text:
\subsection{IMEX time discretization}
\label{sec:3.2}

In this section, we will formulate the IMEX Runge-Kutta (RK) time discretizations for the semi-discrete schemes introduced in Section \ref{sec:3.1}. With similarity and for notational clarity,  this will be presented only for Scheme II in its integral form. Scheme I and the nodal form of Scheme II can be  discussed similarly. Note that Scheme II in the integral form can be given compactly as follows. Find $U_h(\cdot,t)\in {\bf Z}_h^K$, $g_h(\cdot,v,t), r_h(\cdot, t)\in Z_h^K$, such that
$\forall \phi, \psi, \varphi\in Z_h^K$ and $\forall i$,
\begin{subequations}
\label{eq: compact}
\begin{align}
(\partial_t U_h, \phi) + F_h(U_h, \phi) &= D_h( \eps(x) g_h, \phi), \label{eq: compact1}\\
(\eps(x) \partial_t g_h, \psi) + b_{h, v}(\eps(x)g_h, \psi) &= s^{(1)}_h(g_h, \psi) + s^{(2)}_{h, v}(U_h, r_h, \psi), \label{eq: compact2}\\
(r_h, \varphi) =  H_h(U_h, \varphi), \label{eq: compact3}
\end{align}
\end{subequations}
where
\begin{subequations}
\label{eq: S2}
\begin{align}
F_h(U_h, \phi) &= - \int_{\Omega_x} F(U_h)\frac{d \phi(x)}{dx} dx - \sum_i \hat{F}_{h, i+\frac12} [\phi]_{i+\frac12},\\
D_h( \eps(x) g_h, \phi) & =  \int_{\Omega_x} \eps(x) \langle vm g_h\rangle\frac{d \phi(x)}{dx} dx + \sum_i \eps(x_{i+\frac12}) \widehat{\langle vmg_h\rangle}_\iR [\phi]_{i+\frac12}, \\
b_{h, v}(\eps(x)g_h, \psi) &= \int_{\Omega_x} ({\mathbf I}-\Pi_{M_h}) \mD_{h,1}(\eps(x) vg_h)\psi dx,\\
s^{(1)}_h(g_h, \psi) &=-\int_{\Omega_x} g_h\psi dx, \quad
s^{(2)}_{h, v}(U_h, r_h, \psi) =  -\int_{\Omega_x} A_h \frac{r_h}{\sqrt{T_h}} M_h \psi dx, \\
H_h(U_h, \varphi)&= {-\left(\int_{\Omega_x} T_h \frac{d\varphi}{dx} dx + \sum_i \hat{T}_{h, i+\frac12} [\varphi]_{i+\frac12}\right)\big |_{T_h=T_h(U_h)}.}
\end{align}
\end{subequations}

To discretize in time for the scheme in \eqref{eq: compact}, we start with a first order IMEX scheme to introduce our implicit-explicit strategy. Given $U_h^n \in {\bf Z}_h^K$ and $g_h^n \in Z_h^K$ that approximating the solutions $U$ and $g$ at $t=t^n$, respectively,
we look for $U_h^{n+1} \in {\bf Z}_h^K$ and $g_h^{n+1}, {r_h^{n+1}} \in Z_h^K$, such that $\forall \phi, \psi, \varphi \in Z_h^K$,

\begin{subequations}
\label{eq: compact_d}
\begin{align}
\left(\frac{U^{n+1}_h-U^n_h}{\Dt}, \phi \right) + F_h(U^n_h, \phi) &= D_h( \eps(x) g^n_h, \phi), \label{eq: compact_d1}\\
\left(\eps(x) \frac{g^{n+1}_h-g^n_h}{\Dt}, \psi\right) + b_{h, v}(\eps(x)g^n_h, \psi) &= s^{(1)}_h(g^{n+1}_h, \psi) + s^{(2)}_{h, v}(U^{n+1}_h, r_h^{n+1},  \psi), \label{eq: compact_d2} \\
{(r_h^{n+1}, \varphi)} &{= H_h(U_h^{n+1}, \varphi).} \label{eq: compact_d3}
\end{align}
\end{subequations}
This fully discrete scheme can be implemented efficiently. Specifically, one can solve the equation \eqref{eq: compact_d1} for macroscopic variables $U_h^{n+1}$ at the updated time level first, then solve for $r_h^{n+1}$ from \eqref{eq: compact_d3}. Finally one can solve for $g_h^{n+1}$ from \eqref{eq: compact_d2}.

This implicit-explicit procedure can be easily extended to high order globally stiffly accurate IMEX schemes, which can be characterized by a double Butcher Tableau
\beq
\label{eq: B_table}
\begin{array}{c|c}
\tilde{c} & \tilde{A}\\
\hline
 & \tilde{b}^ t\end{array} \ \ \ \ \
\begin{array}{c|c}
{c} & {A}\\
\hline
 & {b^t} \end{array},
\eeq
where $\tilde{A} = (\tilde{a}_{ij})$ is an $s\times s$  lower triangular matrix with zero diagonal for an explicit scheme, and $A = (a_{ij})$ is an $s\times s$ lower triangular matrix with the diagonal entries not all being zero for a diagonally implicit RK (DIRK) method. The coefficients $\tilde{c}$ and $c$ are given by the standard relations
\begin{eqnarray}\label{eq:candc}
\tilde{c}_i = \sum_{j=1}^{i-1} \tilde a_{ij}, \ \ \ c_i = \sum_{j=1}^{i} a_{ij},
\end{eqnarray}
and vectors $\tilde{b} = (\tilde{b}_j)$ and $b = (b_j)$ represent the quadrature weights for internal stages of the RK method. The IMEX RK scheme is defined to be {\em globally stiffly accurate} if $\tilde{c}_s = c_s = 1$ and {$a_{sj} = b_j$, $\tilde{a}_{sj} = \tilde{b}_j$}, $\forall j=1, \cdots, s$. The fully-discrete scheme using the Butcher notation can be written as
follows. Given $U_h^n\in {\bf Z}_h^K$ and $g_h^n \in Z_h^K$, we look for $U_h^{n+1}\in {\bf Z}_h^K$ and $g_h^{n+1} \in Z_h^K$, such that $\forall \phi, \psi \in Z_h^K$,
\begin{subequations}
\label{eq: compact_d_rk}
\begin{align}
\left(U^{n+1}_h, \phi \right) &= \left(U^{n}_h, \phi \right) -\Dt \sum_{l=1}^{s} \tilde{b}_l \left(F_h(U^{(l)}_h, \phi) -D_h( \eps(x) g^{(l)}_h, \phi)\right), \label{eq: compact_d_rk1}\\
\left(\eps(x) g^{n+1}_h, \psi\right) & =\left(\eps(x) g^{n}_h, \psi\right)- \Dt \sum_{l=1}^{s} \tilde{b}_l b_{h, v}(\eps(x)g^{(l)}_h, \psi) + \Dt \sum_{l=1}^{s} b_l \left(s^{(1)}_h(g^{(l)}_h, \psi) + s^{(2)}_{h, v}(U^{(l)}_h, r^{(l)}_h, \psi)\right). \label{eq: compact_d_rk2}
\end{align}
\end{subequations}
Here the approximations at the internal stages of one RK step, $U_h^{(l)}\in {\bf Z}_h^K$ and $g_h^{(l)}, {r_h^{(l)}} \in Z_h^K$ with $l=1, \cdots, s$, satisfy
\begin{subequations}
\label{eq: compact_rk}
\begin{align}
\left(U^{(l)}_h, \phi \right) &= \left(U^{n}_h, \phi \right) -\Dt \sum_{j=1}^{l-1} \tilde{a}_{lj} \left(F_h(U^{(j)}_h, \phi) -D_h( \eps(x) g^{(j)}_h, \phi)\right), \label{eq: compact_rk1}\\
\left(\eps(x) g^{(l)}_h, \psi\right) & =\left(\eps(x) g^{n}_h, \psi\right)- \Dt \sum_{j=1}^{l-1} \tilde{a}_{lj} b_{h, v}(\eps(x)g^{(j)}_h, \psi) + \Dt \sum_{j=1}^{l} a_{lj} \left(s^{(1)}_h(g^{(j)}_h, \psi) + s^{(2)}_{h, v}(U^{(j)}_h, r^{(j)}_h, \psi)\right), \label{eq: compact_rk2}\\
{(r_h^{(l)}, \varphi)} &{= H_h(U_h^{(l)}, \varphi),} \label{eq: compact_rk3}
\end{align}
\end{subequations}
for any $\phi, \psi, \varphi \in Z_h^K$. Similar to the first order IMEX scheme, in a stage-by-stage fashion for {$l=1, \cdots, s$}, one can first solve $U^{(l)}_h$ explicitly from the equation \eqref{eq: compact_rk1}, then plug $U^{(l)}_h$ into \eqref{eq: compact_rk3} to solve $r_h^{(l)}$, and finally solve $g^{(l)}_h$

from \eqref{eq: compact_rk2}.

The third order IMEX scheme we use in our simulations is the globally stiffly accurate ARS(4, 4, 3) scheme \cite{ascher1997implicit} with a double Butcher Tableau
\beq
\label{eq: ars443}

\begin{array}{c|c c c c c}
0 & 0&0&0& 0 & 0\\
1/2 &1/2&0&0& 0 &0\\
2/3 &11/18&1/18&0& 0 &0\\
1/2 &5/6&-5/6&1/2& 0 &0\\
1 &1/4&7/4&3/4& -7/4 &0\\
\hline
&1/4&7/4&3/4& -7/4 &0\\
 \end{array} \ \ \ \ \
\begin{array}{c|c c c c c}
0 & 0&0&0& 0 & 0\\
1/2 &0&1/2&0&0 &0\\
2/3 &0&1/6&1/2& 0 &0\\
1/2 &0&-1/2&1/2& 1/2 &0\\
1 &0&3/2&-3/2& 1/2 &1/2\\
\hline
 &0&3/2&-3/2& 1/2 &1/2\\
 \end{array}
\eeq

\subsection{Formal asymptotic analysis}
\label{sec:3.3}

It is straightforward to see that when $\eps$ goes to $0$, the proposed Schemes I and II
will give rise to a DG method satisfied by $U_h$ with the Lax-Friedrichs flux for the Euler system (see \eqref{cns-bgk} with $\eps=0$), and in fact the DG scheme in the limit is essentially the same as the one in \cite{cockburn1989tvb2}. This can be easily shown for both the semi-discrete and fully-discrete versions of the methods in their integral or nodal form.

In this subsection, we will focus on the formal asymptotic analysis to capture the compressible Navier-Stokes limit for small $\eps$. It is assumed that Knudsen number $\eps$ is constant, and the $v$-direction is continuous without being discretized.
Though both Schemes I and II perform well numerically for a wide range of Knudsen number (see Section \ref{sec:numerical}), only Scheme II can be formally shown to have the correct compressible Navier-Stokes limit. More specifically, in Proposition~\ref{prop: asymptotics} below, we will establish a formal asymptotic analysis, showing that for $0<\eps\ll1$, the fully-discrete DG-IMEX Scheme II in its nodal form will become a local DG method in its nodal form with some explicit RK time discretization, up to $\mathcal{O}(\eps^2)$,  for the compressible Navier-Stokes system.
The limiting scheme is in a similar spirit to the highly cited work proposed by Bassi and Rebay in 1997 \cite{bassi1997high} which was later generalized and analyzed in \cite{cockburn1998local}.

To establish such connection, we first formulate a local DG method with an explicit RK time discretization for the compressible Navier-Stokes system \eqref{cns-bgk2}, which in one dimension is given as follows,
\begin{equation}
\begin{split}
\del_t \left(
\begin{array}{c} \rho\\ \rho u \\ E
\end{array}
 \right)+
 \del_x \left(
 \begin{array}{c}
 \rho u \\ \rho u^2 +p I  \\ (E+p)u
 \end{array}
 \right) = \ep  \del_x \left(
 \begin{array}{c} 0 \\
 0   \\
 \frac32 \rho T \partial_x T
\end{array}
 \right).

\end{split}\label{cns-bgk-1d}
\end{equation}
The semi-discrete local DG method is to find $U_h(\cdot, t) \in {\bf Z}_h^K$ and  $r_h(\cdot, t) \in Z_h^K$, such that $\forall \phi, \varphi \in Z_h^K$
\begin{subequations}
\label{eq:scheme:cns}
\begin{align}
\label{eq: compact_cns}
\left({\partial_t}U_h, \phi \right) + F_h(U_h, \phi) &= \eps F^{(vis)}_h(U_h, r_h, \phi),\\
\label{eq: q}
(r_h, \varphi)&=
-\sum_i\left(\int_{I_i} T_h \frac{d\varphi}{dx} dx + \widehat{T}_{h,i+\frac12} [\varphi]_{\iR}\right).
\end{align}
\end{subequations}
Here $F^{(vis)}_h(U_h, r_h, \phi) = (0, 0,   f^{(vis)}_{E,h})^t$ with
\begin{align}
f^{(vis)}_{E,h}
& = -\frac32\sum_i \left(\int_{I_i}  \rho_h T_h r_h \frac{d\phi}{dx} dx +  \widehat{(\rho_h T_h r_h)}_\iR  [\phi]_{\iR} \right), \label{eq: vis_flux}
\end{align}
where $T_h$ is obtained from $U_h$ based on \eqref{eq: T}.
Note that the right hand side of \eqref{eq: q} is just $H_h(U_h, \varphi)$ defined in \eqref{eq: S2}, and it is explicitly written here to emphasize that  \eqref{eq: q} is to approximate the auxiliary variable $r := \partial_x T$.
The numerical fluxes in \eqref{eq: q} and \eqref{eq: vis_flux} can be taken to be either alternating or central fluxes.
Via the method of line approach, the local DG scheme \eqref{eq:scheme:cns}
can be evolved in time by an explicit RK method characterized by a Butcher table $\tilde{A}$, $\tilde{b}$ and $\tilde{c}$ in \eqref{eq: B_table} as follows: find $U^{n+1}_h(\cdot, t), U^{(l)}_h \in {\bf Z}_h^K$ and  $r^{(l)}_h(\cdot, t) \in Z_h^K$ with $l=1, \cdots, s$, such that $\forall \phi, \varphi \in Z_h^K$,
\beq
\label{eq: compact_cns_rk0}
\left(U^{n+1}_h, \phi \right) = \left(U^{n}_h, \phi \right) -\Dt \sum_{l=1}^{s} \tilde{b}_l \left(F_h(U^{(l)}_h, \phi) -\eps F^{(vis)}_h(U^{(l)}_h, r^{(l)}_h, \phi)\right),
\eeq
with
\beq
\label{eq: compact_cns_rk}
\left(U^{(l)}_h, \phi \right) = \left(U^{n}_h, \phi \right) -\Dt \sum_{j=1}^{l-1} \tilde{a}_{lj} \left(F_h(U^{(j)}_h, \phi) -\eps F^{(vis)}_h (U^{(j)}_h,  r^{(j)}_h, \phi)\right),
\eeq
and
\beq
\label{eq: q_rk}
(r_h^{(l)}, \varphi)=
-\sum_i\left(\int_{I_i} T_h^{(l)} \frac{d\varphi}{dx} dx + \widehat{T}^{(l)}_{h,i+\frac12} [\varphi]_{\iR}\right).
\eeq
The method above can be further given in its nodal form similarly as in Section \ref{sec:3.1}, and this is omitted for brevity.
Next we will state and show a formal asymptotic analysis for the proposed Scheme II in its nodal form. {We consider the third order IMEX scheme ARS(4, 4, 3) as an example to illustrate the analysis, while the result can be extended to general IMEX methods of type A, CK or ARS (for definitions of these types see \cite{boscarino2008error}).}

