\xi|}}{2|\xi|}\widehat
D(|\xi|,\xi) \gamma^0\widehat\psi_0(\xi) +
\frac{e^{-it|\xi|}}{2|\xi|}\widehat
D(|\xi|,-\xi)\gamma^0\widehat\psi_0(\xi) \\
&\quad + \int_0^t \frac{e^{i(t-s)|\xi|}}{2|\xi|}\widehat
D(|\xi|,\xi)i\widetilde
G(s,\xi) \,ds + \int_0^t \frac{e^{-i(t-s)|\xi|}}{2|\xi|}\widehat
D(|\xi|,-\xi)i\widetilde G(s,\xi) \,ds.
\end{aligned} \label{e1.3}
\end{equation}
Rewriting the inhomogeneous terms in \eqref{e1.3} gives
\begin{equation}
\begin{aligned}
\widetilde\psi(t,\xi) &= \big[\frac{e^{it|\xi|}}{2|\xi|}\widehat
D(|\xi|,\xi) + \frac{e^{-it|\xi|}}{2|\xi|}\widehat
D(|\xi|,-\xi)\big]\gamma^0\widehat\psi_0(\xi)    \\
&\quad + \int \big[\frac{e^{it\tau} -
e^{it|\xi|}}{2|\xi|(\tau-|\xi|)} \widehat D(|\xi|, \xi) +
\frac{e^{it\tau} - e^{-it|\xi|}}{2|\xi|(\tau+|\xi|)}\widehat
D(|\xi|,-\xi)\big] \widehat G(\tau,\xi) d\tau.
\end{aligned} \label{e1.4}
\end{equation}

 Now we split the function $  \widehat G$ into several parts in the
following manner. Consider $  \widehat a(\tau)$ a cut-off function equals
1 if $  |\tau| \leq \frac12$ and equals 0 if $|\tau| \geq 1$,
$\widehat a_6(\tau)=\widehat a(\frac\tau6)$, and denote by $h(\tau)$ the
Heaviside function. For simplicity, let us write
\begin{gather*}
\widehat{G}_\pm(\tau,\xi):=
      h(\pm\tau)\widehat{a}(\tau\pm|\xi|)\widehat{G}(\tau,\xi),\\
      %\label{e1.5a}\\
\widehat{G}_f(\tau,\xi):=
\widehat{G}(\tau,\xi)-\big(\widehat{G}_+(\tau,\xi)+
\widehat{G}_-(\tau,\xi)\big),\\
%\label{e1.5b}\\
\widehat{D}_\pm:=
      \widehat{D}(|\xi|,\pm\xi).%\label{e1.5c}
\end{gather*}
Note that $  \widehat{G}_\pm$ are supported in the regions
$\{(\tau,\xi):\pm\tau>0,|\tau\mp|\xi||\leq1\}$ respectively. Using
the decomposition of the forcing term
$ \widehat{G}=\widehat{G}_f+\widehat{G}_++\widehat{G}_-$, the
inhomogeneous term in
\eqref{e1.4} can be written as
\begin{equation}
\begin{aligned}
&\int\big[\frac{e^{it\tau}-e^{it|\xi|}}{2|\xi|(\tau-|\xi|)}
\widehat{D}(|\xi|,\xi)+
\frac{e^{it\tau}-e^{-it|\xi|}}{2|\xi|(\tau+|\xi|)}\widehat{D}(|\xi|,-\xi)
\big]\widehat{G}_f(\tau,\xi)d\tau\\
&=\int{e}^{it\tau}\frac{\widehat{D}(\tau,\xi)}{\tau^2-|\xi|^2}
\widehat{G}_f d\tau-
e^{it|\xi|}\frac{\widehat{D}_+}{2|\xi|}\int
\frac{\widehat{G}_f}{\tau-|\xi|}d\tau
-e^{-it|\xi|}\frac{\widehat{D}_-}{2|\xi|}\int
\frac{\widehat{G}_f}{\tau+|\xi|}d\tau,
\end{aligned} \label{e1.6a}
\end{equation}

\begin{equation}
\begin{aligned}
&\int\frac{e^{it\tau}-e^{it|\xi|}}{2|\xi|(\tau-|\xi|)}\widehat{D}_+
(\widehat{G}_++\widehat{G}_-)d\tau\\
&=e^{it|\xi|}\frac{\widehat{D}_+}{2|\xi|}\int\frac{e^{it(\tau-|\xi|)}-1}
{\tau-|\xi|}
(\widehat{G}_++\widehat{a}(\tau)\widehat{G}_-)d\tau\\
&\quad +\int{e}^{it\tau}\frac{(1-\widehat{a}(\tau))\widehat{D}_+
\widehat{G}_-}{2|\xi|(\tau-|\xi|)}
d\tau - e^{it|\xi|}\frac{\widehat{D}_+}{2|\xi|}
\int\frac{(1-\widehat{a}(\tau))\widehat{G}_-}{\tau-|\xi|}d\tau,
\end{aligned}\label{e1.6b}
\end{equation}

\begin{equation}
\begin{aligned}
&\int\frac{e^{it\tau}-e^{-it|\xi|}}{2|\xi|(\tau+|\xi|)}\widehat{D}_-
(\widehat{G}_++\widehat{G}_-)d\tau\\
&= e^{-it|\xi|}\frac{\widehat{D}_-}{2|\xi|}\int\frac{e^{it(\tau+|\xi|)}-1}
{\tau+|\xi|}
(\widehat{a}(\tau)\widehat{G}_++\widehat{G}_-)d\tau\\
&\quad +\int{e}^{it\tau}\frac{(1-\widehat{a}(\tau))\widehat{D}_-
\widehat{G}_+}{2|\xi|(\tau+|\xi|)}
d\tau - e^{-it|\xi|}\frac{\widehat{D}_-}{2|\xi|}
\int\frac{(1-\widehat{a}(\tau))\widehat{G}_+}{\tau+|\xi|}d\tau,
\end{aligned}\label{e1.6c}
\end{equation}
Combining \eqref{e1.4}-\eqref{e1.6c}, we can give a formula for $
\widehat\psi$, namely
\begin{equation}
 \widehat\psi(\tau,\xi)=\sum_{k=0}^\infty\big(
\delta_+^{(k)}(\tau,\xi)\widehat{A}_{+,k}(\xi) +
\delta_-^{(k)}(\tau,\xi)\widehat{A}_{-,k}(\xi)
\big)+\widehat{K}(\tau,\xi), \label{e1.8}
\end{equation}
where $\delta_\pm(\tau,\xi)$ are the delta functions supported on
$\{\tau=\pm|\xi|\}$ respectively, $ \delta^{(k)}$ mean
derivatives of the delta function, and
\begin{gather}
\widehat{K}(\tau,\xi) := \frac{\widehat{D}(\tau,\xi)}
{\widehat{W}(\tau,\xi)}\widehat{G}_f
 + \frac{(1 - \widehat{a}_6(\tau))\widehat{D}_+\widehat{G}_-}
 {2|\xi|(\tau-|\xi|)} +
\frac{(1 - \widehat{a}_6)\widehat{D}_-\widehat{G}_+}
{2|\xi|(\tau+|\xi|)},
\label{e1.9a}\\
\widehat{A}_{\pm,0}(\xi):=\frac{\widehat{D}_\pm}{2|\xi|}
\big[\gamma^0\widehat\psi_0-
\int\frac{\widehat{G}_f+(1-\widehat{a}_6(\lambda))\widehat{G}_\mp}
{\lambda\mp|\xi|}d\lambda\big],
\label{e1.9b}\\
\widehat{A}_{\pm,k}(\xi):=\frac{\widehat{D}_\pm(-1)^k}{2|\xi|k!}
\int(\lambda\mp|\xi|)^{k-1} \big[\widehat{G}_\pm+
\widehat{a}_6(\lambda)\widehat{G}_\mp\big]
d\lambda. \label{e1.9c}
\end{gather}

