x}f(y)dy$. Such a thing cannot be bounded by
%a square function $g$ with a favorable weight $2^{\delta m}$,
%and there must be a problem at the level of our applications
%to operators, since almost everywhere convergence doesn't hold
%for general orthogonal series, and so forth.
\begin{proposition} \label{prop:numericalbound}
There exists $C<\infty$ such that
for any martingale structure $\{E^m_j\}$,
any $\delta\ge 0$,
any $f_1,\dots,f_n\in L^1(\reals)$,
and any $n\ge 2$,
\begin{equation}
M_n(f_1,\dots,f_n)
\le \frac{C^{n+1}}{\sqrt{n!}}
g_{\delta}(f_1)\cdot g_\delta(\{f_k:k\ge 2\})^{n1}.
\end{equation}
Moreover for any $\delta'>\delta \ge 0$,
there exists $C<\infty$ such that for all $\{f_i\}$ and all $n\ge 2$,
\begin{equation}
M_n^*(f_1,\dots,f_n)
\le \frac{C^{n+1}}{\sqrt{n!}}
g_{\delta}(f_1)\cdot g_{\delta'}(\{f_k:k\ge 2\})^{n1}.
\end{equation}
\end{proposition}
In previous work \cite{christkiselevfiltrations}
we proved the simpler analogue
with $\delta=0$ and with the bound (for $M_n$)
$C^{n+1}(n!)^{1/2}g(\{f_k: k\ge 1\})^n$,
and applied it to the analysis of generalized eigenfunctions,
which can be expanded as sums over $n$ of such
iterated multiple integrals, where $f_k$ is
essentially $\exp(\pm 2i\lambda x)V(x)$, $V$ is the potential,
and $\lambda^2$ is a spectral parameter;
$g(f)$ is then a function of $\lambda$.
In the present work, we need a refinement in which
$f_1$ is essentially $\exp(\pm i\lambda x)h(x)$,
and $h$ is an arbitrary $L^2$ function, unrelated to $V$.
The quantity $g(h)$ is not appropriately bounded for
$h\in L^2$, forcing the introduction of the mollifying
factors $2^{\delta m}$ in its definition. This in turn
forces compensating factors of $2^{+\delta m}$,
leading to the above formulation.
\begin{proof}
It is proved in \cite{christkiselevfiltrations}
that there exist positive constants $b_n$ satisfying
$b_n\le C^{n+1}/\sqrt{n!}$
and $n^{1/2}b_{n+1}/b_n\to C$ as $n\to\infty$,
such that for all nonnegative real numbers $x,y$,
\begin{equation} \label{eq:modifiedbinomial}
b_n y^n + \sum_{i=2}^{n2} b_i b_{ni}x^iy^{ni} + b_n x^n
\le b_n (x^2+y^2)^{n/2}.
\end{equation}
It is also shown in \cite{christkiselevfiltrations} that
\begin{equation} \label{eq:onefunctionversion}
\begin{split}
M_n(f,\dots,f) &\le b_n g_0(f)^n
\\
M_n^*(f,\dots,f) &\le C_\e^n b_n g_\e(f)^n
\end{split}
\end{equation}
for every $\e>0$. Moreover,
for distinct functions $f_i$,
\begin{equation} \label{eq:modifiedonefunctionversion}
\begin{split}
M_n(f_1,\dots,f_n) &\le b_n g_0(\{f_i\})^n
\\
M_n^*(f_1,\dots,f_n)&\le C_\e^n b_n g_\e(\{f_i\})^n.
\end{split}
\end{equation}
Although this bound is not explicitly formulated in
\cite{christkiselevfiltrations}, it follows directly from exactly
the argument given there.
For $n\ge 2$ define $\tilde b_n = R^{n} b_{n2}$,
where $R$ is a sufficiently large positive constant, to be
determined later in the proof.
In order to simplify notation, we will prove
the result in the special case $f_2=f_3=\cdots =f_n=f$,
and will write $f_1=\tilde f$.
The proof will be by induction on $n$. First we will treat only
the case where $n\ge 6$, assuming the result for
all $n\le 5$, and at the end will discuss the modification
for small $n$.
We write
$f^m_j = \chi_{E^m_j}\cdot f$ and $\tilde f^m_j
= \chi_{E^m_j}\cdot \tilde f$,
where $\chi_{E}$ denotes always the characteristic function of a set $E$.
\begin{lemma}
If $R$ is chosen to be sufficiently large then for all $n\ge 2$
and any $\delta\ge 0$,
\begin{equation}
M_n(\tilde f,f,\dots,f)\le \tilde b_n g_{\delta}(\tilde f)
\cdot g_\delta(f)^{n1}.
\end{equation}
\end{lemma}
\begin{proof}
By inequality (4.6) of \cite{christkiselevfiltrations},
\begin{equation}\begin{split} \label{eq:inductivestep}
M_n(\tilde f,f,\dots,f)
\le
&M_n(\tilde f^1_1,f^1_1,\dots,f^1_1)
\\
& + M_{n1}(\tilde f^1_1,f^1_1,\dots,f^1_1)\cdot
{\textstyle\int} f^1_2
\\
& + \sum_{i=2}^{n2} M_{ni}(\tilde f^1_1,f^1_1,\dots,f^1_1)
\cdot M_i(f^1_2,\dots,f^1_2)
\\
& + {\textstyle\int}\tilde f^1_1\cdot M_{n1}(f^1_2,\dots,f^1_2)
+ M_n(\tilde f^1_2,f^1_2,\dots,f^1_2)
\\
\le
& M_n(2^{\delta}\tilde f^1_1,2^\delta f^1_1,\dots,2^\delta f^1_1)
\\
& + M_{n1}(2^{\delta} \tilde f^1_1,
2^\delta f^1_1,\dots,2^\delta f^1_1)
\cdot {\textstyle\int} 2^\delta f^1_2
\\
& + \sum_{i=2}^{n2} M_{ni}( 2^{\delta}\tilde f^1_1,2^\delta f^1_1,\dots,
2^\delta f^1_1)
\cdot M_i(2^\delta f^1_2,\dots,2^\delta f^1_2)
\\
& + {\textstyle\int} 2^{\delta}\tilde f^1_1
\cdot M_{n1}(2^\delta f^1_2,\dots,2^\delta f^1_2)
\\
& + M_n(2^{\delta}\tilde f^1_2,2^\delta f^1_2,\dots,2^\delta f^1_2) .
\end{split}\end{equation}
We have assumed that $n\ge 2$ and $\delta\ge 0$ to ensure that
at least one factor of $2^{\delta}$ offsets the factor of $2^{\delta}$.
The first and last terms in the preceding bound involve $M_n$ itself,
but the former involves only the restrictions of
$\tilde f,f$ to $E^1_1$, while the latter involves only their restrictions
to $E^1_2$; thus these expressions are in a sense simpler than
the original expression $M_n$. We will therefore use as
part of our induction hypothesis the desired inequalities for
$M_n(\tilde f^1_1,f^1_1,\cdots,f^1_1)$ and for
$M_n(\tilde f^1_2,f^1_2,\cdots,f^1_2)$.
For the justification of this method of reasoning see
the two paragraphs immediately following inequality (4.12) of
\cite{christkiselevfiltrations}.
The collection of all those sets $E^m_j$
with $m\ge 1$ and $j\le 2^{m1}$
forms a martingale structure on $E^1_1$;
however, when it is considered as such, the index $m$ should be
replaced by $m1$. Thus
the induction hypothesis, for the first term on the righthand
side of the preceding bound,
may be stated as
\begin{multline}
M_n(2^{\delta} \tilde f^1_1,2^{\delta} f^1_1,\cdots, 2^\delta f^1_1)
\\
\le \tilde b_n
\sum_{m=2}^\infty
2^{m\delta}
(\
\sum_{j=1}^{2^{m1}}
{\textstyle\int} \tilde f^m_j^2
)^{1/2}
\cdot
\Big[
\sum_{m=2}^\infty
2^{m\delta}
(\
\sum_{j=1}^{2^{m1}}
{\textstyle\int} f^m_j^2
)^{1/2}\Big]^{n1}\ .
\end{multline}
There is a corresponding bound for
$M_n(2^{\delta}\tilde f^1_2,2^\delta f^1_2,\cdots,2^\delta f^1_2)$,
with the inner sum running instead over $2^{m1} 0$
\[ u_1(x,E)m(E+i0)+u_2(x,E)=\frac{u(x,E)}{u(0,E)}= \gamma(E)
e^{i \xi(x,E)}(1+o(1)) \]
as $x \rightarrow \infty$
(there is no absolutely continuous spectrum for $E<0$).
The relation between $\gamma$ and $m(E+i0)$
follows by comparing the Wronskians of the left
and right hand sides (taken with their complex conjugates).
\end{proof}
%\noindent \it Remark. {\bf to be deleted later}
%\rm Notice that the whole proof used very few specific
%properties of $\xi(x,z),$ namely \eqref{lbph}, \eqref{ph1}, \eqref{ph2}
% and \eqref{ph3}.
%In particular, it can be readily adapted
%to the case of slowly varying potentials or Stark operators.
Now we are going to rewrite the spectral representation in a manner
convenient for the scattering theory. Notice that
\[ u_1(x,E) = \frac{1}{2i \Im m(E+i0)}
\left( \gamma(E)u(x,E)  \ov{\gamma}(E)\ov{u}(x,E)
\right). \]
Introduce
\begin{equation}
\psi(x,\lambda) = \lambda \ov{\gamma}(\lambda^2) u_1(x,\lambda^2)
= \frac{1}{2i}\Big(
u(x,\lambda^2)
 \frac{{\ov{\gamma}(\lambda^2)}}{\gamma(\lambda^2)}
\cdot \ov{u}(x,\lambda^2) \Big).
\end{equation}
where $u(x,E)$ continue to denote the generalized eigenfunctions
whose existence was established in Theorem~\ref{thm:complexeigenfunctions},
now for $E\in\reals^+$.
Then the results of this subsection may be summarized as follows.
\begin{proposition}
Suppose that $V\in L^1+L^p(\reals^+)$ for some $1 From the preceding section recall the generalized eigenfunctions
$\psi(x,\lambda) = (2i)\rp(u  e^{i\omega(\lambda)}\ov{u}))$,
where $\exp(i\omega(\lambda)) = \ov\gamma(\lambda^2)/\gamma(\lambda^2)$.
Recall the unitary bijection $U_V:L^2(\reals^+,d\lambda)\mapsto
\scripth_{\text{ac}}$ defined
by $U_Vf(x) = \sqrt{2/\pi}\int_0^\infty f(\lambda)\psi(x,\lambda)\,d\lambda$.
Finally, recall that
$e^{iH_Vt}(U_V f)(x) = \sqrt{2/\pi}\int_0^\infty e^{i\lambda^2 t}f(\lambda )\psi(x,\lambda)\,
d\lambda$.
Fix a martingale structure $\{E^m_j\}$ on $\reals^+$ that is
adapted to $V$, in the sense that $\int_{E^m_j}V^p
= 2^{m}\int_{\reals^+}V^p$ for all $m\ge 0$ and all $j$.
For any sufficiently small $\delta>0$, recall the functional
\begin{equation*}
g_\delta(f)(\lambda)
= \sum_{m=1}^\infty 2^{m\delta}
\Big(
\sum_{j=1}^{2^m}
\big\int_{E^m_j} e^{2i\phi(x,\lambda)}f(x)\,dx\big^2
+
\big\int_{E^m_j} e^{2i\phi(x,\lambda)}f(x)\,dx\big^2
\Big)^{1/2}.
\end{equation*}
We have shown that for all sufficiently small $\delta_0$,
$g_{\delta_0}(V)(\lambda)<\infty$ for almost every $\lambda\in\reals^+$.
