text:
\section{Introduction}

Given a function $g\in L^2(\bR)$ and a lattice $\Lambda \subset \mathbb{R} ^2$,
we study the frame property of the set $\{e^{2\pi i \lambda _2 t}
g(t-\lambda _1):  \lambda = (\lambda _1,\lambda _2) \in \Lambda\}$.
 Precisely,  write $\pi _\lambda g = e^{2\pi i \lambda _2 t}
g(t-\lambda _1)$ for the time-frequency shift\ by $\lambda=(\lambda_1,\lambda_2) \in \mathbb{R} ^2$. Then we call
the set $\mathcal{G} (g,\Lambda ) = \{ \pi  _\lambda g : \lambda \in \Lambda
\}$ a \emph{Gabor frame} or \emph{Weyl-Heisenberg frame}, whenever
there  exist  constants $A,B >0$ such that,
for all $f\in L^2(\bR)$,
\begin{equation}
\label{Eq:2.8} A\|f\|^2_{L^2(\bR)} \leq \sum_{\lambda \in
\Lambda}|\langle f, \pi_\lambda g\rangle_{L^2(\bR)}|^2 \leq
B\|f\|^2_{L^2(\bR)}\, .
\end{equation}
  Gabor frames  originate in quantum mechanics through J.~von
Neumann and in information theory through D.~Gabor~\cite{Ga}   and
nowadays have many applications in signal processing. A large body
of results describes the structure of Gabor frames and provides
sufficient conditions of a qualitative nature for  $\mathcal{G} (g,\Lambda )$
to form a (Gabor) frame, see~\cite{daubechies90,Charlybook} for
details and references.

We will also study vector-valued Gabor frames. In this
case the Hilbert space is
$\mathcal{H}=L^2(\mathbb{R}, \mathbb{C}^n)$ consisting of all vector-valued functions
$\bof(t)= \big( f_1(t), \ldots \ , f_n(t) \big)$ with the natural inner product
\begin{equation}
\label{product}
\langle \bof, \bog \rangle = \sum_{j=1}^n
      \int_{-\infty}^\infty f_j(t) \overline{g_j(t)}dt=
               \sum_{j=1}^n \langle f_j, g_j \rangle .
\end{equation}
 The {\em time-frequency shifts} $\pi _z$, $z= (x,\xi )$ act coordinate-wise by
\begin{equation}
  \label{eq:c1}
\pi _z\bof (t) =  e^{2i\pi \xi t}\bof(t-x) \, .
\end{equation}
The vector-valued Gabor system
$\mathcal{G}(\bog, \Lambda)= \{ \pi_\lambda \bog : \lambda \in \Lambda\}   $
is a frame for $L^2(\mathbb{R} , \mathbb{C} ^n)$, if there exist constants $A,B>0$  such that
\begin{equation}
\label{frameinequalitya}
A \|\bof\|^2_{L^2(\mathbb{R}, \mathbb{C}^n)} \leq \sum_{\lambda \in \Lambda}
 |\langle \bof , \pi_\lambda \bog \rangle |^2 \leq B\|\bof \|^2_{L^2(\mathbb{R}, \mathbb{C}^n)},
 \quad \forall \bof \in  L^2(\mathbb{R}, \mathbb{C}^n).
 \end{equation}

Vector-valued Gabor frames were introduced under the name
``superframes'' in signal processing by
R.~Balan~\cite{balan1} in the context of ``multiplexing'' and studied in
~\cite{balan2,han-larsen} for their own sake.  The idea of
``multiplexing'' is to encode $n$ independent signals
(functions) $f_j\in L^2(\mathbb{R} ), j=1,\dots , n,$ as a single sequence that captures the
time-frequency information of each $f_j$. Fixing suitable windows $g_j
\in L^2(\mathbb{R} )$ and using vector-valued notation $\bof = (f_1, \dots ,
f_n)$, one then considers the sequence of numbers
 $\langle \bof , \pi_\lambda \bog \rangle =
  \sum _{j=1}^n \langle f_j, \pi_\lambda g_j\rangle $
  for $\lambda \in \Lambda $, i.e., the inner product in
$L^2(\mathbb{R} , \mathbb{C} ^n) $. Roughly speaking, these numbers then measure the
time-frequency content of the whole  $\bof$ at the point $\lambda $ in the
time-frequency plane.

 Now one  requires that
$\bof $ is completely determined by these inner products and that
there exists a stable reconstruction. This requirement leads to the
definition of a vector-valued Gabor frame~\eqref{frameinequalitya}.

The general problem of characterizing \emph{all} lattices $\Lambda $
for which $\mathcal{G} (g, \Lambda )$ is a frame seems to be extremely
difficult. In fact, this problem is solved only for three classes of
basis functions, namely for the Gaussians  $H_0(t) = e^{-at^2},
a>0,$ in ~\cite{lyu,sw}, for the hyperbolic secant $g(t) =
(\cosh a t )^{-1} $ in~\cite{JS02}, and for  the one-sided exponential function $g(t) =
e^{-a|t|} \chi _{[0,\infty )}(t)$~\cite{janssen96}.  For the Gaussian,
 our understanding is  based on the  connection between the frame property
 of $\mathcal{G} (g,\Lambda )$  and a  classical interpolation problem in the
 Bargmann-Fock  space of entire functions. The case of the hyperbolic
 secant is reduced to the case of the Gaussian.

 For other choices of the $g$, where the connection to complex
 analysis is missing, the   conditions for $\mathcal{G}(g,a\bZ\times b\bZ)$ to form a frame
 become very different and often rather intriguing~\cite{janssen02b}.

Even  less is known about Gabor superframes. Necessary density
conditions were studied in~\cite{balan2}, sufficient density
conditions can be derived from coorbit theory~\cite{fg89jfa} and from
the sampling theory on the Heisenberg group~\cite{fuehr07}. In particular, these
results  imply
that for any  sufficiently dense lattice $\Lambda $ and a mild condition
on the vector $\bog $ the Gabor system $\mathcal{G} (\bog , \Lambda )$ is a
Gabor superframe.

In this article we study the frame property of $\mathcal{G}(\boh , \Lambda)$
in  the case when $\boh $ is the vector of the first $n+1$   Hermite function
 $\boh =(H_0, \dots , H_n)$. In the scalar-valued case, the study of
 Gabor frames with the Gaussian window $H_0$ is natural, because the
 Gaussian minimizes the uncertainty principle. Likewise, for
 vector-valued frames, the study of Gabor superframes with the Hermite
 window $\boh $ is natural, because the first Hermite functions are
 the unique orthonormal set $\{f_0, \dots , f_n\}$ in $L^2(\mathbb{R} )$ of size $n+1$
 satisfying the normalizations $\int_{-\infty}^\infty t |f_j(t)|^2 dt
 =0$ and $ \int_{-\infty}^\infty \xi |\hat{f_j}(\xi)|^2 d\xi =0$,
 $j=0, \dots , n$,   such that  the  uncertainty
 \begin{equation}
 \label{sum_uncertainty}
 \sum _{j=0}^n \left ( \int_{-\infty}^\infty t^2 |f_j(t)|^2 dt +
 \int_{-\infty}^\infty \xi^2 |\hat{f_j}(\xi)|^2 d\xi  \right )
 \end{equation}
is minimized.
Another motivation comes again from signal processing where Hermite
functions are used,  see~\cite{matz}.

The case of the Hermite vector window has been already investigated by
 F\"uhr by employing techniques related to sampling in the space of
bandlimited functions on the Heisenberg group.  He   proved the
following result.

\noindent{\bf Theorem}\cite{fuehr07}.  {\em Let $\boh = (H_0 , \dots ,
  H_n)$.  There exists a constant $C>0$
  with the following property: If the diameter of the (smallest)
  fundamental domain of $\Lambda $ is less that $C(n+1)^{-1/2}$, then
$\mathcal{G}(\boh,\Lambda)$ is a frame for $L^2(\mathbb{R},\mathbb{C}^{n+1})$.}

 Unfortunately nothing can be said about the constant $C$ within such
 an approach. F\"uhr uses the so-called ``oscillation method'' which
 is used for existence results, but  in general does not yield sharp
 results.

In this paper, we give  a complete characterization of vector-valued
frames with Hermite functions. If the lattice is given as $\Lambda =
A\bZ ^2$ for some invertible, real-valued $2\times 2$-matrix $A$, let
$s(\Lambda )= |\det A |$ be the area  of the fundamental domain of
$\Lambda $. Our main result can be formulated as follows.

 \begin{thm}
 \label{mainchar1}
Let $\boh = (H_0 , \dots , H_n)$ be the vector of the first $n+1$
Hermite functions. Then   $\mathcal{G}(\boh,\Lambda)$ is   a frame for
$L^2(\mathbb{R},\mathbb{C}^{n+1})$, if and only if\
 \begin{equation}
 \label{main_estimate}
 s(\Lambda)< \frac{1}{n+1}.
 \end{equation}
\end{thm}

 By specializing to the $n$-th coordinate of $\boh$, we obtain a
  condition for scalar-valued Gabor frames with
Hermite functions. The following result was already announced in~\cite{glyu}.

\begin{prop}
\label{simple_hermite}
 If $s(\Lambda ) < \frac{1}{n+1}$, then $\mathcal{G} (H_n , \Lambda )$ is a
 frame for $L^2(\mathbb{R} )$.
\end{prop}

  For $n=0$, the case of the
 Gaussian, we recover  (the lattice part of) the results
 in~\cite{lyu,sw}. Our method of proof yields several new results
 about the dual window. On the one hand, we construct a dual window
 for $\mathcal{G}(H_n, \Lambda )$ with Gaussian decay in time and
 frequency; on the other hand, we derive a new  estimate for the lower
 frame bound of $\mathcal{G}(H_0, \Lambda )$.
Furthermore, we discuss  an example
 for $n=1$ which   suggests that the
 sufficient conditions of Proposition~\ref{simple_hermite} are sharp for all $n$.

In order to prove these results we combine the   techniques of
Gabor analysis and complex-analytic methods. The (now) classical
structural results related to  the scalar Gabor frame systems can
be formulated for the vector case as well.  They lead  to an interpolation
problem in the Fock space of entire functions.

 This problem is not  "purely holomorphic":   the values of
 linear combinations of  functions from the Fock space
  and their derivatives are prescribed in the lattice points; however,  the
 coefficients  of such combination  are antiholomorphic polynomials.
The classical methods of complex analysis cannot be applied for problems
of this kind.  Fortunately,  in our particular case, i.e.,  for the
lattices of sufficiently
small density,  one may use well-developed machinery of the elliptic functions.

The paper is organized as follows. In  Section~2 we collect the
necessary facts related to vector-valued frames, in particular, the
construction of the  frame operator and
the vector-valued  version of
the structural theorems:  Janssen's representation   and the  Wexler-Raz
biorthogonality criteria. The complex analytic tools  are presented in
Section~3, they include basics about Fock spaces and the growth
estimates of the  Weierstrass  $\sigma$-function. Section~4 contains
the proofs of Theorem  \ref{mainchar1} and Proposition
\ref{simple_hermite}.  We also prove a result (Proposition \ref{counter}) indicating that
the density condition   in Proposition \ref{simple_hermite}  might be
sharp. The rest of Section~4 contains estimates of dual windows for
the Hermitian frames and of the lower frame bound.

\section{General Gabor Superframes}

  \subsection{Scalar case}

To derive the criterion for Gabor superframes from Hermite  functions
(Theorem~\ref{mainchar1}), we need some general notion and results
related  to  Gabor superframes
such as  Janssen's  representation of the frame operator and  Wexler-Raz
identities.  Although these are among the fundamental
results about Gabor frames, they have not yet been formulated  for
Gabor superframes. For convenience and  later reference we formulate
them explicitly for Gabor frames on $\mathbb{R} ^d $ (instead of $\mathbb{R} $) in this
section. Since the vector-valued case is   a simple consequence of the
scalar-valued case, we defer the (easy) proofs to the Appendix.