  Now we split $\psi$ in a different manner. Consider the cut-off
function $ \widehat{b}(\tau)$ equals 1 if  $|\tau|<R$, and equals
0 if $|\tau|>2R$. Let $ \widehat{b}(\tau) + \widehat{c}(\tau) =
1$. Applying \eqref{e1.4}, we can give the following formula for $
\widehat\psi$. \begin{equation} \widehat\psi(\tau,\xi)=\sum_{k=0}^\infty
\big(\delta_+^{(k)}(\tau,\xi)\widehat{U}_{+,k}(\xi) +
\delta_-^{(k)}(\tau,\xi)\widehat{U}_{-,k}(\xi)\big)+\widehat{U}(\tau,\xi),
\label{e1.12}
\end{equation}
where
\begin{equation}
\begin{gathered}
\widehat{U}_{\pm,0}(\xi):=\frac{\widehat{D}_\pm}{2|\xi|}
\big[\gamma^0\widehat\psi_0-\int\frac{\widehat{c}(\lambda\mp|\xi|)}
{\lambda\mp|\xi|}\widehat{G}d\lambda
\big], \\ %\label{e1.12a}\
\widehat{U}_{\pm,k}(\xi):=\frac{\widehat{D}_\pm(-1)^k}{2|\xi|k!}
\int(\lambda\mp|\xi|)^{k-1}\widehat{b}(\lambda\mp|\xi|)\widehat{G}d\lambda,
\\
%\label{e1.12b}\\
\widehat{U}(\tau,\xi):=
\big[\frac{\widehat{D}_+\widehat{c}(\tau-|\xi|)}{2|\xi|(\tau-|\xi|)}
+
\frac{\widehat{D}_-\widehat{c}(\tau+|\xi|)}{2|\xi|(\tau+|\xi|)}\big]
\widehat{G}(\tau,\xi). %\label{e1.12c}
\end{gathered}
\end{equation}
  Consider the wave equation,
\begin{equation}
\begin{gathered}
\square\phi=F, \quad (t,x)\in\mathbb{R}^1\times\mathbb{R}^1,\\
\phi(0)=\phi_0, \quad  \phi_t(0)=\phi_1.
\end{gathered}\label{e1.13}
\end{equation}
Taking Fourier transform on \eqref{e1.13} and solving the resulting ODE
gives
\begin{gather}
\widetilde\phi(t,\xi)=\cos{t|\xi|}\widehat\phi_0(\xi) +
\frac{\sin{t|\xi|}}{|\xi|}\widehat\phi_1(\xi) +
\int_0^t\frac{\sin{(t-s)|\xi|}}{|\xi|}\widetilde{F}(s,\xi)ds.
\label{e1.14a}
\end{gather}
Thus we can rewrite it as follows.
\begin{gather}
\begin{aligned}
\widetilde\phi(t,\xi)
&=\frac{e^{it|\xi|}+e^{-it|\xi|}}{2}\widehat\phi_0(\xi) +
\frac{e^{it|\xi|}-e^{-it|\xi|}}{i2|\xi|}\widehat\phi_1(\xi) \\
&\quad +\frac{-1}{2|\xi|}
\int\frac{e^{it\tau}-e^{it|\xi|}}{\tau-|\xi|}\widehat{F}(\tau,\xi)d\tau+
\frac{1}{2|\xi|}
\int\frac{e^{it\tau}-e^{-it|\xi|}}{\tau+|\xi|}\widehat{F}(\tau,\xi)d\tau.
\end{aligned}\label{e1.14b}
\end{gather}
For convenience, we define the following
\begin{equation}
\begin{gathered}
\widehat{v}_{\pm,0}(\xi) :=
\frac12\widehat\phi_0(\xi) \pm \frac1{i2|\xi|}\widehat\phi_1(\xi) \pm
\frac1{2|\xi|} \int
\frac{\widehat{c}(\lambda \mp |\xi|)}{\lambda\mp|\xi|}
\widehat{F}(\lambda,\xi)d
\lambda, \\
%\label{e1.17a}\\
\widehat{v}_{\pm,k}(\xi):=\mp\frac{(-1)^k}{2|\xi|k!}
\int(\lambda\mp|\xi|)^{k-1}\widehat{b}(\lambda\mp|\xi|)
\widehat{F}(\lambda, \xi)d
\lambda, \\
%\label{e1.17b}\\
\widehat{v}(\tau,\xi):=\frac{-1}{2|\xi|}
\big[\frac{\widehat{c}(\tau-|\xi|)}{\tau-|\xi|} -
\frac{\widehat{c}(\tau+|\xi|)}{\tau+|\xi|}\big]
\widehat{F}(\tau,\xi). %\label{e1.17c}
\end{gathered} \label{e1.17}
\end{equation}
Combining \eqref{e1.14b} and \eqref{e1.17}, and invoking the
cut-off function, we have \begin{equation} \widehat\phi(\tau,\xi) =
\sum_{k=0}^\infty
\big(\widehat{V}_{+,k}(\tau,\xi)+\widehat{V}_{-,k}(\tau,\xi)\big)+
\widehat{V}(\tau,\xi)+
\sum_{k=0}^\infty \widehat{N}_{k}(\tau,\xi)+\widehat{N}(\tau,\xi),
\label{e1.18}
\end{equation}
where
\begin{gather*}
\widehat{V}_{\pm,k}(\tau,\xi)=
\big(1-\widehat{a}(\xi)\big)\delta^{(k)}_\pm(\tau,\xi)
\widehat{v}_{\pm,k}(\xi),\\
%\label{e1.19a}\\
\widehat{V}(\tau,\xi)=\big(1-\widehat{a}(\xi)\big)\widehat{v}(\tau,\xi),\\
%\label{e1.19b}\\
\widehat{N}_{k}(\tau,\xi)= \widehat{a}(\xi)
\big[\delta^{(k)}_+(\tau,\xi)\widehat{v}_{+,k}(\xi)+
\delta^{(k)}_-(\tau,\xi)\widehat{v}_{-,k}(\xi)\big],\\
%\label{e1.19c}\\
\widehat{N}(\tau,\xi) = \widehat{a}(\xi)\widehat{v}(\tau,\xi).
%\label{e1.19d}
\end{gather*}

\subsection*{Remark}
We need to localize the solutions for Dirac
equation and wave equation due to the presence of the delta
function.

\section{Estimates}

  To localize the solution in time, let $\varphi(t)$ be a cut-off
function such that $\varphi(t)$ equals 1 if $|t|\leq 1/2$, and
equals 0 if $|t|>1$, and $  \varphi_T(t) = \varphi(t/T)$. Notice
that, for an arbitrary function $f(t,x)$, we have
\begin{equation}
\|\widehat\varphi_T\ast\widehat{f}\|_{L^2} = \|\varphi_Tf\|_{L^2}
\leq\|\varphi_T\|_{L^\infty}\|f\|_{L^2}.\label{e2.1}
\end{equation}
For the Dirac equation \eqref{e1.2}, using \eqref{e1.12}, we
define
\begin{equation} \widehat\Psi_T(\tau,\xi)=
\widehat\varphi_T\ast\sum_{k=0}^\infty
\big(\delta_+^{(k)}\widehat{U}_{+,k}+\delta_-^{(k)}
\widehat{U}_{-,k}\big)(\tau,\xi)
+\widehat{U}(\tau,\xi), \label{e2.2}
\end{equation}

\begin{lemma} \label{lem2.1}
Let $\epsilon > 0$ and $TR \sim 1$. If $
\psi_0 \in H^{-\alpha}$, then we have
\begin{equation}
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat\Psi_T\big\|
_{L^2(\mathbb{R}^1\times\mathbb{R}^1)}
\leq C\big(\|\psi_0\|_{H^{-\alpha}}+ T^\epsilon
\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}\big). \label{e2.3}
\end{equation}
\end{lemma}

We will only outline the proof. For more details, please see \cite{FG}.

\begin{proof} Applying formulae \eqref{e1.12},
we can derive the following bounds:
\begin{gather*}
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat\varphi_T
\ast \big(\delta_\pm\widehat{U}_{\pm,0}\big)\big\|_{L^2}
 \leq {C}\big(\|\psi_0\|_{H^{-\alpha}}  +
 T^\epsilon\big\|\frac{\widehat{G}}
{\widehat{M}^{\alpha}\widehat{S}^{\frac12-\epsilon}}\big\|_{L^2} \big) ,\\
%\label{e2.4a}\\
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat\varphi_T\ast
\big(\delta_\pm^{(k)}\widehat{U}_{\pm,k}\big)\big\|_{L^2}
\leq{C}\frac{(4RT)^k}{k!}T^\epsilon\big\|\frac{\widehat{G}}
{\widehat{M}^{\alpha}\widehat{S}^{\frac12-\epsilon}}\big\|_{L^2},\\
%\label{e2.4b}\\
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat{U}
\big\|_{L^2(\mathbb{R}^1\times
\mathbb{R}^1)} \leq
CT^\epsilon\big\|\frac{\widehat{G}}
{\widehat{M}^{\alpha}\widehat{S}^{\frac12-\epsilon}}\big\|_{L^2},
%\label{e2.4c}
\end{gather*}
Combining these  estimates, we have \eqref{e2.3}.
\end{proof}


  Consider two Dirac equations,
\begin{equation}
\begin{gathered}
\mathcal{D}\psi_j=G_j,\quad{j}=1,2,\\
\psi_j(0)= \psi_{0j}.
\end{gathered} \label{e2.6}
\end{equation}
For the solutions of this system, we have the following key estimate
whose proof will be presented in the last section.

\begin{lemma}[Null Form Estimate] \label{lem2.2}
Let $\epsilon > 0$, and $ \psi_1$, $ \psi_2$ be the solutions for
\eqref{e2.6}.
If $ \psi_{0j} \in H^{-\alpha}$, we have
\begin{equation}
\begin{aligned}
&\big\|\frac{(\widehat{\varphi_T\overline\psi_1\psi_2})}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}
& \leq C(T)\big(\|\psi_{01}\|_{H^{-\alpha}} +
\big\|\frac{\widehat{G}_1}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}\big)\big(\|\psi_{02}\|_{H^{-\alpha}} +
\big\|\frac{\widehat{G}_2}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}
\big).
\end{aligned} \label{e2.7}
\end{equation}
\end{lemma}

  For the wave equation \eqref{e1.13}, we define
\begin{equation}
\widehat\Phi_T(\tau,\xi)= \widehat\varphi_T\ast\sum_{k=0}^\infty
(\widehat{V}_{+,k}+\widehat{V}_{-,k})(\tau,\xi)+\widehat{V}(\tau,\xi)+
\widehat\varphi_T\ast\sum_{k=0}^\infty\widehat{N}_k(\tau,\xi)+
\widehat{N}(\tau,\xi).
 \label{e2.9}
\end{equation}
Thus we have the following estimate.

\begin{lemma} \label{lem2.3}
Let $\epsilon>0$, $\delta\geq0$, $TR\sim 1$, and $\phi$ be the
solution of \eqref{e1.13}. If $ \phi_0 \in H^{\frac12+\delta}$ and
$ \phi_1 \in H^{-\frac12+\delta}$, then \begin{equation}
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat\Phi_T\big\|_{L^2}\leq
C\big(\|\phi_0\|_{H^{\frac12+\delta}}+\|\phi_1\|_{H^{-\frac12+\delta}}+
T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}\big). \label{e2.10}
\end{equation}
\end{lemma}

\begin{proof} Applying formula \eqref{e1.18}, we can derive the
following bounds:
\begin{gather*}
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat\varphi_T\ast\widehat{V}_{\pm,0}\big\|_{L^2}
 \leq C\big(\|\phi_0\|_{H^{\frac12+\delta}}+
 \|\phi_1\|_{H^{-\frac12+\delta}}+
T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}\big), \\
%\label{e2.11a}\\
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat\varphi_T\ast\widehat{N}_{0}\big\|_{L^2}
 \leq C\big(\|\phi_0\|_{H^{\frac12+\delta}}+
 \|\phi_1\|_{H^{-\frac12+\delta}}+
T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}\big), \label{e2.11b}
\\
\big\|\widehat{S}^\frac12
\widehat{M}^{\frac12+\delta}\widehat\varphi_T\ast\widehat{V}_{\pm,k}
\big\|_{L^2}
\leq C\frac{(4RT)^k}{k!} T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}, \\
%\label{e2.11c}\\
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat\varphi_T\ast\widehat{N}_{k}
\big\|_{L^2} \leq C\frac{(4RT)^k}{k!} T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2},\\
%\label{e2.11d}\\
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat{V}\big\|_{L^2}
\leq{C}T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}, \\ %\label{e2.11e}\\
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat{N}\big\|_{L^2}
\leq{C}T^{\epsilon}
\big\|\frac{\widehat{F}}{\widehat{M}^{\frac12-\delta}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}. %\label{e2.11f}
\end{gather*}
 Combining the above inequalitites, we complete the proof. \end{proof}

