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sparse potentials, singular continuous spectrum, quantum dynamics
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\begin{document}
\begin{center}
{\bf DYNAMICAL ANALYSIS OF SCHRODINGER OPERATORS}
\end{center}
\begin{center}
{\bf WITH GROWING SPARSE POTENTIALS}
\end{center}
\begin{center}{\bf Serguei Tcheremchantsev}
\end{center}
\begin{center}{UMR 6628-MAPMO, Universit\'e d'Orl\'eans, B.P. 6759,}
\end{center}
\begin{center} {F-45067
Orl\'eans Cedex, France}
\end{center}
\begin{center}
{E-mail:
 Serguei.Tcherem@labomath.univ-orleans.fr}
\end{center}




\vskip 2 cm
\begin{center}
\section{Introduction}
\end{center}
\setcounter{equation}{0}
\vskip 1 cm
\par
Consider the discrete Schr\"odinger operators in $l^2(\Z^+)$:
\be\label{in01}
H_\theta \psi (n)=\psi(n-1)+\psi(n+1)+V(n)\psi (n) , 
\ee
where $V(n)$ is some real function, with boundary condition
\be\label{in02}
\psi (0) {\rm cos} \theta +\psi (1) {\rm sin} \theta=0, \ \
\theta \in (-\pi/2, \pi/2). 
\ee
We shall consider the case of sparse potentials. Namely,
$V(n)=V_N$, if $n=L_N$ and $V(n)=0$ else, where $L_N$ is monotone rapidly
increasing sequence. Such potentials were studied, in particular,  
in \cite{G, P, S, SSP, SST, JL, K, KR, Z}.
Their interest lies in the fact that the spectrum on
$(-2,2)$ may be singular continuous with nontrivial Hausdorff
dimension. In the present paper we will be interested by a particular
case of such potentials, considered by Jitomirskaya and
Last \cite{JL}. We consider a slightly more general model. Let  

\be\label{in03}
V(n)=\sum_{N=1}^{\infty} L_N^{\eta - 1 \over 2 \eta} \delta_{L_N,n}+Q(n)
\equiv S(n)+Q(n), 
\ee
where  $L_N$ is some very fast growing sequence such that
$$
L_1 L_2 {\cdots} L_{N-1}=L_N^{\alpha_N},
\ \ {\lim}_{N \to \infty} \alpha_N = 0,
$$
$\eta \in (0,1)$ is a parameter, and $Q(n)$ is any compactly supported real
function  (i.e. $Q(n)=0$ for all $n \ge n_0$). 
It is well known that the study of operator defined by
(\ref{in01})-(\ ref{in02}) 
is equivalent to the study of
operator with Dirichlet boundary condition $\psi(0)=0$
and potential
$$
V_1(n)=V(n)-{\rm tan} \theta \delta_{1,n}. 
$$
It is clear that $V_1(n)=S(n)+Q_1(n)$, where $Q_1$ is another
compactly supported function.
Thus, without loss of generality, we may consider
only operators with Dirichlet boundary condition and potentials
given by (\ref{in03}). We shall denote by $H$ the corresponding
operator.
\par
For such model, it is known \cite{SST}
that $(-2,2)$ belongs to the singular
continuous spectrum of $H$, and there may exist some discrete
point spectrum outside of $[-2,2]$.  It was shown in
\cite{JL} that the Hausdorff dimensionality of the spectrum
in $(-2,2)$ lies between $\eta$ and ${2 \eta \over 1+\eta}$
for all boundary conditions (they consider $Q(n)=0$ in our notations). 
Moreover, for Lebesgue a.e. $\theta$, the spectrum on $(-2,2)$ is
of exact dimension $\eta$. Combes and Mantica \cite{CM} showed that
the parking dimension of the spectral measure restricted to $(-2,2)$
is equal to 1. 
These spectral results imply 
dynamical lower bounds in a usual way \cite{L}, \cite{GSB}.  However,
for the considered model this is only partial dynamical information.   
Some dynamical upper bounds were obtained by Combes and Mantica
\cite{CM} (in our proofs we use some ideas of their paper).  
Krutikov and Remling \cite{KR},
\cite{K} studied the behaviour of the Fourier transform of the
spectral measure at infinity. 
\par
The main motivations of the present paper are the following:
\par\noindent
1. To give a rather complete description of
the (time-averaged) dynamical behaviour
of the considered model related to the
singular continuous part of the spectrum
(and some strong results in the case of more general initial states). 
This is the first example of this kind where dynamics is studied in
such a detailed manner. 
Although this model is simple enough, the results suggest what
could be done in more complicated cases, namely,
for Fibonacci potentials, bounded sparse barriers, random decaying
potentials or random polymers. 
\par\noindent
2. For some initial states $\psi$ we find the exact expression of
the intermittency function (see definition below)
$\beta_\psi^-(p)$ which is nonconstant in $p$. To the best of our
knowledge, this is the first model where such phenomenon of
"quantum intermittency" is rigorously proven. 
\par\noindent
3. Throughout the paper,
we use many different methods to study dynamics
and show how their combination gives stronger results.
In particular, we further develop the method for proving
lower bounds based on Parseval formula \cite{DT}, allowing more
general initial states $\psi$ than $\delta_1$. 
We hope that these ideas will be useful in many other cases. 
\par\noindent
4. For a long time the priority was given to
the spectral analysis of operators with s.c. spectrum
rather than to the
analysis of the corresponding 
dynamics (and most dynamical bounds were obtained as a
consequence of spectral
results). 
In the present paper we show how it is possible
to study {\it direclty} dynamics without virtually any knowledge of the
spectral properties. Indeed, the only information 
we need in our
considerations is that $(-2,2) \in \sigma (H)$.
Although
we prove that the spectral measure is of exact 
Hausdorff dimension  $\eta$ on $(-2,2)$
for {\it all} boundary conditions 
(improving the result of Jitomirskaya and Last), this is just 
a particular simple consequence of our dynamical results. 

\vskip 0.5 cm




\par
Let $\psi \in l^2(\Z^+)$
be some initial state (in particular, $\psi=\delta_1$).
The time evolution is given by 
$$
\psi (t)=\exp (-itH)\psi, 
$$
where $\exp(-itH)$ is the unitary group. We shall be interested by
the time-averaged quantities like
$$
a_\psi (n,T)={1 \over T} \int_0^\infty dt \exp (-t/T) |\psi (t,n)|^2.  
$$
This definition of time-averaging is virtually equivalent to
the Cesaro average, but is more convinient for technical reasons.  
We consider the time averaging because of the rather
irregular behaviour in time of $|\psi (t,n)|^2$ in the case of
singular continuous spectrum. For the sparse barriers model
one can see it from numerical simulations in \cite{CM}.
Moreover, effective analytical methods exist to study
time-averaged quantities. 
Mention that upper bounds for the return probability as $t \to \infty$
without time-averaging  are obtained in \cite{K}, \cite{KR}, and this
is difficult technically.  
\par
We shall study the inside and outside time-averaged probabilities
defined as 
$$
P_\psi (n \le M,T)=\sum_{n \le M} a_\psi (n,T)
$$
and
$$
P_\psi (n \ge M,T)=\sum_{n \ge M} a_\psi (n,T)
$$
respectively. Here $M>0$ are some numbers which may depend on $T$
(growing with $T$).
The quantity $P_\psi (n \le M, T)$ can be interpreted as the time-averaged
probability to find a system inside an interval $[0,M]$ and similarly
for outside probabilities. 
The obtained results are of the form
\be\label{in05}
P_\psi (n \ge M_1(T), T) \ge c>0, \ \ P_\psi (n \le M_2 (T), T) \ge c>0, 
\ee
or
\be\label{in06}
P_\psi (n \ge M_3(T), T) \ge h(T),  \ \ P_\psi (n \le M_4(T), T) \ge g(T), 
\ee
and similarly for the upper bounds, 
where $M_i(T)\to +\infty $ are some growing functions, and
$h(T), g(T)$ tends to 0 not faster then polynomially. Thus, we control
the essential parts of the wavepacket (\ref{in05}), as well as polynomially
small parts of the wavepacket (\ref{in06}) (such bounds for outside
probabilities imply lower bounds for the moments of the position operator). 
\par
We also consider the more traditional quantities:
$$
\la |X|_\psi^p \ra (T)=\sum_{n>0} n^p a_\psi (n,T), \ \ p>0, 
$$
called time-averaged moments of order $p$ of the position operator, as
well as their growth exponents $\beta_\psi^\pm (p)$
(both functions non decreasing in $p$): 
$$
\beta_\psi^-(p)={1 \over p} {\rm limsup}_{T \to \infty}
{ {\rm log} \la |X|_\psi^p \ra (T) \over {\rm log} T}, \ \ p>0,   
$$
and similarly for $\beta_\psi^+(p)$. Since
$$
\la |X|_\psi^p \ra (T) \ge M^p P_\psi (n \ge M,T)
$$
for any $M>0$, it is clear that probabilities and moments are related. 
\par
We shall also study the time-averaged return probability:
$$
J_\psi (1/T, \R)={1 \over T} \int_0^\infty dt \exp (-t/T)
|\la \psi (t), \psi \ra |^2. 
$$


\par



Let us present the main results. Assume first that $\psi$ belongs
to the subspace of continuous spectrum of $H$. Then due to RAGE
theorem, the system escape with time (after time-averaging)
from any finite interval $[1,M]$ and thus the quantum particle
goes to infinity. Since the barriers are very sparse, the picture
of motion is rather obvious. 
If the main part of the wavepacket is far enough
from the barriers: $L_{N-1} << n << L_N$ for some $N$, then the propagation
is ballistic (as in the case of the free particle). When the
wavepacket reaches a barrier $V(L_N)$ (at time $T$ of order $L_N$) 
the motion is slowed down and the process
of tunneling through the high barrier begins. The time necessary for the
essential part of the wavepacket to go through is
about $L_NV^2(L_N)=L_N^{1/\eta}$. During this time the main part
of the wavepacket is confined in the interval $[1, L_N]$.
For $T >> L_N^{1/\eta}$ most of the wavepacket is on
$ [L_N+1, \infty)$ and a new period of ballistic motion begins. 
It is clear that given a large value of $T$, it is crucial to locate
it with respect to the $L_N$. Thus, throughout the paper for any 
$T$ we shall denote by $N$ (depending on $T$ and $N \to \infty$
if $T \to \infty$) the unique value such that
$L_N/C \le T <L_{N+1}/C$ with some $C>1$.  
We prefer considering $L_N/C \le T<L_{N+1}/C$ rather than
$L_N \le T <L_{N+1}$ for the following reason: if $L_N/C \le T \le L_N$,
the far tail of the wavepacket is already
approaching the barrier $V(L_N)$ and the tunneling begins.
For simplicity, we take $C=4$ (of course, one could
take any other value). 
\par
Let $\psi=f(H)\delta_1 \ne 0$, where $f$ is a  complex function from
$f\in C_0^\infty ([-2+\nu, 2-\nu])$ for some $\nu \in (0,1)$.
The operator $f(H)$ is given by spectral theorem.
We shall call these $\psi$ initial states of the first kind. 

The followng bounds are proven:
\par\noindent
For $T: \ L_N/4 \le T \le 2 L_N$,
\be\label{in21}
C_1 L_N^{1-1/\eta-\alpha_N} \le P_\psi (n \ge 2 L_N, T) \le 
C_2 L_N^{1-1/\eta}, 
\ee
where $\alpha_N \to 0$ as $N \to \infty$ (i.e. as $T \to \infty$)

These bounds describe the beginning of tunneling.
\par\noindent
For $T: \ 2 L_N \le T \le L_N^{1/\eta}$, 
\be\label{in22}
C_1 TL_N^{-1/\eta-\alpha_N} \le P_\psi (n \ge T, T) \le
P_\psi (n \ge 2 L_N, T) \le 
C_2 TL_N^{-1/\eta}, \ \alpha_N \to 0.
\ee
These bounds describe the main part of the tunneling process. 

In particular, for  $T: \ L_N/4 \le T \le L_N^{1/\eta}$
\be\label{in23}
P_\psi (n \le 2 L_N, T) \ge ||\psi||^2-C TL_N^{-1/\eta}. 
\ee
Thus, for $T: \ L_N/4 \le T \le c L_N^{1/\eta}$ with $c$ small
enough, the main part of the wavepacket is located in $[1,2L_N]$. 



\par\noindent
Moreover, for $T: \ L_N/4 \le T \le L_N^B$ with some $B>0$,
\be\label{in24}
P_\psi (L_N/4 \le n \le L_N, T) \ge C(B) L_N^{-\alpha_N}. 
\ee
\par\noindent
For $T: \ L_N/4 \le T \le L_N^{1/\eta}$ the following bounds hold
for the time-averaged moments of position operator:
\be\label{in10}
C_1 L_N^{-\alpha_N} (L_N^p+T^{p+1}L_N^{-1/\eta}) \le \la |X|_\psi^p \ra (T)
\le C_2 (L_N^p +T^{p+1}L_N^{-1/\eta}). 
\ee
The bounds (\ref{in21})-(\ref{in10}) are proved in Theorem~\ref{upprob}
and Theorem~\ref{probext}. 
The upper bound in  (\ref{in10}) for the moments averaged over
the boundary condition $\theta$ (\ref{in02}) was proved by Combes and Mantica
in \cite{CM} for $p \le 2$. Our result holds for all $p>0$ and
all compact potential $Q$ (in particular, for
{\it all} boundary conditions).