\medskip

Let $\Lambda = A \bZ ^{2d}$,  be a lattice in $\mathbb{R} ^{2d}$, where $A$
is a non-singular real  $2d \times 2d$-matrix. Let $s(\Lambda )
= |\det| A$ be  the volume of a fundamental domain of $\Lambda$.
The  {\em adjoint lattice} is defined by the commutant property  as
\begin{equation}
  \label{eq:c2}
  \Lambda ^\circ = \{ \mu \in \mathbb{R} ^{2d}: \pi (\lambda ) \pi (\mu ) =
  \pi (\mu ) \pi (\lambda ) \,\, \text{ for all } \, \lambda \in
  \Lambda \}\, .
\end{equation}
 If $\Lambda = \alpha \bZ ^d \times \beta \bZ ^d$, then $\Lambda
^\circ = \beta ^{-1} \bZ ^d \times \alpha ^{-1} \bZ ^d$. In general, for
$\Lambda = A \bZ ^{2d}\subset \mathbb{R}^{2d}$, then
\begin{equation}
  \label{eq:cn1}
  \Lambda ^\circ =
\mathcal{J}(A^T)^{-1} \bZ ^{2d} \, ,
\end{equation}
 where $A^T $ is the transpose of $A$
and $\mathcal{J} =
\begin{pmatrix}
  0 & I \\ -I  & 0
\end{pmatrix}$ (consisting of $d\times d$ blocks) is the
matrix defining the standard symplectic
form~\cite{feichtinger-kozek98}.

 Furthermore, from \eqref{eq:cn1} we see that
\begin{equation}
  \label{eq:c7}
  s(\Lambda ^\circ ) = s(\Lambda ) ^{-1} \, .
\end{equation}
The density of $\Lambda $ is defined as $d(\Lambda ) = s(\Lambda )^{-1}
$, so that $d(\Lambda )$ coincides with the usual notions of density.

\smallskip

In order to make the exposition self-contained  we recall the (now)
standard results about Gabor frames.  We refer the reader to Ch.7 in
\cite{Charlybook} for the complete proofs.

  Given two functions (windows)
$ g, \gamma  \in L^2(\mathbb{R} ^d)$,  the associated Gabor {\em frame
operator} is defined to be
\begin{equation}
\label{foperator_scalar}
Sf=S^\Lambda_{g,\gamma}f=
      \sum_{\lambda \in \Lambda} \langle f,\pi_\lambda g \rangle \pi_\lambda \gamma .
\end{equation}

For arbitrary $ g, \gamma  \in L^2(\mathbb{R} ^d)$ the right-hand side in \eqref{foperator_scalar}
defines a continuous operator from $\mathcal{S} (\mathbb{R} ^d )$
to $\mathcal{S} ' (\mathbb{R} ^d )$ with weak$^*$-convergence of the sum.

 Under  slight conditions on $g$ and $\gamma$,
 for example if both $g$ and $\gamma$  belong to the Feichtinger algebra
 $M^1$,    the series in \eqref{foperator_scalar} converges in the usual
 $L^2(\mathbb{R} ^d)$-norm.  Recall that a function $g$ on $\mathbb{R} ^d $  belongs to the Feichtinger algebra
 $M^1(\mathbb{R} ^d)$, if
 \begin{equation}
 \label{m_one}
\|g \|_{M^1} :=  \int _{\mathbb{R}^{2d} } |
\langle g, \pi _z \varphi  \rangle | \, dz < \infty \, ,
\end{equation}
where $\varphi (t) = 2^{d/4}\, e^{-\pi t^2}$ is the $L^2$-normalized
Gaussian.
This condition is met if, for example,  both $g$ and its Fourier transform
$\hat{g}$, decay sufficiently fast,
 see  \cite{Charlybook}, Ch.~7 and also \cite{ grstudia} for
 discussion and the proofs.

The convergence of \eqref{foperator_scalar} follows from the following
lemma taken from \cite{Charlybook} (the statement is not stated as
explicitly as we want it, but follows from combining
Propositions~11.1.4, 12.1.11, and 12.2.1).

\begin{lemma} \label{bessel}
  If $g\in M^1(\mathbb{R} ^d )$, then, for each lattice $\Lambda$
  \begin{eqnarray}
    \label{eq:cn2}
 &    &     \sum _{\lambda \in \Lambda } |\langle f, \pi _\lambda g\rangle |^2
    \leq n(\Lambda ) \, \|g\|_{M^1} ^2 \, \|f\|_2^2 \\
\mathrm{and} \,\,\,\,\, &   &  \|\sum _{\lambda \in \Lambda } c_\lambda  \pi _\lambda
g\|_2 \leq n(\Lambda ) ^{1/2} \, \|g\|_{M_1} \, \|c\|_2 \, ,  \label{eq:cn3}
  \end{eqnarray}
where $n(\Lambda )  = \kappa \max _{k\in \bZ ^{2d}} \mathrm{card}\, (\Lambda \cap (k+[0,1]^{2d}))$ for some absolute constant $\kappa >0$.
\end{lemma}

In other words if $g\in M^1(\mathbb{R}^d)$, then  for each choice of $\Lambda$  the sequence $\mathcal{G}(g,\Lambda )$
is a {\em  Bessel sequence}, i.e.
\[
 \sum _{\lambda \in \Lambda } |\langle f, \pi _\lambda g\rangle |^2 \leq
  \text{Const}\  \|f\|_2^2.
\]
 Moreover, the
upper frame bound of $\mathcal{G}(g,\Lambda )$ can be estimated
by $n(\Lambda )\, \|g\|_{M^1}^2$.

By a theorem of Rieffel \cite{rieffel88} and Janssen \cite{janssen95} the frame operator can be represented as a sum of time-frequency shifts over the adjoint lattice:
\begin{equation}
\label{janssen_scalar}
Sf=S^\Lambda_{g,\gamma}f=
    s(\Lambda)^{-1}  \sum_{\mu \in \Lambda^\circ} \langle \gamma,\pi_\mu g\rangle
                    \pi_\mu f .
\end{equation}

For $g,\gamma \in M^1(\mathbb{R} ^d)$ the series in the right-hand side converges
both  in  $L^2(\mathbb{R} ^d)$-norm and also in operator norm in   $L^2(\mathbb{R} ^d)$.

Given a frame  $\mathcal{G}(g,\Lambda)$ we say that $\gamma \in L^2(\mathbb{R} ^d)$ is a
{\em  dual window }
if $S_{g,\gamma}^\Lambda = I$

Based on  representation \eqref{janssen_scalar}   one  obtains a criterion for   the system
$\mathcal{G}(g,\Lambda)$ to   form a frame in $L^2(\mathbb{R} ^d)$:

\begin{prop}[Wexler-Raz biorthogonality relation]
\label{WR_scalar}
Assume that both  $\mathcal{G}(g,\Lambda)$ and $\mathcal{G}(\gamma,\Lambda)$  are Bessel sequences in $L^2(\mathbb{R} ^d)$. Then $\mathcal{G}(g,\Lambda)$  is a frame in $L^2(\mathbb{R} ^d)$ with dual window $\gamma$ if and only if
\begin{equation}
\label{biorthogonality_scalar}
\frac 1 {s(\Lambda)} \langle \gamma, \pi_\mu g\rangle = \delta_{0,\mu},
\    \text{for} \ \mu \in \Lambda^\circ.
\end{equation}
\end{prop}

We remark that for $g,\gamma \in M^1(\mathbb{R} ^d)$ the sequences $\mathcal{G}(g,\Lambda)$ and $\mathcal{G}(\gamma,\Lambda)$ always are Bessel sequences for any choice of
 $\Lambda$ by Lemma~\ref{bessel}.

The Wexler-Raz relations yield an estimate for the lower frame bound
that deserves to be better known.

\begin{cor} \label{lowerbound}
  Assume that $\mathcal{G}(g,\Lambda )$ is a frame for $L^2(\mathbb{R} ^d)$ with some dual
  $\gamma \in M^1(\mathbb{R} ^d)$. Then the optimal lower frame bounded
  $A_{\mathrm{opt}} = \|S_{g,g}^{-1} \|_{L^2\to L^2}^{-1}$ satisfies
  \begin{eqnarray}
    \label{eq:cn5}
    A_{\mathrm{opt}} \geq n(\Lambda )^{-1} \,\|\gamma \|_{M^1} ^{-2} \, .
  \end{eqnarray}
\end{cor}

\begin{proof}
  Since $f= S^\Lambda_{g,\gamma }f = \sum _{\lambda \in \Lambda }
  \langle f, \pi _\lambda g\rangle \,  \pi _\lambda \gamma$,
  Lemma~\ref{bessel} implies that
$$
\|f\|_2^2 \leq n(\Lambda ) \|\gamma \|_{M^1}^2 \sum _{\lambda \in
  \Lambda } |\langle f, \pi _\lambda g\rangle |^2 \, ,
$$
whence the estimate for the lower frame bound follows.
\end{proof}

 \subsection{Structure of Vector-Valued Gabor Frames }

For the case of  vector-valued Gabor system the frame-type operator
is defined as
 \begin{equation}
\label{foperator}
S\bof=S^\Lambda_{\bog,\bs{\gamma}}\bof= \sum_{\lambda \in \Lambda}
     \langle \bof, \pi_\lambda \bog \rangle \pi_\lambda \bs{\gamma}.
 \end{equation}
The convergence properties are the same as in the scalar case.

\begin{prop}
\label{janssen_prop}
 {\rm (Janssen representation for Gabor superframes)} Let the
windows $\bs{\gamma}=(\gamma_j)_1^n, \bog=(g_j)_1^n\in L^2(\mathbb{R} ^d, \mathbb{C}^n)$ be such that
$\gamma_j, g_j\in M^1$, $j=1,2, \ldots \ ,n.$  Then  the frame type
operator $S_{\bog , \bs{\gamma} }$ associated to $(\bs{\gamma}, \bog; \Lambda)$
  can be written as
  \begin{equation}
 \label{janssen}
S\bof = S^\Lambda_{\bs{\gamma},\bog}\bof =
 \sum_{\mu \in \Lambda^\circ} \Gamma(\mu)\pi_\mu \bof,
\end{equation}
where $\Gamma(\mu)$ is the $n\times n $ matrix with entries
\begin{equation}
\label{gamma_mu}
\Gamma(\mu)_{kl}= s(\Lambda)^{-1}\langle \gamma_k, \pi_\mu  g_l \rangle \,
\end{equation}
and the sum converges in the operator norm on $L^2(\mathbb{R} , \mathbb{C} ^n) $.
 \end{prop}

Given a Gabor (super)frame  $\mathcal{G} (\bog, \Lambda )$ with $\bog \in
L^2(\mathbb{R} ^d, \mathbb{C}^n)$,  we say that  $\bs{\gamma} \in  L^2(\mathbb{R} ^d, \mathbb{C}^n)$ is a
{\em dual window}  (with respect to the lattice $\Lambda$) if
\begin{equation}
\label{dual}
        S^\Lambda_{\bog, \bs{\gamma} }=   S^\Lambda_{\bs{\gamma},\bog} = I.
\end{equation}

Dual windows always exist, a special choice is given by the
\emph{canonical dual window}
$\bs{\gamma} ^\circ :=  \left (S_{\bog   , \bog }^\Lambda \right )  ^{-1} \bog $;
as in the scalar case the invertibility of
   $ S_{\bog    , \bog }^\Lambda  $
   is an easy consequence of the frame property.