  We will also need some technical lemmas.

\begin{lemma}[Hardy-Littlewood-Polya] \label{lem2.4}
Let ${r}=2-\frac1p-\frac1q$. Then we have
\begin{equation}
 \int_{\mathbb{R}^1 \times \mathbb{R}^1} \frac{f(s)g(t)}{|s-t|^r} ds dt
 \leq
C\|f\|_{L^p}\|g\|_{L^q}. \label{e2.13}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem2.5}
Let $f(t, x)$ and $g(t, x)$ be any functions such that $  f \in
L^q( L^2(\mathbb{R}^n))$ and $  \widehat{S}^{\beta}\widehat{g} \in
L^2(L^2(\mathbb{R}^n))$. Assume that $\epsilon
> 0$, $  \frac1q = 1 - \epsilon$, $
\frac1r=\frac12-\beta$, and $  2\leq{r}<\infty$. Then we have
\begin{gather}
\big\|\frac{\widehat{f}}{\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2(L^2(\mathbb{R}^n))}
\leq{C}\|f\|_{L^q(L^2(\mathbb{R}^n))},\label{e2.14a} \\
\|g\|_{L^r(L^2(\mathbb{R}^n))}
\leq{C}\|\widehat{S}^{\beta}\widehat{g}\|_{L^2(L^2(\mathbb{R}^n))}.
\label{e2.14b}
\end{gather}
\end{lemma}

\begin{proof} The proofs for \eqref{e2.14a} and \eqref{e2.14b} are
analogous. Therefore, we will only prove the case of \eqref{e2.14b}.
Taking the inverse Fourier transform in the time variable over the
identity
\begin{equation}
\widehat{g}=\frac1{\widehat{S}^\beta}\widehat{S}^{\beta}\widehat{g}
\label{e2.18}
\end{equation}
 gives
\begin{equation}
\widetilde g (t, \xi) = \int\frac{e^{\pm i(t-s)|\xi|}}{|t-s|^{1-\beta}}
\mathcal{F}^{-1}_\tau(\widehat S^{\beta} \widehat g)(s, \xi) \,ds.
\label{e2.19}
\end{equation}
Then we use duality and Hardy-Littlewood-Polya inequality to
compute
\begin{equation}
\begin{aligned}
\big|\big<g,\varphi\big>\big|&
=\big|\big<\widetilde{g},\widetilde\varphi\big>\big|
=\big|\iint\int\frac{e^{\pm i(t-s)|\xi|}}{|t-s|^{1-\beta}}
\mathcal{F}^{-1}_\tau(\widehat{S}^{\beta}\widehat{g})(s,\xi) ds
\overline{\widetilde\varphi}(t,\xi) dtd\xi\big|\\
&\leq \int \frac{\|\mathcal{F}^{-1}_\tau(\widehat{S}^{\beta}\widehat{g})(s)
\|_{L^2}\|\widetilde\varphi(t)\|_{L^2}}{|t-s|^{1-\beta}}dsdt \\
&\leq C\|\mathcal{F}^{-1}_\tau(\widehat S^{\beta}\widehat{g})\|_{L^2}
\|\widetilde\varphi\|_{L^{r'}(L^2)}
=C\|\widehat{S}^{\beta}\widehat{g}\|_{L^2}\|\varphi\|_{L^{r'}(L^2)}.
\end{aligned}\label{e2.20}
\end{equation}
This completes the proof of \eqref{e2.14b}. \end{proof}

\section{Local and Global Existence}

  Now we are ready to prove the local existence for the (DKG)
equations.

\begin{proof}[Proof of Theorem \ref{thm0.1}]
 Consider the DKG equations
\begin{equation}
\begin{gathered}
\mathcal{D}\psi=\varphi_T\phi\psi,\\
\square\phi=\varphi_T\overline\psi\psi,
\end{gathered} \label{e3.1}
\end{equation}
 and the map $\mathcal{T}(\psi,\phi) = (\Psi_T,\Phi_T)$. %\label{e3.2a}
We want to show that $\mathcal{T}$ is a contraction under the norm
\begin{equation}
N(\psi,\phi)=\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}
\widehat\psi\big\|_{L^2}
+\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}\widehat\phi
\big\|_{L^2}.
\label{e3.2b}
\end{equation}
For convenience, we define
\begin{equation}
J(0)=\|\phi_0\|_{H^{\frac12+\delta}}+\|\phi_1\|_{H^{-\frac12+\delta}}+
\|\psi_0\|^2_{H^{-\alpha}} + 1. \label{e3.3}
\end{equation}
First we apply \eqref{e2.10} and \eqref{e2.7} to compute
\begin{equation}
\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}\widehat\Phi_T
\big\|_{L^2} \leq{C}\big(J(0)+T^\epsilon\big\|
\frac{\widehat{\varphi_T\overline\psi\psi}}
{\widehat{M}^{\frac12-\delta}\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}\big)
\leq{C}\big(J(0)+T^\epsilon\big\|\frac{\widehat{\varphi_T\phi\psi}}
{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}\big\|^2_{L^2}\big).
\label{e3.4}
\end{equation}
To bound the term above, we first compute
\begin{equation}
\begin{aligned}
\big\|\widehat{M}^{-\alpha}{\widetilde{\phi\psi}(t)}\big\|_{L^2}\sim
\|G_\alpha\ast(\phi\psi)(t)\|_{L^2}
&\leq \|\phi(t)\|_{L^\infty}\|G_\alpha\ast\psi(t)\|_{L^2}\\
&\leq \|\phi(t)\|_{H^{\frac12+\delta}}\|\psi(t)\|_{H^{-\alpha}},
\end{aligned}\label{e3.5}
\end{equation}
 where $  G_\alpha(x)$ is an $L^1$-function such that
\begin{equation}
\widehat{G_\alpha}(\xi)\sim(1+|\xi|)^{-\alpha},\label{e3.6}
\end{equation}
see \cite{S}. Then we invoke \eqref{e2.14a}, \eqref{e2.14b} and obtain
\begin{equation}
\begin{aligned}
\big\|\frac{\widehat{\varphi_T\phi\psi}}{\widehat{M}^{\alpha}
\widehat{S}^{\frac14}}
\big\|_{L^2}
&\leq{C}\big\|\frac{\widetilde{\varphi_T\phi\psi}}{\widehat{M}^\alpha}
\big\|_{L^q(L^2)}\\
&\leq\|\phi\|_{L^{2q}(H^{\frac12+\delta})}\|\psi\|_{L^{2q}(H^{-\alpha})}\\
&\leq \big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat\phi\big\|_{L^2}
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat\psi\big\|_{L^2}.
\end{aligned}\label{e3.7}
\end{equation}
Thus we get
\begin{equation}
\big\|\frac{\widehat{\varphi_T\phi\psi}}{\widehat{M}^{\alpha}
\widehat{S}^{\frac12-\epsilon}}
\big\|_{L^2}\leq CN^2(\psi,\phi).\label{e3.8}
\end{equation}

  Next we want to bound the term involving $ \widehat\Psi_T$. The
estimate \eqref{e2.3} implies
\begin{equation}
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat\Psi_T
\big\|_{L^2(\mathbb{R}^1\times\mathbb{R}^1)}
\leq{C}\big(\|\psi_0\|_{H^{-\alpha}}+T^\epsilon
\big\|\frac{\widehat{\varphi_T\phi\psi}}{\widehat{M}^\alpha
\widehat{S}^{\frac14}}
\big\|_{L^2}\big). \label{e3.9}
\end{equation}
Hence, using \eqref{e3.3}, \eqref{e3.8}, and \eqref{e3.5}, we have
\begin{equation}
N(\mathcal{T}(\psi,\phi))\leq{C}\big(J(0)+T^\epsilon{N}^4(\psi,\phi)\big).
\label{e3.10}
\end{equation}
Choosing sufficiently large $L$, for suitable $T$, we have
\begin{equation}
N(\psi,\phi)\leq{L} \Longrightarrow
N\big(\mathcal{T}(\psi,\phi)\big)\leq{L}, \label{e3.11}
\end{equation}
provided that
$C(J(0)+T^\epsilon{L}^4)\leq{L}$. %\label{e3.12}

