text:
\section{Introduction}
\label{sec:1}
In this paper, we investigate the global behavior of
certain quantum dynamical systems in the presence of magnetic field.
To describe the relevant equations,
introduce the magnetic Laplace  operator
$$
\Delta_{\vec{A}}= \sum\limits_{j=1}^n (i \partial_j+A_j)^2
$$
The magnetic Schr\"odinger  equation is
\begin{equation}
\label{eq:nac}
\left|
\begin{array}{l}
u_t - i (\Delta_{\vec{A}}+V) u=F,\\
u(0,x)=f(x)\in L^2({\mathbf R}^n)
\end{array}
\right.
\end{equation}
In the physically
important case of a real valued $\vec{A}$ and $V$, one has
conservation of charge, $\norm{u(t,\cdot)}{L^2}=\norm{u(0,\cdot)}{L^2}$.
More generally,  by  a result of Leinfelder and Simader, \cite{Leinfelder} if
$A\in L^4_{loc}({\mathbf R}^n)$ and $div(A)\in L^2_{loc}({\mathbf R}^n)$,   $V$ is
relatively bounded with bound less than one
 with respect to $\Delta$ , one has
that the operator $\Delta_{\vec{A}}+V$ is essentially
self-adjoint on $C_0^\infty({\mathbf R}^n)$.
In particular, the  spectrum is real and
one can define functional calculus.

In this paper, we shall be concerned mainly with the case of
{\it time dependent vector
potentials } $\vec{A}$, which are small and real-valued.
This  is
dictated by certain  partial differential equations,   appearing  naturally in
geometry and physics. More specifically, we have
in mind
 the Schr\"odinger map equation in Hodge
gauge \cite{Shatah}, \cite{Stefanov}, the Ishimori system,
\cite{KPV00}, \cite{nahmod}, the Maxwell-Schr\"odinger system,
\cite{Ginibre2},\cite{nakamura},  \cite{tsutsumi}  and
several  other models, related to
the Landau-Lifshitz theory of electromagnetism.

In the case of magnetic-free field ($\vec{A}=0$),
great progress has been made to address the question for global/local
existence and uniqueness
for solutions of
\eqref{eq:nac}, \cite{Cazenave}, \cite{Sulem}. In particular,
when $V$ is small and $n\geq 3$,
one can use the standard Strichartz estimates to
show by a perturbation
argument that the corresponding equation has an unique global
solution
under reasonable assumptions on
the right hand-side and the data $f$. In the same spirit,
one can obtain local well-posedness results for large $V$.

In the magnetics-free case, the Strichartz estimates
are well-known  and play a fundamental role in proving
the existence and uniqueness results alluded to  above. \\
Introduce
$$
\norm{u}{L^q_t L^r}:= \left(\int\limits_0^\infty \left(\int\limits_{{\mathbf R}^n} |u(t,x)|^r dx\right)^{q/r} dt\right)^{1/q}.
$$
for every pair $q,r\geq 1$ and similarly  the mixed Lebesgue spaces
$L^q_t L^{r_1}_{x_1}\ldots L^{r_n}_{x_n}$.

We say that  a pair of indices $(q,r)$  is {\it Strichartz admissible} if
$2\leq q,r \leq \infty$, $2/q+n/r=n/2$ and
$(q,r,n)\neq (2,\infty, 2)$.  Then, by a classical result of
Strichartz, \cite{Strichartz}, later improved by
Ginibre-Velo, \cite{Ginibre} and finally Keel-Tao, \cite{KeelTao}, we have
\begin{eqnarray}
\label{eq:200}
& & \norm{e^{ i t \Delta} f}{L^qL^r}\leq C\norm{f}{L^2}\\
& &
\label{eq:202} \|\int_0^t e^{ i (t-s) \Delta}
F(s,\cdot) ds\|_{L^qL^r}\leq C\norm{F}{L^{\tilde{q}'}L^{\tilde{r}'}},
\end{eqnarray}
where $(\tilde{q}, \tilde{r})$ is another Strichartz admissible pair and
$q'=q/(q-1)$.

Clearly \eqref{eq:200} and  \eqref{eq:202} are equivalent
to $
\norm{u}{L^qL^r}\leq C \norm{f}{L^2}+ \norm{F}{L^{\tilde{q}'}L^{\tilde{r}'}},
$
whenever $u$ is a solution to the free Schr\"odinger equation with initial data $f$ and forcing term $F$.

Another equivalent formulation is that there exists a constant $C$, so that
for all test functions
$\psi:$
$$
 \norm{\psi}{L^qL^r}\leq C(\norm{\psi(0,\cdot)}{L^2}+
\norm{(\partial_t\pm i\Delta)\psi}{L^{\tilde{q}'}L^{\tilde{r}'}}).
$$
 In the sequel, we will  make
extensive use of all these
points of view.

For  the case of small (but non-zero) potential $\vec{A}$, we have
$\Delta_{\vec{A}} =-\Delta+ 2 i \vec{A}\cdot \nabla +
(i div(\vec{A})+\sum_j A_j^2)\cdot=-\Delta+2 i \vec{A}\cdot\nabla  +
\textup{small potential}$, we can effectively treat the
 the magnetic Schr\"odinger equation in the form
\begin{equation}
\label{eq:1}
\left|
\begin{array}{l}
\partial_t u - i \Delta u+\vec{A}(t,x) \cdot \nabla u = F \quad (t,x)\in
{\mathbf R}^+\times {\mathbf R}^n\\
u(0,x)=f(x)
\end{array}
\right.
\end{equation}
where the terms in the form $(i div(\vec{A})+\sum\limits_j A_j^2)u$
are subsumed in the right hand side.

Next, we explain the relevance of the magnetic Strichartz estimates to the
Cauchy problem for
\begin{equation}
\label{eq:477}
\partial_t u- i \Delta u+\vec{A}(u)\cdot  \nabla u= F(u).
\end{equation}
If $\vec{A}=0$, we can clearly use  \eqref{eq:200}, \eqref{eq:202} to
set up an iteration scheme for the {\it semilinear problem }\eqref{eq:477}
in a ball $B=B(0,R)$ in the  ``Strichartz space''
$\cap_{(q,r)}  L^q(0,T) L^r$ to solve for arbitrary $L^2$
data, provided one can show $\norm{F(u)}{L^1(0,T) L^2}\leq
T^\delta M(\norm{u}{L^q(0,T)L^r})$ for some bounded
function $M$.  Choosing $R\sim \norm{f}{L^2}$ and
$T: T^\delta M(R)<< R$ makes such scheme successful to show that the solution
exists for some time $T=T(\norm{f}{L^2})$. There are of course,
remaining unresolved by this approach, including
globality of such solutions\footnote{Note that such approach usually
guarantees the global existence of small solutions.},
smoothness of the solution  etc.

Clearly, to study \eqref{eq:477} with $\vec{A}\neq 0$, one cannot use
the standard Strichartz estimates, by the obvious derivative loss.
One of the goals of this paper is to derive global scale invariant Strichartz
estimates under appropriate smallness assumptions on the vector
potential $\vec{A}$. The pioneering work of Barcelo-Ruiz-Vega,
\cite{barcelo} has addressed some of these issues\footnote{Strictly speaking,
the
results in \cite{barcelo} yield scale invariant
 smoothing estimates, but standard methods allows one to derive
Strichartz estimates from the results there.}, but was
restricted to (essentially) radial vector potentials $\vec{A}$.
To the best of our knowledge, the results in Theorem \ref{theo:pisa} below
 are the first
global estimates of such type for Schr\"odinger equations,
that work for general non-radial potentials $\vec{A}$.

Let us explain the general scheme for applying such Strichartz
estimates to concrete quasilinear PDE's.
Suppose, we have such estimates for the {\it linear gradient
Schr\"odinger equation} \eqref{eq:1}, provided $\norm{\vec{A}}{Y_T}\leq \varepsilon$,
 for some concrete
Banach space $Y_T$ appearing in Section \ref{sec:09}.
 We apply the magnetic
Strichartz estimates  to the nonlinear
equations of the type \eqref{eq:477} as follows. For small
initial data $f$, run an iteration scheme in the ball
$B_X(0,R)$ in an  appropriate
Strichartz space\footnote{Usually one solves the equations
\eqref{eq:477} for data $f$ in some smooth Sobolev space $H^s$  and
very often
in Besov variants of the Strichartz space.}  $X$,
see Section \ref{sec:11} for precise definitions.
This is
possible if
\begin{itemize}
\item one can ensure {\it a priori} the smallness condition
$\norm{\vec{A}(u)}{Y_T}\leq \varepsilon$ for {\it all functions} $u$, that are solutions to \eqref{eq:1}
satisfying $\norm{u}{X}<R$ and for all times $T\leq T_0=T_0(R)\leq \infty$.
\item $\norm{F(u)}{L^1(0,T) L^2}\leq R/2$, whenever $u$ is a function with
$\norm{u}{X}<R$ and for all times $T\leq T_0=T_0(R)\leq \infty$.
\end{itemize}

\subsection{Strichartz estimates for the magnetic Schr\"odinger operator}
\label{sec:09}
The existence and uniqueness problem for \eqref{eq:1} has been studied
extensively by
many authors in the mathematics literature.
We should first point to the pioneering work of Doi, \cite{Doi1}, \cite{Doi2},
who has devised a
method to obtain solutions via energy estimates. The approach then relies
on cleverly exploiting
the properties of pseudodiferential operators of order zero
to obtain {\it a priori} control of
$\norm{u(t,\cdot)}{L^2}$ in terms of
$\norm{f}{L^2}$ and $\norm{F}{L^1_T L^2}$. \\
 We also mention the far
reaching generalization of
Doi's results, due to Kenig-Ponce-Vega, \cite{KPV98}\footnote{In fact, in the
proof of the Strichartz
estimates for \eqref{eq:1}, we shall need a particular
local existence result from \cite{KPV98}, see Section \ref{sec:4}.}.
The authors have considered more general equations and
were  able to
derive {\it a priori} estimates for the $L^2$ norms of the solution as well
as the
validity of
a local smoothing effect, phenomenon well-known for the potential free case.

Note that \eqref{eq:1} has the important
scaling invariance
$u\to u^\lambda(t,x)=u(\lambda^2 t, \lambda x)$,
$A\to A^\lambda(t,x)= \lambda A(\lambda^2 t, \lambda x)$,
$F\to F^\lambda(t,x)=\lambda^2 F(\lambda^2 t, \lambda x)$.  That is,
whenever $(u,A, F)$ satisfy \eqref{eq:1}, so does
$(u^\lambda, A^\lambda, F^\lambda)$ with initial data
$f^\lambda(x)=f(\lambda x)$. \\
We describe the space $Y$ of  vector potentials $\vec{A}$, so
that the corresponding magnetic \\ Schr\"odinger operator satisfies the
Strichartz estimates. Let $SU({\mathbf R}^n)$ be the special unitary group acting on ${\mathbf R}^n$ and $x(t):\mathbf R^1_+\to {\mathbf R}^n$ be arbitrary measurable function. Define
\begin{eqnarray*}
& &\norm{\vec{A}}{Y_0} := \norm{\nabla \vec{A}}{L^1L^\infty}+
\norm{\vec{A}}{L^2L^\infty} +
(\sum\limits_l
2^{2l(1+h)}\norm{A_l}{L^1 L^{n/h}}^2)^{1/2}\\
& &\norm{\vec{A}}{Y_1}:=  \sum\limits_k 2^{k(n-1)} \norm{A_k}{L^\infty_t L^1_x}\\
& &\norm{\vec{A}}{Y_2}:= \sum\limits_k 2^{k(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ \sum\limits_k 2^{k(n-5)/2}\sup\limits_{U\in SU(n), x(t)}
\| (|\partial^2 A_{k}|+|\partial_t A_k|)(t,x(t)+Uz)\|_{L^\infty_t L^2_{z_2,
\ldots, z_n}L^1_{z_1}})\\
& &
\norm{\vec{A}}{Y_3} := \sum\limits_k 2^{k(n+3)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| A_{k}(t,x+ Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+\\
& &+
\sum\limits_k 2^{k(n-1)/2}\sup\limits_{U\in SU(n), x(t)}
\| (|\partial^2 A_{k}|+|\partial_t A_k|)(t,x(t)+Uz)\|_{L^1_t L^2_{z_2,
\ldots, z_n}L^1_{z_1}}).
\end{eqnarray*}
In the case $n\geq 4$, we can replace $Y_1$ by a
bigger space $\tilde{Y_1}$
(with the smaller norm)
$$
\norm{A}{\tilde{Y}_1}=\sum\limits_k 2^{k(n-1)/p_0}
\sup\limits_{U\in SU(n), x}\|A_k(t,x+Uz)\|_{L^\infty_t L^{p_0}_{z_2,
\ldots, z_n}L^1_{z_1}},
$$
for some $p_0: p_0<(n-1)/2$.
\begin{theorem}
\label{theo:pisa}
Let $n\geq 2$. Then, there exists an $\varepsilon=\varepsilon(n)>0$, so that whenever
$\vec{A}:\mathbf R^1\times{\mathbf R}^n\to {\mathbf R}^n$ is a
real-valued
vector potential with $\|\vec{A}\|_{Y_0\cap Y_1\cap Y_2\cap Y_3}
\leq \varepsilon$, (which can be relaxed to
$\|\vec{A}\|_{Y_0\cap \tilde{Y}_1\cap Y_2\cap Y_3}
\leq \varepsilon$, if $n\geq 4$)  \eqref{eq:1}
 has an unique global solution, whenever the initial data
$f\in L^2({\mathbf R}^n)$ and the forcing term $F\in L^1L^2$.
In addition, there exists a constant $C=C(n)$,
so that the {\it a priori} estimate
\begin{equation}
\label{eqn:478}
\sup\limits_{q,r - Str.}\norm{u}{L^qL^r}\leq
C(\norm{f}{L^2}+
\norm{F}{L^1L^2}).
\end{equation}
holds true\footnote{For the two dimensional case, the constant $C$ does
also  depend
on $(q,r)$. More specifically the constant blows up as $(q,r)\to
(2,\infty)$.}. Moreover, for every $\psi\in\mathcal S$
\begin{equation}
\label{eq:1009}
\sup\limits_{q,r - Strichartz admissible} \norm{\psi}{L^qL^r}\leq
C(\norm{\psi(0,\cdot)}{L^2}+\norm{(\partial_t - i
\Delta+\vec{A}\cdot \nabla) \psi}{L^1L^2}).
\end{equation}
One also
 has the $l^2$ Besov space version
\begin{equation}
\label{eq:1011}
(\sum\limits_k \norm{\psi_k}{L^qL^r}^2)^{1/2}
\leq C(\norm{\psi(0,\cdot)}{L^2}+
\norm{(\partial_t - i \Delta+\vec{A}\cdot \nabla) \psi}{L^1L^2}).
\end{equation}
\end{theorem}
{\bf Remark}
\begin{itemize}
\item Note  that for all $Y_j$, $j=0,1,2,3 $ we have that
$\|\vec{A}^\lambda\|_{Y_j}=\|\vec{A}\|_{Y_j}$, that is the spaces are scale
invariant with respect to the natural scaling $A\to A^\lambda$.
\item In the case $n=1$,  the
theorem holds as well.
Our results are
however far from optimal, as shown recently by Burq-Planchon, \cite{Burq}.
It seems that in the one dimensional case one only
needs to require
$\sup_t \norm{A(t, \cdot)}{L^1(\mathbf R^1)}<\infty$, if $A$
is a real-valued potential.
\item If $A:\mathbf R^1\times \mathbf R^1\to \mathcal C$ is complex valued,
and satisfies  \\ $\sup_{t,x}|\int_{-\infty}^x A(t,y)dy|<\varepsilon$ and
$\|\int_{-\infty}^x (\partial_t - i \partial_y^2) A(t,y)dy\|_{L^1_t L^\infty_x}<\varepsilon$,
one has the results of Theorem \ref{theo:pisa}. This is
shown in \cite{Stefanov2}, together with some applications and uniqueness
issues,  consult \cite{Stefanov2} for more details.
Note that by recent examples on  ill-posedness for
derivative Schr\"odinger equations
in $\mathbf R^1$ (due to M. Christ, \cite{Crist20}), some
smallness assumptions are necessary even for a local well-posedness.
\end{itemize}
\subsection{Some corollaries}
We present some corollaries of Theorem \ref{theo:pisa}.
Observe that by Bernstein inequality (Lemma \ref{le:yut}), one can bound
\begin{eqnarray*}
& & \norm{A}{Y_0\cap Y_1\cap Y_2\cap Y_3}\leq C_n
(\sum\limits_k 2^{k(n-1)} \norm{A_k}{L^\infty L^1}+
2^{k(n-3)} \norm{\partial_t A_k}{L^\infty L^1})+ \\
& &+C_n (\sum\limits_k 2^{k(n+1)} \norm{A_k}{L^1 L^1}+
2^{k(n-1)} \norm{\partial_t A_k}{L^1 L^1}).
\end{eqnarray*}
We thus have
\begin{corollary}
\label{cor:90}
There exists a small positive $\varepsilon>0$, so that whenever
a real valued vector potential  $\vec{A}$ satisfies
$$
\sum\limits_k 2^{k(n-1)} \norm{A_k}{L^\infty L^1}+
2^{k(n-3)} \norm{\partial_t A_k}{L^\infty L^1}+  2^{k(n+1)} \norm{A_k}{L^1 L^1}+
2^{k(n-1)} \norm{\partial_t A_k}{L^1 L^1}\leq \varepsilon,
$$
the conclusions \eqref{eqn:478}, \eqref{eq:1009} and \eqref{eq:1011} hold true.
\end{corollary}
\noindent For the case of {\it time independent magnetic  potential} $\vec{A}$, we can formulate the following immediate corollary of Theorem \ref{theo:pisa}.
\begin{corollary}
\label{cor:1}
Let $n\geq 2$. Then there exists an
$\varepsilon=\varepsilon(n)$, so that
the magnetic Schr\"odinger equation \eqref{eq:1} has an
unique global solution, provided $\vec{A}=\vec{A}(x)$ is real-valued
vector function  and
\begin{eqnarray*}
& &\sum\limits_k 2^{k(n+1)} \norm{A_k}{L^1}
<\infty \\
& & \sum\limits_k 2^{k(n-1)} \norm{A_k}{L^1_x}\leq \varepsilon.
\end{eqnarray*}
Moreover, the solution satisfies
$$
\sup\limits_{q,r - Str.}\norm{u}{L^q(0,T) L^r}\leq
C(T,n)(\norm{f}{L^2}+\norm{F}{L^1 L^2}).
$$
\end{corollary}
\noindent Corollary \ref{cor:1} follows easily by just applying Corollary
\ref{cor:90} with a vector potential of the form $\tilde{A}(t,x):=
\vec{A}(x) \chi(t/\delta)$
for some appropriate smooth cutoff function $\chi$ and a small $\delta$.
This will produce a solution in
 a small time interval, say $(0,\delta/2)$, which is iterated and so on.
 The smallness of the potential $\tilde{A}$ is achieved by the smallness of
$\delta$ (used to satisfy the requirements $A\in Y_0\cap Y_2\cap Y_3$)
and by Sobolev embedding and the condition
$\sum\limits_k 2^{k(n-1)} \norm{A_k}{L^1_x}\leq \varepsilon$ in the case $A\in Y_1$.
Such a  proof provides an upper bound $C(t,n)\sim C^{T/c\varepsilon}$,
which is not  optimal in general.
\subsection{Strichartz estimates with derivatives}
The Strichartz estimates described in Theorem \ref{theo:pisa}
can be of course  extended to control the norms of the
solution $u$ in Besov type norms involving derivatives. One way to do that
is  considering the Littlewood-Paley reduction of the equation to
a fixed frequency $k$, applying the regular Strichartz estimates
(either \eqref{eq:1009} or \eqref{eq:1011} with
appropriate $p_1, p_2, q_1, q_2$), then multiplying by the
corresponding power of $2^{ks}$ and square summing in $k$. The result is
\begin{theorem}
\label{theo:3}
Let $n\geq 2$ and $\vec{A}$ satisfies
the assumptions in Theorem \ref{theo:pisa}. Then there exists a
constant $C=C(n)$ ($C=C(n,q,r)$, if $n=2$), so that for
every $s>0$, initial data $f\in \dot{H}^s$ and
forcing term $F\in L^1_t\dot{H}^s$, the global solution $u$
of \eqref{eq:1} satisfies
\begin{equation}
\label{i:1}
(\sum\limits_k 2^{2ks} \norm{u_k}{L^q_t L^r_x}^2)^{1/2}\leq C (\norm{f}{\dot{H}^s}+
\norm{F}{L^1_t\dot{H}^s})+C \norm{\nabla u}{L^2_t L^{n}_x}
(\sum\limits_k 2^{2ks} \norm{A_k}{L^2_t L^{2n/(n-2)}_x}^2)^{1/2}.
\end{equation}
for every Strichartz admissible pair $(q,r)$.
\end{theorem}
\subsection{Strichartz estimates in
 $L^2_t L^{2(n-1)/(n-3)}_{x_2, \ldots, x_n}
L^2_{x_1}$}

We  present an extension of Theorem \ref{theo:pisa},
which allows us  to control a larger set of norms.
\begin{proposition}
\label{prop:3}
Let $n\geq 4$ and $\vec{A}$ satisfies
the assumptions in Theorem \ref{theo:pisa}.
Then there exists a constant $C_n$, so that
the solutions of \eqref{eq:1}, satisfy
\begin{equation}
\label{e:7}
(\sum\limits_k \sup\limits_{U\in SU({\mathbf R}^n), x(t)}
\norm{u_k(t,x(t)+U z)}{L^2_tL^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}
L^2_{z_1}}^2)^{1/2}
\leq C(\norm{f}{L^2}+
\norm{F}{L^1L^2}).
\end{equation}
For $n=3$, take any $(q,r):1/q+1/r=1/2$ and $q>2$. Then  there exists
a constant $C_q$ (which may blow up as $q\to 2$), so that
$$
(\sum\limits_k \sup\limits_{U, x(t)}
\norm{u_k(t,x(t)+U z)}{L^q_t L^{r}_{z_2,z_3}
L^2_{z_1}}^2)^{1/2}
\leq C_q(\norm{f}{L^2}+
\norm{F}{L^1L^2}).
$$
For $n=2$,
$$
(\sum\limits_k \sup\limits_{U, x(t)}
\norm{u_k(t,x(t)+Uz)}{L^4_t L^{\infty}_{z_2}
L^2_{z_1}}^2)^{1/2}
\leq C(\norm{f}{L^2}+
\norm{F}{L^1L^2}).
$$
\end{proposition}
We also have a generalization of Proposition \ref{prop:3} to
the setting of Theorem \ref{theo:3}, that is involving derivatives.
Namely, say when $n\geq 4$, one has the {\it a priori} estimate
\begin{eqnarray*}
& &
(\sum\limits_k \sup\limits_{U\in SU({\mathbf R}^n), x(t)}
2^{2ks}\norm{u_k(t,x(t)+U z)}{L^2_tL^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}
L^2_{z_1}}^2)^{1/2}
\leq C(\norm{f}{\dot{H}^s}+
\norm{F}{L^1\dot{H}^s})+ \\
& &+C \norm{\nabla u}{L^2_t L^{n}}
(\sum\limits_k 2^{2ks} \norm{A_k}{L^2_t L^{2n/(n-2)}_x}^2)^{1/2}.
\end{eqnarray*}

Such results are needed to connect the ``solutions space''
with the space of admissible  vector potentials $Y$. More precisely,
in the applications, we have $A=A(u)$, where
the relation is usually in the form
$A=Q(u, \bar{u})$, where $Q$ is a bilinear (or multilinear)
form acting on the solution and its conjugate. For example, in the
Schr\"odinger map case (see Section \ref{sec:smaps} below),
we have schematically $A=|\nabla|^{-1} (u \bar{u})$
and for the Maxwell-Schr\"odinger system (see Section \ref{sec:MS} below)
we have $A=\square^{-1}(u \nabla u)$.

Ignoring the derivatives (and the invertibility of $\square$ in the MS case)
for a second,
we see that by H\"older's inequality, to have estimates of the form
$$
\norm{A(u)}{Y_2\cap Y_3}\leq C \norm{u}{X}^2
$$
where $X$ is the solutions space, we must rely on mixed Lebesgue
estimates like the one in Theorem \ref{theo:3}. Moreover, $X$ must be intersected with a space given by one of the norms involved  in the left hand side of \eqref{e:7}.

\subsection{Applications to Schr\"odinger maps}
\label{sec:smaps}
In this section, we present a global regularity type result for the
so-called Modified Schr\"odinger map system (MSM), which was derived in
\cite{Stefanov}.  According to Theorem 2.2, \cite{Stefanov},
the Schr\"odinger map problem, with with target $\mathbf S^{n-1}$,
was shown to be equivalent (modulo a Lorentz type gauge transformation),
to a {\it overdetermined} system of Schr\"odinger equations
with attached consistency conditions.

