%LaTeX file: coherent-states.tex
\documentstyle[12pt]{article}
\title{Incompleteness of Sparse Coherent States}
\author{Jayakumar Ramanathan\\
Department of Mathematics\\
Eastern Michigan University\\
Ypsilanti MI 48197
\and
Tim Steger\\
Department of Mathematics\\
University of Georgia\\
Athens GA 30602}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
\newtheorem{example}{Example}
\newtheorem{remark}{Remark}
\newtheorem{conjecture}{Conjecture}
\newtheorem{definition}{Definition}
\begin{document}
\maketitle
\section{Introduction}
The Schr\"odinger reprentation of the Heisenberg group is the unitary
representation defined by
$$
\rho(p,q,t) f(x) = e^{\pi\imath pq} e^{2\pi\imath t} e^{2\pi\imath qx}
f(x+p)
$$
where $p,q,x \in {\bf R}^n$, $t \in {\bf R}$ and $f \in L^2({\bf
R}^n)$. The group law of the Heisenberg group is reflected in the
formula
\[
\rho(p,q,t) \rho(p',q',t') = \rho(p+p',q+q',t+t'+\frac 12
(pq'-p'q)).
\]
The unitary operator $\rho(p,q)f := \rho(p,q,0)f$ may be thought of as a
phase-space translate of the function $f$. This paper is concerned with
the completeness properties of the set
\begin{equation}
\label{eqn:ph_sp_trans}
{\cal S}_{f,\Lambda} = \left\{ \rho(p,q)f: (p,q) \in \Lambda \right\}
\end{equation}
where $f \in L^2({\bf R}^n)$ and $\Lambda$ is a discrete set in ${\bf
R}^{2n}$. It is a well known that the completeness
of such sets depends on the density of the underlying set of phase
space translates. In particular, if the density is sufficiently small, the
set ${\cal S}_{f,\Lambda}$ should be incomplete. Examination of known special
cases leads to the conjecture, made precise in the next paragraph, that the
threshold density should be one.
Let $\Lambda$ be a discrete subset of ${\bf R}^N$.
Choose $E$, a measurable set of measure one. Denote by $\nu^+(r,E)$ the
maximum of the number of
points of $\Lambda$ occurring in any translate of $rE$. The quantity
$\nu^-(r,E)$ is the minimum of the number of points of $\Lambda$ occurring
in any translate of $rE$. The uniform upper and lower density
of the set $\Lambda$ are defined by
\begin{equation}
\label{eqn:upperdensity}
D^+(\Lambda) = \limsup_{r \to \infty} \frac {\nu^+(r,\Lambda)}{r^N}
\end{equation}
and
\begin{equation}
\label{eqn:lowerdensity}
D^-(\Lambda) = \liminf_{r \to \infty} \frac {\nu^-(r,\Lambda)}{r^N}
\end{equation}
respectively. If the two quantities above are the same, the set
$\Lambda$ is said to have {\em uniform density} given by
\[
D(\Lambda) = D^+(\Lambda) = D^-(\Lambda).
\]
It is a theorem of H.J. Landau \cite{Landau} that these uniform
densities are independent of the particular choice of set $E$ with measure one.
\begin{conjecture}
Let $\Lambda \subset {\bf R}^{2n}$ be a discrete set in phase space
and $f \in L^2({\bf R}^n)$
If $D^+(\Lambda) < 1$ then the set ${\cal S}_{f,\Lambda}$
is incomplete in $L^2({\bf R}^n)$.
\label{conj:the_conjecture}
\end{conjecture}
In the case where $f(x)$ is a gaussian, the validity of this
conjecture follows from the work of K. Seip and K. Seip and
Wallst\'en;
see \cite{Seip1} and \cite{Seip2}. The main thrust of the conjecture is that the imcompleteness phenomena is
claimed for an arbitrary function $f \in L^2$.
When $\Lambda$ has the additional structure of a lattice in ${\bf R}^n$
this result follows from work of M. Rieffel \cite{Rieffel}. His approach
relies on the computation of the coupling constant of the von Neumann algebra
generated by $\rho(p,q)$ where $(p,q) \in \Lambda$. Special cases
of this result have been established without recourse to the
machinery of operator algebras. One such approach, using the Zak
transform, has been given by I. Daubechies \cite{Daub} under the additional
assumption that the
density of the lattice is rational. In a recent preprint, A.J.E.M.