Now we consider the difference
$ \mathcal{T}(\psi,\phi)-\mathcal{T}(\psi',\phi')$. Base on the
observations
\begin{equation}
\begin{gathered}
\overline\psi\psi-\overline{\psi'}\psi'=
\frac12(\overline{\psi-\psi'})(\psi+\psi')+
\frac12(\overline{\psi+\psi'})(\psi-\psi'), \\ %\label{e3.13a}\\
\phi\psi-\phi'\psi'= \frac12(\phi-\phi')(\psi+\psi')+
\frac12(\phi+\phi')(\psi-\psi'), %\label{e3.13b}
\end{gathered}  \label{e3.13}
\end{equation}
Using \eqref{e2.10}, \eqref{e2.7}, \eqref{e3.13}, and \eqref{e3.8},
we first calculate
\begin{equation}
\begin{aligned}
&\big\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\mathcal{F}(\Phi_T-\Phi_T')
\big\|_{L^2}\\
&\leq CT^\epsilon\big(\|\frac{\mathcal{F}((\overline{\psi-\psi'})
(\psi+\psi'))}
{\widehat{M}^{\frac12-\delta}\widehat{S}^{\frac12-\epsilon}}\|_{L^2}+
\|\frac{\mathcal{F}((\overline{\psi+\psi'})(\psi-\psi'))}
{\widehat{M}^{\frac12-\delta}\widehat{S}^{\frac12-\epsilon}}\|_{L^2}\big)\\
&\leq  CT^\epsilon \big(
\big\|\frac{\mathcal{F}((\phi-\phi')(\psi+\psi'))}
{\widehat{M}^\alpha\widehat{S}^{\frac14}}\big\|_{L^2} +
\big\|\frac{\mathcal{F}((\phi+\phi')(\psi-\psi'))}
{\widehat{M}^\alpha\widehat{S}^{\frac14}}\big\|_{L^2} \big)\\
&\quad\times \big(J(0)+\big\|\frac{\mathcal{F}(\phi\psi+\phi'\psi')}
{\widehat{M}^\alpha\widehat{S}^{\frac14}}\big\|_{L^2}\big)\\
&\leq CT^\epsilon
\big(\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}\widehat{\phi-\phi'}
\|_{L^2}+
\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat{\psi-\psi'}
\|_{L^2}\big)L(J(0)+L^2)\\
&\leq CT^\epsilon
{L}^3\big(\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}
\widehat{\phi-\phi'}
\|_{L^2}+\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat{\psi-\psi'}
\|_{L^2}\big)
\end{aligned}
\label{e3.15}
\end{equation}
Analogously, we get
\begin{equation}
\big\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat{(\Psi_T-\Psi_T')}
\big\|_{L^2}
\leq{C}T^\epsilon{L}
\big(\|\widehat{S}^\frac12\widehat{M}^{-\alpha}\widehat{\psi-\psi'}
\|_{L^2}+
\|\widehat{S}^\frac12\widehat{M}^{\frac12+\delta}\widehat{\phi-\phi'}
\|_{L^2}\big).
\label{e3.16}
\end{equation}
Combining \eqref{e3.15}and \eqref{e3.16}, we have
\begin{equation}
N\big(\mathcal{T}(\psi-\psi', \phi-\phi')\big) \leq
C T^\epsilon {L}^3 N(\psi-\psi', \phi-\phi'). \label{e3.17}
\end{equation}
Therefore for suitable $T$, we obtain
\begin{equation}
N\big(\mathcal{T}(\psi-\psi', \phi-\phi')\big) \leq\frac12
N(\psi-\psi', \phi-\phi'), \label{e3.18}
\end{equation}
provided that $CT^\epsilon L^3 \leq \frac12$. %\label{e3.19}
We can conclude that the map $\mathcal{T}$ is indeed a contraction with
respect to the norm $N$, thus it has a unique fixed point.
\end{proof}


  We now prove existence of a gobal solution.

\begin{proof}[Proof of Theorem \ref{thm0.2}]
>From the law of conservation of charge, we have
\begin{equation}
\sup_{[0,T]}\|\psi(t)\|_{L^2} = \|\psi_0\|_{L^2}.\label{e3.20}
\end{equation}
To bound $\phi$ we apply the formula \eqref{e1.14a},
\begin{equation}
2\phi(t,x)=(\phi_0(x+t)+\phi_0(x-t))+
\int_{x-t}^{x+t}\phi_1(y) dy +
\int_0^t\int_{x-t+s}^{x+t-s}\overline\psi\psi(s,y)dyds. \label{e3.21}
\end{equation}
First we write $\phi=\phi_L+\phi_N$, the homogeneous and inhomogeneous
parts of the solution, then we obtain
\begin{equation}
\|\phi_L(t)\|_{L^\infty}\leq \|\phi_L(t)\|_{H^{\frac12+\delta}}
\leq \|\phi_0\|_{H^{\frac12+\delta}}+\|\phi_1\|_{H^{-\frac12+\delta}}
\leq{J}(0),  \label{e3.22}
\end{equation}
and
\begin{equation}
\|\phi_N(t)\|_{L^\infty}
\leq\int_0^t\int_{x-t+s}^{x+t-s}\big|\overline\psi\psi(s,y)\big|dy\,ds
\leq{C}T\|\psi_0\|_{L^2}^2. \label{e3.23}
\end{equation}
 Combining \eqref{e3.22} and \eqref{e3.23}, we obtain
\begin{equation}
 \|\phi(t)\|_{L^\infty}\leq{C}(T,J(0)).\label{e3.24}
\end{equation}
Taking Fourier transform of the solution $\phi(t)$, we have
\begin{equation}
\widetilde\phi(t,\xi)=\cos{t|\xi|}\widehat\phi_0(\xi) +
\frac{\sin{t|\xi|}}{|\xi|}\widehat\phi_1(\xi) +
\int_0^t\frac{\sin{(t-s)|\xi|}}{|\xi|}
\widetilde{\varphi_T\overline\psi\psi}(s,\xi)ds.
\label{e3.25}
\end{equation}
 Then we invoke \eqref{e2.7}, \eqref{e2.14a}, (for $\epsilon=\frac14$),
and \eqref{e3.23} to compute
\begin{equation}
\begin{aligned}
 \|\phi(t)\|_{H^{\frac12+\delta}}
&\leq \|\phi_0\|_{H^{\frac12+\delta}}+\|\phi_1\|_{H^{-\frac12+\delta}}+
\int_0^t\|\varphi_T\overline\psi\psi(s)\|_{H^{-\frac12+\delta}}ds\\
&\leq J(0)+T^\frac12
\big\|\frac{\widehat{\varphi_T\overline\psi\psi}}
{\widehat{M}^{\frac12-\delta}}\big\|_{L^2}\\
&\leq J(0)+T^\frac12 \|\widehat{\varphi_T\overline\psi\psi}\|_{L^2}\\
&\leq J(0)+T^\frac12
\big\|\frac{\widehat{\varphi_T\phi\psi}}{\widehat{S}^{\frac14}}
\big\|_{L^2}^2\\
&\leq J(0)+T^\rho\|\varphi_T\phi\psi\|_{L^2}\\
&\leq J(0)+T^\rho\int_0^T\|\phi(t)\|_{L^\infty}^2\|\psi(t)\|_{L^2}^2dt\\
&\leq C\big(T,J(0)\big),
\end{aligned}\label{e3.26}
\end{equation}
where $\rho$ is some positive number. The calculation for
$\|\phi_t(t)\|_{H^{-\frac12+\delta}}$ is analogous. Thus the above
bounds ensure us to proceed the construction of solution beyond $T$.
\end{proof}

\section{Null Form Estimate} %sec 4

In this section, we demonstrate the key estimate in Lemma \ref{lem2.2}:
 Let $\epsilon > 0$ and $\psi_1$, $\psi_2$ be the solutions for the
 Dirac equations \eqref{e2.6}. If the initial data
 $ \psi_{0j}\in H^{-\frac14+\epsilon}$, $j=1, 2$, then we have
\begin{equation}
\big\|\frac{\widehat{\varphi_T\overline\psi_1\psi_2}}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2} \leq
C(T)\big(\|\psi_{01}\|_{H^{-\alpha}} +
\big\|\frac{\widehat{G}_1}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}\big)
\big(\|\psi_{02}\|_{H^{-\alpha}} +
\big\|\frac{\widehat{G}_2}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}\big).\label{e4.1}
\end{equation}


