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# Invertibility of Positive Toeplitz Operators and Associated Uncertainty
Principle

A. Walton Green and Mishko Mitkovski

###### Abstract

We study invertibility and compactness of positive Toeplitz operators
associated to a continuous Parseval frame on a Hilbert space. As applications,
we characterize compactness of affine and Weyl-Heisenberg localization
operators as well as give uncertainty principles for the associated
transforms.

##  1 Introduction

The uncertainty principle in harmonic analysis is a fundamental principle
roughly saying that a function and its Fourier transform cannot be
simultaneously “well localized”. There are many non-equivalent ways to make
this statement precise by imposing different types of “localization”
conditions. One such form of the uncertainty principle is the Benedicks’
theorem [8] which says that a function $f\in L^{2}(\mathbb{R}^{d})$ and its
Fourier transform $\hat{f}$ cannot be both supported on sets of finite
Lebesgue measure. The following is the quantitative form of this result as
obtained by Amrein and Berthier [1]: If $E,F\subseteq\mathbb{R}^{d}$ are sets
of finite Lebesgue measure then there exists $c>0$ such that

$c\left\|f\right\|_{L^{2}(\mathbb{R})}\leq\left\|f\right\|_{L^{2}(E^{c})}+\|\hat{f}\|_{L^{2}(F^{c})},$

for all $f\in L^{2}(\mathbb{R})$. Using the Kohn-Nirenberg quantization one
can restate this result in the following form: Let $P$ be the operator

$Pf(x):=\int_{\mathbb{R}^{d}}1_{(E\times F)^{c}}(x,\xi)\hat{f}(\xi)e^{2\pi
ix\cdot\xi}\,d\xi.$

Then $\langle Pf,f\rangle\geq c\|f\|^{2}$. It was shown by the first author
[18] that a similar result continues to hold for all subsets of
$\mathbb{R}^{2d}$ with finite Lebesgue measure, including the ones which are
not of the form $E\times F$. A natural analog of this result can be obtained
by replacing the Kohn-Nirenberg quantization with the anti-Wick quantization.
In this case we have the following statement for the short-time Fourier
transform: Let $E\subset\mathbb{R}^{2d}$ have finite Lebesgue measure. There
exists $c>0$ such that

$\int_{E^{c}}|V_{\phi}f(p,q)|^{2}\,dp\,dq\geq c\|f\|^{2}$ (1)

for all $f\in L^{2}(\mathbb{R}^{d})$, where $V_{\phi}f$ is the short-time
Fourier transform defined by $V_{\phi}f(p,q)=\int f(x)e^{2\pi
ipx}\phi(x-q)\,dx$. This result was proved independently and almost
simultaneously by Jaming [22], Janssen [24], and Wilczok [35] . Using similar
reasoning Wilczok also showed the following analog for Wavelet transform: Let
$E\subset(0,\infty)\times\mathbb{R}$ with finite affine measure
($\int_{E}a^{-1}dadb<\infty$). There exists $c>0$ such that

$\int_{E^{c}}|W_{\psi}f(a,b)|^{2}\dfrac{dadb}{a}\geq c\|f\|^{2}$ (2)

for all $f\in L^{2}(\mathbb{R}^{d})$ where the wavelet transform $W_{\psi}f$
is defined by
$W_{\psi}f(a,b)=a^{-1/2}\int_{\mathbb{R}}f(x)\psi(a^{-1}x-b)\,dx$ for some
wavelet $\psi$.

In the classical cases, when the window function $\phi$ or the mother wavelet
function $\psi$ are specially chosen, $V_{\phi}L^{2}(\mathbb{R}^{d})$ is a
Fock space of analytic functions and $W_{\psi}L^{2}(\mathbb{R}^{d})$ is a
Bergman space on the upper half space. In both cases, these inequalities hold
for the more general class of so called relatively dense sets, which is known
to be the optimal such class of sets. This has been known since the 80s, and
follows from the work of Luecking [28, 29] on the Bergman space. For the Fock
space, one can consult the work of Janson–Peetre–Rochberg [25], Ortega-Cerdà
[32], and the recent work of Ascensi [3]. However, it can be easily shown (see
section 5 below as well as [3, 23]) that relative density is a too weak
condition for these results to continue to hold for general $L^{2}$ window and
wavelet functions.

Still, it is natural to ask whether there exists a larger class of sets $E$
(of infinite measure) for which (1) and (2) continue to hold for a more
general class of window and wavelet functions. For the short-time Fourier
transform some sufficient conditions can be found in [3, 23] for windows with
varying degrees of regularity. However, for general window functions $\phi\in
L^{2}(\mathbb{R}^{d})$, Fernández and Galbis [15] showed that (1) holds for
all sets $E$ satisfying a certain thinness condition. The main goal of this
paper is to show that (2) also continues to hold for the appropriate analog of
thin sets. Furthermore, we also extend the result of Fernández and Galbis in
the context of short-time Fourier transforms on more general LCA groups. We
actually prove an uncertainty principle of Amrein-Berthier type for more
general Berezin-type quantizations which include both short-time Fourier and
wavelet inequalities as special cases.

To state our result we now introduce the above mentioned Berezin-type
quantization. Let $\mathcal{H}$ be a Hilbert space and $(X,d,\mu)$ be a metric
measure space with a metric $d$ and a Borel measure $\mu$. We assume that the
metric $d$ is proper, i.e., every ball with respect to this metric is
precompact. A collection of vectors $\\{k_{x}\\}_{x\in X}$ indexed by $X$
forms a normalized continuous Parseval frame for $\mathcal{H}$ if

$f=\int_{X}\langle f,k_{x}\rangle k_{x}\,d\mu(x)$ (3)

holds for each $f\in\mathcal{H}$. We will call the continuous map
$k:X\to\mathcal{H}$ given by $k(x)=k_{x}$ the Berezin quantization of $X$. We
will assume that $X$ is a homogeneous space in the sense that some locally
compact group $G$ acts transitively on $X$ in a way that both $\mu$ and $d$
are invariant under this group action ($\mu(gE)=\mu(E)$ and
$d(gx,gy)=d(x,y)$). The group action needs also to respect the inner product
on $\mathcal{H}$ in the following way $|\langle k_{gx},k_{gy}\rangle|=|\langle
k_{x},k_{y}\rangle|$ for all $x,y\in X$, $g\in G$. Under all these
assumptions, we will call the tuple $(\mathcal{H},X,G,k,\mu,d)$ a Berezin
quantization.

Very often, the metric measure space $X$ and its quantization
$k:X\to\mathcal{H}$ of this type arise naturally whenever we are given a
locally compact, second countable topological group $G$ and some of its
nontrivial irreducible, square-integrable, unitary representations (if such
exists), $\pi:G\to\mathcal{U}(\mathcal{H})$. Then $G$ itself can be equipped
with a left-invariant Haar measure $\mu$, and left-invariant metric $d$
(inducing the topology on $G$) which is proper, and for any unit vector
$k\in\mathcal{H}$, the collection $\\{k_{x}\\}_{x\in G}$ defined by
$k(x)=\pi(x)k$ will form a normalized continuous Parseval frame for
$\mathcal{H}$ such that $\langle k_{gx},k_{gy}\rangle=\langle
k_{x},k_{y}\rangle$ for all $x,y,g\in G$. Actually we can obtain a
quantization with essentially the same properties using the same procedure
even when we only have a projective irreducible square-integrable unitary
representation, $\pi:G\to\mathcal{U}(\mathcal{H})/\mathbb{C}$.

The following inequality is the obvious analog of (1) and (2) for Berezin
quantization.

