# Fourier-like multipliers and applications for integral operators

Saifallah Ghobber  LR11ES11 Analyse Mathématiques et Applications, Faculté des
Sciences de Tunis, Université de Tunis El Manar, 2092 Tunis, Tunisie
<EMAIL_ADDRESS>
###### Abstract.

Timelimited functions and bandlimited functions play a fundamental role in
signal and image processing. But by the uncertainty principles, a signal
cannot be simultaneously time and bandlimited. A natural assumption is thus
that a signal is almost time and almost bandlimited. The aim of this paper is
to prove that the set of almost time and almost bandlimited signals is not
excluded from the uncertainty principles. The transforms under consideration
are integral operators with bounded kernels for which there is a Parseval
Theorem. Then we define the _wavelet_ multipliers for this class of operators,
and study their boundedness and Schatten class properties. We show that the
wavelet multiplier is unitary equivalent to a scalar multiple of the phase
space restriction operator. Moreover we prove that a signal which is almost
time and almost bandlimited can be approximated by its projection on the span
of the first eigenfunctions of the phase space restriction operator,
corresponding to the largest eigenvalues which are close to one.

###### Key words and phrases:

Multiplier, localization operator, uncertainty principle, Nash inequality,
Carlson inequality.

###### 1991 Mathematics Subject Classification:

81S30, 94A12, 45P05, 42C25, 42C40

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Notation
    2. 2.2 Generalities
    3. 2.3 Fourier-like Multipliers
  3. 3 Uncertainty principles by means of the frequency limiting operator
    1. 3.1 Uncertainty principles on the space $L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$
    2. 3.2 Uncertainty principles on the space $L^{1}\cap L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$
  4. 4 The wavelet multiplier
    1. 4.1 Boundedness
    2. 4.2 Schatten class properties
    3. 4.3 An uncertainty relation
  5. 5 Uncertainty principles for orthonormal sequences
    1. 5.1 The phase space restriction operator
    2. 5.2 Shapiro–type uncertainty principles
      1. 5.2.1 Localization theorem
      2. 5.2.2 Quantitative dispersion inequality for orthonormal sequences
      3. 5.2.3 Strong uncertainty principle for orthonormal bases

##  1\. Introduction

Timelimited functions and bandlimited functions are basic tools of signal and
image processing. Unfortunately, the simplest form of the uncertainty
principle tells us that a signal cannot be simultaneously time and
bandlimited. This leads to the investigation of the set of almost time and
almost bandlimited functions, which has been initially carried through Landau,
Pollak [17, 18] and then by Donoho, Stark [9]. In the current paper, we review
the uncertainty principles on this set and present and compare different
measure of localization. We made use of compositions of time and bandlimiting
operators and considered the eigenvalue problem associated with these
operators. The resulting operators yield an orthonormal set of eigenfunctions
(well-known as prolate spheroidal functions) which satisfy some optimality in
concentration in a region in the time-frequency domain. We prove a
characterization of functions that are approximately time and bandlimited in
the region of interest, and we obtain approximation inequalities for such
functions using a finite linear combination of eigenfunctions.

The aim of this paper is to continue the study of the uncertainty principle to
a very general class of integral operators, which has been started in [11,
12]. The transforms under consideration are integral operators ${\mathcal{T}}$
with bounded kernels ${\mathcal{K}}$ and for which there is a Parseval
Theorem. This class includes the usual Fourier transform, the Fourier-Bessel
(or Hankel) transform, the Dunkl transform and the deformed Fourier transform
as particular cases. A version of Hardy’s and Donoho-Stark’s uncertainty
principles for integral operators has been proved in [7, 8]. In this paper, we
consider results of a different nature on the subspaces of functions that are
essentially timelimited on $S$ and bandlimited on $\Sigma$, or functions that
are essentially concentrated on $S$ and bandlimited on $\Sigma$, where $S$ and
$\Sigma$ are general subsets of finite measure.

Let us now be more precise. Let $\Omega$ be a convex cones in
${\mathbb{R}}^{d}$ (i.e. $\lambda x\in\Omega$ if $\lambda>0$ and $x\in\Omega$)
with non-empty interior, endowed with the Borel measure $\mu$. The Lebesgue
spaces $L^{p}(\Omega,\mu)$, $1\leq p\leq\infty$, are then defined in the usual
way, where $\|\cdot\|_{\infty}$ is the usual essential supremum norm and form
$1\leq p<\infty$,

$\|f\|_{p,\mu}^{p}=\int_{\Omega}|f(x)|^{p}\,{\mathrm{d}}\mu(x).$

We assume that the measure $\mu$ is absolutely continuous with respect to the
Lebesgue measure and has a polar decomposition of the form
$\mbox{d}\mu(r\zeta)=r^{2a-1}\,\mbox{d}r\,Q(\zeta)\,\mbox{d}\sigma(\zeta)$
where $\mbox{d}\sigma$ is the Lebesgue measure on the unit sphere
${\mathbb{S}}^{d-1}$ of ${\mathbb{R}}^{d}$ and $Q\in
L^{1}({\mathbb{S}}^{d-1},\mbox{d}\sigma)$, $Q\not=0$. Then $\mu$ is
homogeneous of degree $2a$ in the following sense: for every continuous
function $f$ with compact support in $\Omega$ and every $\lambda>0$,

$\int_{\Omega}f\left(\frac{x}{\lambda}\right)\,{\mathrm{d}}\mu(x)=\lambda^{2a}\int_{\Omega}f(x)\,{\mathrm{d}}\mu(x).$
(1.1)

One can then define the integral operator ${\mathcal{T}}$ on
${\mathcal{S}}(\Omega)$ by

${\mathcal{T}}f(\xi)=\int_{\Omega}f(x){\mathcal{K}}(x,\xi)\,{\mathrm{d}}\mu(x),\quad\xi\in\Omega,$
(1.2)

where ${\mathcal{K}}:\Omega\times\Omega\longrightarrow{\mathbb{C}}$ is a
kernel such that:

  1. (1)

${\mathcal{K}}$ is continuous,

  2. (2)

${\mathcal{K}}$ is bounded: $|{\mathcal{K}}(x,\xi)|\leq c_{{\mathcal{K}}}$,

  3. (3)

${\mathcal{K}}$ is homogeneous: ${\mathcal{K}}(\lambda
x,\xi)={\mathcal{K}}(x,\lambda\xi)$.