Fix some $\delta<\delta_0$.
\begin{definition}
A compact set $\Lambda\Subset(0,\infty)$
is said to be a {\em set of uniformity} if
$g_\delta(V)$ is a bounded function of $\lambda\in\Lambda$,
and if
$u(x,\lambda)e^{i\phi(x,\lambda)}\to 0$
as $x\to+\infty$, uniformly for all $\lambda\in\Lambda$.
\end{definition}
\begin{lemma}
For any $f\in L^2(\reals^+)$,
for any $R<\infty$,
\begin{equation*}
\e^{itH_V}U_V f\_{L^2([0,R])} \to 0
\ \ \text{as } t\to\infty.
\end{equation*}
\end{lemma}
\begin{proof}
Since $\exp(itH_V)$ is unitary, it suffices to prove this
for all $f$ in some dense subspace of $L^2(\reals^+)$;
we choose the subspace consisting of all $f$ supported on
some set of uniformity $\Lambda$ (which depends on $f$).
The functions $\psi(x,\lambda)$ are uniformly bounded on
$[0,\infty)\times\Lambda$, as are their derivatives with respect
to $x$, from which it follows that $U_V$ is a compact mapping
from $L^2(\Lambda)$ to $L^2([0,R])$. Thus it suffices to establish
weak convergence to zero.
Now for any $h\in L^2([0,R])$,
\begin{equation}
\langle e^{itH_V}U_V f,h\rangle = \int_\Lambda
e^{i \lambda^2 t }f(\lambda)U_V^*h(\lambda)\,d\lambda,
\end{equation}
which converges to zero by the RiemannLebesgue lemma,
since $U_V^*h\in L^2(\reals^+)$.
\end{proof}
Define
\begin{align}
\psi_0(x,\lambda) & = (2i)\rp\big( e^{i\phi(x,\lambda)}
e^{i\omega(\lambda)}e^{i\phi(x,\lambda)}\big)
\\
U_V^\dagger f(x) & = \sqrt{2/\pi} \int_0^\infty f(\lambda) \psi_0(x,\lambda)\,d\lambda;
\end{align}
these are approximations to $\psi,U_V$, respectively.
The next lemma is the key step in showing that only the
leadingorder approximation $\psi_0$ to $\psi$ contributes to the
longtime asymptotics of the wave group.
\begin{lemma} \label{truekeylemma}
For any set of uniformity $\Lambda$ and any $R>0$,
there exists $C(R,\Lambda)<\infty$ such that for all
$f\in L^\infty(\Lambda)$,
\begin{equation}
\(U_VU_V^\dagger)f\_{L^2([R,\infty))}
\le C(R,\Lambda)\f\_{L^\infty},
\ \text{ where }\ C(R,\Lambda)\to 0 \text{ as } R\to\infty.
\end{equation}
\end{lemma}
\begin{proof}
It suffices to prove this with $\psi(x,\lambda)\psi_0(x,\lambda)$ replaced by
\begin{multline}
\Psi(x,\lambda)  e^{i\phi(x,\lambda)}
\\
= e^{i\phi(x,\lambda)}\sum_{n=1}^\infty S_{2n}(V,V,\dots,V)(x,\lambda)

e^{i\phi(x,\lambda)}\sum_{n=0}^\infty S_{2n+1}(V,V,\dots,V)(x,\lambda),
\end{multline}
for the conclusion for the other terms follows from this by complex conjugation.
We argue by duality; let $h$ be an arbitrary function in
$L^2([R,\infty))$ of norm $1$, and consider
\begin{multline}
\int_R^\infty e^{i\phi(x,\lambda)}h(x)S_{2n}(V,V,\dots,V)(x,\lambda)
\\
= \iint_{R\le t_0\le t_1\le\cdots\le t_{2n}}
e^{i\phi(t_0,\lambda)}h(t_0)\,dt_0\,
\prod_{k=1}^{2n}e^{\pm_k 2i\phi(t_k,\lambda)} V(t_k)\,dt_k.
\end{multline}
Here $\pm_k$ denotes a plus or minus sign depending on $k$ in any manner;
in fact these signs alternate in our expansion, but that is of no
importance here.
Introduce
\begin{equation}
g_{\delta}(h)(\lambda)
= \sum_{m=1}^\infty 2^{m\delta}
\Big(
\sum_{j=1}^{2^m}
\big\int_{E^m_j} e^{i\phi(x,\lambda)}h(x)\,dx\big^2
+
\big\int_{E^m_j} e^{i\phi(x,\lambda)}h(x)\,dx\big^2
\Big)^{1/2}.
\end{equation}
By Proposition~\ref{prop:numericalbound},
\begin{equation}
\Big\iint_{R\le t_0\le t_1\le\cdots\le t_{2n}}
e^{i\phi(t_0,\lambda)}h(t_0)\,dt_0\,
\prod_{k=1}^{2n} e^{\pm_k 2i\phi(t_k,\lambda)}V(t_k)\,dt_k
\Big
\le \frac{C^{n+1}}{\sqrt{2n!}}
g_{\delta}(h)(\lambda)g_\delta(V_R)^{2n}(\lambda)
\end{equation}
where $V_R(x) = V(x)\chi_{[R,\infty)}(x)$;
there is a corresponding bound for the terms arising from the multilinear
expressions $S_{2n+1}$.
We may dominate $\sup_R g_\delta(V_R)(\lambda)^2$
by $C' g_{\delta'}(V)(\lambda)^2$ for any $\delta'>\delta$;
to simplify notation we replace $\delta'$ again by $\delta$.
Consequently by summing over $n$ and comparing with the Taylor expansion
for the exponential function,
\begin{multline}
\Big
\langle \int_\Lambda \int_R^\infty f(\lambda)\left( u(x,\lambda)
e^{i \phi(x,\lambda)}\right) h(x)\,dx\,d\lambda
\rangle\Big
\\
\le
\int_\Lambda
Cg_{\delta}(h)(\lambda)
g_\delta(V_R)(\lambda)
e^{Cg_\delta(V)(\lambda)^2}
f(\lambda)\,d\lambda.
\end{multline}
Recall that
for any $\delta>0$, $\g_{\delta}(h)\_{L^2(\Lambda)}
\le C_{\Lambda,\delta}\h\_{L^2}\le C_{\Lambda,\delta}<\infty$,
for any compact $\Lambda\Subset(0,\infty)$, uniformly over all
martingale structures, not necessarily adapted to $h$.
The factor $\exp(Cg_\delta(V)(\lambda)^2)$
is bounded uniformly for $\lambda\in\Lambda$,
by definition of a set of uniformity.
Thus it suffices to show that
$\g_\delta(V_R)(\lambda)\_{L^2(\Lambda,d\lambda)}\to 0$ as $R\to\infty$.
Now in the sum \eqref{eq:defnofgdelta}
defining $g_\delta(V)$, the $\ell^2$ sum
over $j$ for fixed $m$
is $\le C2^{\e m}\V_R\_{L^2}$ in $L^2(\Lambda)$
for some $\e>0$,
so it suffices to show that
$\int_{E^m_j}e^{2i\phi(x,\lambda)}V_R(\lambda)\,dx
\to 0$ in $L^2(\Lambda)$. This holds by Lemma~\ref{lemma:parseval},
since $V_R\to 0$ in $L^1+L^p$ norm.
\end{proof}
\begin{corollary} \label{cor:dropmultilinearterms}
Let $\rho>0$.
For any $f\in L^2(\reals^+,d\lambda)$ supported on $[\rho,\infty)$,
\begin{equation}
\e^{itH_V}U_V f\sqrt{2/\pi}
\int_0^\infty e^{i\lambda^2 t}f(\lambda)\psi(x,\lambda)
\,d\lambda\_{L^2(\reals^+)}
\to 0
\text{ as } t\to\infty.
\end{equation}
\end{corollary}
The restriction on the support of $f$ comes about because
of the factor of $\lambda\rp$ in the definition of $\phi$,
which causes difficulties as $\lambda\to 0$.
\begin{proof}
It is straightforward to show that
$\int_0^\infty e^{i\lambda^2 t}f(\lambda)\psi(x,\lambda) \,d\lambda \to 0$
in $L^2([0,R])$ for any finite $R$.
Let $\e>0$.
Choose a set of uniformity $\Lambda\subset[\rho,\infty)$
and a function $F\in L^\infty (\Lambda)$
such that $\fF\_{L^2(\reals^+)}<\e$.
Then $\e^{itH_V}(U_V fU_V F)\_{L^2(\reals^+)}<\e$ for every
$t$, as well.
By the preceding lemma,
there exists $R<\infty$ such that
$\e^{itH_V}U_V F\sqrt{2/\pi}
\int_0^\infty e^{i\lambda^2 t}F(\lambda)\psi_0(x,\lambda)
\,d\lambda\_{L^2([R,\infty))}<\e$ for all $t\in\reals$.
Fixing such an $R$,
there exists $T<\infty$ such that
both $e^{itH_V}U_V F$ and $\sqrt{2/\pi}\int_0^\infty e^{i\lambda^2 t}F(E)\psi_0(x,\lambda)
\,d\lambda$ have $L^2([0,R])$ norms $<\e$, for all $t\ge T$.
Thus
\begin{equation}
\e^{itH_V}U_V f\sqrt{2/\pi}\int_0^\infty e^{i\lambda^2 t}F(\lambda)\psi_0(x,\lambda)
\,d\lambda\_{L^2(\reals^+)}<4\e\text{ for all } t\ge T.
\end{equation}
Finally, note that
$\\int_\rho^\infty g(\lambda) u_0(x,\lambda)\,d\lambda\_{L^2(\reals^+)}
\le C_\rho \g\_{L^2([\rho,\infty))}$
for all $g$, where $C_\rho<\infty$ for all $\rho>0$;
this follows from the proof of Lemma~\ref{lemma:parseval}.
Applying this to $g=fF$ allows us to replace $F$ again by $f$
in the preceding inequality, at the expense of replacing $4\e$
by $C_\rho\e$.
Since $\e>0$ was arbitrary, this completes the proof.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{A timedependent phase correction} \label{section:phasereduction}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The goal of this section is to convert the leadingorder asymptotics
$\exp(\pm i\lambda x \mp i(2\lambda)\rp\int_0^x V)$
to a more standard form,
$\exp(\pm i\lambda x \mp i(2\lambda)\rp\int_0^{2\lambda t} V)$.
\begin{lemma} \label{lemma:easyballisticapproximation}
Let $V\in L^1+L^2$. For any $f\in L^2$ supported in a compact
subinterval of $(0,\infty)$,
\begin{multline} \label{eq:twoeasyterms}
\Big\\int_0^\infty e^{i\lambda^2 t+i\lambda x
i(2\lambda)\rp\int_0^x V}
f(\lambda)\,d\lambda
\Big\_{L^2(\reals^+)}
\\
+
\Big\\int_0^\infty e^{i\lambda^2 t+i\lambda x
i(2\lambda)\rp\int_0^{2\lambda t} V}
f(\lambda)\,d\lambda
\Big\_{L^2(\reals^+)}
\to 0 \text{ as } t\to \infty.
\end{multline}
An analogous statement holds as $t\to+\infty$, and follows by
taking complex conjugates.
\end{lemma}
\begin{proof}
Since the mapping $f\mapsto \int_0^\infty e^{i\lambda^2 t+i\lambda x}
e^{i(2\lambda)\rp\int_0^x V} f(\lambda)\,d\lambda$
is bounded from $L^2([\rho,\infty))$ to $L^2(\reals^+)$ for every
$\rho>0$, and since the variant defined by replacing
$\int_0^x V$ by $\int_0^{2\lambda t}V$ is unitary,
it suffices to prove this merely for $f\in C^\infty_0$, which we
assume henceforth.