Janssen \cite{Janssen} has shown that $S_{f,\Lambda}$ can never be
a frame (see section $3$ for the definition) provided
$\Lambda$ is a lattice with density less than one.
In this paper, we shall give a direct proof
of the conjecture when $\Lambda$ is a lattice --- our techniques do not use
sophisticated concepts from operator algebras but rather an elementary
analysis of the \lq phase space content\rq\ of the functions in
${\cal S}_{f,\Lambda}$. It should be noted that the arguments presented
also yield a necessary upper bound on the uniform lower density of a
discrete set of coherent states that is also an Riesz basis for
$L^2$. In addition, certain variants, where $\Lambda$ is
allowed to be more irregular, are also discussed. Necessary
conditions on the density of $\Lambda$ in order for
$S_{f,\Lambda}$ to be a frame, have been given by Landau \cite{Landau2} provided both
$f$ and its Fourier transform have faster than linear decay. This
type of result is shown to hold without the decay condition in theorem
\ref{thm:frames_have_hap}. It is hoped that the work
presented here will provide the basis for proving the above conjecture.
\section{A Comparison Theorem}
The statement of the main result in this section requires the
following definition.
\begin{definition} Suppose the set ${\cal S}_{f,\Lambda}$ is
complete for some $f \in L^2({\bf R}^n)$ and $\Lambda$ a discrete
subset of ${\bf R}^{2n}$. ${\cal S}_{f,\Lambda}$ has the
{\em homogeneous approximation property} if given $g \in L^2({\bf R}^n)$
and $\epsilon > 0$, there is a $R > 0$ such that
for any $(p_0,q_0) \in {\bf R}^{2n}$, $\rho(p_0,q_0)g$ can be
approximated to within $\epsilon$ in the $L^2$-norm by a vector in
\[
\{\rho(p,q)f: (p,q) \in \Lambda \cap B_R(p_0,q_0)\}.
\]
\label{def:hap}
\end{definition}
\begin{theorem}
Let $\Gamma$ and $\Lambda$ be discrete sets. Assume that there are
two functions $\phi,f \in L^2({\bf R}^n)$ such that ${\cal
S}_{\phi,\Gamma}$ forms a Riesz bases for $L^2({\bf R}^n)$ and
that ${\cal S}_{f,\Lambda}$ is complete and has the
homogeneous approximation property.
\begin{enumerate}
\item If $\Gamma$ and $\Lambda$ have uniform upper densities given by
$D^+(\Gamma)$ and $D^+(\Lambda)$ then
\[
D^+(\Gamma) \le D^+(\Lambda).
\]
\item If $\Gamma$ and $\Lambda$ have uniform lower densities given by
$D^-(\Gamma)$ and $D^-(\Lambda)$ then
\[
D^-(\Gamma) \le D^-(\Lambda).
\]
\end{enumerate}
\label{thm:hap_and_density}
\end{theorem}
\noindent{\sc Proof:}
Let $(a,b) \in {\bf R}^{2n}$. Define $V_r$ to be the span of
$$
\{\rho(r_\alpha,s_\alpha)\phi: (r_\alpha,s_\alpha) \in B_r(a,b)
\cap \Gamma\}
$$
and let $W_r$ be the span of
$$
\{\rho(p_i,q_i)f: (p_i,q_i) \in B_r(a,b) \cap \Lambda\}.
$$
A consequence of the preceding discussion is that any $\rho(p,q)\phi$ with
$(p,q) \in B_r(a,b)$ can be approximated to within $\epsilon$ by a function
in $W_{r+R}(a,b)$. Consider the self-adjoint operator from $T:V_r \to V_r$ defined as a
composition of projections:
$$
T = P_{V_r} \circ P_{W_{r+R}}.
$$
Since ${\cal S}_{\phi,\Gamma}$ is an
Riesz basis, there are functions $\psi_\alpha \in L^2({\bf R}^n)$ such
that
$$
\langle \rho(p_\alpha,q_\alpha)\phi,\psi_\beta\rangle =
\delta_{\alpha,\beta}.
$$
The above arguments imply that,
$$
{\rm Tr}(T) = \sum_{(r_\alpha,s_\alpha) \in B_r(a,b) \cap \Gamma}
\langle T(\rho(r_\alpha,s_\alpha)\phi),\psi_\alpha\rangle
\ge (1 - 2 \epsilon)\ {\rm card}\left(B_r(0) \cap
\Gamma\right).