Then ${\mathcal{T}}$ extends into a continuous operator from
$L^{1}(\Omega,\mu)$ to the space of bounded continuous functions
${\mathcal{C}}(\Omega)$, with

${\left\|{{\mathcal{T}}f}\right\|}_{\infty}\leq
c_{\mathcal{K}}{\left\|{f}\right\|}_{1,\mu}.$ (1.3)

Further, if we introduce the dilation operator ${\mathcal{D}}_{\lambda}$,
$\lambda>0$ by:

${\mathcal{D}}_{\lambda}f(x)=\frac{1}{\lambda^{a}}f\left(\frac{x}{\lambda}\right).$

Then the homogeneity of ${\mathcal{K}}$ implies

${\mathcal{T}}{\mathcal{D}}_{\lambda}={\mathcal{D}}_{\frac{1}{\lambda}}{\mathcal{T}}.$
(1.4)

The integral operators under consideration will be assumed to satisfy some of
the following properties that are common for Fourier-like transforms:

  1. (1)

${\mathcal{T}}$ has an Inversion Formula: When both $f\in L^{1}(\Omega,\mu)$
and ${\mathcal{T}}f\in L^{1}(\Omega,\mu)$ we have $f\in{\mathcal{C}}(\Omega)$
and for almost every $x\in\Omega$:

$f(x)=\int_{\Omega}{\mathcal{T}}f(\xi)\overline{{\mathcal{K}}(x,\xi)}\,{\mathrm{d}}\mu(\xi).$
(1.5)

  2. (2)

${\mathcal{T}}$ satisfies Parseval’s Theorem: for every
$f,g\in{\mathcal{S}}(\Omega)$,

${\left\langle{{\mathcal{T}}f,{\mathcal{T}}g}\right\rangle}_{\mu}={\left\langle{f,g}\right\rangle}_{\mu},$
(1.6)

where ${\left\langle{\cdot,\cdot}\right\rangle}_{\mu}$ is the inner product
defined on the Hilbert spaces $L^{2}(\Omega,\mu)$ by

${\left\langle{f,g}\right\rangle}_{\mu}=\int_{\Omega}f(x)\overline{g(x)}\,{\mathrm{d}}\mu(x).$

In particular, ${\mathcal{T}}$ extends to an unitary transform from
$L^{2}(\Omega,\mu)$ onto $L^{2}(\Omega,\mu)$, such that

${\left\|{{\mathcal{T}}f}\right\|}_{2,\mu}={\left\|{f}\right\|}_{2,\mu},$
(1.7)

and for all $f\in L^{2}(\Omega,\mu)$,

${\mathcal{T}}^{-1}f(\xi)=\overline{{\mathcal{T}}\bar{f}(\xi)},\quad\xi\in\Omega.$
(1.8)

This family of transforms includes for instance the Fourier transform (see
e.g. [20]), the Hankel transform (see e.g. [13]), the the Dunkl transform (see
e.g. [14]), the $G$-transform (see e.g. [29]), the deformed Fourier transform
(see e.g. [5]),…

The inversion formula gives us back a signal $f$ via (1.5), and this is the
basis for pseudo-differential operators on $\Omega$. Indeed if $\sigma$ be a
suitable function on $\Omega$, then we define the pseudo-differential operator
$F_{\sigma}$ by

$F_{\sigma}f(x)=\int_{\Omega}\sigma(\xi){\mathcal{T}}f(\xi)\overline{{\mathcal{K}}(x,\xi)}\,{\mathrm{d}}\mu(\xi).$

Pseudo-differential operators $F_{\sigma}$ are known as the multiplier, and in
the case where $\sigma=\chi_{A}$ is a characteristic function, the operator
$F_{\sigma}$ is known as the frequency limiting operator on $\Omega$, we
simply denote it by $F_{A}$. Now if $\sigma$ is identically equal to $1$, then
$F_{\sigma}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ is the identity operator
in view of (1.8).

Our starting point is the following general form of Heisenberg–type
uncertainty inequality (see [11, 12]).

######  Theorem 1.1.

Let $s,\;\beta>0$. Then

  1. (1)

there exists a constant $C=C(s,a,\beta)$ such that for all $f\in
L^{2}(\Omega,\mu)$,

$\||x|^{s}f\|^{\beta}_{2,\mu}\;\||\xi|^{\beta}{\mathcal{T}}f\|^{s}_{2,\mu}\geq
C{\left\|{f}\right\|}^{s+\beta}_{2,\mu},$ (1.9)

  2. (2)

there exists a constant $C=C(s,a,\beta)$ such that for all $f\in
L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$,

$\||x|^{s}f\|_{1,\mu}^{a+\beta}\;\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{a+s}\geq
C\,{\left\|{f}\right\|}_{1,\mu}^{a+s}\,{\left\|{f}\right\|}_{2,\mu}^{a+\beta}.$
(1.10)

The proof of Inequality (1.10) can be obtained by combining a Nash-type
inequality[12, Proposition 2.2] and a Carlson-type inequality[12, Proposition
2.3], while the proof of Inequality (1.9) can be obtained from either the
Faris-type local uncertainty inequalities [11, Theorem A], or from the fact
that the Benedicks-Amrein-Berthier uncertainty inequality [11, Theorem B].

Theorem 1.1 can be refined for orthonormal sequence in $L^{2}(\Omega,\mu)$. In
particular an orthonormal sequence in $L^{2}(\Omega,\mu)$ cannot have uniform
time-frequency localization. This is a consequence of the following Shapiro-
type uncertainty principles.

Theorem A. Let $s>0$ and let $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ be an
orthonormal sequence in $L^{2}(\Omega,\mu)$.

  1. (1)

There exists a positive constant $C$ such that, for every $N\geq 1$,

$\displaystyle\sum_{n=1}^{N}\left(\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu}+\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{2}_{2,\mu}\right)\geq
C\,N^{1+\frac{s}{2a}}.$ (1.11)

  2. (2)

If $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ is an orthonormal basis for
$L^{2}(\Omega,\mu)$, then

$\sup_{n}\left(\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}\,\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}\right)=\infty.$
(1.12)

Notice that the homogeneity of the kernel ${\mathcal{K}}$ plays a key role
only in the proof of Inequality (1.12). Moreover the proof of Theorem A is
inspired from the classical result for the Fourier transform in [20], where
the author prove that Inequality (1.11) is sharp and the equality cases are
attained for the sequence of Hermite functions, (see also [19]). For the
Hankel transform [13] this inequality is also optimal and the optimizers are
the sequence of Laguerre functions.

One would like to find nonzero functions $f\in L^{2}(\Omega,\mu)$, which are
timelimited on a subset $S\subset\Omega$ (i.e. ${\operatorname{supp}}f\subset
S$) and bandlimited on a subset $\Sigma\subset\Omega$ (i.e.
${\operatorname{supp}}{\mathcal{T}}f\subset\Sigma$). Unfortunately, such
functions do not exist, because if $f$ is time and bandlimited on subsets of
finite measure, then $f=0$ (see [11]). As a result, it is natural to replaced
the exact support by the essential support, and to focus on functions that are
essentially time and bandlimited to a bounded region like $S\times\Sigma$ in
the time-frequency plane. To do this, we introduce the time limiting operator

$E_{S}f=\chi_{S}f,\quad f\in L^{1}(\Omega,\mu)\cup L^{2}(\Omega,\mu),$

and from [1, 9, 12] we recall the following definition.