  The proof of this estimate is based on the duality argument and
it will be given in a number of steps. Without loss of generality,
we assume that $ \psi_1=\psi_2$, and prove: if $\psi$ is a
solution of the Dirac equation \eqref{e1.2}, then
\begin{equation}
\big\|\frac{\widehat{\varphi_T\overline\psi\psi}}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2} \leq
C(T)\big(\|\psi_{0}\|_{H^{-\alpha}} +
\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}\big)^2. \label{e4.2}
\end{equation}
Recall the notation:
\begin{gather*}
\widehat{E}(\tau,\xi):=|\tau|+|\xi|+1,\quad
\widehat{S}(\tau,\xi):=\big||\tau|-|\xi|\big|+1,\\ %\label{e4.3a}
\widehat{W}(\tau,\xi):=\tau^2-|\xi|^2,\quad
\widehat{D}(\tau,\xi):=\gamma^0\tau+\gamma^1\xi,\\
\widehat{D}_+:=\widehat{D}(|\xi|,+\xi),\quad
\widehat{D}_-:=\widehat{D}(|\xi|,-\xi), %\label{e4.3c}
\end{gather*}
The formula for $ \widehat\psi$, as in \eqref{e1.8}, for the Dirac
equation \eqref{e1.2} is given by
\begin{equation}
\widehat\psi(\tau,\xi)=\sum_{k=0}^\infty\big(
\delta_+^{(k)}(\tau,\xi)\widehat{A}_{+,k}(\xi) +
\delta_-^{(k)}(\tau,\xi)\widehat{A}_{-,k}(\xi)
\big)+\widehat{K}(\tau,\xi), \label{e4.4}
\end{equation}
where $\delta_\pm(\tau,\xi)$ are the delta functions supported on
$\{\tau=\pm|\xi|\}$ respectively, $ \delta^{(k)}$ mean
derivatives of the delta function, and
\begin{gather}
\widehat{K}(\tau,\xi) := \frac{\widehat{D}(\tau,\xi)}
{\widehat{W}(\tau,\xi)}\widehat{G}_f
 + \frac{(1 - \widehat{a}_6(\tau))\widehat{D}_+\widehat{G}_-}
 {2|\xi|(\tau-|\xi|)} +
\frac{(1 - \widehat{a}_6)\widehat{D}_-\widehat{G}_+}
{2|\xi|(\tau+|\xi|)},
\label{e4.5a}\\
\widehat{A}_{\pm,0}(\xi):=\frac{\widehat{D}_\pm}{2|\xi|}
\big[\gamma^0\widehat\psi_0-
\int\frac{\widehat{G}_f+(1-\widehat{a}_6(\lambda))
\widehat{G}_\mp}{\lambda\mp|\xi|}d\lambda\big],
\label{e4.5b}\\
\widehat{A}_{\pm,k}(\xi):=
\frac{\widehat{D}_\pm(-1)^k}{2|\xi|k!}\int(\lambda\mp|\xi|)^{k-1}
\big[\widehat{G}_\pm+\widehat{a}_6(\lambda)\widehat{G}_\mp\big] d\lambda.
\label{e4.5c}
\end{gather}
Moreover we write
\begin{equation}
\widehat{A}_{\pm,k}(\xi):=\frac{\widehat{D}_\pm}{2|\xi|}
\widehat{f}_{\pm,k}(\xi),
\label{e4.6}
\end{equation}
and set $ \widehat{K}=\widehat{K}_1+\widehat{K}_2$, where
\begin{equation}
\widehat{K}_1:=\frac{\widehat{D}(\tau,\xi)}{\widehat{W}(\tau,\xi)}
\widehat{G}_f; \quad
\widehat{K}_2:=
\frac{b_1\widehat{D}_+\widehat{G}_-+b_2\widehat{D}_-\widehat{G}_+}
{\widehat{E}\widehat{S}},
\label{e4.7}
\end{equation}
and $ b_1$, $  b_2$ are bounded functions. The Fourier
transform of the quadratic expression,
$ \widehat{\overline\psi\psi}=\widehat{\overline\psi}\ast\widehat{\psi}$,
can be written as
the sum of the following terms.
\begin{gather}
\sum_{k,l}\big(\delta_\mp^{(k)}\widehat{\overline{A}}_{\pm,k}\big)\ast
\big(\delta_\pm^{(l)}\widehat{A}_{\pm,l}\big), \label{e4.8a}\\
\sum_{k,l}\big(\delta_\mp^{(k)}\widehat{\overline{A}}_{\pm,k}\big)\ast
\big(\delta_\mp^{(l)}\widehat{A}_{\mp,l}\big), \label{e4.8b}\\
\sum_{k}\big(\delta_\mp^{(k)}\widehat{\overline{A}}_{\pm,k}\big) \ast
\big(\widehat{K}_1+\widehat{K}_2\big) + \big(\widehat{\overline{K}}_1+
\widehat{\overline{K}}_2\big)
 \ast \sum_{k}\big(\delta_\pm^{(k)}\widehat{{A}}_{\pm,k}\big), \
 label{e4.8c}\\
\widehat{\overline{K}}_1\ast\widehat{K}_1 +
\widehat{\overline{K}}_1\ast\widehat{K}_2 +
\widehat{\overline{K}}_2\ast\widehat{K}_1 +
\widehat{\overline{K}}_2\ast\widehat{K}_2.
\label{e4.8d}
\end{gather}
Note that
\begin{gather}
\widehat{A^\dagger_{\pm,k}}(\xi)=\widehat{A}^\dagger_{\pm,k}(-\xi),\quad
\widehat{f^+_{\pm,k}}(\xi)=\widehat{f}^\dagger_{\pm,k}(-\xi),\label{e4.9a}\\
\widehat{\overline{A}_{\pm,k}}(\xi)=\widehat{f}^\dagger_{\pm,k}(-\xi)
\frac{\widehat{D}_\pm}{|\xi|}\gamma^0,\quad
\widehat{\overline{K}}(\tau,\xi)=\widehat{K}^\dagger(-\tau,-\xi)\gamma^0,
\label{e4.9b}
\end{gather}
and
\begin{equation}
\widehat{\overline\psi}(\tau,\xi)=\sum_{k=0}^\infty\big(
\delta_-^{(k)}(\tau,\xi)\widehat{\overline{A}}_{+,k}(\xi) +
\delta_+^{(k)}(\tau,\xi)\widehat{\overline{A}}_{-,k}(\xi)
\big)+\widehat{\overline{K}}(\tau,\xi), \label{e4.10}
\end{equation}

\begin{lemma} \label{lem4.1}
Let $ \alpha<1/4$. Then the following estimate holds
\begin{align*}
&\big\|\frac{\widehat\varphi_T\ast\big(\delta_\mp^{(k)}
\widehat{f^\dagger_{\pm,k}}
\frac{\widehat{D}_\pm}{|\xi|}\gamma^0\big)
\ast\big(\delta_\mp^{(l)}\frac{\widehat{D}_\mp}{|\xi|}
\widehat{f}_{\mp,l}\big)}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2} \\
&\leq C(k+l+1)T^{k+l-\frac12}
\|{f}_{\pm,k}\|_{H^{-\alpha}}\|{f}_{\mp,l}\|_{H^{-\alpha}}.
\end{align*}  % \label{e4.11}
\end{lemma}

\begin{proof} Let
\begin{equation}
 \widehat{Z}_{\pm,k}\equiv \delta_\pm^{(k)}\frac{\widehat{D}_\pm}{2|\xi|}
 \widehat{f}_{\pm,k}
=\delta_\pm^{(k)}\widehat{A}_{\pm,k}. \label{e4.12}
\end{equation}
Using duality, we demonstrate the case $(-,+)$, while the case
$(+,-)$ is being similar. We first compute the fractional term
\[
\frac{\widehat{D}(|\xi|,-\xi)\gamma^0\widehat{D}(|\eta|,\eta)}{|\xi||\eta|}
=\begin{cases}
0, &\text{if }   \xi\eta>0,\\
2\gamma^0\pm2\gamma^1, &\text{if }  \xi\eta<0. \end{cases} %\label{e4.13}
\]
Throughout elementary analysis we have the bound:
\begin{equation}
\frac{(1+|\xi|)^\alpha(1+|\eta|)^\alpha}{(1+|\xi+\eta|)^{\alpha}
\big(\big|\big||\xi|+|\eta|\big|-|\xi+\eta|\big|+1\big)^{2\alpha}}\leq{C},
\label{e4.14}
\end{equation}
for $\xi\eta<0$. Thus
\begin{align*}
&\big|\left<\varphi_T\overline{Z}_{-,k}Z_{+,l},g\right>\big|\\
=&\big|\int\widehat{f}^\dagger_{-,k}(-\xi)
\frac{\widehat{D}(|\xi|,-\xi)\gamma^0\widehat{D}(|\eta|,\eta)}{|\xi||\eta|}
\widehat{f}_{+,l}(\eta)\overline{\widehat{t^{k+l}\varphi_Tg}}(
|\xi|+|\eta|,\xi+\eta)
d\xi d\eta\big|\\
&\leq {C}\|{f}_{-,k}\|_{H^{-\alpha}}\|{f}_{+,l}\|_{H^{-\alpha}}
\|\widehat{M}^{\alpha}\widehat{S}^{2\alpha}\widehat{t^{k+l}\varphi_Tg}
\|_{L^2},
\end{align*} %\label{e4.15a}
 and through some computations, we have
 \begin{equation}
\|\widehat{M}^{\alpha}\widehat{S}^{2\alpha}\widehat{t^{k+l}\varphi_Tg}
\|_{L^2} \leq
C(k+l+1)T^{k+l-\frac12}\|\widehat{M}^{\alpha}\widehat{S}^{2\alpha}
\widehat{g}\|_{L^2}.
\label{e4.15b}
\end{equation}
 This completes the proof.
 \end{proof}

\begin{lemma} \label{lem4.2}
Let $ \alpha<1/4$. The following estimate holds
\begin{align*}
&\big\|\frac{\widehat\varphi_T\ast\big(\delta_\mp^{(k)}
\widehat{f^\dagger_{\pm,k}}
\frac{\widehat{D}_\pm}{|\xi|}\gamma^0\big)
\ast\big(\delta_\pm^{(l)}\frac{\widehat{D}_\pm}{|\xi|}
\widehat{f}_{\pm,l}\big)}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2} \\
&\leq C(k+l+1)T^{k+l-\frac12}
\|{f}_{\pm,k}\|_{H^{-\alpha}}\|{f}_{\pm,l}\|_{H^{-\alpha}}.
\end{align*}  %\label{e4.16}
\end{lemma}

\begin{proof} Using duality, we demonstrate the case $(+,+)$,
while the case $(-,-)$ is being similar. We first compute the
fractional term
\[
\frac{\widehat{D}(|\xi|,\xi)\gamma^0\widehat{D}(|\eta|,\eta)}{|\xi||\eta|}=
\begin{cases}
0, &\text{if }   \xi\eta<0,\\
2\gamma^0\mp2\gamma^1,&\text{if }  \xi\eta>0. \end{cases} %\label{e4.17}
\]
Throughout elementary analysis we have the bound:
\begin{equation}
\frac{(1+|\xi|)^\alpha(1+|\eta|)^\alpha}{(1+|\xi+\eta|)^{\alpha}
\big(\big|\big|-|\xi|+|\eta|\big|-|\xi+\eta|\big|+1\big)^{2\alpha}}\leq{C}
\label{e4.18}
\end{equation}
for $\xi\eta>0$. Thus
\begin{align*}
&\big|\left<\varphi_T\overline{Z}_{+,k}Z_{+,l},g\right>\big|\\
&=\big|\int\widehat{f}^\dagger_{+,k}(-\xi)
\frac{\widehat{D}(|\xi|,\xi)\gamma^0\widehat{D}(|\eta|,\eta)}{|\xi||\eta|}
\widehat{f}_{+,l}(\eta)\overline{\widehat{t^{k+l}\varphi_Tg}}
(-|\xi|+|\eta|,\xi+\eta)
d\xi d\eta\big|\\
&\leq {C}\|{f}_{+,k}\|_{H^{-\alpha}}\|{f}_{+,l}\|_{H^{-\alpha}}
\|\widehat{M}^{\alpha}\widehat{S}^{2\alpha}\widehat{t^{k+l}\varphi_Tg}
\|_{L^2}.
\end{align*} %\label{e4.19}
This together with \eqref{e4.15b} complete the proof.\end{proof}


\begin{lemma} \label{lem4.3}
With the notation above, the following two estimates hold
\begin{gather}
\|{f}_{\pm,0}\|_{H^{-\alpha}}\leq{C}\big(\|\psi_0\|_{H^{-\alpha}}+
\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}\big),
\label{e4.20}\\
\|{f}_{\pm,k}\|_{H^{-\alpha}}\leq{C}\frac{1}{k!}
\|\frac{\widehat{G}_\pm}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}\|_{L^2}.
\label{e4.21}
\end{gather}
\end{lemma}

The proof for the Lemma \ref{lem4.3} is straight forward so that we skip it.
Notice that, in the \eqref{e4.21}, $ \widehat{S}\sim 1$ on the support of
$ \widehat{G}_\pm$.