Let $\Lambda\Subset\reals^+$ be the support of $f$.
By integrating by parts once with respect to $\lambda$,
integrating $\exp(i[\lambda^2 t+\lambda x])$ and differentiating
the rest, and by moving absolute value signs inside the integral,
we obtain a pointwise in $(x,t)$ bound
\begin{multline}
\eqref{eq:twoeasyterms}
\le C\int_\Lambda x2\lambda t\rp
\big(1+
{\textstyle\int}_0^x V
+ {\textstyle\int}_0^{2\lambda t}V
+ tV(2\lambda t) \big)
\,d\lambda
\\
\le C(x+t)\rp
(\big(1+ {\textstyle\int}_0^x V + {\textstyle\int}_0^{C t}V \big)
\end{multline}
for the integrands in \eqref{eq:twoeasyterms}.
For $t\le 1$,
$x2\lambda t\rp\sim (x+t)\rp$, which tends to zero
in $L^2(\reals^+,dx)$.
Thus the first term behaves as desired. The second
and third terms do also, if $V\in L^1$.
For $V\in L^2$, $x\rp\int_0^x V\in L^2$ as well, so
the term $(x+t)\rp \int_0^x V$ is an $L^2$ function times
$x/(x+t)$, hence $\to 0$ in $L^2$ norm.
Lastly, $\int_0^{Ct}V=o(t^{1/2})$,
while $t^{1/2}/(x+t)$ is $O(1)$ in $L^2(\reals^+)$.
\end{proof}
\begin{lemma} \label{lemma:ballisticapproximation}
Let $V\in L^1+L^2$. For any $f\in L^2$ supported in a compact
subinterval of $(0,\infty)$,
\begin{equation} \label{differenceofcorrections}
\Big\
\int_0^\infty e^{i\lambda^2 t+i\lambda x}
\Big[ e^{i(2\lambda)\rp\int_0^x V}
e^{i(2\lambda)\rp\int_0^{2\lambda t} V}
\Big]
f(\lambda)\,d\lambda
\Big\_{L^2(\reals^+)}\to 0 \text{ as } t\to +\infty.
\end{equation}
\end{lemma}
\begin{proof}
As in the preceding lemma, it suffices to prove this for
$f\in C^\infty_0(\reals^+)$.
Fix a cutoff function $\eta\in C^\infty_0(\reals)$ supported on
$(2,2)$ and $\equiv 1$ on $[1,1]$, with $0\le\eta\le 1$.
Let $\epsilon$ be a strictly positive function, such that
$\epsilon(t)\to 0$ as $t\to+\infty$, at a rate to be specified below.
Set
\begin{equation}
\eta(\lambda,t,x) = \eta(\epsilon(t) t^{1/2} (x2\lambda t)).
\end{equation}
Consider the function of $(x,t)$
\begin{equation} \label{eq:functionofxandt}
\int_0^\infty e^{i\lambda^2 t+i\lambda x}
\Big[ e^{i(2\lambda)\rp\int_0^x V}
e^{i(2\lambda)\rp\int_0^{2\lambda t} V}
\Big]
f(\lambda)\,\eta(\lambda,t,x)\,d\lambda\ .
\end{equation}
We have
\begin{equation}\begin{split}
\eqref{eq:functionofxandt}
&\le C \int \int_{y2\lambda t\le 2\epsilon(t)\rp t^{1/2}} V(y)\,dy
\chi_{x2\lambda t\le 2\epsilon(t)\rp t^{1/2}} \,d\lambda
\\
& \le C \int \int_{yx\le 4\epsilon(t)\rp t^{1/2}} V(y)\,dy
\chi_{x2\lambda t\le 2\epsilon(t)\rp t^{1/2}} \,d\lambda
\\
&\le C\epsilon(t)\rp t^{1/2}
\int_{yx\le 4\epsilon(t)\rp t^{1/2}} V(y)\,dy
\\
&\le C\epsilon(t)^{2}MV(x),
\end{split}\end{equation}
where $M$ is the maximal function of Hardy and Littlewood.
Observe that the restriction $\eta(\lambda,t,x)\ne 0$ for
some $\lambda\in\Lambda$ implies that $ct\le x\le Ct$
for some $c,C\in\reals^+$ depending only on $\Lambda$,
provided $\epsilon(t)\ll t^{1/2}$,
and moreover that $V$ may be replaced by its restriction to
such an interval. If $V\in L^2$ then the $L^2$ norm of
$M$ applied to the restriction of $V$ to such an interval
tends to zero as $t\to\infty$, and by choosing $\epsilon(t)$
to tend to zero sufficiently slowly we find that
$\eqref{eq:functionofxandt}\to 0$ in $L^2$.
For $V\in L^1$ we have the pointwise bound
$C\epsilon(t)\rp t^{1/2}\int_{ct}^{Ct}V$,
which is $o(1)\cdot\epsilon(t)\rp$ in $L^2(x\sim t)$;
this tends to zero provided $\epsilon(t)$ does so sufficiently slowly.
In the general case $V\in L^1+L^2$, we decompose
\eqref{eq:functionofxandt}
into two parts, and estimate them separately.
There remains the contribution of $1\eta(\lambda,t,x)$:
\begin{equation}
\int_0^\infty e^{i\lambda^2 t+ i\lambda x}
f(\lambda)
\Big(e^{i(2\lambda)\rp \int_0^{x}V}
e^{i(2\lambda)\rp \int_0^{2\lambda t}V}
\Big)
(1\eta)(\lambda,t,x)
\,d\lambda.
\end{equation}
We would like to apply to it the method
of stationary phase. However, the phase function
$t\lambda^2+\lambda x(2\lambda)\rp\int_0^{2\lambda t}V$ is not
well behaved; its partial derivative with respect to $\lambda$ is
$x2\lambda t
+(2\lambda)^{2}(\int_0^{2\lambda t}V)
(2\lambda)\rp 2tV(2\lambda t)$,
and the final term is not well under control
unless one assumes $V(x) = O(x^{1/2})$.
Instead, we integrate by parts,
integrating $\exp(i[\lambda x\lambda^2 t])$
and differentiating the rest, to obtain
\begin{equation} \label{eq:abigjob}
i\int_0^\infty e^{i\lambda^2 t+ i\lambda x}
f(\lambda)
\frac{\p}{\p\lambda}
\Big[(x2\lambda t)\rp
\Big(e^{i(2\lambda)\rp \int_0^{x}V}
e^{i(2\lambda)\rp \int_0^{2\lambda t}V}
\Big)
(1\eta)(\lambda,t,x)
\Big] \,d\lambda.
\end{equation}
When the derivative is expanded according to Leibniz's rule,
various terms result. The main terms are those in which
$\partial/\partial\lambda$ acts on either of the two exponentials
$\exp( i(2\lambda)\rp(\int_0^* V) )$, and we discuss these first.
One such term is a constant multiple of
\begin{multline}
\int_0^\infty e^{i\lambda^2 t+ i\lambda x
i(2\lambda)\rp \int_0^{2\lambda t}V}
f(\lambda)
(x2\lambda t)\rp
(1\eta)(\lambda,t,x)
\lambda\rp tV(2\lambda t)\,d\lambda
\\
=
c
\int_0^\infty e^{i(y^2/4t)+ i(xy/2t)
it y\rp \int_0^{y}V}
\tilde f(y/2t)
(xy)\rp
(1\eta)(\lambda,t,x)
v(y)
\,dy
\end{multline}
where
we have substituted $y = 2\lambda t$ and written
$\tilde f(\lambda) = \lambda\rp f(\lambda)\in C^\infty_0$,
and where $v=V$ (for the present moment only).
The integral operators with kernels $\exp(iAxy)(xy)\rp
\eta(\delta(xy))$
are bounded on $L^2(\reals)$, uniformly in $A,\delta\in\reals^+$.
Thus the $L^2(dx)$ norm of this last expression is $O(\v\_{L^2})$.
Moreover, since $f$ is supported where $\lambda\ge\rho>0$,
only the restriction of $v$ to $[2\rho t,\infty)$ comes into play, so we
obtain a bound of $C\v\_{L^2([2\rho t,\infty))}$. This holds uniformly in
all realvalued functions $V$ appearing in the exponent.
There is also an easy alternative bound $O(t^{1/4}\v\_{L^1})$,
obtained directly by inserting absolute values inside the integral.
Thus the general case $V\in L^1+L^2$ may be treated by decomposing $v$
as a sum, and estimating the two terms separately.
Another term arising from \eqref{eq:abigjob} differs
only in that $tV(2\lambda t)$
is replaced by $c\lambda^{2}\int_0^{2\lambda t}V$;
in terms of the new variable $y$,
$v(y)$ is replaced by $t\rp\int_0^y v$.
Since $\lambda$ ranges over a compact interval $\Lambda$,
$y$ ranges over an interval $[ct,Ct]$ where $0 0$,
so asymptotic completeness in the relatively weak sense in which we have defined it
implies a stronger form.
Let us say that the halfaxis operator $H_V^\beta$ has boundary condition $\beta$ at the origin if
$u(0)+\beta u'(0)=0$ for any function $u$ in the domain of $H_V^\beta.$
\begin{theorem}\label{bcascom}
Let $H_V^\beta$ be a Schr\"odinger operator defined on a halfaxis with the boundary condition $\beta.$
Assume that $V \in L^1+L^p$ for some $1 0$ for a.e. $E.$ The variation of the boundary condition can
be regarded as rank one perturbation \cite{SVan}, Section I.6. The standard rank one perturbation theory
then implies that for any $\beta$, the singular
part of the spectral measure $\mu_{\rm{s}}^{\beta}$
can only be supported on a fixed set of energies $S \subset \reals^+$ of zero Lebesgue measure
\cite{SVan}, Theorem II.2. But then again by rank one theory for almost every
$\beta$ the singular spectrum on $\reals^+$ is empty \cite{SVan}, Theorem I.8.
Since the potential belongs to $L^1+L^p,$ the spectrum below zero is
discrete (with only $0$ as a possible accumulation point).
\end{proof}
Next, let us consider the following random model:
\begin{equation}\label{rex}
V(x) = \sum\limits_{n=1}^\infty a_n(\omega)g(n) f(xn),
\end{equation}
where $f(x) \in C_0^\infty (0,1),$ $g \in l^p$ for some $p<2$,
and $a_n(\omega)$ are independent identically
distributed bounded random variables with zero expectation. We have
\begin{theorem}\label{ranascom}
Let $H_V$ be a one dimensional Sch\"odinger operator with random potential \eqref{rex}. Then
with probability one, the wave operators $\Omega_\pm$ exist and are asymptotically complete.
\end{theorem}
Indeed, Theorem 9.1 of \cite{KLS} shows that almost surely,
the spectrum of the Schr\"odinger operator with
potential \eqref{rex} is purely absolutely continuous on $\reals^+.$
Notice that our assumptions on the potential easily imply
that the improper integral $\int_0^\infty V$ exists almost surely.
Theorem~\ref{thm:waveoperators}
and decay of the potential then imply asymptotic completeness.
This illustrates another type of situation
where asymptotic completeness holds. We remark that the result holds in a variety of more general
random models for which \eqref{rex} is just an illustration.
For a more general setting, see \cite{KLS}.
\smallskip
{\it Discussion.\/}
The above is only one of various possible definitions of asymptotic completeness.