######  Definition 1.2.

Let $0\leq\varepsilon<1$ and let $S,\Sigma\subset\Omega$. Then

  1. (1)

a nonzero function $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-concentrated on
$S$ if $\|E_{S^{c}}f\|_{2,\mu}\leq\varepsilon\|f\|_{2,\mu}$,

  2. (2)

a nonzero function $f\in L^{1}(\Omega,\mu)$ is $\varepsilon$-timelimited on
$S$ if $\|E_{S^{c}}f\|_{1,\mu}\leq\varepsilon\|f\|_{1,\mu}$,

  3. (3)

a nonzero function $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-bandlimited on
$\Sigma$ if $\|F_{\Sigma^{c}}f\|_{2,\mu}\leq\varepsilon\|f\|_{2,\mu}$,

  4. (4)

a nonzero function $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-localized with
respect to an operator $L:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ if
${\left\|{Lf-f}\right\|}_{2,\mu}\leq\varepsilon\|f\|_{2,\mu}.$

Here $A^{c}=\Omega\backslash A$ is the complement of $A$ in $\Omega$. It is
clear that, if $f$ is $\varepsilon$-bandlimited on $\Sigma$ then by Inequality
(1.7), ${\mathcal{T}}f$ is $\varepsilon$-concentrated on $\Sigma$. Notice also
that, the $\varepsilon$-concentration measure was introduced in [17, 18, 9],
and the idea of $\varepsilon$-localization has been recently introduced in
[1], which arises from the concept of pseudospectra of linear operators.

If $\varepsilon=0$ in the $\varepsilon$-concentration measures, then $S$ and
$\Sigma$ are respectively the exact support of $f$ and ${\mathcal{T}}f$,
moreover when $\varepsilon\in(0,1)$, $S$ and $\Sigma$ may be considered as the
_essential_ support of $f$ and ${\mathcal{T}}f$ respectively. On the other
hand, a function $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-localized with
respect to an operator $L$ is an eigenfunction of $L$ corresponding to the
eigenvalue $1$, if $\varepsilon=0$, otherwise $f$ is called an
$\varepsilon$-approximated eigenfunction (or
$\varepsilon$-pseudoeigenfunction) of $L$ with pseudoeigenvalue $1$, see [17,
24]. For example, since on $L^{2}(\Omega,\mu)$, we have
$F_{\Sigma}+F_{\Sigma^{c}}=E_{S}+E_{S^{c}}=I$, then a nonzero function $f\in
L^{2}(\Omega,\mu)$ is $\varepsilon$-concentrated on $S$ (resp.
$\varepsilon$-bandlimited on $\Sigma$), if and only if, $f$ is
$\varepsilon$-localized with respect to $E_{S}$ (resp. $\varepsilon$-localized
with respect to $F_{\Sigma}$).

Let $\varepsilon_{1},\varepsilon_{2}\in(0,1)$ and let $S$, $\Sigma$ two
measurable subsets of $\Omega$ such that $0<\mu(S),\,\mu(\Sigma)<\infty$. We
denote by $L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ the subspace of
$L^{2}(\Omega,\mu)$ consisting of functions that are
$\varepsilon_{1}$-concentrated on $S$ and $\varepsilon_{2}$-bandlimited on
$\Sigma$ (clearly $L^{2}(0,0,S,\Sigma)=\emptyset$). We denote also by
$L^{1}\cap L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ the subspace of
$L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$ consisting of functions that are
$\varepsilon_{1}$-timelimited on $S$ and $\varepsilon_{2}$-bandlimited on
$\Sigma$.

As a first result, we can remark that the essential supports $S$ and $\Sigma$
cannot be too small, and this is a simple consequence of the following Donoho-
Stark type uncertainty principle (see [8, 11, 12]):

  1. (1)

If $f\in L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ such that
$\varepsilon_{1}^{2}+\varepsilon_{2}^{2}<1$, then

$\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-2}\left(1-\sqrt{\varepsilon_{1}^{2}+\varepsilon_{2}^{2}}\right)^{2}.$
(1.13)

  2. (2)

If $f\in L^{1}\cap L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then

$\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-2}(1-\varepsilon_{1})^{2}(1-\varepsilon_{2}^{2}).$ (1.14)

The second Inequality (1.14) is stronger than (1.13), since it is true for all
$\varepsilon_{1},\varepsilon_{2}\in(0,1)$, and since in (1.14) we can give
separately a lower bound for $\mu(S)$ and $\mu(\Sigma)$, (see Inequality
(3.22).

It is natural to ask if there is a Heisenberg-type uncertainty inequalities
for functions in the subspaces
$L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ and $L^{1}\cap
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, with constant which depends
on $\varepsilon_{1},\varepsilon_{2},S,\Sigma$. In Section 3, we use the local
uncertainty principles for functions either in $L^{2}(\Omega,\mu)$ or in
$L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$ to obtain an uncertainty
inequalities comparing the support and the essential support with the time
dispersion or the frequency dispersion.

Theorem B. Let $\varepsilon_{1},\varepsilon_{2}\in[0,1)$.

  1. (1)

If $s,\beta>a$, then

    1. (a)

there exists a constant $C$ such that for all function $f\in
L^{2}(\Omega,\mu)$ which is $\varepsilon_{1}$-concentrated on $S$,

$\mu(S)^{\frac{\beta}{a}}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{1}^{2}\right)^{\frac{\beta}{a}}\|f\|_{2,\mu}^{2},$
(1.15)

    2. (b)

there exists a constant $C$ such that for all function $f\in
L^{2}(\Omega,\mu)$ which is $\varepsilon_{2}$-bandlimited on $\Sigma$,

$\mu(\Sigma)^{\frac{s}{a}}\,{\left\|{|x|^{s}f}\right\|}_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{2}^{2}\right)^{\frac{s}{a}}\|f\|_{2,\mu}^{2}.$ (1.16)

  2. (2)

If $s,\beta>0$, then

    1. (a)

there exists a constant $C$ such that for all function $f\in
L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$, which is
$\varepsilon_{1}$-timelimited on $S$,

$\mu(S)^{\frac{a+\beta}{2a}}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}\geq
C\left(1-\varepsilon_{1}\right)^{\frac{a+\beta}{a}}{\left\|{f}\right\|}_{1,\mu},$
(1.17)

    2. (b)

there exists a constant $C$ such that for all function $f\in
L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$, which is
$\varepsilon_{2}$-bandlimited on $\Sigma$,

$\mu(\Sigma)^{\frac{a+s}{2a}}\,{\left\|{|x|^{s}f}\right\|}_{1,\mu}\geq
C\left(1-\varepsilon_{2}^{2}\right)^{\frac{a+s}{2a}}{\left\|{f}\right\|}_{2,\mu}.$
(1.18)

Of course, if $\varepsilon_{1}=\varepsilon_{2}=0$, then
$S={\operatorname{supp}}f$ and $\Sigma={\operatorname{supp}}{\mathcal{T}}f$.
Combining the inequalities in Theorem A, we obtain the following Heisenberg-
type uncertainty principles, which can be viewed as the
$\varepsilon$-concentration version of Theorem 1.1:

  1. (1)

If $s,\beta>a$, then there exists a constant $C$ such that for all $f\in
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$,

${\left\|{|x|^{s}f}\right\|}_{2,\mu}^{\beta}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{s}\geq
C\left(\frac{(1-\varepsilon_{1}^{2})(1-\varepsilon_{2}^{2})}{\mu(S)\mu(\Sigma)}\right)^{\frac{s\beta}{2a}}{\left\|{f}\right\|}_{2,\mu}^{s+\beta}.$
(1.19)

  2. (2)

If $s,\beta>0$, then there exists a constant $C$ such that for all $f\in
L^{1}\cap L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$,

${\left\|{|x|^{s}f}\right\|}_{1,\mu}^{a+\beta}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{a+s}\geq
C\left(\frac{(1-\varepsilon_{1})^{2}(1-\varepsilon_{2}^{2})}{\mu(S)\mu(\Sigma)}\right)^{\frac{(a+s)(a+\beta)}{2a}}{\left\|{f}\right\|}_{1,\mu}^{a+s}{\left\|{f}\right\|}_{2,\mu}^{a+\beta}.$
(1.20)

Notice that Inequalities (1.15), (1.16) and (1.19) hold also for
$0<s,\beta\leq a$, but not necessarily with the same constants. Notice also
that results in Theorem B are stronger than Inequalities (1.19) and (3.35),
since in Theorem B we have a lower bounds for the measures of the time and
frequency dispersions separately, this give more information than a lower
bound of the product between them.