\par
The next bounds describe the beginning and the end of ballistic regime.
If $T: \ L_N^{1/\eta} \le T \le L_N^{1/\eta+\delta}$ or
$T: \ L_{N+1}^{1-\delta} \le T <L_{N+1}/4$ for some $\delta>0$, then
\be\label{in25}
CL_N^{-\alpha_N} \le P_\psi (n \ge T,T) \le P_\psi (n \ge 2 L_N, T)
\le ||\psi||^2,  
\ee
and for the moments
\be\label{in26}
C_1 T^p L_N^{-\alpha_N} \le \la |X|_\psi^p \ra (T) \le C_2T^p. 
\ee
\par\noindent
These bounds are proved in Theorem~\ref{probext}. 
\par
Finally, if $T: \ L_N^{1/\eta+\delta} \le T \le L_{N+1}^{1-\delta}$,
the motion is exactly ballistic. Namely, for any $\theta>0$ there
exists $\tau>0$ small enough (independent of $T$) such that
\be\label{in27}
||\psi||^2-\theta \le P_\psi (n \ge \tau T, T) \le ||\psi||^2, 
\ee
for $T$ large enough, and
\be\label{in28}
C_1 T^p \le \la |X|_\psi^p \ra (T) \le C_2T^p. 
\ee
Moreover, for  $T: \ L_N^{1/\eta}\le T <L_{N+1}/4 $,
\be\label{in29}
P_\psi (n \le 2 L_N, T) \le C L_N^{1/\eta+\alpha_N}T^{-1}.  
\ee
The bounds (\ref{in27})-(\ref{in28}) are proved in
Theorem~\ref{ballistic}, and (\ref{in29}) follows from (\ref{upsi1})
of Lemma~\ref{upsingle}. 

\par
For the time-averaged return probability for any
$T: \ L_N/4 \le T <L_{N+1}/4$ the bounds hold (Theorem~\ref{retpro}):
\be\label{in30}
{C \over L_N (1+TL_N^{-1/\eta})} \le J_\psi (1/T, \R)
\le   {C L_N^{\alpha_N} \over L_N (1+TL_N^{-1/\eta})}. 
\ee
A related result (Lemma~\ref{upsingle}) states that
\be\label{in31}
a_\psi (n,T)=\la |\psi|^2 (t,n) \ra (T) 
\le   {C L_N^{\alpha_N} \over L_N (1+TL_N^{-1/\eta})}
\ee
for any $n$. 
\par
As a particular corollary of our bounds for the time-averaged moments,
we obtain the exact expression for the functions $\beta_\psi^\pm (p)$
(the result for $\beta_\psi^+(p)$ follows also from 
${\rm dim}_P (\mu_\psi)=1$ proved in \cite{CM}):
\be\label{in11}
\beta_\psi^-(p)={p+1 \over p+1/\eta}, \ \beta_\psi^+(p)=1, \ p>0. 
\ee
Thus, the upper bound for $\beta_\psi^-(p)$, obtained in \cite{CM}
for $p \le 2$ and for a.e. boundary conditions, gives in fact the
exact expression of $\beta_\psi^-(p)$ for all $p>0$ and all boundary
conditions, as it was conjectured in \cite{CM}.  The result (\ref{in11})
is important from two points of view:
\par\noindent
1. This is the first example where nontrivial (i.e. nonconstant)
function $\beta_\psi^-(p)$ is rigorously calculated.
\par\noindent
2. It implies (Corollary~\ref{betas})
that the restriction of the spectral measure
on $(-2,2)$ is of exact Hausdorff dimension $\eta$. 
This result holds for all compact potentials $Q$ and thus,
in particular, for all
boundary conditions $\theta$ in (\ref{in02}). This improves
the result of \cite{JL}, where it was proven
only for Lebesgue-a.e. $\theta$.
\vskip 0.5 cm

Consider now more general initial states $\psi$, for example,
$\psi=\delta_1$. The problem is that we have no control of the discrete
spectrum outside $(-2,2)$. Thus, it is possible that some part of the
wavepacket remains well localized at any time. On the other hand,
it is also possible that the part of the wavepacket related to the
discrete spectrum moves quasiballistically (the well known
example is the one of \cite{DRJLS}).

As a consequence, we cannot prove
nontrivial upper bounds for the outside probabilities and for the moments,
and we cannnot prove that all the wave function escapes from
$[1,L_N]$ as $T>>L_N^{1/\eta}$. 


However, the part of the wavepacket
corresponding to the continuous spectrum (if nonzero)
behaves in the same manner.
It escapes from any
interval $[1,M]$, moves ballistically between the barriers, tunnels
through the barrier etc. Therefore, we are able to prove nontrivial
lower bounds for outside probabilities and for the moments.
\par
Consider $\psi=f(H)\delta_1\ne 0$, where $f$ is some
bounded Borel complex function such that for some interval
$S=[E_0-\nu, E_0+\nu] \subset [-2+\nu, 2-\nu]$, $f$ is $C^\infty$
on $S$ and $|f(x)| \ge c>0$ on $S$. We call these $\psi$
initial states of the second kind. 
In particular, previously
considered $\psi$ and $\psi=\delta_1$ verify this condition.

 
\par
For $\psi$ described above, the following bounds hold
(proved essentially in Theorem~\ref{probext} and Theorem~\ref{ballistic}): 
\par\noindent
The first bound in (\ref{in21}), and the first and the second
bounds in (\ref{in22}) remain true.  Instead of (\ref{in23}) we prove that
for some $\delta>0$ small enough and
$T: \ L_N/4 \le T \le \delta L_N^{1/\eta}$, 
$$
P_\psi (n \le 2 L_N, T) \ge c_1>0. 
$$
The bound (\ref{in24}) remains true as well
as the first bound in (\ref{in10}). The bound (\ref{in25}) and the
first bound in (\ref{in26}) hold (we do not have a priori ballistic
upper bound for the considered $\psi$, except the case
 where $f$ is smooth, in particular,  $\psi=\delta_1$).
 Instead of (\ref{in27}), one has
the bound
$$
P_\psi (n \ge \tau T, T) \ge c_1>0. 
$$
The first bound in (\ref{in28}) follows. For the time-averaged return
probability the lower bound in (\ref{in30}) holds (Theorem~\ref{retpro}). 
 For the functions
$\beta_\psi^\pm(p)$, one has lower bounds
$$
\beta_\psi^-(p) \ge {p +1 \over p+1/\eta}, \ \beta_\psi^+(p) \ge 1. 
$$
\par
One can ask whether the smoothness condition on $f$ is relevant.
As for the upper bounds for moments and outside probabilities, 
it seems essential.
Some results, namely,
Lemma~\ref{m1}, Corollary~\ref{essen}, Lemma~\ref{upsingle}
and Theorem~\ref{retpro}, 
hold for nonsmooth $f$. 
Probably, lower bounds for outside probabilities and for the moments
(for both kinds of $\psi$) can be proved
without smoothness of $f$.
\par
The paper is organized as following. In Section 2 we first prove
upper bounds for the transfer matrices with complex energies
$T(n,0;z)$
associated to the equation $Hu=zu$.
With this result we obtain some lower
bounds for probabilities and for the moments 
(Theorem~\ref{m2}) using Parseval formula.   
The combination of this  method with the traditional approach
going back to Guarneri, 
 allows us to obtain some control of the essential part of the
 wavepacket (Corollary~\ref{essen}) 
 as well as
better lower bound for the time-averaged moments
(Corollary~\ref{moments} and Theorem~\ref{betamo}). 
 The approach of Section 2
can be applied to a more general class of models, where
the transfer matrix 
has nontrivial upper bound like
$$
||T(n,0; E+i\e)|| \le g_\Delta (n),  \ \ E \in \Delta, \ \e \in (0,1).  
$$
Here $\Delta$ is any compact interval in $(-2,2)$, 
and the function $g_\Delta (n)$, growing not too fast, 
does not depend on $E \in \Delta, \ \e \in (0,1)$.
In particular, $g_\Delta=C(\Delta)n^\alpha$ with some $\alpha>0$
is possible (Theorem~\ref{power}).  
\par
The bounds of Theorem~\ref{m2} show the importance of the integrals
$$
I(\Delta,\e)=\e \int_\Delta dE {\rm Im}^2 F(E+i\e), \
\Delta \subset (-2,2),
$$
where $F$ denotes the Borel transform of spectral measure. 
Good lower bounds for $I(\Delta, \e)$ imply better lower bounds
for probabilities and thus for the moments. These integrals are closely
related to the time-averaged return probabilities and to the
correlation dimensions of the spectral measure restricted to $(-2,2)$. 
\par
In Section 3, which is specific to the considered model with growing
sparse potentials, we obtain upper bounds for
inside (Lemma~\ref{upsingle}) and outside probabilities and 
moments (Theorem~\ref{upprob}). These results are proved for
$\psi=f(H)\delta_1$ with $f$ compactly supported on $(-2,2)$
(and moreover $f \in C_0^\infty$ in Theorem~\ref{upprob}).
When considering the inside probabilities,
we obtain some upper bound for ${\rm Im} F(x+i\e),
\ x \in (-2,2)$. 
It implies a very simple proof of the fact
that for any $\delta>0, \nu>0$
the spectral measure is uniformly $\eta-\delta$-H\"older continuous
on $[-2+\nu, 2-\nu]$ (the result which follows also from the proofs of
\cite{JL}). 
\par
In Section 4 we first use
the obtained upper bounds for outside probabilities to obtain lower
bound for the integrals $I(\Delta, \e)$ which is virtually optimal
(Corollary~\ref{lowI}). Together with Theorem~\ref{m2}, it
implies better lower bounds for probabilities and for the moments
(which are optimal for $\psi$ of the first kind
up to the factors like $L_N^{\alpha_N}$, where
$\alpha_N \to 0$). 
 It implies also bounds for the time-averaged
return probabilities (Theorem~\ref{retpro}). 
 The upper bounds of Section 3 are also used
(Theorem~\ref{ballistic})
to control the essential part of the wavepacket on
$[1, 2 L_N]$ and on $[\tau T, +\infty)$ with some $\tau>0$.  
Finally, we show that the obtained upper bounds for the
moments imply that the restriction of spectral measure on $(-2,2)$
is of exact Hausdorff dimension $\eta$. 
\par\noindent
{\bf Acknowledgments}.  I would like to thank F. Germinet for useful
discussions. 

\vskip 2 cm
\section{Direct lower bounds for probabilities and moments}
\setcounter{equation}{0}
\vskip 1 cm
\par
Define the time-averaged
quantities (which we call probabilities) of the form
$$
P_\psi(n \ge M,T)=\sum_{n \ge M} \la |\psi (t,n)|^2 \ra (T) \equiv
\sum_{n \ge M} {1 \over T}
\int_0^{+\infty} dt e^{-t/T} |\exp(-itH)\psi(n)|^2 
$$
and similarly for $P_\psi (n \le M, T), \ P_\psi (L \le n \le M, T)$,
where $M,L$ may depend on $T$. 
We shall call $P_\psi (n \ge M, T)$ outside and
$P_\psi (n \le M, T)$ inside probabilities respectively. 