We will not discuss here, whether the MSM and
 the Schr\"odinger map problem are equivalent,
and how the (properties of the) solutions to one
relate to the solutions of the other,
with the acknowledgement that these are by no means unimportant
or irrelevant issues. We will concentrate instead on the
question of existence of solutions for MSM, which is
mathematically more tractable. \\
Consider the MSM, which  takes the form
\begin{eqnarray*}
\pp{u_j}{t} &=& i\Delta u_j-2\sum\limits_{k=1}^n  a_k \pp{u_j}{x_k}-
\left(\sum\limits_{k=1}^n  a_k^2\right)u_j+ \\
&+& 2(\sum\limits_{k=1}^n \ Im (\overline{u_j}u_k)u_j) -i a_0u_j \quad\quad\quad j=1,\ldots,n;
\end{eqnarray*}
where
\begin{eqnarray*}
a_k &=& \sum\limits_{l=1}^n \pp{\kappa_{lk}}{x_l};\\
d\ \kappa &=& 0;\\
\Delta \kappa_{kj} &=& -4\ Im(u_k\overline{u_j}) \quad\quad\begin{array}{l}
j=0,1,\ldots,n\\
k=1,\ldots,n,
\end{array}; \\
\Delta a_0 &=& -4 \sum\limits_{j=1}^n\sum\limits_{k=1}^n \left[\pp{}{x_k}\pp{}{x_j}
Re(u_k\overline{u_j})-\f{1}{2}\left(\pp{}{x_k}\right)^2u_j\overline{u_j}
\right].
\end{eqnarray*}
In short, we will consider
the following system of Schr\"odinger equations
\begin{equation}
\label{eqnm:1}
\left|
\begin{array}{l}
\partial_t u - i \Delta u+ A(u)\cdot \nabla u =N(u) \\
u(0,x)=f,
\end{array}\right.
\end{equation}
where $A$ is a real-valued vector potential
($A=(a_1, \ldots, a_n)$ in MSM),
\begin{eqnarray*}
& & A(u)=\partial^{-1} Q_1(u,\bar{u}) \\
& & a_0(u)= \partial^2\Delta^{-1} Q_2(u, \bar{u})\\
& & N(u)= Q_3(u,u, \bar{u})+Q_4(A, A, u)+Q_5(a_0, u).
\end{eqnarray*}
Here, $Q_1, \ldots, Q_5$ are multilinear forms of their arguments, i.e. \\
$Q_j(u^1, \ldots, u^r)=\sum_j  c_{k_1, \ldots, k_r}^j u^{k_1}...
u^{k_r}$ for some constants $c$.
We have also  used the notation $\partial^s$ to denote a multiplier type operator,
whose smooth symbol satisfies  $|s(\xi)|\sim |\xi|^s$. \\
All the results that we obtain for \eqref{eqnm:1} cover the MSM system,
which is our main motivation.
\begin{theorem}(Global regularity of MSM in high dimensions)
\label{theo:MSM} Let $n\geq 6$, $s_0=n/2-1$ and $s\geq (n+1)/2$.
Then, there exists
$\varepsilon>0$, so that whenever $g\in H^s$, with
$\norm{g}{\dot{H}^{s}}\leq 1$,
the solution to \eqref{eqnm:1} with initial data $f=\varepsilon g$
exists globally and satisfies
\begin{eqnarray*}
& &\sup\limits_{t} \norm{u(t, \cdot)}{H^s}\leq C \varepsilon
\end{eqnarray*}
for some constant $C$ depending only on the dimension and $s$.

For $n=5$, there is an appropriate Besov spaces analogue.
\end{theorem}

\subsection{Maxwell-Schr\"odinger system}
\label{sec:MS}
Another system of nonlinear PDE's, for which the Strichartz estimates of
 Theorem \ref{theo:pisa} are applicable is the Maxwell-Schr\"odinger
system. That is
\begin{equation}
\label{eqnm:2}
\left|
\begin{array}{l}
i \partial_t u + \Delta_A u= g(|u|^2) u \\
\square A ={\mathbf P} Im(\bar{u} \nabla_A u) ,
\end{array}\right.
\end{equation}
where ${\mathbf P}$ is the
Leray projection onto the divergence free vector fields and
 $g(|u|^2)$ is either the Hartree
interaction
$g(|u|^2):= \int G(x-y) |u|^2 (y)dy$ or\footnote{Here $G$ is the
Green function in $n$ dimension.}  simply $g(|u|^2)=|u|^2$.

 This system was studied  by Tsutsumi,  \cite{tsutsumi}, where
he  constructs the wave operator on
a class of small scattered states. In particular,
Tsutsumi showed global existence for a particular class of small data.

Recently, Nakamura and Wada,  \cite{nakamura} have considered the
MS system \eqref{eqnm:2}as well. They have obtained local well-posedness
with data $u_0 \in H^{5/3}(\mathbf R^3)$  by using energy estimates approach.
For related results and recent developments for \eqref{eqnm:2}, one might consult the recent work of Ginibre and Velo, \cite{Ginibre2}.

We do not state any precise results here for the
Maxwell-Schr\"odinger system, but it is clear that a variant of Proposition \ref{prop:3} for solutions of the linear wave equation (among other things)
will be required.
These issues will be addressed in a forthcomming paper, \cite{Stefanov3}.

A short outline of the paper is as follows. In Section \ref{sec:prelim},
we give some definitions from harmonic analysis as well as some
facts from the abstract Strichartz estimates theory
due to Keel and Tao, \cite{KeelTao}.
In Section \ref{sec:4},
we give some classical energy estimates and Littlewood-Paley reductions,
which reduce the problem to the existence of parametrix
construction.
In Section \ref{sec:par},
we motivate and construct the parametrix and then we prove some of
its main properties.
In Section \ref{sec:98}, we describe an important
angular decomposition for the phase of the parametrix and as a corollary
we show the  crucial pointwise estimates,
which are used throughout in the sequel.
In Section \ref{sec:6}, we show that the parametrix satisfies
 $L^q_t L^r_x$ estimates.
In Section \ref{sec:7}, we show that the
parametrix almost satisfy the magnetic Schr\"odinger equation.
In Section \ref{sec:8}, we prove the
Strichartz estimates stated in Proposition \ref{prop:3}. In Section \ref{sec:11}, we show the global regularity for the modified Schr\"odinger maps.
Some of the
technical lemmas used in Sections \ref{sec:par} and \ref{sec:7}
are formulated  and proved in  the Appendix.
\section{Preliminaries}
\label{sec:prelim}
\subsection{Fourier transform and Littlewood-Paley projections}
\noindent
Define the Fourier transform  by
$$
\hat{f}(\xi)=\int\limits_{{\mathbf R}^n} f(x)e^{-2\pi i x\cdot  \xi} dx
$$
and hence
$$
 f(x)=\int\limits_{{\mathbf R}^n} \hat{f}(\xi)e^{2\pi i x\cdot  \xi} d\xi
$$
Introduce a
positive, smooth and  even function $\chi:\mathbf R^1\to \mathbf R^1$, supported in
 $\{\xi:|\xi|\leq 2\}$ and $\chi(\xi)=1$ for all
$|\xi|\leq 1$.
 Define $\varphi(\xi)=\chi(\xi)-\chi(2\xi)$,
which is supported in the annulus $1/2\leq |\xi|\leq 2$. Clearly
$\sum_{k\in \mathcal Z} \varphi(2^{-k} \xi)=1$ for all $\xi\neq 0$. \\
The $k^{th}$ Littlewood-Paley projection is defined as a multiplier
type operator by \\
$\widehat{P_k f}(\xi)=\varphi(2^{-k}\xi) \hat{f}(\xi)$.
Note that the kernel of $P_k$ is integrable, smooth and real
valued for every $k$.
In particular, it commutes with differential operators. \\
Also of interest will be the properties
of products under the action of $P_k$. We have that
for any two (Schwartz )
functions $f, g$
\begin{eqnarray*}
P_k( f g) &=& \sum\limits_{l\geq k-2} P_k ( f_l g _{l-2\leq \cdot \leq l+2}) +
\textup{symmetric term}+ \\
&+& P_k( f_{\leq k-4} g_{k-1\leq \cdot\leq  k+1}) +\textup{symmetric
term}=\\
&=& f_{\leq k-4} g_k + [P_k, f _{\leq k-4}] g_{k-1\leq \cdot k+1} +
\textup{symmetric terms} \\
&+& \sum\limits_{l\geq k-2} P_k ( f_l g _{l-2\leq \cdot \leq l+2}) +
\textup{symmetric term}.
\end{eqnarray*}
\begin{lemma}(Bernstein inequality)
\label{le:yut}
Let $f$ be Fourier supported in a rectangle  $Q\subset {\mathbf R}^n$.
Then for every $1\leq p\leq q\leq \infty$, one has
$$
\norm{f}{L^q}\leq C_n|Q|^{1/p-1/q}\norm{f}{L^p}
$$
If $Q={\xi: |\xi|\sim 2^k}$, one can extend to
 mixed $L^pL^q$ spaces. Suppose $p_1>p_2\geq r$.
Then
$$
\norm{f}{L^{p_1}_{x_2, \ldots, x_n} L^{r}_{x_1}}\leq
C_n 2^{k(n-1)(1/p_2-1/p_1)}
\norm{f}{L^{p_2}_{x_2, \ldots, x_n} L^{r}_{x_1}}.
$$
\end{lemma}
\begin{proof}
The first statement is standard. \\
For the second statement, it is equivalent to the boundedness of \\
$P_k:L^{p_2}_{x_2, \ldots, x_n} L^{q_2}_{x_1}\to L^{p_1}_{x_2,
\ldots, x_n} L^{q_1}_{x_1}$ with bound $ C_n 2^{k(n-1)(1/p_2-1/p_1)}$.
One can rescale to the case $k=0$,
since these estimates are scale invariant. Since $P_0$ has integrable
kernel, we have $P_0:L^q_{x_2, \ldots, x_n} L^r_{x_1}\to L^q_{x_2,
\ldots, x_n} L^r_{x_1}$ for every $1\leq q,r\leq \infty$.

On the other hand, an application of the Bernstein inequality
in the $(n-1)$ variables $x_2, \ldots, x_n$
\begin{eqnarray*}
& & \norm{P_0 u}{L^\infty_{x_2, \ldots, x_n} L^1_{x_1}}\leq
\norm{P_0 u}{ L^1_{x_1} L^\infty_{x_2, \ldots, x_n}}\leq
\norm{P_0 u}{ L^1_{x_1} L^1_{x_2, \ldots, x_n}}\leq
\norm{u}{  L^1_{x_2, \ldots, x_n}L^1_{x_1}}.
\end{eqnarray*}
A complex interpolation between the last estimate and
$P_0:L^qL^1\to L^qL^1$ (for every $1\leq q\leq \infty$)
yields $P_0:L^pL^1\to L^{\tilde{p}}L^1$, whenever
$1\leq \tilde{p}<p\leq\infty$. Interpolation between the last estimate and
$p_0:L^\infty L^\infty\to L^\infty L^\infty$ yields
$P_0: L^{rp}_{x_2, \ldots, x_n} L^r_{x_1}\to
 L^{r\tilde{p}}_{x_2, \ldots, x_n} L^r_{x_1} $ for every
$r\leq \tilde{p}<p\leq\infty 1$. This is the second
statement  of Lemma \ref{le:yut} for an appropriate choice of $p, \tilde{p}$.
\end{proof}
We  also need the following technical lemma in the sequel
\begin{lemma}
\label{le:sequences}
Let $\{a_l\}$, $\{b_l\}$ are two sequences and $h>0$. Then
$$
\left(\sum\limits_k 2^{2 h  k} \left(\sum\limits_{l\geq k-2} 2^{-h l} a_l b_l \right)^2\right)^{1/2}\leq
C_h \norm{a}{l^\infty} \norm{b}{l^2}.
$$
\end{lemma}
\begin{proof}
Fix the sequence $\{a_l\}$ and consider the linear operator
(mapping sequence into a sequence)
$$
(Tb)_k := 2^{\varepsilon(k-l)} \sum\limits_{l\geq k-2} a_l b_l
$$
We will show that $T:l^1\to l^1$ and $T:l^\infty\to l^\infty$. Indeed,
\begin{eqnarray*}
& &\norm{Tb}{l^1}\leq \sum\limits_l |a_l|| b_l| \sum\limits_{k\leq l+2} 2^{\varepsilon(k-l)}\lesssim \varepsilon^{-1}
\norm{a}{l^\infty}\norm{b}{l^1},\\
& & \norm{Tb}{l^\infty}\leq \sup\limits_k \sup\limits_l|a_l|  \sup\limits_l|b_l|
\sum\limits_{l\geq k-2} 2^{\varepsilon(k-l)}\lesssim \varepsilon^{-1}
\norm{a}{l^\infty}\norm{b}{l^\infty}.
\end{eqnarray*}
It follows that for $1\leq p\leq \infty: T:l^p\to l^p$ with norm no bigger
 than $C_\varepsilon\norm{a}{l^\infty}$,
hence the statement of the lemma.
\end{proof}
\subsection{Keel-Tao theory}
It is well-known that decay and energy estimates
imply Strichartz estimates in the context of various  dispersive
equations. We would like to state an abstract result due to M. Keel
and T. Tao, \cite{KeelTao}, which proved out to be
very usefull in this context. Let us recall, that this method
in conjunction with the Hausdorf-Young inequality was
used by Ginibre and Velo in  their proof of the Strichartz
estimates for the linear
Schr\"odinger equation away from the endpoint.

The abstract version
of Keel and Tao  has the (somewhat)
unexpected consequence that the endpoint Strichartz estimate follows only
from decay and energy estimates.

Let $H$ be a Hilbert space, and $(X, dx)$ be a measure space and
$U(t):H\to L^2(X)$ be a bounded operator. Suppose that $U$ satisfies
\begin{eqnarray}
\label{e:100}
\norm{U(t) f}{L^2_x}\leq C\norm{f}{H} \\
\label{e:101} \norm{U(t) U(s)^*f}{L^\infty_x}\leq  C
|t-s|^{-\sigma}\norm{f}{L^1}.
\end{eqnarray}
Suppose also that $(q,r)$ are $\sigma$ admissible, that is
 $(q,r): q,r\geq 2; 1/q+\sigma/r=\sigma/2$ and
$(q,r,\sigma)\neq (2,\infty,1)$.
\begin{proposition}(Keel-Tao, \cite{KeelTao})
\label{prop:12} Let $(q,r)$ and $(\tilde{q}, \tilde{r})$
be both $\sigma$ admissible
 and $U(t)$ obeys \eqref{e:100} and \eqref{e:101}. Then
\begin{eqnarray}
\label{e:102}
\norm{U(t)f}{L^q_t L^r_x}\leq C\norm{f}{L^2} \\
\label{e:103} \norm{\int\limits_0^t U(t)(U(s)^*)
F(s,\cdot) ds}{L^qL^r}\leq C \norm{F}{L^{\tilde{q}'}_t L^{\tilde{r}'}_x}.
\end{eqnarray}
\end{proposition}
{\bf Remark} Note that by the Ginibre-Velo original argument
(see also \cite{KeelTao}), the Strichartz estimates \eqref{e:102},
\eqref{e:103} follows only assuming the energy bound \eqref{e:100}
and the ``modified decay bound''
\begin{equation}
\label{e:115}
\norm{U(t) U(s)^*f}{L^{p'}_x}\leq C_p |t-s|^{-(2/p-1)\sigma}\norm{f}{L^p},
\end{equation}
for any $\sigma: (1-2/p)\sigma >1$. Note that \eqref{e:115} follows by interpolation between
\eqref{e:100} and \eqref{e:101}, and so it is in general easier
 to establish. In fact, we use the modified decay bound,
instead of the $L^1\to L^\infty$ decay bound
\eqref{e:101} in order to reduce the smoothness
assumptions\footnote{We show \eqref{e:101} under the condition $A\in Y_1$ and then interpolate with \eqref{e:100} to obtain \eqref{e:115} under the less restrictive condition
$A\in \tilde{Y}_1$.} on our vector potentials,
see Section \ref{sec:6}.\\
\noindent We  need an extension of Proposition \ref{prop:12},
which follows from the same proof as in \cite{KeelTao}.
\begin{proposition}
\label{prop:14}
Let $W(t)$ is an operator defined on all Schwartz
functions on ${\mathbf R}^n$ and it  satisfies
\begin{eqnarray*}
& & \norm{W(t) f}{L^2_{x_2,
\ldots, x_n} L^2_{x_1}}=\norm{W(t) f}{L^2_x}\leq C\norm{f}{L^2_x} \\
& & \norm{W(t)W(s)^* f}{L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}}\leq C |t-s|^{-\sigma}
\norm{f}{L^1_{x_2,
\ldots, x_n} L^2_{x_1}}.
\end{eqnarray*}
Then
\begin{equation}
\label{eq:op}
\norm{W(t) f}{L^q_t  L^r_{x_2, \ldots, x_n}L^2_{x_1}}\leq C
\norm{f}{L^2},
\end{equation}
for all  $\sigma$ admissible pairs $(q,r)$.
The usual averaging argument then implies the ``retarded estimate''
\begin{eqnarray*}
& & \norm{\int\limits_0^t W(t)W(s)^*
F(s, \cdot) ds}{L^q_t  L^r_{x_2, \ldots, x_n}L^2_{x_1}}\leq
C\norm{F}{L^1 L^2}.
\end{eqnarray*}
\end{proposition}
\noindent Indeed,  for the ``retarded estimate'', assume \eqref{eq:op} to get
\begin{eqnarray*}
& & \norm{\int\limits_0^t W(t)W(s)^*
F(s, \cdot) ds}{L^q_t  L^r_{x_2, \ldots, x_n}L^2_{x_1}} \leq
\norm{\int\limits_0^t \norm{W(t)W(s)^*
F(s, \cdot)}{ L^r_{x_2, \ldots, x_n}L^2_{x_1}} ds}{L^q_t}\lesssim \\
& & \lesssim  \int\limits_0^\infty \norm{W(t)[W(s)^*
F(s, \cdot)]}{ L^q_t L^r_{x_2, \ldots, x_n}L^2_{x_1}} ds \lesssim
\int\limits_0^\infty  \norm{W(s)^* F(s, \cdot)}{L^2} ds\leq C \norm{F}{L^1_t L^2_x},
\end{eqnarray*}
where  we have used \eqref{eq:op} in the form
$\norm{W(t)G(s,\cdot) }{L^q_t  L^r_{x_2, \ldots, x_n}L^2_{x_1}}\lesssim
\norm{G(s, \cdot)}{L^2}$ as well as  the energy estimate
$\norm{W(s)^* F}{L^2}\leq C \norm{F}{L^2_x}$.