\smallskip

From Janssen's representation we obtain a criteria for the system
$\mathcal{G}(\bog, \Lambda)  $ forms a frame in $L^2(\mathbb{R} ^d,\mathbb{C}^n)$ with dual window $\bs{\gamma}$,
that is a vector analog of the Wexler-Raz condition.

\begin{prop}
\label{wexler_raz}
{\rm (Wexler-Raz biorthogonality).}
Assume that both $\mathcal{G}(\bog, \Lambda) $ and $   \mathcal{G}(\bs{\gamma}, \Lambda) $
are Bessel sequences in $L^2(\mathbb{R} ^d, \mathbb{C}^n)$. Then  $\mathcal{G}(\bog, \Lambda) $ is a vector-valued frame in
$ L^2(\mathbb{R} ^d, \mathbb{C}^n)$ with dual window $\bs{\gamma}$ if and only if
\begin{equation}
\label{wr_orthogonality}
\frac 1 {s(\Lambda)} \langle \gamma_l, \pi_\mu g_j \rangle =
       \delta_{0, \mu} \delta_{l,j} \quad \text{for} \quad
       \mu \in \Lambda^\circ, \ j, l = 1,2, \ldots , n \, .
\end{equation}
\end{prop}

As in the scalar case we obtain Balan's  necessary density condition~\cite{balan2} by
adjusting an argument of Janssen~\cite{janssen94}.

\begin{prop} \text{(Density theorem.)}
\label{density}
 If $\mathcal{G}(\bog, \Lambda)$ is a frame for
$L^2(\mathbb{R} ^d, \mathbb{C}^n)$, then $s(\Lambda)\leq n^{-1}$.
\end{prop}

\begin{proof}
  Let $\bs{\gamma} ^\circ= S_{\bog , \bog } ^{-1} \bog$   be
 the canonical dual window. Then $\bof = \sum_{\lambda \in \Lambda}
 \langle \bof, \pi (\lambda ) \bs{\gamma} ^\circ \rangle \pi_\lambda \bog
 $ holds for every $\bof \in L^2(\mathbb{R} , \mathbb{C} ^n) $. If $\bof $ has also the representation
$    \bof = \sum_{\lambda \in \Lambda} c_\lambda \pi_\lambda \bog$,
  then  by \cite{DS52}
 \[
 \sum_{\lambda \in \Lambda} |\langle \bof, \pi_\lambda \bs{\gamma} \rangle |^2 \leq
      \sum_{\lambda \in \Lambda} |c_\lambda|^2 \,.
    \]
    We apply this argument to the trivial expansion
    $g= 1\cdot \bog+\sum_{\lambda \neq 0} 0\cdot \pi_\lambda \bog$. Thus we obtain
 \[
 |\langle \bog, \bs{\gamma} \rangle |^2\leq \sum_{\lambda \in
   \Lambda}|\langle\bog,\pi_\lambda\bs{\gamma} \rangle |^2 \leq 1.
 \]
The Wexler-Raz identities \eqref{wr_orthogonality} yield
\[
\langle \bog, \bs{\gamma} \rangle = \sum_{j=1}^n \langle g_j, \gamma_j \rangle = ns(\Lambda),
\]
and the result follows.
\end{proof}

The following duality result is not needed in the sequel, but is
included for completeness. It answers a question of our engineering
colleague G.~Matz, see e.g. \cite{matz}.

\begin{thm}[Janssen-Ron-Shen duality] \label{ron-shen}
  The Gabor system $\mathcal{G} (\mathbf{g}, \Lambda )$ is a frame for
  $L^2(\mathbb{R}^d, \mathbb{C}^n)$,

  if and only if\ the union of Gabor systems $\bigcup _{j=1}^n \mathcal{G} (g_j, \Lambda
  ^\circ )$  is a Riesz sequence for $L^2(\rd ) $.
\end{thm}

 \section{Complex Methods:  Fock space and the Weierstrass
   sigma-function}

The complex analytic techniques  we use   are concentrated around
 the Fock space   of  entire functions and precise estimates for
 the Weierstrass
$\sigma$-function. These topics are closely related: roughly speaking
$\sigma$-functions deliver examples of functions of "maximal possible growth"
in  the Fock space.

\subsection{Fock space}

In this subsection we recall the basic properties of the Fock space
which are important for our applications. We  refer the reader to \cite{Folland},
\cite{Charlybook} for detailed proofs and also for discussion of numerous application
of this space in signal analysis and quantum mechanics.

 \begin{defin}
 \label{Fock}
 The Fock space $\mathcal{F}$ is the Hilbert space of all entire functions such that
 \begin{equation}
 \label{Fock_definition}
 \|F\|^2_\mathcal{F}=\int_ \mathbb{C} |F(z)|^2 e^{-\pi |z|^2} dm_z <\infty,
 \end{equation}
 where $dm$ is the planar Lebesgue measure.
\end{defin}
The natural inner product in $\mathcal{F}$ is  denoted by  $\langle \cdot,\cdot \rangle_\mathcal{F}$.

Below we list the properties of the Fock space, which will be used in the sequel.

(a)  Each  point evaluation is a bounded linear functional in $\mathcal{F}$,
 and  the  corresponding reproducing kernel is the function $w\mapsto
 e^{\pi \bar{z} w }$, precisely,
\begin{equation}
\label{reproducing}
F(z) = \langle F(w), e^{\pi \bar{z} w } \rangle_\mathcal{F} \, ,  \qquad  \forall  F\in \mathcal{F}.
\end{equation}

(b) The  functions from  $  \mathcal{F}$ grow at most  like the  Gaussian
  (see e.g. \cite{LyuSe}),  more precisely,
\begin{equation}
\label{growth_estimate}
|F(z)| = o(1) e^{\frac \pi 2 |z|^2}, \ \text{as} \ z \to \infty, \ F\in \mathcal{F}.
\end{equation}

(c) The collection  of monomials
\begin{equation}
\label{monomials}
e_n(z)=\Big(\frac{\pi^n}{n!}\Big)^{1/2} \, z^n, \ n=0,1,\ldots
\end{equation}
forms an orthonormal basis in $\mathcal{F}$.

(d) Define the Bargmann transform \ \ of a function $f\in L^2(\mathbb{R})$  by
\begin{equation}
 \label{bt}
  f \mapsto \cB f(z)=F(z)= 2^{1/4}e^{-\pi z^2 /2}\int_\mathbb{R} f(t)e^{-\pi
    t^2}e^{2\pi t z} dt.
  \end{equation}

\begin{prop}
\label{unitary}
The Bargmann transform \  is a unitary mapping between $L^2(\mathbb{R})$ and $\mathcal{F}$.
\end{prop}

(e) The  Hermite functions \  \  are defined by
\begin{equation}
\label{HFS}
H_n(t)= c_ne^{\pi t^2} \frac {d^n}{dt^n} e^{-2\pi t^2}, \ n=0,1,2, \dots
, .
\end{equation}
where the coefficients $c_n$ are chosen in order to have $\|H_n\|_2=1$.
It  is a classical result that  the set of Hermite functions
$\{H_n\}_{n=0}^\infty$ forms an  orthonormal basis in  $L^2(\mathbb{R})$.

Their image under the Bargmann transform \  is $\{e_n\}_{n=0}^\infty$  -- the natural orthonormal basis in $\mathcal{F}$:
\begin{equation}
\label{image}
\cB H_n (z) = e_n(z), \ n=0, 1, \ldots \ .
\end{equation}

(f) In what follows we  identify $\mathbb{C}$ and $\mathbb{R}^2$. In particular
for each $\zeta=\xi +i\eta \in \mathbb{C}$ we write
$\pi_\zeta=\pi_{(\xi,\eta)}$.
 Define the shift $\beta_\zeta: \mathcal{F} \to \mathcal{F}$ in the Fock space \ \  by
\begin{equation}
\label{fock_shift}
 \beta_\zeta F(z) = e^{i\pi \xi \eta} e^{-\pi |\zeta|^2/2}e^{\pi {\zeta} z} F(z-\bar{\zeta}).
 \end{equation}
Then $\beta _\zeta $ is unitary on $\mathcal{F} $,  and the   Bargmann transform \  intertwines
the Fock space \  shift and the time-frequency shift:
\begin{equation}
\label{intertwines}
\beta_\zeta \cB = \cB \pi_\zeta.
\end{equation}

Based on this proposition one can easily obtain inner product of a
function $f\in L^2(\mathbb{R})$ with the time-frequency   shifts of a
Hermite function as in~\cite{BS93}. This quantity is    the
short-time Fourier
transform with respect to a  Hermite function.

\begin{prop}
\label{hermite_stft}
Let $f\in L^2(\mathbb{R})$ and  $F(z)=\cB f(z)$. Then, for all $\zeta \in \mathbb{C}$,
\begin{equation}
\label{brcmn_stft}
\langle f, \pi_\zeta H_n \rangle_{L^2(\mathbb{R})}=
               \frac 1{\sqrt{\pi^n n!}}e^{-i\pi \xi \eta} e^{-\pi |\zeta|^2/2}
                    \sum_{k=0}^n \binom{n}{k}  (-\pi {\zeta})^k F^{(n-k)}(\bar{\zeta}) .
\end{equation}
\end{prop}
\begin{proof}
  Using the intertwining property~\eqref{intertwines}, we have
\begin{eqnarray*}
  \langle f, \pi_\zeta H_n \rangle_{L^2(\mathbb{R})}&=&  \langle F, \beta_\zeta \cB
  H_n \rangle_\mathcal{F} \\
&=&     \Big(\frac {\pi^n}{n!} \Big)^{1/2} \, e^{- i\pi \xi \eta}\,  e^{-\pi |\zeta|^2/2}
      \langle F(z), e^{\pi {\zeta} z} (z-\bar{\zeta})^n\rangle \\
&=&  \Big  (\frac {\pi^n}{n!} \Big  )^{1/2} e^{-i\pi \xi \eta}
e^{-\pi |\zeta|^2/2}\sum_{k=0}^n \binom{n}{k}   (- {\zeta})^k
\langle F(z), z^{n-k}e^{\pi {\zeta} z}\rangle      \\
&=&   \Big (\frac {\pi^n}{n!} \Big )^{1/2} e^{-i\pi \xi \eta} e^{-\pi |\zeta|^2/2}\sum_{k=0}^n \binom{n}{k} (-  {\zeta})^k  \pi^{-n+k} \frac{d^{n-k}}{d\bar{\zeta}^{n-k}}  \langle F(z),e^{\pi {\zeta} z}\rangle.
\end{eqnarray*}
  It remains to apply relation  \eqref{reproducing}.
  \end{proof}
\smallskip

Finally we give in this section a description of the space  $M^1$ in
the Bargmann transform  terms.  Since $|\langle f , \pi _{\bar{z}} H_0
\rangle | = |\mathcal{B}f(z)| \, e^{-\pi |z|^2/2}$
(e.g.,\cite{Charlybook} or \cite{Folland}), we obtain
the following.

\begin{prop}
\label{p_one_barg}
A function $f\in L^2(\mathbb{R})$  belongs to $M^1$ if and only if
its Bargmann transform $F(z)=\cB f$ satisfies
\begin{equation}
\label{m_one_barg}
\|f\|_{M^1} = \int_\mathbb{C} |F(z)|e^{-\pi |z|^2 / 2} dm_z < \infty.
\end{equation}
\end{prop}
This is just a reformulation of condition \eqref{m_one}.

\subsection{Estimates of $\sigma$-function}
\label{estimates}

 In this section we collect  definitions and known facts
 about  the Weierstrass functions.  They will be used in the next section.

 Given two numbers   $\omega_1, \omega_2 \in \mathbb{C} $ such that
  \begin{equation}
 \label{positive}
 \Im  \Big (  \omega_2/\omega_1 \Big ) >0.
 \end{equation}
 we consider the  lattice $\Lambda=\{m_1\omega_1+m_2\omega_2; m_1,m_2 \in
 \bZ\}\subset \mathbb{C}$. The numbers    $\omega_1, \omega_2 \in \mathbb{C} $ are
 called {\em periods }  of the lattice $\Lambda$.

   In the previous section we used   lattices of  the form
 $\Lambda = A \bZ^2 \subset \mathbb{R}^2$, where $A$ is an invertible 2$\times$2 matrix.
   The  two constructions coincide after the natural identification of
$\mathbb{R}^2$ and $\mathbb{C} $,  if  $\det A >0$. In this case    $\omega_1, \omega_2$
correspond to the columns of  $A$.

Let $Q=\{(x,y); 0\leq x <1, 0\leq
y <1\}$.
The parallelogram $\Pi _\Lambda = AQ$ with based on
$\omega_1$ and $\omega_2$ is a fundamental domain  for  $\Lambda$
(also called the {\em period    parallelogram}).
 Its area can be  expressed through the periods as
\begin{equation} \label{area}
s(\Lambda ) = \text{Area}(\Pi _\Lambda)= \det A=  \Im (\bar {\omega}_1\omega_2)
= -\frac {i} 2 (\bar{\omega}_1\omega_2- \bar{\omega}_2\omega_1).
  \end{equation}

Next  consider the following Weierstrass functions:
 \begin{equation}
 \label{Pfunction}
\mathcal{P}(z)= \frac 1{z^2} + \sum_{\omega \in \Lambda'}   \left \{
      \frac 1 {(z-\omega)^2} - \frac 1 {\omega^2}  \right \},
\end{equation}
\begin{equation}
\label{zeta}
\zeta(z)= \frac 1 z + \sum_{\omega \in \Lambda'}
     \left \{ \frac 1 {z-\omega} + \frac 1 \omega + \frac z {\omega^2 } \right \},
\end{equation}
and
\begin{equation}
\label{sigma}
\sigma(z)= z \prod_{\omega \in \Lambda'}  \left ( 1 - \frac z \omega \right )
    e^{\frac z \omega + \frac {z^2}{2\omega^2}},
\end{equation}
where  $\Lambda'= \Lambda \setminus \{0\}$. The lattice
$\Lambda$ plays a special role for the Weierstrass functions: $\mathcal{P}$  has poles
of order two precisely on $\Lambda $,  $\zeta$ has  poles
of order one, and  $\sigma$ has simple zeros precisely on $\Lambda $.