Now let $\phi$ and $\psi$ two bounded functions in $L^{2}(\Omega,\mu)$ such
that $\|\phi\|_{2,\mu}=\|\psi\|_{2,\mu}$ and
$\|\phi\|_{\infty}\|\psi\|_{\infty}=1$ . The first aim of Section 4 is to make
precise the definition of the pseudo-differential operator (known as the
wavelet multiplier) $\bar{\psi}F_{\sigma}\phi:L^{2}(\Omega,\mu)\to
L^{2}(\Omega,\mu)$, where $\sigma$ is a symbol in $L^{p}(\Omega,\mu)$, $1\leq
p\leq\infty$, and to prove that the resulting bounded linear operator is in
the Schatten-von Neumann class $S_{p}$. On the other hand, we use the
$\varepsilon$-localization measure introduced in [1] to state a new
uncertainty inequality involving the wavelet multiplier. More precisely we
establish the following results.

Theorem C. Let $\sigma\in L^{p}(\Omega,\mu)$, $1\leq p\leq\infty$, and let
$\varepsilon_{1},\varepsilon_{2}\in(0,1)$ such that
$\varepsilon_{1}+\varepsilon_{2}<1$.

  1. (1)

The wavelet multiplier $\bar{\psi}F_{\sigma}\phi:L^{2}(\Omega,\mu)\to
L^{2}(\Omega,\mu)$ is in $S_{p}$ and

$\|\bar{\psi}F_{\sigma}\phi\|_{S_{p}}\leq
c_{\mathcal{K}}^{\frac{2}{p}}\|\sigma\|_{p,\mu}.$ (1.21)

  2. (2)

If a nonzero function $f\in L^{2}(\Omega,\mu)$ is $\varepsilon_{1}$-localized
with respect to $(\bar{\psi}F_{S}\phi)$ and $\varepsilon_{2}$-localized with
respect to $(\bar{\psi}F_{\Sigma}\phi)$, then

$\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-4}(1-\varepsilon_{1}-\varepsilon_{2}).$ (1.22)

Here we denote by $S_{\infty}$ for the space of bounded operators from
$L^{2}(\Omega,\mu)$ into itself. Notice also that the related results of
Theorem B and Theorem C for the Hankel transform has been studied by the
author in [15].

In Section 5, we will prove that the wavelet multiplier is unitary equivalent
to a scalar multiple of the phase space restriction operator
$L_{S,\Sigma}=E_{S}F_{\Sigma}S_{S}$ on $L^{2}(\Omega,\mu)$ arising from the
Landau-Pollak theory in signal analysis [17, 18]. This leads to a compact
self-adjoint operator with spectral representation:

$L_{S,\Sigma}f=\sum_{n=1}^{\infty}\lambda_{n}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}.$
(1.23)

In the classical setting, these eigenfunctions are known as the prolate
spheroidal wave functions. In particular,

$\|L_{S,\Sigma}\varphi_{n}-\varphi_{n}\|_{2,\mu}={\left\langle{\varphi_{n}-L_{S,\Sigma}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}=1-\lambda_{n}<1,$
(1.24)

then each eigenfunction $\varphi_{n}$ is $(1-\lambda_{n})$-localized with
respect to $L_{S,\Sigma}$, and a simple computation shows also that each
function $f$ in $L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ is
$(2\varepsilon_{1}+\varepsilon_{2})$-localized with respect to $L_{S,\Sigma}$.
Moreover, if a function $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-localized
with respect to $L_{S,\Sigma}$, then it satisfies

${\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}\leq
2\varepsilon\|f\|_{2,\mu}^{2}.$ (1.25)

Conversely, if we denote by

$L^{2}(\varepsilon,S,\Sigma)=\left\\{f\in L^{2}(\Omega,\mu)\ :\
{\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}\leq\varepsilon\|f\|_{2,\mu}^{2}\right\\},$
(1.26)

then, each function $f\in L^{2}(\varepsilon,S,\Sigma)$ is
$\sqrt{\varepsilon}$-localized with respect to $L_{S,\Sigma}$, and each $f\in
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ is in
$L^{2}(2\varepsilon_{1}+\varepsilon_{2},S,\Sigma)$ (see Proposition 5.3 for
more details).

Now let $n(\varepsilon,S,\Sigma)={\operatorname{card}}\\{n\ :\ \lambda_{n}\geq
1-\varepsilon\\}$ the number of eigenvalues which are close to one. In [18],
Landau and Pollak gave an asymptotic estimate for $n(\varepsilon,S,\Sigma)$,
when ${\mathcal{T}}$ is the Fourier transform and $S,\Sigma$ are real
intervals. This result can be interpreted as follows: there exist, up to a
small error, $\frac{|S||\Sigma|}{2\pi}$ independent functions that are
$\varepsilon_{1}$-concentrated on $S$ and $\varepsilon_{2}$-bandlimited on
$\Sigma$, these functions are the so-called prolate spheroidal wave functions.
The last estimate has been recently refined in [1], where the authors instead
of counting the number of eigenfunctions with eigenvalue close to one, they
count the maximum number of orthogonal functions that are
$\varepsilon$-localized with respect to $L_{S,\Sigma}$. In [2], Abreu and
Pereira noted that the sharp asymptotic number of these orthogonal functions
is $\approx(1-\varepsilon)^{-1}\frac{|S||\Sigma|}{2\pi}$. On the other hand,
we establish the following results, which characterize functions that are in
$L^{2}(\varepsilon,S,\Sigma)$, and approximate almost time and bandlimited
functions.

Theorem D. Let $f_{\mathrm{ker}}$ denote the orthogonal projection of $f$ onto
the kernel of $L_{S,\Sigma}$.