\begin{prop}
\label{prop: asymptotics}
For the BGK equation based on the micro-macro formulation \eqref{eq:mmdc}, we consider the fully discrete DG-IMEX scheme \eqref{eq: compact_d_rk}-\eqref{eq: compact_rk}, with operators specified in \eqref{eq: S2} for Scheme II in its nodal form, and {the third order globally stiffly accurate IMEX time discretization ARS(4,4,3) characterized by the double Butcher table \eqref{eq: B_table} and given in \eqref{eq: ars443}}.
 For $0<\eps\ll 1$, the scheme is asymptotically equivalent, up to $\mathcal{O}(\eps^2)$,
 to the local DG method \eqref{eq: compact_cns_rk0}-\eqref{eq: q_rk} in its nodal form for compressible Navier-Stokes equations \eqref{cns-bgk-1d}, coupled with the explicit RK time discretization  characterized by the Butcher table $\tilde{A}$, $\tilde{b}$ and $\tilde{c}$ in \eqref{eq: B_table}.
\end{prop}
\begin{proof}
It is sufficient to prove the equivalence of the discretizations, namely, the nodal forms of the $D_h$ term in \eqref{eq: compact_d_rk}-\eqref{eq: compact_rk} and the $\eps F^{(vis)}_h$ term in \eqref{eq: compact_cns_rk0}-\eqref{eq: compact_cns_rk}, for the viscous term in \eqref{cns-bgk-1d}
for sufficiently small $\eps$.
Consider the fully discrete scheme \eqref{eq: compact_d_rk}-\eqref{eq: compact_rk} in its nodal form (e.g. see equations  \eqref{eq:SDG1:a}, \eqref{eq:SDG2:c}, \eqref{eq:SDG2:d}),
when $0<\eps\ll1$ and for {the RK stage, $l = 2, \cdots, s$},
we have
\beq
\label{eq: ap_g}
(g_i^k)^{(l)} = -(A_i^k)^{(l)} (r_i^k)^{(l)} (M_i^k)^{(l)}/\sqrt{(T_i^k)^{(l)}} + \mathcal{O}(\eps), \quad \forall i=1, \cdots, N_x, \quad k = 0, \cdots, K
\eeq
with $(r_i^k)^{(l)}$ obtained from the equation \eqref{eq:SDG2:d}.
Applying $\langle vm \cdot \rangle$ to the above equation gives
\begin{align}
\langle vm(g_i^k)^{(l)} \rangle & = - \langle vm (A_i^k)^{(l)} (M_i^k)^{(l)} \rangle {(r_i^k)^{(l)}}/{\sqrt{(T_i^k)^{(l)}}} + \mathcal{O}(\eps), \nonumber\\
& = (0, 0,  \frac32 (\rho_i^k))^{(l)} (T_i^k)^{(l)} (r_i^k)^{(l)})^t +\mathcal{O}(\eps), \quad \forall i=1, \cdots, N_x, \quad k = 0, \cdots, K.
\label{eq: vmg}
\end{align}
Here the second equality is obtained along the same line of deriving the viscous term in the equation \eqref{cns-bgk-1d}.
{Since the IMEX method is globally stiffly accurate, $(g_i^k)^{n+1}= (g_i^k)^{(s)}$, hence \eqref{eq: ap_g} holds also at $t^{n+1}$. Such property is needed to justify \eqref{eq: ap_g} for $l=1$ when $a_{11} = 0$ for the next time step evolution, see the Butcher table \eqref{eq: ars443}.}
Now one can plug \eqref{eq: vmg} into the last term in \eqref{eq: compact_d_rk1} and in \eqref{eq: compact_rk1} in the nodal form (see e.g. \eqref{eq:SDG1:a}), and up to $\mathcal{O}(\eps^2)$, this will result in  $\eps F^{(vis)}_{h} (\cdot, \cdot, \cdot)$ in \eqref{eq: compact_cns_rk0}-\eqref{eq: compact_cns_rk} in its nodal form. This, in addition to \eqref{eq:SDG2:d}, determines the discretization of the viscous term up to $\mathcal{O}(\eps^2)$. The explicit part of the IMEX time discretization will be naturally carried over.
\end{proof}
\begin{rem}
The local DG method in \eqref{eq: compact_cns_rk0}-\eqref{eq: q_rk} is in a similar spirit to, yet different from the one proposed in \cite{bassi1997high}. In \cite{bassi1997high}, auxiliary variables $S := {\partial_x U}$ are introduced and approximated together with $U$  by local DG discretizations with central fluxes, while here only one auxiliary variable $r := {\partial_x T}$ is introduced {for one-dimensional problems}.
\end{rem}


text:
\begin{document}
\baselineskip=14.4pt

\title{$L^p -L^{\acute{p}}$ Estimates for the Schr\"{o}dinger Equation on the
Line and Inverse Scattering for the Nonlinear Schr\"{o}dinger Equation
with
 a Potential \thanks{{\sc ams} classification 35P, 35Q, 35R, 34B and 81U. }}

\author{Ricardo Weder\thanks{Fellow Sistema Nacional de Investigadores.}
\thanks{On leave of absence from IIMAS--UNAM. Apartado Postal 20--726.
M\'exico D.F. 01000. E--Mail: weder@servidor.unam.mx }\\
Instituto de F\'{\i}sica de Rosario,\\
Consejo Nacional de Investigaciones Cient\'{\i}ficas y T\'ecnicas,
\\Bv. 27 de febrero 210 bis, 2000 Rosario, Argentina.\\
E-Mail:weder@ifir.ifir.edu.ar}
\date{}
\maketitle

\begin{center}
\begin{minipage}{5.75in}
\centerline{{\bf Abstract}}\bigskip
In this paper I prove a $L^{p}-L^{\acute{p}}$ estimate for the solutions of
the
one--dimensional Schr\"{o}dinger equation with a potential in $L^{1}_{\gamma}$
where in the {\it generic case} $\gamma > 3/2$ and in the
{\it exceptional case}
 (i.e. when
there is a half--bound state of zero energy) $\gamma > 5/2$. I use this
estimate
 to construct the scattering operator for the nonlinear Schr\"{o}dinger
 equation with a potential. I prove moreover, that the low--energy limit of
 the scattering operator
 uniquely determines the potential and the nonlinearity using a method that
  allows as well for the reconstruction of the potential and of the
   nonlinearity.

\end{minipage}
\end{center}
\newpage

\section{Introduction}\sss
Let us consider the Schr\"{o}dinger equation (LS)
\beq
i\frac{\partial}{\partial t} u(t,x)= H_0 u(t,x),\, u(0,x)=\phi (x)
\label{1.1}
\ene
where $H_0$ is the self--adjoint realization of $-\Delta$ in
$L^2\left(\ER^n \right), n \geq 1,$

\beq
H_0 := -\sum^{n}_{j=1}\frac{\partial^2}{\partial x^2_j }.
\label{1.2}
\ene
 The domain of $H_0, D(H_0 ),$ is the Sobolev space $W_2$. The solution to
(\ref{1.1}) is given by $e^{-itH_0} \phi$, where the strongly continuous
unitary group $e^{-itH_0}$ is defined by the functional calculus of
 self--adjoint operators. The kernel of $e^{-itH_0}$ is given by ( see
 Example 3 in page 59 of \cite{6} ) $(4 \pi it )^{-n/2} e^{i|x-y|^2 /4 t}$.
 From this explicit expression for the kernel it follows that the restriction
 of $e^{-it H_0}$ to $L^2 (\ER^n )\cap L^p (\ER^n )$ extends to a bounded
 operator from $L^p (\ER^n ) $ into $L^{\acute{p}} (\ER^n )$ such that
 \beq
 \left\| e^{-it H_0}\right\|_{{\cal B}\left( L^p (\ER^n ), L^{\acute{p}}(\ER^n )
 \right)} \leq \frac{C}{t^{n(\frac{1}{p}-\frac{1}{2})}},\, t > 0,
 \label{1.3}
 \ene
 for some constant $C$, $1 \leq p \leq 2,$ and $\frac{1}{p}
 +\frac{1}{\acute{p}}=1,$ and where for any pair of Banach spaces $X,Y$ we
 denote by ${\cal B}(X, Y )$ the Banach space of all bounded linear operators
 from $X$ into $Y$. In the case when $X=Y$ we use the notation ${\cal B}(X)$.
 Estimate (\ref{1.3}) expresses the dispersive nature of the solutions to
 (\ref{1.1}) and it is a fundamental tool in the study of the nonlinear
 Schr\"{o}dinger equation:

 \beq
 i\frac{\partial}{\partial t}u= H_0 u + f(u)
 \label{1.4}
 \ene
 since it allows to control the nonlinear behaviour of the solutions to
 (\ref{1.4}), that is produced by $f(u)$, in terms of the dispersion that is
 produced by the linear term $H_0 u$. See for example \cite{6},  \cite{20},
 \cite{21}, \cite{22}, \cite{15},
 \cite{16}, \cite{18}, \cite{19},
  \cite{33},
 \cite{27} and \cite{32}.

 In the case of a linear Schr\"{o}dinger equation with a potential (LSP):
 \beq
 i\frac{\partial}{\partial t} u(t,x)=(H_0 +V)u(t,x), \, u(0,x)=\phi,
 \label{1.5}
 \ene
 where $V$ is a real--valued function defined on $\ER^n$ such that the operator
 $H:= H_0 +V$ is self--adjoint on $D(H_0 )$, Journ\'e, Soffer and Sogge
 \cite{10}
 proved
 that for $n \geq 3$

 \beq
\left\|e^{-it H_0} P_c \right\|_{{\cal B}\left(L^p (\ER^n ), L^{\acute{p}}
(\ER^n )\right)} \leq \frac{C}{t^{n(\frac{1}{p}-\frac{1}{2})}},
 \label{1.6}
 \ene
 for $1\leq p \leq 2,\,\frac{1}{p}+\frac{1}{\acute{p}}=1$ and where $P_c$ is
 the orthogonal projector onto the continuous subspace of $H$. Note that
 (\ref{1.6}) can not hold for the pure point subspace of $H$. Estimate
 (\ref{1.6}) is the natural extension of (\ref{1.3}) to the case with a
 potential. Besides conditions on the regularity and the decay of $V$ (see
 equation (1.6) of \cite{10}) Journ\'e, Soffer and Sogge require that zero is
 neither a bound state nor a half--bound state for $H$. The proof given by
 \cite{10} consists of a high--energy estimate that is always true and of a
 low--energy estimate where the condition that zero is neither a bound state nor a
 half--bound state was used. The low--energy estimate of \cite{10} was obtained
 by studying the behaviour near zero of the spectral family of $H$. For this
 purpose Journ\'e, Soffer and Sogge \cite{10} used the estimates on the
 behaviour near
 zero of the resolvent of $H$ obtained by Jensen and Kato \cite{34}, \cite{30}
 and \cite{31} for $n\geq 3$. It is actually here that the restriction $n \geq
 3 $ appears in the result of \cite{10}. One way to understand the
 reasons for the restriction to $n \geq 3$ is to look to the kernel of the
 free resolvent, $\left(H_0 -z \right)^{-1}$. For $n=3$ this kernel is given by

 \beq
 \frac{1}{4\pi} \frac{e^{i\sqrt{z}\, |x-y|}}{|x-y|}.
 \label{1.7}
 \ene
 Note that (\ref{1.7}) behaves nicely as $z\rightarrow 0$. In the case $n \geq
 4$ the kernel of the free resolvent has also a nice behaviour as
 $z \rightarrow 0$. This fact is the starting point of the analysis of Jensen
 and Kato in \cite{34}, \cite{30} and \cite{31}, who use perturbation theory to
 estimate the behaviour near zero of the resolvent of $H$. In the case $n=1$
 the kernel of $\left( H_0 -z \right)^{-1}$ is given by (see Theorem 9.5.2 in
 page 160 of \cite{13})
 \beq
 \frac{i}{2\sqrt{z} } e^{i\sqrt{z}|x-y|}.
 \label{1.8}
 \ene
 The kernel ({\ref{1.8}) is singular as $z\rightarrow 0$ and an approach as in
 \cite{10}, \cite{34}, \cite{30} and \cite{31} does not appears to be
 convenient. We take in Section 2 below  a different point of view. We
 base our analysis of the low--energy behaviour of the spectral family of $H$
 on the generalized Fourier maps that are constructed from the scattering
 solutions $\Psi_+ (x,k), x,k \in \ER$. The crucial issue here is that for
 $n=1$ the construction of the scattering solutions can be reduced to the
 solution of Volterra integral equations. More precisely, the scattering
 solution is given in terms of the Jost solutions, $f_j (x,k), j=1,2,$ as
 follows:
\beq
\Psi_+ (x,k)=\cases{\frac{T(k)}{\sqrt{2\pi}} f_1 (x,k), & $k \geq 0,$\cr\cr
\frac{T(-k)}{\sqrt{2\pi}} f_2 (x,-k),& $k\leq 0,$}
\label{1.9}
\ene
where $T(k)$ is the transmission coefficient. The $f_j$ are solutions to
Volterra integral equations that are obtained by iteration as uniformly
convergent series. See \cite{25}, \cite{26}, \cite{1} and \cite{28}. This fact
allows for a detailed analysis of the low--energy behaviour of the spectral
family of $H$ that coupled with a high--energy estimate allows us to prove in
Section 2 an estimate like (\ref{1.6}) in the case $n=1$.

Since in what
follows we only consider the case $n=1$ we denote below by $L^p ,1 \leq p \leq
\infty,$ the space $L^p \left(\ER^1 \right)$. For any $s\in \ER$ let us denote
by $L^1_s$ the space of all complex--valued measurable functions, $\phi $,
defined on $\ER$ such that
\beq
\|\phi\|_{L^1_s}:=\int_{\ER} |\phi (x)| (1+|x|)^s dx < \infty.
\label{1.10}
\ene
$L^1_s$ is a Banach space with the norm (\ref{1.10}).
Below we always assume that $V\in L^1_1$. It follows from the existence of the
Jost solutions and since the eigenvalues of $-\frac{d^2}{d x^2}+V(x)$ are
simple (see \cite{1}) that the differential expression
$\tau:= -\frac{d^2}{d x^2}+V(x)$ is in the limit point case at $\pm \infty$.
Then by the Weyl criterion (see \cite{12}) $\tau$ is essentially
self--adjoint on the domain
\beq
D(\tau ):=\left\{ \phi \in L^2_C : \phi \,\hbox{and}\, \acute{\phi}
\hbox{ are absolutely
continuous and $\tau \phi \in L^2$ }\right\},
\label{1.11}
\ene
where we denote by $\acute{\phi}(x)= \frac{d}{d x}\phi(x)$
and by $L^2_C$ the set of all $\phi \in L^2$ that have compact support.
We denote by $H$ the unique self--adjoint realization of $\tau $. It is known
that the absolutely continuous spectrum of $H$ is given by $\sigma_{ac}(H)=
[0, \infty ),$ that $H$ has no singular continuous spectrum, that $H$ has no
eigenvalues that are positive or equal to zero and that $H$ has a finite
number, $N$, of negative eigenvalues that are simple and that we denote by
$-\beta^2_N < \beta^2_{N-1} < \cdots < -\beta^2_1 < 0$. Let $F$ denotes the
Fourier
transform as a unitary operator on $L^2$
\beq
F\phi (k)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}
e^{-ikx}\phi (x) dx.
\label{1.12}
\ene
We will also use the notation $\hat{\phi }(k):= F\phi (k)$.
For any
$\alpha \in  \ER$ let us denote by $W_{\alpha}$ the Sobolev space consisting
of the completion of the Schwartz class in the norm
\beq
\left\|\phi \right\|_{\alpha}:=\left\|(1+k^2)^{\alpha /2}
\hat{\phi }(k)\right\|_{L^2}
.
\label{1.13}
\ene
We denote by $h$ the following quadratic form
\beq
h(\phi ,\psi):= (\acute{\phi}, \acute{\psi})_{L^2} + (V\phi , \psi )_{L^2},
\label{1.14}
\ene
with domain $D(h)= W_1 $. Since $V\in L^1_1 \subset L^1_0 \equiv L^1$ it
follows from Theorem 8.42 in page 147 of \cite{13} and from
the remarks above Theorem 9.14.1 in page 183 of \cite{13} that $h$ is
closed and
bounded from below and that the associated operator, $H_h$, is self--adjoint
with domain, $D(H_h )\subset W_1$. Since $D(\tau ) \subset W_1 $ it follows
that $H_h$ is a self--adjoint extension of $\tau$ and as $\tau$ is essentially
self--adjoint we have that $H=H_h$ and then
$
D(|H|)=W_1
$.
For $u, v$ any pair of solutions to the stationary Schr\"{o}dinger equation:
\beq
-\frac{d^2}{d x^2}u + V u=k^2 u,\,\, k\in \ER,
\label{1.16}
\ene
let $[u,v]$ denotes the Wronskian of $u$ and $v$:
\beq
[u,v]:= \acute{u} v -u \acute{v}.
\label{1.17}
\ene
A potential $V$ is said to be {\it generic } if the Jost solutions at zero
energy satisfy \linebreak$[f_1 (x,0) ,f_2 (x,0)]\neq 0$ and $V$ is said
to be {\it
exceptional }if $[f_1 (x,0), f_2 (x,0)]=0$. If the potential $V$ is exceptional
there is a bounded solution (a half--bound state ) to the equation
(\ref{1.16}) with $k=0$. See \cite{2} for these definitions and a discussion of
related issues. Let $P_c$ denotes the projector onto the continuous subspace
of $H$. Note that $P_c =I-P_p$, where $P_p$ is the projector onto the
finite dimensional subspace of $L^2$ generated by the eigenvectors
corresponding to the $N$ eigenvalues of $H$.

Our mail result is the following theorem that we prove in Section 2.
\begin{theorem}(The $L^1 - L^{\infty}$ estimate ).
Suppose that $V\in L^1_{\gamma}$ where in the {\rm generic case} $\gamma > 3/2$
and in the {\rm exceptional case} $\gamma > 5/2$. Then for some constant $C$
\beq
\left\|e^{-it H}P_c \right\|_{{\cal B}\left(L^1 ,L^{\infty}\right)}
\leq \frac{C}{\sqrt{t}}, \, t > 0.
\label{1.18}
\ene
\end{theorem}
\begin{corollary} (The $ L^p -L^{\acute{p}}$ estimate).
 Suppose that the conditions of Theorem 1.1 are satisfied. Then for
$1 \leq p \leq 2$ and $\frac{1}{p}+\frac{1}{\acute{p}}=1$
\beq
\left\|e^{-itH} P_c \right\|_{{\cal B}\left( L^p ,L^{\acute{p}}\right)}
\leq \frac{C}{t^{(\frac{1}{p}-\frac{1}{2})}}, \, t > 0.
\label{1.19}
\ene
\end{corollary}
\begin{corollary} (The espace--time estimate). Suppose that the conditions
of Theorem 1.1 are satisfied. Then

\noindent(a)
\beq
e^{-itH}P_c \in {\cal B}\left(L^2 ,L^6 (\ER \times \ER )\right).
\label{1.20}
\ene

\noindent (b) If moreover, $H$ has no negative eigenvalues and
\beq
i \frac{\partial}{\partial t}u(t,x) = H u(t,x) +g(t,x), \, u(0,x)=f(x),
\label{1.21}
\ene
then
\beq
\left\|u(t,x)\right\|_{L^6 (\ER \times \ER )} \leq C \left[ \|f\|_{L^2}
+\|g\|_{L^{6/5} (\ER \times \ER )}\right].
\label{1.22}
\ene
\end{corollary}
In the case $V=0$ and $n \geq 1$ Theorem 1.1 and Corollaries 1.2 and 1.3
were proven by Strichartz in \cite{11}. They were proven in \cite{10}
for $n \geq 3$ and $V$ satisfying appropriate conditions on regularity and
decay (see \cite{10}, equation (1.6)). In \cite{10} it was assumed moreover,
that zero is neither a bound state nor a half--bound state. Note that we do not
have to assume that zero is not a half--bound state for Theorem 1.1 and
Corollaries 1.2 and 1.3 to hold. In our case it is enough to require that
$V$ has a
slightly  faster decay at infinity when there is a half--bound state at zero.