$$
Let
$$d = d((a,b);r) = {\rm card}\left(B_r(a,b) \cap \Gamma\right)$$
and denote the
eigenvalues of $T$ by $\lambda_1,\cdots,\lambda_d$. Since $T$ is a composition
of projections, $\lambda_i \in [0,1]$. It follows that the rank of
$T$ dominates the trace
\[
\lambda_1 + \cdots + \lambda_d
\]
The observation that the rank of $T$ is no greater than the
dimension of $W_{r+R}$ produces the following estimate
$$
d (1-2\epsilon) \le \mbox{Tr}(T) \le {\rm dim\ }
W_{r + R}(a,b).
$$
The obvious upper bound for the dimension of $W_{r +
R}(a,b)$ then produces the estimate
$$
(1 - 2\epsilon) {\rm card}\left(B_r(a,b) \cap \Gamma\right) \le
{\rm card}\left(B_{r+R}(a,b) \cap \Lambda\right).
$$
Clearly, this is equivalent to
\begin{equation}
(1 - 2\epsilon) \frac {{\rm card}\left(B_r(a,b) \cap \Gamma\right)}
{{\rm vol}\left(B_r(a,b)\right)}
\le
\frac {{\rm card}\left(B_{r+R}(a,b) \cap \Lambda\right)}
{{\rm vol}\left(B_{r+R}(a,b)\right)} \frac {(r + R)^{2n}}{r^{2n}}.
\label{eqn:estimate_at_point}
\end{equation}
Since the last term tends to one as $r \to \infty$, the two
conclusions in the theorem are now straight-forward. Consider item
1. Fix $\delta >
0$, and note that for all $r > r_0$
\[
\frac {{\rm card}\left(B_{r+R}(a,b) \cap \Lambda\right)}
{{\rm vol}\left(B_{r+R}(a,b)\right)} \le
(1+\delta) D^+(\Lambda)
\]
Now choose $(a,b)$ and $r > r_0$ so that
\[
(1 - \delta) D^+(\Gamma) \le \frac {{\rm card}\left(B_r(a,b) \cap \Gamma\right)}
{{\rm vol}\left(B_r(a,b)\right)}.
\]
By putting these estimates together with equation
\ref{eqn:estimate_at_point}, we have
\[
(1 - \delta)(1 - 2\epsilon) D^+(\Gamma) \le (1 + \delta)
D^+(\Lambda).
\]
Since $\epsilon,\delta$ were arbitrary positive numbers the first
item in the theorem follows. The second item is established in an
entirely similar manner.$\diamondsuit$
\begin{theorem} Suppose ${\cal S}_{f,\Lambda}$ is complete for
some $f \in L^2({\bf R}^n)$ and some lattice $\Lambda \in {\bf
R}^{2n}$. Then ${\cal S}_{f,\Lambda}$ has the homogeneous
approximation property.
\label{thm:lattices_have_hap}
\end{theorem}
\noindent{\sc Proof:} Let $\Lambda = \{(p_i,q_i)\}$ and let $D$ be a fundamental domain for the lattice
$\Lambda$ that has the origin in its boundary. Fix $\phi \in L^2$.
Fix $\epsilon > 0$ and note, by virtue of the completeness of ${\cal
S}_{f,\Lambda}$,
that for any $(p,q) \in \bar D$ there is a ball $B$ centered at
the origin (with radius perhaps depending on $(p,q)$)
such that $\rho(p,q)\phi$ can be approximated to within $\epsilon/2$ in
$L^2$-norm by a function in the span of
$$
\{\rho(r,s)f: (r,s) \in B \cap \Lambda\}.
$$
Denote the radius of this ball $R(p,q)$.
The compactness of $\bar D$ and the $L^2$ continuity of the Schr\"odinger
representation implies that there is a finite open cover of $\bar D$ by
open sets $U_1,\cdots,U_k$ such that if $(a,b),(c,d) \in U_i$,
$$
\|\rho(a,b)\phi - \rho(c,d)\phi\| < \epsilon/2.
$$
Choose $(p_i,q_i) \in U_i$ and set $R = \max R(p_i,q_i)$.
It is easy to check that {\em any} $\rho(p,q)\phi$ with
$(p,q) \in \bar D$ can be approximated to within $\epsilon$ by a function
in the span of
$$
\{\rho(p,q)f: (p,q) \in B_R(0) \cap \Lambda\}.
$$
Now any $(r,s) \in {\bf R}^{2n}$ can be expressed as
\[
(r,s) = (p',q') + (r',s') \qquad (r',s') \in \Lambda, (p',q') \in \bar D.