\section{Proof of Theorem \ref{theo:pisa}}
\label{sec:4}
\subsection{Energy estimates and Littlewood-Paley reductions}
\label{sec:3}
To start our argument, we shall need the following $L^2$
existence result, due to
Kenig-Ponce-Vega, \cite{KPV98}, which generalizes
an earlier work of Doi, \cite{Doi1}, \cite{Doi2}. We state it here
only in the particular case of interest to us, namely  Schr\"odinger equation with
first order perturbations.
\begin{proposition}(Kenig-Ponce-Vega, \cite{KPV98})
\label{prop:KPV}
For the equation
$$
\left|
\begin{array}{l}
\partial_t u-i \Delta u+\vec{b}(t,x)\cdot \nabla u= F\\
u(x,0)=f
\end{array}
\right.
$$
there is an unique global solution, provided $\vec{b}\in C^N$,
$|Im\ \vec{b}|\leq <x>^{-m}$ for some large integers $N,m$.
Moreover the solutions
are smooth, provided $f, F$ are smooth  and for
every $T>0$
$$
\norm{u}{L^\infty(0,T)L^2}\leq C_T (\norm{f}{L^2}+\norm{F}{L^1(0,T)L^2}).
$$
\end{proposition}
\noindent In the case, when $\vec{b}$ is real valued and
$\|\vec{b}\|_{L^1L^\infty}<1/2$,
one can derive {\it a priori} estimates for
$\norm{u}{L^\infty(0,T)L^2}$ that are $T$  independent. \\
This is very standard energy estimate.
Indeed, multiply both sides by $\bar{u}$, integrate
in the spatial variable and take real
part\footnote{By the smoothness of  the solutions all the operations are justified.}.
We obtain
\begin{eqnarray*}
\partial_t \int |u|^2 dx+ \int \vec{b}\cdot\nabla |u|^2dx=\f{1}{2}\int (F \bar{u}+\bar{F} u)dx
\end{eqnarray*}
Integrate by parts and then integrate in $(0,T)$ to get
\begin{eqnarray*}
\norm{u(T,\cdot)}{L^2}^2-\norm{f}{L^2}^2&=&\int\limits_0^T \int div(\vec{b})
|u|^2dxdt+ \int\limits_0^T \int (F \bar{u}+\bar{F} u)dx dt
\leq\\
&\leq& \|\nabla \vec{b}\|_{L^1L^\infty}\sup\limits_{0\leq t\leq T}\norm{u(t,\cdot)}{L^2}^2
+\norm{F}{L^1L^2}\sup\limits_{0\leq t\leq T}\norm{u(t,\cdot)}{L^2}.
\end{eqnarray*}
whence since $\|\nabla\vec{b}\|_{L^1L^\infty}<1/2$,
$$
\norm{u(T,\cdot)}{L^\infty L^2}\leq C( \norm{f}{L^2}+\norm{F}{L^1L^2}).
$$
Thus, we have shown the following
\begin{proposition}
\label{prop:5}
Let $f, F$ be smooth functions. Let also
$\vec{A}$ be a smooth, real-valued potential with $\|\nabla\vec{A}\|_{L^1L^\infty}<1/2$.
Then the Schr\"odinger equation
$$
\left|
\begin{array}{l}
\partial_t u-i \Delta u+\vec{A}(t,x)\cdot \nabla u= F\\
u(x,0)=f
\end{array}
\right.
$$
has an unique global solution and moreover there exists an absolute constant $C$,
so that for every $T>0$
\begin{equation}
\label{eq:2300}
\norm{u(T,\cdot)}{L^\infty L^2}\leq C( \norm{f}{L^2}+\norm{F}{L^1L^2}).
\end{equation}
\end{proposition}
We may restate  Proposition \ref{prop:5} in a
slightly different manner. Namely,
the linear operators $U_A(t,s):L^2({\mathbf R}^n)\to L^2({\mathbf R}^n)$,
where  $U_{A}(t,s)f$ is the unqiue solution $u$ of
$$
\left|
\begin{array}{l}
\partial_t u-i \Delta u+\vec{A}(t,x)\cdot \nabla u= 0\\
u(s,x)=f
\end{array}
\right.
$$
are well-defined. Moreover, by uniqueness and
since the equation is time reversible, we can define $U_A(s,t):=U_A(t,s)^{-1}$, which is the solution
operator to the same equation with data at time $t$ backwards in time to $s$.\\
The Duhamel's formula may be used to write the unique solution to \eqref{eq:1} as
\begin{equation}
\label{eq:duh}
u(t,x)=U_A(t,0)f+\int\limits_0^t U_A(t,s) F(s,\cdot) ds.
\end{equation}
\noindent
Next, we take a Littlewood-Paley projections  of \eqref{eq:1}.
We get the equations
\begin{equation}
\label{eq:260}
\left|
\begin{array}{l}
\partial_t u_k - i \Delta u_k + \vec{A}_{\leq k-4} \cdot \nabla u_k= F_k + E^k\\
u_k(0)=f_k
\end{array}
\right.
\end{equation}
where $E^k$ is the error term
\begin{eqnarray}
\label{eq:1201}
E^k &=&   [P_k, \vec{A}_{\leq k-4}] \nabla u_{k-1\leq \cdot k+1} +
\sum\limits_{l\geq k-2} P_k(\vec{A}_l \cdot\nabla u_{l-2\leq \cdot\leq l+2})+\\
&+& \nonumber \sum\limits_{l\geq k-2} P_k(\vec{A}_{l-2\leq\cdot\leq l+2}  \cdot\nabla u_{l})+
P_k ( \vec{A}_{k-1\leq\cdot\leq k+1}\cdot \nabla u_{\leq k-4}).
\end{eqnarray}
Denote
\begin{eqnarray*}
& & \mathcal L \psi:= \psi_t - i \Delta \psi+A\cdot \nabla \psi, \\
& & \mathcal L^k\psi:= \psi_t - i \Delta \psi+A_{\leq k-4} \cdot \nabla \psi.
\end{eqnarray*}
Our next observation is a naive Strichartz estimate for $\mathcal L$,
which will be the  starting point in a continuity argument later on.
\begin{proposition}
\label{prop:7}
For a fixed integer  $k_0$, there exists a time
$T_0=T_0(k_0)\leq \infty$, so that
whenever $0<T< T_0$, and for every $\psi\in \mathcal S$
\begin{equation}
\label{eq:1800}
\norm{P_{<k_0}\psi}{L^q(0,T)L^r}\leq C(T,k_0)(\norm{\psi(0,\cdot)}{L^2}+
\norm{\mathcal L\psi}{L^1L^2}).
\end{equation}
Moreover, $C(T,k_0)$ depends on  $T$ in a continuous way.
\end{proposition}
\begin{proof}
The proof is based on the standard Strichartz estimates for the
linear \\ Schr\"odinger equation, as we treat the term
$\vec{A}\nabla\psi$ as a perturbation.
By H\"older's inequality and since
$\norm{\nabla P_{<k_0}\psi}{L^p}\lesssim 2^{k_0}\norm{\psi}{L^p}$
\begin{eqnarray*}
& &\norm{P_{<k_0}\psi}{L^q(0,T)L^r}\leq C(\norm{\psi(0)}{L^2}+ \|\vec{A}
\nabla P_{<k_0}\psi\|_{L^1(0,T)L^2}+
\norm{\mathcal L P_{<k_0}\psi}{L^1L^2}) \\
&\leq&  C_1(\norm{\psi(0)}{L^2}+\sqrt{T}2^{k_0}\|\vec{A}\|_{L^2_t L_x^\infty}
\norm{\psi}{L^\infty(0,T)L^2})+\\
&+& C_1(
\norm{P_{<k_0}\mathcal L\psi}{L^1L^2}
+\norm{[P_{<k_0},\vec{A}]\nabla \psi}{L^1L^2})\leq \\
& &\leq C_1(\norm{\psi(0)}{L^2}+\norm{\mathcal L\psi}{L^1L^2})+ \\
& &+
C_1(\sqrt{T}2^{k_0}\|\vec{A}\|_{L^2_t L_x^\infty}+
\norm{\nabla \vec{A}}{L^1L^\infty})\norm{\psi}{L^\infty(0,T)L^2} \\
& & \leq C_1(\norm{\psi(0)}{L^2}+\norm{\mathcal L\psi}{L^1L^2})+
C_1\varepsilon(\sqrt{T}2^{k_0}+1)\norm{\psi}{L^\infty(0,T)L^2}) .
\end{eqnarray*}
According to \eqref{eq:2300}, the last expression is controlled by
$$
C_1(1+C \varepsilon (\sqrt{T}2^{k_0}+1)) (\norm{\psi(0)}{L^2}+\norm{\mathcal L\psi}{L^1L^2}).
$$
Hence, \eqref{eq:1800} holds with $C(T,k_0)= C_1(1+C \varepsilon (\sqrt{T}2^{k_0}+1))$.
\end{proof}
\noindent Fix $k_0$ and a small $\varepsilon=\varepsilon(n)$ (to be chosen later).
Set  $0<T^*\leq \infty$ to be
the maximum time,  so that for all
 $0<T<T^*$ and for all $\psi\in \mathcal S$:
\begin{equation}
\label{eq:iteration}
\norm{P_{<k_0}\psi}{L^q(0,T)L^r}\leq \varepsilon^{-1}
(\norm{\psi(0,\cdot)}{L^2}+\norm{\mathcal L \psi}{L^1L^2}).
\end{equation}
If $T^*=\infty$, \eqref{eq:iteration} holds
for the fixed $k_0$. \\
We will show that  $T^*<\infty$ leads to a
contradiction, provided $\varepsilon=\varepsilon_n$ was
chosen suitably small.  We need to consider  separately the homogeneous
and inhomogeneous problems.
\subsection{Homogeneous problem for $\mathcal L$}
Consider
$$
\left|
\begin{array}{l}
\mathcal L g=0\\
g(0,x)=f(x),
\end{array}
\right.
$$
where $g$ is constructed as $g(t,x)=U_{\vec{A}}(t,0)f$.
We will show that there exists a
constant $C$,
depending on the dimension\footnote{The constant $C$
may also depend on the Strichartz pair $(q,r)$ in dimension two,
although we suppress that dependence.} $n$, but not on $T$, $k_0$, or $f$,
 so that for any $T<T^*$ and all $k\leq k_0$
\begin{equation}
\label{eq:3099}
(\sum\limits_k \norm{P_{<k_0}P_k g}{L^q(0,T)L^r}^2)^{1/2}\leq C \norm{f}{L^2}.
\end{equation}
An elementary computation shows that $g_k$ solves
\begin{equation}
\label{eq:1020}
\left|
\begin{array}{l}
\mathcal L g_k=
-[P_k , \vec{A}] \nabla g, \\
g_k (0,x)=f_k.
\end{array}
\right.
\end{equation}
\subsection{A priori $L^2$ estimates for \eqref{eq:1020}}
First, we show an {\it a priori} estimate
for \\ $(\sum\limits_k \norm{g_k}{L^\infty L^2}^2)^{1/2}$ which generalizes
 \eqref{eq:2300}. \\
Apply \eqref{eq:2300} for the solutions of \eqref{eq:1020}. We get
\begin{eqnarray*}
& &\norm{g_k}{L^\infty L^2}\leq C(\norm{f_k}{L^2}+
\norm{[P_k , \vec{A}] \nabla g}{L^1L^2}).
\end{eqnarray*}
To tackle the term $\norm{[ P_k , \vec{A}] \nabla g}{L^1L^2}$,
we split $\vec{A}$ in low and high frequencies portions
respectively. We have
$$
\sum\limits_k \norm{[ P_k , \vec{A}] \nabla g}{L^1L^2}^2\leq
\sum\limits_k \norm{[ P_k , \vec{A}_{\leq k-4}] \nabla g}{L^1L^2}^2+
\sum\limits_k \norm{[ P_k , \vec{A}_{>k-4}] \nabla g}{L^1L^2}^2.
$$
Since $[ P_k , \vec{A}_{\leq k-4}] \nabla g=
[ P_k , \vec{A}_{\leq k-4}]P_{k-2<\cdot<k+2} \nabla g$,
we estimate this  portion by
\begin{eqnarray*}
\sum\limits_k \norm{[P_k,\vec{A}_{\leq k-4}] \nabla g}{L^1L^2}^2\lesssim
\norm{\nabla \vec{A}}{L^1L^\infty}^2 \sum\limits_k
\norm{g_{k-2<\cdot<k+2}}{L^\infty L^2}^2\lesssim  \varepsilon^2
\sum\limits_k
\norm{g_k}{L^\infty L^2}^2.
\end{eqnarray*}
For the high-frequency portion, we have
$$
[ P_k , \vec{A}_{>k-4}] \nabla g=-\vec{A}_{>k-4}\nabla g_k+
P_k(\vec{A}_{>k-4} \nabla g).
$$
 The first term is trivially estimated by
$$
\sum\limits_k \norm{\vec{A}_{>k-4}\nabla g_k}{L^1L^2}^2\lesssim
\norm{\nabla \vec{A}}{L^1L^\infty}^2\sum\limits_k\norm{g_k}{L^\infty L^2}^2.
$$
The rest is estimated
 by
\begin{eqnarray*}
 & & \sum\limits_k \norm{P_k(\vec{A}_{>k-4} \nabla g)}{L^1L^2}^2\leq
\sum\limits_k \norm{P_k(\vec{A}_{k-4<\cdot<k+4} \nabla g)}{L^1L^2}^2+\\
& &+
\sum\limits_k \norm{P_k(\vec{A}_{>k+4} \nabla g)}{L^1L^2}^2
\lesssim \sum\limits_k \norm{\vec{\nabla A}_{k-4<\cdot<k+4}}{L^1L^\infty}^2
\norm{g}{L^\infty L^2}^2+\\
& &+
 \sum\limits_k\left(\sum\limits_{l\geq k+4} \norm{P_k( \vec{A}_l \nabla
g_{l-2\leq \cdot \leq l+2})}{L^1L^2}\right)^2\lesssim\\
& &\lesssim \varepsilon^2\norm{g}{L^\infty L^2}^2+
\sum\limits_k\left(\sum\limits_{l\geq k+4} 2^{h k} \norm{P_k( \vec{A}_l \nabla
g_{l-2\leq \cdot \leq l+2})}{L^1L^{2n/(n+2h)}}\right)^2\lesssim \\
& &\lesssim \varepsilon^2\norm{g}{L^\infty L^2}^2+\sum\limits_k 2^{2 h k}
\left(\sum\limits_{l\geq k+4} 2^l \|\vec{A}_l\|_{L^1 L^{n/h}}
\norm{g_{l-2\leq \cdot \leq l+2}}{L^\infty L^2}\right)^2\lesssim \\
& &\lesssim \varepsilon^2\norm{g}{L^\infty L^2}^2+ \sum\limits_k 2^{2 h k}
\left(\sum\limits_{l\geq k+4} 2^{-h l} \|\vec{A}_l\|_{L^1 W^{n/h, 1+h}}
\norm{g_{l-2\leq \cdot \leq l+2}}{L^\infty L^2}\right)^2
\end{eqnarray*}
for some $h: 0<h<1/4$.
By Lemma \ref{le:sequences}, we get that the second
 expression above is bounded by
$
C\sup_l \norm{g_{l-2\leq \cdot \leq l+2}}{L^\infty L^2}^2
\sum\limits_l \|\vec{A}_l\|_{L^1 W^{n/h, 1+h}}^2.$\\
Altogether, we obtain the estimate
\begin{eqnarray*}
& & \sum\limits_k \norm{g_k}{L^\infty L^2}^2\leq C \norm{f}{L^2}^2+ C\varepsilon^2
\sum\limits_k \norm{g_k}{L^\infty L^2}^2,
\end{eqnarray*}
whence for $\varepsilon$ small enough
\begin{equation}
\label{eq:1021}
(\sum\limits_k \norm{g_k}{L^\infty L^2}^2)^{1/2}\leq 2 C \norm{f}{L^2},
\end{equation}
which generalizes \eqref{eq:2300}, since $\norm{g}{L^\infty L^2}\leq
(\sum\limits_k \norm{g_k}{L^\infty L^2}^2)^{1/2}$.

We  also need the following crucial lemma,
whose proof we postpone for  Section \ref{sec:par}.
\begin{lemma}(existence of parametrix)
\label{le:parametrix}
Given a  potential
$\norm{\vec{A}}{Y_1\cap Y_2\cap Y_3}\leq \varepsilon$ and integer $k$ and $T>0$,
and for every function $f_k\in L^2({\mathbf R}^n)$ with $supp\ \widehat{f_k}
\subset \{|\xi|\sim 2^k\}$,
one can find a
function
$v_k:[0,T)\times {\mathbf R}^n\to \mathcal C$, so that $supp\
\widehat{v_k}\subset \{|\xi|\sim 2^k\}$ and
\begin{eqnarray*}
& &\norm{v_k(0,x)-f_k}{L^2}\leq C\varepsilon \norm{f_k}{L^2} \\
& &\norm{v_k}{L^q(0,T)L^r}\leq C\norm{f_k}{L^2}\\
& &\norm{\mathcal L v_k}{L^1L^2}\leq C\varepsilon \norm{f_k}{L^2},
\end{eqnarray*}
for some $C$ independent of $f, k, T$.
\end{lemma}

\noindent Assuming for the moment the validity
 of Lemma \ref{le:parametrix}, we show \eqref{eq:3099}.
Take $v_k$ as in  Lemma \ref{le:parametrix}.
Applying the {\it a priori} estimate \eqref{eq:iteration} yields
\begin{eqnarray*}
& &\norm{P_{<k_0} P_k g}{L^q(0,T)L^r}\leq
\norm{P_{<k_0}(P_k g-v_k)}{L^q(0,T)L^r}+\norm{P_{<k_0} v_k}{L^q(0,T)L^r}
\leq \\
& & \leq \varepsilon^{-1}(\norm{f_k-v_k(0,x)}{L^2}+\norm{\mathcal L P_k g}{L^1L^2}+
\norm{\mathcal L v_k}{L^1L^2})+C\norm{v_k}{L^q(0,T)L^r}\leq \\
& &\leq \varepsilon^{-1}(C\varepsilon\norm{f_k}{L^2}+
\norm{[ P_k , \vec{A}]
\nabla g}{L^1L^2})+C\norm{f_k}{L^2}.
\end{eqnarray*}
For the proof of \eqref{eq:1021}, we have already estimated
$$
\sum\limits_k \norm{[ P_k , \vec{A}]
\nabla g}{L^1L^2}^2\leq C\varepsilon^2 \sum\limits_k \norm{g_k}{L^\infty L^2}^2,
$$
which by \eqref{eq:1021} is bounded by $C \norm{f}{L^2}^2$.
Squaring and summing the estimates
for \\ $\norm{P_{<k_0} P_k g}{L^q(0,T)L^r}$ yields
$$
\norm{P_{<k_0} g}{L^q(0,T)L^r}\leq
(\sum\limits_k \norm{P_{<k_0} P_k g}{L^q(0,T)L^r}^2)^{1/2}\leq C\norm{f}{L^2}.
$$
This proves
\eqref{eq:3099}.
 Rewriting \eqref{eq:3099} in terms of the operators $U$ yields
$$
(\sum\limits_k \norm{P_{<k_0} P_k U(t,0)f}{L^q(0,T)L^r}^2)^{1/2}\leq C\norm{f}{L^2},
$$
or more generally since
$U(t,s)=U(t,0)U(0,s)$
\begin{equation}
\label{eq:3100}
(\sum\limits_k \norm{P_{<k_0} P_k U(t,s)f_s}{L^q(0,T)L^r}^2)^{1/2}\leq
C\norm{U(0,s)f_s}{L^2}\leq C \norm{f_s}{L^2}.
\end{equation}
which is sometimes more convenient to use. In particular, we obtain
\begin{equation}
\label{eq:3105}
\norm{P_{<k_0} U(t,s)f_s}{L^q(0,T)L^r}\leq
C \norm{f_s}{L^2}.
\end{equation}
\subsection{The inhomogeneous problem for $\mathcal L$}
We derive estimates similar to \eqref{eq:3099} for the
inhomogeneous problem associated with  $\mathcal L$. Namely, we consider
$$
\left|
\begin{array}{l}
\mathcal L w=G\\
w(0,x)=0,
\end{array}
\right.
$$
We show that the solution $w$
(constructed by the Duhamel's formula \eqref{eq:duh}) satisfy
\begin{equation}
\label{eq:3101}
(\sum\limits_k \norm{P_{<k_0}P_k w}{L^q(0,T)L^r}^2)^{1/2}
\leq C\norm{G}{L^1(0,T)L^2},
\end{equation}
whenever $T<T^*$ and the constant $C$ is dependent only on  $n$.\\
Our first step as in the homogeneous case is to project the inhomogeneous
equation by the Littlewood-Paley operator $ P_k$. We have that
\begin{equation}
\label{eq:3102}
\mathcal L  P_k w= P_k G - [ P_k, A] \nabla w,
\end{equation}
whence exactly as in the homogeneous case
(and by using the smallness assumptions on $\vec{A}$),
we conclude
\begin{eqnarray*}
& &
(\sum\limits_k \norm{P_k w}{L^\infty L^2}^2)^{1/2}
\leq C (\sum\limits_k \norm{G_k}{L^1L^2}^2)^{1/2}\leq C \norm{G}{L^1L^2},\\
& & (\sum\limits_k \norm{[ P_k, A] \nabla w}{L^1L^2}^2)^{1/2}\leq C \varepsilon
(\sum\limits_l  \norm{P_l w}{L^\infty L^2}^2)^{1/2}\leq C\varepsilon
\norm{G}{L^1L^2}.
\end{eqnarray*}

By the Duhamel's
formula  applied to \eqref{eq:3102} and \eqref{eq:3105}
\begin{eqnarray*}
& &(\sum\limits_k \norm{P_{<k_0} P_k w}{L^q(0,T)L^r}^2)^{1/2}=
(\sum\limits_k\left\|\int_0^t P_{<k_0} U(t,s) ( P_k
G(s,\cdot)) ds\right\|_{L^q(0,T)L^r}^2)^{1/2} +\\
& & (\sum\limits_k\left\|\int_0^t P_{<k_0} U(t,s)( [P_k,\vec{A}]
\nabla w)ds\right\|_{L^q(0,T)L^r}^2)^{1/2}\lesssim \\
& &\lesssim
(\sum\limits_k (\int\limits_0^T  \norm{P_k
G(s,\cdot)}{L^2}
ds)^2)^{1/2}+ (\sum\limits_k (\int\limits_0^T  \norm{[P_k,\vec{A}]
\nabla w(s,\cdot))}{L^2}
ds)^2)^{1/2}\lesssim \\
& &
\lesssim (\sum\limits_k \norm{G_k}{L^1L^2}^2)^{1/2}+
(\sum\limits_k \norm{[P_k, A] \nabla w}{L^1L^2}^2)^{1/2}
\lesssim \norm{G}{L^1L^2}.
\end{eqnarray*}
In particular,
\begin{equation}
\label{eq:3106}
\norm{P_{<k_0} w}{L^q(0,T)L^r}\leq C \norm{G}{L^1L^2}.
\end{equation}
Combining \eqref{eq:3105},
\eqref{eq:3106} yields that the solution to
$\mathcal L \psi =G, \quad \psi(0,x)=f$, satisfies
$$
\norm{P_{<k_0}\psi}{L^q(0,T)L^r}\leq C_n(\norm{f}{L^2}+\norm{G}{L^1L^2}),
$$
which is a contradiction with the maximality of $T^*$
(see \eqref{eq:iteration}), provided $\varepsilon < 1/C_n$.
Thus $T^*=\infty$ and one has the inequality
$$
\norm{P_{<k_0}\psi}{L^qL^r}\leq C_n(\norm{\psi(0,\cdot)}{L^2}+
\norm{\mathcal L \psi}{L^1L^2}).
$$
for all $k_0$. Taking a limit $k_0\to \infty$ establishes
the Strichartz estimate \eqref{eq:1009}. Moreover,
we have established a more general Besov  type estimate
$$
(\sum\limits_k \norm{\psi_k}{L^q L^r}^2)^{1/2}
\leq C_n(\norm{\psi(0,\cdot)}{L^2}+
\norm{\mathcal L \psi}{L^1L^2}).
$$
\section{The construction of the parametrix}
\label{sec:par}
\noindent In this section, we show the existence of approximate solution
(in  sense of Lemma \ref{le:parametrix}) to the
equation $\mathcal L g=0, \quad g(x,0)=f_k$. \\
We first make some reductions.
Our first  reduction is that
it will suffice to show that there exists
$v_k$ with the properties as listed in Lemma \ref{le:parametrix},
where $v_k$ satisfies $\norm{\mathcal L^k v_k}{L^1 L^2}\leq C\varepsilon\norm{f_k}{L^2}$
instead of $\norm{\mathcal L v_k}{L^1 L^2}\leq C\varepsilon \norm{f_k}{L^2}$. Indeed,
suppose there is $v_k$ with $\norm{\mathcal L^k v_k}{L^1 L^2}\leq
C\varepsilon\norm{f_k}{L^2}$. Then
\begin{eqnarray*}
& &
\norm{\mathcal L v_k}{L^1L^2}\leq
\norm{\mathcal L^k v_k}{L^1 L^2}+\norm{\vec{A}_{>k-4}}{L^1L^\infty}
\norm{\nabla v_k}{L^\infty L^2}\leq\\
& &\leq \norm{\mathcal L^k v_k}{L^1L^2}+C \norm{\partial \vec{A}}{L^1L^\infty}
\norm{v_k}{L^\infty L^2}\leq C\varepsilon \norm{f_k}{L^2}
\end{eqnarray*}
Suppose that one has already  a
function  $v_k$ as in Lemma
\ref{le:parametrix} without
$supp\ \widehat{v_k}\subset \{|\xi|\sim 2^k\}$.
Our claim is that $\tilde{P}_k v_k$ will satisfy
all conditions in Lemma
\ref{le:parametrix},
with $\mathcal L$ replaced by $\mathcal L^k$ according to our first
reduction. \\
By construction $supp\ \widehat{\tilde{P}_k v_k}
\subset \{|\xi|\sim 2^k\}$.
Since  $\tilde{P}_k f_k=f_k$,
$$
\norm{\tilde{P}_k v_k(0,x)-f_k}{L^2}=\norm{\tilde{P}_k
(v_k(0,x)-f_k)}{L^2}\leq
\norm{v_k(0,x)-f_k}{L^2}\leq \varepsilon\norm{f}{L^2}.
$$
Next,
$$
\|\tilde{P}_k v_k\|_{L^qL^r}\leq \norm{v_k}{L^qL^r}\leq C\norm{f_k}{L^2}.
$$
Finally, since $\mathcal L^k \tilde{P}_k v_k=\mathcal L^k v_k-
[ \tilde{P}_k, \vec{A}_{\leq k-4}] \nabla v_k$
\begin{eqnarray*}
& &
\norm{\mathcal L \tilde{P}_k v_k}{L^1L^2}\leq \norm{\mathcal L v_k}{L^1L^2}+
\norm{[ \tilde{P}_k, \vec{A}_{\leq k-4}] \nabla v_k}{L^1L^2}\leq \\
& &
\leq C\varepsilon \norm{f_k}{L^2}+
\norm{\partial \vec{A}}{L^1L^\infty}\norm{v_k}{L^\infty L^2}\leq
C\varepsilon \norm{f_k}{L^2}.
\end{eqnarray*}

Next, since all our estimates will be
scale invariant, we may rescale and assume
without loss of generality  $k=0$. Thus,
matters  are reduced to the following
\begin{lemma}
\label{le:parametrixmod}
Let $\varepsilon>0$ and $\norm{\vec{A}}{Y_1\cap Y_2\cap Y_3}<\varepsilon$ be a potential with \\
$supp\ \widehat{A}(t,\xi)\subset \{|\xi|\lesssim 1\}$.
Then for every $T>0$ and
for every function $f\in L^2({\mathbf R}^n)$ with $supp\
\widehat{f}\subset \{|\xi|\sim 1 \}$,
one can find a
function
$v:[0,T)\times {\mathbf R}^n\to \mathcal C$, so that
\begin{eqnarray}
\label{eq:902}
& &\norm{v(0,x)-f}{L^2}\leq C\varepsilon \norm{f}{L^2} \\
\label{eq:903}
& &\norm{v}{L^q(0,T)L^r}\leq C\norm{f}{L^2}\\
\label{eq:904}
& &\norm{\partial_t v- i \Delta v+ \vec{A}\cdot \nabla v}{L^1L^2}\leq
C\varepsilon \norm{f}{L^2},
\end{eqnarray}
for some $C$ independent on $f,  T, \varepsilon$.
\end{lemma}