  1. (1)

A function $f$ is in $L^{2}(\varepsilon,S,\Sigma)$ if and only if,

$\
\sum_{n=1}^{n(\varepsilon,S,\Sigma)}(\lambda_{n}+\varepsilon-1){\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}\geq(1-\varepsilon)\|f_{\mathrm{ker}}\|_{2,\mu}^{2}+\sum_{n=1+n(\varepsilon,S,\Sigma)}^{\infty}(1-\varepsilon-\lambda_{n}){\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}.$

  2. (2)

If $f$ is in $L^{2}(\varepsilon,S,\Sigma)$, then

${\left\|{f-\sum_{n=1}^{n(\varepsilon_{0},S,\Sigma)}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}}\right\|}_{2,\mu}\leq\sqrt{\frac{\varepsilon}{\varepsilon_{0}}}\;\|f\|_{2,\mu}.$
(1.27)

Using the above comparison, it follows that, if $f\in
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then

${\left\|{f-\sum_{n=1}^{n(\varepsilon_{0},S,\Sigma)}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}}\right\|}_{2,\mu}\leq\sqrt{\frac{2\varepsilon_{1}+\varepsilon_{2}}{\varepsilon_{0}}}\;\|f\|_{2,\mu},$
(1.28)

and if $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-localized with respect to
$L_{S,\Sigma}$, then

${\left\|{f-\sum_{n=1}^{n(\varepsilon_{0},S,\Sigma)}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}}\right\|}_{2,\mu}\leq\sqrt{\frac{2\varepsilon}{\varepsilon_{0}}}\;\|f\|_{2,\mu}.$
(1.29)

##  2\. Preliminaries

###  2.1. Notation

Throughout this paper we denote by ${\left\langle{\cdot,\cdot}\right\rangle}$
the usual Euclidean inner product in ${\mathbb{R}}^{d}$, we write for
$x\in{\mathbb{R}}^{d}$,
${\left|{x}\right|}=\sqrt{{\left\langle{x,x}\right\rangle}}$ and if $A$ is a
measurable subset in ${\mathbb{R}}^{d}$, we will write $A^{c}$ for its
complement in $\Omega$.

For $\xi\in\Omega$, we denote by ${\mathcal{K}}_{\xi}:\Omega\to\Omega$ the
kernel defined by ${\mathcal{K}}_{\xi}(x)={\mathcal{K}}(x,\xi)$, and for
$r>0$, we denote by $B_{r}$ the closed ball in $\Omega$ centred at $0$ and of
radius $r$.

We will write along this paper $C$ for a constant that depends on the
parameters $a$, $s$ and $c_{\mathcal{K}}$ defined above (and may be depends
also on some other parameter $\beta$, $\varepsilon$, $\ldots$). This constant
may changes from line to line.

###  2.2. Generalities

Let X be a separable and complex Hilbert space (of infinite dimension) in
which the inner product and the norm are denoted by
${\left\langle{\cdot,\cdot}\right\rangle}$ and $\|\cdot\|$ respectively. Let
${\mathcal{A}}:X\to X$ be a compact operator for which we denote by
${\mathcal{A}}^{*}:X\to X$ its adjoint. Then the linear operator
$|{\mathcal{A}}|=\sqrt{{\mathcal{A}}^{*}{\mathcal{A}}}:X\to X$ is positive and
compact. The singular values of ${\mathcal{A}}$ are the eigenvalues of the
self-adjoint operator $|{\mathcal{A}}|$. For $1\leq p<\infty$, the Schatten
class $S_{p}$ is the space of all compact operators whose singular values lie
in $\ell_{p}$. In particular, $S_{2}$ is the space of Hilbert-Schmidt
operators, and $S_{1}$ is the space of trace class operators. Moreover from
[22, Section VI.6] and [26, Proposition 2.6], we have the following criterion
for a bounded linear operator to be in the trace class.

######  Proposition 2.1.

Let ${\mathcal{A}}:X\to X$ be a bounded linear operator such that, for all
orthonormal bases $\\{\varphi_{n}\\}_{n=1}^{\infty}$ for $X$,

$\sum_{n=1}^{\infty}{\left|{{\left\langle{{\mathcal{A}}\varphi_{n},\varphi_{n}}\right\rangle}}\right|}<\infty,$
(2.1)

then ${\mathcal{A}}:X\to X$ is in the trace class $S_{1}$ with,

${\mathrm{tr}}({\mathcal{A}})=\sum_{n=1}^{\infty}{\left\langle{{\mathcal{A}}\varphi_{n},\varphi_{n}}\right\rangle},$
(2.2)

where $\\{\varphi_{n}\\}_{n=1}^{\infty}$ is any orthonormal basis for $X$.

If, in addition ${\mathcal{A}}$ is positive, then (see [26, Proposition 2.7]),

${\left\|{{\mathcal{A}}}\right\|}_{S_{1}}={\mathrm{tr}}({\mathcal{A}}).$ (2.3)

Moreover from [26, Proposition 2.8], we have the following criterion for a
bounded linear operator ${\mathcal{A}}:X\to X$ to be in the Hilbert-Schmidt
class $S_{2}$.

######  Proposition 2.2.

Let ${\mathcal{A}}:X\to X$ be a bounded linear operator such that, for all
orthonormal bases $\\{\varphi_{n}\\}_{n=1}^{\infty}$ for $X$,

$\sum_{n=1}^{\infty}\|{\mathcal{A}}\varphi_{n}\|^{2}<\infty,$ (2.4)

then ${\mathcal{A}}:X\to X$ is in the Hilbert-Schmidt class $S_{2}$ with,

$\|{\mathcal{A}}\|_{S_{2}}^{2}=\sum_{n=1}^{\infty}\|{\mathcal{A}}\varphi_{n}\|^{2},$
(2.5)

where $\\{\varphi_{n}\\}_{n=1}^{\infty}$ is any orthonormal basis for $X$.

Finally, if the compact operator ${\mathcal{A}}:X\to X$ is Hilbert-Schmidt,
then the positive operator ${\mathcal{A}}^{*}{\mathcal{A}}$ is in the space of
trace class $S_{1}$ and

${\left\|{{\mathcal{A}}}\right\|}_{HS}^{2}:={\left\|{{\mathcal{A}}}\right\|}_{S_{2}}^{2}={\left\|{{\mathcal{A}}^{*}{\mathcal{A}}}\right\|}_{S_{1}}={\mathrm{tr}}({\mathcal{A}}^{*}{\mathcal{A}})=\sum_{n=1}^{\infty}{\left\|{{\mathcal{A}}\varphi_{n}}\right\|}^{2},$
(2.6)

for any orthonormal basis $\\{\varphi_{n}\\}_{n=1}^{\infty}$ for $X$.

For consistency, we define $S_{\infty}:=B\left(X\right)$ to be the space of
bounded operators from $X$ into $X$, equipped with norm,

${\left\|{{\mathcal{A}}}\right\|}_{S_{\infty}}=\sup_{f\ :\
{\left\|{f}\right\|}\leq 1}{\left\|{{\mathcal{A}}f}\right\|}.$ (2.7)

It is obvious that $S_{p}\subseteq S_{q}$, $1\leq p\leq q\leq\infty$.