$\int_{X\backslash E}|\left\langle f,k_{x}\right\rangle|^{2}\,d\mu(x)\geq
c\|f\|^{2}$ (4)

for all $f\in\mathcal{H}$. By choosing the group $G$ appropriately (see
Section 5 for more details on this) one obtains both (1) and (2) as special
cases.

The inequality (4) can be viewed as a statement about invertibility of a
certain Toeplitz operator. For a non-negative, bounded, measurable function
$\sigma:X\to\mathbb{R}$, the Toeplitz operator
$T_{\sigma}:\mathcal{H}\to\mathcal{H}$ with symbol $\sigma$ is defined by

$T_{\sigma}f=\int_{X}\sigma(x)\langle f,k_{x}\rangle k_{x}\,d\mu(x).$

It is easy to see that each such operator is bounded, self-adjoint, and
positive. We can easily recast (4) in the following way:

$\langle T_{1_{X\backslash E}}f,f\rangle\geq c\|f\|^{2}.$ (5)

This is a statement about boundedness from below, and hence invertibility, of
the Toeplitz operator $T_{\sigma}$ with an indicator symbol. It is clear that
each Toeplitz operator whose symbol is bounded away from zero must be bounded
from below (and hence invertible). However, to study (5), we must broaden this
trivial result since an indicator function is only non-negative. So we would
like to characterize the degree to which a non-negative symbol
$\sigma:X\to\mathbb{R}$ can vanish and still generate an invertible Toeplitz
operator $T_{\sigma}$.

The organization of the paper closely follows the main steps in the proof. The
proof outline that we use goes back at least to Havin and Joricke [19].
Namely, to prove (5) it suffices to show that the operator $T_{1_{E}}$ is
compact with a trivial eigenspace. The later is proved by showing that from
each nontrivial element of this eigenspace, using small translates, one can
construct an infinite dimensional eigenspace of a slightly “bigger” compact
self-adjoint operator of the same form. In Section 2 we address the
compactness problem for positive Toeplitz operators. We show that compactness
can be characterized in terms of the vanishing property of its Berezin
transform, a condition which in turn is closely connected to the thinness
condition of Fernández and Galbis. In Section 3 we examine the linear
independence problem for translations. We provide a fairly general condition
(usually easy to check) on the group which guarantees linear independence of
the translations. In Section 4 we combine the previous conclusions and state
and prove our main result (Theorem 2). Finally, we show how our main result
translates to more familiar settings, proving in particular (2) for thin sets
and a fairly wide class of mother wavelet functions.

##  2 Compactness of positive Toeplitz operators

In this section we deal with the following question: Which non-negative
symbols $\sigma:X\to\mathbb{R}$ generate compact Toeplitz operators
$T_{\sigma}$? We show that, under suitable assumptions on the continuous
Parseval frame $\\{k_{x}\\}_{x\in X}$, a necessary and sufficient condition
for $T_{\sigma}$ to be compact is its “diagonal” $\left\langle
T_{\sigma}k_{x},k_{x}\right\rangle$ to vanish at infinity. We now make this
precise.

For a given bounded operator $T:\mathcal{H}\to\mathcal{H}$ we define its
Berezin transform $\tilde{T}:X\to\mathbb{C}$ by $\tilde{T}(x)=\left\langle
Tk_{x},k_{x}\right\rangle$. In the case when $T$ is a Toeplitz operator with
symbol $\sigma$ we denote its Berezin transform by $\tilde{\sigma}$, i.e.,
$\tilde{\sigma}(x)=\left\langle T_{\sigma}k_{x},k_{x}\right\rangle$. In what
follows, by $y\to\infty$, we will mean that $d(z,y)\to\infty$ for some
(equivalently, all) $z\in X$.

######  Definition 1.

A measurable function $\sigma:X\to\mathbb{C}$ is said to be thin if its
Berezin transform vanishes at infinity, i.e.,

$\tilde{\sigma}(y):=\int_{X}\sigma(x)|\langle
k_{x},k_{y}\rangle|^{2}\,d\mu(x)\to 0\quad\mbox{as }y\to\infty$ (6)

We will also say that a set $E\subset X$ is thin if the indicator function of
$E$ satisfies (6).

The connection between compactness of a Toeplitz operator and the Berezin
transform vanishing at infinity (called thinness here) is another quite old
question [9, 31, 37]. The breakthrough results in this area were due to Axler
and Zheng [4] for the Bergman space on the disk, $A^{p}(\mathbb{D})$, and to
Suárez [34] for $A^{p}(\mathbb{B}_{n})$. Subsequently, these results been
generalized to many different settings [6, 12, 33], even going beyond the
realm of analytic function spaces [5, 7, 10, 14, 36]. Since we are only
dealing with nonnegative symbols, we are able to avoid many difficulties in
these aforementioned results.

The following proposition gives a somewhat more explicit alternative
characterization of thinness.

######  Proposition 1.

Let $\sigma:X\to[0,1]$. The following are equivalent

  * (i)

$\sigma$ is thin.

  * (ii)

$\displaystyle\lim\limits_{y\to\infty}\int_{B(y,R)}\sigma(x)\,d\mu(x)=0$ for
some $R>0$.

  * (iii)

$\displaystyle\lim\limits_{y\to\infty}\int_{B(y,R)}\sigma(x)\,d\mu(x)=0$ for
all $R>0$.

###### Proof.

Fix an element $e\in X$ which is arbitrary, but will be treated as the origin.
Since $X$ is homogeneous, for each $x\in X$, there exists $g_{x}\in G$ (not
necessarily unique) such that $g_{x}e=x$. For each $x$ we pick one such
$g_{x}$ and fix it throughout the proof. By the invariance of $\mu$ and $d$,

$\int_{B(y,R)}\sigma(x)\,d\mu(x)=\int_{B(e,R)}\sigma(g_{y}x)\,d\mu(x).$

Using our initial assumption, the metric $d$ is proper, i.e., every ball is
precompact. So, the ball $B(e,R)$ can be covered by finitely many balls of a
fixed radius. Therefore, we have (ii) implies (iii).

Next, consider the function $h(x)=\langle k_{e},k_{x}\rangle$.
$h(e)=\|k_{e}\|^{2}>0$ and $h$ is continuous so there exists $\delta>0$ such
that $|h(x)|\geq h(e)/2$ for $x\in B(e,\delta)$. Thus,

$\dfrac{h(e)^{2}}{4}\int_{B(e,\delta)}\sigma(g_{y}x)\,d\mu(x)\leq\int_{B(e,\delta)}\sigma(g_{y}x)|h(x)|^{2}\,d\mu(x)$
$\leq\int_{X}\sigma(x)|h(g_{y}^{-1}x)|^{2}\,d\mu(x)=\int_{X}\sigma(x)|\langle
k_{y},k_{x}\rangle|^{2}\,d\mu(x).$

Therefore (i) implies (ii). To show (iii) implies (i), let $\varepsilon>0$. We
can find $R>0$ such that $\int_{B(e,R)^{c}}|h(x)|^{2}\,d\mu(x)<\varepsilon/2$.
For this $R$, by (iii), there exists $N>0$ such that for $d(e,y)\geq N$,
$\int_{B(e,R)}\sigma(g_{y}x)\,d\mu(x)\leq\varepsilon/(2\|k_{e}\|^{4})$.
Therefore,

$\int_{X}\sigma(x)|\langle
k_{y},k_{x}\rangle|^{2}\,d\mu(y)=\int_{B(e,R)}+\int_{B(e,R)^{c}}\sigma(g_{y}x)|h(x)|^{2}\,d\mu(x)\leq\varepsilon,$

for all $y\in X$ with $d(e,y)\geq N$. Here we used the fact that
$|h(x)|\leq\|k_{e}\|\cdot\|k_{x}\|$ but $\|k_{x}\|=\|k_{e}\|$. Indeed,

$\|k_{x}\|^{2}=|\langle k_{x},k_{x}\rangle|=|\langle
k_{e},k_{e}\rangle|=\|k_{e}\|^{2}.$

∎

We are now ready to characterize compactness of Toeplitz operators when the
Parseval frame satisfies some additional decay conditions.