We list some  basic properties of the Weierstrass  functions. See,
e.g.,  \cite{Ahiezer} for detailed proofs.

\begin{itemize}
\item
The $\mathcal{P}$-function  is  elliptic, i.e.,
\begin{equation}
\label{P1}
\mathcal{P}(z)=\mathcal{P}(z+\omega_k), \quad z\in \mathbb{C}, \ k=1,2.
\end{equation}
\item
The function $\zeta$ changes by a constant when its argument changes by a lattice point:
\begin{equation}
\label{eta}
\eta_k:=\zeta(z+\omega_k)- \zeta(z)= \text{Const}\ , \ k=1,2.
\end{equation}
\item The constants $\eta_k$
satisfy the  Legendre relation:
\begin{equation}
\label{legendre}
\eta_1 \omega_2 - \eta_2 \omega_1 = 2\pi i .
\end{equation}
\item
The Weierstrass $\sigma$-function  is quasi-periodic, it satisfies the
following relation
\begin{equation}
\label{sirel}
\sigma(z+\omega_k)= - \sigma(z) e^{\eta_kz + \frac 12 \eta_k \omega_k}
\qquad k=1,2 \, .
\end{equation}
\end{itemize}

\medskip
In order to understand  the growth properties  of the Weierstrass
$\sigma $-function in dependence of the lattice
$ \Lambda$, we follow an elegant argument of Hayman~\cite{Hayman}. He   realized
that, after a proper normalization and growth
compensation, the  absolute value of $\sigma_\Lambda $ becomes a doubly
periodic function.  We include   here  this arguments  in order to make our presentation self-contained.
 \begin{prop}
 \label{lemmasigma}
 Set
 \begin{equation}
 \label{sconstants}
 \alpha(\Lambda)=  \frac{i\pi}{\bar{\omega}_1\omega_2- \bar{\omega}_2\omega_1}=
 \frac{\pi}{2 s(\Lambda)},
 \quad
 a(\Lambda)= \frac 12 \frac {\eta_2\bar{\omega}_1-\eta_1 \bar{\omega}_2}
                      {\omega_1\bar{\omega}_2-\omega_2 \bar {\omega}_1},
\end{equation}
and
\begin{equation}
\label{sigmaa}
\sigma_\Lambda(z)= \sigma(z)e^{a(\Lambda)z^2}.
\end{equation}
Then the function $|\sigma_\Lambda(z)|e^{-\alpha(\Lambda) |z|^2}$ is  periodic  with periods
$\omega_1$ and $\omega_2$.
\end{prop}

 \medskip
The periodicity implies a growth estimate for the
modified sigma-function $\sigma _\Lambda$.

\begin{prop}
\label{growth}
  Set
  \begin{equation}
  \label{C}
  c(\Lambda )= \sup _{z\in \Pi _\Lambda }
  |\sigma_\Lambda(z)| e^{-\alpha(\Lambda) |z|^2}.
  \end{equation}
   Then
\begin{equation}
\label{estimate}
|\sigma_\Lambda(z)| \leq c(\Lambda ) \, e^{\frac{ \pi}{ 2 s(\Lambda)}|z|^2},
\quad \forall z\in \mathbb{C}\, .
\end{equation}
For  each  $\epsilon >0$ we have
\footnote{Here and in what follows the sign $\asymp$ means that the ratio of the left- and right-
hand sides is located between two positive constants.}
\begin{equation}
\label{asymp}
|\sigma_\Lambda(z)|\asymp e^{\alpha(\Lambda ) |z|^2}, \quad \text{whenever dist}(z,\Lambda)>\epsilon.
\end{equation}
\end{prop}

\begin{proof}  The function
$|\sigma_\Lambda(z)|e^{-\alpha(\Lambda) |z|^2}$ is bounded in
$\Pi_\Lambda$ and has its only zeros at the vertices of $\Pi _\Lambda
$. Hence it is bounded away from $0$ on every compact subset of $\Pi
_\Lambda$,  which does not contain its vertices
and relation \eqref{asymp}  follows now by the
periodicity.
\end{proof}
\medskip

\noindent {\em Proof of Proposition \ref{lemmasigma}.} It follows from \eqref{sirel} that
\[
\Big |  \sigma(z+\omega_k) e^{a(\Lambda)(z+\omega_k)^2}   \Big |
e^{-\alpha(\Lambda)|z+\omega_k|^2}=
\Big |  \sigma(z) e^{a(\Lambda)z^2}\Big |e^{-\alpha(\Lambda) |z|^2} e^{\Re A_k(z)}, \ k=1,2,
\]
where
 \begin{equation}
\label{ak}
A_k(z)=  z\Big (\eta_k+2a(\Lambda)\omega_k -
          2\alpha(\Lambda) \bar{\omega}_k\Big )  +
                 \Big ( \frac 12 \eta_k \omega_k  +a(\Lambda)\omega_k^2-
                                    \alpha(\Lambda) |\omega_k|^2 \Big ),   \ k=1,2.
\end{equation}
An explicit calculation, based on relations \eqref{sconstants}  and \eqref{legendre}
shows that
\[
\eta_k+2a(\Lambda)\omega_k -  2\alpha(\Lambda) \bar{\omega}_k= 0, \ k=1,2,
\]
and also
\[
\Re \Big (\frac 12 \eta_k \omega_k  +a(\Lambda)\omega_k^2-
        \alpha(\Lambda) |\omega_k|^2\Big )= 0, \
k=1,2.
\]
Proposition \ref{lemmasigma} now follows.
\hfill $\Box$

\bigskip

\section{Frames with Hermite functions.}

\subsection{Hermite superframes}

In this section we prove  Theorem  \ref{mainchar1}   and thus give a complete
characterization of Gabor superframes with Hermite functions.

We divide the proof into several steps.
First  we will use the matrix form of the Wexler-Raz  biorthogonality relations
  (\ref{wr_orthogonality}) and  translate these relations  to
  an interpolation problem on  Fock space.

Let, as earlier,  $\boh=(H_j)_{j=0}^n$ be the vector-valued window consisting  of the
first $n+1$ Hermite functions.
  For each
$\bs{\gamma}=(\gamma_j)_{j=0}^n\in L^2(\mathbb{R},\mathbb{C} ^{n+1})$ denote
\label{dual_gamma}
\begin{equation} G_j=\cB\gamma_j,  \ j=0,1,\ldots \ ,n.
\end{equation}
Taking $\bog=\boh$ in the biorthogonality relations
\eqref{wr_orthogonality} and using
Proposition \ref{hermite_stft},   we can rewrite these relations as an
interpolation problem  for functions $G_j\in \mathcal{F}$:

\begin{eqnarray}
\label{new_relations}
\langle \gamma_j, \pi_\mu H_l\rangle_{L^2(\mathbb{R})} &=&  e^{-i\pi \Im \mu \Re \mu}e^{-\pi|\mu|^2/2}
       \frac 1 {\sqrt{\pi^l l!}}\sum_{k=0}^l \binom lk
       (-\pi{\mu})^kG_j^{(l-k)}(\bar{\mu})\\
&=&     s(\Lambda)\delta_{\mu,0}\delta_{j,l}, \quad j,l=0,1,\ldots \
              ,n;  \ \mu\in \Lambda^\circ.    \nonumber
\end{eqnarray}

Thus $\mathcal{G}(\boh, \Lambda)$ is a frame for $L^2(\mathbb{R}, \mathbb{C} ^{n+1})$, if
and only if there exist functions   $G_j\in \mathcal{F}$ satisfying \eqref
{new_relations} and if $\mathcal{G} (\bs{\gamma} ,\Lambda )$ is a Bessel sequence in
$\L^2(\mathbb{R}, \mathbb{C}^{n+1}) $.

This interpolation problem can be rewritten in a  simpler way. Indeed,
for each
$\mu\in \Lambda^\circ\setminus \{0\}$  and    $j\in
\{0,1,\ldots \ ,n\}$,
the system \eqref {new_relations}   is a triangular linear system in
the variables
$\{G_j^{(m)}(\bar{\mu})\}_{m=0}^n$ with non-zero diagonal coefficients and
zero right-hand side. Clearly, it has just zero solutions. Thus if
$\boG=(G_j)_{j=0}^n$ satisfies
\eqref {new_relations}, then $G_j^{(m)}(\bar{\mu})=0, \ $
$j,m=0,1,\ldots  ,n, \ $ $\mu\in \Lambda^\circ\setminus \{0\}$. For
$\mu=0$, the   relations  \eqref {new_relations} take the form
\[
\frac 1 {\sqrt{\pi^l l !}} G_j^{(l)}(0)= s(\Lambda) \delta_{j,l}, \ j,l=0,1, \ldots \ ,n.
\]
By adjusting the normalization of the $G_j$,  we obtain the following
statement.

\begin{prop}
\label{delta_prop}
     The Gabor system  $\mathcal{G} (\mathbf{h} , \Lambda ) $ is a  Gabor superframe for
$L^2(\mathbb{R} , \mathbb{C} ^{n+1})$ if and only if\ there exist $n+1$
 functions $\gamma _j \in L^2(\rd ) $  such that $\mathcal{G}(\gamma _j, \Lambda )$ is a
 Bessel sequence and   $G_j=B\gamma _j, j=0, \dots , n$,  satisfy the
 (Hermite) interpolation problem
\begin{equation}
  \label{eq:herm}
G_j^{(\ell )} (\bar{\mu}) = \delta _{\mu ,0} \, \delta _{j,\ell} \qquad
\text{ for } \, \mu \in \Lambda ^{\circ}, j,\ell = 0, \dots, n \, .
\end{equation}
 \end{prop}

\smallskip

\begin{proof}[Proof of Theorem~\ref{mainchar1}]
  \noindent{\em Sufficiency } of \eqref{main_estimate}.
We use the Weierstrass functions described in  Section \ref{estimates}.
 Construct the Weierstrass functions corresponding the lattice
 $\overline{\Lambda^\circ}$, as  given by \eqref{eq:c2}.  Let the constants
 $\alpha(\overline{\Lambda^\circ}), a(\overline{\Lambda^{\circ}})$
 be defined by  \eqref{sconstants}  and the
 function $\sigma_{\overline{\lambda^\circ}}$ be defined by \eqref{sigmaa},
 where the $\sigma$-function is constructed for the lattice
 $\overline{\Lambda^\circ}$. Consider the function
 \begin{equation}
 \label{Sfunction}
 S(z)=\big ( \sigma_{\overline{\Lambda^\circ}}(z)  \big )^{n+1}=
         \sigma   (z)^{n+1}
e^{a(\overline{\Lambda^\circ})(n+1)z^2}.
 \end{equation}

 The zero set of $S$ is  $\overline{\Lambda^\circ}$,  and  the
 multiplicity of each zero is precisely
 $n+1$.
In our case $\alpha(\overline{\Lambda^{\circ}})
 =  \frac{\pi}{2 s(\Lambda ^\circ )} =  \frac \pi 2
s(\Lambda)$ and the growth estimates \eqref{estimate} and
\eqref{asymp} can be rewritten as
 \begin{eqnarray}
\label{Sestimate}
|S(z)| &\leq & e^{\frac \pi 2 (n+1) s(\Lambda)|z|^2}, \  \text{for
  all} \ z\in \mathbb{C}, \\
   |S(z)| &\asymp &  e^{\frac \pi 2 (n+1) s(\Lambda)|z|^2},
  \quad \text{ if  }   \text{dist}\ (z,\overline{\Lambda^\circ})>\epsilon.
  \label{Sestimate2}
  \end{eqnarray}

Now assume that $s(\Lambda ) < (n+1)^{-1} $. Then the functions
 \begin{equation}
 \label{Sj}
 S_m(z) = \frac 1 {z^{n+1-m}} S(z), \ m=0,1,\ldots \ ,n
 \end{equation}
 belong to $\mathcal{F}$. They have zero of order $n+1$ at each $\mu \in
 \overline{\Lambda^\circ} \setminus \{0\}$,  and also
 \[
 S^{(l)}_m(0)=0, \  \text{for} \ 0\leq l < m-1, \ \text{and} \ S^{(m)}_m(0)\neq 0.
 \]

 The solutions $G_j$ to the interpolation problem \eqref{eq:herm} can
 be now found in the form
 \begin{equation}
 \label{solutions}
 G_j=\sum_{m=j}^n c_{m,j}S_m, \,\,\,\,  j=0,1,\ldots \ ,n.
 \end{equation}

  These functions have a zero of multiplicity $n+1$ at all points from
  $\overline{\Lambda^\circ} \setminus \{0\}$, while, for each $j$,  the  condition
  $G_j^{(l)}(0)=\delta_{j,l}$ leads one to a triangular system of linear equations
  with respect to the coefficients $ \{c_{m,j}\}_{m=j}^n$ with non-zero diagonal entries.
  Clearly this system has a (unique) solution.