###  2.3. Fourier-like Multipliers

For $\sigma\in L^{\infty}(\Omega,\mu)$, we define the linear operator
$F_{\sigma}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ by

$F_{\sigma}f={\mathcal{T}}^{-1}\left[\sigma\,{\mathcal{T}}f\right].$ (2.8)

In the case of the Fourier transform, this operator is known as the Fourier
multiplier. Clearly, if $\sigma=1$, then $F_{\sigma}=I$, where $I$ is the
identity operator. Moreover, from the formula (1.7), it is clair that
$F_{\sigma}$ is bounded with

$\|F_{\sigma}\|_{S_{\infty}}\leq\|\sigma\|_{\infty}.$ (2.9)

######  Definition 2.3.

Let $\sigma\in L^{1}(\Omega,\mu)\cup L^{\infty}(\Omega,\mu)$ and let
$\phi,\psi\in L^{\infty}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$ such that
$\|\phi\|_{2,\mu}=\|\psi\|_{2,\mu}=1$. We define the linear operator
$P_{\sigma,\phi,\psi}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ by

${\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}={\left\langle{\sigma{\mathcal{T}}(\phi
f),{\mathcal{T}}(\psi g)}\right\rangle}_{\mu}.$ (2.10)

In the case of the Fourier transform, this operator is known as the _wavelet_
Fourier multiplier, which can be viewed as a variant of a localization
operator with respect to the symbol $\sigma$ and the admissible wavelets
$\phi$ and $\psi$, see the book [26] and the reference therein. Notice that,
if $\sigma=\chi_{A}$ is the characteristic function on the subset
$A\subset\Omega$, then we write $F_{\sigma}$ as $F_{A}$. In this case, we also
write $P_{\sigma,\phi,\psi}$ as $P_{A,\phi,\psi}$ if $\phi\neq\psi$ and
$P_{A,\phi}$, if $\phi=\psi$. The linear operator $F_{A}:L^{2}(\Omega,\mu)\to
L^{2}(\Omega,\mu)$ is a self-adjoint projection, it is known as the _frequency
limiting operator_ on $L^{2}(\Omega,\mu)$ and has many applications in time-
frequency analysis. Moreover we will prove in the last section that
$P_{A,\phi}$ can be viewed as the _phase space_ (or _time frequency_)
_limiting operator_.

The next proposition shows that $P_{\sigma,\phi,\psi}:L^{2}(\Omega,\mu)\to
L^{2}(\Omega,\mu)$ and $\bar{\psi}F_{\sigma}\phi:L^{2}(\Omega,\mu)\to
L^{2}(\Omega,\mu)$ are unitary equivalent.

######  Proposition 2.4.

Let $\sigma\in L^{1}(\Omega,\mu)\cup L^{\infty}(\Omega,\mu)$ and let
$\phi,\psi\in L^{\infty}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$ such that
$\|\phi\|_{2,\mu}=\|\psi\|_{2,\mu}=1$. Then

${\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}={\left\langle{\bar{\psi}F_{\sigma}\phi,g}\right\rangle}_{\mu}.$
(2.11)

###### Proof.

From (2.8) and Parseval’s formula (1.6), we have

$\displaystyle{\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}$
$\displaystyle=$ $\displaystyle{\left\langle{\sigma{\mathcal{T}}(\phi
f),{\mathcal{T}}(\psi g)}\right\rangle}_{\mu}$ $\displaystyle=$
$\displaystyle{\left\langle{{\mathcal{T}}F_{\sigma}(\phi f),{\mathcal{T}}(\psi
g)}\right\rangle}_{\mu}$ $\displaystyle=$
$\displaystyle{\left\langle{F_{\sigma}(\phi f),\psi g}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle{\left\langle{(\bar{\psi}F_{\sigma}\phi)f,g}\right\rangle}_{\mu}.$

The proof is complete. ∎

##  3\. Uncertainty principles by means of the frequency limiting operator

We will need to introduce the following time limiting operator, defined by

$E_{S}f=\chi_{S}f,\quad\quad f\in L^{1}(\Omega,\mu)\cup L^{2}(\Omega,\mu),$

where $S\subset\Omega$. Clearly $E_{S}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$
is a self-adjoint projection.

###  3.1. Uncertainty principles on the space
$L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$

The first known result for functions in
$L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ is the following Donoho-
Stark type uncertainty inequality, see [11, Inequality (3.4)].

######  Theorem 3.1.

Let $\varepsilon_{1},\varepsilon_{2}\in(0,1)$ such that
$\varepsilon_{1}^{2}+\varepsilon_{2}^{2}<1$. Then if $f\in
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ we have

$\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-2}\left(1-\sqrt{\varepsilon_{1}^{2}+\varepsilon_{2}^{2}}\right)^{2}.$
(3.1)

In the case of the Fourier transform, it goes back to Donoho and Stark [9].
This inequality implies that the essential support of $f$ and ${\mathcal{T}}f$
cannot be too small. Moreover, we recall the following local uncertainty
principle, see [11].

######  Theorem 3.2.

  1. (1)

If $\,0<s<a$, then there exists a constant $C$ such that for all $f\in
L^{2}(\Omega,\mu)$ and all measurable subset $\Sigma\subset\Omega$ of finite
measure $0<\mu(\Sigma)<\infty$,

${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}^{2}\leq
C\,\mu(\Sigma)^{\frac{s}{a}}\,\|{\left|{x}\right|}^{s}f\|_{2,\mu}^{2}.$ (3.2)

  2. (2)

If $\,s>a$, then there exists a constant $C$ such that for all $f\in
L^{2}(\Omega,\mu)$ and all measurable subset $\Sigma\subset\Omega$ of finite
measure $0<\mu(\Sigma)<\infty$,

${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}^{2}\leq
C\,\mu(\Sigma)\,\|f\|_{2,\mu}^{2-\frac{2a}{s}}\|{\left|{x}\right|}^{s}f\|_{2,\mu}^{\frac{2a}{s}}.$
(3.3)

Next, take $s=a$. Then, if we apply the first inequality (3.2) with
$a(1-\varepsilon)$, $\varepsilon\in(0,1)$, replacing $s$ and then apply the
following classical inequality

$\||x|^{a-a\varepsilon}f\|_{2,\mu}\leq
C{\left\|{f}\right\|}_{2,\mu}^{\varepsilon}{\left\|{|x|^{a}f}\right\|}_{2,\mu}^{1-\varepsilon},$
(3.4)

we obtain for all $\varepsilon\in(0,1)$,

${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}^{2}\leq
C\,\mu(\Sigma)^{1-\varepsilon}\,\|f\|_{2,\mu}^{2\varepsilon}\,\|{\left|{x}\right|}^{a}f\|_{2,\mu}^{2-2\varepsilon}.$
(3.5)

Consequently we conclude the following first corollary comparing the support
of ${\mathcal{T}}f$ and the generalized time dispersion $\||x|^{s}f\|_{2,\mu}$
for function in the range of $F_{\Sigma}$:

$\operatorname{Im}(F_{\Sigma})=\\{f\in L^{2}(\Omega,\mu)\ :\
{\operatorname{supp}}{\mathcal{T}}f\subset\Sigma\\}.$

######  Corollary 3.3.

Let $s>0$. Then there exists a constant $C$ such that for all
$f\in\operatorname{Im}(F_{\Sigma})$,

$\mu\left({\operatorname{supp}}{\mathcal{T}}f\right)\|{\left|{x}\right|}^{s}f\|_{2,\mu}^{\frac{2a}{s}}\geq
C{\left\|{f}\right\|}_{2,\mu}^{\frac{2a}{s}}.$ (3.6)

###### Proof.

Let $s>0$ and $f\in\operatorname{Im}(F_{\Sigma})$. Then $f=F_{\Sigma}f$, and
we apply (3.2), (3.3), (3.5) to obtain the desired result. ∎

Notice that, if $\mu\left({\operatorname{supp}}{\mathcal{T}}f\right)$ is
finite, then $\mu\left({\operatorname{supp}}f\right)$ is infinite, because $f$
and ${\mathcal{T}}f$ cannot be simultaneously supported on subsets of finite
measure, see [11, Corollary 3.7]. This result is known as the Benedicks-
Amrein-Berthier uncertainty principle.