######  Theorem 1.

Let $(\mathcal{H},X,G,k,\mu,d)$ be a Berezin quantization. Suppose that the
continuous Parseval frame $\\{k_{x}\\}_{x\in X}$ satisfies the following:
there exists a weight $w:X\to(0,\infty)$ and $M>0$ such that

  * (i)

$\displaystyle w(y)^{-1}\int_{X}|\langle k_{x},k_{y}\rangle|w(x)\,d\mu(x)\leq
M$ for all $y\in X$.

  * (ii)

$\displaystyle\lim_{R\to\infty}\sup_{y\in X}w(y)^{-1}\int_{B(y,R)^{c}}|\langle
k_{x},k_{y}\rangle|w(x)\,d\mu(x)=0$.

  * (iii)

$|\langle k_{x},k_{y}\rangle|\to 0$ as $d(x,y)\to\infty$.

Then $\sigma\in L^{\infty}(X)$ is thin if and only if $T_{\sigma}$ is compact.

######  Remark 1.

In the case $X=G$, it is often easier to check that $x\mapsto\langle
k_{1},k_{x}\rangle w(x)$ is in $L^{1}(X)$ for some submultiplicative weight
$w$ ($w(xy)\leq Cw(x)w(y)$). Then, (i) and (ii) follow from the group
invariance of $|\langle k_{x},k_{y}\rangle|$. In this way, these conditions
are related to the so-called analyzing vectors for the coorbit spaces of
Gröchenig and Feichtinger [13].

###### Proof.

To prove necessity we only need to use (iii). First, we show that (iii)
implies $k_{y}{\overset{\mbox{\tiny w}}{\to}}0$ as $y\to\infty$. We have
$\langle k_{y},k_{x}\rangle\to 0$ for each $x\in X$ as $d(e,y)\to\infty$.
Moreover, since $\\{k_{x}\\}_{x\in X}$ is a continuous Parseval frame we have
$\left\|f\right\|^{2}=\int_{X}\left|\left\langle
f,k_{x}\right\rangle\right|^{2}d\mu(x)$, and hence
$\overline{\operatorname{span}}\\{k_{x}\\}=\mathcal{H}$. Let $f\in\mathcal{H}$
and $\varepsilon>0$. There exists
$f_{\varepsilon}\in\operatorname{span}\\{k_{x}\\}$ such that
$\|f-f_{\varepsilon}\|\leq\varepsilon/(2\|k_{e}\|)$. Moreover, there exists
$M$ such that if $d(e,y)>M$ then $|\langle
k_{y},f_{\varepsilon}\rangle|\leq\varepsilon/2$. Thus,

$|\langle k_{y},f\rangle|\leq\|k_{e}\|\cdot\|f-f_{\varepsilon}\|+|\langle
k_{y},f_{\varepsilon}\rangle|\leq\varepsilon$

whenever $d(e,y)>M$. Thus, if $T_{\sigma}$ is compact we have
$T_{\sigma}k_{y}\to 0$ as $y\to\infty$, and hence
$\tilde{\sigma}(y)=\left\langle T_{\sigma}k_{y},k_{y}\right\rangle\to 0$ as
$y\to\infty$.

We now prove sufficiency. For this we use (i) and (ii). Let $\varepsilon>0$.
There exists $R>0$ such that

$w(y)^{-1}\int_{B(y,R)^{c}}|\langle
k_{y},k_{z}\rangle|w(z)\,d\mu(z)\leq\varepsilon$

for all $y\in X$. First we estimate the “tails” using the Schur property of
$\\{k_{x}\\}$. For any $f\in\mathcal{H}$,

$\int_{X}\left|\int_{B(x,R)^{c}}\sigma(y)\langle f,k_{y}\rangle\langle
k_{y},k_{x}\rangle\,d\mu(y)\right|^{2}\,d\mu(x)$
$\leq\|\sigma\|_{\infty}^{2}\int_{X}\int_{B(x,R)^{c}}|\langle
f,k_{y}\rangle|^{2}|\langle
k_{y},k_{x}\rangle|w(y)^{-1}\,d\mu(y)\int_{B(x,R)^{c}}|\langle
k_{y},k_{x}\rangle|w(y)\,d\mu(y)\,d\mu(x)$
$\leq\varepsilon\|\sigma\|_{\infty}^{2}\int_{X}|\langle
f,k_{y}\rangle|^{2}\int_{X}w(x)|\langle
k_{y},k_{x}\rangle|w(y)^{-1}\,d\mu(x)\,d\mu(y)$ $\leq
M\|\sigma\|_{\infty}^{2}\varepsilon\|f\|^{2}.$

Now, since $\sigma$ is thin and bounded, by Proposition 1, there exists $S>0$
such that for $d(e,y)\geq S$,
$\int_{B(y,R)}|\sigma|^{2}\,d\mu\leq\varepsilon$. Thus,

$\int_{B(e,S)^{c}}\left|\int_{B(x,R)}\sigma(y)\langle f,k_{y}\rangle\langle
k_{y},k_{x}\rangle\,d\mu(y)\right|^{2}\,d\mu(x)$
$\leq\varepsilon\int_{X}\int_{X}|\langle f,k_{y}\rangle|^{2}|\langle
k_{x},k_{y}\rangle|^{2}\,d\mu(y)\,d\mu(x)=\varepsilon\|k_{e}\|^{2}\|f\|^{2}.$

Now, let $f_{n}{\overset{\mbox{\tiny w}}{\to}}0$ be arbitrary. Then,
$\|f_{n}\|$ is uniformly bounded.

$\displaystyle\|T_{\sigma}$ $\displaystyle f_{n}\|^{2}=\int_{X}|\langle
T_{\sigma}f_{n},k_{x}\rangle|^{2}\,d\mu(x)$
$\displaystyle=\int_{X}\left|\int_{B(x,R)}+\int_{B(x,R)^{c}}\sigma(y)\langle
f_{n},k_{y}\rangle\langle k_{y},k_{x}\rangle\,d\mu(y)\right|^{2}\,d\mu(x)$
$\displaystyle\leq\int_{B(e,S)}+\int_{B(e,S)^{c}}2\left|\int_{B(x,R)}\sigma(y)\langle
f_{n},k_{y}\rangle\langle
k_{y},k_{x}\rangle\,d\mu(y)\right|^{2}\,d\mu(x)+C\varepsilon$
$\displaystyle\leq 2\int_{B(e,S)}\left|\int_{B(x,r)}\sigma(y)\langle
f_{n},k_{y}\rangle\langle
k_{y},k_{x}\rangle\,d\mu(y)\right|^{2}\,d\mu(x)+C\varepsilon.$

Since the integrand goes to zero pointwise, applying the dominated convergence
theorem we obtain

$\limsup_{n\to\infty}\|T_{\sigma}f_{n}\|^{2}\leq C\varepsilon.$

But $\varepsilon$ is arbitrary so $\lim\limits_{n\to\infty}T_{\sigma}f_{n}=0$.
Therefore $T_{\sigma}$ is compact.