\begin{proof}(Lemma \ref{le:parametrixmod})
We construct $v$ in the form
$$
v(t,x)=\Lambda f(t,x)= \int e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi,
$$
where the $\Omega$ is a smooth cut-off of the annulus $|\xi|\sim 1$ and the
phase correction $\sigma$ is to be selected momentarily. We have
\begin{eqnarray*}
& &\mathcal L v= \partial_t v - i \Delta v+\vec{A}\cdot \nabla v =\\
& &=\int (i \partial_t \sigma+\Delta \sigma+ 2\pi i(\dpr{\nabla\sigma}{\xi}+
 \vec{A}\cdot \xi)+ i
[(\nabla\sigma)^2+\vec{A}\cdot \nabla \sigma])  \times \\
& &\times
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
\end{eqnarray*}
We first comment on possible choices for $\sigma$. Clearly,
since we are trying to almost solve $\mathcal L v=0$,
we should choose $\sigma$ in a way, so that the main terms are
resolved  in  the formula for $\mathcal L v$.
We see that since the potential $\vec{A}$ is
supported in the low frequencies and is small, the main terms are
those, that are either  linear in $\vec{A}$ or linear in $\nabla \sigma$.
It seems then  reasonable to choose $\sigma$, so that
$$
\dpr{\nabla\sigma}{\xi}+ \vec{A}\cdot\xi=0.
$$
However, it turns out that when $|\dpr{\xi/|\xi|}{\eta}|\lesssim |\eta|^2$
(here $\eta$ is the Fourier variable for $\sigma$),
one has that $\Delta\sigma$ is actually ``bigger''
compared to $\dpr{\nabla\sigma }{\xi}$. We therefore modify our
choice as follows.\\
Set $\sigma=\sigma^0+\sigma^1$, where
\begin{eqnarray*}
& &
\sigma^0(t,x,\xi)=\sum\limits_{k\leq -2}   \int\limits_0^\infty \vec{A}_k(t, x+z\xi/|\xi|)
\cdot \f{\xi}{|\xi|} \chi(2^{2k}z) dz,\\
& &
\sigma^1(t,x,\xi)= 2\pi i \sum\limits_{k\leq -2} 2^{2k}
\int_0^\infty \Delta^{-1}\vec{A}_k(t, x+z\xi/|\xi|)\cdot \xi\chi'(2^{2k}z) dz.
\end{eqnarray*}
It is  elementary to see that
$$\dpr{\nabla\sigma^0}{\xi}+
\vec{A}\cdot\xi=- \sum\limits_k 2^{2k}\int\limits_0^\infty
\dpr{\vec{A}_k(t, x+z\xi/|\xi|)}{\xi}
\chi'(2^{2k}z) dz,
$$
 whence
$$
\Delta \sigma^1+2\pi i(\dpr{\nabla\sigma^0}{\xi}+ \vec{A}\cdot\xi)=0.
$$
Denote $\vec{\tilde{A}}_k:=2^{2k}\Delta^{-1} \vec{A}_k$. Then,
we rewrite
$$
\sigma^1(t,x,\xi)= 2\pi i \sum\limits_{k\leq -2}
\int_0^\infty \vec{\tilde{A}}_k(t, x+z\xi/|\xi|)\cdot \xi\chi'(2^{2k}z) dz.
$$
According to the choice of $\sigma^0, \sigma^1$, we get
\begin{equation}
\label{eq:501}
\begin{array}{l}
\mathcal L (\Lambda f) (x,t) = \int (i \partial_t (\sigma^0+\sigma^1)
+\Delta \sigma^0 + 2\pi i \dpr{\nabla\sigma^1}{\xi}+ i
[(\nabla\sigma)^2+\dpr{\vec{A}}{\nabla \sigma}])  \times \\
\times
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
\end{array}
\end{equation}
Now every term in the formula for $\mathcal L (\Lambda f)$
except $\dpr{\nabla\sigma^1}{\xi}$ either has a
two spatial derivatives or one time  derivative acting on it (recall
that in our scaling time derivatives are worth two spatial derivatives) or is
quadratic in $\nabla\sigma$ ( since $\nabla \sigma \sim A$ ). \\
However by our choice of $\sigma^1$, we have
\begin{eqnarray}
& & \nonumber
\dpr{\nabla\sigma^1}{\xi} =  2\pi i
\sum\limits_{k\leq -2}
\int\limits (\sum\limits_{j, m=1}^n \partial_m \vec{\tilde{A}}^j_k(x+z \xi/|\xi|) \xi_j \xi_m )
\chi'(2^{2k} z) dz=\\
\nonumber
& &=2\pi i
\sum\limits_{k\leq -2}  \int\limits\left( \f{d}{d z}
\vec{\tilde{A}}_k(x+z\xi/|\xi|) \cdot \xi\right)|\xi|  \chi'(2^{2k} z) dz=\\
\label{e:1}
& &= - 2\pi i
\sum\limits_{k\leq -2}  2^{2k}\int\limits
\vec{\tilde{A}}_k(x+z\xi/|\xi|) \cdot \xi |\xi|  \chi''(2^{2k}z) dz.
\end{eqnarray}
In this last expression, one has
 multiplication by $2^{2k}$, which behaves
like two spatial derivatives on
$\vec{\tilde{A}}_k$. \\
Now that we have made our selection of
 $\sigma$, we go back to the proof of Lemma \ref{le:parametrixmod}.
First, expanding the exponential $e^{i \sigma}$ in
Taylor series yields the representation \\
$\Lambda = \sum\limits_{\alpha\geq 0} i ^{\alpha}(\alpha! )^{-1}
\Lambda^{\alpha}$, where
$$
\Lambda^{\alpha}f(t,x)=\int (\sigma(t,x,\xi))^\alpha
e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
$$
and similar for the expression for $\mathcal L(\Lambda f)$. It is clear now that in
this formulation, it is convenient to think that  $\sigma$ is in the form
$$
\sigma(t,x,\xi) \sim
\sum\limits_{k\leq -2}   \int\limits_0^\infty A_k(t, x+z\xi/|\xi|)
 \chi(2^{2k}z) dz
$$
for some $C^\infty_0$
function
$\chi$ supported in $(-2,2)$.
 This is done by subsuming  the harmless terms $\xi/|\xi|$, $\xi$ and
$\xi|\xi|$ in the multiplier $\Omega(\xi)$ and by considering the resulting
expressions componentwise. This will be a good
strategy for
all estimates involving $L^2_x$
norms.\footnote{We will omit the interval of integration $(0,\infty)$
in the formula for $\sigma$. In other words, we will tacitly replace
$\chi(2^{2k}z)$ by $\chi_+(2^{2k}z):=\chi(2^{2k}z)\chi_{(0,\infty)}(z)$.
This is not going to make any difference in the $L^2$ estimates, since
no smoothness of the amplitude $\chi_+(2^{2k}z)$ is  needed.}

We will however need to show also
decay estimates for $\Lambda^\alpha f$,  in which case,  it is better to
think of $\sigma$ in the form
$$
\sigma(t,x,\xi)\sim  \sum\limits_{k\leq -2} \sum\limits_{l\leq -2k}
\int\limits_0^\infty A_k(t, x+z\xi/|\xi|)
 \varphi(2^{-l}z) dz.
$$
This is so since $\sum_{l\leq -2k} \varphi(2^{-l} z)=\chi(2^{2k}z)$ and
$\chi'(2^{2k}z)$ which enters in $\sigma^1$ has
essentially the same form as $\varphi(2^{2k}z)$. Note that since
$\varphi(2^{-l}z)$ has support away from the endpoints of the
interval of integration, we can also  write
$$
\sigma(t,x,\xi)\sim  \sum\limits_{k\leq -2} \sum\limits_{l\leq -2k}
\int\limits A_k(t, x+z\xi/|\xi|)
 \varphi(2^{-l}z) dz,
$$
with some smooth compactly supported function $\varphi$ with $supp\ \varphi
\subset (1/2,4)$.
\section{Pointwise estimates on $\sigma$}
\label{sec:98}
 To start us off, we will need an additional
angular decomposition for $\sigma$, which we describe next. Note that in this section, we completely ignore the $t$ dependence, since it is
irrelevant in that setting. This is so, because in the spaces of
interest to us (i.e. the Strichartz space and its dual)
 the $L^r_x$ norm always comes first.

To put ourselves in the desired framework, let us fix a family of unit
vectors \\
$\{\theta_j^m\}_{j\in [1, c2^{m(n-1)}],
m\in [-2, \infty)}$  with the property:
 For fixed $m$,
the family of balls
$\{B(\theta_j^m, 2^{-m})\}=:\{B_j^m\}$ forms a
covering of $\mathbf S^{n-1}$
with every ball in the family intersecting at most a
fixed number
(depending only on the dimension n)
 of other balls in the family.

In other words for fixed $m$, $\theta^m_j$ forms a $2^{-m}$ net over $\mathbf S^{n-1}$ and
the distance between any two $\theta^m_j, \theta^m_s$ is bounded below by
$c2^{-m}$.

For the locally finite covering $\{B(\theta_j^m, 2^{-m})\}$ of the annulus
$\{\xi: 1-2^{-m-5}\leq |\xi|\leq 1+2^{-m-5}\}$, we find a smooth
partition of unity subordinated to the covering.
That is, there exists a familly
$\{\psi_{j,m}\}\subset C^\infty_0$ with
$$
\sum\limits_j \psi_{j,m}(2^m(\xi/|\xi|-\theta_j^m))=1.
$$
Moreover,  the functions $\psi_{j,m}$
satisfy uniform bound on their derivatives,
i.e.
$\sup\limits_{j,m}|\partial^\gamma \psi_{j,m}(x)|\leq C_n^\gamma$.

Fix $k,l: l>-k$ For any  function $H_k$,
expand $H_k(x+z\xi/|\xi|)$ in Taylor series around $x+z\theta^{l+k}_j$, when
$|\xi/|\xi|-\theta^{l+k}_j|\lesssim 2^{-l-k}$.
We get
\begin{eqnarray*}
& &
 H_k( x+z \xi/|\xi|)\psi_{j,l+k}(2^{l+k}(\xi/|\xi|-\theta_j^{l+k}))= \\
& &=
\sum\limits_{\gamma\geq 0} \f{1}{\gamma!}
\partial^\gamma H_k( x+z \theta_j^{l+k})z^{|\gamma|} (\xi/|\xi|-\theta^{l+k}_j)^\gamma
\psi_{j,l+k}(2^{l+k}(\xi/|\xi|-\theta_j^{l+k}))
=\\
& &=\sum\limits_{\gamma\geq 0} \f{z^{|\gamma|}}{2^{l|\gamma|}\gamma!}
H^\gamma_k( x+z \theta_j^{l+k})\psi^\gamma_{j,l+k}(2^{l+k}(\xi/|\xi|-\theta_j^{l+k}))
\end{eqnarray*}
where $\psi_{j,m}^\gamma (\mu)=\psi_{j,m}(\mu)\mu^{|\gamma|}$.
We will also adopt the convention of naming a  function
$H_k^\gamma$, whenever  $H^\gamma_k$ has the same Fourier support properties as
$H_k$ and $\|H_k^\gamma\|_{Z}\leq C^{|\gamma|}
\|H_k\|_{Z}$ for all Banach spaces $Z$ used
throughout the paper.
Using the formula  for \\ $H_k(x+z \xi/|\xi|)\psi_{j,l+k}(2^{l+k}
(\xi/|\xi|-\theta_j^{l+k}))$,
we arrive at
\begin{equation}
\label{eq:decompose}
\begin{array}{l}
\int H_k(x+z\xi/|\xi|)\psi_{j,l+k}(2^{l+k}(\xi/|\xi|-\theta_j^{l+k}))
\varphi(2^{-l}z) dz= \\
=\sum\limits_{\gamma\geq 0} (\gamma!)^{-1}
(\int H^\gamma_k( x+z \theta^{l+k}_j)\varphi^\gamma
(2^{-l}z)dz) \psi_{j,l+k}^\gamma(2^{l+k}(\xi/|\xi|-\theta_j^{l+k}))
\end{array}
\end{equation}
If $l\leq -k$,
the above decomposition trivializes in the sense that we can write
\begin{eqnarray*}
& & \int H_k(x+z\xi/|\xi|)
\varphi(2^{-l}z) dz= \\
& & =\sum\limits_{\gamma\geq 0} (\gamma!)^{-1}
(\int H^\gamma_k( x+z e_1)\varphi^\gamma
(2^{-l}z)dz)(2^{l+k}(\xi/|\xi|-e_1))^\gamma,
\end{eqnarray*}
that is, just the vector $e_1$ would suffice in that situation.

For the $L^2$ estimates, we use the decomposition
\begin{equation}
\label{eq:54}
\begin{array}{l}
\int H_k(x+z\xi/|\xi|)\psi_{j,-k}(2^{-k}(\xi/|\xi|-\theta^{-k}_j))
\chi(2^{2k} z)
  dz= \\
 = \sum\limits_{\gamma\geq 0} (\gamma!)^{-1}
(\int H^\gamma_k( x+z \theta^{-k}_j)\chi^\gamma(2^{2k}z) dz)
\psi_{j,-k}^\gamma(2^{-k}(\xi/|\xi|-\theta^{-k}_j)),
\end{array}
\end{equation}
which is derived similar to \eqref{eq:decompose}.

\noindent
We have the following
pointwise bound on  functions in the form  $\int H_k(x+\theta_j^{l+k})
\varphi(2^{-l}z)dz$.
\begin{lemma}
\label{le:pointwise}
Let $k$ be fixed integer and $\varphi$ be  a
fixed Schwartz function with $supp \ \varphi\subset (1/2, 4)$.
Then
\begin{equation}
\label{eq:254}
\sup\limits_x \sum\limits_{l>-k} \sum\limits_j \int |H_k(x+z\theta_j^{l+k})| \varphi(2^{-l}z) dz
\leq C_n
2^{k(n-1)}
\norm{H_k}{L^1_x}.
\end{equation}
In the case $l\leq -k$, we trivially have
$$
\sup\limits_x \sum\limits_{l\leq -k}
\sum\limits_j \int |H_k(x+z e_1)| \varphi(2^{-l}z) dz\leq C 2^{-k}\norm{H_k}{L^\infty}\leq C_n 2^{k(n-1)}\norm{H_k}{L^1_x}.
$$
\end{lemma}
\begin{proof} We concentrate on the case $l>-k$, since the other inequality
is obvious.

Represent $H_k(x) =2^{kn} \int
\varsigma(2^{2k}|x-y|^2) H_k(y) dy$ with some suitable Schwartz  function
$\varsigma:\mathbf R^1\to \mathbf R^1$. Clearly
\begin{eqnarray*}
& &2^{2k}|x+z\theta_j^{l+k}-y|^2=2^{2k}(|x-y|-z)^2+\\
& &+
2(2^{2k})(|x-y|z)(1-\dpr{\theta_j^{l+k}}{(y-x)/|y-x|}.
\end{eqnarray*}
We thus have
$$
|\varsigma(2^{2k}|x+z\theta_j-y|^2)|\leq \f{C_N}{(1+2^{k}||x-y|-z|)^N
(1+2^{2k}(|x-y|z)\gamma^2)^N},
$$
where $\gamma$ is the angle between the unit vectors $\theta_j^{l+k}$ and
$(y-x)/|y-x|$ and $N$ is arbitrary integer. It follows
that
\begin{eqnarray*}
& & \sum\limits_{l>-k} \sum\limits_j \int |H_k(x+z\theta_j)|
\varphi(2^{-l}z) dz\leq \\
& &\leq \sum\limits_{l>-k} \sum\limits_j  \int
\f{C_N 2^{k n}|H_k(y)|\varphi(2^{-l}z) dz}{(1+2^{k}||x-y|-z|)^N
(1+2^{2k}(|x-y|z)\gamma^2)^N}  dydz=\\
& &=\sum\limits_{l>-k} \sum\limits_j  \int
\f{C_N 2^{k n}
|H_k(x+r\theta)|r^{n-1}\varphi(2^{-l}z) dz}{(1+2^{k}|r-z|)^N
(1+2^{2k}(r z)|\theta_j^{l+k}-\theta|^2)^N}  drd\theta dz
\end{eqnarray*}
The main term in the expression above is when the integration is
over \\
$r\sim 2^{l}$, $|z-r|\leq 2^{-k}$ and $|\theta_j^{l+k}-\theta|\leq 2^{-k-l}$,
with the corresponding decay away from this set. We  estimate by
\begin{eqnarray*}
& & C \sum\limits_{l>-k} \sum\limits_j 2^{k(n-1)} \int_{r\sim 2^l}
\int_{|\theta_j^{l+k}-\theta|\leq 2^{-l-k}} |H_k(x+r\theta)||r|^{n-1} dr
\end{eqnarray*}
Clearly the summation in $j$ is
sum of integrals over  (almost)  disjoint subsets of $\mathbf S^{n-1}$ and as a result
it  gives the integration over the whole $\mathbf S^{n-1}$
 (since $\{\theta_j^{l+k}\}$ were chosen to be a $2^{-l-k}$ net of $\mathbf S^{n-1}$). \\
We get
\begin{eqnarray*}
& & \sum\limits_{l>-k} \sum\limits_j \int |H_k(x+z\theta_j^{l+k})|
\varphi(2^{-l}z) dz\lesssim 2^{k(n-1)} \int |H_k(x+y)|
dy\leq 2^{k(n-1)} \norm{H_k}{L^1}.
\end{eqnarray*}
\end{proof}

\section{$L^qL^r$ estimates for the parametrix}
\label{sec:6}
We show that the parametrix is close to the initial
data at $t=0$ and stays
 in the Strichartz space $L^qL^r$.
Taking into account that $\Lambda^{0}f = e^{i t \Delta} f$, it is clear
that \eqref{eq:902}, \eqref{eq:903} will follow from
\begin{equation}
\label{eq:906}
\norm{\Lambda^{\alpha}f}{L^qL^r}\leq C_{n}^\alpha
\norm{A}{\tilde{Y}_1}^{\alpha} \norm{f}{L^2}.
\end{equation}
The case $\alpha=0$
corresponds to the case of free solutions, which are  in $L^qL^r$ by the
standard Strichartz estimates.
We prove  \eqref{eq:906} by showing that
 $\Lambda^{\alpha}$ satisfies appropriate
 decay and energy estimates.

\noindent We will show that for a fixed $s,t$
\begin{eqnarray}
\label{eq:90}
& &\norm{\Lambda^{\alpha}f(t,\cdot)}{L^2}\leq
C_n^\alpha\left(\sum\limits_k \sup\limits_{x\in{\mathbf R}^n,\theta\in\mathbf S^{n-1}}
\int |A_k(t,x+z\theta)|dz\right)^\alpha
\norm{f}{L^2}\\
\label{eq:91}
& & |\Lambda^{\alpha}(t)\Lambda^{\alpha}(s)^*f(x)|\leq C_n^{2\alpha}
(\sum\limits_k 2^{k(n-1)}\norm{A_k}{L^\infty L^{1}})^{2 \alpha}
 |t-s|^{-n/2} \norm{f}{L^1},
\end{eqnarray}
whence by the abstract Strichartz estimates of
Keel and Tao,  \cite{KeelTao}(see also
Proposition \ref{prop:12}), one gets \eqref{eq:906}.

Note that by the Bernstein inequality
\begin{eqnarray*}
& & \sum\limits_k \sup\limits_{x\in{\mathbf R}^n,\theta\in\mathbf S^{n-1}}\int |A_k(t,x+z\theta)|dz|
= \sum\limits_k \sup\limits_{U\in SU(n), x}\|A_k(t,x+Uz)\|_{L^\infty_t
L^\infty_{z_2, \ldots, z_n}L^1_{z_1}} \lesssim \\
& &\lesssim  \sum\limits_k 2^{k(n-1)}\|A_k\|_{L^\infty L^1_x}=\norm{A}{Y_1}
\end{eqnarray*}
and therefore \eqref{eq:90} and \eqref{eq:91} hold for $A\in Y_1$.

If however $n\geq 3$, we have by  complex multilinear
 interpolation between
\eqref{eq:90} and \eqref{eq:91} that  the ``modified decay estimate''
(see \eqref{e:115})
\begin{eqnarray*}
& &
\|\Lambda^{\alpha}(t)\Lambda^{\alpha}(s)^*f(x)\|_{L^{p'}}\leq C_n^{2\alpha}
(\sum\limits_k 2^{k(n-1)/p}\sup\limits_{U\in SU(n),x}
\norm{A_k(t,x+Uz)}{L^\infty_t L^{p}_{z_2, \ldots, z_n} L^1_{z_1}})^{2 \alpha}\times \\
& & \times
 |t-s|^{-n/(2p)} \norm{f}{L^p}= C_n^\alpha \norm{A}{\tilde{Y}_1}^{2\alpha} \norm{f}{L^p},
\end{eqnarray*}
which suffices for \eqref{eq:906}. Thus,
 \eqref{eq:90} and \eqref{eq:91} imply \eqref{eq:906}.
\subsection{Energy estimates: Proof of \eqref{eq:90}} It is
 more convenient to show
$(\Lambda^{\alpha})^*:L^2\to L^2$, which is equivalent to  \eqref{eq:90}.
Clearly
$$
(\Lambda^{\alpha})^*f(x,t)=\int\limits
 e^{4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
\Omega(\xi) f(y) \overline{\sigma^{\alpha}(t,y,\xi)} dy d\xi.
$$
Having in mind the specific form of $\sigma$, matters reduce to
\begin{lemma}
\label{le:34}
Let $\alpha$ be an positive integer and $k_1,  k_2 \ldots,  k_\alpha$
are integers. Let
$\{F^\mu_{k_\mu}\}$ be a collection of functions with
$supp \widehat{F^\mu_{k_\mu}}\subset \{\xi:|\xi|\sim 2^{k_\mu}\}$. Then
for the multilinear operator
\begin{eqnarray*}
& &\Xi^{k_1, \ldots, k_\alpha}_{F^1, \ldots, F^\alpha} f(t,x)=
\int
e^{4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
\Omega(\xi)  d\xi f(y) \prod_{\mu=1}^\alpha
(\int F^\mu_{k_\mu}(t,y+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz )dy,
\end{eqnarray*}
there is the estimate
\begin{equation}
\label{a:11}
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F^1, \ldots, F^\alpha}
 f(t, \cdot)}{L^2_x}\leq C_n^\alpha \norm{f}{L^2}\prod_{\mu=1}^\alpha
\sup\limits_{x,\theta} \int |F^\mu_{k_\mu}(t, x+z\theta)| dz.
\end{equation}
\end{lemma}
The Lemma is applied to $(\Lambda^\alpha)^*$ in an obvious way. That is,
 write \\
$\sigma=\sum\limits_{k\leq -2} \int A_k(t,x+z\xi/|\xi|) \chi(2^{2k}z) dz$ and
\begin{eqnarray*}
& &
(\Lambda^\alpha)^* f(t,x)= \\
& & =\sum\limits_{k_1, \ldots, k_\alpha}
\int e^{4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
\Omega(\xi)  f(y) \prod_{\mu=1}^\alpha
(\int A_{k_\mu}(t,y+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz )dy d\xi =\\
& &= \sum\limits_{k_1, \ldots, k_\alpha}
 \Xi_{A, \ldots, A}^{k_1, \ldots k_\alpha} f(t,x)
\end{eqnarray*}
It follows that
\begin{eqnarray*}
& & \norm{\Lambda^\alpha f}{L^2}\leq \sum\limits_{k_1, \ldots, k_\alpha}
\norm{\Xi_{A, \ldots, A}^{k_1, \ldots k_\alpha} f}{L^2}\leq C_n^\alpha \norm{f}{L^2}
(\sum\limits_{k} \sup\limits_{x, \theta} \int |A_k(t,x+z\theta)|dz)^\alpha,
\end{eqnarray*}
as claimed.
\begin{proof}(Lemma \ref{le:34})
Let us first present to the proof in the case $\alpha=1$,
since the proof in the general case follows similar scheme,
with somewhat cumbersome notations.

The basic idea is to ``pretend''
that $\int F_{k_\mu}(t,y+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz$ is independent
of $\xi$.
We  show that this is almost true,
modulo the angular  decomposition, that we have described earlier.

For the given $k$,  introduce the partition of unity
$$
\sum\limits_j \psi_{j, -k}(2^{-k}(\xi/|\xi|-\theta_j^{-k}))=1
$$
Next, expand $F_k(y+z\xi/|\xi|)$ around $y+z\theta^{-k}_j$. We get
\begin{eqnarray*}
& &
F_k(t,y+z\xi/|\xi|) \chi(2^{2k}z)
\psi_{j, -k}(2^{-k}(\xi/|\xi|-\theta_j^{-k}))= \\
& &=
\sum\limits_{\gamma\geq 0} (\gamma!)^{-1}
F^\gamma_k(t, y+z\theta_j^{-k}) \chi^\gamma(2^{2k}z)
\psi^\gamma_{j, -k}(2^{-k}(\xi/|\xi|-\theta_j^{-k}))
\end{eqnarray*}
We drop the $\gamma$'s, since in the end, we always
add up with the help of $(\gamma!)^{-1}$. We have
\begin{equation}
\label{eq1}
\Xi f(t,x)=
e^{- i t \Delta} \sum\limits_{j}
P^{j}
[f(\cdot)\int F_k(t,\cdot+z\theta_j^{-k}) \chi(2^{2k}z)dz ],
\end{equation}
where $\widehat{P^{j}g}(\xi)= \hat{g}(\xi)
\psi_{j, -k}(2^{-k}(\xi/|\xi|-
\theta_{j}^{-k})) $. Note that the function \\
$y\to \int F_k(y+z\theta_j^{-k}) \chi(2^{2k}z)dz$ has a
Fourier transform supported in $\{\xi: |\xi|\sim 2^k\}$. It follows
\begin{equation}
\label{eq2}
P^{j}
[f(\cdot)\int F_k(t,\cdot+z\theta_j^{-k}) \chi(2^{2k}z)dz ]=P^{j}
[(\int F_k(t,\cdot+z\theta_j^{-k}) \chi(2^{2k}z)dz )\tilde{P}^j(f)],
\end{equation}
for some $\tilde{P}^j: \tilde{P}^jP^j=P^j$ and $\tilde{P}^j$
has  multiplier with support inside $|\xi/|\xi|-
\theta_{j}^{-k}|\leq 2^{2k+2}$.

This implies
\begin{eqnarray*}
& & \norm{\Xi f(t,x)}{L^2_x}\lesssim \left(\sum\limits_j \norm{P^{j}
[f(\cdot)\int F_k(t,\cdot+z\theta_j^{-k})
\chi(2^{2k}z)dz ]}{L^2}^2\right)^{1/2}= \\
& &= \left(\sum\limits_j \norm{P^j[(\int F_k(t,\cdot+z\theta_j^{-k})
\chi(2^{2k}z)dz) \tilde{P}^j f]}{L^2}^2\right)^{1/2} \lesssim \\
& &\lesssim  \left(\sum\limits_j \norm{(\int F_k(t,\cdot+z\theta_j^{-k})
\chi(2^{2k}z)dz) \tilde{P}^j f}{L^2}^2\right)^{1/2} \lesssim\\
& &\lesssim \left(\sum\limits_j \norm{\int F_k(t,\cdot+z\theta_j^{-k})
\chi(2^{2k}z)dz}{L^\infty}^2\norm{\tilde{P}^j f}{L^2}^2\right)^{1/2}
\lesssim \\
& &\leq C_n  \sup\limits_{y, \theta} |\int F_k(t,y+z\theta)
\chi(2^{2k}z)dz| \norm{f}{L^2}.
\end{eqnarray*}
Note that the previous calculation requires slightly augmenting $P^j$ to
$\tilde{P}^j$ for an additional constant $C_n$ comming
on the account of the extra overlap of the supports of different
$\tilde{P}^j$.