\par
Throughout the paper we shall consider two kinds of initial states
$\psi$:
\par\noindent
1. $\psi=f(H)\delta_1$, where $f \in C_0^\infty ([-2+\nu, 2-\nu])$
for some $\nu>0$ and $f(x_0) \ne 0$ for some $x_0$. 
We shall call these $\psi$ initial states of the
first kind. 
\par\noindent
2. $\psi=f(H)\delta_1$ where $f: \R \to \C$
is a bounded Borel function such that
for some $[E_0-\nu,E_0+\nu] \subset [-2+\nu,2-\nu]$, with $\nu>0$,  
\be\label{condf}
f \in C^{\infty} ([E_0-\nu, E_0+\nu])
\ {\rm and}  \ |f(x)| \ge c>0, \
x \in [E_0-\nu, E_0+\nu]. 
\ee 
In particular, one can take $\psi=\delta_1$.  We shall call these
$\psi$ initial states of the second kind. One can observe that
any $\psi$ of the first kind belongs to the second kind. 
\par
In the case of any $\psi$ we shall denote by
$\mu_\psi$ the corresponding spectral 
measure, and by
 $\mu\equiv \mu_{\delta_1}$ the measure of the state $\delta_1$. Observe
that
$$
d \mu_{\psi} (x)= |f(x)|^2 d \mu (x).
$$
\par
Let $\psi$ be any vector and $\mu_\psi$ its spectral measure. For any
Borel set $\Delta$ and $\eps >0$
define the following integrals:
$$
J_\psi (\eps, \Delta)=\int_\Delta d \mu_\psi (x) \int_{\R}
d \mu_\psi (y) R((x-y)/\eps), 
$$
where $R(w)=1/(1+w^2)$. These quantities will play an important role
in the sequel. 
What one can observe is the following
identity (which can be easily proved using spectral theorem):
\be\label{return}
{1 \over T} \int_0^\infty dt \exp (-t/T) |\la \psi (t), \psi \ra |^2=
J_\psi (\eps, \R), \ \ \eps=1/T.  
\ee
Thus, $J_\psi (\eps, \R)$ coincides with the time-averaged return
probability. 
\par
The first statement is of a rather general nature, and holds in fact for any
self-adjoint operator $H$. 
\begin{lem}\label{m1}
{\sl
Let $H$ be some self-adjoint operator in $l^2(\N)$ and $\psi$ any vector
such that $c_1=\mu_{\psi}(\Delta)>0$, where $\Delta$ is some Borel set. 
Let  $M(T)=c_1^2/(16 J_{\psi}(T^{-1}, \Delta))$. Then
$$
P_\psi (n \ge M(T), T) \ge c_1/2>0.
$$
}
\end{lem}
Proof. The result follows rather directly from \cite{T} and
is obtained using  
the traditional approach developed by Guarneri-Combes-Last. 
For the sake of completeness we shall give the main lines of the 
proof.
Define
$$
\rho=X_{\Delta} \psi, \ \chi=\psi-\rho, 
$$
where $X_S$ is the spectral projector of the operator $H$ on the set $S$. 
One has $\rho \ne 0$ since $\| \rho \| ^2= \mu_\psi (\Delta)=c_1>0$.
One shows \cite{T} that for any $M>0$,
\be\label{p03}
P_\psi(n \ge M,T) \ge ||\rho||^2-2|D(M,T)|, 
\ee
where 
$$
D(M,T)={1 \over T} \int_0^{+\infty} dt  \exp(-t/T)
\sum_{n<M} \psi (t,n) {\overline {\rho(t,n)}}.
$$
Lemma 2.1 of \cite{T} (with $h(u)=\exp(-u), \ u>0$) implies 
\be\label{p04}
|D(M,T)| \le \int_{\Delta} d \mu_\psi (x) 
\sqrt{ b(x,T) S_M(x)},
\ee
where 
\be\label{bimf}
b(x,T)=\int_{\R} d \mu_\psi (u) R((T(x-u))
=\eps {\rm Im} F_{\mu_\psi} (x+i\eps), \ \eps={1 \over T}, 
\ee
$F_{\mu_\psi}$ is the Borel transform of spectral measure,  
$ S_M(x)=\sum_{n<M}|u_\psi(n,x)|^2$, and $u_\psi(n,x)$ are generalised 
eigenfunctions associated to the state $\psi$.
Since
$$
\int_{\R} d \mu_\psi (x) |u_\psi(n,x)|^2 \le 1  
$$
for any $n$, the bound (\ref{p04}) and Cauchy-Schwarz inequality yield
\be\label{p05}
|D(M,T)|^2 \le M J_{\psi}(\e, \Delta), \ \e=1/T.
\ee
Let us take $M=||\rho||^4 (16 J_{\psi}(\e, \Delta))^{-1}$. 
It follows from (\ref{p03}),
(\ref{p05}) that
$$
P_\psi(n \ge M,T)\ge ||\rho||^2/2=c_1/2>0.
$$
The proof is completed.
\par
In the sequel we shall also
need the following integrals:
$$
I_\psi (\e, \Delta)=\e \int_\Delta dE {\rm Im}^2 F_\psi (E+i\e)=
\e^3 \int_\Delta dE \left( \int_\R {d \mu_\psi (u) \over
\e^2+(E-u)^2} \right) ^2,  
$$
where $\psi$ is some state and $F_\psi$ denotes the Borel transform
of its spectral measure. 
In fact, the integrals $I_\psi (\e,\Delta)$ and $J_\psi (\e, \Delta)$ are
closely related. 

\begin{lem}\label{IJ}
{\sl
Let 
$0<\e<1$, $\Delta=[a,b]$ some bounded interval. 
  The uniform estimate holds:
\be\label{p10}
J_{\psi}(\e, \Delta) \le C(\Delta) I_{\psi} (\e, \Delta).
\ee
}
\end{lem}
Proof. For simplicity we shall omit the dependence on $\psi$ in the proof.   
The definition of $I$ implies
$$
I(\e, \Delta)=\e^3 \int_{\R} d \mu (x) \int_{\R} d \mu (u)
\int_\Delta  
{ dE \over ((u-E)^2+\e^2) ((x-E)^2+\e^2) }. 
$$
Thus
\be\label{p11}
I(\e, \Delta) \ge \int_\Delta d \mu (x) 
\int_{\R} d \mu (u) f(x,u,\e), 
\ee
where 
$$
f(x,u,\e)=\e^3 \int_a^b {dE \over 
((u-E)^2+\e^2)((x-E)^2+\e^2)}, \ \Delta=[a,b].  
$$
One changes the variable $t=(E-x)/\e$ in the integral over $E$: 
$$ 
f(x,u,\e)=\int_A^B {dt \over (t^2+1)((t+s)^2+1)}, 
$$
where $A=(a-x)/\e, B=(b-x)/\e, \ s=(x-u)/\e$.
Since one integrates in (\ref{p11}) over $x \in [a,b]$,
and $0<\e<1$, one can easily see that 
$$
f(x,u,\e) \ge c/(s^2+1)
$$
with uniform positive constant. The bound (\ref{p11}) yields
\be\label{p20}
I(\e, \Delta) \ge c \int_\Delta  d \mu (x) 
\int_{\R} d \mu (u) R((x-y)/\eps)=cJ(\eps, \Delta). 
\ee
\par


\par
As a basis of our proofs we shall use the following
\begin{lem}\label{transfer}
{\sl
Let $x \in [-2+\nu, 2-\nu]$ with some $\nu>0$, $\eps \in [0,1)$.
The following uniform
bounds hold under condition $n \eps \le K$ for some $K>0$.   
\par\noindent a)
If $n<L_N$, then
\be\label{tr1}
||T(n,0; x+i\eps)|| \le C(K, \nu) L_N^{\alpha_N}, \ \ \alpha_N \to 0.   
\ee
If $n: \ L_N \le n < L_{N+1} $,  
then
\be\label{tr2}
||T(n,0; x+i \eps)|| \le C(K, \nu) L_N^{{1 - \eta \over 2 \eta} +\alpha_N}, \ \
\alpha_N \to 0. 
\ee
}
\end{lem}
Proof. 
Assume first that $Q(n) \equiv 0$. Then one can easily see that
for any $n: \ L_m \le n <L_{m+1}$ with some $m \ge 1$, 
\be\label{tr3}
T(n,0; z)=T_0(n-L_m, z) A(L_m, z) T_0(L_m -L_{m-1}+1, z)
A(L_{m-1}, z) \cdots A(L_1,z) T_0(L_1-1, z). 
\ee
Here $T_0(k,z)=A_0(z)^k$ is the free transfer matrix with
\be\label{a0z}
A_0(z)=\left( \begin{array}{cr} z & -1 \\  1 & 0 \end{array}
\right) 
\ee
and
$$
A(L_k,z)=\left(
\begin{array}{cr} z-L_k^{1 - \eta \over 2 \eta} & -1 \\
 1 & 0 \end{array} \right).
$$
For real $x \in [-2+\nu, 2-\nu]$ one can show by direct calculations
that $||T_0(k,x)|| \le C$ uniformly in $x,k$. For complex
$z=x+i \eps $ one represents
$A_0(z)=A_0(x)+i \eps D$ with
$$
D=\left( \begin{array}{cr} 1 & 0 \\ 0 & 0 \end{array} \right). 
$$
Developing $A_0(z)^k$, one sees
that one has still
\be\label{tr4}
||T_0(k,z)||=||A_0(z)^k || \le C
\ee
while $k \eps \le K$. As to $A(L_k,z)$, it can be bounded by
\be\label{tr5}
||A(L_k, z)|| \le C L_k^{1 - \eta \over 2\eta}, 
\ee
since $x \in [-2+\nu, 2-\nu], \ \eps \in [0,1)$. The statement of the
Lemma follows directly from the bounds (\ref{tr3})-(\ref{tr5}) and
the sparseness condition:
$$
L_1 L_2 \cdots L_m \equiv L_{m+1}^{\nu_{m+1}}, \ \ \nu_m \to 0. 
$$
For more details see the similar proof in \cite{JL}.
If one add the finite range perturbation $Q(n)$, it is clear that
the norms $||T(n,0;z)||$ remain bounded by the same expressions
(\ref{tr1}), (\ref{tr2}) with different constants. 
 The proof is completed. 
\par
In the next statement we shall use notation
$I(\e, \Delta)=I_{\delta_1} (\e, \Delta)$. In all statements of the paper
$\alpha_N$ denotes sequences such that $\lim \alpha_N=0$
(not necessarily the same). 

\begin{teo}\label{m2}
{\sl Assume that $\psi$ is of the second kind 
(in particular, $\psi=\delta_1$). 
Let $\Delta=[E_0-\nu/2, E_0+\nu/2]$, where $\nu$ comes from
(\ref{condf}). 
\par\noindent
1. Let $L_N \le T <L_{N+1}/4$ for some $N$. The uniform in
$T$ bound holds:
\be\label{p06}
P_\psi(n \ge T,T) \ge cT L_N^{{\eta -1 \over \eta} -\alpha_N} 
(I(1/T, \Delta)+{1 \over T}) \ge c L_N^{{\eta -1 \over \eta}-\alpha_N}.  
\ee
\par\noindent
2. Let $L_N/4 \le T \le 4 L_N$. Then
\be\label{p06a}
P_\psi(n > L_N,T) \ge c L_N^{2 -{1 \over \eta} -\alpha_N} 
(I(1/T, \Delta)+{1 \over T}) \ge c L_N^{{\eta -1 \over \eta}-\alpha_N}.   
\ee
\par\noindent

3. Let $L_N/4 \le T \le L_N^B$ with some $B>1$. Then the uniform bound holds:
\be\label{p07}
P_\psi(L_N/4 \le n \le L_N, T) \ge c_B L_N^{1-\alpha_N} 
(I(1/T, \Delta)+{1 \over T})\ge c_B L_N^{1-\alpha_N} T^{-1}.  
\ee
In all bounds (\ref{p06})-(\ref{p07}), $c>0$ and
$\lim_{N \to \infty} \alpha_N=0$.

}
\end{teo}
Proof. We shall follow the ideas of
\cite{DT}. The starting point is the Parseval formula:
\be\label{p08}
\la |\psi(t,n)|^2 \ra (T)={\e \over \pi} \int_{\R} 
dE |(R(E+i\e) \psi) (n)|^2, \ \e=(2T)^{-1}, 
\ee
where $R(z)=(H-zI)^{-1}$.
\par\noindent
a) We begin with $\psi=\delta_1$. 
Let $u(n,z)=(R(z)\delta_1)(n)$. It is well 
known  \cite{KKL}  that 
\be\label{p080}
(u(n+1,z),u(n,z))^T=T(n,0,z) (F(z), -1)^T, \ n \ge 0, 
\ee
where $T(n,0,z)$ is the transfer matrix associated to the equation
$Hu=zu$ and $F$ is the Borel transform of the spectral measure
$F(z)=\int_{\R} {d \mu_{\delta_1} (x) \over x-z}$. 
Let $E \in [-2+\delta, 2-\delta]$ with some $\delta \in (0,1)$,
$z=E+i\e, \ \e=(2T)^{-1}$. 
Assume first that $L_N \le T \le L_{N+1}/4$. 
The bound (\ref{tr2})
of Lemma~\ref{transfer} and (\ref{p080}) 
imply (since $||T||=||T^{-1}||$)
for any $L_N \le n \le 2T$
$$
|u(n+1,z)|^2 +|u(n,z)|^2
$$
\be\label{p09}
\ge ||T(n,0,z)||^{-2} (|F(z)|^2+1) \ge
a(\delta) L_N^{{\eta -1 \over \eta} -2 \alpha_N} (({\rm Im}^2 F(z)+1),
\ee
where $\alpha_N \to 0$. 
Summation in (\ref{p09}) over $n: T \le n \le 2T$ and integration over 
$E\in [-2+\delta,2-\delta]$ in (\ref{p08}) yields (\ref{p06}) with
$\Delta=[-2+\delta, 2-\delta]$. We have used a simple bound
$I(u/2, \Delta) \ge 1/8 I(u, \Delta)$, which directly follows
from the definition of integrals $I$. 
If $L_N/4 \le T\le 4L_N$, one considers
$n: 2L_N \le n \le 3 L_N$ to get (\ref{p06a}). 
The bound (\ref{p07}) is proved in a similar manner using the
bound (\ref{tr1}) of Lemma~\ref{transfer} and summating over
$n: L_N/4 \le n <L_N-1$.
\par\noindent b)
Assume now that $\psi$ is such that $\psi=g(H)\delta_1$, 
$g(x)\in C_0^\infty (S)$, where
$S=[E_0-\nu,E_0+\nu] \subset [-2+\nu,2-\nu]$ for some $\nu \in (0,1)$. 
Assume also that $g(x) \equiv 1, \ x \in [E_0-3\nu/4,E_0+3\nu/4]$.   
Consider the decomposition
$$
\delta_1=\psi+\chi, \ \psi=g(H)\delta_1, \ \chi=(1-g(H)) \delta_1.
$$
Let $L_N \le T \le L_{N+1}/4$. 
Since
$$
|R(z)\psi (n)|^2 \ge 1/2 |R(z)\delta_1(n)|^2 -|R(z)\chi (n)|^2,
$$
integration over $\Delta=[E_0-\nu/2,E_0+\nu/2]$ and summation over
$n: T \le n \le 2T$ yields (using the proof of a)):
\be\label{gk2}
P_\psi(n \ge T,T) \ge c T L_N^{{\eta-1 \over \eta}-\alpha_N}
(I(1/(2T), \Delta)+1/T) -c/T \int_\Delta
dE \sum_{n \ge T} |R(E+i \e)\chi (n)|^2
\ee