Moreover, we can also obtain an inequality comparing the essential support of
${\mathcal{T}}f$ and the generalized time dispersion $\||x|^{s}f\|_{2,\mu}$
for functions that are $\varepsilon_{2}$-bandlimited on $\Sigma$.

######  Corollary 3.4.

Let $s>0$.

  1. (1)

If $\,0<s<a$, then there exists a constant $C$ such that for all function $f$
which is $\varepsilon_{2}$-bandlimited on $\Sigma$,

$\mu(\Sigma)^{\frac{s}{a}}\,{\left\|{|x|^{s}f}\right\|}_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{2}^{2}\right)\|f\|_{2,\mu}^{2}.$ (3.7)

  2. (2)

If $s>a$, then there exists a constant $C$ such that for all function $f$
which is $\varepsilon_{2}$-bandlimited on $\Sigma$,

$\mu(\Sigma)^{\frac{s}{a}}\,{\left\|{|x|^{s}f}\right\|}_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{2}^{2}\right)^{\frac{s}{a}}\|f\|_{2,\mu}^{2}.$ (3.8)

  3. (3)

For all $\varepsilon\in(0,1)$, there exists a constant $C$ such that for all
function $f$ which is $\varepsilon_{2}$-bandlimited on $\Sigma$,

$\mu(\Sigma)\,{\left\|{|x|^{a}f}\right\|}_{2,\mu}^{2}\geq
C(1-\varepsilon_{2}^{2})^{\frac{1}{1-\varepsilon}}\|f\|_{2,\mu}^{2}.$ (3.9)

###### Proof.

Since $f\in L^{2}(\Omega,\mu)$ is $\varepsilon_{2}$-bandlimited on $\Sigma$,
then

$\|F_{\Sigma}f\|_{2,\mu}^{2}=\|f\|_{2,\mu}^{2}-\|F_{\Sigma^{c}}f\|_{2,\mu}^{2}\geq(1-\varepsilon_{2}^{2})\|f\|_{2,\mu}^{2}.$

For the first result, we use the local inequalities (3.2). Analogously, for
the second inequality, we use (3.3), and finally, for the third inequality, we
use (3.5). ∎

Now, since,
${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}={\left\|{E_{\Sigma}{\mathcal{T}}f}\right\|}_{2,\mu}$,
then by interchanging the roles of $f$ and ${\mathcal{T}}f$ in Theorem 3.2,
Corollary 3.3 and Corollary 5.8 , we obtain the following results involving
the time limiting operator instead of the frequency limiting operator, and the
frequency dispersion instead of the time dispersion.

######  Theorem 3.5.

Let $\beta>0$.

  1. (1)

If $\,0<\beta<a$, then

    1. (a)

there exists a constant $C$ such that for all $f\in L^{2}(\Omega,\mu)$ and all
measurable subset $S\subset\Omega$ of finite measure $0<\mu(S)<\infty$,

${\left\|{E_{S}f}\right\|}_{2,\mu}^{2}\leq
C\mu(S)^{\frac{\beta}{a}}\|{\left|{x}\right|}^{\beta}{\mathcal{T}}f\big{\|}_{2,\mu}^{2},$
(3.10)

    2. (b)

there exists a constant $C$ such that for all function $f$ which is
$\varepsilon_{1}$-concentrated on $S$,

$\mu(S)^{\frac{\beta}{a}}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{1}^{2}\right)\|f\|_{2,\mu}^{2}.$ (3.11)

  2. (2)

If $\beta>a$, then

    1. (a)

there exists a constant $C$ such that for all $f\in L^{2}(\Omega,\mu)$ and all
measurable subset $S\subset\Omega$ of finite measure $0<\mu(S)<\infty$,

${\left\|{E_{S}f}\right\|}_{2,\mu}^{2}\leq
C\,\mu(S)\|f\|_{2,\mu}^{2-\frac{2a}{\beta}}\|{\left|{\xi}\right|}^{\beta}{\mathcal{T}}f\|_{2,\mu}^{\frac{2a}{\beta}},$
(3.12)

    2. (b)

there exists a constant $C$ such that for all function $f$ which is
$\varepsilon_{1}$-concentrated on $S$,

$\mu(S)^{\frac{\beta}{a}}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{1}^{2}\right)^{\frac{\beta}{a}}\|f\|_{2,\mu}^{2}.$
(3.13)

  3. (3)

For all $\varepsilon\in(0,1)$,

    1. (a)

there exists a constant $C$ such that for all $f\in L^{2}(\Omega,\mu)$ and all
measurable subset $S\subset\Omega$ of finite measure $0<\mu(S)<\infty$,

${\left\|{E_{S}f}\right\|}_{2,\mu}^{2}\leq
C\,\mu(S)^{1-\varepsilon}\,\|f\|_{2,\mu}^{2\varepsilon}\,\|{\left|{\xi}\right|}^{a}{\mathcal{T}}f\|_{2,\mu}^{2-2\varepsilon},$
(3.14)

    2. (b)

there exists a constant $C$ such that for all function $f$ which is
$\varepsilon_{1}$-concentrated on $S$,

$\mu(S)\,{\left\|{|\xi|^{a}{\mathcal{T}}f}\right\|}_{2,\mu}^{2}\geq
C\left(1-\varepsilon_{1}^{2}\right)^{\frac{1}{1-\varepsilon}}\|f\|_{2,\mu}^{2}.$
(3.15)

  4. (4)

There exists a constant $C$ such that for all
$f\in\operatorname{Im}(E_{S})=\\{f\in L^{2}(\Omega,\mu)\ :\
{\operatorname{supp}}f\subset S\\}$,

$\mu\left({\operatorname{supp}}f\right)\|{\left|{\xi}\right|}^{\beta}{\mathcal{T}}f\|_{2,\mu}^{\frac{2a}{\beta}}\geq
C{\left\|{f}\right\|}_{2,\mu}^{\frac{2a}{\beta}}.$ (3.16)

Finally we can formulate our new Heisenberg-type uncertainty inequalities for
functions in $L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, with constants
that depend on $\varepsilon_{1},\,\varepsilon_{2},\,S$ and $\Sigma$.

######  Theorem 3.6.

Let $s,\beta>0$. Then for all $f\in
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$:

  1. (1)

if $0<s,\beta<a$,

${\left\|{|x|^{s}f}\right\|}_{2,\mu}^{\beta}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{s}\geq
C\;\frac{(1-\varepsilon_{1}^{2})^{s/2}(1-\varepsilon_{2}^{2})^{\beta/2}}{\left(\mu(S)\mu(\Sigma)\right)^{\frac{s\beta}{2a}}}{\left\|{f}\right\|}_{2,\mu}^{s+\beta},$
(3.17)

  2. (2)

if $s,\beta>a$,

${\left\|{|x|^{s}f}\right\|}_{2,\mu}^{\beta}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{s}\geq
C\left(\frac{(1-\varepsilon_{1}^{2})(1-\varepsilon_{2}^{2})}{\mu(S)\mu(\Sigma)}\right)^{\frac{s\beta}{2a}}{\left\|{f}\right\|}_{2,\mu}^{s+\beta},$
(3.18)