Unfortunately, we have to be extra careful
for the case of $\alpha>1$, because the constants (following this argument)
are estimated by $\alpha^\alpha$, whereas we need (and can manage) constants of
magnitude  $C_n^\alpha$. This problems occurs, when too many of the
frequencies $k_1, \ldots , k_\alpha$ are equal.

Thus, we start our considerations for $\alpha>1$
by ordering these frequencies. Without loss of generality
let us assume
$k_1=\ldots=k_{s_1}<k_{s_1+1}=\ldots=k_{s_2}<\ldots
k_{s_{m-1}+1}=\ldots=k_{s_m}=k_\alpha$, where we have
set $s_0=0$ for convenience.

 Since for every $1\leq r\leq m$,  and
$\mu\in[s_{r-1}+1,s_r]$
$$
\sum\limits_{j_r} \psi_{j_r, -k_{\mu}}(2^{-k_{\mu}}(\xi/|\xi|-
\theta_{j_r}^{-k_{\mu}}))=1,
$$
we expand  $F^\mu_{k_{\mu}}(t, y+z\xi/|\xi|)$ around
$y+z \theta_{j_r}^{-k_{\mu}}$. Just as in the case $\alpha=1$, this
 allows us to write
\begin{equation}
\label{eqn12}
\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f(t,x)= e^{- i t \Delta}
\sum\limits_{j_1, \ldots, j_m}
\sum\limits_{\gamma_1, \ldots
\gamma_\alpha}
\f{ P^{j_1}\ldots P^{j_m} [
f(\cdot)\prod_{\mu=1}^\alpha g_\mu^\gamma(t,\cdot)]}{(\gamma_1! \ldots
\gamma_\alpha! )},
\end{equation}
where for $\mu\in[s_{r-1}+1,s_r]$
\begin{eqnarray*}
& & \widehat{P^{j_r}g}(\xi)= \hat{g}(\xi)
\psi_{j_r, -k_{\mu}}^{\gamma}(2^{-k_\mu}(\xi/|\xi|-
\theta_{j_r}^{-k_\mu})) ,\\
& & g_\mu^\gamma(t,y)=\int F^{\mu, \gamma}_{k_\mu}(t, y+z\theta_{j_r}^{-k_\mu})
\chi(2^{2k_\mu}z)dz.
\end{eqnarray*}
As always we drop the $\gamma$'s and concentrate on the case
$\gamma_1=\ldots=\gamma_\alpha=0$.

Next,  observe that since $e^{-i t \Delta}$ is an isometry on $L^2$,
we can dispose of it immediately. We have
$$
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2
\leq \sum\limits_{j_1, \ldots, j_m} \norm{P^{j_1}\ldots
P^{j_m} [f(\cdot)\prod_{\mu=1}^\alpha g_\mu^\gamma(t,\cdot)]}{L^2}^2.
$$
For technical reasons, it is more convenient to
replace $P^{j_r}$ by ``rough'' versions of the same. Namely,
 introduce the Fourier restriction operators $Q^{j_r}$,
which act  via \\
$\widehat{Q^{j_r} g}(\xi)= \hat{g}(\xi) 1_{\{|\xi/|\xi|-
\theta_{j_r}^{k_{s_r}}|\leq 2^{k_{s_r}}\}}$. More generally, for any $a>0$,
$$
\widehat{Q^{j_r}_a g}(\xi)= \hat{g}(\xi) 1_{\{\xi: |\xi/|\xi|-\theta_{j_r}^{k_{s_r}}|\leq a\}}
$$
Since the Fourier supports of the multipliers of $P^{j_r}$ are in
$\{\xi: |\xi/|\xi|-\theta_{j_r}^{k_{s_r}}|\leq 2^{k_{s_r}}\}$,  we have
\begin{equation}
\label{eqn:7}
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2
\leq  \sum\limits_{j_1, \ldots, j_m} \norm{Q^{j_1}\ldots
Q^{j_m} [f(\cdot)\prod_{\mu=1}^\alpha g_\mu(t,\cdot)]}{L^2}^2.
\end{equation}
Fix $j_1$. Clearly the summation in any $j_r$ runs only on
$j_r: |\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}$, since  otherwise $Q^{j_1} Q^{j_r}=0$. Thus,
\begin{equation}
\label{eqn:1}
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2
\leq \sum\limits_{j_1}
\sum\limits_{j_2,  \ldots, j_m:|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}}
\norm{Q^{j_1} [f(\cdot)\prod_{\mu=1}^\alpha g_\mu(t,\cdot)]}{L^2}^2.
\end{equation}
Note that since $supp\ \hat{g_\mu}\subset \{\xi: |\xi|\leq 2^{k_\mu+1}\}$,
we have that $supp\ \widehat{g_1\ldots g_{s_1}}
\subset \{\xi: |\xi|\leq 2 s_1 2^{k_{s_1}}\}$. It follows that
\begin{equation}
\label{eqn14}
\begin{array}{l}
Q^{j_1} [f(\cdot)\prod_{\mu=1}^\alpha g_\mu(t,\cdot)]=
Q^{j_1} [(g_1\ldots g_{s_1})
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))]=\\
= Q^{j_1} [(g_1\ldots g_{s_1})
Q^{j_1}_{2^{k_{s_1}}+2s_1
2^{k_{s_1}}}(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))]
\end{array}
\end{equation}
Plugging this back in \eqref{eqn:1} and taking into account
$\norm{Q^{j_1}}{L^2\to L^2}=1$ yields
\begin{eqnarray*}
& &
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2
\leq  \\
& &\leq \sum\limits_{j_1}
\sum\limits_{j_2,  \ldots, j_m:|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}}
 \norm{g_1 \ldots g_{s_1} Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))}{L^2}^2\leq \\
& &
\leq \sum\limits_{j_1}
\sum\limits_{j_2,  \ldots, j_m:|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}}
\norm{g_1 \ldots g_{s_1}}{L^\infty}^2
 \norm{Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))}{L^2}^2\leq \\
& &\leq \prod_{\mu=1}^{s_1}
\sup_{y, \theta} |\int F^\mu_{k_\mu}(t,y+z\theta) \chi(2^{2k_\mu}z) dz|^2
\times \\
& &\times \sum\limits_{j_1}
\sum\limits_{j_2,  \ldots, j_m:|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}}
 \norm{Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))}{L^2}^2.
\end{eqnarray*}
It is now time to reintroduce the $Q^{j_r}$ multipliers.
Since $|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}$, we conclude that for every $r: 1\leq r\leq m$,
$Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}=Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
Q^{j_r}_{2^{k_{s_r}}+(2s_1+2)2^{k_{s_1}}}$ and we have
\begin{eqnarray*}
& & \sum\limits_{j_1}
\sum\limits_{j_2,  \ldots, j_m:|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}}
 \norm{Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))}{L^2}^2\leq \\
& &\leq \sum\limits_{j_2,  \ldots, j_m}\sum\limits_{j_1}
\norm{ Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
Q^{j_2}_{2^{k_{s_2}}+(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}} (f(\cdot)
\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))}{L^2}^2
\end{eqnarray*}
Note that
\begin{equation}
\label{eqn17}
\sum\limits_{j_1} \norm{Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}} g}{L^2}^2\leq C_n
\left(\f{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}{2^{k_{s_1}}}\right)^n
\norm{g}{L^2}^2=C_n(1+2s_1)^n \norm{g}{L^2}^2 ,
\end{equation}
because of the extra overlap created by passing from $Q^{j_1}$ to
$Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}$. It follows that
\begin{eqnarray*}
& &\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2  \leq
C_n \prod_{\mu=1}^{s_1}
\sup_{y, \theta} |\int F^\mu_{k_\mu}(t,y+z\theta)
\chi(2^{2k_\mu}z) dz|^2 (1+2s_1)^n \times \\
& & \times \sum\limits_{j_2,  \ldots, j_m}
\norm{Q^{j_2}_{2^{k_{s_2}}+(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}} (f(\cdot)
\prod_{\mu=s_1+1}^\alpha g_\mu(t,\cdot))}{L^2}^2.
\end{eqnarray*}
This is very similar to \eqref{eqn:7}, except that the sum in $j_1$ is taken care of and
$Q^{j_2}\ldots Q^{j_m}$ gets
replaced by the slightly larger
$Q^{j_2}_{2^{k_{s_2}}+(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}}$. Continuing in this fashion yields
the estimate
\begin{eqnarray*}
& &\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2  \leq
C_n^m \prod_{\mu=1}^{\alpha}
\sup_{y, \theta} |\int F^\mu_{k_\mu}(t,y+z\theta)
\chi(2^{2k_\mu}z) dz|^2 \norm{f}{L^2}^2 (1+2s_1)^n  \times \\
& & \times  \prod_{r=2}^m
\left(\f{2^{k_{s_r}}+(2s_1+2) 2^{k_{s_1}}+ (2(s_2-s_1)+2) 2^{k_{s_2}}+\ldots +
(2(s_r-s_{r-1})+2)2^{k_{s_r}}}{2^{k_{s_r}}}\right)^n.
\end{eqnarray*}
Note that for every $1\leq i \leq r$, there is
 $k_{s_r}-k_{s_i}\geq r-i$, since
there are $r-i$ strict inequalities in the chain  $k_{s_i}<\ldots<k_{s_r}$.
We therefore need an estimate for
$$
G=\prod_{r=1}^m (1+\sum\limits_{i=1}^r (2(s_i-s_{i-1})+2)2^{i-r})^n
$$
For $\ln(G)$, note that
 since $\ln(1+x)\leq x$ and $m\leq \alpha$,  we can  estimate
$$
\ln(G)\leq n \sum\limits_{r=1}^m \sum\limits_{i=1}^r (2(s_i-s_{i-1})+2)2^{i-r}
\leq 4n (m+\sum\limits_{i=1}^{m} (s_i - s_{i-1}))=4n(m+\alpha)\leq  8n\alpha.
$$
It follows that $G\leq e^{8 n\alpha}$ and
$$
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}^2 \leq  C_n^\alpha
\prod_{\mu=1}^{\alpha}
\sup_{y, \theta} |\int F^\mu_{k_\mu}(t,y+z\theta)
\chi(2^{2k_\mu}z) dz|^2 \norm{f}{L^2}^2
$$

\end{proof}

\subsection{Dispersive estimates: Proof of \eqref{eq:91}}
For the dispersive estimate,  write
\begin{eqnarray*}
& &\Lambda^{\alpha}(t)\Lambda^{\alpha}(s)^*f = \int\limits   \sigma^{\alpha}(t,x,\xi)
 \overline{\sigma^{\alpha}(s,y,\xi)}
 e^{-4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
\Omega^2(\xi) f(y) dy d\xi
\end{eqnarray*}
Clearly for $\alpha=0$, we have
\eqref{eq:91} by the decay  estimates for the free solution.
Consider the case $\alpha=1$ for simplicity, the general case to
be addressed momentarily.\\
Expand $\sigma(t,x,\xi)$ to get
\begin{eqnarray*}
& &\Lambda^{1}(t)\Lambda^{1}(s)^*f= \sum\limits_{k_1 \leq -2}
\sum\limits_{l_1\leq -2k_1}
\sum\limits_{\gamma_1\geq 0} (\gamma_1!)^{-1}\sum\limits_{j_1}
(\int A^{\gamma_1}_k(t, x+z \theta^{l_1+k_1}_{j_1})\varphi^{\gamma_1}
(2^{-l_1}z)dz) \times \\
& &\times \int
\psi_{j_1,l_1+k_1}^\gamma(2^{l_1+k_1}(\xi/|\xi|-\theta_{j_1}^{l_1+k_1}))
 e^{-4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
\Omega^2(\xi) f(y) \overline{(\sigma)(s,y,\xi)} dy d\xi=\\
& &=\sum\limits_{k_1\leq -2}
\sum\limits_{l_1\leq -2 k_1}
\sum\limits_{\gamma_1\geq 0} (\gamma_1!)^{-1}\sum\limits_{j_1}
(\int A^{\gamma_1}_{k_1}(t, x+z \theta^{l_1+k_1}_{j_1})\varphi^{\gamma_1}
(2^{-l_1}z)dz)\Gamma^1_{j_1, l_1+k_1} f,
\end{eqnarray*}
where\footnote{In the case $l_1\leq -k_1$ the summation in $j_1$ collapses to
a single term and $\theta_{j_1}^{l_1+k_1}=e_1$ as pointed out in the previous section.}
$$
\Gamma^\alpha_{j, l+k} f(x)= \int\limits
 \overline{\sigma^{\alpha}(s,y,\xi)}
 e^{-4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
\psi_{j,l+k}^\gamma(2^{l+k}(\xi/|\xi|-\theta_{j}^{l+k}))
\Omega^2(\xi) f(y) dy d\xi
$$
It is easy to see that
\begin{eqnarray*}
& &\norm{\Lambda^{1}(t)\Lambda^{1}(s)^*f}{L^\infty_x}\lesssim \\
& &\lesssim  \sup\limits_x
(\sum\limits_{k_1\leq -2}
\sum\limits_{l_1\leq -2 k_1} \sum\limits_{\gamma_1\geq 0} (\gamma_1!)^{-1}\sum\limits_{j_1}
\int |A^{\gamma_1}_{k_1}(t, x+z \theta^{l_1+k_1}_{j_1})|\varphi^{\gamma_1}
(2^{-l_1}z)dz )\times \\
& &\times \sup\limits_{j_1, l_1, k_1} \norm{\Gamma^1_{j_1, l_1+k_1} f}{L^\infty}.
\end{eqnarray*}
Bythe pointwise estimates of
 Lemma \ref{le:pointwise} and $A\in Y_3$, the last expression is
bounded by $C\norm{A}{Y_1}^2 |t-s|^{-n/2}$, provided one can show
$\norm{\Gamma^1_{j_1, l_1+k_1} f}{L^\infty}\leq C\norm{A}{Y_1}|t-s|^{-n/2}\norm{f}{L^1}$.

\noindent
More generally, it is easy to see that by
iterating the argument above, we have
\begin{eqnarray*}
& &\Lambda^{\alpha}(t)\Lambda^{\alpha}(s)^*f(x)= \\
& &=
\prod_{\mu=1}^\alpha (\sum\limits_{k_\mu\leq -2}
\sum\limits_{l_\mu\leq -2k_\mu}
\sum\limits_{\gamma_\mu} (\gamma_\mu!)^{-1}
\sum\limits_{j_\mu}
\int A^{\gamma_\mu}_{k_\mu}(t, x+z \theta^{l_\mu+k_\mu}_{j_\mu})\varphi^{\gamma_\mu}
(2^{-l_\mu}z)dz )
\Gamma^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f,
\end{eqnarray*}
where
\begin{eqnarray*}
& & \Gamma^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f(x)  =
\int\limits
 \overline{\sigma^{\alpha}(s,y,\xi)}
 e^{-4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
 \times \\
& & \times  \prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}^{\gamma_\mu}
(2^{l_\mu+k_\mu}(\xi/|\xi|-\theta_{j_\mu}^{l_\mu+k_\mu}))
\Omega^2(\xi) f(y) dy d\xi.
\end{eqnarray*}
Thus,
\begin{eqnarray*}
& &\norm{\Lambda^{\alpha}(t)\Lambda^{\alpha}(s)^*f}{L^\infty}\leq \\
& & \leq  \prod_{\mu=1}^\alpha \sup\limits_x[
\sum\limits_{k_\mu\leq -2}
\sum\limits_{l_\mu\leq -2k_\mu}
\sum\limits_{\gamma_\mu\geq 0} (\gamma_\mu)^{-1}  \sum\limits_{j_\mu}
\int |A^{\gamma_\mu}_{k_\mu}(t, x+z \theta^{l_\mu+k_\mu}_{j_\mu})|\varphi^{\gamma_\mu}
(2^{-l_\mu}z) dz )]\times \\
& & \times \sup\limits_{j_1,l_1,k_1;\ldots; j_\alpha, l_\alpha,k_\alpha}
\norm{\Gamma^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f}{L^\infty}.
\end{eqnarray*}
By Lemma \ref{le:pointwise}, we have
\begin{eqnarray*}
& &\norm{\Lambda^{\alpha}(t)\Lambda^{\alpha}(s)^*f}{L^\infty}
\leq C_n^{2 \alpha} \norm{A}{Y_1}^{2\alpha} |t-s|^{-n/2}\norm{f}{L^1},
\end{eqnarray*}
provided one can show
\begin{equation}
\label{eq:340}
 \sup\limits_{j_1,l_1,k_1;\ldots; j_\alpha, l_\alpha,k_\alpha}
\norm{\Gamma^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f}{L^\infty}
\leq C_n^\alpha \norm{A}{Y_1}^\alpha |t-s|^{-n/2}\norm{f}{L^1}
\end{equation}
Note that $\Gamma^\alpha_{j_1, l_1+k_1; \ldots ; j_\alpha, l_\alpha+k_\alpha} f$
looks somewhat like $\Lambda(s)^* f$, with the important difference
that it has the  ``multiplier''
$\prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}(2^{l_\mu+k_\mu}(\xi/|\xi|-
\theta_{j_\mu}^{l_\mu+k_\mu}))$ in its definition. \\
Dualizing \eqref{eq:340} leads us to showing that
\begin{eqnarray*}
& & [(\Gamma^\alpha_{j_1, l_1+k_1; \ldots;
j_\alpha, l_\alpha+k_\alpha})^* f] (x) = \int\limits
 \sigma^{\alpha}(s,x,\xi)
 e^{ 4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
 \times \\
& & \times \prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}^{\gamma_\mu}
(2^{l_\mu+k_\mu}(\xi/|\xi|-\theta_{j_\mu}^{l_\mu+k_\mu}))
\Omega^2(\xi) f(y) dy d\xi.
\end{eqnarray*}
maps $L^1\to L^\infty$ with norm no bigger than
$C_n^\alpha \norm{A}{Y_1}^\alpha |t-s|^{-n/2}$. Expand again to get
\begin{eqnarray*}
& & [(\Gamma^\alpha_{j_1, l_1+k_1; \ldots;
j_\alpha, l_\alpha+k_\alpha})^* f](x)= \\
& &=
\prod_{\nu=1}^\alpha (\sum\limits_{\tilde{k}_\nu \leq -2}
\sum\limits_{\tilde{l}_\nu\leq -2\tilde{k}_\nu}
\sum\limits_{\gamma_\nu} (\gamma_\nu!)^{-1} \sum\limits_{j_\nu}
\int A^{\gamma_\nu}_{\tilde{k}_\nu}(s,
x+z \theta^{\tilde{l}_\nu+\tilde{k}_\nu}_{j_\nu})\varphi^{\gamma_\nu}
(2^{-\tilde{l}_\nu}z)dz )\times \\
& &\times \int\limits
 e^{ 4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-y}}
  \prod_{\nu=1}^\alpha \psi_{j_\nu,\tilde{l}_\nu+\tilde{k}_\nu}^{\gamma_\nu}
(2^{\tilde{l}_\nu+\tilde{k}_\nu}(\xi/|\xi|-
\theta_{j_\nu}^{\tilde{l}_\nu+\tilde{k}_\nu})) \times \\
& &\times
\prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}^{\gamma_\mu}
(2^{l_\mu+k_\mu}(\xi/|\xi|-\theta_{j_\mu}^{l_\mu+k_\mu}))
\Omega^2(\xi) f(y) dy d\xi.
\end{eqnarray*}
By the Krein-Milman
theorem, $span[\delta_b]$ is  $w^*$ dense in
$M({\mathbf R}^n)\supset L^1({\mathbf R}^n)$. This, together with
Lemma \ref{le:pointwise} allows us to estimate
\begin{eqnarray*}
& & \norm{(\Gamma^\alpha_{j_1, l_1+k_1; \ldots;
j_\alpha, l_\alpha+k_\alpha})^* f}{L^\infty_x}\leq  C_n^\alpha (\norm{A}{Y_1})^\alpha
\times \\
& & \sup\limits_{x,b\in {\mathbf R}^n; m_1, \ldots, m_{2\alpha}\geq 0;\tilde{\theta}_1,
\ldots, \tilde{\theta}_{2\alpha} \in \mathbf S^{n-1}}|\int\limits
 e^{ 4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x-b}}
\prod_{\mu=1}^{2\alpha} \psi_{j_\mu,m_\mu}
(2^{m_\mu}(\xi/|\xi|-\tilde{\theta}_{\mu}))
\Omega^2(\xi)d\xi|.
\end{eqnarray*}
The oscillatory integral above is bounded by $C_n^\alpha |t-s|^{-n/2}$ by
  Lemma \ref{le:osc} in the Appendix, and the required
dispersive estimates hold.
\end{proof}

\section{The parametrix almost satisfies the equation}
\label{sec:7}
In this section, we  show
 that the parametrix satisfies \eqref{eq:904}.
We have several types of terms that arise according to
 \eqref{eq:501}.

First, we take on the terms
$$
\int (i \partial_t (\sigma^0+\sigma^1)+\Delta \sigma^0+2\pi i \dpr{\nabla\sigma^1}{\xi})
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
$$
These are {all terms linear in either $\sigma^0$ or $\sigma^1$}
with either a time derivative or two spatial derivative acting on them
(recall from the expression
\eqref{e:1} that $\dpr{\nabla\sigma^1}{\xi})$ is also of this form).
\subsection{Main result}
\begin{lemma}
\label{le:10}
 Let $\Xi^{k_1, \ldots,k_\alpha}_{F^1, \ldots, F^\alpha}$
 be as in Lemma \ref{le:34}, i.e.
\begin{eqnarray*}
& &\Xi^{k_1, \ldots,k_\alpha}_{F^1, \ldots, F^\alpha}f(t,x)=
\int \prod_{\mu=1}^\alpha
(\int F^\mu_{k_\mu}(t,x+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz )
e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
\end{eqnarray*}
Then  there exists a constant $C_n$, so that
\begin{eqnarray*}
& &
\norm{\Xi^{k_1, \ldots,k_\alpha}_{F^1, \ldots, F^\alpha}f}{L^1_tL^2_x}\leq
C_n^\alpha \norm{f}{L^2} \times \\
& &\times
\prod_{\mu=1}^{\alpha-1}
(2^{k_\mu(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_t \sup\limits_x
\| F^\mu_{k_\mu}(t,x+ Uz)\|_{L^2_{z_2, \ldots, z_n}L^1_{z_1}}+
\\
& &+ 2^{k_\mu(n+3)/2}\sup\limits_{U\in SU(n)} \| \sup\limits_x
\| F^\mu_{k_\mu}(t,x+Uz)\|_{L^2_{z_2, \ldots, z_n}L^1_{z_1}}\|_{L^1_t})
 \times \\
& &(2^{k_\mu(n-1)/2}\sup\limits_{U\in SU(n)} \| \sup\limits_x
\| F^\alpha_{k_\alpha}(t,x+Uz)\|_{L^2_{z_2, \ldots, z_n}L^1_{z_1}}\|_{L^1_t}+ \\
& &+
2^{k_\mu(n-5)/2}\sup\limits_{U\in SU(n)} \sup\limits_t \sup\limits_x
\| F^\alpha_{k_\alpha}(t,x+ Uz)\|_{L^2_{z_2, \ldots, z_n}L^1_{z_1}}).
\end{eqnarray*}
In a different form,
\begin{eqnarray*}
& &
\norm{\Xi^{k_1, \ldots,k_\alpha}_{F^1, \ldots, F^\alpha}f}{L^1_tL^2_x}\leq
C_n^\alpha \norm{f}{L^2} \times \\
& &\times
\prod_{\mu=1}^{\alpha-1}
(2^{k_\mu(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| F^\mu_{k_\mu}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k_\mu(n+3)/2}\sup\limits_{U\in SU(n), t\to x(t)}
\| F^\mu_{k_\mu}(t,x(t)+Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}})
 \times \\
& &(2^{k_\mu(n-1)/2}\sup\limits_{U\in SU(n), t\to x(t)}
\| F^\alpha_{k_\alpha}(t,x(t)+Uz)\|_{L^1_t
L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &+
2^{k_\mu(n-5)/2}\sup\limits_{U\in SU(n)}\sup\limits_x
\| F^\alpha_{k_\alpha}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}),
\end{eqnarray*}
where $\sup\limits_{t\to x(t)}$ is taken over all measurable
 functions $x(\cdot): \mathbf R^1\to {\mathbf R}^n$.
\end{lemma}
\subsection{Application of Lemma \ref{le:10} to various terms of $\norm{\mathcal L v}{L^1L^2}$.}
\noindent Assuming the validity of Lemma \ref{le:10}, one can easily handle the first type of terms.