The notion of asymptotic
completeness is intended to describe a situation where the Hilbert space is split into
two orthogonal subspaces \cite{RS3}: $\scripth_{\text{pp}}(H_V)$ and the range of wave operators,
$\scripth_{\text{ac}}(H_V).$
On $\scripth_{\text{ac}}(H_V)$ the perturbed dynamics is close to the modified free evolution
at large times, and corresponds to the scattering states. On $\scripth_{\text{pp}}(H_V)$ the dynamics
is supposed to be bounded in some sense. However, the intuitive physical assumption
that pure point
spectrum leads to dynamics which is bounded needs to be clarified,
and in recent years there have appeared examples with very nontrivial transport on the
pure point component. A
%reasonable and
widely accepted way to calibrate transport properties is to consider evolution of
the averaged moments of coordinate operator:
\begin{equation}\label{moments}
\langle \langle X^m \rangle_\phi \rangle_T = \frac{1}{T} \int\limits_0^T  \langle e^{iHt}\phi, X^m e^{iHt}\phi
\rangle  \, dt,
\end{equation}
where $\phi$ is the initial state, $\langle \phi_1, \phi_2 \rangle$ is the inner product and $X^m$
the operator of multiplication by $(x+1)^m$ in coordinate representation.
The paper \cite{DJLS} contains an example of a (discrete) Schr\"odinger operator $h$ with pure point spectrum and
exponentially decaying eigenfunctions, such that
\begin{equation}\label{pointran}
\limsup_{t \rightarrow \infty}
\langle \langle X^2 \rangle_{\delta_0} \rangle_T /T^{\alpha} = \infty
\end{equation}
for any $\alpha <2.$ Here the initial state is $\delta_0,$ the vector localized at the origin. Given that the rate
of growth $T^2$ for the second moment corresponds to ballistic motion, as for the free Laplace
operator, this example shows that in some sense
the transport associated with point spectrum can be
very fast.
%Because of such examples, it is clear that the definition given above does not completely %correspond to the
%intuitive picture we would like to capture.
However, there is still an important difference between transport
associated with point and singular continuous spectrum. Namely, let $B_R$ denote the ball of radius $R$
centered at the origin and $B_R^c$ its complement. Let $P_{{\rm pp}}$ and $P_{{\rm c}}$ be
the orthogonal projections
on $\scripth_{\text{pp}}(H_V)$ and the continuous subspace $\scripth_{\text{c}}(H_V)$
respectively.
%Then if the projection of the vector $\phi$
%on $\scripth_{\text{pp}}(H_V)$ is nonzero,
Then
%we have that
for any $\epsilon >0$ there exists $R_\epsilon$ such that
\begin{equation}\label{ppbound}
\e^{iH_Vt} P_{{\rm pp}}\phi \^2_{L^2(B^c_{R_\epsilon})} < \epsilon
\end{equation}
for all $t.$ The growth of the moments in \eqref{pointran}
is achieved not because of the motion
of the whole wavepacket, but because of thin tails escaping to infinity.
On the other hand,
we have
\begin{equation}\label{probfin}
\frac{1}{T} \int\limits_0^T \ e^{iH_Vt} P_{{\rm c}}
\phi \^2_{L^2(B_R)} \,dt \stackrel{T \rightarrow \infty}{\longrightarrow} 0
\end{equation}
%converges as $T \rightarrow \infty$ to a positive number
for any finite $R.$
%which is sufficiently large.
%That it is, the average probability to find a particle in a finite ball is greater than a
%fixed positive number.
Equation \eqref{probfin} is one of the statements of
the RAGE theorem (see, e.g. \cite{CFKS}) and is
basically a corollary of Wiener's theorem
on Fourier transforms of measures.
%with pure point component,
%which also implies that conversely,
%if the projection of $\phi$ on $\scripth_{\text{pp}}(H_V)$ is zero,
%then \eqref{probfin} tends to zero as $T \rightarrow \infty.$
Morever, there exist examples of Schr\"odinger operators \cite{KL} in which
the dynamics corresponding to the singular continuous subspace is almost ballistic
in a sense that the whole wavepacket is moving to infinity at a fast rate:
for any $\rho>0$ there exists $C_\rho$ such that
\[ \frac{1}{T} \int\limits_0^T \ e^{iH_Vt}
\phi \^2_{L^2(B_{C_\rho T^{1\epsilon}})} \, dt < \rho \]
for all $T$ and $\phi$ lying in the singular continuous subspace of $H_V$.
\section{Potentials in $\ell^p(L^1)$}
Following \cite{christkiselevdecaying}, we sketch here the small modifications needed
to extend the analysis of potentials in $L^p$ to those in $L^1+L^p$,
and indeed those in the larger class $\ell^p(L^1)$.
A locally integrable function $f$ is said to belong to the amalgamated space
$\ell^p(L^1)$ if
\begin{equation*}
\sum_{n\in\integers} \big(\int_n^{n+1}f(x)dx \big)^p<\infty.
\end{equation*}
The norm $\f\_{\ell^p(L^1)}$ is the $p$th root of this expression.
For any $1\le p<\infty$, this defines the
Banach space $\ell^p(L^1)$, which contains $L^1+L^p$.
A martingale structure $\{E^m_j\}$ is said to be adapted to $f$ in
$\ell^p(L^1)$ if
\begin{equation}
\f\cdot\chi_{E^m_j}\_{\ell^p(L^1)}^p
\le 2^{m} \f\_{\ell^p(L^1)}^p
\end{equation}
for all $m,j$.
For any $f \in\ell^p(L^1)$, there does exist an adapted martingale
structure \cite{christkiselevdecaying}.
Lemma~\ref{lemma:parseval} extends to $\ell^p(L^1)$: this Banach space is mapped boundedly
to $L^{p'}(\Lambda)$ for any compact interval $\Lambda\Subset(0,\infty)$,
for any $1\le p\le 2$. The proof is essentially unchanged; see
the analogous proof of Proposition~3.5 of \cite{christkiselevdecaying}.
Corollary~\ref{cor:nontangentialmaxbound} may now be refined by replacing $\f\_{L^p}$
by $\f\_{\ell^p(L^1)}$ on the righthand side of each inequality.
With these bounds for the operator $G$ in hand, the remainder of the proof is unchanged.
\medskip
{\bf Acknowledgement} We thank Barry Simon for useful discussions. \\
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\end{document}\end
0110161419415
0$ be a constant to be specified.
By Proposition~\ref{prop:numericalbound},
\begin{equation} \label{eq:boundfromtheorem1}
\scriptt_n^*(f_1,\dots,f_n)(\lambda)
\le \frac{C^{n+1}}{\sqrt{n!}}
\tilde G(\lambda)
G(\lambda)^{n1}
\end{equation}
where
\begin{align}
\tilde G(\lambda)
& = \sum_{m=1}^\infty 2^{\delta m}
(\sum_j T_1(f_1\cdot \chi^m_j)(\lambda)^2)^{1/2}
\\
G(\lambda)
& = \sum_{m=1}^\infty 2^{\delta m}
(\sum_j \max_iT_i(f_i\cdot \chi^m_j)(\lambda)^2)^{1/2}.
\end{align}
Now
\begin{multline}
\\tilde G\_{L^2} \le \sum_m 2^{\delta m}
(\sum_j\T_1(f_1\cdot\chi^m_j)\_{L^2}^2)^{1/2}
\\
\le \sum_m 2^{\delta m}
\T_1\_{2,2}(\sum_j\f_1\cdot\chi^m_j)\_{L^2}^2)^{1/2}
\le C_\delta
\T_1\_{2,2}\f_1\_{L^2}
\end{multline}
provided that $\delta>0$.
As for $G$, we may majorize
\begin{equation}
G(\lambda)
\le \sum^*_i \sum_{m=1}^\infty 2^{\delta m}
(\sum_j T_i(f_i\cdot \chi^m_j)(\lambda)^2)^{1/2}.
\end{equation}
It is shown on page 413 of \cite{christkiselevfiltrations}
(see also the proof of Corollary~\ref{cor:nontangentialmaxbound} below)
that since $p<2\le q$, there exists $\e>0$ such that
\begin{equation}
\(\sum_j T_i(f_i\cdot \chi^m_j)^2)^{1/2}\_{L^q}
\le C2^{\e m},
\end{equation}
under the condition that $\f_i\chi^m_j\_{L^p}^p\le C' 2^{m}$
for all $j,m$.
Choosing $\delta<\e$ and summing over $m$ gives
\begin{equation}
\G\_{L^q} \le C<\infty,
\end{equation}
of course under the hypothesis that $\f_i\_{L^p}=1$ for all $i\ge 2$.
An application of H\"older's inequality concludes the proof.
\end{proof}
Our main theorems are based on estimates for such multilinear operators;
however, we require not the conclusion of
Theorem~\ref{thm:multilinearoperators}, but rather the
more detailed information contained in \eqref{eq:boundfromtheorem1}
together with the norm bounds for $\tilde G,G$.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Nontangential convergence and the maximal function}\label{nontan}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Throughout the paper we write $\complex^+=\{z\in\complex: \Im(z)>0\}$.
Consider the cones
\begin{align*}
\Gamma_{\alpha}(w) &=\{z\in\complex^+: \Re(z)w<\alpha\Im(z)\},
\\
\Gamma_{\alpha,\delta}(w)
&=\{z\in\complex^+\cap\Gamma_\alpha(w): \Im(z)<\delta\}.
\end{align*}
A function $f(z)$ defined on $\complex^+$
is said to converge to a limit $a$ as $z\to w$ nontangentially
if for every $\alpha<\infty$,
$f(z)\to a$ as $z\to w$ within the cone $\Gamma_\alpha(w)$.
The nontangential maximal function of $f$ is defined by
\begin{equation}
Nf(w) =N_{\alpha,\delta}f(w)= \sup_{z\in \Gamma_{\alpha,\delta}(w)}f(z).
\end{equation}
We will need the following local variant of a standard property of
functions holomorphic in the whole half plane $\complex^+$.
\begin{lemma}\label{analyticf}
Let $\delta>0$. Let $\Lambda$ be an open subinterval of $\reals$,
and let $\Lambda'\Subset\Lambda$ be a relatively compact subinterval.
Let $0\le q\le\infty$. Let $B$ be a Banach space.
Suppose that $F$ is a holomorphic $B$valued function
in $\{\lambda+i\e: 0<\e<\delta,\ \lambda\in\Lambda\}$.
Then for any $\alpha<\infty$,
there exist $C<\infty$ and $\delta'>0$
depending on $\alpha,\Lambda,\Lambda'$, but not on $F$, such that
\begin{equation}
\N_{\alpha,\delta'}F(\cdot)\_{L^q(\Lambda')} \le
C \sup_{0<\e<\delta}
\F(\cdot+i\e)\_{L^q(\Lambda)} .
\end{equation}
\end{lemma}
By a $B$valued holomorphic function we mean a continuous function $f$ from
an open subset of $\complex$ to $B$, such that $z\mapsto \ell(f)(z)$
is holomorphic for every bounded linear functional $\ell$ on $B$.
\begin{proof}
Suppose first that $f$ is continuous on the closed rectangle
$\Lambda+i[0,\delta]$, and that $1
0$. Fixing any $0
s$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Generalized eigenfunctions associated to complex
spectral parameters} \label{section:complex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%{\bf Remark for us.}
%Main results of paper should generalize to slowly varying case where
%some derivative is in $L^p$ \dots
%\smallskip
Consider the generalized eigenfunction equation
\begin{equation} \label{eigenfneqn}
u'' + V(x)u = zu,
\end{equation}
where $z$ is permitted to be complex.
In earlier work we have analyzed the solutions to this equation for $z$
real and positive, and have shown that for almost every such $z$ there
exists a solution with certain asymptotic behavior as $x\to+\infty$.
Our present purpose is to analyze solutions for complex $z$ and to show
that they tend almost everywhere to the solutions for real $z$,
as $z$ approaches the positive real axis.
We restrict attention to parameters $z\in\complex^+\cup \reals^+$,
and choose a branch of $\sqrt{z}$ which has nonnegative imaginary part
for such $z$.
Define the phases
\begin{equation}
\xi(x,z) = \sqrt{z}x(2\sqrt{z})\rp\int_0^x V.