It remains to mention that each  system
$\mathcal{G}(\gamma _j, \Lambda )$ is a  Bessel sequence.  Each $G_j$
inherits its growth from the functions $S_j$ and from $S$, thus
\eqref{Sestimate} and Proposition~\ref{p_one_barg} imply that $\gamma
_j \in M^1(\mathbb{R} ) $ for $j=0, \dots , n$. Now apply  Proposition~\ref{bessel}.

  \smallskip

  \noindent{\em Necessity.}
 Let now $s(\Lambda)\geq (n+1)^{-1}$. The growth estimate~
  \eqref{Sestimate2} implies that
  \begin{equation}
  \label{belowestimate}
   |S(z)| \geq \text{Const}\   e^{\frac \pi 2  |z|^2},
 \quad \text{whenever} \,  \text{dist}\ (z,\overline{\Lambda^\circ})>\epsilon.
\end{equation}

Now assume that  $\mathcal{G}(\boh, \Lambda)$ is a frame for $L^2(\mathbb{R},\mathbb{C}
^{n+1})$. Then  there exists a system of functions
$G_0, G_1,\ldots \ ,G_n\in \mathcal{F}$ which satisfy the interpolation
problem~\eqref{eq:herm}. The function
$G_n$ has zeros of multiplicity $n+1$ at $\Lambda^\circ\setminus \{0\}$ and
a zero of multiplicity $n$ at the origin. Therefore
\[
\Phi(z)= \frac {zG_n(z)}{S(z)}
\]
is an entire function. Estimates \eqref{belowestimate}
and~\eqref{growth_estimate}  yield
\[
|\Phi(z)| = o(|z|), \ z \to \infty, \ \text{dist}\ (z, \overline{\Lambda^\circ}) >\epsilon.
\]
By the maximum principle, the  restriction
$ \text{dist}\ (z, \Lambda^\circ) >\epsilon$   can be removed,
and $\Phi $ is in fact a bounded entire function. The
Liouville theorem implies that  $\Phi$ is a constant, or
\[
G_n(z)= C \frac{S(z)}z.
\]
But  $z^{-1}S(z) \not \in \mathcal{F} $, this follows again from \eqref {belowestimate}. Therefore
$C=0$ and hence $G_n$ is identically zero in contradiction to  \eqref{eq:herm}. This completes the proof of necessity.
\end{proof}

\noindent{\em Remark.} It follows from the density theorem
(Proposition \ref{density} ) that $\mathcal{G} (\boh , \Lambda )$ cannot be a frame for
$s(\Lambda ) > \frac{1}{n+1}$. Our proof shows that we must also
exclude the case of the critical density $s(\Lambda ) =
\frac{1}{n+1}$. This is  a Balian-Low type phenomenon (see the
review~\cite{BHW95}). We conjecture that  in general, if $g_j \in M^1,
j=1, \dots , n$ and $\mathcal{G} (\bog , \Lambda )$ is a Gabor frame, then
$s(\Lambda ) < \frac{1}{n}$,
but we will not pursue this question
further in this work.

\subsection{Hermite frames}
Next  we consider scalar-valued Gabor frames with Hermite
functions.

We remark that if for some $\bog \in L^2(\mathbb{R}, \mathbb{C}^{n+1})$ and lattice
$\Lambda \subset \mathbb{R}^2$ the system
 $\mathcal{G}(\bog, \Lambda)$ is a vector-valued frame in
$L^2(\mathbb{R}, \mathbb{C}^{n+1})$, then, trivially,  for each $j=0,1,\ldots \ ,n$, the Gabor system
$\mathcal{G}(g_j, \Lambda)=\{\pi_\lambda g_j, \lambda \in \Lambda\}$ is a frame for
$L^2(\mathbb{R})$. More generally, for each
 $\mathbf{c}\in \mathbb{C} ^{n+1}$,  $ \mathbf{c}\neq 0$, the system
 $\mathcal{G}(h, \Lambda)=\{\pi_\lambda h, \lambda \in \Lambda\}$  with
$h = \sum _{j=0}^n c_j g_j$  also is a frame in  $L^2(\mathbb{R})$.
By applying this observation to the window $\bog = (H_0,\ldots \ , H_n)$
we obtain Proposition \ref{simple_hermite}, which gives
a condition for the one dimensional system with the Hermite window to be a frame for  $L^2(\mathbb{R})$. Actually a    slightly more general result   related to   linear
combinations of Hermite functions holds true

\begin{prop}
  \label{hermitegeneral}
 Let $n\in \bZ , n\geq 0$ and $h = \sum _{k=0}^n c_k H_k$, \, $\sum_0^n |c_k|\neq 0$.   If
 $s(\Lambda ) < (n+1)^{-1}$,    then
$\mathcal{G} (h , \Lambda ) $ is a frame for $L^2(\mathbb{R} )$.
\end{prop}

 \noindent{\em Remark.} Clearly the estimate above is not optimal in case
 $c_n=0$.
 \smallskip

Amazingly enough the sufficient density of Theorem~\ref{hermitegeneral} might be
sharp, as is suggested by the following counter-example.

\begin{prop}
  \label{counter}
If $\Lambda = a \bZ \times b \bZ $ and $ab = 1/2$, then
$\mathcal{G} (H_{2n+1}, \Lambda )$ is not a Gabor frame for $L^2(\mathbb{R})$ for all
integers $n\geq 0$.
\end{prop}
\begin{proof}
We use a Zak transform argument. For $a>0$ the Zak transform is defined as
\begin{equation}
\label{zak}
Z_a f(x,\xi)=\sum_{k=-\infty}^\infty f(x-ak)e^{2i\pi a k \xi}.
\end{equation}
We refer the reader to  \cite[Sec.~8.3]{Charlybook} for a detailed
discussion of  its properties.   In particular it is a unitary mapping
between  $L^2(\mathbb{R})$
 and $L^2(Q_{a,1/a})$ -- space of square summable functions in the  rectangle
 $Q_{a,1/a}=[0,a)\times[0,1/a)$.
For the case $ab=1/2$ (more generally for the case $(ab)^{-1}\in \bZ$)
the Gabor   frame operator is unitarily equivalent to a
multiplication operator  on $
L^2(Q_{a,1})$ by means of  the Zak transform. Precisely,
\[
Z_a S_{H_n,H_n} f(x,\xi) =
     \big ( |Z_aH_n(x, \xi)|^2 + |Z_a H_n (x-\frac a2, \xi)|^2 \big )
               Z_af(x,\xi), \quad f \in   L^2(\mathbb{R}).
\]
Consequently  $\mathcal{G} (H_n, \Lambda )$ is a frame for  $L^2(\mathbb{R})$ if and only if\
\begin{multline*}
 \quad 0< \inf_{Q_{a,1}} \{ ( |Z_aH_n(x, \xi)|^2 + |Z_a H_n (x-\frac
 a2, \xi)|^2 \big )  \}   \\
\leq \sup_{Q_{a,1}} \{ ( |Z_aH_n(x, \xi)|^2 + |Z_a H_n (x-\frac a2, \xi)|^2 \big )  \} <\infty,
         \quad
 \end{multline*}
 see relation (8.21) in~\cite{Charlybook}.

Clearly,   $Z_aH_n(x,\xi)$ is
 a continuous function.  Since $H_{2n+1}$ is an odd function, its Zak
 transform   satisfies $Z_aH_{n+1}(0,0)=Z_aH_{n+1}(\frac a2, 0)=0$. This
 contradicts  the above criterium, therefore $\mathcal{G} (H_{2n+1}, \Lambda
 )$ cannot be a frame.
 \end{proof}

In~\cite[Prop.~3.3]{glyu} we showed that $\mathcal{G}(H_n, \Lambda )$ is a
frame, if and only if there exists a $\gamma_n \in L^2(\mathbb{R} )$, such
that $\mathcal{G}(\gamma _n, \Lambda )$ is a Bessel system and the
Bargmann transform $G_n = \mathcal{B}\gamma _n$ is in $\mathcal{F}$ and  satisfies the interpolation
problem
\begin{equation}
  \label{Eq:3.2}
  \sum _{k=0}^n   \binom{n}{k} \,
         (-\pi \mu )^k G_n^{(n-k)} (\bar{\mu })   = \delta _{\mu , 0} \, \qquad
         \,\, \forall \mu \in \Lambda ^\circ \, .
\end{equation}

Note that \eqref{Eq:3.2} coincides with   \eqref{new_relations} for $j=l=n$
and that for $s(\Lambda ) < 1/(n+1)$ the function
\begin{equation}
  \label{eq:dualcc}
  G_n(z) = S_n(z) = c z^{-1}\sigma_{\overline{\Lambda^\circ}}^{n+1}(z)
\end{equation}
is a solution of \eqref{Eq:3.2} in $\mathcal{F}$.

If $s(\Lambda ) \geq 1/(n+1)$, then we do not know whether
\eqref{Eq:3.2} has any solution in $\mathcal{F}$. The difficulty is
that this interpolation problem is not entirely holomorphic, and
the standard complex variable methods do not seem sufficient to investigate this
problem. In the light of Proposition~\ref{counter} it is conceivable
that the sufficient condition  $s(\Lambda) < 1/(n+1) $ in
Proposition~\ref{hermitegeneral} is also necessary.

\subsection{Norms and Decay of the Dual Window}

In the rest of this section we assume that $ s(\Lambda)<\frac 1 {n+1}$,
and  we estimate  the $L^2$ and $M^1$-norms of the
dual window for the frame
$\mathcal{G} (H_n , \Lambda )$.  Though the estimates are  simple, they  seem
to be new even for the Gaussian case  $n=0$.

 \begin{lemma} \label{norms}
Assume that $ (n+1)s(\Lambda)<1$ and let $\gamma _n$ be the dual
window with Bargmann transform $G_n$ defined in~\eqref{eq:dualcc}.
Setting  $\kappa = \frac{1-(n+1)s(\Lambda)}{3-(n+1)s(\Lambda)}$, then
\begin{equation}
  \label{eq:cne2}
    |\gamma_n(t)|+ |\widehat{\gamma _n}(t)| \leq C e^{-\pi \kappa t^2}
    \, .
\end{equation}
Furthermore,
\begin{equation}
\label{m_one_norm}
\|\gamma\|_{M_1} \leq  4  c({\overline{\Lambda ^\circ}})^{n+1}  \frac{
  1}{1-(n+1)s(\Lambda   )} ,
\end{equation}
and
 \begin{equation}
\label{focknorm}
\|\gamma _n\|_{L^2(\mathbb{R})}^2 \asymp    \log
\frac{1}{1-(n+1)s(\Lambda )},
\end{equation}
where $c(\Lambda )  $ is defined in~\eqref{C}.
\end{lemma}
\begin{proof}
The estimates involve a routine calculation with Gaussians and are a
consequence of the growth estimate~\eqref{estimate} of Proposition~\ref{growth}.
Set $\rho = 1-(n+1)s(\Lambda )$.