To bound from above $|R(E+i \eps) \chi (n)|, \ \ E \in \Delta$, 
we shall use now the following result
from \cite{GK}:
\be\label{z5}
|(u(H) \delta_m )(n)| \le C _k |||u|||_{k+2} (1+|n-m|^2)^{-k/2},
\ee
for any integer $k>0$, where $u$ is some smooth complex function,
$$
|||u|||_k=\sum_{r=0}^k \int_{\R} dx |u^{(r)} (x)| (1+|x|^2)^{(r-1)/2},
$$
and the constants in (\ref{z5}) are independent of $u$ and $H$.
Although the result of \cite{GK} is stated in the continuous case,
one can easily see that the result holds in the
discrete case for any self-adjoint operator $H$.
\par We shall take
$u_{E+i\eps} (x)={\chi (x) \over x-E-i \eps}$, where
$\chi(x)=1-g(x)$, and $z=E+i \eps$
is considered as a parameter. Thus, 
$R(E+i\eps) \chi (n)=(u_{E+i\eps} (H)\delta_1)(n)$. 
The definition of $f$ implies that $\chi (x)=0$ for any
$x \in [E_0-3 \nu/4, E_0+3 \nu/4]$.  
One can easily show that $|||u_{E+i \e}|||_k \le C(k,\nu)$ for any $k$
and any $E \in \Delta, \ \e>0$ with uniform constants. 
Thus, (\ref{z5}) implies
$|R(E+i\e)\chi (n)| \le C(k) n^{-k}$ and 
\be\label{gk4}
\sum_{n \ge T} |R(E+i\e)\chi (n)|^2 \le C(k)T^{-k}. 
\ee
Taking $k$ large enough, we see that (\ref{gk2}),
(\ref{gk4}) 
imply the same bound (\ref{p06}), since $T \ge L_N$ and thus the 
integral in (\ref{gk2}) is small with respect to the first term.
The bounds (\ref{p06a}) and (\ref{p07}) can be proved in the same manner.

\par\noindent c) Let now $\psi$ be any vector of the second kind. 
Let $g$ be some function verifying conditions of the part b), that
is, $g \in C_0^\infty (S)$, $S \equiv [E_0-\nu, E_0+\nu]$,  
$g(x)=1, \ x \in [E_0-3\nu/4,E_0+3\nu/4]$.
One can write
$$
g(x)=l(x) f(x), 
$$
where $l(x)=0$ if $|x-E_0| > \nu$ and $l(x)=g(x)/f(x)$, \
$|x-E_0| \le \nu$.
The facts that $f \in C^\infty(S), \
g \in C_0^\infty(S)$ and
$|f(x)| \ge c>0$ on $S$ imply that
$l \in C_0^\infty (S)$.  Again, due to (\ref{z5}), 
the kernel of $l(H)$ is fast decaying in $|n-m|$, so that for any $k>0$,
$$
|R(E+i\eps) g(H) \delta_1 (n)|^2 \le \sum_m {C_k \over 1+|n-m|^k}
|R(E+i\eps) f(H)\delta_1(m)|^2. 
$$
Therefore, for any $L>0$, 
$$
A(2L,T) \equiv 1/T \sum_{n \ge 2L} \int_{\Delta} dE
|R(E+i\eps) g(H)\delta_1(n)|^2 \le
$$
$$
1 /T \int_{\Delta}
dE \sum_m h_k(m,T) |R(E+i\eps) f(H)\delta_1(m)|^2,
$$
where
$$
h_k(m,T)=\sum_{n \ge 2L} {C_k \over 1+|n-m|^k}. 
$$
Let us split the sum over $m$ into two with $m<L$ and $m \ge L$. 
One observes that $h_k(m,T) \le C_k L^{1-k}$ in the first case and
one uses trivial bound $h_k(m,T)\le C_k $ in the second case. Thus, we get
\be\label{p311}
A(2L,T) \le C_k L^{1-k} +C_k /T \int_{\Delta} 
dE \sum_{m \ge L} |R(E+i\eps)f(H)\delta_1 (m)|^2, 
\ee
where we used the fact that
$$
\eps \sum_m \int_\R dE |R(E+i\eps) \psi (m)|^2 = \pi ||\psi||^2. 
$$
Let us assume first that $L_N \le T \le L_{N+1}/4$. 
One can easily see from the proofs of the part b) that 
the quantity $A(2T, T)$ is bounded from below by the r.h.s. of (\ref{p06})
(only the constant changes). 
Taking $k>1/\eta$, using (\ref{p311}) and 
Parseval equality, we get (\ref{p06}) for $\psi=f(H)\delta_1$.
For (\ref{p06a}) the proof is similar with $L=2L_N$. 
To prove (\ref{p07}), one considers
$$
A(T) \equiv 1/T \sum_{L_N/2 \le n \le 3L_N/4} \int_{\Delta} dE
|R(E+i\eps) g(H)\delta_1(n)|^2 \le
$$
\be\label{p3111}
1 /T \int_{\Delta}
dE \sum_m h_k(m) |R(E+i\eps) f(H)\delta_1(m)|^2 
\ee
with $h_k(m)=\sum_{L_N/2 \le n \le 3L_N/4} C_k (1+|n-m|^k)^{-1}$. 
Splitting the sum over $m$ in (\ref{p3111}) into three with
$m <L_N/4, \ m >L_N$ and $L_N/4 \le m \le L_N$, one shows that
the first two are bounded from above
by $C_k L_N^{1-k}$ and the third by
$$
C/T \int_\Delta \sum_{L_N/4 \le m \le L_N} |R(E+i\e)f(H)\delta_1(m)|^2. 
$$
On the other hand, $A(T)$ is bounded
from  below by the r.h.s. of (\ref{p07}) (the proof is identical
with the one of the part b), only the constants change).
Since $T \le L_N^B$, taking $k$ large enough we get
the bound (\ref{p07}) for $\psi=f(H)\delta_1$. 
The proof is completed.
\par\noindent


\begin{cor}\label{twoint}
{\sl
Let $\Delta=[-2+\nu,2-\nu]$ with some $\nu>0$.   
Let $\eps>0$ and $N$ be such that $L_N/4 \le T\equiv 1/\eps <L_{N+1}/4$.
The following
estimate holds:
\be\label{ij1}
J(\eps,\Delta) 
\le C I(\eps,\Delta) \le {C L_N^{\alpha_N} \over 
L_N+TL_N^{\eta-1 \over \eta} } 
\ee
with uniform in $T$ constants and $\lim \alpha_N =0$. Here the integrals
$J,I$ correspond to $\psi=\delta_1$. 
}

\end{cor}
Proof. We shall use the bounds of Theorem~\ref{m2} for $\psi=\delta_1$.
In this case, as it follows from the part a) of the proof,
(\ref{p06}), (\ref{p06a}),
(\ref{p07}) hold with $\Delta=[-2+\nu,2-\nu]$. Moreover, (\ref{p07})
holds for all $L_N/4 \le T \le L_{N+1}/4$ without restriction. 
On the other hand,
all the quantities $P_\psi(n \ge T, T), P_\psi (n\ge 2L_N, T)$,
$P_\psi(L_N/4 \le n \le L_N, T)$ are bounded from above by $1$.
We thus obtain the last inequality in (\ref{ij1}).
The first inequality is that of Lemma~\ref{IJ}. 
The proof is completed.   


\begin{cor}\label{essen}
{\sl
Let $\psi$ be any vector of the second kind (but $f$ is not necessarily
smooth on $\Delta=[E_0-\nu, E_0+\nu]$). 
Then
\be\label{p31}
P_\psi (n \ge M(T),T) \ge c>0,\  {\rm for}\
M(T)=CL_N^{-\alpha_N} \left( L_N+TL_N^{\eta-1 \over \eta} \right), 
\ee
where again $L_N/4 \le T <L_{N+1}/4$ and $\alpha_N \to 0$.  
}

\end{cor}
Proof. 
Since $|f(x)| \ge c>0, \ x \in \Delta$ and
$\Delta \subset (-2,2) \subset \sigma (H)$,
it is clear that $\mu_\psi (\Delta) \ge
c^2 \mu(\Delta)>0$. On the other hand, since $f$ is bounded, by 
(\ref{ij1}), 
\be\label{pila}
J_\psi (\eps, \Delta) \le C J(\eps, \Delta) \le C {L_N^{\alpha_N}
\over L_N+TL_N^{\eta-1 \over \eta} }. 
\ee
The result now follows from (\ref{pila}) and Lemma~\ref{m1}. 
The proof is completed. 
\par
Generally speaking, to obtain better lower bound
for $M(T)$, one should
better estimate from above the integrals $J(\eps,\Delta)$. Similarly,
to get better lower bounds for probabilities (Theorem~\ref{m2}),
one should bound from below the integrals $I(\eps,\Delta)$. 
These quantities are both closely related to the correlation dimensions
$D^\pm (2)$  \cite{T}
of the spectral measure restricted to
$\Delta$. To get
good bounds for $I,J$, one should have a rather 
good knowledge of the fine structure of the spectral measure.
In the Section 4 we shall use the obtained upper bound for the outside
probabilities to obtain optimal lower bounds for $I(\eps,\Delta)$.
The idea is the following: upper bound on outside probabilities \ $=>$
upper bound on $M(T)$ such that $P_\psi (n \ge M(T), T) \ge c>0$
\ $=>$ lower bound on  $J$ $=>$ lower bound on $I$.
This method, however, is specific to the considered model with 
unbounded sparse potentials.

\par
Consider now applications of the obtained results for
probabilities to
the time-averaged moments of the position operator:
$$
\la |X|^p_\psi \ra (T)\equiv \sum_n |n|^p \la |\psi(t,n)|^2 \ra (T),
\ \ p>0. 
$$
An immediate consequence of Lemma~\ref{m1} and Theorem~\ref{m2} is the
following. 


\par
\begin{cor}\label{moments}
{\sl
Let $\psi$ be of the second kind,    
$p>0, T: L_N/4 \le T <L_{N+1}/4$ for some $N$. 
The bounds hold:
\be\label{moij}
\la |X|_{\psi}^p \ra (T) \ge C 
J(\eps, \Delta)^{-p} +
C\left( L_N^{p+1-\alpha_N}+
T^{p+1} L_N^{{\eta-1 \over \eta} -\alpha_N} \right) 
(I(\eps, \Delta)+1/T) 
\ee
\be\label{moij2}
\ge C (L_N+TL_N^{\eta -1 \over \eta})^p L_N^{-p\alpha_N}+
C T^p L_N^{{\eta -1 \over \eta} -\alpha_N}, 
\ee

where $\eps=1/T$ and $\alpha_N \to 0$.

}
\end{cor}
Proof. One observes that
$\la |X|_{\psi}^p \ra (T) \ge M^p P_\psi(n \ge M,T)$ for any $M,T$.
The bound (\ref{moij}) now follows directly from 
Lemma~\ref{m1},
Theorem~\ref{m2} and Lemma~\ref{IJ}. The bound (\ref{moij2})
follows directly from (\ref{moij}) and
(\ref{pila})
(since $T \ge L_N$, the term with
$L_N^{p+1}/T$ is smaller than $L_N^p$, so we don't keep it in (\ref{moij2})). 
The proof is completed. 
\par

What is interesting is the
following observation:
even if one has no additional information about integrals $I,J$,
one can obtain the bound better than (\ref{moij2}), 
optimising (\ref{moij}) as a sum of two related via
Lemma~\ref{IJ} terms.


\begin{teo}\label{betamo}
{\sl Let $\psi$ be of the second kind.  
Let $p>0, \ T: L_N/4 \le T <L_{N+1}/4$. The uniform in $T$ estimate holds:
\be\label{momain}
\la |X|^p_{\psi} \ra (T) \ge C L_N^{-\alpha_N}
\left( L_N^p+ T^p L_N^{{p \over p+1}{\eta-1 \over \eta}} \right), 
\ee
where $\alpha_N \to 0$. 
In particular, 
\be\label{beta-}
\beta_\psi^-(p) \ge {(p+1) \over p+1/\eta}, \ \ \ 
\beta_\psi^+(p)\ge 1.
\ee
}
\end{teo}
Proof.  
The bound (\ref{moij}) of
Corollary~\ref{moments} and Lemma~\ref{IJ} imply
$$
\la |X|_{\psi}^p \ra (T) \ge C \left(
z^{-p} +L_N^{-\alpha_N} 
(L_N^{p+1}+T^{p+1}L_N^{{\eta-1 \over \eta}} )z \right), 
$$
where $z=I(\eps, \Delta)$. 
The function $f(z)=z^{-p}+K z, \ z>0$, is bounded from below by
$c(p)K^{p \over p+1}$. The bound (\ref{momain}) follows. To prove the second
statement, define
$s={p(1-\eta) \over (p+1) \eta}$. Considering 
$T: L_N/4 \le T \le L_N^{p+s \over p}$ and 
$T: L_{N+1}/4>T \ge L_N^{p+s \over p}$,
one can easily see from (\ref{momain}) that in both cases
$$
\la |X|_{\psi}^p \ra (T) \ge c L_N^{-\alpha_N} T^{p^2 \over p+s} \ge 
cT^{-\alpha_N+{p^2 \over p+s}}. 
$$
The first bound of (\ref{beta-}) follows. To see that $\beta^+(p)\ge 1$
for any $p>0$, it is sufficient to take the sequence $T_N=L_N$ in
(\ref{momain}). 
The proof is completed.
\par\noindent
Remark 1. A priori we don't have upper bounds for the moments. However,
if $\psi$ is such that ballistic upper bound holds, then
$\beta_\psi^+(p)=1$ for any $p$. 