\]
We know $\rho(p',q')\phi$ can be approximated to within $\epsilon$
by a sum of the form
\[
\sum_{n=1}^N c_n \rho(a_n,b_n)f
\]
where $(a_n,b_n) \in B_R(0) \cap \Lambda$. Upon applying the
unitary operator $\rho(r',s')$ and using the group law for the
Heisenberg group, we get a sum of the form
\[
\sum_{n=1}^N c_n' \rho(a_n+r',b_n+s')f
\]
as an $\epsilon$ approximant to $\rho(r',s')\rho(p'q')\phi$.
Now, in view of the group law for the Heisenberg group,
$$
\rho(r',s')\rho(p'q')\phi = \exp(\pi\imath(r'q'-s'p'))
\rho(r,s)\phi.
$$
Therefore
\[
\sum_{n=1}^N \exp(-\pi\imath(r'q'-s'p'))c_n' \rho(a_n+r',b_n+s')f
\]
approximates $\rho(r,s)\phi$ to within $\epsilon$ in $L^2$-norm.
Note that the points $(a_n+r',b_n+s')$ are in $\Lambda$ and within
distance $R$ of $(r,s)$ and the required property is established.$\diamondsuit$
Two corollaries follow immediately.
\begin{corollary}
Let $\Lambda$ be a lattice in ${\bf R}^{2n}$ with density smaller than
one and $f \in L^2({\bf R}^n)$. The set ${\cal S}_{f,\Lambda}$ is
incomplete.
\end{corollary}
\noindent {\sc Proof:} Take $\phi(x)$ to be the characteristic
function of the unit cube $[0,1]^n$ and $\Gamma$ to be the integer
lattice. The set ${\cal S}_{\phi,\Gamma}$ is an orthonormal basis
for $L^2$. If ${\cal S}_{f,\Lambda}$ is complete, theorems
\ref{thm:lattices_have_hap} and \ref{thm:hap_and_density} yield the result
that
\[
1 \le D^+(\Lambda).\diamondsuit
\]
A proof, using von Neumann algebra theory,
of the preceding result follows from the work of M. Rieffel
\cite{Rieffel}. A proof using
the Zak transform has been given by I. Daubechies \cite{Daub} in the
case where the density of $\Lambda$ is a rational number larger
than one. A recent preprint of A.J.E.M. Janssen \cite{Janssen} gives a proof of
the weaker statement that ${\cal S}_{f,\Lambda}$ can never be a
frame under these conditions.
\begin{corollary}
Let $\Gamma$ be a discrete set in ${\bf R}^{2n}$ with $D^+(\Gamma) > 1$. Then
there is no $\phi \in L^2({\bf R}^n)$ for which ${\cal S}_{\phi,\Gamma}$
is an Riesz basis of $L^2({\bf R}^n)$.
\end{corollary}
\noindent {\sc Proof:} Suppose a $\Gamma$ and $\phi$ have the
property that ${\cal S}_{\phi,\Gamma}$ is a Riesz basis. Then let
$f$ be the characteristic function of the unit cube and $\Lambda$
the integer lattice. Then ${\cal S}_{f,\Lambda}$ is complete and
has the homogeneous approximation property by theorem
\ref{thm:lattices_have_hap}. Upon applying theorem
\ref{thm:hap_and_density},
\[
D^+(\Gamma) \le 1.\diamondsuit
\]
\section{Density of Frames}
Recall that a sequence of vectors $v_n$ in a Hilbert space $H$ is
a {\em frame} if there are constants $A,B > 0$ such that
\[
A \|v\|^2 \le \sum_n \langle v,v_n\rangle^2 \le B \|v\|^2 \qquad
\mbox{for all $v \in H$.}
\]
This {\em frame inequality} insures that a frame is always a complete set.
In this section, we will actually consider a
somewhat weaker condition.
\begin{definition} A sequence $v_n$ in a Hilbert space $H$ is a
{\em quasi-frame} with respect to a dense subspace $D$ of $H$ if
\[
A\|v\|^2 \le \sum_n \langle v,v_n\rangle^2 < \infty
\]
for every $v \in D$. We will call the constant $A$ the {\em
quasi-frame constant}.