Theorem 1.1 and Corollaries 1.2 and 1.3 open the way to the study of the
 scattering theory for the nonlinear Schr\"{o}dinger equation
with a potential (NLSP):
\beq
i\frac{\partial}{\partial t}u = H u + f(|u|) \frac{u}{|u|}.
\label{1.23}
\ene
As a first application we study in this paper the low--energy scattering for
the NLSP and we prove that the low--energy limit of the scattering
operator uniquely determines the
potential and the nonlinearity. For this purpose we proceed as in \cite{23}
were the case $n \geq 3 $ was considered. Let us assume that $H$ has no
negative eigenvalues. Then $H >0$ and since $D(\sqrt{H})=W_1$ the operators
$\sqrt{H+1}\, (-\Delta +1)^{-1/2}$ and $\sqrt{-\Delta +1}\, (H+1)^{-1/2}$ are
bounded in $L^2$. It follows that the norm associated to the following
scalar product
\beq
 (\phi ,\psi )_{X}:=\left( \sqrt{H+1} \,\phi , \sqrt{H+1} \,\psi \right)_{L^2},
 \label{1.24}
 \ene
 is equivalent to the norm of $W_1$. We denote by $X$ the Sobolev space $W_1$
 endowed with the scalar product (\ref{1.24}). The space $X$ is a Hilbert
 space. Clearly, $e^{-itH}$ is a strongly continuous group of unitary operators
 on $X$. For any $\delta > 0$ we denote:

 \beq
 X(\delta ):=\left\{ \phi \in X : \|\phi \|_{X} < \delta \right\}.
 \label{1.25}
 \ene
 Let us denote  $X_3 := L^{p+1}$ and $r=(p-1)/(1-d)$ with
 $d:=\frac{1}{2} (p-1)/
 (p+1)$ and $5 \leq p < \infty$. In what follows for functions $u(t,x)$ defined
 on $\ER \times \ER$ we write $u(t)$ for $u(t,\cdot )$.

 \begin{theorem} (Low--energy scattering). Suppose that $V \in L^1_{\gamma}$
 where in the {\rm generic case} $ \gamma > 3/2$ and in the
 {\rm exceptional case} $\gamma
 > 5/2$ and that $H$ has no negative eigenvalues. Assume moreover, that the
  function $f$ in (\ref{1.23}) is defined on $\ER$, that it is real--valued
  and $C^1$. Furthermore, $f(0)=0$ and
  \beq
  \left| \frac{d}{d \mu}f(\mu )\right| \leq C |\mu |^{p-1},
  \label{1.26}
  \ene
  for some  $ 5 \leq p < \infty $. Then there is a $\delta > 0$ such that for
  every $\phi_- \in X(\delta )$ there is a unique solution to the NLSP,
  $u(t,x)$, such that $u\in C(\ER , X)\cap L^r (\ER , X_3 )$ and
  \beq
  \lim_{t \rightarrow - \infty} \left\| u(t) -e^{-itH} \phi_- \right\|_X =0.
  \label{1.27}
  \ene
  Moreover, there exists a unique $\phi_+ \in X$ such that
  \beq
  \lim_{t \rightarrow \infty}\left\|u(t)- e^{-itH}\phi_+ \right\|_X=0.
  \label{1.28}
  \ene
  For all $t\in \ER$
  \beq
  \frac{1}{2} \|u(t)\|^2_X +\int_{\ER} F(|u(t)|) dx =\frac{1}{2}
  \|\phi_- \|^2_X
  =\frac{1}{2}\|\phi_+\|^2_X,
  \label{1.29}
  \ene
  where $F$ is the primitive of $f$ such that $F(0)=0$. In addition the
  nonlinear scattering operator $S_V : \phi_- \rightarrow \phi_+$ is a
  homeomorphism from $X(\delta )$ onto $X(\delta )$.
  \end{theorem}
 Theorem 1.4 is proven in Section 3 using Theorem 1.1, Corollaries 1.2 and 1.3
 and the abstract low--energy scattering theory of Strauss \cite{15},
 \cite{16}.
The scattering operator $S_V$ compares solutions of the NLSP (\ref{1.23}) with
 solutions to the LSP (\ref{1.5}). To reconstruct $V$ we consider below the
 scattering operator, $S$, that compares solutions to the NLSP with solutions
 to the LS (\ref{1.1}). For this purpose let us  consider the wave operators
 \beq
 W_{\pm}:= \hbox{s}-\lim_{t \rightarrow \pm \infty} e^{itH}\, e^{-itH_0}.
 \label{1.30}
 \ene
 The $W_{\pm}$ are unitary on $L^2$ (note that $H$ has no eigenvalues). The
 existence of the strong limits in (\ref{1.30}) is well known (see Theorem
 9.14.1 in page 183 of\cite{13}). Moreover, by the intertwining relations,
 $\sqrt{H} W_{\pm}= W_{\pm}\sqrt{H_0}$ and as $D(\sqrt{H})=W_1$, we have that
 $W_{\pm}$ and $W^{\ast}_{\pm}$ belong to ${\cal B}(W_1)$ and for
 $0 < \delta_1
 < \delta$
  they send $X(\delta_1 )$ into $X(\delta )$ if $\delta_1$ is small
 enough. Let us define:
 \beq
 S:= W^{\ast}_+ S_V W_- .
 \label{1.31}
 \ene
 Take $\delta_1$ so small that $W_- X(\delta_1 )\subset X(\delta )$ with
 $\delta$ as in Theorem 1.4 and then $\delta_2$ so large that $W^{\ast}_+
 X(\delta ) \subset X( \delta_2 )$. Then $S$ sends $X(\delta_1 )$ into
 $X(\delta_2 )$. Moreover, for any $\psi_- \in X(\delta_1 )$ let us take in
 Theorem 1.4
 $\phi_- \equiv W_- \psi_-$  and let $u(t,x)$ and $\phi_+$ be as in
 Theorem 1.4.
 Let us denote $\psi_+ :=S \psi_- = W^{\ast}_+ \phi_+$. Then by Theorem 1.4 and
 (\ref{1.30})
 \beq
 \lim_{t \rightarrow \pm \infty}\left\|u(t,x)-e^{-itH_0}\psi_{\pm}
 \right\|_{L^2}=0.
 \label{1.32}
 \ene
 That is to say, $S$ sends the initial data at $t=0,\psi_-  ,$ of the
 incoming solution to LS to the initial data at $t=0, \psi_+ ,$ of the
 outgoing solution to LS. Let us denote by $S_L$ the linear scattering operator
 corresponding to the LS and the LSP:
 \beq
 S_L := W^{\ast}_+ \,W_-.
 \label{1.33}
 \ene
 In Theorem 1.5 below, $S_L$ is reconstructed from the low--energy limit
 of $S$.
 \begin{theorem}
 Suppose that the assumptions of Theorem 1.4 are satisfied. Then for every
 $\phi ,\psi \in X$
 \beq
 \lim_{\epsilon \downarrow 0}\left(S \epsilon \phi , \psi \right)_{L^2}=
 \left( S_L \phi ,\psi \right)_{L^2}.
 \label{1.34}
 \ene
 \end{theorem}
 Since, as is well known, from $S_L$ we can uniquely reconstruct $V$ we
 obtain the following Corollary.
 \begin{corollary}
 Suppose that the assumptions of Theorem 1.5 are satisfied. Then the scattering
 operator, S, uniquely determines the potential $V$.
 \end{corollary}
 In the case where $f(u)= \lambda |u|^p$, we can also uniquely reconstruct the
 coupling  constant $\lambda$.
 \begin{corollary}
Suppose that the assumptions of Theorem 1.4 are satisfied and that moreover,
$f(u)=\lambda |u|^p $, for some constant $\lambda$. Then the scattering
operator, S, uniquely determines the potential $V$ and the coupling constant
$\lambda$. Furthermore, for all $0\neq \phi \in X \cap L^{1+\frac{1}{p}}$:
\beq
\lambda =\lim_{\epsilon \downarrow 0} \frac{1}{\epsilon^p} \frac{\left(
(S_V -I)\epsilon \phi ,\phi \right)_{L^2}}{\int^{\infty}_{-\infty}
\left\|e^{-itH} \phi \right \|^{1+p}_{L^{1+p}}}.
\label{1.35}
\ene
\end{corollary}
Remark that by Sobolev's imbedding theorem \cite{14}, $X \subset L^{1+p}$. Then
by (\ref{1.19})
\beq
0 < \int^{\infty}_{-\infty} \left\| e^{-itH} \phi \right\|^{1+p}_{L^{1+p}} dt
< \infty.
\label{1.36}
\ene
 Theorem 1.5 and Corollaries 1.6 and 1.7 are proven
 as in \cite{23} (see Section 3).

 We use below the letter $C$ to denote any
 positive constant whose particular value is not relevant.

\section{The $L^p -L^{\acute{p}}$ Estimate}\sss
We assume that $V\in L^1_1$. For any complex number, $k$,
 we denote by $\Re k$ and $\Im k$, respectively, the real and the imaginary
 parts of $k$. The Jost solutions $f_j (x,k), j=1,2,$
are solutions
to the stationary Schr\"{o}dinger equation
\beq
-\frac{d^2}{d x^2} f_j (x,k) +V(x) f_j (x,k)= k^2 f_j (x,k)
\label{2.1}
\ene
 were $\Im k \geq 0$. To construct the Jost solution we define $ m_1 (x,k):=
 e^{-ikx} f_1 (x,k)$ and $m_2 (x,k) :=e^{ikx} f_2 (x,k)$. They are,
 respectively, solutions of the following equations:

 \beq
 \frac{d^2}{d x^2} m_1 (x,k) +2ik \frac{d}{d x} m_1 (x,k)= V(x) m_1 (x,k),
 \label{2.2}
 \ene
 \beq
 \frac{d^2}{d x^2} m_2 (x,k) -2ik \frac{d}{d x} m_2 (x,k)= V(x) m_2 (x,k).
 \label{2.3}
 \ene
The $m_j (x,k), j=1,2,$ are the unique solutions of the Volterra integral
equations
\beq
m_1(x,k)= 1 + \int_x^{\infty} D_k (y-x) V(y) m_1 (y,k) dy,
\label{2.4}
 \ene
 \beq
 m_2 (x,k)= 1+\int^x_{-\infty} D_k (x-y) V(y) m_2 (y,k) dy,
 \label{2.5}
 \ene
 where
 \beq
 D_k (x):=\int_0^x e^{2iky}dy =
 \cases{ \frac{1}{2ik} (e^{2ikx}-1),& $k\neq 0,$ \cr\cr
 x,& $k=0.$}
 \label{2.6}
 \ene
 Note that $f_1 (x,k) \sim e^{ikx}$ as $ x \rightarrow \infty$ and that
 $f_2 (x,k)\sim e^{-ikx}$ as $ x\rightarrow -\infty$. A detailed study of the
 properties of the $m_j (x,k),j=1,2,$ was  carried over in \cite{1}. Here we
  state a number of results from \cite{1} that we need. In what follows we
  denote by $C$ any positive constant whose specific value is not relevant to
  us and by $\dot{g}(x,k):= \frac{\partial}{\partial k}g(x,k)$.
  For each fixed $x \in \ER$ the $m_j (x,k)$ are analytic in $k$ for $\Im k > 0$
  and continuous in $\Im k \geq 0$ and
  \beq
  |m_1 (x,k) -1 | \leq C\,\, \frac{1+\hbox{max} (-x ,0)}{1+|k|},
  \label{2.7}
  \ene
  \beq
  |m_2 (x,k) -1| \leq C\,\, \frac{1+\hbox{max}(x,0)}{1+|k|}.
  \label{2.8}
  \ene
  Moreover, $\dot{m}_j (x,k),j=1,2,$
  exits for $\Im k\geq 0, k\neq 0$, $ k\dot{m}_j (x,k)$ is continuous in $k$ for
   each
  fixed $x\in \ER $ and for each fixed $x_0\in \ER$ there is a constant
  $C_{x_0}$  such that
  \beq
  |\dot{m}_1 (x,k)|\leq C_{x_0} \,\,\frac{1}{|k|}, x \geq x_0,
  \label{2.10}
  \ene
  \beq
  |\dot{m}_2 (x,k)| \leq C_{x_0} \,\,\frac{1}{|k|}, x \leq x_0 .
  \label{2.11}
  \ene
  In the Lemma below we slighly improve the estimates (\ref{2.10}) and
  (\ref{2.11}) under the assumption that $V \in L^1_{\gamma}$ for
  $1 < \gamma \leq 2$.
  \begin{lemma}
  Suppose that $V \in L^1_{\gamma}$ for some $1 \leq \gamma \leq 2$. Then for
  each $x_0 \in \ER$ there is a constant $C_{x_0}$ such that
  \beq
|\dot{m}_1 (x,k)| \leq C_{x_0}\, \frac{|k|^{\gamma}}{|k|^2 (1+|k|)^{\gamma-1}}
, x \geq x_0,
\label{2.12}
\ene
\beq
|\dot{m}_2 (x,k)| \leq C_{x_0} \,\frac{|k|^{\gamma}}{|k|^2 (1+|k|)^{\gamma -1}}
, x \leq x_0 .
\label{2.13}
\ene
\end{lemma}
\noindent{\it Proof :}\, We give the proof in the case of $\dot{m}_1 (x,k)$.
The case of $\dot{m}_2 (x,k)$ follows similarly. It follows from (\ref{2.6})
that for $k\neq 0$
\beq
\left|\dot{D}_k (x)\right|=\left|\frac{1}{k}\int_0^{x} y \left(
\frac{\partial}{\partial y}e^{2iky}\right) \, dy \right|\leq 2 \frac{|x|}{|k|},
\label{2.14}
\ene
and that
\beq
\left|\dot{D}_k (x)\right| \leq |x|^2.
\label{2.15}
\ene
By (\ref{2.14}) and (\ref{2.15}) for any $1\leq \gamma \leq 2$
\beq
|\dot{D}_k (x)|\leq \frac{2^{2-\gamma} |x|^{\gamma}}{|k|^{2-\gamma}}.
\label{2.15b}
\ene
Since (\ref{2.4}) is a Volterra integral equation, $m_1 (x,k)$ is obtained by
iteration \cite{1}:
\beq
m_1 (x,k)=\lim_{n \rightarrow \infty} m_{1,n}(x,k),
\label{2.16}
\ene
where $m_{1,0}(x,k)=1$ and for $n=1,2,\cdots$
\beq
m_{1,n}(x,k)=1+\sum_{l=1}^{n}g_l (x,k),
\label{2.17}
\ene
where
\beq
g_l (x,k)=\int_{x\leq x_1 \leq x_2 \leq \cdots \leq x_l}D_k (x_1 -x)
D_k (x_2 -x_1 ) \cdots D_k (x_l -x_{l-1}) V(x_1 )\cdots V(x_l )dx_1
\cdots dx_l.
\label{2.18}
\ene
Moreover, the $m_{1,n}$ satisfy the following equation for $n=0,1,\cdots$
\beq
m_{1,n+1}(x,k)=1+\int_x^{\infty}D_k (y-x) V(y) m_n (y,k) dy.
\label{2.19}
\ene
Then,
\beq
\dot{m}_{1,n+1}(x,k)=\int_x^{\infty}\dot{D}_k (y-x) V(y) m_n (y,k) dy
+\int_x^{\infty}D_k (y-x)V(y) \dot{m}_n (y,k) dy.
\label{2.20}
\ene
Furthermore, since by (\ref{2.6})
\beq
|D_k (x)|\leq |x|,
\label{2.21}
\ene
 it follows from (\ref{2.18}) that
 \beq
 |g_l (x,k)| \leq \frac{1}{l!} \left(\int_x^{\infty} (y-x) V(y) dy \right)^l,
 \label{2.22}
 \ene
 and then by (\ref{2.17}) for $x \geq x_0$

 $$
 |m_{1,n}(x,k)|\leq 1+ \sum_{l=1}^n \frac{1}{l!} \left(\int_x^{\infty} (y-x)
 |V(y)| dy\right)^l
$$
 \beq
 \leq  e^{(\int_x^{\infty} (|x_0 |+|y|) |V(y)| dy)}, \,
 x \geq x_0.
 \label{2.23}
 \ene
 We can now estimate the first integral in the right--hand side of
 (\ref{2.20}) as follows

$$
\left| \int_x^{\infty}\dot{D}_k (y-x) V(y) m_n (y,k) dy \right|\leq
$$
\beq
\frac{2^{2-\gamma}}{|k|^{2-\gamma}}\int_x^{\infty} |y-x|^{\gamma}\, |V(y)|
e^{ \int_x^{\infty}(|x_0 |+|y|) |V(y)| dy}\leq C \,\frac{1}{|k|^{2-\gamma}},
x \geq x_0,
\label{2.24}
\ene
 where we used (\ref{2.15b}). Then using again (\ref{2.20}) and (\ref{2.21})
 we obtain that
 \beq
 |\dot{m}_{1,n+1}(x,k)|\leq \, \frac{C}{|k|^{2-\gamma}} + \int_x^{\infty}
 |y-x| |V(y)| |\dot{m}_n (y,k)| dy.
 \label{2.25}
 \ene
 Since $m_0 (y,k) \equiv 1$ it follows from (\ref{2.25}) with $n=0$ that
 \beq
 |\dot{m}_{1,1}(x,k)|\leq \frac{C}{|k|^{2-\gamma}}.
 \label{2.26}
 \ene
 Then by (\ref{2.25}) we have that
 \beq
 |\dot{m}_{1,2}(x,k)|\leq\, \frac{C}{|k|^{2-\gamma}} (1+ q(x)), x \geq x_0,
 \label{2.27}
 \ene
 where
 \beq
 q(x):=  \int_{x}^{\infty} (|x|_0 +|y|) |V(y)| dy,
 \label{2.28}
 \ene
 and then, iterating (\ref{2.25}) $n-1$ more times we prove that
 \beq
 |\dot{m}_{1,n+1}(x,k)|\leq \frac{C}{|k|^{2-\gamma}}\sum_{l=0}^n
 \frac{(q(x))^l}{l!}.
 \label{2.29}
 \ene
 Taking the limit as $n \rightarrow \infty$ in (\ref{2.29}) we prove that
 \beq
 |\dot{m}_1 (x,k)| \leq \frac{C}{|k|^{2-\gamma}} e^{q(x)}, x \geq x_0.
 \label{2.30}
 \ene
 Since $V\in L^1_{\gamma} \subset L^1_1 $, we can take $\gamma =1$ in
 (\ref{2.30}) and then