\par\noindent
Remark 2.  
In somewhat paradoxal manner,
one can obtain better {\it lower} bounds for the moments,
if one has good {\it upper} bounds. This can be done in the following way. 
Assume that
$$
\la |X|_\psi^r \ra (T) \le h_r(T), \ \ r>0, 
$$
with some nontrivial $h_r(T)$ (that is, better than ballistic).
Then the bound (\ref{moij}) implies some nontrivial lower bound 
$J \ge A(r,\eps) $ and upper bound  $I \le B(r, \eps)$. 
 The result of
Lemma~\ref{IJ} yields $I \ge C A(r, \eps)$ and $J \le C B(r, \eps)$. 
These two bounds (with any values $r=r_1$ and $r=r_2$ respectively)
can be inserted into (\ref{moij}). Finally, one
can optimise the obtained bound (for a given $p>0$), 
choosing appropriate values of $r_1,r_2$. Most probably, one
should take $r_1$ small and $r_2$ large. 

\par
The methods developed in this section, as it was mentioned in Introduction,
can be applied to more general models. For example, one can prove the
following statement. It can be applied, in particular, to the operators f
with bounded sparse potentials considered in \cite{Z}, \cite{GKT} and gives
a better result. 
\begin{teo}\label{power}
{\sl
Let $H$ be any operator in $l^2(\Z^+)$ such that the corresponding
transfer matrix verifies the condition:
$$
||T(n,0; E+i \e)|| \le C(\delta) n^\alpha, \ \ \alpha>0,
$$
for any $E \in [-2+\delta, 2-\delta], \ \e \in (0,1)$
and $n$ such that $n \e \le K, \ K>0$. 
Let $\psi $ be of the second kind (in particular, $\psi=\delta_1$).
For any $T$ the bounds hold:
\be\label{kwa1}
P_\psi (n \ge T, T)\ge T^{1-2 \alpha} (I(1/T, \Delta) +1/T)
\ge CT^{-2 \alpha},   
\ee
\be\label{kwa2}
\la |X|_\psi^p \ra (T) \ge C I^{-p} (1/T, \Delta)+
T^{p+1-2 \alpha} I(1/T, \Delta) \ge C(p) T^{p-2p\alpha/(p+1)}.
\ee
Thus,
$$
\beta_\psi^-(p) \ge 1- {2 \alpha \over p+1}
$$
(this bound is nontrivial only beginning from $p>2 \alpha-1$). 
}
\end{teo}
Proof. The bound (\ref{kwa1}) is obtained following the proof
of Theorem~\ref{m2}. The first inequality in (\ref{kwa2}) follows from
the proof of Corollary~\ref{moments}, and the second from the proof of
Theorem~\ref{betamo}. 



\vskip 2 cm
\begin{center}
\section{Dynamical upper bounds}
\end{center}
\setcounter{equation}{0}
\vskip 1 cm
\par
In this section we shall establish some upper bounds for the inside
and outside probabilities and the moments. 
 It is clear that one cannot consider the same
class of initial states $\psi$ as in the previous section. The problem is
that we do not have dynamical control of the possible pure point
spectrum outside $(-2,2)$.
Thus, we shall consider only $\psi=f(H)\delta_1$ such that
${\rm supp} f \subset (-2,2)$. Moreover, to control the decay at
infinity (when considering outside probabilities), 
we shall assume that the function $f$ is infinitely smooth
(recall that we call these $\psi$ initial states of the first kind). 
\par
We begin with the inside probabilities. 
Let $\psi=f(H)\delta_1$, where $f$ is a bounded Borel function
such that ${\rm supp} f \subset \Delta=[-2+\nu,2-\nu]$ for some $\nu>0$.
Following the proof of Lemma~\ref{m1}, one can show that
for any $K,M>0$,
\be\label{zo1}
\sum_{n=K}^{n=K+M} \la |\psi (t,n)|^2 \ra (T) 
 \le C \sqrt{M J(\eps, \Delta)} \le
C \sqrt{{M L_N^{\alpha_N} \over L_N+TL_N^{\eta -1 \over \eta}}}. 
\ee
In fact, a slightly better result can be obtained
using the upper bound for the imaginary part of the Borel
transform of spectral measure.
Such a bound represent an independent interest since
it provides an upper bound for the measure of intervals and thus
a lower bound for Hausdorff and packing dimensions
of the spectral measure. 

\begin{lem}\label{meup}
{\sl
Let $\mu$ be the spectral measure of the state $\psi=\delta_1$
and $F(z)$ its Borel transform.  For
any $\nu \in (0,1)$ there exists constant $C(\nu)$ such that
for all $x \in [-2+\nu,2-\nu]$ and
$\e: {4 \over L_{N+1}} < \e \le {4 \over L_N}$ the bound holds:
\be\label{s01}
{1 \over 2\e} \mu([x-\e, x+\e]) \le 
{\rm Im} F(x+i\eps) \le
C(\nu)  L_N^{\alpha_N}
\left( \e L_N+L_N^{\eta-1 \over \eta} \right)^{-1},
\ee
where $\alpha_N \to 0$.
}
\end{lem}
Proof. It is well known that
$$
{\rm Im} F(z)={\rm Im} z ||R(z)\delta_1||^2={\rm Im} z
\sum_{n=1}^{\infty} |u(n,z)|^2,
$$
where $F(z)$ is the Borel transform of $\mu$.
The first inequality in (\ref{p09}) implies
\be\label{s02}
{\rm Im} F(z) \ge c {\rm Im} z ({\rm Im}^2 F(z)+1)
\sum_{n=1}^{\infty} ||T(n,0,z)||^{-2}.
\ee
Let $x \in [-2+\nu,2-\nu], \ \e \in (4/L_{N+1}, 4/L_N]$,
$z=x+i \e$. We can summate over $n: 1 \le n <L_N$ and over
$n: L_N \le n \le K/\e$ with suitable $K$ ($K=8$ for
$1/(2L_N) \le \e \le 4/L_N$ 
and $K=1$ for $\e<1/(2L_N)$, for example)
using the upper bounds for $||T||$ of
Lemma~\ref{transfer}. 
Thus, we obtain from (\ref{s02}):
$$
{\rm Im} F(z) \ge C(\nu) \e {\rm Im}^2 F(z) L_N^{-2 \alpha_N}
(L_N+\e^{-1} L_N^{\eta-1 \over \eta}).
$$
Since ${\rm Im} F(x+i\e) \ge 1/(2\e) \mu([x-\e, x+\e])$, the result
follows. 
\par\noindent
Remark.
The proof is rather simple because we have from the very beginning the
upper bound for $||T(n,0,z)||$ for complex $z$. In most applications,
however, one has such bounds only for real $z$, and one should proceed in a
more complicated way using the Jitomirskaya-Last method \cite{JL}. 
 \par
As a first direct consequence of this result, one can obtain the
already known upper bounds (\ref{ij1}) on $I(\e, \Delta), J(\e, \Delta)$. 
Indeed, for $\Delta=[-2+\nu, 2-\nu]$, 
$$
J(\e, \Delta)=\int_\Delta  d \mu (x) b(x,T),
$$
where
\be\label{sq01}
b(x,T)=\e {\rm Im} F(x+i\e)\le
C(\nu) \e L_N^{\alpha_N}
\left( \e L_N+L_N^{\eta-1 \over \eta} \right)^{-1}
\ee
due to (\ref{s01}). The bound for $J(\e, \Delta)$ follows. 
Next, 
$$
I(\e, \Delta)=\eps \int_\Delta dE {\rm Im}^2 F(E+i\e) =
\int_\Delta dE b(E,T) {\rm Im} F(x+i\e)  
$$
The bound (\ref{sq01}) and
$$
\int_\R d E {\rm Im} F(x+i\e)=\mu (\R)=1
$$
imply the bound for $I(\e, \Delta)$. 
\par
Before stating the next corollary, let us recall the definition
of the lower and upper Hausdorff dimension of Borel measure:
$$
{\rm dim}_* (\mu)={\rm inf} \{ {\rm dim} (S) \ | \ \mu(S)>0 \}, 
$$
$$
{\rm dim}^* (\mu)={\rm inf} \{ {\rm dim} (S) \ | \ \mu(S)=\mu(\R) \}, 
$$
where ${\rm dim} (S)$ denotes Hausdorff dimension of the set $S$.
Thus, the measure gives zero weight to any set $S$ with
${\rm dim} (S)<{\rm dim}_*(\mu)$ and for any $\e>0$ is supported by 
some set $S$ with ${\rm dim} (S)<{\rm dim}^* (\mu)+\e$.  
The measure is of exact Hausdorff dimension if
${\rm dim}_* (\mu)={\rm dim}^*(\mu)$. It is known (see \cite{T} for
the referencies) that 
\be\label{ha}
{\rm dim}_* (\mu)=\mu - {\rm essinf} \gamma^-(x)=
{\rm sup} \{ \alpha \ | \ \gamma^-(x) \ge \alpha \ \mu-a.s. \}, 
\ee
\be\label{ha2}
{\rm dim}^* (\mu)=\mu - {\rm esssup} \gamma^-(x)=
{\rm inf} \{ \alpha \ | \ \gamma^-(x) \le \alpha \ \mu-a.s. \}.  
\ee
Here $\gamma^- (x)$ is the lower local exponent of $\mu$:
$$
\gamma^-(x)=\liminf _{\e \to 0}  { {\rm log} \mu ([x-\e,x+\e])
\over {\rm log} \e }. 
$$
For the packing dimension similar formulae hold (see \cite{T} for
details). 



\begin{cor}\label{dimh}
{\sl
1. For any $\delta \in (0,1), \ \nu>0$ the spectral measure
$\mu$ of the state
$\psi=\delta_1$ is uniformly $\eta-\delta$-H\"older continuous on
$[-2+\nu,2-\nu]$. In particular, for $\mu'$, the restriction of $\mu$
on $(-2,2)$,  
${\rm dim}_*(\mu') \ge \eta$.
\par\noindent
2. The packing dimension of $\mu$ is 1.

}
\end{cor}
Proof. Let $\e \in (4/L_{N+1}, 4/L_N]$ for some $N$.
One can easily see that
$$
\e L_N+L_N^{\eta-1 \over \eta} \ge \e^{1-\eta}.
$$
Therefore, Lemma~\ref{meup} implies
$$
\mu ([x-\e,x+\e]) \le C(\delta)\e^{\eta}L_N^{\alpha_N} \le
C_1(\delta) \e^{\eta-\alpha_N}
$$
for any $x \in [-2+\nu,2-\nu]$.
Since $\lim \alpha_N=0$, the uniform $\eta-\delta$-continuity of
$\mu$ restricted to $[-2+\nu,2-\nu]$ follows.
As a particular consequence, $\gamma^-(x) \ge \eta$ for all
$x \in (-2,2)$. The equality (\ref{ha}) implies
${\rm dim}_* (\mu')\ge \eta$. 