\end{equation}
\begin{theorem} \label{thm:complexeigenfunctions}
Let $1\le p<2$ and assume that $V\in L^1+L^p$.
For each $z\in\complex^+$ there exists a (unique) solution $u(x,z)$ of
the generalized eigenfunction equation \eqref{eigenfneqn}
satisfying
\begin{equation} \label{wkbasymptotics}
u(x,z)  e^{i\xi(x,z)} \to 0
\text{ and }
\partial u(x,z)/\partial x  i\sqrt{z}e^{i\xi(x,z)} \to 0
\qquad\text{as } x\to +\infty.
\end{equation}
$u(x,z)$ is continuous as a function on $\complex^+\times\reals$,
and is holomorphic with respect to $z$ for each fixed $x$.
Likewise, there exists such a (unique) solution
for almost every $z\in\reals^+$.
For almost every $E\in\reals$,
$u(x,z)$ converges to $u(x,E)$ uniformly for all $x$ in any
interval bounded below, as $z\to E$ nontangentially.
\end{theorem}
Here ``almost every'' means with respect to Lebesgue measure.
The existence of such solutions for almost every $z\in\reals^+$ is proved
in \cite{christkiselevdecaying}. For $z\in\complex^+$ it is well known
under weaker hypotheses on $V$.
The new point here is the convergence as $z\to\reals^+$, and in particular,
the fact that it is globally uniform in $x$.
It suffices to prove the theorem for $x\in [0,\infty)$, since the
conclusion for $x\in [\rho,\infty)$, for any $\rho>\infty$,
then follows via the eigenfunction equation \eqref{eigenfneqn}.
By rewriting \eqref{eigenfneqn} as a firstorder system, performing
a couple of algebraic transformations, reducing to an integral equation,
and solving it by iteration,
one arrives \cite{christkiselevdecaying} at a formal
series representation for solutions of \eqref{eigenfneqn}:
\begin{equation} \label{formalsolutionseries}
\begin{pmatrix} u(x,z) \\ u'(x,z) \end{pmatrix}
=
\begin{pmatrix}
e^{i\xi(x,z)} & e^{i\xi(x,z)} \\
i\sqrt{z} e^{i\xi(x,z)} & i\sqrt{z} e^{i\xi(x,z)}
\end{pmatrix}
\cdot
\begin{pmatrix}
\sum_{n=0}^\infty T_{2n} (V, \dots, V)(x,z)
\\
\sum_{n=0}^\infty T_{2n+1} (V, \dots, V)(x,z)
\end{pmatrix}
\end{equation}
where
\begin{equation}
T_n (V_1, \dots, V_n)(x,z)= (2\sqrt{z})^{n}
\int_{x \leq t_1 \leq \dots \leq t_n} \prod_{j=1}^n
e^{2i (1)^{nj} \xi(t_j,z)} V_j(t_j)\,dt_j,
\end{equation}
with the convention $T_0(\cdot) \equiv 1$.
To prove Theorem~\ref{thm:complexeigenfunctions}
we will show that each multilinear expression $T_n$
is welldefined for all $z\in\complex^+$,
that $T_n(\cdot)(x,z)\to 0$ as $x\to+\infty$ for all $n\ge 1$,
that they have the natural limits as $z\to\reals^+$ nontangentially,
and that these expressions satisfy bounds sufficiently strong to enable
us to sum the infinite series to obtain the desired conclusions.
Substitute $\zeta=\sqrt{z}$ and write $\zeta = \lambda+i\e$,
noting that $\e>0$.
Also write
$\phi(x,\zeta) = \xi(x,z)$, $S_n(V_1,\dots)(x,\zeta)
= T_n(V_1,\dots)(x,z)$.
Let $\{E^m_j\}$ be a martingale structure on $\reals^+$.
Denote by $t_{m,j}^\pm$ respectively the right ($+$) and left ($$)
endpoints of the interval $E^m_j$.
The real part of the exponent
$2i\sum_{j=1}^n (1)^{nj} \xi(t_j,z)$
is to leading order
\begin{equation*}
2\Im(\sqrt{z}[(t_nt_{n1})+(t_{n2}t_{n3})+\cdots])
= 2\e\cdot [(t_nt_{n1})+(t_{n2}t_{n3})+\cdots],
\end{equation*}
which is nonnegative (for $x\ge 0$)
for all $z\in\complex^+$ since $t_1\le t_2\cdots$; the exponential
factor decays rapidly as
$[(t_nt_{n1})+(t_{n2}t_{n3})+\cdots]\to\infty$.
This accounts for the difference between $z\in\complex^+$
and $z\in\reals^+$.
Define
\begin{equation}\begin{split}
G_m(V)(\zeta)
&=
\Big(
\sum_{j=1}^{2^m}
s^{m,}_j(V,\zeta)^2 +  s^{m,+}_j(V,\zeta)^2\Big)^{1/2}
\\
G(V) &= \sum_{m=1}^\infty mG_m(V)
\end{split}\end{equation}
where
\begin{equation} \label{emjintegrals} \begin{split}
s^{m,}_j(V,\zeta)
&= \int_{E^m_j} e^{2i[\phi(t,\zeta)\phi(t^_{m,j},\zeta)]}V(t)\,dt
\\
s^{m,+}_j(V,\zeta) &= \int_{E^m_j} e^{2i[\phi(t^+_{m,j},\zeta)\phi(t,\zeta)]}V(t)\,dt.
\end{split}\end{equation}
When $j=2^m$, and only then, the right endpoint of $E^m_j$ is infinite.
To simplify notation we make the convention that
for $j=2^m$, the second term
$s^{m,+}_j(V,\zeta)^2$
is always to be omitted, in the definition of $G$ and anywhere
else that the quantities $\phi(t^+_{m,j},\zeta)$ arise.
The following definitions will allow us to regard $G$ as a linear operator,
and hence to exploit properties of holomorphic functions.
\begin{definition}
$\scriptb$ denotes the Banach space consisting of all
sequences $\{\complex^2\owns s^m_j: m\ge 0,\ 1\le j\le 2^m\}$, with the norm
$\s\_\scriptb = \sum_m m\big(\sum_{j=1}^{2^m} s^m_j^2\big)^{1/2}$.
$\frakG: L^p\mapsto\scriptb$ denotes the operator
\begin{equation}
\frakG(V)(\zeta) = \{(s^{m,+}_j(V,\zeta),s^{m,}_j)(V,\zeta): 1\le m<\infty, 1\le j\le 2^m\}.
\end{equation}
\end{definition}
Thus
\begin{equation*}
G(V)(\zeta) = \\frakG(V)(\zeta)\_{\scriptb}.
\end{equation*}
Likewise we may write $G_M(V) = \\frakG_M(V)\_{\scriptb}$
with the analogous definition of $\frakG_M$.
The real parts of $i\phi(t,\zeta)i\phi(t^_{m,j},\zeta)$
and $i\phi(t^+_{m,j},\zeta)i\phi(t,\zeta)$
are bounded above uniformly for $0<\Im(\zeta)\le 1$,
and are $\le c\Im(\zeta)(tt^_{m,j})$ and
$\le c\Im(\zeta)(t^+_{m,j}t)$, respectively;
see \eqref{realpartnonpositive}.
Therefore for $V\in L^1+L^\infty$ and $\zeta\in\complex^+$,
each of these integrals converges absolutely, and each defines a holomorphic
scalarvalued function of $\zeta\in\complex^+$.
Thus $\frakG(V)$ may be regarded as a $\scriptb$valued holomorphic function\footnote{
By this we mean simply that it is a continuous mapping into $\scriptb$
with respect to the norm topology, and that each $s^{m,\pm}_j$ is
a holomorphic scalarvalued function.}
in any open set where it can be established that the series
defining $\\frakG(V)\_\scriptb$ converges uniformly.
We will also use the following
variant. Given a collection of functions $(V_1,\dots,V_n)$, we define
\begin{equation}
G(\{V_i\})(\zeta)
= \sum_{m=1}^\infty m\cdot
\Big(
\sum_{j=1}^{2^m}
\sum^*_i
s^{m,+}_j(V_i,\zeta)^2+s^{m,}_j(V_i,\zeta)^2
\Big)^{1/2},
\end{equation}
where $\sum\limits^*$ indicates that the sum is taken over a maximal
set of indices $i$ for which the functions $V_i$ are all distinct.
% ??? do we need this paragraph at all?
\begin{lemma} \label{lemma:Gmajorizationwithmodifiedphases}
For all $V,n$ and all $\zeta\in \complex^+\cup\reals^+$,
\begin{equation}
\sup_{x\in\reals^+}S_n(V,V,\dots,V)(x,\zeta)
\le \frac{C^{n+1}}{\sqrt{n!}}
G(V)(\zeta)^n.
\end{equation}
More generally,
for all $n$ and all $\{V_1,\dots,V_n\}$,
\begin{equation}
\sup_{x\in\reals^+}S_n(V_1,V_2,\dots,V_n)(x,\zeta)
\le \frac{C_k^{n+1}}{\sqrt{n!}}
G(\{V_i\})(\zeta)^n,
\end{equation}
provided that $\{V_i\}_{i=1}^n$ has cardinality $\le k$.
\end{lemma}
Write $p'=p/(p1)$.
\begin{lemma} \label{lemma:parseval}
For any compact interval $\Lambda\Subset(0,\infty)$,
there exists $C<\infty$ such that
for any $1\le p\le 2$,
for all $t'\in\reals$ and $f\in L^p(\reals)$,
for every $\e\ge 0$,
\begin{align}
&\\int_{t\ge t'} e^{2i[\phi(t, \lambda+i\e)  \phi(t', \lambda+i\e)]}
f(t)\,dt\_{L^{p'}(\Lambda,d\lambda)}
\le C\f\_{L^p}
\\
&\\int_{t\le t'} e^{2i[\phi(t', \lambda+i\e)  \phi(t, \lambda+i\e)]}
f(t)\,dt\_{L^{p'}(\Lambda,d\lambda)}
\le C\f\_{L^p}.
\end{align}
\end{lemma}
For $\e=0$ these integrals need not converge absolutely, hence require
interpretation. They are initially welldefined for compactly
supported $f$, and the lemma asserts an {\em a priori} bound for
such functions. Then they are defined for general $f\in L^p$
by approximating in $L^p$ norm by compactly supported functions,
and passing to the limit in $L^{p'}$ norm.
Let an exponent $p<\infty$ be specified.
Recall that a martingale structure is said to be adapted to $f$ in $L^p$
if $\int_{E^m_j}f^p = 2^{m}\intf^p$ for all $m,j$.
Recall from the preceding section the definitions of the cones $\Gamma_{\alpha,\delta}$
and associated nontangential maximal functions $N_{\alpha,\delta}$.
\begin{corollary} \label{cor:nontangentialmaxbound}
Let $\alpha<\infty$, let $1\le p \le 2$, and let
$\Lambda\Subset(0,\infty)$ be any compact subinterval.
Then there exist $C<\infty,\delta>0$ such that for any
$f\in L^p(\reals)$ and for any martingale structure
$\{E^m_j\}$ on $\reals^+$
% adapted to $f$,
\begin{equation} \label{nonadaptedbound}
\N_{\alpha,\delta} G_m(f)(\lambda)\_{L^{p'}(\Lambda,d\lambda)}
\le C\f\_{L^p}.
\end{equation}
Moreover for each $1\le p<2$
there exists $\rho>0$ such that for any $f\in L^p$ and
for any martingale structure adapted
to $f$ in $L^p$,
\begin{equation} \label{GMdecaybound}
\N_{\alpha,\delta} G_m(f)(\lambda)\_{L^{p'}(\Lambda,d\lambda)}
\le C2^{\rho m}\f\_{L^p}.