For the decay property, we use the inversion formula for the
short-time Fourier transform (e.g.,~\cite[Prop.~3.2.3]{Charlybook})
which states
$$
\gamma _n(t) = \int _{\mathbb{C} } \langle \gamma _n , \pi _z H_0\rangle \,
\pi _z H_0(t) \, dm_z
$$
with absolute convergence of the integral for each $t\in\mathbb{R} $. Since
by \eqref{estimate}
$$
|\langle \gamma _n , \pi _{\overline{z}} H_0\rangle| =
|\mathcal{B}\gamma _n(z) |  e^{-\pi |z|^2/2} \leq C e^{-\pi \rho
  |z|^2/2}
$$
and since $\pi _{x+i\xi} H_0(t) = e^{2\pi i \xi t} H_0(t-x)$, we obtain
that
$$
|\gamma _n (t) | \leq C \int _{\mathbb{R} } e^{-\pi \rho \xi ^2/2}\, d\xi \int _{\mathbb{R}
} e^{-\pi \rho x ^2/2} e^{-\pi (t-x)^2} \, dx \, .
$$
The convolution of the two Gaussians in this integral is a multiple of
the Gaussian $e^{-\pi \frac{\rho}{2+\rho}t^2}$ by the semigroup
property of Gaussians with respect to
convolution~\cite[Lemma~4.4.5]{Charlybook}.

The $M^1$-norm of  $\gamma _n$ is readily estimated by
\begin{eqnarray*}
  \|\gamma _n\|_{M^1} &=& \int _{\mathbb{C} } |G_n(z) | \, e^{-\pi
    |z|^2/2}\, dm_z \\
&\leq & c(\overline{\Lambda ^\circ})^{n+1}  \int _{\mathbb{C} } \frac{1}{|z|}
\, e^{-\pi \rho     |z|^2/2}\, dm_z  \\
&=& 2\pi c(\overline{\Lambda ^\circ})^{n+1} \int _0^\infty  e^{-\pi
  \rho r/2}\, dr = 4 c(\overline{\Lambda ^\circ})^{n+1}
\frac{1}{\rho}\, .
\end{eqnarray*}

For the estimate of $\|\gamma _n \|_2^2$ we have similarly
\[
\|\gamma\|_{L^2(\mathbb{R})}^2 = \|G\|^2_\mathcal{F} =
\int_{\mathbb{C}} |z|^{-2} |\sigma_{\overline{\Lambda^\circ}} (z)|^{2(n+1)} e^{-\pi |z|^2} dm_z.
\]
When $|z|\leq 1$  the integrand is bounded uniformly,
 so this
part does not bring essential contribution into the whole norm. It
follows now from
\eqref{asymp} that
\begin{multline*}
\|\gamma\|_{L^2(\mathbb{R})}^2 \asymp
\int_{|z|>1}|z|^{-2} e^{-\pi \rho  |z|^2 } dm_z  \asymp
 \int_1^\infty  \frac 1 r e^{-\pi \rho r^2} dr
    \asymp  - \log (1-(n+1)s(\Lambda ))\, .
 \end{multline*}
This is the desired inequality.
\end{proof}
\noindent{\em Remarks.} 1. In general, the dual window $\gamma _n$ is not
the canonical dual window (given by $S_{H_n,H_n}^{-1}
H_n$). Janssen~\cite{janssen96} observed that for $n=0$ and $\Lambda = a\bZ
\times (aN)^{-1} \bZ $ with a positive integer $N$ the canonical dual
window decays at most exponentially, whereas the dual window
constructed by means of the $\sigma $-function possesses Gaussian
decay.

2. The Gaussian decay~\eqref{eq:cne2} of the dual window of
$\mathcal{G}(H_n, \Lambda )$ is usually expressed by saying that
$\gamma _n$ belongs to the Gelfand-Shilov space $S^{1/2}_{1/2}$. This
is considered the smallest reasonable space of test functions in
time-frequency analysis. An important consequence of this observation
is the existence of universal Gabor frames, i.e., the frame
$\mathcal{G}(H_n, \Lambda )$  is a (Banach) frame for all reasonable
modulation spaces. We refer to~\cite{gro07c} for a precise statement
and further details.

\vspace{ 3mm}

Finally let us briefly describe how the lower frame bound of the  Gabor
frame $\mathcal{G}(H_0, \Lambda )$ with Gaussian window behaves, when
the  lattice approaches  the critical size $s(\Lambda ) =1$.
Since the constants in Lemma~\ref{norms} depend on (the excentricity
of) the lattice $\Lambda $,   we fix a lattice $\Lambda $ with size
$s(\Lambda )=1$ and study the
behavior of lower frame bound of $\mathcal{G}(H_0, q \Lambda )$,
as $q $ tends to $1$. As long as $q <1$,
$\mathcal{G}(H_0, q \Lambda )$ is a frame by the classical results
in~\cite{lyu,sw}, however, when $q  =1$, then
$\mathcal{G}(H_0, q \Lambda )$ cannot be a frame by the Balian-Low
theorem~\cite{BHW95}.  The upper frame bound can be controlled
uniformly with Proposition~\ref{bessel}, therefore    the lower frame
bound must $A$
converge to $0$ as $q \to 1$.

\begin{prop}
  Assume that $s(\Lambda )=1$ and $q   <1$.
   Let  $A_q$ denote the optimal lower
  frame bound of   $\mathcal{G}(H_0,q \Lambda )$. Then
$$
A_q \geq c (1-q ^2)^2 = c (1-s(q \Lambda ))^2
$$
for some constant independent of $q $.
\end{prop}

\begin{proof}
Let $\gamma _0$ be the dual window defined by ~\eqref{eq:dualcc}.
 Corollary~\ref{lowerbound}  of  the
  Wexler-Raz relations  and Lemma~\ref{norms} imply that
  $$
A_q \geq (n(q \Lambda ) \|\gamma _0\|_{M^1}^2)^{-1} \geq
 \Big(16 n(q \Lambda) c(\overline{(q \Lambda )^\circ})^2 \Big)^{-1}
 (1-s(q \Lambda ))^2
$$
Thus we need to show that the two constants $n$ and $c$ are bounded,
as long as $q $ is bounded away from $0$, $q \geq 1/2$, say.
Clearly the constant $n(q \Lambda )$, which measures the maximal number
of lattice points in a unit cube (Lemma~\ref{bessel}), can be bounded uniformly.
As for $c$, which is the supremum of the $\sigma $-function over the
fundamental parallelogram~\eqref{C}, we note that
$(q \Lambda )^\circ= q ^{-1} \Lambda ^\circ$
and that
$\sigma _{q ^{-1}  \Lambda }(z) = q^{-1} \sigma _{\Lambda} (q z)$ by
\eqref{sigma}, \eqref{legendre} and \eqref{sconstants}. Consequently,
$\sup _{1/2\leq q \leq 1} c(q^{-1} \overline{\Lambda ^\circ})
< \infty $, and we are done.
\end{proof}

\section{Appendix}

In this appendix we show how the structural results about Gabor
superframes can be derived from the corresponding well-known results
for scalar Gabor frames.

\begin{proof}[Proof of Janssen's representation, Theorem~\ref{janssen_prop}]
  We look at the $l$-th component of $S\bof$. Using definition
\eqref {foperator} it can be written explicitly as
 \[
 (S\bof)_l= \sum_{\lambda \in \Lambda} \langle \bof, \pi_\lambda \bog \rangle
                               \pi_\lambda \bs{\gamma} _l =
  \sum_{j=1}^n    \sum_{\lambda \in \Lambda}  \langle f_j, \pi_\lambda g_j \rangle
                               \pi_\lambda \gamma_l =
\sum_{j=1}^n S^\Lambda_{g_j,\gamma_l}f_j.
\]
If all $g_j$, $\gamma_j$ are in $M^1$, then Janssen's representation~\eqref{janssen}
can be applied to each of the frame-type operators occurring in the
above sum and we obtain that
\begin{equation}
(S\bof)_l= s(\Lambda)^{-1}\sum_{\mu \in \Lambda^\circ} \sum_{j=1}^n
\langle \gamma_l, \pi_\mu g_j\rangle \pi_\mu f_j=
 \sum_{\mu \in \Lambda^\circ} \big ( \Gamma(\mu) \pi_\mu \bof \big )_l,
\end{equation}
as was to be shown.
\end{proof}

\smallskip

\begin{proof}[Proof of the Wexler-Raz biorthonality conditions,
  Prop.~\ref{wexler_raz}]
 We remark first that time-frequency shift s on a lattice are linearly
independent in the following sense:  if $c=(c_\mu )_{\mu \in \Lambda ^\circ } \in
\ell ^\infty (\Lambda ^\circ )$ and $\sum _{\mu \in \Lambda ^\circ}
c_m \pi (\mu ) = 0$ as an operator from $M^1$ to $(M^1)^* $, then
$c_\mu = 0$ for all $\mu \in \Lambda ^\circ $. See, e.g., \cite{gro07a,
rieffel88}.
If $S_{\mathbf{g}, \mathbf{\gamma}} = \mathrm{I}$, then by Janssen's
representation
$$
f_{\ell } = (S\mathbf{f})_{\ell} =  s(\Lambda)^{-1} \, \sum
       _{\mu \in \Lambda ^\circ } \sum _{j=1}^n \langle \gamma _{\ell } , \pi _\mu
       g_j\rangle \pi _\mu f_j \, ,
$$
and so the linear independence forces $ s(\Lambda)^{-1} \langle \gamma
_{\ell } , \pi _\mu
       g_j\rangle = \delta _{\mu , 0} \delta _{j,\ell }$, or in short
       notation,
$s(\Lambda )^{-1} \Gamma (\mu ) = \delta _{\mu ,0} \mathrm{I}$.

Conversely, if the biorthogonality condition $s(\Lambda )^{-1} \Gamma
(\mu ) = \delta _{\mu ,0} \mathrm{I}$  holds, then obviously
$S_{\mathbf{g}, \mathbf{\gamma}} = \mathrm{I}$.
\end{proof}
\medskip

\begin{proof}[Proof of the Ron-Shen duality, Theorem~\ref{ron-shen}]
Assume first that  $\mathcal{G} (\mathbf{g}, \Lambda )$ is a frame for
$L^2(\mathbb{R}^d, \mathbb{C} ^n)$. Then the
canonical dual is given by $\bs{\gamma} ^\circ  = S^{-1} \mathbf{g}$, and by
general frame theory $\bs{\gamma}  ^\circ $ satisfies $S_{\mathbf{g},
 \bs{\gamma} ^\circ} = \mathrm{I}$. Furthermore, $\mathcal{G} (\bs{\gamma} ^\circ ,
\Lambda )$ is again a frame for
$L^2(\mathbb{R}^d, \mathbb{C}^n)$,  in particular it is a Bessel
sequence. If
\begin{equation}
  \label{eq:vv14}
  f= \sum _{j=1}^n \sum _{\mu \in  \Lambda ^\circ }
c_{j,\mu  } \pi _\mu g_j
\end{equation} for some sequence $(c_{j,\mu })
\in \ell ^2(\{1, \dots , n\}\times \Lambda )$,
then by
the Bessel property of $\mathcal{G} (g_j, \Lambda ^\circ)$ for each $j$, we obtain
that
$$
\|f \|_2^2 = \|\sum _{j=1}^n \sum _{\mu \in \Lambda ^\circ }
c_{j,\mu } \pi _\mu g_j \| \leq  B \sum _{j=1} ^n \sum
_{\mu  \in \Lambda ^\circ } |c_{j,\mu }|^2_2 = B \|c\|_2^2 \, .
$$
For the converse inequality we use the Wexler-Raz relations. If $f \in
L^2(\rd ) $ is given by \eqref{eq:vv14}, then the coefficients are uniquely
determined by
$$
c_{j,\mu } = \langle f, \pi _\mu \gamma _j \rangle  \, .
$$
Again, since $\mathcal{G} (\gamma _j, \Lambda ^\circ )$ possesses the Bessel
property, we find that $\|c\|_2^2 \leq A' \|f\|_2^2$. Altogether we
have shown that the set $\bigcup _{j=1}^n \mathcal{G} (g_j , \Lambda ^\circ )$ is
a Riesz sequence.