\begin{lemma} \label{lem4.4}
With the notation above, the following estimate holds
\begin{equation}
\big\|\frac{\widehat\varphi_T\ast\widehat{\overline{K}}_1\ast\widehat{K}_1}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}
\leq{C}\big\|\frac{\widehat{G}_f}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|^2_{L^2}. \label{e4.22}
\end{equation}
\end{lemma}

\begin{proof} For simplicity, we write $ \widehat{G}:=\widehat{G}_f$ and
$ \widehat{K}:=\widehat{K}_1$. We use dyadic decomposition to handle
this case.
Assume that
\begin{equation}
 \widehat{G}=\sum_{k=1}^\infty\widehat{G}_{\pm,\pm,k},\label{e4.23}
\end{equation}
where $ \widehat{G}_{\pm,\pm,k}(\tau,\xi)$ is supported in one of the
following four types of regions:
\begin{equation}
\begin{gathered}
\Sigma_{+,+}:=\{(\tau,\xi):\tau>0,+2^{k-1}<\tau-|\xi|<+2^{k+1}\},\\
%\label{e4.24a}\\
\Sigma_{+,-}:=\{(\tau,\xi):\tau>0,-2^{k+1}<\tau-|\xi|<-2^{k-1}\},\\
%\label{e4.24b}\\
\Sigma_{-,+}:=\{(\tau,\xi):\tau<0,+2^{k-1}<\tau+|\xi|<+2^{k+1}\},\\
%\label{e4.24c}\\
\Sigma_{-,-}:=\{(\tau,\xi):\tau<0,-2^{k+1}<\tau+|\xi|<-2^{k-1}\}.\\
%\label{e4.24d}
\end{gathered} \label{e4.24}
\end{equation}
The decomposition of $ \widehat{G}$ induces a decomposition for
$ \widehat{K}$, namely
\begin{equation}
\widehat{K}_{\pm,\pm,k}=\frac{\widehat{D}}{\widehat{W}}
\widehat{G}_{\pm,\pm,k}.
\label{e4.25a}
\end{equation}
To compute the convolution in \eqref{e4.22},
\begin{equation}
\begin{aligned}
\widehat{\overline{K}}_{\pm,\pm,k}\ast\widehat{K}_{\pm,\pm,l}(-\tau,-\xi)
&=\int\widehat{\overline{K}}_{\pm,\pm,k}(-\tau-\sigma,-\xi-\eta)
\widehat{K}_{\pm,\pm,l}(\sigma,\eta) d\sigma{d}\eta\\
&=\int\widehat{{K}}^\dagger_{\pm,\pm,k}(\tau+\sigma,\xi+\eta)\gamma^0
\widehat{K}_{\pm,\pm,l}(\sigma,\eta) d\sigma{d}\eta,
\end{aligned}\label{e4.25b}
\end{equation}
we have 16 cases resulted from \eqref{e4.24} and \eqref{e4.25b} as follows.
\begin{gather*}
\{(\tau,\sigma,\xi,\eta) :\tau+\sigma > 0,\sigma > 0,
\tau+\sigma-|\xi+\eta| \sim \pm2^k,\sigma-|\eta| \sim \pm2^l\}\\
%\label{e4.26a}\\
\{(\tau,\sigma,\xi,\eta) :\tau+\sigma < 0,\sigma < 0,
\tau+\sigma+|\xi+\eta| \sim \pm2^k,\sigma+|\eta| \sim \pm2^l\}\\
%\label{e4.26b}\\
\{(\tau,\sigma,\xi,\eta) :\tau+\sigma < 0,\sigma > 0,
\tau+\sigma+|\xi+\eta| \sim \pm2^k,\sigma-|\eta| \sim \pm2^l\}\\
%\label{e4.26c}\\
\{(\tau,\sigma,\xi,\eta) :\tau+\sigma > 0,\sigma < 0,
\tau+\sigma-|\xi+\eta| \sim \pm2^k,\sigma+|\eta| \sim \pm2^l\}\\
%\label{e4.26d}
\end{gather*}
We label them as
$\Sigma_{k,l}[(\pm ,\pm);(\pm ,\pm)]$, % \label{e4.27}
and denote by $ \Sigma_{k,l}$ without specifying which one precisely.
We also use $ \widehat{K}_{k}$ for abbreviation of $
\widehat{K}_{\pm,\pm,k}$
and $ \widehat{G}_{k}$ for $ \widehat{G}_{\pm,\pm,k}$ .

  Let $g$ be an arbitrary function. We first compute
\[
\big[\gamma^0(\tau+\sigma)-\gamma^1(\xi+\eta)\big]\gamma^0
\big[\gamma^0\tau+\gamma^1\eta\big]
=\gamma^0\big[(\tau+\sigma)\sigma-(\xi+\eta)\eta\big]+
\gamma^1\big[(\tau+\sigma)\eta-\sigma(\xi+\eta)\big]. %\label{e4.28}
\]
Thus, we have
\begin{align*}
\big|\left<\widehat{\overline{K}}_k\ast\widehat{K}_l,\widehat{g}\right>\big|
&=\Big|\int\widehat{G}_k^\dagger(\tau+\sigma,\xi+\eta)
\frac{\gamma^0(\tau+\sigma)-\gamma^1(\xi+\eta)}
{(\tau+\sigma)^2-(\xi+\eta)^2}\gamma^0
\frac{\gamma^0\sigma+\gamma^1\eta}{\sigma^2-\eta^2}
\widehat{G}_l(\sigma,\eta)\\
&\quad\times \overline{\widehat{g}}(-\tau,-\xi)d\sigma \,d\eta\, d\tau \,
d\xi\Big|\\
&\leq C\|\frac{\widehat{G}_k}{\widehat{M}^\alpha}\|_{L^2}\|
\frac{\widehat{G}_l}{\widehat{M}^\alpha}\|_{L^2}
\Big(\int{I}_{k,l}(\tau,\xi)
\big|\widehat{g}(-\tau,-\xi)\big|^2d\tau\,d\xi\Big)^{1/2},
\end{align*} %\label{e4.29a}
where
\begin{gather}
{I}_{k,l}(\tau,\xi):=\int_{D_{k,l}}
\frac{\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)
Q(\tau,\sigma,\xi,\eta)}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma d\eta,
\label{e4.29b} \\
 Q(\tau,\sigma,\xi,\eta):=
\big[(\tau+\sigma)\sigma-(\xi+\eta)\eta\big]^2+
\big[(\tau+\sigma)\eta-\sigma(\xi+\eta)\big]^2, \label{e4.29c}
\end{gather}
and $ {D_{k,l}(\tau,\xi)}$ is a slice of $ \Sigma_{k,l}$ for
fixed $(\tau,\xi)$; i.e.,
\begin{equation}
{D_{k,l}(\tau,\xi)}:=\{(\sigma,\eta):(\tau,\sigma,\xi,\eta)\in
\Sigma_{k,l}\}.
\label{e4.29d}
\end{equation}

We need to sort the cases into two sets,
\begin{equation}
\Sigma_{k,l}[(\pm,\cdot);(\pm,\cdot)]\quad\text{and}\quad
\Sigma_{k,l}[(\pm ,\cdot);(\mp ,\cdot)], \label{e4.30}
\end{equation}
due to the fact that the computation for the 8 cases in each set is
similar. For simplicity, we will assume $k \geq l$, while the other
case is similar.

\subsection*{Cases H} We have the following estimate
\begin{gather}
\big\|\frac{\widehat{\overline{K}}_{+,\cdot,k}\ast\widehat{K}_{+,\cdot,l}}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}\leq
\frac{C}{2^\frac{l}2}\frac1{2^{(\frac12-\alpha)k}}
\big\|\frac{\widehat{G}_{+,\cdot,k}}{\widehat{M}^\alpha}\big\|_{L^2}
\big\|\frac{\widehat{G}_{+,\cdot,l}}{\widehat{M}^\alpha}\big\|_{L^2},
\label{e4.31a}\\
\big\|\frac{\widehat{\overline{K}}_{-,\cdot,k}\ast\widehat{K}_{-,\cdot,l}}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}\leq
\frac{C}{2^\frac{l}2}\frac1{2^{(\frac12-\alpha)k}}
\big\|\frac{\widehat{G}_{-,\cdot,k}}{\widehat{M}^\alpha}\big\|_{L^2}
\big\|\frac{\widehat{G}_{-,\cdot,l}}{\widehat{M}^\alpha}\big\|_{L^2},
\label{e4.31b}
\end{gather}
In these cases, we have $(\tau+\sigma)\sigma>0$. Throughout some algebraic
manipulation, the expression $Q$ can be written as
\begin{equation}
\begin{aligned}
2Q&=(\tau+\sigma-|\xi+\eta|)^2(\sigma+|\eta|)^2 +
(\tau+\sigma+|\xi+\eta|)^2(\sigma-|\eta|)^2\\
&\quad +8(\tau+\sigma)\sigma\big[|\xi+\eta||\eta|-(\xi+\eta)\eta\big].
\end{aligned}\label{e4.32}
\end{equation}
Take the case of
\begin{equation}
\widehat{\overline{K}}_{+,+,k}\ast\widehat{K}_{+,+,l},\label{e4.33}
\end{equation}
 as an example and in which
$D_{k,l}=\{(\eta,\sigma):\tau+\sigma-|\xi+\eta|\sim 2^k,
\sigma-|\eta|\sim2^l,(\tau,\sigma,\xi,\eta)\in\Sigma_{k,l}[(+,+);(+,+)]\}$.
In this case $\tau+\sigma>0$ and $\sigma>0$. In the $\eta\sigma$-plane,
this is
the region of the intersection of two forward cones. One has the
thickness of $2^k$ and the translation of $(-\xi,-\tau)$, while the
other has thickness of $2^l$. It is bounded mostly, except for the
extreme case which is when one cone moves along the other cone such
that the intersection region is unbounded.