\end{equation}
Consequently under these additional hypotheses,
\begin{equation}
\N_{\alpha,\delta} G(f)(\lambda)\_{L^{p'}(\Lambda,d\lambda)}
\le C\f\_{L^p}.
\end{equation}
Moreover, for almost every $\lambda\in\Lambda$,
\begin{equation}
\\frakG(f)(\zeta)\frakG(f)(\lambda)\_{\scriptb} \to 0
\end{equation}
as $\zeta\to\lambda$ nontangentially.
\end{corollary}
We postpone the proofs of Lemmas~\ref{lemma:Gmajorizationwithmodifiedphases},
\ref{lemma:parseval} and Corollary~\ref{cor:nontangentialmaxbound}
until the end of the section.
For any $t\ge t'$ and any $\zeta=\lambda+i\e$ with $\e\ge 0$,
\begin{equation} \label{realpartnonpositive}
\Re(i\phi(t,\lambda+i\e)i\phi(t',\lambda+i\e))
= \e(tt') \e(\lambda^2+\e^2)\int_{t'}^t V,
\end{equation}
and $\int_{t'}^t V\le tt'^{1/p'}\V\_{L^p}$.
Therefore
% ?? phi, xi depend on V; need to explain/deal with this ... ??
\begin{equation}
e^{i[\phi(t,\lambda+i\e)\phi(t',\lambda+i\e)]}
\le Ce^{c\ett'}
\end{equation}
where $C,c\in\reals^+$ are constants which depend only on the $L^p$ norm
of $V$.
Hence for all $n\ge 1$,
\begin{align}
S_{2n}(V,V,\dots,V)(x,\lambda+i\e)
& \le \iint_{x\le t_1\le\cdots\le t_{2n}}
e^{c\e(t_{2n}t_{2n1}+t_{2n2}\cdots)}
\prod_j V(t_j)\,dt_j
\notag
\\
& \le \frac{C^n}{n!}
\Big(\iint_{x\le t'\le t} e^{c\e(tt')}V(t')V(t)\,dt'\,dt\Big)^n
\notag
\\
& \le \frac{C^n}{n!}
\e^{2n(1p\rp)} \V\_{L^p([x,\infty)}^{2n}.
\end{align}
In the same way, for $x\ge 0$, one obtains for $n\ge 0$
\begin{equation}
S_{2n+1}(V,V,\dots,V)(x,\lambda+i\e)
\le \frac{C^n}{n!}
\e^{(2n+1)(1p\rp)} \V\_{L^p([x,\infty)}^{2n+1}.
\end{equation}
Let $z$ belong to any compact subset of $\complex^+$.
Then $\e=\Im(\sqrt{z})$ has a strictly positive lower bound.
Therefore the individual terms of the series \eqref{formalsolutionseries}
defining a formal solution of \eqref{eigenfneqn} do define uniformly bounded
functions of $(x,\zeta)$ for $x\ge 0$ and $\zeta$ in any compact subset
of $\complex^+$, and moreover, the
series are uniformly absolutely convergent.
As in Lemma~4.2 of \cite{christkiselevdecaying},
it follows that the sums of these series define solutions of the
generalized eigenfunction equation \eqref{eigenfneqn} for all such $z$.
Because the $L^p$ norm of $V$ over $[x,\infty)$ tends to zero
as $x\to+\infty$, only $T_0(x)$ contributes in that limit,
so these solutions do have the desired WKB asymptotics $\exp(i\xi(x,z))$.
Clearly each summand depends holomorphically on $\zeta$, hence so do
the sums.
Existence of a (unique) solution for almost every $z\in\reals^+$
is proved in \cite{christkiselevdecaying}.
We come now to the main step, where we relate
$z\in\complex^+$ to $z\in\reals^+$.
For compactly supported $f\in L^1$,
the quantities $s^{m,\pm}_j(f,\zeta)$
are clearly holomorphic functions of $\zeta$ where
$\Im(\zeta)>0$, and are continuous at $\e=0$ for each $0\ne \lambda\in\reals$.
The same holds for $S_n(f,f,\dots,f)(x,\zeta)$, for each $x\in\reals$.
$\frakG(f)$ is likewise a $\scriptb$valued holomorphic function
of $\zeta$, continuous on $\complex^+\cup\reals^+$,
for compactly supported $f\in L^1$.
This follows by combining the holomorphy of the individual terms
\eqref{emjintegrals} with the rapid convergence bound \eqref{GMdecaybound}.
\begin{lemma} \label{lemma:Snconvergence}
Let $1\le p<2$ and $V\in L^p(\reals)$.
For almost every $E\in\reals$,
for every $n\ge 1$,
$T_n(V,V,\dots,V)(x,z)\to T_n(V,V,\dots,V)(x,E)$
uniformly for all $x\ge 0$
as $\complex^+\owns z\to E$ nontangentially.
\end{lemma}
\begin{proof}
It is equivalent to show that
\begin{equation*}
\sup_{x\ge 0}\big
S_n(V,V,\dots,V)(x,\zeta)S_n(V,V,\dots,V)(x,\lambda)\big \to 0
\end{equation*}
as $\zeta\to\lambda$ nontangentially, for almost all $\lambda\in\Lambda$,
for any fixed compact interval $\Lambda\subset\reals^+$.
For $V=W\in L^1$ with compact support, we have already established
convergence uniformly in $x,\lambda$, as $\complex^+\owns\zeta\to \lambda$
unrestrictedly (rather than merely nontangentially),
since the phases $\phi(t,\zeta)$ converge uniformly to $\phi(t,\lambda)$
for $t,\zeta,\lambda$ in any compact set.
Let $V$ be given, and remain fixed for the remainder of this proof.
Set $\phi_V(t,\zeta)=\zeta t  (2\zeta)\rp\int_0^t V$.
Whenever we write $S_n(f_1,f_2,\dots,f_n)$, it is defined
in terms of the phases $\phi(t_i,\zeta) = \phi_V(t_i,\zeta)$,
independent of $f_1,f_2,\dots$. Thus the $S_n$ are here
genuine multilinear operators.
Let $\e>0$ be arbitrary, and fix a martingale structure $\{E^m_j\}$
adapted to $V$ in $L^p$ on $\reals^+$.
Decompose $V=W+(VW)$
where $W(x) =V(x)\chi_{(0,R]}(x)$, with $R$ chosen so
that $\VW\_{L^p}<\e$
and moreover so that $\NG(VW)\_{L^{p'}(\Lambda)}<\e$.
Such a choice is possible, since
\begin{equation*}
\NG_M(V\chi_{[R,\infty)})\_{L^{p'}(\Lambda)}
\le C\min(2^{M\delta}\V\_{L^p},\,\ V\chi_R\_{L^p}).
\end{equation*}
Then
\begin{multline}
S_n(V,V,\dots,V)(x,\zeta)S_n(V,V,\dots,V)(x,\lambda)
\\
\le
S_n(W,W,\dots,W)(x,\zeta)S_n(W,W,\dots,W)(x,\lambda)
\\
+
S_n(V,V,\dots,V)(x,\zeta)S_n(W,W,\dots,W)(x,\zeta)
\\
+
S_n(V,V,\dots,V)(x,\lambda)S_n(W,W,\dots,W)(x,\lambda) .
\end{multline}
The first term on the right tends to zero, in the sense desired.
Majorize the second by
\begin{multline}
S_n(V,V,\dots,V)(x,\zeta)S_n(W,W,\dots,W)(x,\zeta)
\\
\le
\sum_{i=1}^n
S_n(V,\dots,V,VW,W,\dots,W)(z,\zeta)
\end{multline}
where in the $i$th summand, the argument of $S_n$ has
$i1$ copies of $V$ and $ni$ copies of $W$.
Fix any aperture $\alpha\in\reals^+$.
Thus as established in the proof of
Proposition~4.1 of \cite{christkiselevdecaying},
\begin{multline}
\sup_{x\ge 0}
\sup_{\zeta\in \Gamma_{\alpha,\delta} (\lambda)}
S_n(V,V,\dots,V)(x,\zeta)S_n(W,W,\dots,W)(x,\zeta)
\\
\le
C_n
\sum_{i=1}^n
\sup_{\zeta\in \Gamma_{\alpha,\delta}(\lambda)}
G(V)^{i1}(\zeta)G(W)^{ni}(\zeta)G(VW)(\zeta)
\\
\le
C_n
\sum_{i=1}^n
NG(V)^{i1}(\lambda)NG(W)^{ni}(\lambda)NG(VW)(\lambda).
\end{multline}
Let $q=p'$.
By Chebyshev's inequality,
for any $\beta>0$,
\begin{equation}\begin{split}
&\{\lambda\in\Lambda:
\sup_{\zeta\in \Gamma_{\alpha,\delta}(\lambda)}\sup_{x\ge 0}
S_n(V,V,\dots,V)(x,\zeta)S_n(W,W,\dots,W)(x,\zeta)>\beta\}
\\
&\qquad\qquad \le
C_n\beta^{q/n} \sum_{i=1}^n
\ NG(V)^{i1}NG(W)^{ni}NG(VW)\_{L^{q/n}(\Lambda)}^{q/n}
\\
&\qquad\qquad \le
C_n\beta^{q/n} \sum_{i=1}^n
(\NG(V)\_{L^q(\Lambda)}^{i1}
\NG(W)\_{L^q(\Lambda)}^{ni}
\NG(VW)\_{L^q(\Lambda)})^{q/n}
\\
&\qquad\qquad \le
C_n\beta^{q/n} \sum_{i=1}^n
(\V\_{L^p}^{n1}\VW\_{L^p})^{q/n}
\\
&\qquad\qquad \le
C_n\beta^{q/n}
\V\_{L^p}^{q(n1)/n}\e^{q/n}.
\end{split}\end{equation}
The same reasoning gives
\begin{multline}
\{\lambda\in\Lambda:
\sup_{x\ge 0}
\bigS_n(V,V,\dots,V)(x,\lambda)S_n(W,W,\dots,W)(x,\lambda)>\beta\}\big
\\
\le
C_n\beta^{q/n}
\V\_{L^p}^{q(n1)/n}\e^{q/n}.
\end{multline}
Consequently
\begin{multline}
\{\lambda\in\Lambda:
\limsup_{\Gamma_\alpha(\lambda)\owns\zeta\to\lambda}\,\,
\sup_{x\ge 0}\,
\bigS_n(V,V,\dots,V)(x,\zeta)S_n(V,V,\dots,V)(x,\lambda)\big
>\beta\}
\\
\le
C_n\beta^{q/n}
\V\_{L^p}^{q(n1)/n}\e^{q/n},
\end{multline}
for all $\beta,\e\in\reals^+$.
Letting $\e\to 0$, we conclude that the $\limsup$ vanishes for almost
every $\lambda$.
\end{proof}
It is now straightforward to sum the series to obtain the same
conclusion regarding convergence of $u(x,z)=u(x,\zeta^2)$
to $u(x,E)=u(x,\lambda^2)$:
\begin{equation}
u(x,\zeta^2)u(x,\lambda^2)
\le
\sum_{n=0}^\infty
S_n(V,V,\dots,V)(x,\zeta)S_n(V,V,\dots,V)(x,\lambda)
\end{equation}
and
\begin{equation}\begin{split}
&\sup_{\zeta\in\Gamma_{\alpha,\delta}(\lambda)}\sup_{x\ge 0}
\sum_{n=M}^\infty
S_n(V,V,\dots,V)(x,\zeta)S_n(V,V,\dots,V)(x,\lambda)
\\
&\qquad\qquad\le
\sup_{\zeta\in\Gamma_{\alpha,\delta}(\lambda)}\sup_{x\ge 0}
\sum_{n=M}^\infty
(S_n(V,V,\dots,V)(x,\zeta)+S_n(V,V,\dots,V)(x,\lambda))
\\
&\qquad\qquad\le
\sum_{n=M}^\infty \frac{C^{n+1}}{\sqrt{n!}}
\big(NG(V)(\lambda)+G(V)(\lambda)\big)^n
\\
&\qquad\qquad\le
\frac{C^{M+1}}{\sqrt{M!}}\big(NG(V)(\lambda)+G(V)(\lambda)\big)^M
\sum_{k=0}^\infty
\frac{C^{k+1}}{\sqrt{k!}}
\big(NG(V)(\lambda)+G(V)(\lambda)\big)^k
\\
&\qquad\qquad\le
\frac{C^{M+1}}{\sqrt{M!}}\big(NG(V)(\lambda)+G(V)(\lambda)\big)^M
\exp(C \big(NG(V)(\lambda)+G(V)(\lambda)\big)^2).