 \beq
 |\dot{m}_1 (x,k)|\leq \frac{C}{|k|} \, e^{q(x)},\, x \geq x_0.
 \label{2.31}
 \ene
 Equation (\ref{2.12}) follows from  (\ref{2.30}) and (\ref{2.31}).

 \begin{corollary}
 Suppose that $V \in L^1_{\gamma}$, for some $1\leq \gamma \leq 2$. Then for each
 $x_0 \in \ER$ there is a constant $C_{x_0}$ such that for all $\Im k \geq 0$
 \beq
 \left|\dot{\acute{m}}_1 (x,k)\right|\leq C_{x_0} \left[ 1+\frac{|k|^{\gamma}}
 {|k|^2 (1+|k|)^{\gamma -1}}\right], x \geq x_0 ,
 \label{2.32}
 \ene
 \beq
 \left|\dot{\acute{m}}_2 (x,k)\right| \leq C_{x_0} \left[1+
 \frac{|k|^{\gamma}}{|k|^2 (1+|k|)^{\gamma -1}}\right], \, x \leq x_0.
 \label{2.33}
 \ene
 \end{corollary}

 \noindent {\it Proof :}\, We prove (\ref{2.32}). The proof of (\ref{2.33})
 is similar. By (\ref{2.4}) and (\ref{2.6})
 \beq
 \acute{m}_1 (x,k)= -\int_{x}^{\infty} e^{2ik(y-x)} V(y) m_1 (y,k)  dy,
 \label{2.34}
 \ene
 and then
 \beq
 \dot{\acute{m}}_1 (x,k) =-\int_x^{\infty} \left[2i e^{2ik(y-x)} (y-x) V(y)
 m_1 (y,k)  + e^{2ik(y-x)} V(y) \dot{m}_1 (y,k)\right] dy.
 \label{2.35}
 \ene
 It follows from (\ref{2.7}), (\ref{2.12}) and (\ref{2.35}) that
 \beq
 \left|\dot{\acute{m}}_1 (x,k)\right| \leq C_{x_0} \left[ 1+
 \frac{|k|^{\gamma}}{|k|^2 (1+|k|^{\gamma -1})}\right], x \geq x_0.
 \label{2.36}
 \ene

 \begin{lemma}
 Suppose that $V\in L^1_{\gamma}$, for some $2\leq \gamma \leq 3$. Then for
 every
 $x_0 \in \ER $ there is a constant $C_{x_0}$ such that
 \beq
 \left|\dot{m}_1 (x,k)-\dot{m}_1 (x,0)\right|\leq C_{x_0} |k|^{\gamma -2},
 x \geq x_0,
 \label{2.37}
 \ene
\beq
\left|\dot{m}_2 (x,k) -\dot{m}_2 (x,0)\right|\leq C_{x_0} |k|^{\gamma -2},
x \leq x_0.
\label{2.38}
\ene
\end{lemma}
\noindent{\it Proof:}\, It follows from the definition of $D_k (x)$ in
(\ref{2.6}) that
\beq
\left|\dot{D}_k (x) -\dot{D}_0 (x)\right|\leq \frac{4}{3} |k| |x|^3,
\label{2.39}
\ene
and that
\beq
\left|\dot{D}_k (x) -\dot{D}_0 (x)\right| \leq 2 |x|^2.
\label{2.40}
\ene
Then for any $2\leq \gamma \leq 3$ there is a constant, $C_{\gamma}$, such that
\beq
\left|\dot{D}_k (x)-\dot{D}_0 (x)\right|\leq  C_{\gamma} |k|^{\gamma-2}
|x|^{\gamma}.
\label{2.41}
\ene
We obtain from (\ref{2.20}) that
$$
\dot{m}_{1,n+1}(x,k)- \dot{m}_{1,n+1}(x,0)=\int_x^{\infty}
\left[\dot{D}_k (y-x)-\dot{D}_0 (y-x)\right] V(y) m_n (y,k) dy +
$$
$$
\int_x^{\infty}\left\{\dot{D}_0 (y-x) V(y) \left[m_n (y,k)-m_n (y,0)\right]+
\left[D_k (y-x)-D_0 (y-x)\right] V(y) \dot{m}_n (y,k)\right\}dy +
$$
\beq
\int_x^{\infty}D_0 (y-x) V(y) \left[\dot{m}_n (y,k)-\dot{m}_n (y,0)\right]
 dy.
\label{2.42}
\ene
Moreover, by (\ref{2.23}) and (\ref{2.41})
\beq
\left|\int_0^{\infty}\left[ \dot{D}_k (y-x) -\dot{D}_0 (y-x)\right] V(y)
m_n (y,k) dy \right| \leq C_{x_0} |k|^{\gamma -2}, x \geq x_0.
\label{2.43}
\ene
By (\ref{2.29}) with $\gamma =2$
\beq
\left|m_{1,n}(x,k)-m_{1,n}(x,0)\right|=\left|\int_0^k \dot{m}_{1,n}(x,s)ds
\right| \leq C_{x_0}\, |k|, x \geq x_0,
\label{2.44}
\ene
and then by (\ref{2.15})
\beq
\left|\int_x^{\infty} \dot{D}_0 (y-x) V(y) \left[m_n (y,k)-m_n (y,0)\right]dy
\right| \leq C_{x_0}\, |k|,\, x \geq x_0.
\label{2.45}
\ene
Moreover, by (\ref{2.6})
\beq
|D_k (y)- D_0 (y)| \leq |k||y|^2,
\label{2.46}
\ene
and it follows from (\ref{2.29}) with $\gamma =2$ that
\beq
\left|\int_x^{\infty}\left[D_k (y-x)-D_0 (y-x)\right] V(y) \dot{m}_n (y,k)dk
\right|\leq C_{x_0} |k|, x \geq x_0.
\label{2.47}
\ene
Then we obtain from (\ref{2.21}), (\ref{2.42}), (\ref{2.43}), (\ref{2.45}) and
(\ref{2.47}) that for $|k| \leq 1$:
$$
\left|\dot{m}_{n+1}(x,k)-\dot{m}_{n+1}(x,0)\right|\leq C_{x_0}|k|^{\gamma-2}+
$$
\beq
\int_x^{\infty} (y-x) |V(y)| \left|\dot{m}_n (y,k)-\dot{m}_n (y,0)\right|dy,
x \geq x_0.
\label{2.48}
\ene
But since $m_0 (x,k) \equiv 1$ it follows from (\ref{2.48}) with $n=0$ that
\beq
\left|\dot{m}_{1,1}(x,k) -\dot{m}_{1,1}(x,0)\right|\leq C_{x_0} |k|^{\gamma-2}.
\label{2.49}
\ene
Iterating (\ref{2.48}) $n$ more times we prove
that
\beq
\left|\dot{m}_{1,n+1}(x,k)-\dot{m}_{1,n+1}(x,0)\right|\leq C_{x_0}
 |k|^{\gamma-2}\left(1+\sum_{l=1}^{n} \frac{(q(x))^l}{l!}\right),
\label{2.50}
\ene
with $q(x)$ as in (\ref{2.28}) and taking the limit as
$n\rightarrow \infty$ we have that
\beq
\left|\dot{m}_1 (x,k)- \dot{m}_1 (x,0)\right| \leq C_{x_0} |k|^{\gamma -2}
e^{q(x)}, x \geq x_0 ,
\label{2.51}
\ene
and this proves (\ref{2.37}). Equation (\ref{2.38}) follows similarly.

\begin{corollary}
 Suppose that $V \in L^1_{\gamma}$ for some $2 \leq \gamma \leq 3$. Then for
 every $x_0 \in \ER$ there is a constant $C_{x_0}$ such that
 \beq
 \left| \dot{\acute{m}}_1 (x,k) -\dot{\acute{m}}_1 (x,0)\right| \leq
 C_{x_0}\, |k|^{\gamma -2}, x \geq x_0 ,
 \label{2.52}
 \ene
 \beq
 \left| \dot{\acute{m}}_2 (x,k)-\dot{\acute{m}}_2 (x,0)\right|\leq C_{x_0}
 \,|k|^{\gamma -2}, x \leq x_0 .
 \label{2.53}
 \ene
 \end{corollary}

 \noindent {\it Proof:} \, We give the proof of (\ref{2.52}). Equation
 (\ref{2.53}) follows in a similar way. By (\ref{2.35})
 $$
 \dot{\acute{m}}_1 (x,k) - \dot{\acute{m}}_1 (x,0)= -\int_x^{\infty}
 dy\, \left[ e^{2ik(y-x)}-1\right] V(y) \left\{2i(y-x) m_1 (y,k)+
 \dot{m}_1 (y,k) \right\}-
  $$
  \beq
  \int_x^{\infty}dy\, V(y) \left[ 2i (y-x) \left(m_1 (y,k) -m_1 (y,0)\right)+
  \dot{m}_1 (y,k) -\dot{m}_1 (y,0)\right] dy.
  \label{2.54}
  \ene
  Then by (\ref{2.7}), (\ref{2.12})with $\gamma =2$ and (\ref{2.37})
\beq
\left|\dot{\acute{m}}_1 (x,k)-\dot{\acute{m}}_1 (x,0)\right| \leq C_{x_0}
\, |k|^{\gamma -2}, x \geq x_0 .
\label{2.55}
\ene

\bull

The Jost solutions, $f_j (x,k), j=1,2,$ are independent solutions to
(\ref{2.1}) for $k\neq 0$ and there are unique functions $T(k)$ and
$R_j (k), j=1,2,$ such that \cite{1}

\beq
f_2 (x,k)= \frac{R_1 (k)}{T(k)} f_1 (x,k) +\frac{1}{T(k)} f_1 (x,-k),
\label{2.56}
\ene
\beq
f_1 (x,k)= \frac{R_2 (k)}{T(k)} f_2 (x,k)+ \frac{1}{T(k)} f_2 (x,-k),
\label{2.57}
\ene
for $k \in \ER \setminus 0$. The function $T(k) f_1 (x,k)$ describes the
scattering from left to right of a plane wave $e^{ikx}$ and $T(k) f_2 (x,k)$
describes the scattering from right to left of a plane wave $e^{-ikx}$. The
function $T(k)$ is the transmission coefficient , $R_2 (k)$ is the reflection
coefficient from left to right and $R_1 (k)$ is the reflection coefficient
from right to left. The relations (\ref{2.56}) and (\ref{2.57}) are expressed
as follows in terms of the $m_j (x,k), j=1,2,$
\beq
T(k) m_2 (x,k)=R_1 (k) e^{2ikx} m_1 (x,k) +m_1 (x,-k),
\label{2.58}
\ene
\beq
T(k) m_1 (x,k)= R_2 (k) e^{-2ik} m_2 (x,k) +m_2 (x,-k).
\label{2.59}
\ene
Moreover, $T(k)$ is meromorphic for $\Im k > 0$ with a finite number of
simple poles,\linebreak
$i\beta_N , i\beta_{N-1},\cdots  ,i\beta_1,\, \beta_j > 0,\,j=1,2,
\cdots ,N,$ on the imaginary axis. The numbers, $-\beta^2_N ,\linebreak
 -\beta^2_{N-1},
\cdots ,-\beta^2_1 ,$ are the simple eigenvalues of $H$. Furthermore, $T(k)$
is continuous in $\Im k \geq 0, k\neq i\beta_1 ,i\beta_2, \cdots i\beta_N $
and $T(k)\neq 0$ for $k \neq 0$. the $R_j (k),j=1,2,$ are continuous for $k
\in \ER$. Moreover, the following formulas hold \cite{1}

\beq
\frac{1}{T(k)}= \frac{1}{2ik} [f_1 (x,k), f_2 (x,k)]=1-\frac{1}{2ik}
\int_{-\infty}^{\infty} V(y)\, m_j (y,k)\, dk,\, j=1,2.
\label{2.60}
\ene
 \beq
 \frac{R_1 (k)}{T(k)}=\frac{1}{2ik} [f_2 (x,k), f_1 (x, -k)]=\frac{1}{2ik}
 \int_{-\infty}^{\infty} e^{-2iky}\,V(y)\, m_2 (y,k)\,dy,
 \label{2.61}
 \ene
 \beq
 \frac{R_2 (k)}{T(k)}=\frac{1}{2ik} [f_2 (x,-k) ,f_1 (x,k)]=\frac{1}{2ik}
 \int_{-\infty}^{\infty} e^{2iky} V(y)\, m_1 (y,k)\, dy.
 \label{2.62}
 \ene
 Furthermore,
 \beq
 T(k)=1+O\left(\frac{1}{|k|}\right),\, |k|\rightarrow \infty , \Im k \geq 0,
 \label{2.63}
 \ene
 \beq
 R_j (k)= O\left(\frac{1}{|k|}\right), |k| \rightarrow \infty , k \in \ER,
 \label{2.64}
 \ene
 and
 \beq
 |T(k)|^2 +|R_j (k)|^2 =1, \, j=1,2, \, k \in \ER .
 \label{2.65}
 \ene
 The behaviour as $ k \rightarrow 0$ is as follows:

 \noindent (a) In the {\it generic case}
 \beq
 T(k)= \alpha k +o(k),\, \alpha \neq 0 ,k \rightarrow 0 ,\Im k \geq 0,
 \label{2.66}
 \ene
 and $R_1 (0)=R_2 (0)= -1$.

 \noindent (b) In the {\it exceptional case}
 \beq
 T(k)= \frac{2a}{1+a^2}+o(1),\, k \rightarrow 0 , \Im k \geq 0,
 \label{2.67}
 \ene
 \beq
 R_1 (k)= \frac{1-a^2}{1+a^2} +o(1), \, k \rightarrow 0 ,\, k\in \ER ,
 \label{2.68}
 \ene
 \beq
 R_2 (k)=\frac{a^2 -1}{1+a^2}+o(1),\, k \rightarrow 0,\, k \in \ER ,
 \label{2.69}
 \ene
 where $a= \lim_{x \rightarrow -\infty}\, f_1 (x,0)\neq 0.$
 For the results above about $T(k)$ and $R_j (k), \,j=1,2,$ see \cite{1},
 \cite{2} and \cite{3}. In particular for the continuity of $T(k)$ and of
 $R_j (k)$  as $k \rightarrow 0$ in the exceptional case for $V \in L^1_1$
 see \cite{3}.

\begin{theorem}Assume that $V\in L^1_{\gamma}$.

\noindent (a)If $V$ is {\rm generic} and $1 \leq \gamma \leq 2$, then
\beq
|\dot{T}(k)|\leq C (1+|k|)^{-1} ,\, \Im k \geq 0,
\label{2.70}
\ene

\beq
R_j (k_1 )-R_j (k_2 )=\cases{ o\left(|k_1 -k_2 |^{\gamma -1}\right),&
$1\leq \gamma <2$, \cr\cr
O(|k_1 -k_2 |), & $\gamma =2$, }
\label{2.72}
\ene
as $ k_1 -k_2 \rightarrow 0$.

\noindent (b) If $V$ is {\rm exceptional} and $ 2 \leq \gamma \leq 3$, then
\beq
|\dot{T}(k)|\leq C \frac{|k|^{\gamma -3}}{(1+|k|)^{\gamma -2}},
\label{2.73}
\ene
\beq
T(k)-T(0)=O(|k|), k \rightarrow 0,
\label{2.74}
\ene

\beq
R_j (k)-R_j (0)=O(|k|),\, k \rightarrow 0,\, j=1,2.
\label{2.76}
\ene
Moreover, if $\gamma > 2$
\beq
R_j (k_1 )-R_j (k_2 )= O\left(|k_1 -k_2 |^{\gamma -2}\right), \,
k_1 -k_2 \rightarrow 0.
\label{2.77}
\ene
\end{theorem}

\noindent{\it Proof:}\, It follows from (\ref{2.7}) and (\ref{2.34}) that
\beq
\left|\acute{m}_1 (x,k)\right| \leq C,\, x\in \ER ,\, \Im k \geq 0.
\label{2.78}
\ene
We similarly prove that
\beq
\left|\acute{m}_2 (x,k)\right| \leq C, \, x\in \ER , \, \Im k \geq 0.
\label{2.79}
\ene
Then (\ref{2.70}) follows from (\ref{2.7}), (\ref{2.8}), (\ref{2.12}),
(\ref{2.13}), (\ref{2.32}), (\ref{2.33}), the first equality in
(\ref{2.60}), (\ref{2.63}),
(\ref{2.66}), (\ref{2.78}) and (\ref{2.79}).