For the first part, we distinguish three cases: $|\xi+\eta|\leq|\eta|$,
$|\xi+\eta|\geq|\eta|$, and the extreme case. For the first two cases,
we have
\begin{equation}
\begin{aligned}
I^1_{k,l}(\tau,\xi)
&:=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma-|\xi+\eta|)^2(\sigma+|\eta|)^2}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma\, d\eta\\
&=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(\tau+\sigma+|\xi+\eta|)^2(\sigma-|\eta|)^2}d\sigma{d}\eta\\
&\sim \frac1{2^{l}}
\int_{\widetilde{D}_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(2^k+|\xi+\eta|)^2}{d}\eta\\
&\leq \frac1{2^{l}}
\int_{\widetilde{D}_{k,l}}\frac1{(2^k+|\xi+\eta|)^{2-2\alpha}}{d}\eta
\widehat{M}^{2\alpha}(\xi)\\
&\leq \frac{C}{2^{(1-2\alpha)k+l}}\widehat{M}^{2\alpha}(\xi)
\widehat{S}^{4\alpha}(\tau,\xi).
\end{aligned} \label{e4.34a}
\end{equation}
For the extreme case, we obtain
\begin{equation}
\begin{aligned}
I^1_{k,l}(\tau,\xi)
&\sim \frac1{2^{l}}
\int_{\widetilde{D}_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(2^k+|\xi+\eta|)^2}{d}\eta\\
&\leq \frac1{2^{l}}
\int_{\widetilde{D}_{k,l}}\frac1{(2^k+|\xi+\eta|)^{2-4\alpha}}{d}\eta\\
&\leq \frac{C}{2^{(1-2\alpha)k+l}}\widehat{M}^{2\alpha}(\xi)
\widehat{S}^{4\alpha}(\tau,\xi).
\end{aligned} \label{e4.34b}
\end{equation}

For the second part, again we distinguish three cases:
$|\xi+\eta|\leq|\eta|$, $|\xi+\eta|\geq|\eta|$, and the extreme case.
For the first two cases, we get
\begin{equation}
\begin{aligned}
 I^2_{k,l}(\tau,\xi)
&:=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma+|\xi+\eta|)^2(\sigma-|\eta|)^2}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma \,d\eta\\
&=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(\tau+\sigma-|\xi+\eta|)^2(\sigma+|\eta|)^2}d\sigma{d}\eta\\
&\sim \frac1{2^{2k-l}}\int_{\widetilde{D}_{k,l}}
\frac{\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)}
{(2^l+|\eta|)^2}{d}\eta\\
&\leq \frac1{2^{2k-l}}\int_{\widetilde{D}_{k,l}}
\frac1{(2^l+|\eta|)^{2-2\alpha}}{d}\eta\widehat{M}^{2\alpha}(\xi)\\
&\leq \frac{C}{2^{k+(1-2\alpha)l}}\widehat{M}^{2\alpha}(\xi)
\widehat{S}^{4\alpha}(\tau,\xi).
\end{aligned} \label{e4.35a}
\end{equation}
For the extreme case, we have
\begin{equation}
\begin{aligned}
 I^2_{k,l}(\tau,\xi)
&\sim \frac1{2^{2k-l}}
\int_{\widetilde{D}_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(2^l+|\eta|)^2}{d}\eta\\
&\leq \frac1{2^{2k-l}}
\int_{\widetilde{D}_{k,l}}\frac1{(2^l+|\eta|)^{2-4\alpha}}{d}\eta
\widehat{M}^{2\alpha}(\xi)\\
&\leq \frac{C}{2^{k+(1-2\alpha)l}}\widehat{M}^{2\alpha}(\xi)
\widehat{S}^{4\alpha}(\tau,\xi).
\end{aligned} \label{e4.35b}
\end{equation}

For the third part, we get
\begin{equation}
\begin{aligned}
I^3_{k,l}(\tau,\xi)
&:=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma)\sigma\big[|\xi+\eta||\eta|-(\xi+\eta)\eta\big]}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma d\eta\\
&\leq \frac{C}{2^{2k+2l}}\int_{D_{k,l}}
\frac{\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma)\sigma|\xi+\eta||\eta|}
{(\tau+\sigma+|\xi+\eta|)^2(\sigma+|\eta|)^2}d\sigma{d}\eta\\
&\leq \frac{C}{2^{2k+2l}}
\int_{D_{k,l}}\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)
d\sigma{d}\eta\\
&\leq \frac{C}{2^{(1-2\alpha)k+l}}\widehat{M}^{2\alpha}(\xi)
\widehat{S}^{4\alpha}(\tau,\xi).
\end{aligned} \label{e4.36a}
\end{equation}
The extreme case will not cause trouble since $\xi+\eta$ and $\eta$ are
of the same sign except on a bounded region, i.e.
$\big[|\xi+\eta||\eta|-(\xi+\eta)\eta\big]=0$ except on a bounded
region. Let us denote the small region by $R$.
\begin{equation}
\begin{aligned}
I^3_{k,l}(\tau,\xi)
&\leq \frac{C}{2^{2k+2l}}\int_{R}
\frac{\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma)\sigma|\xi+\eta||\eta|}
{(\tau+\sigma+|\xi+\eta|)^2(\sigma+|\eta|)^2}d\sigma{d}\eta\\
&\leq \frac{C}{2^{2k+2l}}
\int_{R}\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)
d\sigma{d}\eta\\
&\leq \frac{C}{2^{2k+2l}}
\int_{R} d\sigma{d}\eta2^{2\alpha{k}}\widehat{M}^{2\alpha}(\xi)\\
&\leq \frac{C}{2^{(1-2\alpha)k+l}}\widehat{M}^{2\alpha}(\xi)
\widehat{S}^{4\alpha}(\tau,\xi).
\end{aligned} \label{e4.36b}
\end{equation}

\noindent\textbf{Cases E} We have the following estimate
\begin{gather}
\big\|\frac{\widehat{\overline{K}}_{+,\cdot,k}\ast\widehat{K}_{-,\cdot,l}}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}\leq
\frac{C}{2^\frac{l}2}\frac1{2^{(\frac12-\alpha)k}}
\big\|\frac{\widehat{G}_{+,\cdot,k}}{\widehat{M}^\alpha}\big\|_{L^2}
\big\|\frac{\widehat{G}_{-,\cdot,l}}{\widehat{M}^\alpha}\big\|_{L^2},
\label{e4.37a}\\
\big\|\frac{\widehat{\overline{K}}_{-,\cdot,k}\ast\widehat{K}_{+,\cdot,l}}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}\leq
\frac{C}{2^\frac{l}2}\frac1{2^{(\frac12-\alpha)k}}
\big\|\frac{\widehat{G}_{-,\cdot,k}}{\widehat{M}^\alpha}
\big\|_{L^2}
\big\|\frac{\widehat{G}_{+,\cdot,l}}{\widehat{M}^\alpha}\big\|_{L^2},
\label{e4.37b}
\end{gather}
In these cases, we have $(\tau+\sigma)\sigma<0$. Throughout some algebraic
manipulation, the expression $Q$ can be written as
\begin{align*}
2Q&=(\tau+\sigma+|\xi+\eta|)^2(\sigma+|\eta|)^2 +
(\tau+\sigma-|\xi+\eta|)^2(\sigma-|\eta|)^2\\
&\quad -8(\tau+\sigma)\sigma\big[|\xi+\eta||\eta|+(\xi+\eta)\eta\big].
%\label{e4.38}
\end{align*}
Take the case of
\begin{equation}
\widehat{\overline{K}}_{-,+,k}\ast\widehat{K}_{+,+,l},\label{e4.39}
\end{equation} as
an example and in which
$D_{k,l}=\{(\eta,\sigma):\tau+\sigma+|\xi+\eta|\sim2^k,
\sigma-|\eta|\sim2^l,(\tau,\sigma,\xi,\eta)\in\Sigma_{k,l}[(-,+);
(+,+)]\}$, In this case $\tau+\sigma<0$ and $\sigma>0$.
In $\eta\sigma$-plane, this
is the region of the intersection of a forward cone with a truncated
backward cone. One has the thickness of $2^k$ and the translation of
$(-\xi,-\tau)$, while the other has thickness of $2^l$. It is bounded
for all cases. We still have the extreme case which is when one cone
moves along the other cone, though the region of intersection can be as
large as possible, nevertheless it is bounded.

Again for the first part, we can estimate
\begin{equation}
\begin{aligned}
I^1_{k,l}(\tau,\xi)
&:=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma+|\xi+\eta|)^2(\sigma+|\eta|)^2}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma d\eta\\
&=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(\tau+\sigma-|\xi+\eta|)^2(\sigma-|\eta|)^2}d\sigma{d}\eta\\
&\leq \frac1{2^{2l}}\int_{D_{k,l}}
\frac{(|\xi+\eta|+1)^{2\alpha}(|\eta|+1)^{2\alpha}}
{(\tau+\sigma-|\xi+\eta|)^2}d\sigma{d}\eta.
\end{aligned} \label{e4.40a}
\end{equation}
To estimate the above integral, we separate the cases for
$|\xi+\eta|\geq|\eta|$, $|\xi+\eta|\leq|\eta|$, and the extreme case.
Throughout some calculations, in each case, we have
\begin{equation}
I^1_{k,l}(\tau,\xi)
\leq\frac1{2^l}\frac1{2^{(1-2\alpha)k}}\widehat{M}^{2\alpha}
\widehat{S}^{4\alpha}.
\label{e4.40b}
\end{equation}