\par
Taking $\e_N=1/L_N$, we obtain from Lemma~\ref{meup} that
$$
\mu ([x-\e_N,x+\e_N]) \le C(\delta) \e_N^{1-\alpha_N}.
$$
Therefore, for the upper local exponents of the measure we have
$$
\gamma^+(x) \equiv {\rm limsup}_{\e \to 0}
{{\rm log} \mu([x-\e,x+\e]) \over {\rm log} \e} \ge 1.
$$
The fact that ${\rm dim}_P(\mu)=1$ follows \cite{T}. 
The proof is completed.
\par\noindent
Remark. These results are not new. The fact that
${\rm dim}_*(\mu') \ge \eta$ is proved in \cite{JL} and
${\rm dim}_P(\mu)=1$ in \cite{CM}. Our proof, however, is more
simple. Moreover, the upper bound (\ref{s01}) contains more information. 
\par
\begin{lem}\label{upsingle}

{\sl  Let $\psi=f(H)\delta_1$, where $f$ is bounded Borel function
such that ${\rm supp} f \subset \Delta=[-2+\nu,2-\nu]$ for some $\nu>0$.
Let $T: \ L_N/4 \le T <L_{N+1}/4$ for some $N$. 
\par\noindent
1. For any $n$ the bound holds:
\be\label{upsi1}
\la |\psi (t,n)|^2 \ra (T) \le C {L_N^{\alpha_N}
\over L_N+TL_N^{\eta -1 \over \eta} }. 
\ee
\par\noindent
2. Define $M(T)=L_N^{-\delta} (L_N+TL_N^{{\eta -1 \over \eta}})$
with some $\delta>0$. Then
$$
P_\psi (n \le M(T), T) \le C L_N^{-\delta/2}  
$$
for $T$ large enough and thus
$$
P_\psi (n \ge M(T), T) \ge ||\psi||^2-CL_N^{-\delta/2}. 
$$
\par\noindent
3. For the time-averaged return probability the bound holds:
\be\label{zret}
{1 \over T} \int_0^\infty dt \exp (-t/T)
|\la \psi(t), \psi \ra |^2 \le C {L_N^{\alpha_N}  \over
L_N+TL_N^{\eta -1 \over \eta} }. 
\ee

}
\end{lem}
Proof. Using spectral Theorem in a standard way (see \cite{T}, for
example), one first shows that
$$
2 \la |\psi (t,n) |^2 \ra (T)=
\int_\R \int _\R d \mu_\psi (x) d \mu_\psi (y) u_\psi (n,x)
{\overline {u_\psi (n,y)}} R(T(x-y)) \le 
$$
\be\label{qw1}
2 \int_\R d \mu_\psi (x) |u_\psi (n,x)|^2 b_\psi (x,T),  
\ee
where
$$
b_\psi (x,T)=\int_\R d \mu_\psi (u) R(T(x-u)) =\e {\rm Im}
F_{\mu_\psi} (x+i\e), \ \e=1/T. 
$$
Since $f$ is bounded and  ${\rm supp} f \subset \Delta$, 
we get
$$
\int_\R d \mu_\psi (x) |u_\psi (n,x)|^2 b_\psi (x,T)
\le C \int_\Delta  d \mu_\psi (x) b(x,T)|u_\psi (n,x)|^2.
$$
The bound (\ref{s01})
and
$$
\int_\R d \mu_\psi (x) |u_\psi (n,x)|^2 \le 1
$$
yield (\ref{upsi1}). The second statement of Lemma directly follows. 
For the return probabilities the result follows from the bound
$$
J_\psi (\eps, \R) \le C J(\eps, \Delta)  
$$
and the established upper bound for $J(\e, \Delta)$
(Corollary~\ref{twoint}).
The proof is completed. 
\par
The situation is more difficult with the upper bounds for outside
probabilities. 
We shall consider the initial state $\psi$ of the form
$\psi=f(H) \delta_1$, where $f \in C_0^{\infty}([-2+\nu,2-\nu])$
with some $\nu \in (0,1/2)$. For smooth $f$ it is well known that 
the function $\psi(n)$
decays at infinity faster than any inverse power and moreover,
for the moments of the time-averaged position operator, the ballistic
upper bound holds:
\be\label{balq1}
\la |X|_\psi^p \ra (T) \le C(p)T^p, \ \ p>0. 
\ee

The following statement holds (where we use some ideas of \cite{CM} in the
proof). 
\begin{teo}\label{upprob}
{\sl Consider $\psi$ of the first kind. 
Let $T$ be such that $L_N/4 \le T \le L_N^{1/\eta}$ for some $N$.
\par\noindent
1. For any $p \ge 0$ the following bound holds
\be\label{zop1}
\sum_{n \ge 2 L_N} n^p \la |\psi(t,n)|^2 \ra (T) \le C(p)
T^{p+1} L_N^{-1/\eta} 
\ee
\par\noindent
In particular, 
\be\label{z1}
P_\psi (n \ge 2 L_N, T) \le C T L_N^{-1/\eta}
\ee
and 
\be\label{z2}
\la |X|_\psi^p \ra (T) \le C L_N^p + CT^{p+1}L_N^{-1/\eta}. 
\ee
\par\noindent
2. Let $T: L_N/4 \le L_{N+1}^{1-\delta}$ with some $\delta>0$.
For $M>2L_N$ and any $A>0$ the uniform bound holds:
\be\label{z001}
P_\psi ( 2 L_N \le n \le M, T) \le
C {M \over T+L_N^{1/\eta}}+ {C_A \over L_N^A}.
\ee
}
\end{teo}
Proof. First of all, observe that the ballistic upper bound (\ref{balq1})
implies
$$
\la |\psi (t,n)|^2 \ra (T) \le C(r) T^r n^{-r} 
$$
for any $r>0$. Therefore, taking $r$ large enough, we obtain
$$
\sum_{n \ge T^2} n^p  \la |\psi (t,n)|^2 \ra (T) \le
C(r,p) T^{2p+2-r} \le C(p,A) T^{-A}
$$
for any $A>0$. Thus, to prove (\ref{zop1}), it is sufficient
to consider the sum over $n: \ 2 L_N \le n \le T^2$. 
We use again the Parseval formula:
\be\label{z3}
\la |\psi (t,n)|^2 \ra (T)={\e \over \pi} \int_{\R} dE
|(R(E+i \e)f(H) \delta_1) (n)|^2, \ \ \e={1 \over 2T}.
\ee
Define $\Delta=[-2+\nu/2, 2-\nu/2]$, where
$f \in C_0^\infty ([-2+\nu, 2-\nu])$. 
We shall denote by $a_1(n,T)$ the integral over $\R \setminus \Delta$
in (\ref{z3}), and by $a_2(n,T)$ the integral over $\Delta$.
Since $f(x)=0, \ |x| \ge 2-\nu$, one can show, as in the proof
of the part b) of Theorem~\ref{m2} (bounds (\ref{gk2})-(\ref{gk4})),  
 that
$$
|R(E+i\eps) f(H) \delta_1(n)| \le {C(k, \nu) \over E(1+|n|^2)^{k/2}}
$$
for
any integer $k>0$ and all
$E \in \R \setminus \Delta$ with uniform in $n,E, \eps$ constants.
Therefore, 
\be\label{z6}
a_1(n,T) \le {C(k,\nu) \over T}  (1+|n|^2)^{-k}
\ee
for any $k>0$. In particular, taking $k$ large enough, we obtain 
\be\label{z61}
\sum_{n \ge 2 L_N} n^p a_1(n,T) \le C(p,A) L_N^{-A}
\ee
for any $A>0$. 
\par
Consider now the term $a_2(n,T)$. Since $R(z)f(H)=f(H)R(z)$, 
one can write it as follows:
\be\label{z7}
a_2(n,T)={\e \over \pi} \int_\Delta dE |(f(H)R(E+i \e) \delta_1)(n)|^2.
\ee
Since $f \in C_0^{\infty} ([-2,2])$, it follows again from 
results of \cite{GK} that for any $\chi \in l^2(\N)$
$$
|(f(H) \chi)(n)|^2 \le C(k) \sum_m (1+|n-m|^2)^{-k} |\chi (m)|^2
$$
Inserting this bound in (\ref{z7}) yields after integration:
\be\label{z8}
a_2(n,T) \le C(k) \e \sum_m (1+|n-m|^2)^{-k} 
\int_\Delta dE |(R(E+i \e)\delta_1)(m)|^2
\ee
for any  $k>0$. 
Denote by $a_{21} (n,T)$ the sum in (\ref{z8}) over $m: m \le L_N$, 
by $a_{22} (n,T)$ the sum over $m: \ L_N < m \le T^2+L_N$
and by $a_{23}(n,T)$ the sum over $m: \ m >T^2+L_N$.  
It is clear that
for any $A>0$, 
\be\label{z9}
\sum_{2 L_N \le n \le T^2} n^p (a_{21} (n,T)+a_{23} (n,T))
 \le C(p,A) L_N^{-A} \e \sum_m \int_\Delta
|(R(E+i \e)\delta_1)(m)|^2. 
\ee
The fact that 
$$
{\e \over \pi} \sum_m \int_{\R} dE |(R(E+i\e)\delta_1)(m)|^2=
||\delta_1||^2=1
$$
and (\ref{z9}) yield
\be\label{z10}
\sum_{2 L_N \le n \le T^2} n^p (
a_{21} (n,T) +a_{23} (n,T)) \le C(p,A) L_N^{-A}.
\ee
The summation over $n$ in the expression of $a_{22} (n,T)$ yields
\be\label{z11}
\sum_{2 L_N \le n \le T^2} n^p a_{22} (n,T) \le C \e
\sum_{L_N < m \le T^2+L_N} m^p  \int_\Delta
dE |(R(E+i\e)\delta_1)(m)|^2.
\ee
To  bound from above the r.h.s. of (\ref{z11}), we shall introduce
in $l^2(\N)$ operator
$$
H_N=H_0+V_N, \ \ V_N(n)=F(n \le L_N)V(n) 
$$
with compactly supported potential
and thus  absolutely continuous spectrum on $(-2,2)$.
We denote by $R(z)$ and $R_N(z)$ the resolvents of $H$ and $H_N$
respectively.
One  can see that for $N$ large enough (so that $Q(n)$ disappear), 
\be\label{z110}
(H-H_N)\phi(n)=\sum_{k=N+1}^{\infty} \delta_{L_k} (n) V(L_k) \phi(L_k).
\ee
The resolvent  equation implies that for any complex $z=E+i\e$,
\be\label{z111}
||R(z)\delta_1 -R_N(z)\delta_1 || \le
{1 \over \e} ||(H-H_N)R_N(z)\delta_1||.
\ee
To bound from above the r.h.s. of (\ref{z11}) and the r.h.s. of (\ref{z111}), 
we need to control $g(n)=R_N(z)\delta_1 (n)$ for $n>L_N$. 
In fact, a rather explicit expression can be obtained.
Since $(H_N-z)g=\delta_1$ and $V_N(n)=0$ for  $n>L_N$,  
$$
g(n-1)+g(n+1)-z g(n)=0, \ \ n>L_N. 
$$
Thus,
\be\label{z501}
(g(n+1), g(n))^T= T_0(n-L_N, z) (g(L_N+1), g(L_N))^T, \ \ n \ge L_N,  
\ee
where $T_0(m,z)=A_0(z)^m$ is the free transfer matrix with
$A_0(z)$ given by (\ref{a0z}).
Since $E \in \Delta=[-2+\nu/2, 2-\nu/2]$,
the matrix $A_0(z)$ has two complex eigenvalues
$$
\lambda_{1,2}={1 \over 2} (z \pm \sqrt{z^2-4}) 
$$
with corresponding eigenvectors $e_{i}=(\lambda_i, 1)^T, \ i=1,2$.
It follows from (\ref{z501})
that
$$
(g(n+1), g(n))^T= C_1 \lambda_1^{n-L_N} e_1+C_2 \lambda_2^{n-L_N} e_2,
\ \ n \ge L_N, 
$$
with some complex $C_1,C_2$. Since ${\rm Im} z=\eps >0$, 
one of the two eigenvalues, say, $\lambda_1$,
is such that $|\lambda_1|<1$ and then $|\lambda_2|>1$. On the other hand,
since $g=R_N(z)\delta_1$, it should be square integrable in $n$. Therefore,
$C_2=0$ and
$$
(g(n+1), g(n))^T=C \lambda_1^{n-L_N} (\lambda_1, 1)^T. 
$$
Finally, we obtain that
\be\label{z115}
g(n) \equiv (R_N(z)\delta_1)(n)=\lambda_1^{n-L_N} g(L_N)
\ee
for any $n \ge L_N$. 
One can see from the expression of $\lambda_1$ that
\be\label{z116}
\exp(-c_1 \e) \le |\lambda_1| \le \exp (-c \e)
\ee
with uniform $c_1, c>0$ for all $E \in \Delta, \ \e \in (0,1)$. 
\par
Let us return to the resolvent $R(z)$. Using the trivial bound
$|g(L_N)| \le 1/\e$, one gets from (\ref{z110})-(\ref{z111}) and 
(\ref{z115})-(\ref{z116}):
\be\label{117}
||R(z)\delta_1-R_N(z)\delta_1||^2 \le \e^{-4}
\sum_{k=N+1}^{\infty} V^2(L_k) \exp (-2c \e(L_k-L_N).
\ee
Since $T={1 \over 2\e} \le L_{N+1}^{1-\delta}$ in all three statements
of the Theorem,  
 \ $V(L_k)=L_k^{(1-\eta)/2 \eta}$,
and $L_k$ is a very fast
growing sequence, it is easy to check that
\be\label{z1180}
||R(z)\delta_1-R_N(z)\delta_1||^2 \le C \exp (- 1/\e^\alpha)
\ee
with some $\alpha>0$ for all $E \in \Delta, \ 
\e \in [L_{N+1}^{\delta-1}, 4 L_N^{-1}]$.
Thus, the bounds (\ref{z11}) and (\ref{z1180}) imply
\be\label{z118}
\sum_{2 L_N \le n \le T^2} n^p  a_{22}(n,T)
\le C/ \e^{2p} \exp (-1 /\e^\alpha)+
C \e \sum_{L_N\le m \le T^2+L_N} m^p
\int_\Delta dE |R_N(E+i\e)\delta_1 (m)|^2.
\ee
It follows from (\ref{z115})-(\ref{z116}) and $\e \le 2/L_N$ that
\be\label{z119}
\sum_{L_N\le m \le T^2+L_N} m^p  |R_N(E+i\e)\delta_1(m)|^2 
\le C \e^{-p-1} |R_N(E+i\e)\delta_1(L_N)|^2.
\ee
To bound $R_N(E+i\e)\delta_1(L_N)$, one can use the result of Lemma 4
in \cite{CM}. For the sake of completeness we shall give here a simple
and slightly different proof. Namely, we shall show that
\be\label{z13}
|(R_N(E+i\e) \delta_1)(L_N)|^2 \le C(\Delta)
{1 \over 1+\e L_N^{1/\eta}} 
{\rm Im} F_N (E+i\e), 
\ee
where $E \in \Delta$ and $F_N$ denotes the Borel
transform of the spectral measure
associated to the state $\delta_1$ and operator $H_N$.
First, it follows from (\ref{z115})-(\ref{z116}) that
$$
{1 \over \e} {\rm Im} F_N(E+i\e)=||R_N(E+i\e)\delta_1||^2 \ge
\sum_{m>L_N} |g(m)|^2 \ge {C \over \e} |g(L_N)|^2.
$$
Therefore,
\be\label{z1300}
|g(L_N)|^2 \le C {\rm Im} F_N(E+i\e). 
\ee
Let $L_{N-1}<n<L_{N+1}$. The definition of $g=R_N(z)\delta_1$ implies
\be\label{z131}
g(n+1)+g(n-1)-z g(n)=0, \ \ n \ne L_N, 
\ee
\be\label{z13100}
g(L_N+1)+g(L_N-1)+(V(L_N)-z) g(L_N)=0.
\ee
It is clear that for $n>L_N$,  
\be\label{z132}
(g(n+1), g(n))^T =T_0(n-L_N,z) (g(L_N+1),g(L_N))^T,
\ee
and for $n<L_N-1$
\be\label{z133}
(g(n+1),g(n))^T=T_0(n-L_N+1,z) (g(L_N), g(L_N-1))^T, 
\ee
where $T_0(m,z)$ is the free transfer matrix. Since $z=E+i\e, E \in \Delta$, 
its norm is uniformly bounded for $|m| \le K/\e$. Using the fact that
$||T^{-1}||=||T||$ and $\e \le 2/L_N$,  we thus get that 
for $2L_N>n>L_N$, 
$$
|g(n+1)|^2+|g(n)|^2 \ge c (|g(L_N+1)|^2+|g(L_N)|^2).
$$
with uniform $c>0$. Summating this bound, one obtains 
\be\label{z134}
c L_N (|g(L_N+1)|^2+|g(L_N)|^2) \le 2 ||g||^2=2/\e {\rm Im} F_N(E+i\e).
\ee
Similarly, summation over $L_{N}/2<n<L_N$ yields (since $L_N>2L_{N-1}$):
\be\label{z135}
c/2 L_N(|g(L_N)|^2+|g(L_N-1)|^2) \le 2/\e {\rm Im} F_N(E+i\e).
\ee
Thus, (\ref{z134})-(\ref{z135}) yield
\be\label{z136}
|g(L_N-1)|^2+|g(L_N+1)|^2 \le {C \over \e L_N} {\rm Im} F_N(E+i\e).
\ee
It follows from (\ref{z13100}) that
$$
|V(L_N)-z|^2 |g(L_N)|^2 \le {C \over \e L_N} {\rm Im} F_N(E+i\e).
$$
Since $|z| \le 3$ and $V(L_N)=L_N^{(1-\eta)/2 \eta}$, 
we obtain
\be\label{z1360}
|g(L_N)|^2 \le C(\Delta)\e^{-1} L_N^{-1/\eta} {\rm Im} F_N(E+i\e). 
\ee
The bound (\ref{z13}) follows from (\ref{z1300}) and (\ref{z1360}). 