\label{def:q-frame}
\end{definition}
In this section $H = L^2({\bf R}^n)$ and $D$ is the dense subspace
consisting of finite linear combinations of functions of the form
\[
\rho(p,q) \phi_0(x)
\]
where $\phi_0(x) = 2^{n/4} \exp( - \pi x^2)$ and $(p,q) \in {\bf
R}^{2n}$. Note that $D$ is invariant under the action of the
Heisenberg group.
We will develop necessary conditions on a
set $\Lambda$ so that $S_{f,\Lambda}$ is a quasi-frame
with respect to the subspace $D$. To this end, we need the
following definition.
\begin{definition}
A {\em uniformly} discrete set $\Lambda$ of ${\bf R}^{N}$ is a discrete
set for which there is a $\delta > 0$ satisfying
\[
|x_1 - x_2| > \delta \qquad \mbox{for all distinct $x_1,x_2 \in \Lambda$.}
\]
The constant $\delta$ is called the {\em separation constant}.
\label{def:unif_dis}
\end{definition}
Let $\Lambda$ be a uniformly discrete subset of ${\bf R}^{2n}$
with separation constant $\delta>0$ and $f \in L^({\bf R}^n)$. We
first show that
\begin{equation}
\sum_{(p_0,q_0) \in \Lambda} \left| \langle
\rho(p_0,q_0)f,g\rangle\right|^2 < \infty
\label{eqn:q-fr_finite}
\end{equation}
for all $g \in D$.
Let
\[
g(x) = \sum_{i = 1}^N c_i \rho(r_i,s_i) \phi_0.
\]
For any function $h \in L^2$
\begin{equation}
|\langle h, g\rangle|^2 \le \left(\sum_{i = 1}^N
|c_i|^2\right) \left(\sum_{i = 1}^N |\langle h,
\rho(r_i,s_i)\phi_0\rangle|^2\right).
\label{eqn:split_apart}
\end{equation}
Now fix $(p_0,q_0) \in \Lambda$ and set $h = \rho(p_0,q_0)f$.
We first claim that there is a constant $C > 0$, depending just on
the separation constant, such that
\begin{equation}
|\langle \rho(p_0,q_0)f, \rho(r,s)\phi_0\rangle|^2 \le C
\int_{B_{\delta/2}(0,0)}
|\langle \rho(p_0+p,q_0+q)f,\rho(s,t)\phi_0\rangle|^2\,dpdq.
\label{eqn:integral_bound}
\end{equation}
To see this, note that the map
\[
F: p + \imath q \mapsto \langle \rho(p_0-s,q_0-t)f, \rho(- p,- q)\phi_0\rangle
\exp\left(\pi/2(p^2+q^2)\right)
\]
is an analytic function in the variable $z = p + \imath q$ (see
\cite{Barg}). An application of a standard integral bound for analytic functions
\cite{Hor} gives:
\[
|F(0)|^2 \le C \int_{B_{\delta/2}(0,0)} |F(z)|^2.
\]
The group law for the Heisenberg group can be used to compute that
\[
|\langle \rho(p_0-s,q_0-t)f, \rho(- p,- q)\phi_0\rangle| =
\langle \rho(p_0+p,q_0+q)f, \rho(s,t)\phi_0\rangle|.
\]
The equation \ref{eqn:integral_bound} follows.
An application of equation \ref{eqn:integral_bound} to each term
in the sum
\[
A = \sum_{(p,q) \in \Gamma} |\langle \rho(p,q)f,\rho(r_i,s_i)\phi_0\rangle|^2
\]
produces the bound
\[
A \le C \int_{{\bf R}^{2n}}
|\langle \rho(p,q)f,\rho(r_i,s_i)\phi_0\rangle|^2\,dpdq
\]
The integral on the right is finite by Moyal's formula. It is
useful to actually glean a little more from this argument. Clearly,
\begin{eqnarray}
\lefteqn{\sum_{(p,q) \in \Gamma \backslash B_{r}}
|\langle \rho(p,q)f,\rho(r_i,s_i)\phi_0\rangle|^2
\le} \nonumber \\
& & C \int_{B_{r - \delta}^c}
|\langle \rho(p,q)f,\rho(r_i,s_i)\phi_0\rangle|^2\,dpdq = I(r)
\label{eqn:contol_far_away}
\end{eqnarray}
with $I(r) \to 0$ as $r \to \infty$. This observation, together
with equation \ref{eqn:split_apart} readily yields the following
lemma.
\begin{lemma} Let $g$ be any finite linear combination of phase
space translates of the gaussian $\phi_0$. Given $\epsilon,\delta
> 0$ there is a conmpact set $K$ such that
\[
\sum_{(p,q) \in \Gamma \backslash K} |\langle
\rho(p,q)f,g\rangle|^2 \le \epsilon
\]
for any uniformly discrete set $\Gamma$ with separation constant
$\delta > 0$.