 For the second part, we derive
\begin{equation}
\begin{aligned}
 I^2_{k,l}(\tau,\xi)
&:=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)
(\tau+\sigma-|\xi+\eta|)^2(\sigma-|\eta|)^2}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma d\eta\\
&=\int_{D_{k,l}}\frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)}
{(\tau+\sigma+|\xi+\eta|)^2(\sigma+|\eta|)^2}d\sigma{d}\eta\\
&\leq \frac{C}{2^{2k}}\int_{D_{k,l}}
\frac{(|\xi+\eta|+1)^{2\alpha}(|\eta|+1)^{2\alpha}}{(\sigma+|\eta|)^2}
d\sigma{d}\eta\\
&\leq \frac{C2^l}{2^{2k}}\int_{\widetilde{D}_{k,l}}
\frac{(|\xi+\eta|+1)^{2\alpha}}{(2^l+|\eta|)^{2+2\alpha}}{d}\eta.
\end{aligned} \label{e4.41a}
\end{equation}
To estimate the above integral, we separate the cases for
$|\xi+\eta|\geq|\eta|$, $|\xi+\eta|\leq|\eta|$, and the extreme case.
Throughout some calculations, in each case, we have
\begin{equation}
I^2_{k,l}(\tau,\xi)
\leq\frac1{2^l}\frac1{2^{(1-2\alpha)k}}\widehat{M}^{2\alpha}
\widehat{S}^{4\alpha}.
\label{e4.41b}
\end{equation}
For the third part, we have
\begin{equation}
\begin{aligned}
I^3_{k,l}(\tau,\xi)
&:=\int_{D_{k,l}}    \frac{\widehat{M}^{2\alpha}(\xi+\eta)
\widehat{M}^{2\alpha}(\eta)
|\tau+\sigma|\sigma\big[|\xi+\eta||\eta|+(\xi+\eta)\eta\big]}
{\widehat{W}^2(\tau+\sigma,\xi+\eta)\widehat{W}^2(\sigma,\eta)}
d\sigma d\eta\\
&\leq \frac{C}{2^{2k+2l}}\int_{D_{k,l}}
\frac{\widehat{M}^{2\alpha}(\xi+\eta)\widehat{M}^{2\alpha}(\eta)
|\tau+\sigma|\sigma|\xi+\eta||\eta|}
{(\tau+\sigma-|\xi+\eta|)^2(\sigma+|\eta|)^2}d\sigma{d}\eta\\
&\leq \frac{C}{2^{2k+2l}}
\int_{D_{k,l}}(|\xi+\eta|+1)^{2\alpha}(|\eta|+1)^{2\alpha}  d\sigma{d}\eta.
\end{aligned} \label{e4.42a}
\end{equation}
To estimate the above integral, we separate the cases for
$|\xi+\eta|\geq|\eta|$, $|\xi+\eta|\leq|\eta|$, and the extreme case.
Notice that for the extreme case, we have
$|\xi+\eta||\eta|+(\xi+\eta)\eta=0$ except on a small part of the
region of the intersection. Throughout some calculations, in each case,
we have
\begin{equation}
 I^3_{k,l}(\tau,\xi)
\leq\frac1{2^l}\frac1{2^{(1-2\alpha)k}}\widehat{M}^{2\alpha}
\widehat{S}^{4\alpha}.
\label{e4.42b}
\end{equation}

  Now we return to the proof of \eqref{e4.22}. Combine the above, we get
\begin{equation}
\begin{aligned}
\big|\left<\overline{K}_kK_l,g\right>\big|\leq
&C\big\|\frac{\widehat{G}_k}{\widehat{M}^\alpha}\big\|_{L^2}
\big\|\frac{\widehat{G}_l}{\widehat{M}^\alpha}\big\|_{L^2}
\big(\int{I}_{k,l}(\tau,\xi)
\big|\widehat{g}(-\tau,-\xi)\big|^2d\tau{d}\xi\big)^{1/2}\\
&\leq \frac{C}{2^\frac{l}2}\frac1{2^{(\frac12-\alpha)k}}
\big\|\frac{\widehat{G}_k}{\widehat{M}^\alpha}\big\|_{L^2}
\big\|\frac{\widehat{G}_l}{\widehat{M}^\alpha}\big\|_{L^2}
\|\widehat{M}^{\alpha}\widehat{S}^{2\alpha}\widehat{g}\|_{L^2}\\
&\leq \frac{C}{2^{\frac14l}}\frac1{2^{\epsilon{k}}}
\big\|\frac{\widehat{G}_k}{\widehat{M}^\alpha\widehat{S}^{\frac14}}
\big\|_{L^2}
\big\|\frac{\widehat{G}_l}{\widehat{M}^\alpha\widehat{S}^{\frac14}}
\big\|_{L^2}
\|\widehat{M}^{\alpha}\widehat{S}^{2\alpha}\widehat{g}\|_{L^2}.
\end{aligned}
\label{e4.43}
\end{equation}
Finally, we have
\begin{equation}
\begin{aligned}
\big\|\frac{\widehat{\overline{K}}\ast\widehat{K}}{\widehat{M}^{\alpha}
\widehat{S}^{2\alpha}}\big\|_{L^2}
&\leq\sum_{k,l}
\big\|\frac{\widehat{\overline{K}}_k\ast\widehat{K}_l}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}
\big\|_{L^2}\\
&\leq \sum_{k,l}\frac{C}{2^{\epsilon{k}+{\frac14l}}}
\big\|\frac{\widehat{G}_k}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}
\big\|\frac{\widehat{G}_l}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}
\leq{C}\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}
\widehat{S}^{\frac14}}\big\|_{L^2}^2.
\end{aligned}\label{e4.44}
\end{equation}
This completes the proof.\end{proof}

  The estimates for the remaining cases are given in the following
Lemma.

\begin{lemma} \label{lem4.5} For $j=1, 2$ and $k=0, 1, 2,
\cdot\cdot\cdot$. The following estimates hold
\begin{gather*}
\big\|\frac{
\widehat\varphi_T\ast\big(\delta_\mp^{(k)}\widehat{f^\dagger_{\pm,k}}
\frac{\widehat{D}_\pm}{|\xi|}\gamma^0\big)
\ast\big(\widehat{K}_j\big)}{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}
\big\|_{L^2}
\leq{C}(k+1)T^{k-\frac12}\|{f}_{\pm,k}\|_{H^{-\alpha}}
\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2},\\
%\label{e4.45a}
\big\|\frac{\widehat\varphi_T\ast\widehat{\overline{K}}_j\ast
\big(\delta_\pm^{(k)}\frac{\widehat{D}_\pm}{|\xi|}\widehat{f}_{\pm,k}\big)}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}
\leq{C}(k+1)T^{k-\frac12}\|{f}_{\pm,k}\|_{H^{-\alpha}}
\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2},\\
% \label{e4.45b}
\big\|\frac{\widehat\varphi_T\ast\widehat{\overline{K}}_1\ast\widehat{K}_2}
{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}\big\|_{L^2}
\leq{C}\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}^2, \\ %\label{e4.45c}
\big\|\frac{\widehat\varphi_T\ast\widehat{\overline{K}}_2
\ast\widehat{K}_j}{\widehat{M}^{\alpha}\widehat{S}^{2\alpha}}
\big\|_{L^2}
\leq{C}\big\|\frac{\widehat{G}}{\widehat{M}^{\alpha}\widehat{S}^{\frac14}}
\big\|_{L^2}^2, %\label{e4.45d}
\end{gather*}
\end{lemma}

The proof of this lemma is a repetition of the arguments presented
in Lemmas \ref{lem4.1}, \ref{lem4.2}, and \ref{lem4.4}, so that we omit it.


\subsection*{Acknowledgement} The author want to express his
gratitude to Manossos Grillakis and Chang-shou Lin for their
encouragement and help.

\begin{thebibliography}{00}

\bibitem{B} J. Bourgain, \emph{Invariant Measures for NLS in
Infinite Volume}, Commun. Math. Phys , Vol. 210 (2000),
605-620.

\bibitem{Ba} A. Bachelot, \emph{Global existence of large
amplitude solutions for Dirac-Klein-Gordon systems in Minkowski
space}, Lecture Notes in Math. , Vol. 1402 (1989), 99-113. Springer, Berlin

\bibitem{Bo} N. Bournaveas, \emph{A new proof of global
existence for the Dirac-Klein-Gordon equations in one space
dimension}, J. Funct. Anal. , Vol. 173 (2000), 203-213.

\bibitem{C} J. Chadam, \emph{Global Solutions of the Cauchy
Problem for the (Classical) Coupled Maxwell-Dirac Equations in one
Space Dimension}, J. Funct. Anal. , Vol. 13 (1973), 173-184.

\bibitem{CG} J. Chadam \& R. Glassey, \emph{On Certain Global
Solutions of the Cauchy Problem for the (Classical) Coupled
Klein-Gordon-Dirac equations in one and three Space Dimensions},
 Arch. Rational Mech. Anal. , Vol. 54 (1974), 223-237.

\bibitem{F1} Yung-fu Fang, \emph{Local Existence for Semilinear
Wave Equations and Applications to Yang-Mills Equations}, Ph.D
dissertation , Vol. (1996),. University of Maryland

\bibitem{F2} Yung-fu Fang, \emph{A Direct Proof of Global
Existence for the Dirac-Klein-Gordon Equations in One Space
Dimension},  (2002), preprint.

\bibitem{FG} Yung-fu Fang \& Manoussos Grillakis,
\emph{Existence and Uniqueness for Boussinesq type Equations on a Circle},
Comm. PDE, Vol. 21 (1996), 1253-1277.

\bibitem{G} V. Georgiev, \emph{Small amplitude solutions of the
Maxwell-Dirac equations}, Indiana Univ. Math. J., Vol. 40,
(1991), 845-883.

\bibitem{GS} R. Glassey \& W. Strauss, \emph{Conservation laws
for the classical Maxwell-Dirac and Klein-Gordon-Dirac equations},
J. Math. Phys. , Vol. 20 (1979), 454-458.

\bibitem{KM} S. Klainerman and M. Machedon, \emph{Space-time
estimates for null forms and the local existence theorem},
Comm. Pure Appl. Math. , Vol. XLVI (1993), 1221-1268.

\bibitem{Ku} Sergej Kuksin, \emph{Infinite-Dimensional
Symplectic Capacities and a Squeezing Theorem for Hamiltonian
PDE's}, Commun. Math. Phys , Vol. 167 (1995), 531-552.

\bibitem{S} E. M. Stein, \emph{Singular Integrals and
Differentiability Properties of Functions},
Princeton University Press, 1970.

\bibitem{Z} Y. Zheng, \emph{Regularity of weak solutions to a
two-dimensional modified Dirac-Klein-Gordon system of equations},
Commun. Math. Phys., Vol. 151 (1993), 67-87.

\end{thebibliography}

\end{document}