Expand  $e^{i \sigma(t,x,\xi)}$ in powers of $\sigma$
as in the previous section yields
\begin{eqnarray*}
& &
\int (i \partial_t (\sigma^0+\sigma^1)+\Delta \sigma^0+2\pi
i\dpr{\nabla\sigma^1}{\xi})
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi =\\
& &=\sum\limits_{\alpha\geq 1} \f{i^{\alpha-1}}{(\alpha-1)!}
\int \sigma(t,x,\xi)^{\alpha-1} (i \partial_t (\sigma^0+\sigma^1)+\Delta \sigma^0+2\pi
i\dpr{\nabla\sigma^1}{\xi})\times \\
& &\times
 e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi =\sum\limits_{\alpha\geq 1} \f{i^{\alpha-1}}{(\alpha-1)!}
 \sum\limits_{k_1, \ldots, k_\alpha}
 \\
& &
=\int \prod_{\mu=1}^{\alpha-1}
(\int A_{k_\mu}(t, x+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz)
(\int \partial^2 A_{k_\alpha}(t, x+z\xi/|\xi|) \chi(2^{2k_\alpha}z) dz)\times \\
& &\times
e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi =\\
& &= \sum\limits_{\alpha\geq 1} \f{i^{\alpha-1}}{(\alpha-1)!}
 \sum\limits_{k_1, \ldots, k_\alpha} \Xi^{k_1, \ldots, k_\alpha}_{A, \ldots,A, \partial^2 A},
\end{eqnarray*}
where we have denoted by
$\int \partial^2 A_{k_\alpha} (t, x+z\xi/|\xi|) \chi(2^{2k}z) dz$
all the terms
$P_{k_\alpha}(i \partial_t (\sigma^0+\sigma^1)+\Delta \sigma^0+2\pi
i\dpr{\nabla\sigma^1}{\xi})$. \\
Thus, an application of Lemma \ref{le:10} yields
\begin{eqnarray*}
& &\norm{\int (i \partial_t (\sigma^0+\sigma^1)+\Delta \sigma^0+2\pi
i\dpr{\nabla\sigma^1}{\xi})
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi}{L^1 L^2} \\
& & \leq C_n \norm{f}{L^2}
exp(C_n(\sum\limits_k 2^{k(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k(n+3)/2}\sup\limits_{U\in SU(n), t\to x(t)}
\| A_{k}(t,x(t)+Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}))\times \\
& &\times \sum\limits_k (2^{k(n-1)/2}\sup\limits_{U\in SU(n), x(t)}
\| (|\partial^2 A_k|+|\partial_t A_k|) (t,x(t)+Uz)\|_{L^1_t L^2_{z_2,
\ldots, z_n}L^1_{z_1}}+ \\
& &+
2^{k(n-5)/2}\sup\limits_{U\in SU(n), x(t)}
\| (|\partial^2 A_k|+|\partial_t A_k|) (t,x(t) + Uz)\|_{L^\infty_t
L^2_{z_2, \ldots, z_n}L^1_{z_1}}).
\end{eqnarray*}

Next, we show how to use Lemma \ref{le:10} to control terms in the form
$$
\int  (\nabla\sigma)^2
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
$$
in $L^1L^2$ norm. We expand the same way as before
$e^{i \sigma}=\sum_{\alpha\geq 2}i^{\alpha-2}/(\alpha-2)! \sigma^{\alpha-2}$ to get
\begin{eqnarray*}
& & \int  (\nabla\sigma)^2
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi = \\
& &= \sum_{\alpha\geq 2} \f{i^{\alpha-2}}{(\alpha-2)!}
 \sum\limits_{k_1, \ldots, k_\alpha} \int
\prod_{\mu=1}^{\alpha-2}
(\int A_{k_\mu}(t, x+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz)\times \\
& &\times
(\int \partial A_{k_{\alpha-1}}(t, x+z\xi/|\xi|) \chi(2^{2k_{\alpha-1}}z) dz)
(\int \partial A_{k_{\alpha}}(t, x+z\xi/|\xi|) \chi(2^{2k_\alpha}z) dz)\times \\
& &\times
e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi= \sum_{\alpha\geq 2} \f{i^{\alpha-2}}{(\alpha-2)!}
 \sum\limits_{k_1, \ldots, k_\alpha} \Xi^{k_1,
\ldots, k_\alpha}_{A, \ldots, A, \partial A, \partial A}.
\end{eqnarray*}
By the symmetry in the last entries $\partial A_{k_{\alpha-1}}$ and $\partial A_{k_\alpha}$,
we will without loss of generality assume $k_{\alpha-1}\leq k_\alpha$.
Applying Lemma \ref{le:10} yields
\begin{eqnarray*}
& & \norm{\int  (\nabla\sigma)^2
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi}{L^1L^2} \leq  \\
& &\leq C_n \norm{f}{L^2}
 exp(C_n(\sum\limits_k 2^{k(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k(n+3)/2}\sup\limits_{U\in SU(n), t\to x(t)}
\| A_{k}(t,x(t)+Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}))\times\\
& & \sum\limits_{k_\alpha}
\sum\limits_{k_{\alpha-1}\leq k_\alpha} 2^{k_{\alpha-1}(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| \partial A_{k_{\alpha-1}}(t,x+
Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k_{\alpha-1}(n+3)/2}\sup\limits_{U\in SU(n), t\to x(t)}
\| \partial A_{k_{\alpha-1}}(t,x(t)+Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}))\times \\
& &\times (2^{k_\alpha(n-1)/2}\sup\limits_{U\in SU(n)} \sup\limits_x
\| \partial A_{k_\alpha}(t,x+ Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k_\alpha(n-5)/2}\sup\limits_{U\in SU(n), t\to x(t)}
\| \partial A_{k_\alpha}(t,x(t)+Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}).
\end{eqnarray*}
We simply now replace (at the expense of a constant)
 $\partial A_{k_{\alpha-1}}$ by $2^{k_{\alpha-1}} A_{k_{\alpha-1}}$,
which is in turn smaller than $2^{k_{\alpha}} A_{k_{\alpha-1}}$
in all the norms above involving $A_{k_{\alpha-1}}$.  We get an estimate
\begin{eqnarray*}
& & \norm{\int  (\nabla\sigma)^2
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi}{L^1L^2} \leq  \\
& &\leq C_n \norm{f}{L^2} exp(C_n(\sum\limits_k 2^{k(n-1)/2}\sup\limits_{U, x}
\| A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k(n+3)/2}\sup\limits_{U,x(t)}
\| A_{k}(t,x(t)+Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}))\times \\
& &
(\sum\limits_{k} 2^{k(n-1)/2}\sup\limits_{U, x}
\| A_{k}(t,x+
Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &\sum\limits_k 2^{k (n+3)/2}\sup\limits_{U,x(t)}
\| A_{k}(t,x(t)+Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}})\times \\
& &\times (\sum\limits_k 2^{k(n+1)/2}\sup\limits_{U,x}
\| \partial A_{k}(t,x+ Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}+ \\
& &
+ 2^{k(n-3)/2}\sup\limits_{U,x(t)}
\| \partial A_{k}(t,x(t)+Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}).
\end{eqnarray*}
\noindent Next,  using the fact that
$\norm{A}{L^2L^\infty}\leq \varepsilon$, one sees that to control
the term
$$
\int (i \dpr{\vec{A}}{\nabla\sigma})
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi.
$$
in $L^1L^2$, we need to control $\|\int \nabla\sigma(t,x,\xi)
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi\|_{L^2_t L^2_x}$.

For the proof of that, we can proceed by interpolation
between the $L^1L^2$ estimates of Lemma \ref{le:10} and the $L^\infty L^2$ estimates
 of Lemma \ref{le:34}.  An even easier way is the following. By inspection of the
proof of Lemma \ref{le:10}, one sees that $L^2_t L^2_x$ estimate  is
much easier to prove (since $L^2_t$ norm
 commutes with $L^2_x$ norm)  and one gets
\begin{eqnarray*}
& & \|\int \nabla\sigma(t,x,\xi)
e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\Omega(\xi) \widehat{f}(\xi) d\xi\|_{L^2_t L^2_x}\leq \\
& & \leq C_n \norm{f}{L^2} exp(C_n\sum\limits_k 2^{k(n-1)/2}\sup\limits_{U, x}
\| A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}} )\times\\
& & \times \sum\limits_k 2^{k(n-1)/2}\sup\limits_{U,x}
\| \partial  A_{k}(t,x+ Uz)\|_{L^2_t L^2_{z_2, \ldots, z_n}L^1_{z_1}}.
\end{eqnarray*}
Note that by the convexity of the norms
\begin{eqnarray*}
& & 2^{k(n-1)/2} \sup\limits_{U,x}
\| \partial  A_{k}(t,x+ Uz)\|_{L^2_t L^2_{z_2, \ldots, z_n}L^1_{z_1}} \leq \\
& &
\leq (2^{k(n-1)/2} \sup\limits_{U,x}
\|A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}})^{1/2}\times \\
& &\times
(2^{k(n-1)/2} \sup\limits_{U,x}
\|\partial^2 A_{k}(t,x+ Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}})^{1/2},
\end{eqnarray*}
whence
\begin{eqnarray*}
& & \sum\limits_k 2^{k(n-1)/2} \sup\limits_{U,x}
\| \partial  A_{k}(t,x+ Uz)\|_{L^2_t L^2_{z_2, \ldots, z_n}L^1_{z_1}} \leq \\
& &\leq (\sum\limits_k 2^{k(n-1)/2} \sup\limits_{U,x}
\|A_{k}(t,x+ Uz)\|_{L^\infty_t L^2_{z_2, \ldots, z_n}L^1_{z_1}})^{1/2}\times \\
& &\times (\sum\limits_k 2^{k(n-1)/2} \sup\limits_{U,x}
\|\partial^2 A_{k}(t,x+ Uz)\|_{L^1_t L^2_{z_2, \ldots, z_n}L^1_{z_1}})^{1/2}
\leq \norm{A}{Y_2\cap Y_3}.
\end{eqnarray*}

\noindent Thus, it remains to prove Lemma \ref{le:10}.

\subsection{Proof of Lemma \ref{le:10}.}
To outline the main ideas,
we start the proof with the simpler case $\alpha=1$.
\subsubsection{The case $\alpha=1$}
We follow the method of Lemma \ref{le:34}.
According to \eqref{eq1}, \eqref{eq2},
$$
\Xi^{k_1}_F f(t,x)= e^{-i t \Delta} \sum\limits_j P^j[
(\int F_k(t,\cdot+z\theta_j^{-k})\chi(2^{2k}z) dz)\tilde{P}^j (f)],
$$
where the summation in $j$ is over the family $\{\theta_j^{-k}\}$.
Taking $L^2_x$ norms and taking into account the (almost)
orthogonality of the $j$ sum, we have
\begin{eqnarray*}
& &\norm{\Xi^{k_1}_F f(t,\cdot)}{L^2}\leq C_n  \left(\sum\limits_j
\norm{(\int F_k(t,\cdot+z\theta_j^{-k})
\chi(2^{2k}z) dz)\tilde{P}^j (f)}{L^2}^2\right)^{1/2}\leq \\
& &\leq C_n \left(\sum\limits_j \sup\limits_x|h_j(t,x)|^2
\norm{\tilde{P}^j f}{L^2}^2\right)^{1/2},
\end{eqnarray*}
where $h_j(t,x)= \int F_k(t,x+z\theta_j^{-k})
\chi(2^{2k}z) dz$.
Take  $L^1_t$ norm. We will show
\begin{equation}
\label{eqn20}
\begin{array}{l}
\norm{\left(\sum\limits_j \sup\limits_x|h_j(t,x)|^2
\norm{\tilde{P}^j f}{L^2}^2\right)^{1/2}}{L^1_t}\leq \\
\leq C_n
2^{k(n-1)/2} \sup\limits_{U\in SU(n)}
\|\sup\limits_x\|F_k(t, x+Uz)\|_{L^{2}_{z_2,\ldots, z_n}
L^1_{z_1}}\|_{L^1_t}\norm{f}{L^2}.
\end{array}
\end{equation}
This follows by complex interpolation between
\begin{equation}
\label{eqn10}
\begin{array}{l}
\norm{\sum\limits_j \sup\limits_x|h_j(t,x)|
\norm{\tilde{P}^j f}{L^2}}{L^1_t}\leq \\
\leq
\sup\limits_{U\in SU(n)}
\norm{\sup\limits_x
\norm{F_k(t,x+Uz)}{L^{\infty}_{z_2,\ldots, z_n}L^1_{z_1}}}{L^1_t}
\sum\limits_j \norm{\tilde{P}^j f}{L^2}
\end{array}
\end{equation}
and
\begin{equation}
\label{eqn11}
\norm{\sup\limits_j[ \sup\limits_x|h_j(t,x)|
\|\tilde{P}^j f\|_{L^2}]}{L^1_t}\leq  C_n
2^{k(n-1)} \norm{F_k}{L^1_t L^1_z}\sup_j \norm{\tilde{P}^j f}{L^2},
\end{equation}
because $\norm{F_k}{L^1_t L^1_x}=\sup_{U\in SU(n)}
\norm{\sup_x
\norm{F_k(t,x+Uz)|}{L^{1}_z}}{L^1_t}$. \\
The $l^1_j$ estimate,  \eqref{eqn10} is straightforward, since
$\tilde{P}^j f$ is indendent of $t$. We have
\begin{eqnarray*}
& & \norm{\sum\limits_j \sup\limits_x|h_j(t,x)|
\norm{\tilde{P}^j f}{L^2}}{L^1_t}=  \sum\limits_j
\norm{\sup\limits_x |h_j(t,x)|}{L^1_t}
 \norm{\tilde{P}^j f}{L^2}\leq \\
& &\leq \sup\limits_j \norm{\sup\limits_x |h_j(t,x)|}{L^1_t}
\sum\limits_j \norm{\tilde{P}^j f}{L^2}
\end{eqnarray*}
It remains to observe that
\begin{eqnarray*}
& & \sup\limits_j \norm{\sup\limits_x |h_j(t,x)|}{L^1_t}\leq
\sup\limits_\theta \norm{\sup\limits_x \int |F_k(t,x+z\theta)|dz}{L^1_t}= \\
& &=
\sup\limits_{U\in SU(n)}
\norm{\sup\limits_x\norm{F_k(t,x+Uz)|}{L^\infty_{z_2, \ldots, z_n}L^1_{z_1}}}{L^1_t}.
\end{eqnarray*}
For the $l^\infty_j$ estimate,  \eqref{eqn11},  write
\begin{eqnarray*}
& & \norm{\sup\limits_j [\sup\limits_x|h_j(t,x)|
\norm{\tilde{P}^j f}{L^2}]}{L^1_t}=  \norm{\sup\limits_x\sup\limits_j |h_j(t,x)| }{L^1_t}
\sup\limits_j \norm{\tilde{P}^j f}{L^2}.
\end{eqnarray*}
Note that
\begin{eqnarray*}
& &
\sup\limits_x \sup\limits_j |h_j(t,x)|\leq \sup\limits_\theta \sup\limits_x
\sum\limits_{l\leq -2k} \int |F_k(t,x+z\theta)|\varphi(2^{-l}z) dz\leq \\
& &\leq
\sum\limits_{l\leq -2k} \sup\limits_\theta \sup\limits_x
 \int |F_k(t,x+z\theta)|\varphi(2^{-l}z) dz.
\end{eqnarray*}
Since for every fixed $\theta$,
$2^{(l+k)(n-1)}\int_{\theta_0\in\mathbf S^{n-1}: |\theta_0-\theta|<2^{-l-k}}
d\theta_0 \gtrsim  1$, write
\begin{eqnarray*}
& &  \int |F_k(t,x+z\theta)|\varphi(2^{-l}z) dz\leq \\
& &\leq C_n
2^{(l+k)(n-1)} \int_{\theta_0\in\mathbf S^{n-1}: |\theta_0-\theta|<2^{-l-k}}
\int |F_k(t,x+z\theta)|\varphi(2^{-l}z) dz
 d\theta_0
\end{eqnarray*}
Expand out as before $F_k(t,x+z\theta)$ around $x+z\theta_0$. We get
\begin{eqnarray*}
& &  \sup\limits_\theta\sup\limits_x  \int |F_k(t,x+z\theta)|\varphi(2^{-l}z) dz\leq \\
& &\leq C_n
2^{(l+k)(n-1)} \sum\limits_{\gamma\geq 0}(\gamma!)^{-1}
\int_{\theta_0\in\mathbf S^{n-1}: |\theta_0-\theta|<2^{-l-k}}
\int |F_k^\gamma(t,x+z\theta_0)|\varphi^\gamma(2^{-l}z) dz d\theta_0\leq \\
& & \leq C_n
2^{(l+k)(n-1)} \sum\limits_{\gamma\geq 0}(\gamma!)^{-1}
\int_{\theta_0\in\mathbf S^{n-1}}
\int |F_k^\gamma(t,x+z\theta_0)|\varphi^\gamma(2^{-l}z) dz d\theta_0\leq \\
& &\leq
C_n
2^{k(n-1)} \sum\limits_{\gamma\geq 0}(\gamma!)^{-1}
\int_{\theta_0\in\mathbf S^{n-1}}
\int |F_k^\gamma(t,x+z\theta_0)||z|^{n-1}\varphi^\gamma(2^{-l}z) dz d\theta_0.
\end{eqnarray*}
Summing the last inequalities in $l$ implies
\begin{equation}
\label{eqn18}
\begin{array}{l}
|\sup\limits_x \sup\limits_j h_j(t,x)|\leq C_n
2^{k(n-1)} \sum\limits_{\gamma\geq 0}(\gamma!)^{-1}  \\\sum\limits_l  \int_{\theta_0\in\mathbf S^{n-1}}
\int |F_k^\gamma(t,x+z\theta_0)||z|^{n-1}\varphi^\gamma(2^{-l}z) dz d\theta_0
\leq
C_n
2^{k(n-1)} \norm{F_k(t, \cdot)}{L^1_x}
\end{array}
\end{equation}
Taking $L^1_t$ norm  yields
$$
 \norm{\sup\limits_x \sup\limits_j |h_j(t,x)|}{L^1_t}\leq C_n
2^{k(n-1)} \norm{F_k}{L^1_{t x}},
$$
as required.

\subsubsection{The case $\alpha>1$}
The strategy here is to start ``peeling off'' the functions
$\int F^\mu_{k_\mu}(t,x+z\xi/|\xi|) \chi(2^{2k_\mu}z) dz$ in a way
 similar to Lemma \ref{le:34}. Recall that the method presented
in Lemma \ref{le:34} starts the ``peeling'' argument with the terms
with the lowest frequency \\
$k_{\min}=\min(k_1, \ldots, k_\alpha)$.

We have here
the extra complication of having to take $L^1_t$. Moreover, we must
measure  the
last  term $\int F^\alpha_{k_\alpha}(t,x+z\xi/|\xi|) \chi(2^{2k_\alpha}z) dz$
in the $L^1_t$ norm, while all the other terms should be
 measured in $L^\infty_t$.

As we have alluded to above, the order of the frequencies is not
insignificant.
The case $\alpha=1$,  considered in the
previous section roughly corresponds to the case when the last  frequency
$k_\alpha$ is maximal, i.e. $k_\alpha=\max(k_1, \ldots, k_\alpha)$.
We consider this case, and then
we indicate the necessary changes
when $k_\alpha<\max(k_1, \ldots, k_\alpha)$. \\
{\bf SubCase 1: $k_\alpha=\max(k_1, \ldots, k_\alpha)$}.\\
By the symmetry of  the terms $1, \ldots, (\alpha-1)$,
assume without loss of generality $k_1\leq \ldots\leq  k_{\alpha-1}$.
In fact, following Lemma \ref{le:34}, let
$k_1=\ldots=k_{s_1}<k_{s_1+1}=\ldots=k_{s_2}<\ldots
k_{s_{m-1}+1}=\ldots=k_{s_m}=k_\alpha$.

According to \eqref{eqn12}, we can write
$\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha}$
as a free solution (with time  depending data). Recall also
\eqref{eqn:7} and \eqref{eqn:1}, which give an
estimate for its $L^2$ norm (for a fixed time $t$).
Take into account \eqref{eqn14}, to get
 \begin{eqnarray*}
& &
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}
\leq  \\
& &\leq C_n  (\sum\limits_{j_1}
\sum\limits_{j_2,  \ldots, j_m:|\theta_{j_r}^{-k_{s_r}}-\theta_{j_1}^{-k_{s_1}}|\leq
2^{k_{s_1}}+2^{k_{s_r}}}
 \norm{g_1^{j_1} \ldots g_{s_1}^{j_1} Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))}{L^2_x}^2)^{1/2},
\end{eqnarray*}
where for $\mu\in [s_{r-1}+1, s_r]$, we have introduced
$$
g_\mu^{j_r}(t,x)= \int F_{k_\mu}^\mu(t,x+z
\theta_{j_r}^{-k_\mu})\chi(2^{2k_\mu}z) dz.
$$
Furthermore, by support considerations
(as discussed in the proof of Lemma \ref{le:34}),
$$
Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}=Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
Q^{j_2}_{2^{k_{s_2}}+(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_r}_{2^{k_{s_r}}+(2s_1+2)2^{k_{s_1}}}.
$$
We conclude
\begin{equation}
\label{eqn16}
\begin{array}{l}
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^2_x}
\leq C_n  (\sum\limits_{j_1, j_2, \ldots, j_m}
 \|g_1^{j_1}(t,x) \ldots g_{s_1}^{j_1}(t,x) \times \\
\times   Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}\ldots
Q^{j_r}_{2^{k_{s_r}}+(2s_1+2)2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu^{j_r(\mu)}(t,\cdot))\|_{L^2_x}^2)^{1/2}.
\end{array}
\end{equation}
Take $L^1_t$ norm. Our  claim is that
$$
\begin{array}{l}
\norm{\Xi^{k_1, \ldots, k_\alpha}_{F_1, \ldots, F_\alpha} f}{L^1_t L^2_x}
\leq C_n (1+2s_1)^n
\prod_{\mu=1}^{s_1} 2^{k_\mu(n-1)/2}\sup\limits_t \sup\limits_{U\in SU(n)}
\sup\limits_x \norm{F_{k_\mu}^\mu(t, x+Uz)}{L^2_{z_2, \ldots, z_n}L^1_{z_1}} \\
\times
 \|(\sum\limits_{j_2,  \ldots, j_m}
\norm{Q^{j_2}_{2^{k_{s_2}}+(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}} (f(\cdot)
\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))}{L^2}^2)^{1/2}\|_{L^1_t}.
\end{array}
$$
On account of \eqref{eqn16}, this follows
by a complex multilinear  interpolation between the $l^1_j$ estimate
\begin{eqnarray*}
& & \|\sum\limits_{j_1, j_2,  \ldots, j_m}
 \|g_1^{j_1} \ldots g_{s_1}^{j_1} Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))\|_{L^2_x}\|_{L^1_t} \leq \\
& &\leq \prod_{\mu=1}^{s_1} (\sup\limits_t \sup\limits_{\theta, x}
\int |F_{k_\mu}^\mu(t,x+z\theta)|dz) \times \\
& &\times \|\sum\limits_{
j_1, j_2,  \ldots, j_m} \| Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
Q^{j_2}_{2^{k_{s_2}}+(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}} (f(\cdot)
\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))\|_{L^2}\|_{L^1_t}.
\end{eqnarray*}
and the $l^\infty_j$ estimate
\begin{eqnarray*}
& &\| \sup\limits_{j_1,j_2, \ldots, j_m}
 \|g_1^{j_1} \ldots g_{s_1}^{j_1} Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))\|_{L^2_x}\|_{L^1_t}
 \leq \\
& &\leq C_n \prod_{\mu=1}^{s_1} 2^{k_\mu(n-1)}
\norm{F_{k_\mu}^\mu}{L^\infty_t L^1_x} \times \\
& &\times
\|\sup\limits_{j_1, j_2, \ldots, j_m}\|Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}} Q^{j_2}_{2^{k_{s_2}}+
(2s_1+2)2^{k_{s_1}}}\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}} (f(\cdot)
\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))\|_{L^2}\|_{L^1_t}.
\end{eqnarray*}
Note that after the complex interpolation, the sum in $j_1$ disappears (at the expense of a constant $C_n(1+2s_1)^n$),
since we have estimated by \eqref{eqn17}.

The proof of the $l^1_j$ estimate is immediate, by just pulling  out \\
$$
\prod_{\mu=1}^{s_1}
\sup_{j_1,t,x} |g_1^{j_1}(t,x)\ldots g_{s_1}^{j_1}(t,x)|\leq
\prod_{\mu=1}^{s_1} \sup\limits_t \sup\limits_{\theta, x}
\int |F_{k_\mu}^\mu(t,x+z\theta)|dz.
$$
For the $l^\infty_j$ estimate, estimate by
\begin{eqnarray*}
& &\| \sup\limits_{j_1,j_2, \ldots, j_m}
 \|g_1^{j_1} \ldots g_{s_1}^{j_1} Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha g_\mu^{j_{r(\mu)}}(t,\cdot))\|_{L^2_x}\|_{L^1_t}
 \leq \\
& &\leq \prod_{\mu=1}^{s_1} \sup\limits_t \sup\limits_{j_1,x} |g_\mu^{j_1}(t,x)| \times \\
& &\times \|\sup\limits_{j_1,j_2, \ldots, j_m}
\|Q^{j_1}_{2^{k_{s_1}}+2s_1 2^{k_{s_1}}}
\ldots
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}}
(f(\cdot)\prod_{\mu=s_1+1}^\alpha
g_\mu^{j_{r(\mu)}}(t,\cdot))\|_{L^2_x}\|_{L^1_t}.
\end{eqnarray*}
By \eqref{eqn18} however,
$$
\sup\limits_t \sup\limits_{j_1,x} |g^{j_1}_\mu(t,x)|\leq C_n
2^{k_\mu(n-1)}\sup\limits_t \norm{F^\mu_{k_\mu}(t,\cdot)}{L^1_x},
$$
as required.