If follows from (\ref{2.19}) that
$$
m_{1,n+1}(x,k_1 )-m_{1,n+1}(x, k_2 )=f_n (x,k_1 ,k_2 )+
\int_x^{\infty}D_{k_2}(y-x) V(y)
$$
\beq
\left[m_{1,n}(y,k_1 )-m_{1,n}(y,k_2 ) \right] dy,
\label{2.80}
\ene
where
\beq
f_n (x,k_1 ,k_2 ):= \int_x^{\infty} \left[D_{k_1}(y-x)-D_{k_2}(y-x)\right]
V(y) m_{1,n}(y,k_1 ) \, dy.
\label{2.81}
\ene
Moreover, by (\ref{2.6})
\beq
\left|D_{k_1}(x)-D_{k_2}(x)\right|\leq 2
\frac{|k_1 -k_2 ||x|}{1+ |k_1 -k_2 ||x|} |x|.
\label{2.82}
\ene
Then by (\ref{2.23}) for $x \geq 0$
\beq
\left|f_n (x,k_1 ,k_2 )\right| \leq f_{\gamma} (k_1 -k_2 ),
\label{2.83}
\ene
where for $1 \leq \gamma \leq 2$
\beq
f_{\gamma}(k)= C\, |k|^{\gamma-1} \int_0^{\infty} \, y^{\gamma} |V(y)|
\left(\frac{|k|y}{1+|k|y}\right)^{2-\gamma}\, dy.
\label{2.84}
\ene
Note that as $k \rightarrow 0$
\beq
f_{\gamma}(k)=\cases{ o\left(|k|^{\gamma -1}\right),& $1 \leq \gamma < 2$,\cr\cr
O\left( |k|\right),& $\gamma =2$.}
\label{2.85}
\ene
Since the function: $\lambda \rightarrow  |k| \lambda (1+|k| \lambda )^{-1}$
is an increasing function
of $\lambda $, for $\lambda \geq 0 $, we have that (see (\ref{2.7}) and
(\ref{2.82})) for all $x\in \ER $
$$
\int_0^{\infty}\left|D_{k_1}(y-x)-D_{k_2}(y-x)\right| |V(y)\, m_{1,n}(y,k_2 )|
\,dy
\leq C \int_0^{\infty} \frac{|k_1 -k_2 |(|x|+|y|)}{1+|k_1 -k_2 | (|x|+|y|)}
$$
$$
(|x|+|y|) |V(y)| dy\leq C\, \frac{|k_1 -k_2 ||x|^2}{1+|k_1 -k_2 ||x|}
+C\, \int_0^{\infty}\frac{|k_1 -k_2 ||y|^2}{1+|k_1 -k_2 ||y|} dy
$$
\beq
\leq C\left[\frac{|k_1 -k_2 ||x|}{1+|k_1 -k_2 ||x|}+f_{\gamma}(k_1-k_2 ).
\right] (1+|x|).
\label{2.86}
\ene
Furthermore, for $x \leq 0$ (see (\ref{2.7}) and (\ref{2.82}))
$$
\int_x^0\left|D_{k_1}(y-x)-D_{k_2}(y-x)\right||V(y)\,m_{1,n}(y,k_2 )|
dy \leq
$$
\beq \int_x^0 \frac{|k_1 -k_2 |\, (|x|+|y|)^2}{1+|k_1 -k_2 |(|x|+|y|)}\,
|V(y)|\,(1+|y|) dy \leq C \frac{|k_1 -k_2 ||x|^2}{1+|k_1 -k_2 ||x|}.
\label{2.87}
\ene

By (\ref{2.86}) and (\ref{2.87}) for $x \leq 0$
\beq
\left|f_n (x,k_1 ,k_2 )\right|\leq g_{\gamma}(x,k_1 -k_2 )
\label{2.88}
\ene
where
\beq
g_{\gamma}(x,k):=C\, \left[ \frac{|k||x|}{1+|k||x|}+f_{\gamma}(k)\right]
(1+|x|).
\label{2.89}
\ene
By (\ref{2.80}) and (\ref{2.83}) we have that for $x \geq 0$
\beq
\left|m_{1,n+1}(x,k_1 )-m_{1,n+1}(x,k_2 )\right|\leq f_{\gamma}(x, k_1 -k_2 )
+\int_x^{\infty} \left|m_{1,n}(y,k_1 )-m_{1,n}(y,k_2 )\right|\,y\, |V(y)| dy.
\label{2.90}
\ene
Since $m_{1,0}(x,k) \equiv 1$, it follows from (\ref{2.80}) and (\ref{2.83})
that
\beq
\left|m_{1,1}(x, k_1 )- m_{1,1}(x, k_2 )\right|\leq f_{\gamma}(x, k_1 -k_2 ),\,
x \geq 0.
\label{2.91}
\ene
Then iterating (\ref{2.90}) we prove that
\beq
\left|m_1 (x,k_1 )-m_1 (x,k_2 )\right| \leq f_{\gamma}(x, k_1 -k_2 )\,
e^{\left(\int_x^{\infty} y \, |V(y)|\, dy\right)},\, x \geq 0.
\label{2.92}
\ene
Moreover, taking the limit as $n \rightarrow \infty$ in
(\ref{2.80}) and using (\ref{2.21}), (\ref{2.88}) and (\ref{2.92})
we obtain that for $x \leq 0$
\beq
\left|m_1 (x,k_1 )-m_1 (x,k_2 )\right|\leq g_{\gamma}(x,k_1 -k_2 )+
 \int_x^0 (|x|+|y|) |V(y)| \left|m_1 (y,k_1 )-m_1 (y,k_2 )\right| dy,
\label{2.93}
\ene
where in the right--hand side of (\ref{2.89}) we take a constant $C$ large
enough. Let us denote
\beq
h(x,k_1 ,k_2 ):= \frac{\left|m_1 (x,k_1 )- m_1 (x,k_2 )\right|}
{g_{\gamma}(x, k_1 -k_2 )}.
\label{2.94}
\ene
Then it follows from (\ref{2.93}) that for $x \leq 0$
\beq
h(x,k_1 ,k_2 )\leq 1 + \int_0^x (1+|y|) |V(y)| h(y, k_1 ,k_2 ) dy,
\label{2.95}
\ene
where we used that $ g_{\gamma}(x,k) /(1+|x|)$ is an increasing function
of $|x|$.
By (\ref{2.95}) and Gronwall's inequality (see page 204 of \cite{7}) we have
that
\beq
h(x, k_1 ,k_2 )\leq e^{\int_0^{\infty} (1+|y|)|V(y)| dy}
\label{2.96}
\ene
 and then taking in (\ref{2.89}) $C$ large enough we obtain that
 \beq
 \left|m_1 (x,k_1 )-m_1 (x,k_2 )\right|\leq g_{\gamma}(x,k_1 -k_2 ).
 \label{2.97}
 \ene
 We similarly prove that
 \beq
 \left|m_2 (x,k_1 )-m_2 (x,k_2 )\right|\leq g_{\gamma}(x, k_1 -k_2 ).
 \label{2.98}
 \ene
 Note that in the proof of (\ref{2.97}), (\ref{2.98}) we only used that
 $V \in L^1_{\gamma},\, 1 \leq \gamma \leq 2$. We now prove (\ref{2.72}).
 It follows from (\ref{2.58}) that
 $$
 R_1 (k_1 )-R_1 (k_2 )=\left( m_1 (x,k_2 )\right)^{-1} \left[ e^{-2ik_1 x}
 T(k_1 )\, m_2 (x,k_1 ) -e^{-2ik_2 x} T(k_2 )\, m_2 (x,k_2 ) +\right.
 $$
 \beq \left.
 e^{-2ik_2 x}
 m_1 (x,-k_2)
 -e^{-2ik_1 x} m_1 (x, -k_1 ) +R_1 (k_1 ) (m_1 (x,k_2 )-m_1 (x,k_1 )) \right].
 \label{2.99}
 \ene
 Then by (\ref{2.4}) and (\ref{2.7}) there is an $x_0 \in \ER$
 such that
 \beq
 |m_1 (x,k)|\geq \frac{1}{2},\, x \geq x_0 , \, k \in \ER.
 \label{2.100}
 \ene
 Then (\ref{2.72}) with $j=1$ follows from (\ref{2.70}), (\ref{2.97})  and
 (\ref{2.98}) taking in (\ref{2.99}) any $x \geq x_0$. Equation (\ref{2.72})
 with $j=2$ is proven in a similar way.
Equation (\ref{2.73}) follows from (\ref{2.7}), (\ref{2.8}), (\ref{2.12}),
(\ref{2.13}), (\ref{2.32}), (\ref{2.33}), (\ref{2.37}),
(\ref{2.38}), (\ref{2.52}), (\ref{2.53}), the first equality in the
right--hand side of (\ref{2.60}) and (\ref{2.65}) and noting that if
$V \in L^1_2 $
\beq
 [f_1 (x,k), f_2 (x,k)]=ik\frac{1+a^2}{a}+O\left(k^2 \right), k \rightarrow 0.
 \label{2.106}
 \ene
 Equation (\ref{2.106}) is proven by the argument given in \cite{3} to prove
 that
 \beq
 [f_1 (x,k),f_2 (x,k)]=ik \frac{1+a^2}{a}+o(k), k \rightarrow 0,
 \label{2.107}
 \ene
 in the case when $V \in L^1_1 $. The fact that in (\ref{2.106}) we have
 $O\left(k^2 \right)$ instead of $o(k)$ follows because we assume that
 $V \in  L^1_{\gamma} , \gamma \geq 2 $ (see (\ref{2.12}) and (\ref{2.13})).
  Equation (\ref{2.74}) follows from the first equality
  in the right--hand side of
 (\ref{2.60}) and by (\ref{2.106}). Also (\ref{2.76}) follows from the
 first equality in the right--hand side of (\ref{2.61}) and (\ref{2.62})and
 observing that
 \beq
 \left[ f_1 (x,k), f_2 (x, -k)\right] =-ik \frac{a^2 -1}{a}
 +O\left(k^2 \right), \, k \rightarrow 0.
 \label{2.101}
 \ene
Equation (\ref{2.101}) is proven as (\ref{2.106}). It follows from (\ref{2.73})
that
 \beq
 T(k_1) -T(k_2 )=O\left(|k_1 -k_2 |^{\gamma -2}\right), k_1 -k_2 \rightarrow 0.
 \label{2.108}
 \ene
 Then (\ref{2.77}) with $j=1$ follows from (\ref{2.97}), (\ref{2.98}),
 (\ref{2.99}) and (\ref{2.108}). Equation (\ref{2.77}) with $j=2$ is proven
 in the same way.

 \bull

The results on the spectral theorem for $H$ that we state below follow from
the Weyl--Kodaira--Titchmarsch theory. See for example \cite{1}. For a
version of the Weyl--Kodaira-- Titchmarsch theory adapted to our situation
see Appendix 1 of \cite{4} and also the proof of Theorem 6.1 in page 78 of
\cite{4}. Let us denote for any $k \in \ER$
\beq
\Psi_+ (x,k):= \cases{ \frac{1}{\sqrt{2 \pi }} T(k) f_1 (x,k),& $ k \geq 0$,\cr\cr
\frac{1}{\sqrt{2\pi}} T(-k) f_2 (x,-k),& $k < 0$,}
\label{2.109}
\ene
and $\Psi_- (x,-k) := \overline{\Psi_+ (x,k)}$. Let ${\cal H}_{ac}(H)$ be the
subspace of absolute continuity of $H$. Then the following limits
\beq
\hat{\phi}_{\pm}(k):= s-\lim_{N \rightarrow \infty}\int_{-N}^{N}
\overline{\Psi_{\pm}(x,k)} \,\phi (x) \, dx
\label{2.110}
\ene
exist in the strong topology in $L^2$ for every $\phi \in L^2 $ and the
operators
\beq
\left(F_{\pm} \phi \right)(k):= \hat{\phi}_{\pm}(k)
\label{2.111}
\ene
are unitary operators from ${\cal H}_{ac}(H)$ onto $L^2$. Moreover, the
${F^{\ast}}_{\pm}$ are given by
\beq
\left({F^{\ast}}_{\pm} \phi \right)(x)= s-\lim_{N\rightarrow \infty}
\int_{-N}^{N} \Psi_{\pm}(x,k) \,\phi (k) \, dk,
\label{2.112}
\ene
where the limits exist in the strong topology in $L^2$. Furthermore, the
operators ${F^{\ast}}_{\pm }F_{\pm}$ are the orthogonal projection onto
${\cal H}_{ac}(H)$. For each eigenvalue of $H$, let $\Psi_j , j=1,2,\cdots ,N$
be the corresponding eigenfunction normalized to one, i.e.
$\|\Psi_j \|_{L^2}=1$. The operators:
\beq
F_j \phi := (\phi ,\Psi_j ) \Psi_j, j=1,2,\cdots , N,
\label{2.113}
\ene
are unitary from the eigenspace generated by $\Psi_j$ onto $C$. The following
operators
\beq
F^{\pm}= F_{\pm} \oplus^N_{j=1} F_j ,
\label{2.114}
\ene
are unitary from $L^2$ onto $L^2 \oplus^N_{j=1} C$ and for any $\phi \in D(H)$
\beq
F^{\pm} H \phi= \left\{ k^2 (F_{\pm} \phi )(k) , -\beta^2_1 F_1 \phi ,
\cdots , - \beta^2_N F_N \phi \right\}.
\label{2.115}
\ene
Moreover, for any bounded Borel function , $ \Phi$, defined on $\ER$
\beq
F^{\pm} \Phi (H) \phi = \left\{ \Phi (k^2) (F_{\pm}\phi )(k), \Phi(-\beta^2_1
)F_1 \phi ,\cdots , \Phi (-\beta^2_N ) F_N \phi \right\}.
\label{2.116}
\ene
The projector, $P_p$, onto the subspace of $L^2$ generated by the eigenvectors
of $H$ is given by
\beq
P_p \phi:= \sum_{j=1}^N  (\phi ,\Psi_j ) \Psi_j .
\label{2.117}
\ene
 Since $H$ has no
singular--continuous spectrum the projector onto the continuous subspace of
$H$ is given by: $P_c := I-P_p$. It follows from (\ref{2.116}) that
 \beq
 e^{-it H}P_c = F^{\pm \ast} e^{-ik^2 t} F^{\pm}.
 \label{2.118}
 \ene
 Equation (\ref{2.118}) is the starting point of our proof of the
  $L^1 - L^{\infty} $ estimate (Theorem 1.1). We divide the proof of the
  $L^1 -L^{\infty}$  estimate into a high--energy estimate and a low--energy
  estimate. For this purpose, let $\Phi$ be any continuous and bounded function
  on $\ER$ that has a bounded derivative and such that $\Phi (k) =0$ for
  $|k| \leq k_1 $ and $\Phi (k)=1$ for $|k| \geq k_2 $ for some $0 < k_1  < k_2$.