∎

##  3 Independence of Translations

The action of $G$ on $X$ naturally induces the following translation action on
$L^{2}(X,d\mu)$. For a given $h\in G$, the translation operator
$\tau_{h}:L^{2}(X,d\mu)\to L^{2}(X,d\mu)$ is defined by
$\tau_{h}F(x)=F(h^{-1}x)$ for $F\in L^{2}(X,d\mu)$. We denote the translated
function by $F_{h}:=\tau_{h}F$. We can view $\mathcal{H}$ as a subspace of
$L^{2}(X)$ by identifying $f\in\mathcal{H}$ with $x\mapsto\langle
f,k_{x}\rangle\in L^{2}(X,d\mu)$. In this way, $\tau_{h}$ induces the
following shift operator $S_{h}:\mathcal{H}\to\mathcal{H}$ defined by

$S_{h}f=\int_{X}\langle f,k_{h^{-1}x}\rangle k_{x}\,d\mu(x)=\int_{X}\langle
f,k_{x}\rangle k_{hx}\,d\mu(x).$

It is easy to check that $S_{h}^{*}=S_{h^{-1}}$ and $S_{h}$ is bounded. While
it is not true in general that $S_{h}k_{x}=k_{hx}$, this does hold if $\langle
k_{hx},k_{hy}\rangle=\langle k_{x},k_{y}\rangle$, and then $S_{h}$ is unitary
so that $S_{h}T_{\sigma}S_{h}^{*}=T_{\sigma_{h}}$. For this reason, we
introduce the subgroup of $G$,

$G_{\tau}=\\{h\in G:\langle k_{hx},k_{hy}\rangle=\langle
k_{x},k_{y}\rangle\mbox{ for all }x,y\in X\\}.$ (7)

In this way, $h\mapsto S_{h}$ is a homomorphism from
$G_{\tau}\to\mathcal{U}(\mathcal{H})$.

We are interested in conditions on the action under which different shifts of
any given non-zero $f\in\mathcal{H}$ form a linearly independent set. Observe
first that this problem can be reduced to the more studied problem of linear
independence of translations. Indeed, for any nonzero $f\in\mathcal{H}$ define
$F(x)=\langle f,k_{x}\rangle$. Then $F\in L^{2}(X)$ and $\langle
S_{h}f,k_{x}\rangle=F(h^{-1}x)$ for $h\in G_{\tau}$. The last relation clearly
implies that the linear independence of
$\\{S_{h_{1}}f,S_{h_{2}}f,\dots,S_{h_{n}}f\\}$ is equivalent to the linear
independence of the corresponding translations of $F$,
$\\{F_{h_{1}},F_{h_{2}},\dots,F_{h_{n}}\\}$.

Note, however, that the linear independence problem for translations is in
general a very difficult problem. Even for some fairly simple group actions,
translations may be not linear independent. Specifically, in the case of the
Affine group, $\chi_{[0,1]}(x)=\chi_{[0,1]}(2x)+\chi_{[0,1]}(2x-1)$ is a
simple example. Moreover, for the Heisenberg Group, the linear independence
problem for time-frequency shifts (known as the HRT conjecture [20]) is still
open and is widely considered to be very difficult.

However, in the case of Abelian groups $G$, a theorem of Edgar and Rosenblatt
[11] says that the translations of any non-zero $F\in L^{2}(G)$ are linearly
independent as long as $G$ has no nontrivial compact subgroups. Our main goal
will be to extend this result to the case of homogeneous spaces $X$.

Let $\Gamma\subset G$. Denote

$\mathbb{C}\Gamma=\\{\sum c_{n}h_{n}:h_{n}\in\Gamma,c_{n}\in\mathbb{C}\\}.$

The elements $\theta\in\mathbb{C}\Gamma$ act on $F\in L^{2}(X)$ by

$\theta F=\sum c_{n}F_{h_{n}}.$

Thus, a collection of translations $\Gamma F=\\{F_{h}:h\in\Gamma\\}$ is
linearly independent if and only if $\theta F\neq 0$ for all
$0\neq\theta\in\mathbb{C}\Gamma$. This leads to the following definition.

######  Definition 2.

We say that $L^{2}(X)$ has linearly independent $\Gamma$-translations if for
all $F\in L^{2}(X)$, $\theta\in\mathbb{C}\Gamma$,

$\theta F=0\mbox{ implies }F=0\mbox{ or }\theta=0.$

We want to study the independence of $\Gamma$-translates of $L^{2}(X)$ by
reducing it to the case $L^{2}(\Gamma)$ where we can use the result of Edgar
and Rosenblatt. The following proposition establishes this reduction. It can
be viewed as a generalization of a result from [27] to non-discrete $\Gamma$.

######  Proposition 2.

Let $(X,\mu)$ be a measure space. Let $\Gamma$ be a unimodular locally compact
group $\Gamma$ acting on $X$ under which $\mu$ is invariant. Then $L^{2}(X)$
has linearly independent $\Gamma$-translations if $L^{2}(\Gamma)$ has linearly
independent $\Gamma$-translations.

In proving this, we will make crucial use of the following identity.

######  Lemma 1.

Let $\Gamma$ and $(X,\mu)$ be as above. Then there exists a measure $\nu$ on
$X/\Gamma$ such that

$\int_{X/\Gamma}\int_{\Gamma}F(\gamma x)\,d\lambda(\gamma)\,d\nu(\Gamma
x)=\int_{X}F(x)\,d\mu(x)$ (8)

holds for all $0\leq F\in L^{1}(X,\mu)$, where $\lambda$ is the Haar measure
on $\Gamma$.

We are not aware of such a fact in the literature. However, in the case where
$X$ is a group and $\Gamma\leq X$, it is well-known [16, 30], so we show at
the end of this section that a slight modification of these proofs yields (8)
in our setting.

###### Proof of Proposition 2.

Let $F\in L^{2}(X)$ be nonzero. Let $E$ denote the set of $\Gamma x\in
X/\Gamma$ with $\int|F(\gamma x)|^{2}\,d\lambda(\gamma)\in\\{0,\infty\\}$. By
formula (8), replacing $F$ with $|F|^{2}$, it must be that $\nu(E^{c})>0$. So,
there exists $\Gamma x\in E^{c}$ such that $g(\gamma):=F(\gamma x)$ satisfies
$g\in L^{2}(\Gamma)$ and $g\neq 0$. Thus, for each
$\theta\in\mathbb{C}\Gamma$, if $\theta F=0$, then $\theta g=0$ which implies
$\theta=0$. ∎

Applying the known result for translations on Abelian groups from [11], we
obtain the following corollary.

######  Corollary 1.

Let $\Gamma$ be an Abelian subgroup of $G_{\tau}$ with no nontrivial compact
subgroups. For any $\\{h_{n}\\}\subset\Gamma$, and $f\in\mathcal{H}$,
$\\{S_{h_{n}}S_{h_{n-1}}\cdots S_{h_{1}}f\\}_{n=1}^{\infty}$ is linearly
independent.

###### Proof.

Let $f\in\mathcal{H}$. Define $F(x)=\langle f,k_{x}\rangle$. $F\in L^{2}(X)$
and $\langle S_{h}f,k_{x}\rangle=F(h^{-1}x)$. Therefore since $L^{2}(\Gamma)$
has linearly independent $\Gamma$ translations by the result of Edgar and
Rosenblatt [11], $L^{2}(X)$ has linearly independent $\Gamma$ translations by
Proposition 2. ∎

Finally, we give the proof of Lemma 1. Let $\lambda$ be the Haar measure for
$\Gamma$. Since $\Gamma$ is unimodular, the Haar measure is both right and
left invariant and $d\lambda(\gamma)=d\lambda(\gamma^{-1})$. Define
$P:C_{c}(X)\to C_{c}(X/\Gamma)$ by

$Pf(\Gamma x)=\int_{\Gamma}f(\gamma x)\,d\lambda(\gamma).$

This is well-defined since if $\Gamma y=\Gamma x$, then $\gamma_{0}y=x$ so
$Pf(\Gamma y)=Pf(\Gamma x)$ by the invariance of $\lambda$.