Conversely, assume that  $\bigcup _{j=1}^n \mathcal{G} (g_j , \Lambda ^\circ )$
is a Riesz sequence that generates a subspace $\mathcal{K} \subseteq L^2(\rd )
$. Then there exists a biorthogonal basis $\{e_{j,\mu }\}$ contained
in $\mathcal{K}$. By the invariance of the Gabor systems $\mathcal{G} (g_j, \Lambda
^\circ)$, we find that the biorthogonal basis must be of the form
$e_{j,\mu } = \pi _\mu \gamma _j$ for some $\gamma _j \in L^2(\rd ) $. By
the general properties of Riesz bases, $\mathcal{G} (\gamma _j, \Lambda ^\circ
)$ is a Bessel sequence for each $j$, and, after some rescaling, the biorthogonality states
that $s(\Lambda )^{-1} \, \langle \pi (\mu ')\gamma _j, \pi _\mu g _\ell \rangle =
\delta _{\mu,\mu '} \delta _{j,\ell }$. According to the Wexler-Raz
relations, this implies that $\mathcal{G} (\mathbf{g}, \Lambda )$ is a frame
for $L^2(\mathbb{R} ^d, \mathbb{C} ^n) $.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
A film about fears, life decisions, doubts and faith. It tells the story Alice who, faced with an important decision that must be made for her life, works through a series of fears and traumas that have accompanied her throughout her life.
The film symbolically portrays the way in which we, as humans, relate to our internal fears and how uncertainty about the future creates the perfect setting for fears to appear and, in too many cases, dominate us.
Fear blocks us. Fear makes us doubt. Fear lies to us.
Can uncertainty be the best school to help us learn to live with our fears?
How to learn to handle uncertainty?
onsdag 25.01.2023 08:33 - 09:17
onsdag, 25. januar 2023 / 07:13 - 07:57

redpajama_set_name: RedPajamaCommonCrawl


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text:
Court hearing appeal by female ski jumpers
VANCOUVER, British Columbia—A British Columbia Court of Appeal heard arguments Thursday about why female ski jumpers should have the right to compete at the 2010 Winter Olympics.
Fourteen women are appealing a British Columbia Supreme Court ruling that found the Vancouver Olympic organizing committee does not have the power to change a decision made by the International Olympic Committee.
The IOC voted in 2006 against including women's ski jumping in the 2010 Games, saying the sport had not developed enough to meet the basic criteria for participation.
The women skiers argue that the Vancouver Olympic organizing committee, VANOC, is subject to Canada's Charter of Rights and Freedoms, and its failure to include women's ski jumping is discrimination since there will be men's ski jumping at the games.
American jumper Lindsey Van said the case is about fighting for the Olympic ideal.
"You have to have hope and show up or you're already defeated," she said.
Ross Clark, the lawyer representing the women, told the panel of three judges that the Canadian government cannot decide that Swiss law applies on Canadian soil.
"(Canada) cannot host the games contrary to the Charter. The IOC cannot force VANOC or anyone else here to violate the laws of Canada," he said.
Clark asked the court to rule that the organizing committee must inform the IOC that the women should be included in the games.
If the court refuses to do so, Clark said his clients might ask the committee to cancel the men's ski jumping events or hold them elsewhere.
VANOC spokeswoman Renee Smith-Valade said the organizing committee will abide by the IOC's decision.
The court appeared sympathetic to the jumpers.
"We know VANOC has done everything it can to include women ski jumpers," said Justice Anne Rowles. "The IOC has turned a deaf ear to it."
Clark said if the jumpers receive the declaration they seek and VANOC and the IOC do not comply, they would seek a court injunction to stop the discrimination.
The group of women will be in court on Friday and hope that a decision will be reached by the end of Friday's proceedings.


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Home Magazines BizTimes Milwaukee Port development plan includes 14 office towers
Port development plan includes 14 office towers
BizTimes Staff
A Minnesota man and three investment partners have bought the former Milwaukee Solvay Coke & Gas Co. site on the city's south side, where they plan to build the largest real estate development in the state's history and change the city's skyline.
Thomas Short and his partners, whom he declined to identify, recently purchased the long-abandoned industrial site at 311. E. Greenfield Ave.
The investors intend to develop a gargantuan, $1.5 billion, mixed-use complex of 14 20-story towers suited for residential, office, retail and marina use, Short said.
The 46-acre site is located within the shadows of the Rockwell Automation clock tower, directly south Greenfield Avenue, east of First Street and north of the Kinnickinnic River.
Short and his investment partners formed Golden Marina Causeway LLC to acquire the property from Cleveland Cliffs Inc. of Ohio. Real estate sources said the limited liability corporation paid approximately $400,000 for the site and agreed to put additional funds into an escrow account to help pay for demolition and site remediation. The property had a total assessed valuation of $929,900 in 2002.
Short is the managing partner of the new corporation, which is gathering the financing to develop the project.
He tentatively refers to the office tower redevelopment as Gas Light Park.
The investors already have some office tenants who have given preliminary commitments to occupy space in the towers, Short said.
Short, who still resides in St. Paul, Minn., said the property and others like it near the Port of Milwaukee are ripe for development. Short said he was involved in similar industrial rehabilitation projects in St. Paul when that city essentially reinvented itself by capitalizing on its waterfront properties along the Mississippi River.
"When I came here from St. Paul, I looked at the Port of Milwaukee and said to myself, 'This is just like St. Paul 20 years ago,'" Short said.
Short said he has reached a verbal agreement with Lyle Balistreri, president of the Milwaukee building and Construction Trades Council, AFL-CIO, for the towers to be built by union labor.
Although the investors have acquired the Solvay property, they will face plenty of obstacles in their path to launch the development.
The first issue to be overcome is the environmental contamination at the site.
The site was first developed as an industrial site in 1902, where metallurgical coke was manufactured for use in the production of steel, foundry coke, coal gas and coal tar, according to historical archives at the University of Wisconsin-Milwaukee.
The US Environmental Protection Agency has not declared the site a Superfund site, but the EPA has documented some contamination at the site, which has been vacant since 1982. Contaminants found in the soil, groundwater and sediment, including arsenic, chromium, cyanide and lead, according to the agency.
The City of Milwaukee has classified the Solvay site as a "mothballed brownfield."
Indeed, the site, which is located between two sets of railroad tracks, is riddled with dilapidated buildings, rubbish, barrels and overgrowth.
"It's a blighting influence and a nuisance," said William Zaferos, spokesman for the Milwaukee Department of City Development.
"We've made a deal with the EPA to clean it up," Short said.
The EPA issued an administrative order by consent for Short to remediate the site. Terms of the agreement were not disclosed, but a cost assessment study conducted by the Milwaukee School of Engineering in 2001 estimated the site remediation costs at $10.4 million.
The environmental remediation will be performed by Eli Cos., an environmental and demolition company, which shares an office with Short's Water Street Holdings LLC near the Port of Milwaukee.
The investors have contracted Earth Tech Inc., an engineering company based in Long Beach, Calif., to be the project manager for the remediation.
"We expect to start cleanup within 30 days and hope to begin development by the end of the year," Short said.
Compared with other types of seeping contamination, soil that is polluted by coal is relatively easy to remove, he said.
In addition to the contamination, the Solvay project also faces the obstacle of gaining the approval of the City of Milwaukee to proceed with development.
The city has not devised a plan for the development of land surrounding the Port of Milwaukee, and Julie Penman, commissioner of the Department of City Development, said the city does not want the port to be developed in "piece meal" fashion.
The Solvay site is zoned by the city for industrial use and had a 2002 property assessment value of $929,900.
A final obstacle for the redevelopment project may be the market. Some real estate officials in Milwaukee question whether the city's office and residential markets could support such a large-scale development.
To put the scope of the Solvay project into perspective, its projected $1.5 billion cost would be more than three times the value of the Miller Park project, which exceeded $400 million, and would be five times the value of the proposed $300 million Pabst City project at the former Pabst Brewery site in downtown Milwaukee.
The only other development project to rival the Solvay project would be the proposed $1 billion Pabst Farms mixed-use project in Oconomowoc.
April 4, 2003 Small Business Times, Milwaukee
A Minnesota man and three investment partners have bought the former Milwaukee Solvay Coke & Gas Co. site on the city's south side, where they plan to build the largest real estate development in the state's history and change the city's skyline.
"When I came here from St. Paul, I looked at the Port of Milwaukee and said to myself, 'This is just like St. Paul 20 years ago,'" Short said.
"It's a blighting influence and a nuisance," said William Zaferos, spokesman for the Milwaukee Department of City Development.
"We've made a deal with the EPA to clean it up," Short said.
The environmental remediation will be performed by Eli Cos., an environmental and demolition company, which shares an office with Short's Water Street Holdings LLC near the Port of Milwaukee.
A final obstacle for the redevelopment project may be the market. Some real estate officials in Milwaukee question whether the city's office and residential markets could support such a large-scale development.


text:
Dr Keith
LLB(Hons); Dip Export; IMD (Hons); M Jur ; PhD
Professor, Associate Dean
Dr Thompson has enjoyed a wide and varied career in the law with experience in property, commercial, international, constitutional, criminal and human rights law. He worked for 11 years in Auckland in commercial and property law, including six years as a partner with Fortune Manning and was then appointed as International Legal Counsel for The Church of Jesus Christ of Latter-day Saints. In that role, Dr Thompson supervised that Church's legal work through the Pacific Area (for 18 years) and then through the African Continent for a further 2 years.
Since returning to Australia in 2011, he has been appointed as Professor and Associate Dean at the University of Notre Dame Australia's Sydney School of Law where he teaches Constitutional Law, Law and Religion as well as some Evidence and Contemporary Legal Issues. He has published a book entitled Religious Confession Privilege and the Common Law and has also published articles on subjects as diverse as Insolvency, Vicarious Liability in Tort, Foreign Corrupt Practices and Freedom of Religion.
e. keith.thompson@nd.edu.au |


text:
Sportscene >
Snooker >
World Snooker Championships
Hendry retires from snooker
Peter Huys Wikimedia Commons
Seven times world champion Stephen Hendry has quit the green baize. The Scot announced his retirement after his quarter-final defeat by fellow-Scot Stephen Maguire at the Crucible.
A combination of other interests, business commitments and a failure to recapture his best form consistently all contributed to Hendry's decision to call it a day. "It was quite an easy decision," he said. "There's a few reasons. The schedule didn't help. The fact that I'm not playing the snooker I want to play didn't help. The fact I'm not enjoying practice doesn't help."
Hendry managed a 147 break in his first round match, and beat champion John Higgins in the previous round, suggesting a return to form. He was never in contention against Maguire though. "The time is right for me," he said. "If I'd have won the title it would have been an even better way to go out. I'm delighted I made a maximum here, that's why I was more animated than normal when making it. I was delighted to do it on my last appearance here."
Hendry, always an icy calm presence on the tables ever since he arrived on the scene as a teenage prodigy, wasn't about to break down and shed a tear at the thought of never playing at the Crucible again. "I've never been the most emotional person even when I win," he said. "It's sad that I won't play here again. I love playing here, but no, it's a relief as much as anything."
World Championship,
John Higgins,
Stephen Hendry,
Higgins crushed by resurgent Hendry.
Hendry claims maximum break despite jet lag
I'm innocent
Hendry takes on Williams in World Championships
O'Sullivan's latest world title could be his last


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St. Mark's Lutheran Church in Delaware is celebrating the 500th anniversary of the Protestant Reformation.
Activities will begin at noon on Saturday, Oct. 28 at the church, located at the corner of William and Union streets.
According to a press release issued by the church, on Oct. 31, 1517, a young Roman Catholic Priest and Augustinian monk named Martin Luther posted his "Disputation on the Power and Efficacy of Indulgences," more commonly known as "The 95 Theses," which was the spark that ignited the Protestant Reformation.
Rue said despite many differences that have arisen between denominations over the years, many churches are still working toward unity.
Rue said the public is invited to the church parking lot from noon to 3 p.m. on Saturday, Oct. 28 for the celebration. German cuisine from Schmidt's sausage truck will be served. Bounce houses will be open to children.
Join in the table talk with Martin and Katie Luther. Learn about the reforms of the church as you participate in games and activities. Stop by the Gifts of Grace booth to pick up a free gift.
For information, contact the church office at 740-362-6771 or office@stmarksdelaware.com.


12pm - Meet at the Saint-Germain-des-Prés Christmas market and dive into the atmosphere.
2pm - Share an educational moment at the Caves du Louvre (Rue de l'Arbre Sec) for an introduction to wine.
3pm - Do some shopping at BHV Marais, where there's something to suit every taste. Wait for the others while sipping a cocktail and enjoying the view from Perchoir on the roof.
5pm - Get your adrenaline rush at the Funfair department of La Villette, called 'Holidays'.
7pm - Stroll around Bercy Village to enjoy the lights and break for a meal.
10.30pm - Enjoy a boozy evening on board a riverboat at Quai de la Gare – and party into the early hours!


Falk and Bennis launch Falk's great new book on Middle East!
This is the recording of the great discussion that Richard Falk, author of "Chaos and Counterrevolution: After the Arab Spring" had with Phyllis Bennis of the Institute for Policy Studies, to mark the book's June 9th launch. A thoughtful seminar with two renowned experts.