\begin{lemma}(The high--energy estimate). Suppose that $V \in L^1_1$. Then
$e^{-itH} \Phi (H)P_c $ extends to a bounded operator from $L^1$ to $L^{\infty}$ and
there is a constant $C$ such that
\beq
\left\| e^{-itH} \Phi (H) P_c \right\|_{{\cal B}\left( L^1 ,L^{\infty}\right)} \leq
 \frac{C}{\sqrt{t}}, \, t > 0.
 \label{2.119}
 \ene
 \end{lemma}
 \noindent {\it Proof:}\,Let us take $\chi \in C^{\infty}, \, \chi (k)=1,
 |k| \leq 1$ and $\chi (k)=0, \, k \geq 2$, and let us denote $ \chi_n (k)=
 \chi (k/n), n=1,2,\cdots $. Then it follows from (\ref{2.118}) that for any
 $f, g \in L^1 \cap L^2 $:
\beq
 \left(e^{-itH}\Phi (H) P_c f, g \right)= \lim_{n \rightarrow \infty}
 \left( e^{-itH} \Phi (H) \chi_n (H) P_c f, g \right)=
\lim_{n \rightarrow \infty}\int dx \, dy \Phi_{t,n}(x,y) f(x) \overline{g(y)},
\label{2.120}
\ene
where
\beq
\Phi_{t,n}(x,y):=\int_{-\infty}^{\infty} e^{-ik^2 t} \chi_n (k^2 ) \Phi (k^2 )
\overline{\Psi_+ (x,k)} \Psi_+ (y,k) dk.
\label{2.121}
\ene
We have that,
\beq
\Phi_{t,n}(x,y) = \Phi^{(0)}_{t,n} (x,y)+\Phi^{(1)}_{t,n}(x,y)
+\Phi^{(+)}_{t,n}(x,y)+\Phi^{(-)}_{t,n}(x,y),
\label{2.122}
\ene
where
\beq
\Phi^{(0)}_{t,n}(x,y):= \int_{-\infty}^{\infty} e^{-ik^2 t} \,
\frac{e^{-ik(x-y)}}{2 \pi} \chi_n (k^2 ) dk,
\label{2.123}
\ene
\beq
\Phi^{(1)}_{t,n}(x,y) :=\int_{-\infty}^{\infty} e^{-ik^2 t} \,
\frac{e^{-ik(x-y)}}{2 \pi}  \chi_n (k^2 ) (\Phi (k^2 )-1) dk,
\label{2.124}
\ene
\beq
\Phi^{(+)}_{t,n}(x,y):= \int_0^{\infty}e^{-ik^2 t}\,
\frac{e^{-i(x-y)}}{2 \pi } \chi_n (k^2 ) m_+ (x,y,k) dk,
\label{2.125}
\ene
\beq
\Phi^{(-)}_{t,n}(x,y):= \int_{-\infty}^0 e^{-ik^2 t}\,
\frac{e^{-ik(x-y)}}{2\pi } \chi_n (k^2 ) m_- (x,y,k) dk,
\label{2.126}
\ene
with
$$
m_{\pm}(x,y,k):= \Phi (k^2 )\left[ \overline{(T(k) m_{j(\pm )}(x,k)-1)} T(k)
m_{j(\pm)}(y,k) + \right.
$$
\beq
\left. T(k) m_{j(\pm)} (y,k) -1 \right],
\, \pm k \geq 0,
\label{2.127}
\ene
where $j(+)=1$ and $j(-)=2$.
Since the inverse Fourier transform of $\frac{1}{\sqrt{2 \pi} }e^{-ik^2 t}$
 is
 \beq
\Phi^{(0)}_t (x):= \frac{1}{\sqrt{4 \pi i t}}\, e^{ix^2 / 4t}
\label{2.127b}
\ene
it follows that
\beq
\lim_{n \rightarrow \infty} \int dx \, dy \, \Phi^{(0)}_{t,n}(x,y) f(x)
\overline{g(y)}= \int dx \, dy\, \Phi^{(0)}_t (x,y) f(x)
\overline{g(y)}.
\label{2.128}
\ene
Changing the coordinates of integration in (\ref{2.124}) to $p=k-k_0$ where
$k_0 = (y-x)/2t$ we obtain that
$$
\Phi^{(1)}_{t,n} (x,y)=\frac{1}{2 \pi}  e^{i(x-y)^2 /4t}
\int_{-\infty}^{\infty} dp e^{-ip^2 t}
\chi_n \left((p+k_0 )^2 \right) \left(\Phi\left((p+k_0 )^2 \right)-1 \right)=
$$
\beq
\frac{1}{2 \pi \sqrt{2it}}\,  e^{i(x-y)^2 /4t} \int_{-\infty}^{\infty}
dp \, e^{i \rho^2 /4t} \, \hat{h}_n (\rho ),
\label{2.131}
\ene
where in the second equality we used the Plancherel theorem and
$\hat{h}_n (\rho )$ is the Fourier transform of the function
$ h_n (\rho )$  defined as
follows
\beq
h_n (\rho ):= \overline{ \chi_n \left((p+k_0 )^2 \right)
\left(\Phi \left((p+k_0 )^2\right)-1 \right)}.
\label{2.132}
\ene
Since,
\beq
\left\| \hat{h}_n \right\|_{L^1} \leq C \left\|h_n \right\|_{W_1} \leq C
\left\|\Phi (p^2 )-1 \right\|_{W_1},
\label{2.133}
\ene
we have that
\beq
\left| \Phi^{(1)}_{t,n}(x,y) \right| \leq \frac{C}{\sqrt{t}}.
\label{2.134}
\ene
Let us denote $h(p):= \overline{ \Phi \left((p+k_0 )^2 \right)-1}$. Then
since $\hat{h}_n (p)$ converges to $\hat{h}(p)$ in the $L^1$ norm, it
follows  from (\ref{2.131})
and the dominated convergence theorem  that
\beq
\lim_{n \rightarrow \infty} \Phi^{(1)}_{t,n} (x,y) =\Phi^{(1)}_t (x,y):=
\frac{1}{2 \pi \sqrt{2it}} \,e^{i(x-y)^2 /4t} \int_{-\infty}^{\infty}
e^{i \rho^2 /4t} \, \hat{h}(\rho )d\rho,
\label{2.135}
\ene
and that
\beq
\left| \Phi^{(1)}_t (x,y)\right| \leq \frac{C}{\sqrt{t}}, \, x,y \in \ER.
\label{2.136}
\ene
Using the dominated convergence theorem again we prove that
\beq
\lim_{n \rightarrow \infty}\int dx \, dy\, \Phi^{(1)}_{t,n}(x,y) f(x)
\overline{g(y)}= \int dx \, dy \,\Phi^{(1)}_t (x,y) f(x) \overline{g(y)}.
\label{2.137}
\ene
We denote
\beq
m_{+,e} (x,y,k):=\cases{ m_+ (x,y,k),& $k \geq 0$, \cr\cr
0,& $k < 0 $.}
\label{2.138}
\ene
Then since $\Phi (k^2 )=0$ for $|k| \leq \sqrt{k_1}$ and $\Phi (k^2 )=1$
for $|k| \geq \sqrt{k_2}$, it follows from (\ref{2.7}), (\ref{2.12}),
 (\ref{2.63}), (\ref{2.70}) and (\ref{2.127}) that for some constant $C$
 \beq
 \left\|m_{+,e}(x,y,\cdot )\right\|_{W_1} \leq C, \,\,x, y \geq 0.
 \label{2.139}
 \ene
 Then, as in the case of $\Phi^{(1)}_{t,n}$ we prove that
 \beq
 \left|\Phi^{(+)}_{t,n}(x,y)\right| \leq \frac{C}{\sqrt{t}}, \,\,x,y \geq 0,
  t > 0.
  \label{2.140}
  \ene
  and that
  \beq
  \lim_{n \rightarrow \infty} \Phi^{(+)}_{t,n}(x,y)=\Phi^{(+)}_t (x,y):=
  \frac{1}{2 \pi \sqrt{2it}}\int_{-\infty}^{\infty}
  e^{i\rho^2 /4t} \,\tilde{m}_{+,e}(x,y,\rho )d \rho,
  \label{2.141}
  \ene
  where $\tilde{m}_{+,e}(x,y,\rho )$ is the Fourier transform of
  $m_{+,e}(x,y,k+k_0 )$, and that
  \beq
  \left|\Phi^{(+)}_{t}(x,y)\right| \leq \frac{C}{\sqrt{t}},\,\, x,y \geq 0,
  t > 0.
  \label{2.142}
  \ene
  Using (\ref{2.58}) we write (\ref{2.126}) as follows
  \beq
  \Phi^{(-)}_{t,n} (x,y)=\sum_{j=2}^5 \Phi^{(j)}_{t,n}(x,y),
  \label{2.143}
  \ene
  where
  \beq
  \Phi^{(j)}_{t,n}(x,y):=\int_{-\infty}^0 e^{-ik^2 t}
  \frac{e^{-i k (lx-ry)}}{2\pi}\, \chi_n (k^2)\,m_j (x,y,k) dk,
  \label{2.144}
  \ene
 where for $j=2,\, l=r=3$, for $j=3, \, l=3, r=1$, for $j=4, \, l=1, r=3$,
 and for $j=5,\, l=r=1$. Moreover, (recall that
 $m_j (x,-k)=\overline{m_j (x,k)}$)
 \beq
 m_2 (x,y,k):= \Phi (k^2 )\left[|R_1 (k)|^2 \overline{m_1 (x,k)} m_1 (y,k)
 \right],
 \label{2.145}
 \ene
 \beq
 m_3 (x,y,k):= \Phi (k^2 ) \overline{R_1 (k^2 ) m_1 (x,k) m_1 (y,k)},
 \label{2.146}
 \ene
 \beq
 m_4 (x,y,k):= \Phi(k^2 )R_1 (k) (\overline{m_1 (x,k)}-1) m_1 (y,k),
 \label{2.147}
 \ene
and
\beq
m_5 (x,y,k):= \Phi (k^2 )\overline{(m_1 (x,k)-1) m_1 (y,k)}.
\label{2.148}
\ene
Then as in the case of $\Phi^{(+)}_{t,n}$ we prove that
\beq
\left|\Phi^{(-)}_{t,n}(x,y)\right|\leq \frac{C}{\sqrt{t}},\, x,y \geq 0,
t > 0,
\label{2.149}
\ene
and that
\beq
\lim_{n\rightarrow \infty}\Phi^{(-)}_{t,n}(x,y)=\Phi^{(-)}_t (x,y),
\, x,y \geq 0, t > 0,
\label{2.150}
\ene
where
\beq
\Phi^{(-)}_t (x,y)=\sum _{j=2}^5 \Phi^{(j)}_t (x,y),
\label{2.151}
\ene
with
\beq
\Phi^{(j)}_t (x,y):= \frac{1}{2 \pi \sqrt{2t}}\int_{-\infty}^{\infty}
e^{i\rho^2 /4t} \tilde{m}_j (x,y,\rho )d\rho,
\label{2.152}
\ene
with $\tilde{m}_j (x,y,\rho )$ the Fourier transform of
$m_j (x,y, p+(ry-lx)/2t)$. We also have that

\beq
\left|\Phi^{(-)}_t (x,y)\right|\leq \frac{C}{\sqrt{t}},\, x,y \geq 0, t > 0.
\label{2.153}
\ene
By the same argument as above and using also (\ref{2.59}) we prove that for
$( x \geq 0 , y \leq 0), (x \leq 0 ,y \geq 0 )$ and $ (x \leq 0 ,y \leq 0)$
\beq
\left|\Phi^{(\pm)}_{t,n} (x,y)\right| \leq \frac{C}{\sqrt{t}},\, t > 0,
\label{2.154}
\ene
and that
\beq
\lim_{n \rightarrow \infty}\Phi^{(\pm)}_{t,n}(x,y)= \Phi^{(\pm)}_t (x,y),
\label{2.155}
\ene
for  functions $\Phi^{(\pm)}_t (x,y)$ that satisfy
\beq
\left|\Phi^{(\pm)}_t (x,y)\right| \leq \frac{C}{\sqrt{t}},\, t > 0.
\label{2.156}
\ene
We can explicitly compute $\Phi^{\pm}_t (x,y)$ as in the case
$(x \geq 0 ,y \geq 0)$.Then (\ref{2.154}), (\ref{2.155}) and (\ref{2.156})
hold for all $x,y \in \ER$ and using (\ref{2.120}), (\ref{2.122}),
(\ref{2.128}), (\ref{2.134}), (\ref{2.137}), \linebreak (\ref{2.154} )
and (\ref{2.155})
we prove that
\beq
\left(e^{-itH}\Phi(H) P_c f, g \right)=\int dx \, dy \left[ \Phi^{(0)}_t (x,y)
+\Phi^{(1)}_t (x,y)+ \Phi^{(+)}_t (x,y)+ \Phi^{(-)}_t (x,y)\right] f(x)
\overline{g(y)}.
\label{2.157}
\ene
Then by (\ref{2.127b}), (\ref{2.136}) and (\ref{2.156})
\beq
\left|\left(e^{-itH} \Phi (H) P_c f, g \right)\right|\leq \frac{C}{\sqrt{t}}
\|f\|_{L^1} \|g||_{L^1},\, t > 0,
\label{2.158}
\ene
for all $f, g \in L^1 \cap L^2$. By continuity this estimate holds for all
$f, g \in L^1$ and (\ref{2.119}) follows.

\bull

Let $\Psi$ be any function on $C^{\infty}_0 \left(\ER \right)$ such that
$\Psi (k)=1, |k| \leq \delta$, for some $\delta > 0$.

\begin{lemma} (The low--energy estimate).
Suppose that $V \in L^1_{\gamma}$ where in the {\rm generic case} $\gamma > 3/2$
and in the {\rm exceptional case} $ \gamma > 5/2$. Then $e^{-itH} \Psi(H) P_c$
extends to a bounded operator from $L^1$ to $L^{\infty}$ and there is a
constant $C$ such that
\beq
\left\|e^{-itH} \Psi (H) P_c \right\| \leq  \frac{C}{\sqrt{t}},\, t > 0.
\label{2.159}
\ene
\end{lemma}
\noindent {\it Proof :}\, As in the proof of Lemma 2.6 it follows from
(\ref{2.118}) that for all $f, g \in L^1 \cap L^2$
\beq
 \left(e^{-itH} \Psi (H) P_c f, g \right)=\int dx \, dy \, \Phi_t (x,y)
 f(x) \overline{g(y)},
 \label{2.160}
 \ene
 where
 \beq
 \Phi_t (x,y)= \Phi^{(+)}_t (x,y) +\Phi^{(-)}_t (x,y),
 \label{2.161}
 \ene
 with
 \beq
 \Phi^{(+)}_t (x,y):= \int_0^{\infty} e^{-ik^2 t}\,
 \frac{e^{-ik(x-y)}}{2 \pi}\, m_+ (x,y,k) \,dk,
 \label{2.162}
 \ene
 \beq
 \Phi^{(-)}_t (x,y);= \int_{-\infty}^0 e^{-ik^2 t}\,
 \frac{e^{-ik(x-y)}}{2 \pi} \, m_- (x,y,k) \, dk,
 \label{2.163}
 \ene
 and
 \beq
 m_{\pm}(x,y,k):= \Psi (k^2 ) q_{\pm}(x,y,k)
 \label{2.164}
 \ene
 with
 \beq
 q_{\pm}(x,y,k):=\overline{T(k) m_{j(\pm)}(x,k)} T(k) m_{j(\pm)}(y,k), \,
 \pm k > 0,
 \label{2.165}
 \ene
 where $ j(+)=1$ and $j(-)=2$.

 Let us consider first the {\it generic case}. In this case it follows from
 (\ref{2.66}) that \linebreak $m_{\pm}(x,y, 0{\pm})=0$. We denote
 \beq
 m_{+,e}(x,y,k):=\cases{m_+ (x,y,k),& $ k \geq 0$,\cr\cr
 0,& $k < 0$.}
 \label{2.166}
 \ene
 Let us denote by $\omega_{+,x,y}(\rho )$ the modulus of continuity
 of $m_{+,e}(x,y)$, i.e.,
 \beq
 \omega_{+,x,y}(\rho ):= \left\|m_{+,e}(x,y,k+\rho )
 -m_{+,e}(x,y,k)\right\|_{L^2}.
 \label{2.167}
 \ene
 Remark that
 \beq
 \omega_{+,x,y}(\rho )\leq 2 \left\|m_{+,e}(x,y,\cdot )\right\|_{L^2}
 \leq C_{x_0}, \, x, y \geq x_0.
 \label{2.168}
 \ene
 Without lossing generality we can assume that $ \gamma \leq 2$. Then
 by (\ref{2.7}), (\ref{2.12}), (\ref{2.70}), (\ref{2.163}) and
 (\ref{2.164}) for $|\rho |\leq 1$
 \beq
 \omega_{+,x,y}(\rho )\leq C_{x_0}\, |\rho |^{\gamma -1}, x, y \geq x_0.
 \label{2.169}
 \ene
It follows from (\ref{2.168}) and (\ref{2.169}) that for any $0 \leq \alpha <
\gamma -1 $
\beq
\int d \rho |\omega_{+,x,y}(\rho )|^2 \frac{1}{|\rho |^{1+2 \alpha}}  < \infty
\label{2.170}
\ene
and then by Proposition 4 in page 139 of \cite{5}
\beq
\left\|m_{+,e}(x,y,\cdot )\right\|_{W_{\alpha}} \leq C_{\alpha ,x_0},
x, y \geq x_0,
\label{2.171}
\ene
for any $0 < \alpha < \gamma -1$. Let us denote $k_0 = (y-x)/2t$. Then
we prove as in Lemma 2.6 that
(\ref{2.134})
\beq
\Phi^{(+)}_t (x,y)=\frac{1}{2\pi \sqrt{2it}}\, \int_{-\infty}^{\infty}
e^{i\rho^2 /4t}\, \tilde{m}_{+,e}(x,y,\rho )\, d\rho,
\label{2.172}
\ene
with $\tilde{m}_{+,e}(x,y,\rho )$ the Fourier transform of
$m_{+,e}(x,y,k+k_0 )$. But since for $ \frac{1}{2} < \alpha < \gamma -1$
\beq
\left\|\tilde{m}_{+,e}(x,y,\cdot ) \right\|_{L^1} \leq C
\left\|(1+\rho^2 )^{\frac{\alpha} {2}} \tilde{m}_{+,e}(x,y, \cdot )
\right\|_{L^2}= C \left\|m_{+,e}(x,y,\cdot )\right\|_{W_{\alpha}} \leq
C , \, x, y \geq 0 ,
\label{2.173}
\ene
we have that
\beq
\left|\Phi^{(+)}_t (x,y)\right| \leq \frac{C}{\sqrt{t}},\, x, y \geq 0
,  t > 0.
\label{2.174}
\ene
Using (\ref{2.7}), (\ref{2.8}), (\ref{2.12}), (\ref{2.13}),
(\ref{2.58}), (\ref{2.59}), (\ref{2.61}) and (\ref{2.72}) we prove in the
same way that
(\ref{2.174}) holds for $(x \geq 0 , y < 0), (x \leq 0, y \geq 0)$
and $(x \leq 0, y\leq 0)$ and that the same is true for $\Phi^{(-)}_t (x,y)$
(see the proof of Lemma 2.6 for a similar argument). Then we have that
\beq
\left|\Phi_t (x, y)\right| \leq \frac{C}{\sqrt{t}}, \,x, y \in \ER,\, t > 0.
\label{2.175}
\ene
Equation (\ref{2.159}) follows from (\ref{2.175}) as in the proof of Lemma 2.6