\end{split}\end{equation}
For almost every $\lambda\in\reals$,
$\big(NG(V)(\lambda)+G(V)(\lambda)\big)<\infty$,
and hence this expression tends to zero as $M\to\infty$.
Coupled with the convergence established for the individual terms
$S_n$ in Lemma~\ref{lemma:Snconvergence}, this completes the proof of
Theorem~\ref{thm:complexeigenfunctions}, modulo the proofs of
Lemmas~\ref{lemma:parseval} and
\ref{lemma:Gmajorizationwithmodifiedphases},
and Corollary~\ref{cor:nontangentialmaxbound}.
\hfill\qed
\begin{proof}[Proof of Lemma~\ref{lemma:parseval}]
The proofs of the two inequalities are essentially the same,
so we discuss only the first.
The exponent here is $i\Phi(t,\lambda+i\e)
= i\phi(t,\lambda+i\e) i\phi(t',\lambda+i\e)
= (i\lambda\e)(tt')  (i\lambda+\e)(\lambda^2+\e^2)\rp(\int_{t'}^t V)$.
Since $t\ge t'$ and $\int_{t'}^t V \le C+Ctt'^{1/2}$,
the real part of $\Phi$ is bounded above, uniformly for all
$\lambda$ in any compact subinterval $\Lambda\Subset\reals\backslash\{0\}$
and $\e\ge 0$. Thus $L^1$ is mapped boundedly to $L^\infty(\Lambda)$,
uniformly in $\e$; by interpolation it suffices to prove the $L^2$
estimate.
Fix any cutoff function $\eta\in C^\infty(\reals\backslash\{0\})$
and consider
\begin{equation}
\int \Big\int_{t>t'}
e^{i\Phi(t,\lambda+i\e)}f(t)\,dt\Big^2\eta(\lambda)\,d\lambda
=
\iint_{s,t\ge t'} f(t)\bar f(s) K(t,s)\,dt\,ds
\end{equation}
where
\begin{equation}
K(t,s)
=\int e^{\Psi(t,s,\lambda+i\e)}
\eta(\lambda)
\,d\lambda
\end{equation}
with
\begin{multline}
\Psi(t,s,\lambda+i\e) = {2i\Phi(t,\lambda+i\e)2i\bar\Phi(s,\lambda+i\e)}
\\
= 2i\big[\lambda(ts)\lambda(\lambda^2+\e^2)\rp{\textstyle\int}_s^t V\big]
\\
2\e\big[(tt')+(st')
+ (\lambda^2+\e^2)\rp({\textstyle\int}_{t'}^t V)
+ (\lambda^2+\e^2)\rp({\textstyle\int}_{t'}^s V)\big].
\end{multline}
We claim that $K(t,s)\le C(1+st)^{2}$,
uniformly in $\e\ge 0$; this would suffice to imply the $L^2$ bound.
The integrand itself is bounded, uniformly in all parameters,
so it suffices to restrict attention to the case where $st\ge C_0$,
where $C_0$ is a sufficiently large constant.
In that case we integrate by parts,
integrating $\exp(2i[\lambda(ts)\lambda(\lambda^2+\e^2)\rp\int_s^t V])$,
and differentiating
$\eta(\lambda)\cdot
\exp(2\e[(tt')+(st')
+ (\lambda^2+\e^2)\rp(\int_{t'}^t V)
+ (\lambda^2+\e^2)\rp(\int_{t'}^s V))$,
noting that $\partial [\lambda(ts)\lambda(\lambda^2+\e^2)\rp\int_s^t V]
/\partial\lambda\ge st/2$ provided $C_0$ is chosen to be
sufficiently large.
Thus we gain a factor of $(st)\rp$.
On the other hand, differentiating the other exponential with respect
to $\lambda$ brings in an unfavorable term
$O(\e \int_{t'}^{\max(s,t)}V)$.
After two integrations by parts, the integrand is
\begin{equation}
O(st^{2})\cdot e^{2\e(st')2\e(tt')}
\cdot O( 1+\e^2( \int_{t'}^{\max(s,t)}V )^2 )
=
O(st^{2}),
\end{equation}
uniformly in $\e$.
\end{proof}
\begin{proof}[Proof of Lemma~\ref{lemma:Gmajorizationwithmodifiedphases}]
The new feature here is the introduction of the modifying factors
$\exp( \pm 2i\phi(t_{m,j}^\pm,\zeta) )$; without these, this is proved
in \cite{christkiselevfiltrations}. We will merely indicate
the modification needed in the argument, referring to
\cite{christkiselevfiltrations} for the rest.
Consider
\begin{equation}
\iint_{x\le t_1\le\cdots\le t_n}e^{2i[\phi(t_n,\zeta)\phi(t_{n1},\zeta)
+ \phi(t_{n2},\zeta)\cdots ]}
f(t_1)f(t_2)\cdots f(t_n)\,dt_1\cdots\,dt_n.
\end{equation}
Decompose the region of integration $\{t=(t_1,\cdots,t_n):
x\le t_1\le\cdots\le t_n\}$ as $\cup_{k=0}^n\Omega_k$
where $\Omega_k=\{t: x\le t_1\le\cdots\le t_k\le t^+_{1,1}
= t^_{1,2}\le t_{k+1}\le\cdots\le t_n\}$.
The total integral is
\begin{equation}\begin{split}
&\sum_k\iint_{\Omega_k}
e^{2i[\phi(t_n,\zeta)\phi(t_{n1},\zeta)+\cdots\pm\phi(t_{k+1},\zeta)
\mp\phi(t^_{1,2},\zeta)]}
\cdot
e^{2i[\pm\phi(t^+_{1,1}
\mp \phi(t_k,\zeta)\pm\phi(t_{k1},\zeta)\mp\cdots]}
\prod_{j=1}^n f(t_j)\,dt_j
\\
& \qquad=
\sum_k
\Big(
\iint_{t^_{1,2}\le t_{k+1}\le\cdots\le t_n}
e^{2i[\phi(t_n,\zeta)\phi(t_{n1},\zeta)+\cdots
+(1)^{nk1}\phi(t_{k+1},\zeta)]}
\prod_{j=k+1}^n f(t_j)\,dt_j
\Big)
\\
&\qquad\qquad\cdot
\Big(
\iint_{x\le t_1\le\cdots\le t_k\le t^+_{1,1} }
e^{2i(1)^{nk}[\phi(t_k,\zeta)  \phi(t_{k1},\zeta) + \cdots
+ (1)^{n1}\phi(t_1,\zeta)]}
\prod_{j=1}^k f(t_j)\,dt_j
\Big).
\end{split}\end{equation}
For each $k$, each of the two factors on the righthand side has
the same form as the multiple integral with which we began,
except that when $nk$ is odd,
an extra factor of $1$ appears in the exponent in
the integral with respect to $dt_k\cdots dt_1$; this minus sign
destroys the bounds we seek, as is clear from
\eqref{realpartnonpositive}.
Therefore when $nk$ is odd, we rewrite the corresponding term
as the modified product
\begin{multline*}
\Big(
\iint_{t^_{1,2}\le t_{k+1}\le\cdots\le t_n}
e^{2i[\phi(t_n,\zeta)\phi(t_{n1},\zeta)+\cdots
+\phi(t_{k+1},\zeta)\phi(t^_{1,2},\zeta)]}
\prod_{j=k+1}^n f(t_j)\,dt_j
\Big)
\\
\cdot
\Big(
\iint_{x\le t_1\le\cdots\le t_k\le t^+_{1,1} }
e^{2i[\phi(t^+_{1,1},\zeta)
\phi(t_k,\zeta) + \phi(t_{k1},\zeta) + \cdots
+ (1)^{n1}\phi(t_1,\zeta)]}
\prod_{j=1}^k f(t_j)\,dt_j
\Big).
\end{multline*}
Suppose now that $n$ is even.
The proof in \cite{christkiselevfiltrations}
is an induction based on a repeated application of this decomposition;
each step of that recursion involves a ``cut point'' $t^+_{m,j}
= t^_{m,j+1}$ playing the
same role as $t^+_{1,1}=t^_{1,2}$ in the above formula.
At each step, the region of integration is decomposed into
subregions as above, and corresponding to each subregion there is
a splitting of the terms in the phase into two subsets.
At any step which results in an odd number of terms appearing
in one (hence both) subsets,
we modify the resulting phases by introducing a factor
$1 = \exp(\pm 2i[\phi(t^+_{m,j},\zeta)\phi(t^_{m,j+1},\zeta)])$,
factoring it as a product of one exponentials, and splitting those
two exponential factors as above. This, together with the argument
in \cite{christkiselevfiltrations}, yields the assertion of the lemma
for even $n$.
For odd $n$ we introduce a factor of
$1 = \exp(2i[\phi(x,\zeta)\phi(x,\zeta)])$,
incorporate $\exp(2i\phi(x,\zeta))$ into the phase, thus
reducing matters again to the case where there an even number of
terms. The remaining factor of $\exp(2i\phi(x,\zeta))$
is bounded above, uniformly for $0\le\Im(\zeta)\le 1$ and $x\ge 0$,
so is harmless.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{cor:nontangentialmaxbound}]
Let $\Lambda\Subset(0,\infty)$ be a compact interval, and let $1< p<2$,
the case $p=1$ being trivial.
Set $q = p'=p/(p1)$.
We discuss only the contributions of terms involving
$\phi(t^_{m,j},\zeta)$
to $G$ and $G_M$; those involving $t^+_{m,j}$ are treated in
exactly the same way.
For any $m\ge 1$ and any $f\in L^p$ we have, since $q/2\ge 1$,
\begin{multline}
\\Big(
\sum_{j=1}^{2^m}
\Big\int_{E^m_j} e^{2i[\phi(t,\lambda+i\e)\phi(t^_{m,j},\lambda+i\e)]}
f(t)\,dt\Big^2
\Big)^{1/2}\_{L^{q}(\Lambda,d\lambda)}^q
\\
\le
\Big(\sum_j
\Big[\int_{\Lambda}
\Big\int_{E^m_j} e^{2i[\phi(t,\lambda+i\e)\phi(t^_{m,j},\lambda+i\e)]}
f(t)\,dt\Big^q
\,d\lambda\Big]^{2/q}\Big)^{q/2},
\end{multline}
by Minkowski's integral inequality.
% marker
By Lemma~\ref{lemma:parseval}, the righthand side is
\begin{equation*}
\le
C \Big(\sum_j \f\cdot \chi_{E^m_j}\_{L^p}^2\Big)^{q/2}.
\end{equation*}
Since $p\le 2$, this is
\begin{equation*}
\le
C\Big(\sum_j \f\_{L^p}^{2p} \f\cdot \chi_{E^m_j}\_{L^p}^p )^{q/2}
= C\f\_{L^p}^q.
\end{equation*}
%where $\rho>0$ if $p<2$, and $\rho=0$ if $p=2$.