\par
We can finish now the proof of the first part of the Theorem.

It follows from (\ref{z118}),(\ref{z119}) and (\ref{z13})
that
$$
\sum_{2 L_N\le n \le T^2} n^p a_{22} (n,T)
\le C/\e^{2p}\exp (-1/\e^\alpha)+
$$
\be\label{z14}
C \e^{-p-1} L_N^{-1/\eta}
\int_\Delta dE {\rm Im} F_N (E+i\e) \le C \e^{-p-1} L_N^{-1/\eta}, 
\ee
since $\e \le 2 L_N^{-1}$ and
$$
\int_\Delta dE {\rm Im} F_N(E+i\e) \le \int_{\R} dE {\rm Im} F_N(E+i\e)=
\pi \mu_N (\R)=\pi.
$$


The bound (\ref{zop1}) of 
Theorem follows from Parseval equality, (\ref{z61}), (\ref{z10})
(one takes $A=1/\eta$)
and (\ref{z14}).   Taking $p=0$, we obtain the bound for outside
probabilities. Since
$$
\la |X|_\psi^p \la (T) \le (2L_N)^p ||\psi||^2+
\sum_{n \ge 2 L_N} n^p \la |\psi (t,n)|^2 \ra (T), 
$$
the upper bound for the moments follows. 

\par
The proof of the second statement is similar. One defines $a_1(n,T)$ and
$a_2(n,T)$ in the same manner. The bound (\ref{z61}) yields
$$
\sum_{2 L_N \le n \le M} a_1(n,T) \le C_A L_N^{-A}. 
$$
Next, one denotes as $a_{21}(n,T), a_{22}(n,T)$
and $a_{23}(n,T)$ the sums in (\ref{z8})
over $m \le L_N, \ m: L_N<m<2M$ and $m: m \ge 2M$ respectively.
The bound (\ref{z10}) yields
$$
\sum_{2 L_N \le n \le M} a_{21} (n,T) \le C_A L_N^{-A}. 
$$
Similarly, one shows that 
$$
\sum_{2 L_N \le n \le M} a_{23} (n,T) \le C_A L_N^{-A}. 
$$
Thus, it is sufficient to bound from above the r.h.s. of 
$$
\sum_{2L_N \le n \le M} a_{22} (n,T)\le C \e
\sum_{L_N <m<2M} \int_{\Delta} dE |(R(E+i\e) \delta_1)(m)|^2. 
$$
The same consideration as in the proof of the part 1 yields
\be\label{z402}
\sum_{2 L_N \le n \le M} a_{22} (n,T) \le C \exp (-1/\e^\alpha)+
C \e \sum_{L_N <m<2M} \int_\Delta dE |(R_N(E+i\e)\delta_1)(m)|^2,  
\ee
where
\be\label{z401}
(R_N (E+i\e) \delta_1)(m)=g(L_N) \lambda_1^{m-L_N}, \ m \ge L_N. 
\ee
The bounds (\ref{z13}) and (\ref{z401}) imply
$$
\e \sum_{L_N <m<2M} |(R_N(E+i\e)\delta_1)(m)|^2 \le C
{M \e \over 1+\e L_N^{1/\eta}} 
{\rm Im} F_N(E+i\e). 
$$
Inserting this bound in (\ref{z402}), we obtain
$$
\sum_{2L_N \le n \le M} a_{22}(n,T)\le C {M \e \over 1+\e L_N^{1/\eta}}.  
$$
The bound (\ref{z001}) follows. 



\vskip 2 cm
\begin{center}
\section{Improved lower bounds}
\end{center}
\setcounter{equation}{0}
\vskip 1 cm
\par

The result of Theorem~\ref{upprob} allows us to have total control
of the integrals $I,J$ (up to factor like $L_N^{\alpha_N}$).  
\begin{cor}\label{lowI}
{\sl Let $\Delta$ be nonempty interval such that
$\Delta \subset [-2+\nu, 2-\nu]$ for
some $\nu>0$. There exist positive constants uniform in $\e$
such that
for all $\eps:  4 L_{N+1}^{-1} < \eps \le 4 L_N^{-1}$,  
\be\label{stable}
{C \eps \over \eps L_N+L_N^{\eta -1 \over \eta}}
\le J(\eps,\Delta) \le C I(\eps,\Delta) \le
 {C L_N^{\alpha_N} \eps \over \eps L_N+L_N^{\eta -1 \over \eta}},  
\ee
where $\alpha_N \to 0$. 
}
\end{cor}
Proof.
Pick a slightly smaller interval $\Delta'\subset \Delta$.
Let $f$ be a function
from $C_0^\infty (\Delta)$ such that $0 \le f(x) \le 1$ and
$f(x)=1$ on $\Delta'$. Define
$\psi=f(H)\delta_1$.
The result of
Lemma~\ref{m1} applied to interval $\Delta'$ yields 
\be\label{z601}
P_\psi (n \ge M_1(T), T) \ge D_1 c_1>0, 
\ee
where $c_1=\mu_\psi(\Delta')=\mu(\Delta')>0$, and
\be\label{z6002}
M_1(T)=c_1^2/(16 J_\psi (\eps, \Delta')), \ \ \eps=1/T. 
\ee
On the other hand, Theorem~\ref{upprob} implies
\be\label{z603}
P_\psi (n \ge 2 L_N, T) \le D_2 TL_N^{-1/\eta}, \ \ T \le L_N^{1/\eta}.  
\ee
If $T \le {D_1 c_1 \over 2 D_2} L_N^{1/\eta}\equiv \gamma L_N^{1/\eta}$, then
(\ref{z601}), (\ref{z603}) imply
$P_\psi ( n \ge 2L_N, T)< P_\psi (n \ge M_1(T),T)$ and
thus $M_1(T) < 2 L_N$. It follows from (\ref{z6002}) that
\be\label{z951}
{C \over L_N} \le J_\psi (\eps, \Delta') \le J_\psi (\eps, \R) 
\ee
for $\eps \in [\e_0, 4L_N^{-1}]$, where
$\eps_0=(\gamma)^{-1} L_N^{-1/\eta}$. 
Recall the equality
\be\label{w02}
J_\psi (\eps, \R)=
\eps \int_0^\infty dt \exp(- \eps t) |\la \psi (t), \psi \ra |^2. 
\ee
The crucial observation is that $J_\psi (\eps, \R)/\eps$ is decreasing
in $\eps$. 
Therefore, it follows from (\ref{z951})-(\ref{w02}) that
\be\label{z952}
J_\psi (\eps, \R) \ge \eps/ \eps_0 J_\psi (\eps_0, \R) \ge
 C \eps L_N^{1-\eta \over \eta}
\ee
for all $\eps \le \eps_0$.   The bounds (\ref{z951}), (\ref{z952})
imply that
$$
J_\psi (\eps, \R) \ge {C \eps \over \eps L_N + L_N^{\eta -1 \over \eta}} 
$$
for all $\eps \in (4L_{N+1}^{-1}, 4L_N^{-1}]$ with suitable constant.
Since $0 \le f(x) \le 1$ and
$f(x)=0$ for $x$ outside of $\Delta$, the definition of $J_\psi, J$
implies                                   
$$
J_\psi(\eps, \R) \le \int_\Delta d \mu (x)\int_\Delta d \mu (y)
R((x-y)/\eps) \le J(\eps, \Delta).
$$
Thus, we get
\be\label{wau}
J(\e, \Delta) \ge \int_\Delta d \mu (x) \int_\Delta d \mu (y) R((x-y)/\e) 
\ge {C \e \over \e L_N +L_N^{\eta -1 \over \eta}}.
\ee
The first inequality
in (\ref{stable}) follows.  
The second and the third inequalities
follow from Lemma~\ref{IJ} and Corollary~\ref{twoint}. 
\par
As a direct consequence of this result, one gets lower bound for
the time-averaged return probabilities $J_\psi (1/T, \R)$. In fact, if
the measure $\mu_\psi$ has a nontrivial point part:
$\mu_\psi(\{ E_0 \} )=\gamma >0$ for some $E_0$, then clearly
$J_\psi (\eps, \R) \ge \gamma^2 >0$ for any $\eps$. The situation is more
interesting if the measure is continuous, in our case if
${\rm supp} \mu_\psi \subset (-2,2)$. 
\begin{teo}\label{retpro}
{\sl
Assume that $\psi=f(H)\delta_1$, where $f$ is a bounded Borel function
\par\noindent
a) supported on $[-2+\nu, 2-\nu]$ for some $\nu>0$
\par\noindent
b) such that
$|f(x)| \ge c>0$ on some interval $\Delta \subset [-2+\nu, 2-\nu]$.
Then 
$$
{C \eps \over \eps L_N+L_N^{\eta -1 \over \eta}} \le J_\psi (1/T, \R) \le
{C \eps L_N^{\alpha_N} \over  \eps L_N+L_N^{\eta -1 \over \eta}}, \
\e=1/T, 
$$
for $T: \ L_N/4 \le T <L_{N+1}/4$.
If only the condition b) is fulfilled, then only the lower
bound for $J_\psi (\e, \R)$ holds. 
}
\end{teo}
Proof. The upper bound is proved in Lemma~\ref{upsingle}. Since
$$
J_\psi (\eps, \R) \ge c^4 \int_\Delta d \mu (x) \int_\Delta d \mu (y)
R((x-y)/\eps),
$$
the second inequality in (\ref{wau}) yields lower bound. 
The proof is completed. 
\par
One observes that the integral
$$
T J_\psi (1/T, \R)=\int_0^\infty dt \exp(-t/T)
| \la \psi (t), \psi \ra |^2
$$
grows linearly for $T: \ L_N \le T \le L_N^{1/\eta}$
and remains stable for $T: \ L_N^{1/\eta} <T<L_{N+1}$ (up to factors
like $CL_N^{\alpha_N}$).
Since the main contribution to the integral comes
from interval $[0,T]$, 
one can conjecture that the return probability
$R_\psi (t)=|\la \psi(t), \psi \ra |^2=|{\hat {\mu}}_\psi (t)|^2  $ is
essentially constant of order $L_N^{-1}$ if $t \in [L_N, L_N^{1/\eta}]$, 
and is small (decaying at least as $1/t$) if
$t \in [L_N^{1/\eta+\delta}, L_{N+1}]$. 