\label{lem:control_far_away}
\end{lemma}
Our objective is the following theorem.
\begin{theorem} Let $\Lambda \in {\bf R}^{2n}$ be a uniformly discrete
subset with $D^+(\Lambda) < \infty$. Suppose that
for $f \in L^2({\bf R}^n)$, ${\cal S}_{f,\Lambda}$ is a quasi-frame.
Then ${\cal S}_{f,\Lambda}$ has the homogeneous approximation
property.
\label{thm:frames_have_hap}
\end{theorem}
When taken together with theorem 1, this yields the following
strengthening of a result of H. J. Landau \cite{Landau2}.
\begin{corollary}
Let $\Lambda \in {\bf R}^{2n}$ be a uniformly discrete set. Suppose that
for $f \in L^2({\bf R}^n)$, ${\cal S}_{f,\Lambda}$ is a quasi-frame.
Then
\[
D^-(\Lambda) \ge 1.
\]
\end{corollary}
We first develop some preliminary results necessary for the proof
of theorem \ref{thm:frames_have_hap}.
\begin{definition} The Frechet distance $[Q,R]$ between two closed sets
$Q,R$ of ${\bf R}^n$ is the smallest $t \ge 0$ such that
\[
Q \subset R_t \qquad \mbox{and} \qquad R \subset Q_t
\]
where $Q_t$ is the set of all points within the distance $t$ of
$Q$. $Q_n$ converges wealky to $Q$ if
\[
[Q_n \cap K, Q \cap K] \to 0
\]
for every compact set $K$.
\label{def:conv_of_sets}
\end{definition}
\begin{lemma} Let $\Lambda$ be a uniformly discrete set and $f \in
L^2({\bf R}^n)$ such that $S_{f,\Lambda}$ is a frame. Let
$\Lambda'$ be a uniformly discrete set that is a weak limit of the
sequence of translates of $\Lambda$. Then $S_{f,\Lambda'}$ is a
frame with the same frame bounds as $S_{f,\Lambda}$.
\label{lem:wk_limits_are_frames}
\end{lemma}
\begin{lemma} Let $\Lambda_n$ be a sequence of translates
of a uniformly discrete set, $\Lambda$ with $D^+(\Lambda)
< \infty$. Then there is a sub-sequence $n_i$ and a uniformly
discrete set $\Lambda'$ such that $D^+(\Lambda') < \infty$ and
$\Lambda_{n_i}$ converges weakly to $\Lambda'$.
\label{lem:wk_compactness}
\end{lemma}
We defer the proofs of these two lemmas till after the proof of
the above theorem.
\noindent{\sc Proof of Theorem \ref{thm:frames_have_hap}:} Suppose
the set $S_{f,\Lambda}$ {\em does not} satisfy the homogeneous
approximation property. Fix $\epsilon > 0$. Then there is a
$g \in L^2({\bf R}^n)$ and sequences $(p_n,q_n)$ and $r_n \to \infty$ such that
the distance of $\rho(p_n,q_n)g$ to the span of
\[
\{\rho(p,q)f: (p,q) \in \Lambda \cap B_{r_n}(p_n,q_n)\}
\]
is at least $\epsilon$. Set $\Lambda_n = \Lambda - (p_n,q_n)$. By
applying the unitary tranformation $\rho(p_n,q_n)$ and using the
group law for the Heisenberg group, we observe that $g$ has
distance at least $\epsilon$ to the span of
\[
\{\rho(p,q)f: (p,q) \in \Lambda_n \cap B_{r_n}(0,0)\}.