This shows the main step in the argument. We continue in this
fashion (and as in Lemma \ref{le:34}, we keep incurring
constants comming from the increased
overlap of the supports of  the operators $Q^j$)
until we reach the maximal frequency $k_\alpha$. In this final step,
we finally have to  take the $L^1_t$ norm on
$\int F_{k_\alpha}(t, x+z\theta)\chi(2^{2k_\alpha}z) dz$. We have
\begin{eqnarray*}
& & \norm{\Xi^{k_1, \ldots, k_\alpha}_{F^1, \ldots, F^\alpha}}{L^1_t L^2_x}\leq \\
& & \leq C_n^\alpha \prod_{\mu=1}^{k_{s_{m-1}}}
2^{k_\mu(n-1)/2}\sup\limits_t \sup\limits_{U\in SU(n)}
\sup\limits_x \norm{F_{k_\mu}^\mu(t, x+Uz)}{L^2_{z_2, \ldots, z_n}L^1_{z_1}}\times \\
& & \times  \|(\sum\limits_{j_m}
\|\prod_{\mu=s_{m-1}+1}^\alpha g_\mu^{j_m}(t,\cdot)
Q^{j_m}_{2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}+
\ldots (2(s_{m}-s_{m-1})+2)2^{k_{s_m}}} [f(\cdot)]\|_{L^2}^2)^{1/2}\|_{L^1_t}.
\end{eqnarray*}
Observe that since $p(\alpha,m)=2^{k_{s_m}}+(2s_1+2)2^{k_{s_1}}+
\ldots (2(s_{m}-s_{m-1})+2)2^{k_{s_m}}\leq 4\alpha 2^{k_{s_m}}$, we have
$$
(\sum\limits_{j_m} \norm{Q^{j_m}_{p(\alpha,m)}g}{L^2}^2)^{1/2}\leq
C_n (p(\alpha,m)/2^{k_{s_m}})^n \norm{g}{L^2}\leq C_n \alpha^{n}\norm{g}{L^2} .
$$
It remains  to show
\begin{eqnarray*}
& & \|(\sum\limits_{j_m}
\|\prod_{\mu=s_{m-1}+1}^\alpha g_\mu^{j_m}(t,\cdot)
Q^{j_m}_{p(\alpha, m)} [f(\cdot)]\|_{L^2}^2)^{1/2}\|_{L^1_t}
\leq C_n^\alpha (\prod_{\mu=s_{m-1}+1}^{\alpha-1} \sup\limits_{t, \theta, x}\int
|F_{k_\mu}^\mu(t,x+z \theta)|dz)\times \\
& &\times   2^{k_\alpha(n-1)/2}
\sup\limits_{U\in SU(n)}
\|\sup\limits_x \|F_{k_\alpha}(t, x+Uz)
\|_{L^2_{z_2, \ldots, z_n}L^1_{z_1}}\|_{L^1_t}\norm{f}{L^2}.
\end{eqnarray*}
This follows as in \eqref{eqn20}, by interpolating between
\begin{eqnarray*}
& & \|\sum\limits_{j_m}
\|\prod_{\mu=s_{m-1}+1}^\alpha g_\mu^{j_m}(t,\cdot)
Q^{j_m}_{p(\alpha,m)} [f(\cdot)]\|_{L^2}\|_{L^1_t}
\leq \\
& &\leq C_n (\prod_{\mu=s_{m-1}+1}^{\alpha-1} \sup\limits_{t, \theta, x}\int
|F_{k_\mu}^\mu(t,x+z \theta)|dz)
 \sum\limits_{j_m}
\|Q^{j_m}_{p(\alpha,m)} [f(\cdot)]\|_{L^2}\times\\
& &\times \sup\limits_{\theta}  \|\sup\limits_x\|\int
|F_{k_\alpha}^\alpha(t,x+z \theta)|dz\|_{L^1_t}
\end{eqnarray*}
and
\begin{eqnarray*}
& & \|\sup\limits_{j_m}
\|\prod_{\mu=s_{m-1}+1}^\alpha g_\mu^{j_m}(t,\cdot)
Q^{j_m}_{p(\alpha,m)} [f(\cdot)]\|_{L^2}\|_{L^1_t}
\leq \\
& &\leq C_n (\prod_{\mu=s_{m-1}+1}^{\alpha-1} 2^{k_\mu(n-1)}
\norm{F_{k_\mu}^\mu}{L^\infty_t L^1_{x}})
 \sup\limits_{j_m}
\|Q^{j_m}_{p(\alpha,m)} [f(\cdot)]\|_{L^2}\times\\
& &\times 2^{k_\alpha(n-1)}
\norm{F_{k_\alpha}^\alpha}{L^1_t L^1_{x}}) .
\end{eqnarray*}
We omit the details, as they are exactly as in the proof of \eqref{eqn20}.
This completes the proof of Subcase I.  \\
{\bf Subcase II: $k_\alpha<\max(k_1, \ldots, k_\alpha)$}\\
Let for some $1\leq r_0 <\alpha$, we have $k_{r_0}=
\max(k_1, \ldots, k_\alpha)$.
The estimates proved in Subcase I yield
\begin{eqnarray*}
& & \norm{\Xi^{k_1, \ldots, k_\alpha}_{F^1, \ldots, F^\alpha}}{L^1_t L^2_x}\leq
 C_n^\alpha
(\prod_{\mu=1,\mu\neq r_0}^{\alpha-1}
2^{k_\mu(n-1)/2}\sup\limits_t \sup\limits_{U\in SU(n)}
\sup\limits_x \norm{F_{k_\mu}^\mu(t, x+Uz)}{L^2_{z_2, \ldots, z_n}L^1_{z_1}})
 \times \\
& & \times 2^{k_\alpha(n-1)/2}\sup\limits_t \sup\limits_{U\in SU(n)}
\sup\limits_x \norm{F_{k_\alpha}^\alpha(t, x+Uz)}{L^2_{z_2, \ldots, z_n}L^1_{z_1}}\times\\
& &\times 2^{k_{r_0}(n-1)/2}
\sup\limits_{U\in SU(n)}
\|\sup\limits_x \|F_{k_{r_0}}^{r_0}(t, x+Uz)\|_{L^2_{z_2, \ldots,
z_n}L^1_{z_1}}\|_{L^1_t}
\norm{f}{L^2}
\end{eqnarray*}
Since $k_\alpha<\max(k_1, \ldots, k_\alpha)=k_{r_0}$, we have $2^{k_\alpha(n-1)/2}2^{k_{r_0}(n-1)/2}\leq 2^{k_\alpha(n-5)/2}2^{k_{r_0}(n+3)/2}$. We get an estimate
\begin{eqnarray*}
& & \norm{\Xi^{k_1, \ldots, k_\alpha}_{F^1, \ldots, F^\alpha}}{L^1_t L^2_x}\leq
 C_n^\alpha
(\prod_{\mu=1,\mu\neq r_0}^{\alpha-1}
2^{k_\mu(n-1)/2}\sup\limits_t \sup\limits_{U\in SU(n)}
\sup\limits_x \norm{F_{k_\mu}^\mu(t, x+Uz)}{L^2_{z_2, \ldots, z_n}L^1_{z_1}})
 \times \\
& & \times 2^{k_\alpha(n-5)/2}\sup\limits_t \sup\limits_{U\in SU(n)}
\sup\limits_x \norm{F_{k_\alpha}^\alpha(t, x+Uz)}{L^2_{z_2, \ldots, z_n}L^1_{z_1}}\times\\
& &\times 2^{k_{r_0}(n+3)/2}
\sup\limits_{U\in SU(n)}
\|\sup\limits_x \|F_{k_{r_0}}^{r_0}(t, x+Uz)\|_{L^2_{z_2, \ldots,
z_n}L^1_{z_1}}\|_{L^1_t}
\norm{f}{L^2}
\end{eqnarray*}
This finishes the Proof of Lemma \ref{le:10}.

\section{Strichartz estimates in
$L^2_t L^{\f{2(n-1)}{(n-3)}}_{x_2, \ldots, x_n}L^2_{x_1}$}
\label{sec:8}
In this section, we establish Proposition \ref{prop:3}. We concentrate
on the case $n\geq 4$, since this is what we use anyway.
On the  other hand,
minor changes are needed in the proofs  for
$n=2$ and $n=3$.

Let us show first
Proposition \ref{prop:3} for the case $A=0$, that is for free solutions.
\begin{lemma}
\label{le:890}
Let $n\geq 4$. Then
\begin{eqnarray}
\label{eq:600}
& & \sup\limits_{U\in SU({\mathbf R}^n), x(t)} \norm{e^{i t \Delta} f(x(t)+U z)}{L^2_t L^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}}\leq C_n \norm{f}{L^2} \\
\label{eq:601}
& &
\sup\limits_{U\in SU({\mathbf R}^n), x(t)} \norm{ \int\limits_0^t e^{i (t-s) \Delta}
F(s,(x(s)+U z)ds)}{L^2_t L^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}}\leq C_n
\norm{F}{L^1L^2}.
\end{eqnarray}
\end{lemma}
\begin{proof}
For a fixed $U\in SU({\mathbf R}^n)$ and fixed measurable function $x(t)$,
denote \\ $W(t) f(y):=(e^{i t \Delta} f)(x(t)+Uy)$. According
to Proposition \ref{prop:14}, we need to verify that $W(t):L^2\to L^2$ and
$W(t)W(s)^*:L^1_{y_2, \ldots, y_n} L^2_{y_1}\to
L^\infty_{y_2, \ldots, y_n} L^2_{y_1}$ with norm no larger than \\
$C_n |t-s|^{(n-1)/2}$. \\
The $L^2$ boundedness is obvious since,
$$
\norm{W(t)f}{L^2}=\norm{(e^{it \Delta}f)(x(t)+U y)}{L^2_y}=
\norm{(e^{it \Delta}f)(x(t)+y)}{L^2_y}=
\norm{e^{it \Delta}f}{L^2_y}=\norm{f}{L^2}.
$$

For the dispersive estimates, note that
$$
W(t)W(s)^* f(y)= (e^{i (t-s)\Delta} f)(y+U^*x(s)-U^* x(t)).
$$
It suffices to verify
the decay estimate for a familly of extreme points, whose convex
span is $w^*$ dense in the unit ball of
$L^1_{y_2, \ldots, y_n} L^2_{y_1}$. Since the $\delta$ functions
provide such a set in $L^1_X$ for any measure space $X$, it will
suffice to take $f(y)=\delta(\bar{y}-b) g(y_1)$, where
$\bar{y}=(y_2, \ldots, y_n)$, $b\in {\mathbf R}^{n-1}$ and $g\in L^2(\mathbf R^1)$.
Fix $s$ and $t$ and denote $z=U^*x(s)-U^* x(t)$, which is a fixed vector in ${\mathbf R}^n$.
Compute $W(t)W(s)^* f$
$$
W(t)W(s)^* f(y)=(e^{i (t-s) \partial_1^2} g)(y_1+z_1)
[e^{i(t-s)\Delta_{n-1}}\delta(\cdot-b)](\bar{y}+\bar{z}).
$$
Immediately,
$\norm{W(t)W(s)^* f(\bar{y})}{L^2_{y_1}}=
|e^{i(t-s)\Delta_{n-1}}\delta(\cdot-b)(\bar{y}+\bar{z})|\|g\|_{L^2_{y_1}}$ and thus
$$
\norm{W(t)W(s)^* f}{L^\infty_{y_2, \ldots, y_n}
L^2_{y_1}}\leq C_n|t-s|^{-(n-1)/2}\|g\|_{L^2_{y_1}}=
C_n|t-s|^{-(n-1)/2}\|f\|_{L^1_{z_2, \ldots, z_n}L^2_{y_1}}.
$$
Note that the constant $C_n$ is independent of $g$, and $z\in {\mathbf R}^n$
and depends only on the dimension $n$.
\end{proof}

For the case $A\neq 0$, we proceed as in the proof of Theorem
\ref{theo:pisa}. We first establish a ``naive'' Strichartz estimate similar to Proposition \ref{prop:7}. Then, we show that to extend the
naive Strichartz estimate indefinitely in time (under the assumption
that $A$ is small), we need an estimate of the parametrix, similar to Lemma \ref{le:parametrix} in the appropriate norms.

We start with the
naive Strichartz estimate. This is done
exactly the same way as Proposition \ref{prop:7},
given that we already have Lemma \ref{le:890}.
\begin{proposition}
\label{prop:8} For a fixed integer  $k_0$, there exists a time
$T_0=T_0(k_0)\leq \infty$, so that
whenever $0<T< T_0$, $\psi\in \mathcal S$
\begin{equation}
\label{eq:1820}
\sup\limits_{U\in SU({\mathbf R}^n), x(t)} \norm{(P_{<k_0}\psi)(t, x(t)+U z)}{L^2(0,T)
L^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}}\leq
C(T,k_0)(\norm{\psi(0,\cdot)}{L^2}+
\norm{\mathcal L\psi}{L^1L^2}).
\end{equation}
Moreover, $C(T,k_0)$ depends on  $T$ in a continuous way.
\end{proposition}
Fix $k_0$ and a small $\varepsilon$.
Set as before $0<T^*\leq \infty$ to be the maximum time,
so that for every $0<T<T^*$, one has
$$
\sup\limits_{U\in SU({\mathbf R}^n), x(t)} \norm{(P_{<k_0}\psi)(t, x(t)+U z)}{L^2(0,T)
L^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}}\leq
\varepsilon^{-1}(\norm{\psi(0,\cdot)}{L^2}+
\norm{\mathcal L\psi}{L^1L^2}),
$$
for all Schwartz functions $\psi$.

The goal would be again to show that for small
enough $\varepsilon$, we have that $T^*=\infty$. This is reduced in a standard way (recall that $P_{<k_0}$ has an integrable kernel and is therefore bounded on
$L^2(0,T)
L^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}$) to the following
estimate for the parametrix, constructed in Section \ref{sec:par}. This
 needs to be compared to Lemma \ref{le:parametrix}.
\begin{lemma}
\label{le:9}
Given
$\norm{\vec{A}}{\tilde{Y}_1\cap Y_2\cap Y_3}\leq \varepsilon$, integer $k$ and $T>0$,
and for every function $f_k\in L^2({\mathbf R}^n)$ with $supp\ \widehat{f_k}
\subset \{|\xi|\sim 2^k\}$,
one can find a
function
$v_k:[0,T)\times {\mathbf R}^n\to \mathcal C$, so that $supp\
\widehat{v_k}\subset \{|\xi|\sim 2^k\}$ and
\begin{eqnarray*}
& &\norm{v_k(0,x)-f_k}{L^2}\leq C\varepsilon \norm{f_k}{L^2} \\
& & \sup\limits_{U\in SU(n), x(t)}
\norm{v_k(t, x(t)+Uz)}{L^2_tL^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}}
\leq C\norm{f_k}{L^2}\\
& &\norm{\mathcal L v_k}{L^1L^2}\leq C\varepsilon \norm{f_k}{L^2},
\end{eqnarray*}
for some $C$ independent of $f, k, T$.
\end{lemma}
\noindent It remains to prove Lemma \ref{le:9}.
\begin{proof}(\ref{le:9})
We reduce as in Section \ref{sec:par} to the case, when $k=0$ and
$supp \hat{A}\subset
\{\xi: |\xi|<<1\}$ and {\it without the condition}
 $supp\ \hat{v}\subset \{\xi: |\xi|\sim 1\}$.
We again use the function $v$ constructed in Section \ref{sec:par}. \\
Since we have already verified
$\norm{v(0,x)-f}{L^2}\leq C\varepsilon \norm{f}{L^2}$ and
$\norm{\mathcal L v}{L^1L^2}\leq C\varepsilon \norm{f}{L^2}$ under the consitions imposed
in Lemma \ref{le:9}, it remains to check
$$
\sup\limits_{U\in SU(n), x(t)}
\norm{v(t, x(t)+Uz)}{L^2(0,T)
L^{2(n-1)/(n-3)}_{z_2, \ldots, z_n}L^2_{z_1}}\leq C\norm{f}{L^2}.
$$
for
$$
v(t,x)=\Lambda f(t,x) = \int e^{i \sigma(t,x,\xi)} e^{-4\pi^2 i t|\xi|^2} e^{2\pi i \dpr{\xi}{x}}\Omega(\xi) \hat{f}(\xi) d\xi.
$$
Expanding $e^{i \sigma}=\sum\limits_\alpha (i^\alpha \sigma^\alpha)/(\alpha!)$ reduces
matters to showing
\begin{equation}
\label{eq:op1}
\sup\limits_{U\in SU(n), x(t)}   \norm{[\Lambda^\alpha f](t,x(t)+Ux)}{L^2(0,T)
L^{2(n-1)/(n-3)}_{x_2, \ldots, x_n}L^2_{x_1}}\leq C_n^\alpha \varepsilon^\alpha
\norm{f}{L^2}.
\end{equation}
Fix $U\in SU(n)$ and a measurable function $x(t)$. Set
$$
W^\alpha(t) f(x)= [\Lambda^\alpha f](t,x(t)+Ux)=
\int \sigma^\alpha(t,x(t)+Ux,\xi) e^{-4\pi^2 i t|\xi|^2}
e^{2\pi i \dpr{\xi}{x(t)+Ux}}\Omega(\xi) \hat{f}(\xi) d\xi.
$$
By Proposition \ref{prop:14}, \eqref{eq:op1} follows from
 the energy estimate
$$
\norm{W^\alpha(t) f}{L^2}\leq C_n^\alpha(\sum\limits_k
\sup\limits_{U,x} \|A_k(t, x+Uz)\|_{L^\infty_t
L^\infty_{z_2, \ldots, z_n}L^1_{z_1}})\alpha
\norm{f}{L^2}
$$
and
the decay estimate
$$
\norm{W^\alpha(t)W^\alpha(s)^* f}{L^{\infty}_{x_2,
\ldots, x_n} L^2_{x_1}}\leq C_n^{\alpha} (\sum\limits_k 2^{k(n-1)}
\|A_k\|_{L^\infty_t L^1_{x}}) ^{2\alpha} |t-s|^{-(n-1)/2}
\norm{f}{L^1_{x_2,
\ldots, x_n} L^2_{x_1}}.
$$
By interpolation between the last two estimates, we obtain the ``modified decay estimate''
\begin{eqnarray*}
& &
\norm{W^\alpha(t)W^\alpha(s)^* f}{L^{p'}_{x_2,
\ldots, x_n} L^2_{x_1}}\leq \\
& &\leq C_n^{\alpha} (\sum\limits_k 2^{k(n-1)/p_0}
\sup\limits_{U,x} \|A_k(t, x+Uz)\|_{L^\infty_t
L^{p_0}_{z_2, \ldots, z_n}L^1_{z_1}}) ^{2\alpha} |t-s|^{- (n-1)/(2p_0)}
\norm{f}{L^p_{x_2,
\ldots, x_n} L^2_{x_1}}= \\
& &= C_n^\alpha |t-s|^{-(n-1)/(2p_0)} \norm{A}{\tilde{Y}_1}^{2\alpha}
\norm{f}{L^p_{x_2,
\ldots, x_n} L^2_{x_1}},
\end{eqnarray*}
which implies \eqref{eq:op1}, as long as $p_0<(n-1)/2$,
according to the Remark before  Proposition \ref{prop:14}.
\subsection{Energy estimates for $W^\alpha$}
Observe that the case $\alpha=0$ is
simply the energy estimate in Lemma \ref{le:890}, while for
the general case
observe\footnote{Recall  that $\Omega$ is a radial  function.}
\begin{eqnarray*}
& &
W^\alpha(t) f(x)= [\Lambda^\alpha f](t,x(t)+Ux)= \\
& &= \int \sigma^\alpha(t,x(t)+Ux,U^* \xi) e^{-4\pi^2 i t|\xi|^2}
e^{2\pi i \dpr{\xi}{U^* x(t)+x-y}}\Omega(\xi) d\xi f(Uy) dy.
\end{eqnarray*}
By a simple translational invariance,
$\norm{f(U\cdot)}{L^2}=\norm{f}{L^2}$ and
$\norm{A(t, x(t)+ U\cdot)}{Y_j}=\norm{A}{Y_j}$,
we have that by
\eqref{eq:906}, the energy estimate $\norm{W^\alpha(t) f}{L^2}\leq C_n^\alpha\varepsilon^\alpha \norm{f}{L^2}$  is satisfied for every $\alpha>0$.
\subsection{Decay estimates for $W^\alpha$}
The case $\alpha=0$ is  the decay estimate in Lemma \ref{le:890}.

For $\alpha\geq 1$, we are following the approach of the dispersive estimates in
Section \ref{sec:6}. Write
\begin{eqnarray*}
& &W^{\alpha}(t)W^{\alpha}(s)^*f(x)= \\
& &=
\prod_{\mu=1}^\alpha (\sum\limits_{k_\mu\leq -2}
\sum\limits_{l_\mu\leq -2k_\mu}
\sum\limits_{\gamma_\mu} (\gamma_\mu!)^{-1}
\sum\limits_{j_\mu}
\int A^{\gamma_\mu}_{k_\mu}(t, x(t)+Ux +z \theta^{l_\mu+k_\mu}_{j_\mu})
\varphi^{\gamma_\mu}
(2^{-l_\mu}z)dz )\times \\
& &\times
H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f(x),
\end{eqnarray*}
where
\begin{eqnarray*}
& & H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f(x)  =
\int\limits
 \overline{\sigma^{\alpha}(s,x(s)+U y,\xi)}
 e^{-4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x(t)-x(s)+Ux-Uy}}
 \times \\
& & \times  \prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}^{\gamma_\mu}
(2^{l_\mu+k_\mu}(\xi/|\xi|-\theta_{j_\mu}^{l_\mu+k_\mu}))
\Omega^2(\xi) f(y) dy d\xi.
\end{eqnarray*}
Hence by Lemma \ref{le:pointwise}, we conclude
\begin{eqnarray*}
& &
\norm{W^{\alpha}(t)W^{\alpha}(s)^*f}{L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}}\leq  \\
& & \leq \sup\limits_x \prod_{\mu=1}^\alpha (\sum\limits_{k_\mu\leq -2}
\sum\limits_{l_\mu\leq -2k_\mu}
\sum\limits_{\gamma_\mu} (\gamma_\mu!)^{-1}
\sum\limits_{j_\mu}
\int A^{\gamma_\mu}_{k_\mu}(t, x(t)+Ux +z \theta^{l_\mu+k_\mu}_{j_\mu})
\varphi^{\gamma_\mu}
(2^{-l_\mu}z)dz ) \times \\
& & \times \sup\limits_{j_1,l_1,k_1;\ldots; j_\alpha, l_\alpha,k_\alpha}
\norm{H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f}{L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}}.\\
& &
\leq C_n^\alpha  (\norm{A}{Y_1})^\alpha
\sup\limits_{j_1,l_1,k_1;\ldots; j_\alpha, l_\alpha,k_\alpha}
\norm{H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f}{L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}}.
\end{eqnarray*}
Dualizing the needed estimate for
$\sup\limits_{j_1,l_1,k_1;\ldots; j_\alpha, l_\alpha,k_\alpha}
\norm{H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha} f}{L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}}$, reduces matters to showing that
\begin{eqnarray*}
& & (H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha})^* f(x)=
\int\limits   \sigma^{\alpha}(s,x(s)+U x,\xi)
 e^{4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x(s)-x(t)+Ux-Uy}}
 \times \\
& & \times  \prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}^{\gamma_\mu}
(2^{l_\mu+k_\mu}(\xi/|\xi|-\theta_{j_\mu}^{l_\mu+k_\mu}))
\Omega^2(\xi) f(y) dy d\xi
\end{eqnarray*}
is a mapping $L^1_{x_2,
\ldots, x_n} L^2_{x_1}\to L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}$ with norm no bigger than $C_n^\alpha \norm{A}{Y_1}^{\alpha}
|t-s|^{-(n-1)/2}$.