Let us now consider the {\it exceptional case}. The new problem is that
 now $m_{\pm}(x, y, 0\pm )\neq 0$. Let us write $\Phi^{(+)}_t$ as
 follows
 \beq
 \Phi^{(+)}_t (x, y)= \Phi^{(1)}_t (x,y) +\Phi^{(2)}_t (x,y),
 \label{2.176}
 \ene
 where
 \beq
 \Phi^{(j)}_t (x,y):=\int_0^{\infty} e^{-ik^2 t}\, \frac{e^{-ik(x-y)}}{2 \pi}
 m^{(j)}(x,y,k)\, dk,\, j=1,2,
 \label{2.177}
 \ene
 with
 \beq
 m^{(1)}(x,y,k):= \Psi (k^2 )\left[q_+ (x,y,k)-q_+ (x,y,0+)\right],
 \label{2.178}
 \ene
 \beq
 m^{(2)}(x,y,k):=\Psi (k^2 )q_+ (x,y,0+).
 \label{2.179}
 \ene
 Then using Theorem 2.5 (b) we prove as in the {\it generic case}\, that
 \beq
 \left|\Phi^{(1)}_t (x,y)\right| \leq \frac{C}{\sqrt{t}},\,x,y \in \ER,\,
 t > 0.
 \label{2.180}
 \ene
 Let $\hat{\Psi}(\lambda ), \lambda \geq 0$, be the cosine transform of
 $\Psi (k^2 )$:
 \beq
 \hat{\Psi}(\lambda ):= \int_0^{\infty} \cos (\lambda k) \Psi(k^2 )dk.
 \label{2.181}
 \ene
 Then integrating by parts we prove that for any $N > 0$ there is a constant
 $C_N$ such that
 \beq
 \left|\hat{\Psi}(\lambda )\right| \leq C_N \, (1+|\lambda |)^{-N}.
 \label{2.182}
 \ene
 Since
 \beq
 \Psi (k^2 )=\frac{2}{\pi} \int_0^{\infty} \cos (\lambda k) \hat{\Psi}(\lambda ) d \lambda ,
 \label{2.183}
 \ene
 we have that
 \beq
 \Phi^{(2)}_t (x,y)=\frac{q_+ (x,y,0+)}{ \pi} \int_0^{\infty} d \lambda
 \hat{\Psi}(\lambda ) \int_0^{\infty} e^{-ik^2 t }e^{-ik(x-y)} \cos(\lambda k)
 \, dk.
 \label{2.184}
 \ene
But

\beq
\left|\int_0^{\infty} e^{-ik^2 t} e^{-ik(x-y)} \cos (\lambda k)\, dk \right|
\leq \frac{C}{\sqrt{t}},\, t > 0.
\label{2.185}
\ene
The estimate (\ref{2.185}) is proven by explicitly evaluating the cosine
transform using the following equations from \cite{8}: 3 in page 7,
1 in page 23, 7 in page 24, 3 in page 63, 1 in page 82 and 3 in page 83.
Then by (\ref{2.182}), (\ref{2.184}) and (\ref{2.185})
\beq
\left|\Phi^{(2)}_t (x,y)\right| \leq \frac{C}{\sqrt{t}},\, x,y \in \ER ,\,
 t > 0.
 \label{2.186}
 \ene
 It follows from (\ref{2.176}), (\ref{2.180}) and (\ref{2.186}) that
 \beq
 \left|\Phi^{(+)}_t (x,y)\right| \leq \frac{C}{\sqrt{t}}, \, x,y \in \ER ,\,
 t > 0.
 \label{2.187}
 \ene

 We prove in the same way that
 \beq
 \left|\Phi^{(-)}_t (x,y)\right|\leq \frac{C}{\sqrt{t}},\, x,y \in \ER ,
 \, t > 0.
 \label{2.188}
 \ene
 Equation (\ref{2.159}) follows from (\ref{2.160}), (\ref{2.161}),
 (\ref{2.187}) and (\ref{2.188}) as in the {\it generic case}.

\noindent {\it Proof of Theorem 1.1:}\, The theorem follows from Corollaries
2.6 and 2.7.

\noindent {\it Proof of Corollary 1.2:}\, Since $H$ is self--adjoint
\beq
\left\|e^{-itH} P_c \right\|_{{\cal B}\left(L^2 \right)} \leq 1.
\label{2.189}
\ene
Then the corollary follows interpolating between (\ref{1.18}) and
(\ref{2.189}) (see the Appendix to \cite{6}).

\noindent{\it Proof of Corollary 1.3:}\, Corollary 1.3 follows from Corollary
1.2 as in the proof of Theorem 4.1 of \cite{10}.

\section{Inverse Scattering}\sss
\noindent {\it Proof of Theorem 1.4:}\, We prove this theorem by verifying
the conditions of the abstract Theorems 1 and 2 of \cite{15} and of
Theorem 16 of \cite{16}. This is done as in Theorem 8 of \cite{15} and Theorem
17 of \cite{16}. We define $X$ and $X_3$ as in the Introduction and
$X_1 :=L^{1+\frac{1}{p}}$. It
 follows from the Sobolev imbedding theorem (see \cite{14}) that
 $X \subset X_3 $, with continuous imbedding. Concerning hyphotesis (V) in page
 113 of \cite{15}: note that since by Sobolev's imbedding theorem
 $W_1 \subset L^{1+p}$; we have that $X_1 \subset W_1$. But as $e^{-itH} \in
 {\cal B}(W_1 )$, it follows by duality that $e^{-itH} \in {\cal B}(W_{-1} )$.
 Then for all $\phi \in X_1$, $e^{-itH}\,\phi \in W_{-1}$ and
  $e^{-itH} \, e^{-isH} \phi =e^{-i(t+s)H} \phi$ for all $t,s \in \ER$.

 To verify hypothesis VII of Theorem 16 of \cite{16}, as in the proof of
 Theorem 8 of \cite{15}, we need the following result. Let $g$ be any
 real--valued $C^2$ function defined on $\ER$ such that $g(0)=0$ and for all
 $u, v \in \ER $:
 \beq
 |g(u)-g(v)|+ |\acute{g}(u)-\acute{g}(v)| \leq C |u-v|,
 \label{3.1}
 \ene
 and
 \beq
 |g(u)|\leq |f(u)|.
 \label{3.3}
 \ene
 For $I$ any interval let us denote by $C(I, X)$ the Banach space of bounded
 and
 continuous functions from $I$ into $X$ with the supremun norm and by
 $B_{\rho} (I,X)$ the ball of center zero and radius $\rho$ in $C(I,X)$.
 Then for any $\phi \in X(\rho /2 )$ and any $s \in \ER$ the equation
 \beq
 u(t)= e^{-itH} \phi +\frac{1}{i} \int_s^t e^{-i(t-\tau )H} P_g (u(\tau ))
 d\tau,
 \label{3.4}
 \ene
 where
 \beq
 P_g (u(\tau )):= g(|u(\tau )|) \frac{u(\tau )}{|u(\tau )|}
 \label{3.5}
 \ene
 has a unique solution $u(t)\in B_{\rho}(\ER ,X)$ and moreover, the $L^2$
 norm and the energy are conserved:
 \beq
 \|u(t)\|_{L^2}=\hbox{constant}
 \label{3.6}
 \ene
 \beq
 E_g := \frac{1}{2} \|\sqrt{H}u(t)\|^2_{L^2}+\int dx G(|u(t)|)=\hbox{constant},
 \label{3.7}
 \ene
 for all $t \in \ER$, where $G$ is the primitive of $g$ such that $G(0)=0$.
 To prove this result we observe that it follows from (\ref{3.1})
 and (\ref{3.3}) that
 \beq
 \| P_g (\phi )-P_g (\psi )\|_X \leq C\left(\|\phi\|_X + \|\psi\|_X \right)
 \|\phi -\psi \|_X ,
 \label{3.8}
 \ene
 for all $\phi ,\psi \in X$. Then by a standard contraction mapping argument
 (\ref{3.4}) as a unique solution on $C([s-\epsilon ,s+\epsilon ],X)$ provided
 that $0 < \epsilon \leq  1/ 3C \rho$ and $ 0 < \epsilon <
 1/2C$. Suppose that (\ref{3.6}) and
 (\ref{3.7}) are true for $t \in [s-\epsilon ,s+\epsilon ]$. Then since
 $|G(\lambda )| \leq C \lambda^2$,
 \beq
 \|u(t)\|_X^2 \leq 2 E_g+ 2 (1+C) \|u(t)\|^2_{L^2} \leq C,\, t \in [s-\epsilon ,
 s+\epsilon ].
 \label{3.9}
 \ene
 Since $\|u(t)\|_X$ remains bounded as $t \rightarrow s \pm \epsilon$ by a
 constant $C$ that depends only on $\|\phi \|_X$ we can extend $u(t)$ into a
 global solution such that (\ref{3.6}), (\ref{3.7}) hold for all $t \in \ER$.
 It remains to prove that (\ref{3.6}), (\ref{3.7}) are true for
 $t \in [s-\epsilon , s+\epsilon ]$. In the constant coefficient case, $V=0$,
 this is accomplished by approximating the local solution in $W_1$ by solutions
 in $W_2$, see \cite{18} and \cite{19} or by regularizing equation (\ref{3.4})
 by taking convolution with a function in Schwartz space, see \cite{20},
 \cite{21} and \cite{22}. This is possible because in the constant
 coefficient case $D(H) = D(\Delta )=W_2$. In our case this is not a convenient
 approach. Since we only assume that $V \in L^1_{\gamma}$ we do not
 have much control over $D(H)$. We only know that $D(H)$ is a dense set in $X$.
 To solve this problem we regularize (\ref{3.4}) multiplying it by an
 appropriate function of $H$. Let us denote
 $r_n (H):= \left(\frac{H}{n}+1 \right)^{-1}, n=1,2,\cdots  $. The regularized
 equation is given by
 \beq
 u_n (t)=e^{-itH} r_n (H) \phi +\frac{1}{i} \int_s^t e^{-i(t -\tau )H}
 r_n (H) P_g \left(r_n (H) u_n (\tau )\right)\, d \tau.
 \label{3.10}
 \ene
 As above we prove that (\ref{3.10}) has a unique solution for
 $t \in [s-\epsilon , s+\epsilon ]$. Note that we can take $\epsilon$
 independent
 on $n$. Moreover, since $H r_n (H) \in {\cal B}(X )$ we have that actually
 $u_n (t) \in C^1 ([s-\epsilon ,s+\epsilon ], X)$. Then
 \beq
 \frac{d}{d t} \|u(t)\|_{L^2}^2 = 2 \Re \left( u_n (t),
 \frac{\partial}{\partial t} u_n (t)\right).
 \label{3.11}
 \ene
 Since $u_n (t)$ is a solution to the equation
 \beq
 i\frac{\partial}{\partial t}u_n (t)=H u_n(t)+
 r_n (H) g\left(|r_n (H) u_n (t)|\right)
 \frac{r_n (H) u_n (t)}{|r_n (H) u_n (t)|}
 \label{3.12}
 \ene
 and since $H$ is self--adjoint, it follows from (\ref{3.11}) that
 \beq
 \frac{d}{d t} \|u_n (t)\|^2_{L^2} = 0.
 \label{3.13}
 \ene
 Furthermore,
 \beq
 \frac{1}{2} \frac{d}{d t} \|\sqrt{H} u_n (t)\|^2_{L^2} = \Re \left( \sqrt{H}
 u_n (t), \sqrt{H} \frac{\partial}{\partial t} u_n (t) \right).
 \label{3.14}
 \ene
 Let us define
 \beq
 Q_n (t):= \int dx \, G\left(|r_n (H) u_n|\right).
 \label{3.15}
 \ene
 Since $|G(\lambda )| \leq C |\lambda |^2$,
 \beq
 |Q_n (t)| \leq C \|u_n (t)\|^2_{L^2}.
 \label{3.16}
 \ene
 Furthermore, since $u_n (t) \in C^1 ([s-\epsilon ,s+\epsilon ], X)$ it follows
 from a simple proof using the fundamental theorem of calculus (see
 the proof of Lemma 3.1 of \cite{20} for a similar argument) that
 \beq
 \frac{d}{d t} Q_n (t)= \Re \left( r_n (H)
 \frac{g(|r_n (H) u_n (t)|)}{|r_n (H) u_n (t)|} r_n (H) u_n (t),
 \frac{\partial}{\partial t} u_n (t) \right).
 \label{3.17}
 \ene
 We define the regularized energy as follows
 \beq
 E_n (t):= \frac{1}{2}\|\sqrt{H} u_n (t) \|^2_{L^2} +Q_n (t).
 \label{3.18}
 \ene
 It follows from (\ref{3.12}), (\ref{3.14}), (\ref{3.17}) and since $H$ is
 self--adjoint that
 \beq
 \frac{d}{d t} E_n (t)=0.
 \label{3.19}
 \ene
 By (\ref{3.13}) and (\ref{3.19}), $\|u_n (t)\|_{L^2}$ and $E_n (t)$ are
 constant for $t \in [s-\epsilon ,s+\epsilon ]$. We prove below that $u_n (t)$
 converges strongly in $X$ to $u(t)$. Since moreover, $r_n (H)$ converges to
 the identity strongly in $X$, equations (\ref{3.6}) and (\ref{3.7}) hold
 for $t \in [s-\epsilon ,s+\epsilon ]$. It only remains to prove
 that
 \beq
 \lim_{n \rightarrow \infty}\|u_n (t) - u(t) \|_X=0.
 \label{3.20}
 \ene
 But by (\ref{3.4}), (\ref{3.8}) and (\ref{3.10})
 $$
 \|u_n (t)-u(t)\|_X \leq \int_s^t d\tau \, \|r_n (H) P_g (r_n (H) u_n )
 -r_n (H) P_g (r_n (H)u)\|_X \,+
 $$
 $$
 \int_s^t d\tau \|r_n (H) P_g  (r_n (H)u)-P_g (u)\|_X \leq
 2 C \epsilon \, \rho \|u_n  -u\|_{C([s-\epsilon ,s+\epsilon ],X)}+
 $$
 \beq
 2 C \epsilon \, \rho \ \int_s^t \|(r_n (H)-1) u(\tau )\|_X \, d\tau .
 \label{3.21}
 \ene
 But since $2 C \epsilon \rho < 2/3$
 \beq
 \|u_n -u\|_{C([s-\epsilon ,s+\epsilon ],X)} \leq  6 C \epsilon \rho
 \int_{s-\epsilon}^{s+\epsilon} \| (r_n (H)-1) u(\tau )\| d\tau \rightarrow 0,
 \label{3.22}
 \ene
 as $n \rightarrow \infty$.
 As in the proof of Theorem 17 of \cite{16} we have to prove that
 $e^{-itH} \in {\cal B}\left( X, L^r \left(\ER , L^{1+p}\right)\right)$.
 Let us denote by ${\cal D}$ the set of points in the
 $(\frac{1}{q}, \frac{1}{r})$ plane, $ 1\leq p,q \leq \infty$, such
 that
 $e^{-itH}\in {\cal B}\left(X,L^r \left( \ER ,L^{q}\right)\right)$.
 We already know that $A:= (\frac{1}{2},0) \in {\cal D}$ because
 $e^{-itH}$ is a unitary  operator on $L^2$. Since $e^{-itH}$ is unitary on
 $X$, we have that $e^{-itH} \in {\cal B}\left( X, L^{\infty}(X)\right)$ and as
 by Sobolev's theorem \cite{14} $X$ is continuously embedded in $L^{\infty}$
 it follows that  $B:= (0,0) \in {\cal D}$. By Corollary 1.3
 $e^{-itH} \in {\cal B}\left( L^2 , L^6 \left(\ER ,L^6 \right)\right)$ and
 then  $C:= (\frac{1}{6},\frac{1}{6}) \in {\cal D}$. Since
 $A, B, C \in {\cal D}$  it follows by interpolation (see \cite{6}) that the
 solid triangle with vertices $A, B, C$ belongs to ${\cal D}$. Let us consider
 the following curve, ${\cal C}$, in the $(\frac{1}{q}, \frac{1}{r})$ plane:
 \beq
 \frac{1}{r}:= \left(\frac{1}{2}+\frac{1}{q}\right)/\left(q-2\right)
 =-\frac{1}{2} +\frac{1}{2-\frac{4}{q}}-\frac{1}{2q},\, 1 \leq q\leq 6.
 \label{3.23}
 \ene
  Note that  ${\cal C}$ goes
 from $B$ to $C$ and that for $0 \leq \frac{1}{q} \leq \frac{1}{6}$ the curve
 ${\cal C}$ is contained in the triangle with vertices $(A, B, C )$. Then
 ${\cal C} \subset {\cal D}$  for $0 \leq \frac{1}{q} \leq \frac{1}{6}$ and
 then taking $q=p-1$, we have that
 $e^{-itH} \in {\cal B}\left( X, L^r \left( L^{p+1}\right)\right)$
 for $5 \leq p \leq \infty$, with $r:= (p-1)/(1-d)$.

 \noindent{ \it Proof of Theorem 1.5 :}\, The proof of Theorem 1.1 of \cite{23}
 applies in our case with no changes.