We claim that $Pf=0$ implies $\int_{X}f\,d\mu=0$. Indeed, take $\phi\in
C_{c}(X)$ such that $P\phi=1$ on $\Gamma\operatorname{supp}f$ (this can be
done by surjectivity of $P$ which we will prove below). Then,

$0=\int_{X}\phi(x)\int_{\Gamma}f(\gamma
x)\,d\lambda(\gamma)\,d\mu(x)=\int_{X}\int_{\Gamma}\phi(\gamma^{-1}x)\,d\lambda(\gamma)\,f(x)\,d\mu(x)$
$=\int_{X}P\phi(\Gamma x)f(x)\,d\mu(x)=\int_{X}f(x)\,d\mu(x).$

Now we are ready to define the measure $\nu$. Define the functional on
$C_{c}(X/\Gamma)$ by $Pf\to\int_{X}f\,d\mu$. This is a well-defined positive
linear functional by the previous claim and Lemma 2 below. By the Riesz
Representation Theorem, there exists $\nu$ such that

$\int_{X/\Gamma}Pf\,d\nu=\int_{X}f\,d\mu$

which is the desired identity (8). It only remains to prove the surjectivity
of $P$.

######  Lemma 2.

Let $\phi\in C_{c}(X/\Gamma)$. There exists $f\in C_{c}(X)$ such that
$Pf=\phi$, $\Gamma(\operatorname{supp}f)=\operatorname{supp}\phi$, and $f\geq
0$ if $\phi\geq 0$.

###### Proof.

Denote by $E=\operatorname{supp}\phi$. Since $E$ is compact, there exists
$\\{g_{k}\\}_{k=1}^{N}\subset G$ such that $E\subset\cup_{k=1}^{N}\Gamma
g_{k}V$ where $V$ is a compact neighborhood of $e$. $V$ has an open
neighborhood $U$ with compact closure. Then, consider the function
$\psi(x)=\max_{k}\\{d(g_{k}^{-1}x,U^{c})\\}$. $\psi\in C_{c}(X)$, $\psi>0$ on
$\cup_{k=1}^{N}g_{k}V$, and for $\Gamma x\in E$,

$P\psi(\Gamma x)=\int_{\Gamma}\psi(\gamma x)\,d\lambda(\gamma)>0.$

Now, setting

$f(x)=\dfrac{\phi(\Gamma x)}{P\psi(\Gamma x)}\psi(x),$

we see that $f\in C_{c}(X)$,
$\Gamma\operatorname{supp}f=\operatorname{supp}\phi$, and $f\geq 0$ if
$\phi\geq 0$. We only need to check

$Pf(\Gamma x)=\int_{\Gamma}\dfrac{\phi(\Gamma\gamma x)}{P\psi(\Gamma\gamma
x)}\psi(\gamma x)\,d\lambda(\gamma)$ $=\dfrac{\phi(\Gamma x)}{P\psi(\Gamma
x)})P\psi(\Gamma x)=\phi(\Gamma x).$

∎

##  4 Thin Toeplitz Operators

In this section, we show that from one compact Toeplitz operator $T_{\sigma}$,
one can create $T_{\rho}$ with $\rho\geq\sup_{n}\sigma_{h_{n}}$ for some
infinite, but small, sequence of translates $\\{h_{n}\\}$, while $T_{\rho}$
remains compact.

The set of positive linear operators $B(\mathcal{H})^{+}$ on a Hilbert space
$\mathcal{H}$ possesses a partial ordering. We say $A\geq B$ if $A-B$ is a
positive operator, i.e $\langle(A-B)f,f\rangle\geq 0$ for all
$f\in\mathcal{H}$. Letting $(\sigma\vee\rho)(x)=\max\\{\sigma(x),\rho(x)\\}$,
we have

  * (i)

$T_{\sigma},T_{\rho}\leq T_{\sigma\vee\rho}$.

  * (ii)

If $T_{\sigma}$ and $T_{\rho}$ are compact, so is $T_{\sigma\vee\rho}$.

  * (iii)

$\|T_{\sigma\vee\rho}\|\leq\max\\{\|\sigma\|_{\infty},\|\rho\|_{\infty}\\}$

Since $\langle T_{\sigma}f,f\rangle=\int\sigma(x)|\langle
f,k_{x}\rangle|^{2}\,d\mu(x)$, ordering of the Toeplitz operators follows from
the ordering of their symbols. (ii) follows from the fact that
$T_{\sigma\vee\rho}\leq T_{\sigma}+T_{\rho}$. (iii) is a consequence of the
trivial estimate that $\|T_{\sigma}\|\leq\|\sigma\|_{\infty}$.

######  Lemma 3.

Let $\sigma$ be thin and $\\{h_{k}\\}\subset G$ such that $h_{k}\to 1$. Then,
there exists $\\{m_{k}\\}$ such that

$\rho(x)=\sup_{k}\sigma_{h_{m_{k}}h_{m_{k-1}}\cdots h_{m_{1}}}(x)$

is thin.

###### Proof.

Let $K_{y}(x)=|\langle k_{x},k_{y}\rangle|^{2}\in L^{1}(X)$. Then,

$\tilde{\sigma}(y)=\int_{X}\sigma(x)K_{y}(x)\,d\mu(x).$

Therefore, by the boundedness of $\sigma$ and the continuity of the
translations on $L^{1}(X)$,

$|\tilde{\sigma}(y)-\tilde{\sigma}_{h}(y)|=\int_{X}\sigma(x)[K_{y}(x)-K_{y}(hx)]\,d\mu(x)\to
0$

as $h\to 1$ for each $y$. Since $h_{m}\to 1$ we can pick a subsequence such
that $\\{\prod_{finite}h_{m}\\}\subset B_{1}$. From this subsequence, we pick
another one. Set $\rho_{0}=\sigma$. Then, since $\sigma$ is thin, there exists
$R_{0}$ such that $\tilde{\rho}_{0}\leq 1$ outside $B_{R_{0}}$. Then, pick
$m_{0}$ such that

$|\tilde{\rho}_{0}-\tilde{\rho}_{0,h_{m_{0}}}|\leq 1$

on $B_{R_{0}+1}$. Then,

$|\tilde{\rho}_{0}-\tilde{\rho}_{0,h_{m_{0}}}|\leq 3.$

Now, having $\rho_{j}$ (thin) and $h_{m_{j}}$ for $j<k$, pick $m_{k}$ in the
same manner as above (finding $R_{k}$ since $\rho_{j}$ are all thin) so that

$|\tilde{\rho}_{j}-\tilde{\rho}_{j,h_{m_{k}}}|\leq 3\cdot 2^{-k}$

for all $j<k$. then set $\rho_{k}=\max\\{\rho_{k-1},\rho_{k-1,h_{m_{k}}}\\}$.