Again, as in Section \ref{sec:6}, we expand
$$
\sigma^\alpha(s, x(s)+Ux,\xi)=\prod_{\nu=1}^\alpha (\sum\limits_{k_\nu\leq -2,
l_\nu\leq -2k_\nu}
\sum_{\gamma_\nu, j_\nu} (\gamma_\nu!)^{-1}
\int A^{\gamma_\nu}_{k_\nu}(t, x(s)+Ux +z \theta^{l_\nu+k_\nu}_{j_\nu})
\varphi^{\gamma_\nu}
(2^{-l_\nu}z)dz )
$$ and
estimate by Lemma \ref{le:pointwise}. We get
\begin{eqnarray*}
& &
 \norm{(H^\alpha_{j_1,l_1+k_1;\ldots; j_\alpha, l_\alpha+k_\alpha})^* f}{L^\infty_{x_2,
\ldots, x_n} L^2_{x_1}}\leq C_n^\alpha\norm{A}{Y_1}^\alpha \times \\
& &\times \int\limits
 e^{4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{x(s)-x(t)+Ux-Uy}}
  \prod_{\mu=1}^\alpha \psi_{j_\mu,l_\mu+k_\mu}^{\gamma_\mu}
(2^{l_\mu+k_\mu}(\xi/|\xi|-\theta_{j_\mu}^{l_\mu+k_\mu})) \times \\
& &\times  \prod_{\nu=1}^\alpha \psi_{j_\nu,l_\nu+k_\nu}^{\gamma_\nu}
(2^{l_\nu+k_\nu}(\xi/|\xi|-\theta_{j_\nu}^{l_\nu+k_\nu}))
\Omega^2(\xi) f(y) dy d\xi
\end{eqnarray*}
By the Krein-Milman theorem, the linear span of
$\{\delta_b(x_2, \ldots, x_n) g(x_1): g\in L^2(\mathbf R^1)\}$ is $w^*$ dense in
$L^1_{x_2, \ldots, x_n} L^2_{x_1}$. Thus,
it will suffice to verify the estimate
\begin{eqnarray*}
& &\sup\limits_{b,\bar{x} \in R^{n-1}, z\in {\mathbf R}^n, \tilde{\theta}_1, \ldots,
\tilde{\theta}_{2\alpha}\in\mathbf S^{n-1}}
\|\int\limits
 e^{4\pi^2 i (t-s)|\xi|^2} e^{2\pi i \dpr{\xi}{z+U(x_1, \bar{x})-U(y_1, b)}}\times \\
& &
  \prod_{\mu=1}^{2\alpha}
\psi_{\mu}(2^{m_\mu}(\xi/|\xi|-\tilde{\theta}_\mu))
\Omega^2(\xi) g(y_1) dy_1 d\xi\|_{L^2_{x_1}} \leq C_n^\alpha |t-s|^{-(n-1)/2}
\norm{g}{L^2(\mathbf R^1)}.
\end{eqnarray*}
Clearly, by rotational invariance, we can assume $U=Id$.
But then the expression above is equal to
$$
\int\limits
 e^{4\pi^2 i (t-s)|\xi_1|^2} e^{2\pi i (\xi_1)(z_1+x_1-y_1)} K(\xi_1)
g(y_1) dy_1= e^{-i(t-s)\partial_1^2}K(\partial_1/2i \pi)[g(\cdot)](z_1+x_1),
$$
where
$$
 K(\xi_1) = \int \int\limits
 e^{4\pi^2 i (t-s)|\bar{\xi}|^2}
e^{2\pi i \dpr{\bar{\xi}}{\bar{z}+\bar{x}-b}}
  \prod_{\mu=1}^{2\alpha}
\psi_{\mu}(2^{m_\mu}(\xi/|\xi|-\tilde{\theta}_\mu))
\Omega^2(\xi) d\xi_2\ldots d \xi_n.
$$
By Lemma \ref{le:osc} (see its second statement), we have
$$
\sup\limits_{\xi_1} |K(\xi_1)|\leq C_n^{\alpha}|t-s|^{-(n-1)/2}.
$$
We get
$$
\sup\limits_{b,\bar{x} \in R^{n-1}, z\in {\mathbf R}^n, \tilde{\theta}_1, \ldots,
\tilde{\theta}_{2\alpha}} \norm{e^{-i(t-s)\partial_1^2}K(\partial_1/2i \pi)[g(\cdot)]}{L^2}
\lesssim \norm{g}{l^2}\sup\limits_{\xi_1} |K(\xi_1)|\leq C_n^{\alpha}|t-s|^{-(n-1)/2}
\norm{g}{l^2},
$$
as required.

\end{proof}
\section{Global regularity for  Schr\"odinger maps}
\label{sec:11}
In this section, we sketch the proof of Theorem \ref{theo:MSM}.
As it was discussed earlier, it will suffice to show that $A$
stays small in the space of vector potentials
 $Y$, given the {\it a priori} information
that $u$ is small in (a portion of )
the solution space to be described below.
Let $\dot{X}^\alpha$ be the completion of all Schwartz functions in the norm
$$
\|u\|_{\dot{X}^\alpha} = (\sum\limits_k 2^{2 \alpha k}\sup\limits_{q,r- Str.}
\|u_k\|_{L^q L^r}^2 +2^{2 \alpha k} \sup\limits_{U\in SU(n), x(t)}
\|u_k(t, x(t)+Ux)\|_{L^2_t L^{2(n-1)/(n-3)}_{x_2,
\ldots, x_n}L^2_{x_1}}^2 )^{1/2}
$$
Let $X^s:=\dot{X}^{s}\cap \dot{X}^{0}$.
We will generally
measure the solution in $X^s$, but
moreover, we will show it is small in $\dot{X}^{s_0}$. Note that
since $s>s_0$,
$\norm{u}{\dot{X}^{s_0}}\lesssim \norm{u}{X^s}$.

The space $Y$ of acceptable vector potentials
is on the level of smoothness of $\dot{X}^{s_0}$.\\
 Fix $\delta>0$, so that
$s>(n+1)/2+\delta$. We will only assume that $\norm{f}{\dot{H}^{(n+1)/2+\delta}}
\norm{f}{\dot{H}^{(n-5)/2-\delta}}<<1$. Clearly\footnote{Note that in the fromulation of the theorem, we have asked for a lot more, namely
$\norm{f}{\dot{H}^s}=\varepsilon\norm{g}{\dot{H}^s}<<1$ for all $s\in[0,(n+1)/2+\delta)$.}
$\norm{f}{\dot{H}^{s_0}}\leq (\norm{f}{\dot{H}^{(n+1)/2+\delta}}\norm{f}{\dot{H}^{(n-5)/2-\delta}})^{1/2}<<1$.  \\
 Under this assumptions, it will suffice to check
\begin{itemize}
\item ($A=A(u)$ is controlled by
$\norm{u}{\dot{X}^{s_0}}$ and $\norm{u}{X^s}$)
\begin{eqnarray}
\label{eqnm:3}
& & \norm{A(u)}{Y}\leq C_n
\norm{u}{\dot{X}^{(n+1)/2+\delta}} \norm{u}{\dot{X}^{(n-5)/2-\delta}}, \\
\label{eqnm:4}
& & (\sum\limits_k 2^{2ks} \norm{A_k}{L^2_t L^{2n/(n-2)}}^2)^{1/2}\leq
C_n \norm{u}{\dot{X}^s}\norm{u}{\dot{X}^{s_0}}\quad\q \textup{for every} \quad s\geq 0.
\end{eqnarray}
\item ($N(u)$ is controlled by $\norm{u}{\dot{X}^{s_0}}$ and $\norm{u}{X^s}$)
\begin{eqnarray}
\label{eqnm:6}
& & \norm{\partial^{s} N(u)}{L^1 L^2} \leq C_n
\norm{u}{\dot{X}^{s}}(\norm{u}{\dot{X}^{s_0}}^2+\norm{u}{\dot{X}^{s_0}}^4)
\end{eqnarray}
\end{itemize}
Le ts us first show how  Theorem \ref{theo:MSM} follows from \eqref{eqnm:3},
\eqref{eqnm:4}, \eqref{eqnm:6}.

To that end, we know that the Strichartz estimates  hold for at
least for some time $T$, so that $\norm{A}{Y_T}\leq \varepsilon$. Fix one such $T$.
We have for every $s\geq 0$, by  \eqref{eqnm:4} and \eqref{eqnm:6}
\begin{eqnarray*}
& & \norm{u}{\dot{X}_T^{s}}\leq
C_n (\norm{f}{\dot{H}^{s}}+
\norm{\partial^s N(u)}{L^1_TL^2})+  C_n \norm{\nabla u}{L^2_T L^n_x}
(\sum\limits_k 2^{2ks}\norm{A_k}{L^2_T L^{2n/(n-2)}}^2)^{1/2}\leq \\
& &\leq C_n \norm{f}{\dot{H}^{s}}+C_n \norm{u}{\dot{X}_T^s}
(\norm{u}{\dot{X}_T^{s_0}}^2+\norm{u}{\dot{X}_T^{s_0}}^4)+
C_n \norm{u}{\dot{X}_T^{s_0}}
(\sum\limits_k 2^{2ks}\norm{A_k}{L^2_T L^{2n/(n-2)}}^2)^{1/2}\leq \\
& & \leq C_n (\norm{f}{\dot{H}^{s}}+
\norm{u}{\dot{X}_T^s}
(\norm{u}{\dot{X}_T^{s_0}}^2+
\norm{u}{\dot{X}_T^{s_0}}^4)).
\end{eqnarray*}
In particular, for $s=s_0$ and the smallness of
$\norm{f}{\dot{H}^{s_0}}$, it follows that
$$
\norm{u}{\dot{X}_T^{s_0}}\leq C_n \norm{f}{\dot{H}^{s_0}}.
$$
This means that $\norm{u}{\dot{X}_T^{s_0}}$ is
small (independently of $T$), which  in
turn implies that
$$
\norm{u}{\dot{X}_T^{s}}\leq C_n \norm{f}{\dot{H}^{s}},
$$
for every $s\geq 0$.
But how far can we really push that? Recall \eqref{eqnm:3}, which
gives us a control of $A$ back in terms of $u$. Namely, since
$$
 \norm{A(u)}{Y_T}\leq C_n
\norm{u}{\dot{X}_T^{(n+1)/2+\delta}} \norm{u}{\dot{X}_T^{(n-5)/2-\delta}}\leq C_n
\norm{f}{\dot{H}^{(n+1)/2+\delta}} \norm{f}{\dot{H}^{(n-5)/2-\delta}}<<1.
$$
This implies  that $\norm{A}{Y_T}$ is small and one could
apply back the Strichartz estimates, which means
that $T$ could be taken to be $\infty$.
Theorem \ref{theo:MSM} follows.

\subsection{Proof of \eqref{eqnm:3}, \eqref{eqnm:4}}
We will not give the full details of
 \eqref{eqnm:3}, \eqref{eqnm:4} , since
these are standard Besov type estimates for products.

Let us for example consider the estimate for  $\norm{A}{Y_3}$.
First, it is not hard to see that the terms containing $\partial_t A$ one  uses the equation
\eqref{eqnm:1}, to write it
like
$$
\partial_t A = \partial_t Q_1(u,\bar{u})\sim  \tilde{Q}_1 (u,u_t)\sim
 \tilde{Q}_1 (u,\Delta u)+\tilde{Q}_1 (u,N(u)).
$$
Thus, everything is reduced to the terms containing
$\partial^2 A$ and $N(u)$, the latter being easy to treat.

So, we concentrate on the terms involving $\partial^2 A$.
For those,
take into account $\partial^2A_k \sim 2^{2k}A_k$ and $A\sim \partial^{-1}Q(u, \bar{u})$, to conclude
\begin{eqnarray*}
& &\norm{A}{Y_3}\sim \sum\limits_k  2^{k(n+3)/2} \sup\limits_{U\in SU(n), x(t)}
\norm{A_k(t, x(t)+Ux)}{L^1_t L^2_{x_2, \ldots, x_n}L^1_{x_1}}\sim \\
& &\sim
\sum\limits_k  2^{k(n+1)/2}
\sup\limits_{U\in SU(n), x(t)}
\norm{(u v)_k (t, x(t)+Ux)}{L^1_t L^2_{x_2, \ldots, x_n}L^1_{x_1}}.
\end{eqnarray*}
Following Lemma 3.1 and Lemma 3.2 in \cite{rodnianski},
we have to split
$(u v)_k$
in\footnote{Here $v$ might be either $u$ or $\bar{u}$.}
two types of terms  - high low interactions $u_{\sim k} v_{<k+5}$
and
high-high interactions
$\sum\limits_{l>k-5}(u_{l} v_{l-2<\cdot< l+2})_k$.

The high-low interactions are more difficult to
handle in this context\footnote{For the high-high interactions, one can actually split the $(n+1)/2$ derivatives between the two terms and get a better more
balanced estimate.}, so let us concentrate on these.
We have by Cauchy-Schwartz and
Bernstein inequalities
\begin{eqnarray*}
& & 2^{k(n+1)/2}\sup\limits_{U\in SU(n), x(t)}
\norm{u_{k} v_{ k-m}
(t, x(t)+Ux)}{L^1_t L^2_{x_2, \ldots, x_n}L^1_{x_1}}\leq \\
& & \leq 2^{k(n+1)/2}
\sup\limits_{U, x(t)}
\norm{u_{ k} (t, x(t)+Ux)}{L^2_t L^{2(n-1)/(n-3)}_{\bar{x}}L^2_{x_1}}
\norm{v_{k-m } (t, x(t)+Ux)}{L^2_t L^{n-1}_{\bar{x}}L^2_{x_1}}\\
& &\leq C_n 2^{-\delta m}
\sup\limits_{U, x(t)} 2^{k((n+1)/2+\delta)}
\norm{u_{k} (t, x(t)+Ux)}{L^2_t L^{2(n-1)/(n-3)}_{\bar{x}}L^2_{x_1}}\times \\
& &
\sup\limits_{U, x(t)} 2^{(k-m)((n-5)/2-\delta)}
\norm{v_{k-m } (t, x(t)+Ux)}{L^2_t L^{2(n-1)/(n-3)}_{\bar{x}}L^2_{x_1}}
\end{eqnarray*}
Note that at this stage to apply the Bernstein inequality in the variables
$x_2, \ldots, x_n$ \\
$\|v_{k-m}\|_{L^{n-1}_{x_2, \ldots, x_n}L^2_{x_1}}\leq C_n
2^{(k-m)(n-5)/2}\|v_{k-m}\|_{L^{2(n-1)/(n-3)}_{x_2, \ldots, x_n} L^2_{x_1}}$,
we needed $n-1>2(n-1)/(n-3)$, which is the dimensional restriction $n>5$. \\
Summing in $m>-5$ yields
\begin{eqnarray*}
& & \sum\limits_k 2^{k(n+1)/2}\sup\limits_{U\in SU(n), x(t)}
\norm{u_{ k} v_{k-m}(t, x(t)+Ux)}{L^1_t
L^2_{x_2, \ldots, x_n} L^1_{x_1}}\leq \\
& & \leq
(\sum\limits_k 2^{k((n+1)+2\delta)} \sup\limits_{U, x(t)}
\norm{u_k(t, x(t)+Ux)}{L^2_t L^{2(n-1)/(n-3)}_{\bar{x}}
L^2_{x_1}}^2)^{1/2}\times \\
& & \times
(\sum\limits_k  2^{k((n-5)-2\delta)}  \sup\limits_{U, x(t)}
\norm{v_k (t, x(t)+Ux)}{L^2_t
L^{2(n-1)/(n-3)}_{\bar{x}}L^2_{x_1}}^2)^{1/2}\leq \\
& & \leq C_n
\norm{u}{\dot{X}^{(n+1)/2+\delta}}\norm{v}{\dot{X}^{(n-5)/2-\delta}}.
\end{eqnarray*}

The proof of \eqref{eqnm:4} is in fact very similar
 and boils down to the same Besov space estimates for products.
\subsection{Proof of \eqref{eqnm:6}}
The estimates for the nonlinearities are the easiest ones.
It basically suffice to apply the Kato-Ponce type-estimates
$$
\norm{\partial^s(u v)}{L^r}\leq C\norm{\partial^s u}{L^p}\norm{v}{L^q}+
C\norm{\partial^s v}{L^p}\norm{u}{L^q}
$$
whenever $1/r=1/p+1/q$. We omit the details.

\section{Appendix}
\subsection{Decay estimates for the free Schr\"odinger equation
with initial data Fourier supported in a small cap}
\begin{lemma}
\label{le:osc}
Let $k_1, \ldots, k_\mu$ be positive integers.
Let also $\{\theta_j\}\in \mathbf S^{n-1}$ and
$\psi_j$ be smooth cutoff  functions adapted to
$\{|\xi|\leq 1\}$, whose smoothness bounds are uniform in $j$.
 Then there exists a constant $C$
depending only on the dimension, so that
\begin{equation}
\label{eq:514}
\sup\limits_{\theta_1, k_1, \ldots, \theta_\mu\in\mathbf S^{n-1}, k_\mu}\sup\limits_x | \int\limits
 e^{-4\pi^2 i t |\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\prod_{j=1}^\mu \psi_{j}(2^{k_j}(\xi/|\xi|-\theta_{j}))
\varphi(\xi) d\xi|\leq  C_n^\mu |t|^{-n/2}.
\end{equation}
Also, if one fixes $\xi_1$ and integrates with respect
to $\xi_2, \ldots, \xi_n$,
$$
\sup\limits_{\theta_1, k_1, \ldots, \theta_\mu \in \mathbf S^{n-1}, k_\mu}
\sup\limits_{x,\xi_1} | \int\limits
 e^{-4\pi^2 i t |\bar{\xi}|^2} e^{2\pi i \dpr{\bar{\xi}}{\bar{x}}}
\prod_{j=1}^\mu \psi_{j}(2^{k_j}(\xi/|\xi|-\theta_{j}))
\varphi(\xi) d\xi_2\ldots d\xi_n|\leq  C_n^\mu |t|^{-(n-1)/2}.
$$
\end{lemma}
\begin{proof}
We prove only \eqref{eq:514}, since the second statement in Lemma
\ref{le:osc} requires  only a slight adjustment of the argument.

This a standard stationary phase argument,
except that we have to keep track of the
derivatives that may pile up from the cutoffs
$\psi_{j}(2^{k_j}(\xi/|\xi|-\theta_{j}))$. Fix $k_1, \ldots, k_\mu$ and let
$k=\max(k_1, \ldots, k_\mu)=k_{j_0}$.
If $2^k\geq \sqrt{t}$, we pass to polar coordinates
and estimate by
\begin{eqnarray*}
& &C \int\limits_{|\theta-\theta_{j_0}|\lesssim  2^{-k}} |\int
e^{-4\pi^2 i t \rho^2} e^{2\pi \rho i \dpr{\theta}{x}}
\varphi(\rho)d\rho|d\theta \leq
C 2^{-k(n-1)} |t|^{-1/2}\leq C |t|^{-n/2},
\end{eqnarray*}
where we have used the decay estimate by $C t^{-1/2}$ for the 1 D Schr\"odinger
equation.

Thus assume $2^k\leq \sqrt{t}$.
We (smoothly) split the region of integration, according to the
 size of the derivative of the phase. If
 $|-8\pi^2 t \xi+2\pi x|\leq \sqrt{t}$,
we estimate by absolute values and obtain the desired estimate
by the volume of the $\xi$ support, which is $C_n |t|^{-n/2}$.

 It remains to show that
\begin{eqnarray*}
& &| \int\limits
 e^{-4\pi^2 i t |\xi|^2} e^{2\pi i \dpr{\xi}{x}}
\varphi(2^{-m} t^{-1/2} (-8\pi^2 t \xi+2\pi x))
\prod_{j=1}^\mu \psi_{j}(2^{k_j}(\xi/|\xi|-\theta_{j}))
\varphi(\xi) d\xi|\\
& & \leq C 2^{-m} C_{n}^\mu |t|^{-n/2},
\end{eqnarray*}
Now that we have $2^{k_j}\leq 2^{k}\leq \sqrt{t}$. Then, the
argument goes as in the classical estimate, that is after $n+1$
integration by parts with the phase function
$\varrho(\xi)=i(-4\pi^2  t |\xi|^2+2\pi  \dpr{\xi}{x})$,
(at  each step one gains at least a
factor of $2^{-m}|t|^{-1/2}$
and  loses a factor of $C_{n} \sqrt{|t|}$ at the most),
we put in absolute values. Taking
into account the volume of the support
$\leq C_n 2^{mn} t^{-n/2}$, we estimate by
$$
 D_{n}2^\mu  2^{-m} |t|^{-n/2}\leq C_n^\mu 2^{-m} |t|^{-n/2},
$$
whence summation by $m\geq 0$ yields the result.
\end{proof}
\subsection{Estimates on the error term}
In this section we give an estimate on the error terms $E^k$, defined in
\eqref{eq:1201}.
\begin{lemma}
Let  $\norm{A}{Y_0}\leq \varepsilon$ and
$s\geq 0$. Let also $1=1/p_1+1/p_2; 1/2=1/q_1+1/q_2$. Then
\begin{eqnarray}
\label{eq:error}
\sum\limits_k 2^{2ks} \norm{E^k}{L^1 L^2}^2\lesssim \varepsilon^2
\sum\limits_k 2^{2ks} \norm{u_k}{L^\infty L^2}^2+
\norm{u}{L^{p_1}L^{q_1}}^2
\sum\limits_k 2^{2k(s+1)} \norm{A_k}{L^{p_2}_t L^{q_2}_x}^2,\\
\label{e:11}
\sum\limits_k 2^{2ks} \norm{E^k}{L^1 L^2}^2\lesssim \varepsilon^2
\sum\limits_k 2^{2ks} \norm{u_k}{L^\infty L^2}^2+
\norm{\nabla u}{L^{p_1}L^{q_1}}^2
\sum\limits_k 2^{2k s} \norm{A_k}{L^{p_2}_t L^{q_2}_x}^2.
\end{eqnarray}
In particular, from \eqref{eq:error}, with
$s=0, q_1=2, p_1=\infty, p_2=1, q_2=\infty$,
and since $\norm{A_k}{L^1L^\infty_x}
\leq 2^{kh} \norm{A_k}{L^1 L^{n/h}}$
\begin{equation}
\label{e:10}
\sum\limits_k \norm{E^k}{L^1 L^2}^2\lesssim \varepsilon^2
\sum\limits_k  \norm{u_k}{L^\infty L^2}^2
\end{equation}
\end{lemma}
\begin{proof}
We estimate on a term-by-term basis in formula \eqref{eq:1201}.

For the first term, we have
\begin{eqnarray*}
\sum\limits_k  2^{2ks} \norm{[P_k,\vec{A}_{\leq k-4}]
\nabla u_{k-1\leq \cdot\leq k+1}}{L^1L^2}^2
\lesssim
\norm{\nabla A}{L^1L^\infty}^2 \sum\limits_k 2^{2ks} \norm{u_k}{L^\infty L^2}^2.
\end{eqnarray*}

The second and third terms in \eqref{eq:1201} are treated in a similar
fashion, so we
concentrate on the second one. For any positive $h\leq 1$, we have
\begin{eqnarray*}
& & \sum\limits_k 2^{2ks} \left(\sum\limits_{l\geq k-2} \norm{P_k( A_l \nabla
u_{l-2\leq \cdot \leq l+2})}{L^1L^2}\right)^2\lesssim\\
& &\lesssim \sum\limits_k 2^{2ks}
\left(\sum\limits_{l\geq k-2} 2^{h k} \norm{P_k( A_l \nabla
u_{l-2\leq \cdot \leq l+2})}{L^1L^{2n/(n+h)}}\right)^2\lesssim \\
& &\lesssim\sum\limits_k 2^{2 k(s+h)} \left(\sum\limits_{l\geq k-2} 2^l
\norm{A_l}{L^1 L^{2 n/h}}
\norm{u_{l-2\leq \cdot \leq l+2}}{L^\infty L^{2}}\right)^2\lesssim \\
& & \lesssim\sum\limits_k 2^{2 k(s+h)} \left(\sum\limits_{l\geq k-2} 2^{-h l}
(2^{l+lh}\norm{A_l}{L^1 L^{n/h}})
\norm{u_l}{L^2L^{2n/(n-2)}}\right)^2
\end{eqnarray*}
One obtains by
Lemma \ref{le:sequences} an estimate by
\begin{eqnarray*}
\sum\limits_l 2^{2l(1+h)}\norm{A_l}{L^1 L^{n/h}}^2
\sum\limits_l 2^{2ls} \norm{u_l}{L^\infty L^2}^2.
\end{eqnarray*}
The last fourth term in \eqref{eq:1201} can be bounded in two ways.
\begin{eqnarray*}
& & \sum\limits_k 2^{2ks} \norm{A_{k-1\leq \cdot\leq k+1}\nabla
u_{\leq k-4})}{L^1 L^2}^2\lesssim \norm{u}{L^{p_1} L^{q_1}}^2
\sum\limits_k 2^{2k (s+1)} \norm{A_k}{L^{p_2} L^{q_2}}^2,\\
& & \sum\limits_k 2^{2ks} \norm{A_{k-1\leq \cdot\leq k+1}\nabla
u_{\leq k-4})}{L^1 L^2}^2\lesssim \norm{\nabla u}{L^{p_1} L^{q_1}}^2
\sum\limits_k 2^{2k s} \norm{A_k}{L^{p_2} L^{q_2}}^2
\end{eqnarray*}

\end{proof}

redpajama_set_name: RedPajamaArXiv