 \noindent {\it Proof of Corollary 1.6 :} \, By Theorem 1.5 $S$ determines
 uniquely $S_L$. Let us denote
 \beq
 \hat{S_L}:= F S_L F^{\ast}
 \label{3.24}
 \ene
 and let $U$ be the following unitary operator from $L^2$ onto $L^2 (\ER^+ )
 \oplus L^2 (\ER^+ ):$
 \beq
 Uf(k):= \left\{ \begin{array}{c}f_1 (k)\\ f_2 (k)\end{array}\right\},
 \label{3.25}
 \ene

 where $f_1 (k):= f(k), k \geq 0 ,$ and $f_2 (k):= f(-k), k \geq 0.$
 Let us denote
 \beq
 \tilde{S}_L := U \hat{S}_L U^{\ast}.
 \label{3.26}
 \ene
 Pearson proved in Section 9.7 of \cite{24} that for $V$ bounded and with fast
 decay:
 \beq
 \tilde{S}_L  \left\{ \begin{array}{c}f_1 (k)\\ f_2 (k)\end{array}\right\}=
 \left[ \begin{array}{lr}T(k)& R_1 (k)\\R_2 (k) & T(k)\end{array}\right]
  \left[ \begin{array}{c}f_1 (k)\\ f_2 (k)\end{array}\right].
  \label{3.27}
  \ene
  Let us assume that $V \in L^1_{\delta}$ for some $\delta > 1$. Let $V_n \in
  C^{\infty}_0 , n=1,2,\cdots $ be such that
  \beq
  \lim_{n \rightarrow \infty}\left\| V_n -V \right\|_{L^1_{\delta}} =0.
  \label{3.28}
  \ene
  Let us denote by $S_{L,n}, T_n (k)$ and $R_{j,n}(k), j=1,2,$ the
  scattering operator, the transmission coefficient and the reflection
  coefficients corresponding to $V_n$. Then by the proof of Lemma 1 of \cite{1}
  and by equations (\ref{2.60}) to (\ref{2.62})
  \beq
  \lim_{n \rightarrow \infty}T_n (k)= T(k), \,\,
  \lim_{n \rightarrow \infty}R_{j,n}(k)=R_j (k), j=1,2.
  \label{3.29}
  \ene
  Moreover, by the stationary formula for the wave operators (see equation
  (12.7.5) of \cite{13}) and from the results in Chapter 12 of \cite{13}

  \beq
  s-\lim_{n \rightarrow \infty}S_{L,n}=S_L ,
  \label{3.30}
  \ene
  where the limit exists in the strong topology in $L^2$.
  Then by continuity (\ref{3.27}) is true also for
  $V\in L^1_{\delta}, \delta > 1$ and it follows that from $S_L$ we obtain
  the transmission coefficient and the reflection coefficients. But since $V$
  has no bound states one of the reflection coefficients uniquely determines
  $V$ ( see for example \cite{25}, \cite{26}, \cite{1}, \cite{me}
  \cite{28} or \cite{29}).

\noindent {\it Proof of Corollary 1.7:}\,The proof of Corollary 1.3 of
\cite{23}
applies in this case with no changes.

config: arxiv


TITLE: Problem with example of $G_\delta$ set
QUESTION [3 upvotes]: It was known that every closed set is a $G_\delta$ set. Now, I am trying to make an example about that.
This is my example:
$$[1,2]=\bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$$
I just can show that by visualization on a real line. But, when I try to proof that mathematically, I get stuck. This is my attempt:

$[1,2]\subseteq \bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$. I think for this part is clear, because $[1,2]\subseteq \bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$ for all $n\in\mathbb{N}$.
Conversely, i.e. $\bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)\subseteq [1,2]$. I try to use contraposition method, take any $x\notin [1,2]$, so $x<1$ or $x>2$. Then I will proof that $x\notin \bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)$, that means $x\in \bigg(\bigcap_{n=1}^\infty\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)\bigg)^c=\bigcup_{n=1}^{\infty}\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)^c$. Because $x \in  \bigcup_{n=1}^{\infty}\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)^c$, there exists $n\in \mathbb{N}$ sucht that $x\in\bigg(1-\frac{1}{n},2+\frac{1}{n}\bigg)^c$, so $x\in\bigg(-\infty,1-\frac{1}{n}\bigg]$ or $x\in\bigg[2+\frac{1}{n},\infty\bigg)$. Take any $\varepsilon>0$, by Archimedian Property, we have $\frac{1}{n}<\varepsilon$, then
$$x\leq 1-\frac{1}{n}<1-\varepsilon<1$$
or $$x\geq2+\frac{1}{n}$$
I get stuck on inequality of $x\geq2+\frac{1}{n}$, because it's impossible that $x\geq2+\frac{1}{n}<2+\varepsilon$.
So, I am not sure about my attempt. Is there another way to prove that? Thanks for any help.

REPLY [1 votes]: Okay so there is a slight mistake in the last steps you made. By the Archimedean property you get that $1/n<\epsilon$ but this means when you switch signs that you get $-\epsilon<-1/n$ and so the sequence of inequalities should really be $x\leq 1-\epsilon<1-\frac{1}{n}<1$. Also it's not clear what purpose $\epsilon$ is supposed to have. I am taking the liberty to assume you meant it to be $\epsilon = min\{\vert x-1 \vert, \vert x-2\vert\}$. Intuitively speaking this is the distance that $x$ has compared to the the set $[1,2]$. Then you need to break the proof down into two cases. The first case is the one in which $x<1$ and the second one when $x>2$. The first case besides the small error works since you showed if $x<1$ then by the definition of $\epsilon$ $x\leq 1-\epsilon<1-\frac{1}{n}$ and so it's for some $n\in\mathbb{N}$ in the the set $(-\infty,1-\frac{1}{n}]$. For the second case you consider the case $x>2$ so by the definition of $\epsilon$ you have $x\geq 2+\epsilon>2+\frac{1}{n}$ and so $x\in [2+\frac{1}{n},+\infty)$. You need to be careful not to mix the inequalities that appear in one of the cases with the inequalities in the other case.


TITLE: Is there a general way to integrate for the convolution sum of two random variables?
QUESTION [1 upvotes]: Say $f_X(x), f_Y(y)$ are two random independent PDF's with domains $D_X$ and $D_Y$.  Let $Z = X + Y$.  I know $f_Z(z) = f_X \star f_Y(z)$ where $\star$ means convolve.  How do I determine how to integrate this given only the domains of the random variables?
I am trying to solve this for $X - U(0,1)$ and $f_Y(y) = a^{-y}$ but I realized I don't understand how to choose the bounds in general.
For my example I would think to do $\int_0^1a^{-(z-x)}dx$ but I don't understand how $z$ would have anything to do with this.
Anyone have any ideas?

REPLY [1 votes]: $f_X(x)=1_{(0,1)}(x)$ and I'll assume $f_Y(y)=a^{-y}\cdot 1_{(0,\infty)}(y)$ and moreover $a=e$.
Then for $z\ge 0$,
$$f_Z(z)=\int_{-\infty}^{\infty} f_X(x)f_Y(z-x)\,dx=\int_0^1 a^{-(z-x)}\cdot 1_{(0,\infty)}(z-x)\,dx$$
$$=\int_0^{\min(z,1)} a^{-(z-x)}\,dx=e^{-z}\int_0^{\min(z,1)}e^{x}\,dx$$
$$=e^{-z}(e^{\min(z,1)}-1)=\begin{cases}1-e^{-z}&\text{if }z<1,\\ e^{-z}(e-1)& \text{if }z\ge 1.\end{cases}$$
So maybe the point that you were missing is how to work with the domains, like
$$\int_{-\infty}^\infty 1_{(a,b)}(x)1_{(c,d)}(x)f(x)\,dx=\int_{-\infty}^\infty 1_{(a,b)\cap(c,d)}(x)f(x)\,dx=\int_{\max(a,c)}^{\min(b,d)}f(x)\,dx$$
if $\max(a,c)<\min(b,d)$.


TITLE: How to sum a "complex" finite geometric series?
QUESTION [0 upvotes]: I'm trying to prove analytically the time complexity of a problem in "Cracking The Coding Interview".  The algorithm in question involves calling an O(1) operation on each node of a binary tree once for each level of nodes above it and every node in the tree is visited.  The cost of such an algorithm, it seems to me would be:
$2^1(1) + 2^2(2) + 2^3(3) + ... + 2^{log_2 n}log_2(n)$ where n is the number of nodes.
I'm not sure how to evaluate the above using the usual equation for finite geometric series: $a_1\frac{(1-r^n)}{(1-r)}$ where $a_1$ is the first value and r is the constant factor by which the series grows, because the series does not seem to grow at a constant factor; r grows with each step of the algorithm.  Does this sort of series have a name?

REPLY [1 votes]: Consider the geometric formula
$$\frac{r^n-1}{r-1} = 1+r+r^2+r^3+...+r^{n-1}$$
We take the derivative on both sides in terms of $r$:
$$\frac{(nr^{n-1})(r-1)-(r^n-1)}{(r-1)^2}=1+2r+3r^2+...+(n-1)r^{n-2}$$
We then multiply both sides by $r$ in order for the coefficient to match the exponent of $r$:
$$r\frac{(nr^{n-1})(r-1)-(r^n-1)}{(r-1)^2}=r+2r^2+3r^3+...+(n-1)r^{n-1}$$
Substituting $r=2$:
$$2[(n2^{n-1})-(2^n-1)]=2(1)+2^2(2)+2^3(3)+...+2^{n-1}(n-1)$$
At this point you're finished, but you can substitute $n=n+1$ so that $n$ equals the number of terms, and we can also simplify the LHS. The final formula is
$$2^{n+1}(n-1)+2$$where $n$ is the number of terms.


TITLE: whether electricity that comes in our home comes directly after generated or first stored and then distributed
QUESTION [3 upvotes]: Please give answer that, whether electricity in our home comes directly after it is generated e.g. at a dam or nuclear plant or first it is stored somewhere and then distributed as it is needed.

REPLY [1 votes]: There are some ways to store energy - http://en.wikipedia.org/wiki/Grid_energy_storage ; hydro storage was mentioned here, but theoretically it is also possible to store energy using flywheels - http://www.beaconpower.com/ (but  that specific company went bust:-)).


TITLE: Zero divisors in the hyperreal numbers
QUESTION [2 upvotes]: I am currently reading this introduction to hyperreal numbers. On the first page, to illustrate the problem with just taking hyperreal numbers to be sequences of reals, the following example is used:
$$ (0,1,0,1,...) \cdot (1,0,1,0,...) = (0,0,0,...),$$
demonstrating that this fails to be a field. Thus, the construction using ultrafilters and taking sequences mod a certain equivalence relation is motivated.
I see how this construction works for convergent series. However, in the example above, I fail to see that this actually solves our problem. It still seems that we have zero divisors. If $(0,1,0,1...) \sim (0,0,0,0...)$ and $(1,0,1,0,...) \sim (0,0,0,0,...)$, then the sets $2\mathbb{N}, 2\mathbb{N}-1$ are contained in the ultrafilter, and since ultrafilters are closed under intersection, their intersection $\emptyset$ is as well. Thus, these two elements must be nonzero and we have zero divisors.
There must be an error in my reasoning, but unfortunately I don’t see where. The introduction cited uses the transfer principle to show that the hyperreal numbers do indeed form a field, but this feels kind of unsatisfactory given this example.

REPLY [4 votes]: As you have noted, $2\mathbb{N}$ and $2\mathbb{N}-1$ are the complement of each other in $\mathbb{N}$. By definition of an ultrafilter, this means that precisely one of these lies in the ultrafilter; say $2\mathbb{N}$ does, for example. You are correct that this means that the equivalence class of $(0,1,0,1,\dots)$ is non-zero. But this also means that the equivalence class of $(1,0,1,0,\dots)$ is equal to the equivalence class of zero (why?). So, even though the product of these two equivalence classes is zero, this does not give an example of non-trivial zero divisors, since one of the elements in question is equal to zero. In short, the fact that $2\mathbb{N}$ and $2\mathbb{N}-1$ are complementary means that precisely one of the elements you describe is non-zero, not that both of them are.

REPLY [3 votes]: If $a$ and $b$ are complementary binary sequences, then
$$\begin{aligned}[]
 [a] = 0 &\iff \{n \mid a_n = 0\} \in\mathcal F
\iff \{n \mid b_n = 1\} \in\mathcal F
\iff [b]=1
\end{aligned}$$
So by design it is guaranteed that either $[a]=0$ and $[b]=1$ or $[a]=1$ and $[b]=0$.


TITLE: Is this right? Topology with closures
QUESTION [3 upvotes]: I want to show that (possibly) $$cl(A-B) = cl(A) - cl(B).$$
I know that $$cl(A-B) \subset cl(A) - cl(B).$$ already, but for the other inclusion I tried this.
Let $x \in  cl(A) - cl(B)$, so that for any neighborhood $U$of $x$, we know $U \cap (A - B) =\emptyset$
However, we also get the inclusion$$U \cap (A - B)  \subset U \cap cl(A-B) \subset U \cap cl(A) - cl(B).$$
Or did I get the other way around? Because I found the counterexample $A = [0,1)$ and $B = (-1,0)$.
So that $cl(A - B) = cl(\emptyset) = \emptyset$ and as well as $$cl(A) - cl(B) = (0,1)$$
So I think I got tihs backwards
Update: As pointed out, equality does'nt really hold and my inclusion argument is incorrect as I have incorrectly applied a techinuqe to the wrong problem. Thus, mynew quesiton is, is there ever inclusion (proper)?

REPLY [1 votes]: Consider $[0,1],\{1\} \subseteq \Bbb R$, where $\Bbb R$ carries the standard topology. Then $cl([0,1] - \{1\}) = [0,1] \neq [0,1) = cl([0,1]) - cl(\{1\})$,
so the result doesn't hold true.
edit: Oh, I somehow missed  doppz's comment. Please let me know whether you want me to delete this answer.
edit: Yes, there is inclusion. In fact: $cl(A)-cl(B) \subseteq cl(A-B)$.
proof:
If $x \not \in cl(A-B)$, there is an open neighborhood $x \in U$, such that $U \cap (A-B) = \emptyset$. On the other hand, if $x \not \in cl(B)$, there is an open neighborhood $x \in V$, such that $V \cap B = \emptyset$. Now observe that
$$U \cap (A-B) = (U \cap A)-(U \cap B)$$, so that $U \cap A \subseteq U \cap B$. This gives
$$V \cap U \cap A  \subseteq V \cap U \cap B \subseteq V \cap B = \emptyset$$, so $x \not \in cl(A)$.
Let's see what we proved:
If $x\not \in cl(A-B)$ and $x \not \in cl(B)$, then $x \not \in cl(A)$. So if $x \in cl(A)\setminus cl(B)$, then we must have that $x \in cl(A-B)$, i.e. $cl(A)-cl(B) \subseteq cl(A)-cl(B)$.
Note that this also shows $cl(cl(A)-cl(B)) \subseteq cl(A-B)$.


TITLE: determinant of matrix in polynomial
QUESTION [1 upvotes]: Suppose $X=\sqrt{k^2 - k_1^2}, Y= \sqrt{k^2 - k_2^2},Z=\sqrt{k^2 - k_3^2}, S=\sqrt{k^2 - k_4^2}, T= \sqrt{k^2 - k_5^2}$. \
And there is $5 \times 5 $ matrix as given below.
\begin{equation}
\begin{vmatrix}
a_{11} & a_{12}  & a_{13} & a_{14}*S & a_{15}*T\\
a_{21}*X & a_{22}*Y  & a_{23}*Z & a_{24} & a_{25}\\
a_{31}*X & a_{32}*Y  & a_{33}*Z & 0 & 0\\
a_{41} & a_{42}  & a_{43} & a_{44}*S & a_{45}*T\\
a_{51}*X & a_{52}*Y  & a_{53}*Z & a_{54}& a_{55}\\
\end{vmatrix} = 0,
\end{equation}
I have to find the determinant in terms of k with inter powers. By usual exapnsion method (squaring the terms), finding the expression for k is getting complex. Is there any other approach to find the expression for above determinant. And that expression should not contain fraction powers of k, means expression should have $X^2, Y^2,Z^2,S^2,T^2$ instead of $X, Y,Z,S,T$.

REPLY [1 votes]: Here is a special case of your matrix
$$
\begin{vmatrix}
1 & 0  & 0 & 0 & 0\\
0& 0  & 0 & 1 & 0\\
0 & 0  &  Z & 0 & 0\\
0 & 1  & 0 & 0 & 0\\
0 & 0  & 0 & 0& 1\\
\end{vmatrix}.
$$
It evaluates to $Z=\sqrt{k^2 - k_3^2}$. (Or is it $-Z$? No matter.) You can't achieve what you are asking for.


TITLE: How many $5$ digit natural numbers have at least one $5$ or one $7$?
QUESTION [2 upvotes]: There are $9 \cdot 10^4=90,000$ possible $5$ digit numbers.
Numbers with no $5$ is $8\cdot 9\cdot 9\cdot 9\cdot 9 = 52,488$.
Numbers with no $7$ is $52,488$.
Numbers with at least one $5$ is $90,000 - 52,488$
Similarly numbers with at least one $7$ is $90,000-52,488$
So answer is $2(37,512)=75,024$
But this answer is wrong. Please provide correct solution..

REPLY [4 votes]: Hint: Count the number of five digit-numbers that contain NO $5$'s and  no $7$'s; that gives us  $7\cdot 8^4$ such five-digit numbers.
The remainder ($90000 - 7\cdot 8^4$) will be the number of five digit numbers with at least one $5$ OR one $7$.