Posted in: Art, Inspiration, Photography, Weekly Photo Challenge.	Tagged: DPchallenge, love, monochrome, Phish, postaday, prayer garden, Release, Twist, Unity Village, Weekly Photo Challenge, Wisdom.
Love – Two hearts are intertwined, signifying eternal harmony and unity.
discerning and resolving — finding a pearl of wisdom in our center.
and bringing to the surface things we need to release.
Wow, these are beautiful statues with so much meaning. Thank you 😊 .
These are beautiful. Like the Hearts best.
This is really a beautiful interpretation of these locked and twisted figures.
Pingback: Twisted – Buttercups | What's (in) the Picture?


Green Bean Casserole is my FAVORITE thanksgiving side dish. But the version I grew up with is loaded with processed foods and carbs. I decided to make a keto green bean casserole to serve at our keto thanksgiving. This keto green bean casserole recipe uses a homemade cream of mushroom and crispy baked onions. If you are looking for keto Thanksgiving side dishes, you should give this gluten free green bean casserole a try. It's the perfect keto side dish for you low carb thanksgiving.
Combine the ingredients for the crispy onions in a bowl. Toss to coat the onions evenly. Spread the onions over a parchment lined baking tray and spray oil over the top of them. Place in the oven to bake for 12 – 15 minutes. Watch them carefully so they don't burn.
Set onions aside to cool and lower oven temp to 400.
While the onions are cooking, you can make the green beans. Bring a gallon or so of water and 2 tablespoons of salt to boil. Add the beans and blanch for 5 minutes. Drain and immediately place them in an ice bath to stop the cooking. Drain and set aside.
Melt the butter in a large skillet over medium-high heat. Add the mushrooms and salt. Cook until the mushrooms give up some of their liquid (about 5 minutes). Add the garlic, nutmeg and pepper. Cook for another minute. Sprinkle arrowroot powder over the mushrooms and stir to combine. Add in the broth and cook until simmering.
Next: Pumpkin Chocolate Chip Bundt Cake (Thanksgiving Dessert Ideas) Fall Season Recipes.


My name is Ken, and I am a PHP developer. I live and work in the city of Charlotte, North Carolina.
This entire site was designed and developed by myself. Its sole purpose is to showcase and demonstrate my skills in programming and web design.
On a more personal note, I enjoy developing websites, learning about new technologies, reading books, watching TV and movies, playing on the Xbox 360, and meeting people with similar interests.
I don't believe in being fake, and I am a man of principle. I am also honest, kind, polite, and thoughtful. At least, those are the values that I strive for each day.
Copyright © 2008-2017, Ken Stanley. All rights reserved.


text:
Apple's 2010 iPad launch was the last great Steve Jobs keynote
On the 10th anniversary of Apple's tablet, an appreciation of a master at work.
Steve Jobs brandishes the first iPad at the Yerba Buena Gardens theater in San Francisco on January 27, 2010. [Photo: Kimberly White/Corbis via Getty Images]
By Harry McCracken 9 minute Read
Ten years ago today, at a press event on January 27, 2010, Apple introduced a netbook competitor called the iSlate. Its starting price was an intimidating $999. The sheer sticker shock—and lack of crucial features such as Adobe's Flash—led to the iSlate becoming one of the company's most ignominious flops.
Oh, fine—that only happened in the alternate universe in which pundits' expectations about Apple's new device panned out, rather than being hilariously inaccurate.
In our world, what Apple cofounder and CEO Steve Jobs unveiled was a $499 tablet called the iPad, which went on sale the following April 3. The device was more an anti-netbook than a rival to the stripped-down mini-laptops that were—briefly—a hot commodity in the Windows world. And even though it ran a custom version of the software then known as iPhone OS, it was much more than a humongous-screened iPhone.
In the best possible way, Apple had built a product that defied expectations and fulfilled needs people didn't realize they had. Though the iPad has had its ups and downs over the years, it's been an enormous, category-defining success by any measure. Apple sold over 400 million of them by late 2018, which was when it stopped breaking out unit sales.
As usual, the invite to Apple's January 2010 product launch was a tease, not a tell. [Image: Courtesy of Apple]
Back in January 2010, I was in the audience at San Francisco's Yerba Buena Center for Jobs's presentation. I've always remembered it as among the best "Stevenotes" I ever witnessed in person. (I didn't get to go to Jobs's legendary iPhone launch three years earlier, and also failed to see him demo the original 1984 Mac and lay out his plan for rescuing Apple in 1997—events I could have attended but idiotically skipped.)
Like plenty of other tech writers who sat through the 2010 iPad event, I was more concerned with live-blogging the proceedings than luxuriating in them myself. So I had my eyes on my MacBook rather than the stage for much of the 90-minute event. That meant that when I recently watched the keynote on YouTube, it wasn't a simple act of nostalgia. I was literally seeing some of it for the first time.
In case you'd like to reacquaint yourself with the keynote—or never saw it at all—here you go:
Reviewing all this in 2020, a bunch of things stood out for me, only some of which I'd appreciated when I experienced it in person a decade earlier. Such as . . .
It was a different Apple. The company seemed remarkably successful at the time, but the numbers involved were dinky by 2020 standards. The first thing Jobs reveals is that the iPod—at the time, still an entirely relevant product in Apple's lineup—has recently sold its 250 millionth unit. (As of late 2018 the company had sold more than 2 billion iPhones.) Jobs also expresses disbelief that Apple has reached $50 billion in annual sales. (In 2019, that figure was $260 billion.)
A sizable percentage of Apple's current omnipresence in our lives was yet to come in 2010—in part because the iPod, iPhone, and Mac didn't add up to an ecosystem anywhere near as sweeping as what we have now.
Apple understood the importance of nailing the software experience from the get-go. These days, the company's launches often dwell on the tech and specs of new products, from the processors it designs itself to the sensors that capture photographs. At the 2010 iPad event, Jobs crammed the hardware section into less than three minutes of the 90-minute running time. He didn't even bother to specify that the iPad was Apple's first device to ship with an Apple-designed CPU, the A4.
The iPad wasn't just a fluffy and superficial entertainment machine.
By fast-forwarding through such details, Jobs and the other presenters had all the time they needed to show that the iPad was neither a scaled-up iPhone nor a stripped-down PC from a software standpoint. The company had devoted copious time to considering how core apps like a web browser, music player, and photo viewer should behave, and it showed. Later, Android and Windows tablets would struggle to match what Apple got right from the start with the iPad—and made the focal point of its launch.
The demos emphasized the iPad's lean-back nature. The classic tech-company keynote involves executives pacing back and forth on a stage. But for much of the iPad launch, Jobs, software chief Scott Forstall, and marketing honcho Phil Schiller plopped themselves in a comfy black leather chair and did their demos as if they were spending quality time with a favorite book. That happened to be a use-case scenario that the device served better than a phone or laptop, so the sit-down presentation made a point even though Jobs didn't play it up. (Late in the event, Schiller did, declaring "I get to sit in the chair!" to nobody in particular.)
A year later, Apple had sold almost 15 million iPads. By that point, the device already felt so familiar that the iPad 2 launch didn't have to acclimate anyone to what made Apple's tablet unique. Jobs stayed upright throughout, and only a couple of demos toward the end, conducted by others, involved the chair.
And yet Apple also played up productivity. Once the iPad shipped, the blogosphere was rife with debate about whether it could be used for serious content creation as well as passive consumption. I said yes, but felt outnumbered by naysayers; remarkably, that debate continues on today.
From the start, however, Apple positioned the iPad as being useful to artists and business people who wanted to create as well as consume. The developer behind an iPhone painting program called Brushes got to show the iPad version onstage, and Schiller devoted more than 12 minutes to demoing Apple's Pages, Numbers, and Keynote—productivity apps that cost 10 bucks apiece. (In 2o13, the company began giving them away to buyers of new devices.)
Apple presumably didn't expect to make a fortune by selling office software. But by building such apps, it conveyed a fact that was critical to the iPad's future: It wasn't just a fluffy and superficial entertainment machine.
In 2010, nobody knew for sure what sort of battery life a tablet like the iPad could or would achieve. [Image via YouTube]
iPhone app compatibility turned out to be a short-term security blanket. In January 2010, the iPhone App Store was a year and a half old and boasted a then-staggering 140,000 apps. One of Forstall's primary contributions to the event was showing that you could run almost all of them on an iPad, either with mammoth black borders or in a slightly pixelated 2X mode. That was a kludge at best, quickly rendered largely superfluous when third-party developers brought out a profusion of real iPad apps. But for the purposes of introducing an unfamiliar platform, the backwards compatibility was a necessary reassurance.
10 hours of battery life was a big deal. I'm always iffy about gauging the reaction to an Apple event by the applause it generated, since at least some of the people wildly clapping are Apple employees. Yet the original iPad's 10-hour battery life—announced to cheers from the audience—did feel like a feat that was key to its appeal. After all, you can't kick back with a device if you're carefully monitoring its battery gauge. Apple must have been pleased with the running time it achieved: It's the only feature from the 2010 iPad that has remained constant through every subsequent model.
The problem is, netbooks aren't better at anything!"
The iPad started as a peripheral for your Mac or Windows PC. Once iCloud came along in October 2011, the iPad gained a fair amount of autonomy. But at first, it was dependent on iTunes for backups. You were also likely to hook up your tablet to a computer via USB just to get music and movies onto it. As iTunes' reputation degraded in later years, such a prospect would have sounded horrifying to many people. But during the 2010 keynote, Apple framed the iPad's iTunes compatibility as a selling point. Which, at the time, it was.
Netbooks made for a convenient whipping boy. When Windows netbooks became a hit starting in 2007, multiple writers helpfully explained that Apple had to make a cheap mini-MacBook of its own. By the time the iPad's release was imminent, this conventional wisdom had morphed into an assumption that an Apple tablet would compete with netbooks. In reality, the iPad—a posh-feeling gadget with a largely new user interface and no keyboard—had little to do with netbooks, which were almost entirely about using off-the-shelf technology to hit a price point.
Rather than ignoring the comparison, however, Jobs ran with it: "The problem is, netbooks aren't better at anything!" It turned out that the world agreed with him, and the fad for them dwindled away even as the iPad took off.
The price was one of the most dramatic reveals of all. It's easy to forget that nobody had ever seen a tablet that much resembled the iPad, which made it tough to guess what it might cost. That's why pundits wasted an infinite number of hours anticipating that Apple's device would cost $1,000 or more. (That seemed plausible to me—in 2005, I'd bought a Toshiba Tablet PC for $1,597.)
The $499 price tag, which few Apple seers had even raised as a possibility, was so low that Jobs held off revealing it until the end of its event. The company also referenced the cost in its amusingly hype-y (though not altogether inaccurate) tagline, "a magical and revolutionary device at an unbelievable price."
Today, the question "How much does an iPad cost?" has many answers, ranging from $329 (the price for the current base model) to $1,699 (for a 12.9″ iPad Pro decked out with cellular connectivity and 1TB of storage, or 62.5 times the original version's starting capacity of 16GB). But in 2010, associating the new device with one price point that was far less than people had expected was atypical for Apple—but highly effective.
Related: Why the iPad is my personal gadget of the decade
Steve Jobs is irreplaceable. That's a truism, not a revelation. Still, rewatching the iPad launch was a powerful reminder of how unmatched he was as an explainer of new products. For all his talk of Apple's tablet being "magical," "revolutionary," and "unbelievable," the meat of his presentation was a clear, convincing explanation of what it was and why you'd want one. And as much as the iPad has evolved in the past 10 years, it remains the device that Jobs introduced, as if he understood where it might go in the years to come. By contrast, the 2014 event where Tim Cook and company unveiled the Apple Watch—the biggest all-new Apple device since the iPad—doesn't hold up nearly as well.
Between the January 27, 2010 event and his death on October 5, 2011, Jobs presided over another five keynotes. None was anywhere near as significant as the iPad's debut. (The one that would have come closest, the iPhone 4 introduction, had much of the suspense drained out of it when a preproduction phone went missing in a Redwood City beer garden.) As Jobs's final epoch-shifting product launch, the iPad introduction was both the start of something big and the end of an era—supremely effective marketing then, and even more meaningful now.
Harry McCracken is the technology editor for Fast Company, based in San Francisco. In past lives, he was editor at large for Time magazine, founder and editor of Technologizer, and editor of PC World.
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