Now we are ready to prove that $\rho=\lim\rho_{k}$ is thin. Let
$\varepsilon>0$ and pick $j$ such that
$3\sum_{k=j}^{\infty}2^{-k}\leq\tfrac{\varepsilon}{2}$. Let $R>0$ such that
$\tilde{\sigma}(y)\leq\tfrac{\varepsilon}{2^{j+1}}$ for $d(1,y)\geq R$. Then,
for $d(1,y)\geq R+1$, $\tilde{\sigma}_{h}(y)\leq\tfrac{\varepsilon}{2^{j+1}}$
for all $h\in B_{1}$. Then, by the fact that $|P_{j}|=2^{j}$ ($P_{j}$ is the
power set of $\\{0,1,2,\ldots,j\\}$), and noticing that $\rho_{k}=\max_{I\in
P_{k}}\sigma_{\prod_{i\in I}h_{m_{i}}}$,

$\tilde{\rho}_{j}(y)\leq\sum_{I\in P_{j}}\tilde{\sigma}_{\prod_{i\in
I}h_{m_{i}}}(y)\leq
2^{j}(\tfrac{\varepsilon}{2^{j+1}})=\tfrac{\varepsilon}{2}.$

By construction of $h_{m_{k}}$, $|\tilde{\rho}_{k}-\tilde{\rho}_{k+1}|\leq
2^{-k}$. Therefore,

$\tilde{\rho}(y)\leq\tilde{\rho}_{j}(y)+\sum_{k=j}^{\infty}|\tilde{\rho}_{k}(y)-\tilde{\rho}_{k+1}(y)|\leq\tfrac{\varepsilon}{2}+3\sum_{k=j}^{\infty}2^{-k}\leq\varepsilon.$

∎

###  4.1 Main Results

We now assemble the pieces from the previous sections, specifically Theorem 1
and Corollary 1 to prove our main results.

######  Theorem 2.

Let $(\mathcal{H},X,G,k,\mu,d)$ be a Berezin quantization. Suppose there
exists $w:X\to(0,\infty)$ and $M>0$ such that

  * (i)

$\displaystyle w(y)^{-1}\int_{X}|\langle k_{x},k_{y}\rangle|w(x)\,d\mu(x)\leq
M$ for all $y\in X$.

  * (ii)

$\displaystyle\lim_{R\to\infty}\sup_{y\in X}w(y)^{-1}\int_{B(y,R)^{c}}|\langle
k_{x},k_{y}\rangle|w(x)\,d\mu(x)=0$.

Suppose also that $G_{\tau}$ contains an Abelian subgroup which has an
accumulation point at $1$ and no nontrivial compact subgroups. Then, if
$\sigma:X\to[0,1]$ is thin, there exists $c>0$ such that

$\langle T_{1-\sigma}f,f\rangle\geq c\|f\|^{2}$

for all $f\in\mathcal{H}$.

###### Proof.

Suppose there exists $f\neq 0$ such that $T_{\sigma}f=f$. By Lemma 3, there
exists a sequence $\\{h_{k}\\}_{k=1}^{\infty}\subset\Gamma$ such that
$\rho(x)=\sup_{k}\sigma_{h_{k}\cdots h_{1}}(x)$ is thin. Define
$f_{k}=S_{h_{k}}\cdots S_{h_{1}}f$. Then,

$1\geq\langle T_{\rho}f_{k},f_{k}\rangle\geq\langle S_{h_{k}}\cdots
S_{h_{1}}T_{\sigma}(S_{h_{k}}\cdots S_{h_{1}})^{*}f_{k},f_{k}\rangle=\langle
T_{\sigma}f,f\rangle=1.$

This implies $T_{\rho}f_{k}=f_{k}$. However, the collection $\\{f_{k}\\}$ is
linearly independent by Corollary 1 since $\Gamma$ is Abelian, has no
nontrivial compact subgroups, and is contained in $G_{\tau}$. Therefore the
dimension of the eigenspace corresponding to the eigenvalue $1$ is infinite.
However, $T_{\rho}$ is compact by Theorem 1 which is a contradiction. ∎

Properties (i) and (ii) often follow from the integrability of $\langle
k_{x},k_{e}\rangle$, see Remark 1. In case the decay of $\langle
k_{x},k_{y}\rangle$ is not sufficient for (i) and (ii), we can obtain
something weaker.

######  Theorem 3.

Let $(\mathcal{H},X,G,k,\mu,d)$ be a Berezin quantiztion and suppose
$G_{\tau}$ contains an Abelian subgroup which has an accumulation point at $1$
and no nontrivial compact subgroups. If $\sigma:X\to[0,1]$ is in $L^{1}(X)$,
then there exists $c>0$ such that

$\langle T_{1-\sigma}f,f\rangle\geq c\|f\|^{2}$

for all $f\in\mathcal{H}$.

###### Proof.

First, let us see that $\sigma\in L^{1}(X)$ implies $T$ is compact. Indeed, if
$f_{n}{\overset{\mbox{\tiny w}}{\to}}0$, then $\langle f_{n},k_{x}\rangle\to
0$ for each $x\in X$. Then, by dominated convergence,

$\langle T_{\sigma}f_{n},f_{n}\rangle=\int\sigma(x)|\langle
f_{n},k_{x}\rangle|^{2}\,d\mu(x)\to 0.$

Secondly, since the translations are continuous on $L^{1}(X)$, one can find
$h_{k}$ such that $\|\sigma_{h_{k}}-\sigma\|_{L^{1}}\leq 2^{-k}$ which implies
$\rho(x)=\sup_{k}\sigma_{h_{k}\cdots h_{1}}(x)$ is still in $L^{1}$. Then we
follow the same proof as Theorem 2. ∎

##  5 Applications

###  5.1 Short-Time Fourier Transform on LCA groups

Theorem 3 is similar to Benedicks Theorem (also called the Amrein-Berthier
Theorem or Qualitative Uncertainty Principle) for the Plancherel groups,
proved in [2], which states that if $f\in L^{2}(G,\mu)$ and $\hat{f}\in
L^{2}(\hat{G},\hat{\mu})$ are each supported on sets of finite $\mu$ and
$\hat{\mu}$ measure respectively, then $f=0$. This uncertainty principle
limits the joint time-frequency distribution of these functions. Namely, it
states that the joint time-frequency support (a set in $\hat{G}\times G$)
cannot be contained in a set of finite $\hat{\mu}\otimes\mu$ measure.

However, looking at other joint time-frequency distributions, such as the
Wigner distribution, Ambiguity function, or Short-Time Fourier transform
(STFT) yields a stronger result. We will only focus on the STFT, which is
defined, for $f,\phi\in L^{2}(G)$,

$V_{\phi}f(p,q)=\int_{G}f(t)\overline{p(q^{-1}t)\phi(q^{-1}t)}\,d\mu(t)=p(q)\hat{(}f\phi_{q})(p)$

for $(p,q)\in\hat{G}\times G$. We will only deal with locally compact groups
$G$ which are Abelian and second countable. The second countability is used to
ensure that the invariant metric $d$ is proper. In this way, by the Plancherel
theorem for $G$, we obtain

$\langle
V_{\phi}f,V_{\psi}g\rangle=\int_{G}\langle\hat{(}f\cdot\overline{\phi_{q}}),\hat{(}g\cdot\overline{\psi_{q}})\rangle_{L^{2}(\hat{G},\hat{\mu})}\,d\mu(q)$
$=\int_{G}\int_{G}f(x)\overline{\phi(q^{-1}x)g(x)}\psi(q^{-1}x)\,d\mu(x)\,d\mu(q)=\langle
f,g\rangle\cdot\langle\psi,\phi\rangle.$

Which shows that $\phi_{p,q}(x)=p(q^{-1}x)\phi(q^{-1}x)$ is a Parseval frame
for $L^{2}(G)$ if $\|\phi\|=1$. This implies
$\|\phi\|\cdot\|f\|=\|V_{\phi}f\|$. In order to apply Theorem 2, we take
${\hat{G}\times G}$ to be the homogeneous space with the measure
$\hat{\mu}\otimes\mu$ acted on by the “Heisenberg” group $H(G):={\hat{G}\times
G}\times\mathbb{T}$ with the operation

$(p^{\prime},q^{\prime},z^{\prime})(p,q,z):=(pp^{\prime},qq^{\prime},zz^{\prime}p(q^{\prime})).$

$H(G)$ acts on ${\hat{G}\times G}$ by
$(p,q,z)(p^{\prime},q^{\prime})=(pp^{\prime},qq^{\prime})$. It can also be
checked that

$|\langle\phi_{(p,q,z)(p^{\prime},q^{\prime})},\phi_{(p,q,z)(p^{\prime\prime},q^{\prime\prime})}\rangle|=|\langle\phi_{p^{\prime},q^{\prime}},\phi_{p^{\prime\prime},q^{\prime\prime}}\rangle|$

and moreover
$\langle\phi_{(1,q,1)(p^{\prime},q^{\prime})},\phi_{(1,q,1)(p^{\prime\prime},q^{\prime\prime})}\rangle=\langle\phi_{p^{\prime},q^{\prime}},\phi_{p^{\prime\prime},q^{\prime\prime}}\rangle$
so that $\\{1_{\hat{G}}\\}\times G\times\\{1\\}\leq H(G)_{\tau}$. If we had
used the more conventional definition of $V_{\phi}f(p,q)$ to be
$\hat{(}f\phi_{q})(p)$ then $\hat{G}\times\\{1_{G}\\}\times\\{1\\}\leq
H(G)_{\tau}$ instead.