If we assume that the martingale structure is adapted to $f$ in $L^p$,
then for $p<2$ we have the improved majorization
\begin{equation*}
\f\cdot \chi_{E^m_j}\_{L^p}^2
\le 2^{m(2p)/p}\f\_{L^p}^{2p}\f\cdot \chi_{E^m_j}\_{L^p}^p,
\end{equation*}
whence the final bound is $2^{\rho m}\f\_{L^p}^q$ for some
$\rho(p)>0$.
Therefore
\begin{equation}
\int_\Lambda \\frakG_M(f)(\lambda+i\e)\_{\scriptb}^q\,d\lambda
\le C2^{\rho M}\f\_{L^p}^q,
\end{equation}
uniformly for all $\e>0$.
The first two conclusions of the Corollary now follow from
Lemma~\ref{analyticf},
since $\Lambda$ is an arbitrary compact interval.
That $\frakG(f)(\zeta)$ converges almost everywhere to
$\frakG(f)(\lambda)$
in the $\scriptb$ norm as $\zeta\to \lambda$ nontangentially, is an
immediate consequence of the bound
$\NG_M(f)\_{L^q}\le C2^{\varepsilon M}\f\_{L^p}$, since
\begin{equation}
\int_{t\ge t'} e^{2i[\phi(t, \zeta)  \phi(t', \zeta)]}
f(t)\,dt
\to
\int_{t\ge t'} e^{2i[\phi(t, \lambda)  \phi(t', \lambda)]}
f(t)\,dt
\end{equation}
almost everwhere as $\zeta\to\lambda$ nontangentially, for all $f\in L^p$.
This holds for all $f$ in the dense subspace $L^1\cap L^p$,
and then follows for general $f$ by Lemma~\ref{analyticf},
Lemma~\ref{lemma:parseval}, and standard reasoning.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Resolvents and spectral projections}
\label{section:limiting absorption}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The halfline case} \label{section:resolvents}
Consider the operator
\begin{equation}\label{so}
H_V = \frac{d^2}{dx^2} +V(x)
\end{equation}
on $\reals^+$, with Dirichlet boundary condition at the origin.
Any other selfadjoint boundary condition can be treated in a similar way.
For each $z\in\complex$, let $u_1(x,z), u_2(x,z)$ be the
unique solutions of
\begin{equation}\label{efeq}
u''+V(x)u = zu
\end{equation}
satisfying the boundary conditions
$u_1(0,z) = (0,1)^{t}, u_2(0,z)=(1,0)^t$;
the superscripts $t$ denote transposes.
The classical theory of second order differential operators
(see e.g. \cite{CL,Tit}) tells us that if \eqref{so} is in the
limit point case, then for any $z \in \complex \setminus \reals$
there exists a unique complex number
$m(z)$, called the Weyl $m$function, such that
\[ f(x,z) = u_1(x,z)m(z)+u_2(x,z) \in L^2(\reals^+). \]
We will always consider potentials which
lead to the limit point situation,
as is the case if $V \in L^1+L^p$ for some $1\le p<\infty$; see for example
\cite{ReSi2}.
The Weyl $m$ function is a Herglotz function, that is, it is analytic
in the upper halfplane and has positive
imaginary part there.
%The simplest way to see this positivity
%is to consider the Wronskian
%$W[f,\ov{f}]$, where in general, $W[f,g]= f'{g}f{g}'$.
%Then $W[f,\ov{f}](0)= 2i \Im m(z)$, $W[f,\ov{f}](\infty)=0$,
%and $W'[f,\ov{f}](x) = 2i \Im z f(x,z)^2$. Hence
%\[ \Im m(z) = \Im z \int_0^\infty f(x,z)^2\,dx. \]
By direct computation, the resolvent $(H_Vz)^{1}g$
for $z\in\complex^+$ is given by
\begin{equation}\label{resol}
(H_Vz)^{1} g(x) = u_1(x,z) \int_x^\infty f(y,z)g(y)\,dy +
f(x,z) \int_0^x u_1(y,z) g(y)\,dy.
\end{equation}
Denote by $P_{(a,b)}$ the spectral projection associated to
$H_V$ and to the interval $(a,b)$.
>From the resolvent formula \eqref{resol} we may derive a
formula for the projections $P_{(a,b)}$. In doing so, we will use the
following three routine facts.
\\
1. The functions $u_1(x,z), u_2(x,z)$ are continuous in $x$ for
each $z$, and are entire holomorphic functions of $z$ for each $x$.
\\
2. The $m$ function (in fact, any Herglotz function)
%satisfying an appropriate moment condition)
has a representation
\begin{equation}
m(z) = C_1 + C_2z+
\int_\reals \left( \frac{1}{tz}  \frac{t}{1+t^2} \right)
d\mu(t)
\end{equation}
for some positive Borel measure $\mu$ satisfying $\int (1+t^2)^{1}
d\mu(t) < \infty,$ see e.g. \cite{AD}.
In the $m$ function context, $\mu$ is often called the
spectral measure. For Dirichlet boundary conditions, as considered here,
it is the spectral measure corresponding to the generalized vector $\delta_0'$,
defined by $\delta_0'(u)=u'(0)$ for any function $u$ in the domain of $H_V.$
The moment condition above corresponds to the fact that
the derivative $\delta'_0$ belongs to the Sobolevlike space $H_{2}(H_V)$ associated
to $H_V$ (see e.g.\ \cite{BeSh} for details on families of Sobolevlike spaces
associated with any selfadjoint operator $A$). \\
3. $\Im m(E+i\epsilon)$ converges weakly to $\pi \mu$
as $\epsilon \rightarrow
0^+$. Moreover, $\Im m(E+i\epsilon)$ has limiting boundary values for
Lebesguealmost every $E$,
and the density of the absolutely continuous part of $\mu$ satisfies
\begin{equation}\label{acpart}
d\mu_{\rm{ac}}(E)= \frac{1}{\pi}\Im m(E+i0) \,dE,
\end{equation}
where $m(E+i0) = \lim_{\e\to 0^+}m(E+i\e)$.
Since the imaginary part of $m$ is simply a Poisson integral of $\mu$,
this is straightforward.
Fix functions $h$ and $g$ with compact support.
We integrate the resolvent element $\langle (H_Vz)^{1}g,h \rangle$
over the contour $\gamma_\epsilon$ in complex plane consisting of two
horizontal intervals $(a \pm i \epsilon, b \pm i \epsilon)$, and
two vertical intervals at the ends
connecting them. In the limit $\epsilon \rightarrow 0^+$,
the contributions of the vertical intervals disappear unless $a$ or $b$
is an eigenvalue (point mass of $\mu$); we will assume this
is not the case.
By the spectral theory (see, e.g. \cite{ReSi1}, Stone formula)
we get then the following
expression for $\langle P_{(a,b)}g,h \rangle$:
\begin{multline}\label{spro}
\langle P_{(a,b)}g,h \rangle
= \frac{1}{2\pi i}\lim_{\epsilon \rightarrow 0}
\int_{\gamma_\epsilon}
\langle (H_Vz)^{1}g,h \rangle \, dz \\
=
\int_a^b \int_\reals \int_\reals
u_1(x,E)u_1(y,E) g(x)\overline{h}(y)\,dx\,dy
\,d\mu(E).
\end{multline}
In passing to the last line,
we have taken into account the resolvent formula \eqref{resol},
the properties of $u_{1},u_2$
(in particular, $u_2$ drops out because of analyticity), and the fact that
$\pi^{1}\Im m(E+i\epsilon)$ converges weakly to $\mu$. The compact
supports of $h,g$ ensure that the integral is welldefined.
Similar formulas can be found in \cite{CL, Tit}.
Since $P_{(a,b)}$ is by its definition an orthogonal projection,
an immediate consequence
of \eqref{spro} is that the mapping $g\mapsto \int_\reals
u_1(x,E)g(x)\,dx$, initially defined for continuous $g$ having compact
support, extends to an orthogonal projection from $L^2(\reals^+,dx)$
to $L^2((a,b),\mu)$.
Dually, from \eqref{spro} we see that for each $g\in L^2(\reals,d\mu)$,
\begin{equation}
U_0 g(x) =\lim_{N \rightarrow \infty} \int_{N}^N
u_1(x,E)g(E) d\mu(E)
\end{equation}
exists in $L^2(\reals^+,dx)$ norm,
and that the linear operator $U_0$ thus defined
is a unitary bijection from $L^2(\reals,d\mu)$ to $L^2(\reals,dx)$
with inverse
\begin{equation}
U_0^{1}
g(E) = \lim_{N \rightarrow \infty} \int_{N}^N u_1(x,E) g(x) dx,
\end{equation}
where the limit is again taken in $L^2$ norm.
With the formula \eqref{spro} for the spectral projections in hand,
we can invoke general spectral theory to find expressions for
the spectral representation and other functions of $H_V$
(see, e.g.\ \cite{BS}).
To $H_V$ and any interval $(a,b)$
are associated a maximal closed subspace of $L^2(\reals^+)$
on which $H_V$ has purely absolutely continuous spectrum,
and spectrum contained in $(a,b)$.
By \eqref{spro} and \eqref{acpart} as well as by
definition of the absolutely continuous part of the spectral projection,
the projection $P^{\rm{ac}}_{(a,b)}$ of $L^2(\reals^+)$
onto this subspace can be written as
\begin{equation}\label{acpro}
P_{(a,b)}^{\rm{ac}}g(x) =
\frac{1}{\pi}
\int_a^b u_1(x,E) \Big( \int_\reals u_1(y,E) g(y) \,dy \Big)
\Im m(E+i0)\,dE.
\end{equation}
The integral over $\reals$ here is generally understood in the
$L^2$limiting sense; we will omit such explanatory remarks in the future.
Consequently the
operator $U$ mapping continuous functions with compact support to
$L^2(\reals, dx)$, defined by
\begin{equation} Uh(x) = \frac{1}{\pi}\int_{\reals}
u_1(x,E) h(E) \Im m(E+i0) \,dE \end{equation}
extends to an isometry of $L^2(\reals,\Im m(E+i0) \,dE)$
onto the absolutely continuous subspace associated to $H_V$.
The unitary evolution operator on the absolutely continuous subspace is
given by
\begin{equation}\label{evol}
e^{iH_V t} g(x) =
\frac{1}{\pi}\int_\reals e^{iEt}u_1(x,E) \tilde{g}(E)
\Im m(E+i0)\,dE,
\end{equation}
where
\[ \tilde{g}(E) =
\frac{1}{\pi}\int_\reals u_1(y,E) g(y)\,dy. \]
Finally, in the case $V=0$ we can compute explicitly
$u_1(x,E) = \sqrt{E}^{1} \sin \sqrt{E}x,$
$m(z) = \sqrt{z},$ and so the evolution operator can be written as
\begin{equation}\label{freeevol}
e^{iH_0 t} g(x) =
\frac{1}{\pi}\int_\reals e^{iEt} \sin(\sqrt{E}x) \hat{g}(E)
\,\frac{dE}{\sqrt{E}},
\end{equation}
where
\[ \hat{g}(E) = \int \sin (\sqrt{E}x) g(x)\,dx. \]
For almost every $E>0$, define the scattering coefficient
$\gamma(E)\in\complex$ by
\begin{equation} \label{reflectioncoefficientdefn}
\gamma(E) = 1/ u(0,E)
\end{equation}
where $u(x,E)$ is the unique generalized eigenfunction
asymptotic to $\exp(i\xi(x,E))$ as $x\to+\infty$,
whose existence was established in Theorem~\ref{thm:complexeigenfunctions}.
The following proposition connects the formulae of this section
with the generalized eigenfunctions analyzed in \S\ref{section:complex}.
\begin{proposition}[Limiting absorption principle]\label{relim}
Assume that $V\in L^1+L^p(\reals)$ for some $1