\]
Lemma \ref{lem:wk_compactness} assures us that, by
passing to a subsequence, $\Lambda_n$ converges
weakly to a uniformly discrete set $\Lambda'$. Fix $r > 0$. Given
$\kappa > 0$, there is an $N > 0$ such that every $(p,q) \in
\Lambda' \cap B_r(0,0)$ is within $\kappa$ distance of a point from
$\Lambda_n$ for all $n \ge N$. Consequently, the unit vectors in
the span of
\[
\{\rho(p,q)f: (p,q) \in \Lambda' \cap B_r(0,0)\}
\]
can be made uniformly close to the span of
\[
\{\rho(p,q)f: (p,q) \in \Lambda_n \cap B_{r_n}(0,0)\}
\]
provided $n$ is sufficiently large. Hence the distance between $g$
and the span of
\[
\{\rho(p,q)f: (p,q) \in \Lambda' \cap B_r(0,0)\}
\]
cannot be less than $\epsilon$. Since this statement is
independent of $r$, $g$ cannot be in the span of
\[
S_{f,\Lambda'} = \{\rho(p,q)f: (p,q) \in \Lambda'\}.
\]
But this last set is a quasiframe and, hence, complete. Contradiction.$\diamondsuit$
\noindent{\sc Proof of lemma \ref{lem:wk_limits_are_frames}:}
Let $\Lambda_n$ be a sequence of translates of
$\Lambda$ that converges weakly to $\Lambda'$. It is clear that
there is a $\delta > 0$ that serves as a separation constant for
both the $\Lambda_n$ and $\Lambda'$. Let $g(x)$ be a linear combination of a
finite number of phase space translates of the gaussian
\[
2^{n/4} \exp(- \pi x^2).
\]
Fix $\epsilon > 0$.
Let $K$ be a compact subset of ${\bf R}^{2n}$ satisfying the
conditions of lemma \ref{lem:control_far_away}. For all $n$
sufficiently large, we must have that
\[
\left| \sum_{(p,q) \in \Lambda' \cap K} |\langle f,
\rho(p,q)g\rangle|^2 - \sum_{(p,q) \in \Lambda_n \cap K} |\langle f,
\rho(p,q)g\rangle|^2 \right| \le \epsilon.
\]
By putting this together with the estimate in lemma
\ref{lem:control_far_away}, we obtain the estimate
\[
\left| \sum_{(p,q) \in \Lambda'} |\langle f,
\rho(p,q)g\rangle|^2 - \sum_{(p,q) \in \Lambda_n} |\langle f,
\rho(p,q)g\rangle|^2 \right| \le 3\epsilon
\]
for all $n$ sufficiently large. The sets $S_{\Lambda_n,f}$ are all
quasi-frames with the same quasi-frame bound. Hence,
$S_{\Lambda',f}$ must also have the same quasi-frame bound.$\diamondsuit$
\noindent{\sc Proof of lemma \ref{lem:wk_compactness}:} Decompose
${\bf R}^{2n}$ as the union of non-overlapping closed unit cubes:
\[
{\bf R}^{2n} = \cup_k Q_k
\]
By passing to a subsequence, if necessary, we may assume that the
number of points of $\Lambda_n$ in a particular cube $Q_{k_0}$
eventually stabilizes. This can be done because of the uniformly
discrete nature of the $\Lambda_n$. A diagonalization argument
shows that --- by passing to a subsequence, if necessary --- for
any $Q_k$ there is an $N = N(k)$ such that the number of points in
$\Lambda_n \cap Q_k$ stabilizes for all $n \ge N$. Each cube
in the decomposition is compact. For
each $k$ and $n \ge N(k)$, list the points in $\Lambda_n \cap Q_k$
by $a_{1k}^{(n)},\cdots,a_{l_kk}^{(n)}$. Note that by passing to a
subsequence, we may assume that each $a_{ik}^{(n)}$ converges as
$n \to \infty$. A diagonalization argument allows us to assume, by
passing to a subsequence if necessary, that this is true for each
$k$. Note by $a_{ik}$ the limit of $a_{ik}^{(n)}$. Set $\Lambda'$
to be the collection of these limit points. It is easy to verify
that $\Lambda'$ is a uniformly discrete set with finite upper
uniform density.$\diamondsuit$
We conclude this paper with the following direct consequence to
corollary 2 and corollary 3.
\begin{corollary} Let $\Lambda$ be a uniformly discrete set and $f
\in L^2({\bf R}^{2n})$ such that $S_{f,\Lambda}$ is a Riesz basis.
Then $\Lambda$ must have a uniform density of one:
\[
D^+(\Lambda) = D^-(\Lambda) = 1.
\]
\label{cor:density_R_basis}
\end{corollary}
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\end{document}