######  Theorem 4.

Let $G$ be a noncompact second countable LCA group. For any $\phi\in
L^{2}(G)$, $\sigma:{\hat{G}\times G}\to[0,1]$ is thin if and only if

$T_{\sigma}f=\int_{\hat{G}\times G}\sigma(p,q)\langle
f,\phi_{p,q}\rangle\phi_{p,q}\,d({\hat{\mu}\otimes\mu})(p,q)$

is compact.

###### Proof.

To apply our compactness characterization, Theorem 1, let us first show that
for $\phi\in L^{2}(G)$,
$\langle\phi_{p,q},\phi_{p^{\prime},q^{\prime}}\rangle\to 0$ as
$d((p,q),(p^{\prime},q^{\prime}))\to\infty$. By invariance of the frame, it
enough to check $\langle\phi,\phi_{p,q}\rangle\to 0$ as $(p,q)\to\infty$.
First, if $p\to\infty$ and $q$ remains bounded, then for each $q$, by the
Riemann-Lebesgue Lemma $\lim_{p\to\infty}\langle\phi,\phi_{p,q}\rangle=0$.
Therefore since $q$ remains in a compact set,
$\langle\phi,\phi_{p,q}\rangle\to 0$. Otherwise, we consider $q\to\infty$. Let
$\varepsilon>0$. Since $\phi\in L^{2}(G)$, there exists $R$ such that

$\int_{B(1_{G},R)^{c}}|\phi|^{2}\,d\mu\leq\varepsilon.$

Then, there exists $M$ such that for $d(q,1)>M$, $B(q^{-1},R)\subset
B(1_{G},R)^{c}$. Then,

$|\langle\phi,\phi_{(p,q)}\rangle|\leq\int_{B(1_{G},R)}+\int_{B(1_{G},R)^{c}}|\phi||\phi(q^{-1}\cdot)|\,d\mu\leq
2\|\phi\|\varepsilon.$

Applying Theorem 1 gives the necessity. To show sufficiency, consider the
Feichtinger algebra

$S_{0}(G):=\\{\phi\in L^{2}(G):V_{\phi}\phi\in L^{1}({\hat{G}\times G})\\}.$

It is known that $S_{0}(G)$ is dense in $L^{2}(G)$, see [21, Lemma 4.19].
First note that $T_{\sigma}$ is compact for any $\phi\in S_{0}(G)$ by Theorem
1. Then,

$\langle T_{\sigma}^{\phi,\psi}f,g\rangle=\int\sigma(p,q)\langle
f,\phi_{p,q}\rangle\langle\psi_{p,q},g\rangle$
$\leq\|\sigma\|_{\infty}\|f\|_{L^{2}(G)}\|g\|_{L^{2}(G)}\|\psi\|_{L^{2}(G)}\|\phi\|_{L^{2}(G)}.$

Therefore, we have
$\|T_{\sigma}^{\phi,\psi}\|\leq\|\sigma\|_{\infty}\|\phi\|_{L^{2}(G)}\|\psi\|_{L^{2}(G)}$.
Then,
$T_{\sigma}^{\phi,\phi}-T_{\sigma}^{\phi^{\prime},\phi^{\prime}}=T_{\sigma}^{\phi-\phi^{\prime},\phi}+T_{\sigma}^{\phi,\phi-\phi^{\prime}}$.
This concludes the proof since compact operators are closed in the operator
norm topology. ∎

Putting this together with Theorem 2, we obtain the following uncertainty
principle (see [15] for the case $G=\mathbb{R}^{d}$ and $\sigma$ an indicator
function).

######  Theorem 5.

Let $G$ be a second countable LCA group containing a subgroup $\Gamma$ such
that $\Gamma$ has an accumulation point at $1$ and $\Gamma$ has no nontrivial
compact subgroups. Let $\|\phi\|_{L^{2}(G)}=1$. If $\sigma:{\hat{G}\times
G}\to[0,1]$ is thin, then there exists $c>0$ such that

$\langle T_{1-\sigma}f,f\rangle=\int_{{\hat{G}\times
G}}(1-\sigma)|V_{\phi}f|^{2}\,d({\hat{\mu}\otimes\mu})\geq c\|f\|^{2}$

for all $f\in L^{2}(G)$.

###### Proof.

The subgroup $\\{1_{\hat{G}}\\}\times\Gamma\times\\{1\\}\leq H(G)$ gives the
independence of translations and the tautology that $\sigma$ is thin if and
only if $T_{\sigma}$ is compact concludes the proof using the argument of
Theorem 2. ∎

######  Corollary 2.

Under the assumptions of Theorem 5, if $E\subset{\hat{G}\times G}$ is thin,
then there exists $c>0$ such that

$\int_{{\hat{G}\times G}\backslash
E}|V_{\phi}f|^{2}\,d{\hat{\mu}\otimes\mu}\geq c\|f\|^{2}$

for all $f\in L^{2}(G)$.

In particular, $\operatorname{supp}V_{\phi}f$ cannot be thin unless $f=0$.

######  Remark 2.

If $\phi$ and $f$ are both supported on compact sets $K_{1}$ and $K_{2}$, then
$\operatorname{supp}V_{\phi}f\subset\hat{G}\times K_{2}K_{1}^{-1}$. This shows
the thinness condition cannot be relaxed from vanishing at infinity to some
smallness at infinity condition.

###  5.2 Wavelet Transform

Next, we apply these results to wavelets, which are a special case of the
affine group acting on $L^{2}(\mathbb{R}^{d})$. For $\phi\in
L^{2}(\mathbb{R}^{d})$, define the following unitary representation of the
wavelet group $G=(0,\infty)\times\mathbb{R}^{d}$ on $L^{2}(\mathbb{R}^{d})$ by

$\phi_{a,b}(x)=a^{-1/2}\phi(a^{-1}x-b)$ (9)

for $a\in(0,\infty)$ and $b\in\mathbb{R}^{d}$. The group operation is then
$(a_{1},b_{1})(a_{2},b_{2})=(a_{1}a_{2},a_{2}^{-1}b_{1}+b_{2})$ and the Haar
measure is $d\lambda(a,b)=a^{-1}\,da\,db$.

We say that a wavelet $\phi\in L^{2}(\mathbb{R}^{d})$ is admissible if

$\int_{0}^{\infty}\dfrac{|\hat{\phi}(x\xi)|^{2}}{\xi}\,d\xi=1$ (10)

for a.e. $x\in\mathbb{R}^{d}$. Due to the invariance of the measure
$a^{-1}\,da$, it is enough to check this for almost every
$x\in\mathbb{S}^{d-1}$. We define the set $\mathcal{A}$ to be the set of all
admissible $\phi$.