\par
The obtained lower bounds for $I(\eps, \Delta)$ imply also improved
lower bounds for probabilities and moments.  


\par
\begin{teo}\label{probext}
{\sl
Let $\psi$ be of the second kind.  Then
\par\noindent
1. For $T: \  L_N \le T \le L_N^{1/\eta}$:
$$
P_\psi(n \ge T,T) \ge C TL_N^{-1/\eta-\alpha_N} 
$$
and for $T: \ L_N^{1/\eta} \le T <L_{N+1}/4$, 
$$
P_\psi (n \ge T,T) \ge CL_N^{-\alpha_N}. 
$$
\par\noindent
2. For $T: \ L_N/4 \le T \le L_N^{1/\eta}$,
$$
P_\psi (L_N/4 \le n \le L_N, T) \ge C L_N^{-\alpha_N}. 
$$
\par\noindent
As a consequence,
for $T: \ L_N/4 \le T \le L_N^{1/\eta}$, 
$$
\la |X|_\psi^p \ra (T) \ge C(L_N^p +T^{p+1} L_N^{-1/\eta})
L_N^{\alpha_N},
$$
and for $T: \ L_N^{1/\eta} \le T <L_{N+1}/4$, 
$$
\la |X|_\psi^p \ra (T) \ge C T^p L_N^{-\alpha_N}. 
$$
}                                                     
\end{teo}
The results for probabilities follow directly from Theorem~\ref{m2} and
Corollary~\ref{lowI}. The bound $\la |X|_\psi^p \ra (T) \ge
M^p P_\psi(n \ge M, T)$ for any $M$ yields the result for the moments. 
The proof is completed. 
\par
The result of the Theorem tells, in particular, that for
$T \ge L_N^{1/\eta}$, some (not too small) part of the wavepacket 
has gone through the barrier and moves ballistically:
\be\label{z955}
P_\psi (n \ge T,T)\ge C L_N^{-\alpha_N}\ge CT^{-\alpha_N}, \ \ \alpha_N \to 0. 
\ee
On the other hand, Corollary 1.6 implies that for
$T \ge L_N^{1/\eta+\delta}$ with some $\delta>0$, 
$$
P_\psi (n>L_N, T) \ge
P_\psi (n \ge T L_N^{{\eta -1 \over \eta}-\alpha_N}, T) \ge
c>0.
$$
Thus, some essential (and not small)
part of the wavepacket is on the right of
$L_N$. This part will
continue to move ballistically
up to the next barrier located at $n=L_{N+1}$. Therefore, one can
expect the bound like
$$
P_\psi (n \ge T,T) \ge c_1>0
$$
for $T>L_N^{1/\eta+\delta}$, which is better than just (\ref{z955}).  
The following statement confirm this conjecture. Slightly modifying the
proof, we show also that
$$
P_\psi (n \le 2 L_N, T) \ge c_2>0
$$
for $T \le \tau L_N^{1/\eta}$ with $\tau>0$ small enough. This
is better than $P_\psi (n \le 2 L_N, T)\ge C L_N^{-\alpha_N}$, which
follows from Theorem~\ref{probext}. 

\begin{teo}\label{ballistic}
{\sl The following statements hold:
\par\noindent
1. Assume that $\psi$ is of the second kind. 
For any $\delta>0$ there exist $\tau>0, c_1>0$
such that for $T: \ L_N^{1/\eta+\delta} \le T <L_{N+1}^{1-\delta}$
with $N$ large enough,
$$
P_\psi (n \ge \tau T, T) \ge c_1>0. 
$$
If $\psi$ is of the first kind, 
for any $\theta>0$ one can choose
$\tau$ so that 
$$
P_\psi (n \ge \tau T, T) \ge ||\psi||^2-\theta. 
$$
 \par\noindent
In both cases, for such $T$,
$$
\la |X|_\psi^p \ra (T) \ge C(p)T^p, \ \ p>0. 
$$
\par\noindent
2. Let $\psi$ be of the second kind. There exists $\tau>0$ small enough
such that
$$
P_\psi (n \le 2 L_N, T) \ge c_2>0
$$
for all $T: \ L_N/4 \le T \le \tau L_N^{1/\eta}$. If $\psi$ is of the
first kind, then better bound holds:
\be\label{zy34}
P_\psi (n \le 2 L_N, T) \ge ||\psi||^2-CTL_N^{-1/\eta}.
\ee

}
\end{teo}
Proof. Recall that $\psi=f(H) \delta_1$, where $f$ is a bounded
Borel function, 
$f \in C^\infty (S)$, $S=[E_0-\nu, E_0+\nu] \subset [-2+\nu,2-\nu]$
and $|f(x)| \ge c>0$ on $S$. Let $h$ be some function such that
$0 \le h(x) \le 1$, 
$h \in C_0^\infty([E_0-\gamma, E_0+\gamma])$
and $h(x)=1, \ x \in [E_0-\theta, E_0+\theta]$, where
$0<\theta<\gamma<\nu$. Define $g(x)=f(x)h(x)$. It is clear that
$g \in C_0^\infty ([E_0-\gamma, E_0+\gamma])$. Let
$$
\rho=g(H)\delta_1, \ \chi=\psi-\rho=(f(H)-g(H)) \delta_1. 
$$
As $|f(x)| \ge c>0$ on $S$, 
$$
\alpha \equiv ||\rho ||^2 \ge c^2 \mu ([E_0-\theta, E_0+\theta]) >0. 
$$
Since
$$
\la \rho, \chi \ra =\int d \mu (x) |f(x)|^2 h(x)(1-h(x)) 
$$
and $f$ bounded, choosing the parameter $\gamma$
in the definition of $h$ close enough
to $\theta$, one can ensure that
\be\label{zy1}
|\la \rho, \chi \ra \le  ||\rho||^2/4=\alpha/4.  
\ee


Let $\rho(t)=\exp(-itH)\rho, \ \chi(t)=\exp(-itH)\chi$
and $\psi(t)=\exp(-itH)\psi$. For any $n \in \N$,
$$
|\psi(t,n)|^2=|\rho(t,n)|^2+|\chi(t,n)|^2+
2 {\rm Re} (\rho(t,n) {\overline {\chi(t,n)}}). 
$$
Let $M>0$. Summation over $n \le M$ and time-averaging yield
for any $T>0$:
$$
\sum_{n \le M} \la |\psi(t,n)|^2 \ra (T) \le
$$
\be\label{zy2}
||\chi ||^2 +\sum_{n \le M} \la |\rho (t,n)|^2 \ra (T)
+ 2 ||\chi|| \left( \sum_{n \le M}
\la |\rho (t,n)|^2 \ra (T) \right)^{1/2}. 
\ee
We have used the fact that $||\chi (t)||=||\chi||$ and the
Cauchy-Schwarz inequality.  The condition (\ref{zy1}) implies
that $||\chi||^2 \le ||\psi||^2-\alpha/2$. Therefore,
(\ref{zy2}) yields
\be\label{zy33}
P_\psi (n \le M, T) \le ||\psi||^2 -\alpha/2+ P_\rho (n \le M,T)+
C (P_\rho (n \le M,T))^{1/2}
\ee
Thus, if $P_\rho (n \le M,T)\le \eta$,
where $\eta$ is small enough (depending on $\alpha$),
then $P_\psi (n \ge M,T) \ge \alpha/4>0$. 
\par
Let $M>2L_N$. To bound from above $P_\rho (n \le M,T)$, we shall write
$$
P_\rho (n \le M,T) = P_\rho (n \le 2 L_N,T)+P_\rho (2L_N<n \le M,T)
$$
Recall that $\rho=g(H)\psi$, where
$g \in C_0^\infty([E_0-\gamma,E_0+\gamma])$ and
$$
[E_0-\gamma,E_0+\gamma] \subset
[E_0-\nu,E_0+\nu] \subset [-2+\nu,2-\nu].
$$
Therefore, all upper bounds
of the previous section hold for $\rho$. 
Since
$T \ge L_N^{1/\eta+\delta}$,
the bound (\ref{upsi1}) of Lemma~\ref{upsingle} yields
\be\label{zy3}
P_\rho (n \le 2 L_N) \le C L_N^{\alpha_N-\delta} \le C L_N^{-\delta/2}
\ee
for $N$ large enough. On the other hand, the bound (\ref{z001}) of
Theorem~\ref{upprob}
implies
\be\label{zy4}
P_\rho (2 L_N \le n \le M, T) \le C {M \over T}+C_A L_N^{-A}
\ee
for any $A>0$. The bounds (\ref{zy3})-(\ref{zy4}) yield
$$
P_\rho (n \le M, T) \le C {M \over T} +\beta_N, \ \ \beta_N \to 0. 
$$
It is clear that taking $M=\tau T$ with $\tau>0$ small enough, 
for $N$ large enough we get $P_\rho (n \le M,T) \le \eta$ and thus
$P_\psi (n \ge M,T) \le \alpha/4>0$.
\par
In the case of $\psi$ of the first kind
the proof is more simple. One can directly estimate $P_\psi (n \le 2 L_N,T)$
and $P_\psi(2 L_N \le n \le M, T)$ as in (\ref{zy3}), (\ref{zy4}).
Taking $\tau$ small enough, one obtains for $T$ large enough that
$P_\psi (n \le \tau T, T)\le \theta$. 
For the moments the bound directly follows. 
\par
To prove the second part of the Theorem, one shows the bound similar to
(\ref{zy33}):
$$
P_\psi (n \ge M, T) \le ||\psi||^2 -\alpha/2+ P_\rho (n \ge M,T)+
C (P_\rho (n \ge M,T))^{1/2}. 
$$
Taking $M=2L_N$ and using the bound (\ref{z1})
of Theorem~\ref{upprob}
for the state $\rho$,
we get
$$
P_\rho (n \ge 2 L_N,T) \le CTL_N^{-1/\eta}.
$$
One sees that for $L_N/4 \le T \le \tau L_N^{1/\eta}$ with
$\tau$ small enough, 
$$
P_\rho (n \ge 2 L_N, T)+C (P_\rho (n \ge 2 L_N, T))^{1/2} \le \alpha/4.
$$
Thus, $P_\psi (n \le 2 L_N, T) \ge \alpha/4>0$.
In the case of $\psi$ of the first kind, the bound (\ref{zy34})
follows directly from the bound (\ref{z1}) of Theorem~\ref{upprob}. 
The proof is completed.


\par




\begin{cor}\label{betas}
{\sl
Let $\psi$ be of the first kind. 
\par \noindent 1. The equalities hold:
$$
\beta_\psi^- (p)={p + 1 \over p+ 1/\eta}, \ \ \beta_\psi^+ (p)=1. 
$$
\par\noindent
2. The measure $\mu_\psi$ and the restriction of
$\mu_{\delta_1}$ to $(-2,2)$,
have exact Hausdorff dimension $\eta$. 
}
\end{cor}
Proof. The first statement is proved using the bounds for the moments
of Theorem~\ref{betamo} and Theorem~\ref{upprob}
 and considering $L_N \le T \le L_N^{\alpha}$
and $L_N^{\alpha} \le T <L_{N+1}$
with suitable $\alpha$ as in the proof of Theorem~\ref{betamo}.
The bound ${\rm dim}_* (\mu_\psi) \ge \eta$ was proved by
Jitomirskaya and Last (see also Corollary~\ref{dimh} for a more simple
proof). 
On the other hand, one has the well known inequality
$\beta_\psi^-(p) \ge {\rm dim}^*(\mu_\psi)$ for all $p>0$
(which follows from the results of \cite{L}). 
Since
$\beta_\psi^-(p)={p+1 \over p+1/\eta}$,  letting $p \to 0$
we obtain the upper bound ${\rm dim}^* (\mu_\psi) \le \eta$. 
Thus, $\mu_\psi$ has exact Hausdorff dimension $\eta$. Since it is
true for any $f$ of the first kind, it is true for the restriction
of $\mu_{\delta_1}$ to $(-2,2)$. The proof is completed. 



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