If $\phi$ is admissible, then $\\{\phi_{g}\\}_{g\in G}$ does form a
generalized Parseval frame with the measure $\lambda$. To see this, define the
wavelet transform $W_{\phi}f(a,b):=\langle f,\phi_{a,b}\rangle$. Then, since
$\int e^{-ib\xi}W_{\phi}f(a,b)\,db=a^{-1/2}\hat{f}(a^{-1}\xi)\hat{\phi}(\xi)$,
by the Plancherel theorem on $\mathbb{R}^{d}$,

$\int_{G}|W_{\phi}f|^{2}\,d\lambda=\int_{0}^{\infty}\int_{\mathbb{R}^{d}}|\hat{f}(a^{-1}\xi)\hat{\phi}(\xi)|^{2}a^{-1}\,d\xi
a^{-1}\,da$
$=\int_{\mathbb{R}^{d}}|\hat{f}(\eta)|^{2}\int_{0}^{\infty}|\hat{\phi}(a\eta)|^{2}\,a^{-1}da\,d\eta=\|f\|^{2}.$

We want to check the Schur condition (i) and (ii) in Theorem 2 so we study the
decay properties of $W_{\phi}\phi$. Define

$\mathcal{B}^{1}_{w}=\\{\phi\in L^{2}(\mathbb{R}^{d}):W_{\phi}\phi\in
L^{1}(w\,d\lambda)\\}$

for a weight $w$. We can show that a very large class of functions is
contained in $\mathcal{B}^{1}_{w}$. Define the translation operator
$\tau_{h}f(x)=f(x+h)$. For $0<\alpha\leq 1$, denote by $\Lambda_{\alpha}$ the
class of $L^{1}$ functions such that $\|\tau_{h}f-f\|_{L^{1}}\leq
Ch^{\alpha}$. $\Lambda_{1}$ contains the Schwarz functions as well as less
smooth functions like indicator functions (thus including the Haar wavelet).

######  Lemma 4.

Let $0<\varepsilon<\alpha\leq 1$ and
$w_{\varepsilon}(a,b)=a^{d/2+\varepsilon}$. Then,

$\Lambda_{\alpha}\cap
L_{0}^{1}(|x|^{\alpha})\subset\mathcal{B}^{1}_{w_{\varepsilon}}$

where $L^{1}_{0}(|x|^{\alpha})=\\{f\in L^{1}(\mathbb{R}^{d}):\int f=0\mbox{
and }\int|f(x)|\,|x|^{\alpha}\,dx<\infty\\}$.

In particular, this weight is multiplicative, so by Remark 1, $\phi_{a,b}$
satisfies the Schur conditions (i) and (ii) in Theorem 2 for any
$\phi\in\Lambda_{\alpha}\cap L^{1}_{0}(|x|^{\alpha})$.

###### Proof.

We split $(0,\infty)=(0,1)\cup(1,\infty)$. On $(0,1)$,

$\int_{\mathbb{R}^{d}}\int_{0}^{1}\left|\int\phi(x)\phi(a^{-1}x-b)\,dx\right|a^{-d/2}w(a)\,\dfrac{da}{a}\,db$
$\leq\|\phi\|_{L^{1}}^{2}\int_{0}^{1}a^{d/2+\varepsilon}\,\dfrac{da}{a^{d/2+1}}<\infty$

On the other hand, using the mean zero property of $\phi$, we estimate

$\int|W_{\phi}\phi(a,b)|\,db=a^{-d/2}\int\left|\int\phi(x)\left[\phi(a^{-1}x-b)-\phi(-b)\right]\,dx\right|\,db$
$\leq a^{-d/2}\int|\phi(x)|\|\tau_{a^{-1}x}\phi-\phi\|_{L^{1}}\,dx.\leq
Ca^{-d/2-\alpha}\int|\phi(x)|\cdot|x|^{\alpha}\,dx$

Therefore,

$\int_{\mathbb{R}^{d}}\int_{1}^{\infty}|W_{\phi}\phi|w\,d\lambda\leq
C\int_{1}^{\infty}\dfrac{w(a)}{a^{d/2+1+\alpha}}\,da\|\phi|x|^{\alpha}\|_{L^{1}}.$

Taking $\varepsilon<\alpha$ ensures that the $a$ integral is finite. ∎

This lemma gives us plenty of information about the space
$\mathcal{B}^{1}_{w}$. It can be verified that the admissibility condition
(10) holds for any radial, normalized mean zero function in $L^{1}\cap
L^{1}(|x|)\cap L^{2}$. From this discussion and the previous lemma, we have

$L^{1}\cap L^{1}_{0}(|x|)\cap L^{2}\cap(\cup_{0<\alpha\leq
1}\Lambda_{\alpha})\cap\\{\phi\mbox{ radial }\\}$
$\subset(\cup_{0<\varepsilon<1}\mathcal{B}^{1}_{w_{\varepsilon}})\cap\mathcal{A}=:\mathcal{A}_{1}.$

Therefore, by Theorem 1, the following compactness result holds for many
wavelets $\phi$. In particular, all Schwarz functions and the Haar function.

######  Theorem 6.

Let $\phi\in\mathcal{A}_{1}$ and $\sigma\in
L^{\infty}((0,\infty)\times\mathbb{R}^{d})$ be thin. Then,

$T_{\sigma}f(x)=\int\sigma(a,b)W_{\phi}f(a,b)\phi_{a,b}(x)\,\dfrac{da}{a}db$

is compact.

The converse also holds if $\phi$ is admissible and Schwartz since
$\langle\phi_{a,b},\phi\rangle\to 0$ as $d((a,b),(1,0))\to\infty$, see for
example [17, Appendix, Lemmas 2 and 4]. This yields the following positivity
result, by taking $\Gamma$ in Theorem 2 to be the subgroup
$\\{1\\}\times\mathbb{R}^{d}$.

######  Theorem 7.

Let $\phi\in\mathcal{A}_{1}$ and
$\sigma:(0,\infty)\times\mathbb{R}^{d}\to[0,1]$ be thin. Then there exists
$c>0$ such that

$\langle
T_{1-\sigma}f,f\rangle=\int_{(0,\infty)\times\mathbb{R}^{d}}(1-\sigma)|W_{\phi}f|^{2}\,d\lambda\geq
c\|f\|^{2}$

for all $f\in L^{2}(\mathbb{R}^{d})$.

As an immediate consequence, we obtain the following uncertainty principle.

######  Corollary 3.

Let $\phi\in\mathcal{A}_{1}$. If $E\subset\mathbb{R}^{d}\times(0,\infty)$ is
thin, then there exists $c>0$ such that

$\int_{(0,\infty)\times\mathbb{R}^{d}\backslash
E}|W_{\phi}f(a,b)|^{2}\dfrac{da\,db}{a}\geq c\|f\|^{2}$

for all $f\in L^{2}(\mathbb{R}^{d})$.

As in the STFT case, this implies $\operatorname{supp}W_{\phi}f$ can only be
thin if $f=0$, and this is sharp in the sense that we cannot improve from
vanishing sets to small ones for general $\phi$. If $f$ and $\phi$ are both
supported in a ball $B$, then $W_{\phi}f$ is supported the region
$\\{(a,b):b\in a^{-1}B-B\\}$ which contains the strip $(0,\infty)\times B$.

We also mention that these results continue to hold for higher dimensional
wavelet transforms such as the shearlet [26], but describing the classes
$\mathcal{B}^{1}_{w}$ and $\mathcal{A}$ is more difficult. However, we mention
that Schwarz functions with Fourier support in a bounded set away from the
$y$-axis are included in $\mathcal{B}^{1}$ without any weight as shown in
[26].

### Acknowledgements

The authors would like to thank Karlheinz Gröchenig for his comments.

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