  3. (3)

for all $\varepsilon\in(0,1)$,

${\left\|{|x|^{a}f}\right\|}_{2,\mu}\,{\left\|{|\xi|^{a}{\mathcal{T}}f}\right\|}_{2,\mu}\geq
C\;\frac{\big{(}(1-\varepsilon_{1}^{2})(1-\varepsilon_{2}^{2})\big{)}^{\frac{1}{2-2\varepsilon}}}{\sqrt{\mu(S)\mu(\Sigma)}}{\left\|{f}\right\|}_{2,\mu}^{2}.$
(3.19)

######  Remark 3.7.

  1. (1)

Notice that Corollary 5.8 and Inequalities (3.11), (3.13) and (3.15) give
separately a lower bounds for the measures of the time dispersion
${\left\|{|x|^{s}f}\right\|}_{2,\mu}$ and the frequency dispersion
$\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}$, which give more information than a
lower bound of the product between them in Theorem 3.6

  2. (2)

On the other hand, from Corollary 5.8 and Inequalities (3.11), (3.13) and
(3.15), we can obtain separately a lower bounds, that depend of the signal
$f\in L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, for the measures of
$\mu(S)$ and $\mu(\Sigma)$, from which we deduce, in the spirit of [6], the
following lower bounds for the product between them:

$\mu(S)\mu(\Sigma)\geq\begin{cases}C.C_{f}(s,a,\beta)\big{(}(1-\varepsilon_{1}^{2})^{\frac{1}{\beta}}(1-\varepsilon_{2}^{2})^{\frac{1}{s}}\big{)}^{a},&0<s,\beta<a,\\\
\\\
C.C_{f}(s,a,\beta)\,(1-\varepsilon_{1}^{2})(1-\varepsilon_{2}^{2}),&s,\beta>a,\\\
\\\
C.C_{f}(a,a,a)\big{(}(1-\varepsilon_{1}^{2})(1-\varepsilon_{2}^{2})\big{)}^{\frac{1}{1-\varepsilon}},&\mathrm{otherwise},\end{cases}$
(3.20)

where $C$ is a constant that depend only on
$s,a,c_{\mathcal{K}},\beta,\varepsilon,$ and

$C_{f}(s,a,\beta)=\left(\frac{{\left\|{f}\right\|}_{2,\mu}^{s+\beta}}{{\left\|{x^{s}f}\right\|}_{2,\mu}^{\beta}\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{s}}\right)^{\frac{2a}{s\beta}}.$
(3.21)

###  3.2. Uncertainty principles on the space $L^{1}\cap
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$

The first known result for functions in $L^{1}\cap
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ is the following Donoho-Stark
type uncertainty inequality, see [12, Proposition 2.6].

######  Theorem 3.8.

Let $\varepsilon_{1},\varepsilon_{2}\in(0,1)$. Then if $f\in L^{1}\cap
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ we have

$\mu(S)\geq\frac{{\left\|{f}\right\|}_{1,\mu}^{2}}{{\left\|{f}\right\|}_{2,\mu}^{2}}(1-\varepsilon_{1})^{2},\quad\mu(\Sigma)\geq
c_{\mathcal{K}}^{-2}\frac{{\left\|{f}\right\|}_{2,\mu}^{2}}{{\left\|{f}\right\|}_{1,\mu}^{2}}(1-\varepsilon_{2}^{2}),$
(3.22)

and then

$\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-2}(1-\varepsilon_{1})^{2}(1-\varepsilon_{2}^{2}).$ (3.23)

Theorem 3.8 is stronger then Theorem 3.1, in the sense that the previous
theorem give a lower bound of $\mu(S)$ and $\mu(\Sigma)$ separately, which is
not possible in Theorem 3.1.

Now we will recall the following Carlson-type and Nash-type inequalities, see
[12, Proposition 2.2, Proposition 2.3].

######  Theorem 3.9.

Let $s,\beta>0$. Then we have:

  1. (1)

_A Carlson-type inequality:_ there exists a constant $C=C(s,a)$ such that for
all $f\in L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$,

${\left\|{f}\right\|}_{1,\mu}^{1+\frac{s}{a}}\leq
C\|f\|_{2,\mu}^{\frac{s}{a}}\;{\left\|{|x|^{s}f}\right\|}_{1,\mu}.$ (3.24)

  2. (2)

_A Nash-type inequality:_ there exists a constant $C=C(\beta,a)$ such that for
all $f\in L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$,

${\left\|{f}\right\|}_{2,\mu}^{1+\frac{\beta}{a}}\leq
C\|f\|_{1,\mu}^{\frac{\beta}{a}}\;\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}.$
(3.25)

Consequently we obtain a lower bounds for the time and frequency dispersions:

${\left\|{|x|^{s}f}\right\|}_{1,\mu}\geq
C\left(\frac{{\left\|{f}\right\|}_{1,\mu}}{\|f\|_{2,\mu}}\right)^{\frac{s}{a}}{\left\|{f}\right\|}_{1,\mu}\quad\mathrm{and}\quad\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}\geq
C\left(\frac{{\left\|{f}\right\|}_{2,\mu}}{\|f\|_{1,\mu}}\right)^{\frac{\beta}{a}}{\left\|{f}\right\|}_{2,\mu}.$
(3.26)

######  Corollary 3.10.

Let $s,\beta>0$. Then

  1. (1)

there exists a constant $C=C(a,\beta,s)$ such that for all $f\in
L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$,

$\||x|^{s}f\|_{1,\mu}^{\beta}\;\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{s}\geq
C\,{\left\|{f}\right\|}_{1,\mu}^{\beta}\,{\left\|{f}\right\|}_{2,\mu}^{s},$
(3.27)

  2. (2)

there exists a constant $C$ such that for all $f\in L^{1}(\Omega,\mu)\cap
L^{2}(\Omega,\mu)$ and all measurable subset of $\Sigma$ of finite measure,

${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}^{2}\leq
C\,\mu(\Sigma){\left\|{f}\right\|}^{\frac{2s}{a+s}}_{2,\mu}\,{\left\|{|x|^{s}f}\right\|}^{\frac{2a}{a+s}}_{1,\mu},$
(3.28)

  3. (3)

there exists a constant $C$ such that for all $f\in L^{1}(\Omega,\mu)\cap
L^{2}(\Omega,\mu)$ and all measurable subset $S$ of finite measure,

${\left\|{E_{S}f}\right\|}_{1,\mu}^{2}\leq
C\,\mu(S){\left\|{f}\right\|}^{\frac{2\beta}{a+\beta}}_{1,\mu}\,\||\xi|^{\beta}{\mathcal{T}}f\|^{\frac{2a}{a+\beta}}_{2,\mu},$
(3.29)

  4. (4)

there exists a constant $C$ such that for all $f\in L^{1}(\Omega,\mu)\cap
L^{2}(\Omega,\mu)$ with ${\operatorname{supp}}{\mathcal{T}}f\subset\Sigma$,

$\mu\big{(}{\operatorname{supp}}{\mathcal{T}}f\big{)}\,{\left\|{|x|^{s}f}\right\|}^{\frac{2a}{a+s}}_{1,\mu}\geq
C\,{\left\|{f}\right\|}^{\frac{2a}{a+s}}_{2,\mu},$ (3.30)

  5. (5)

there exists a constant $C$ such that for all $f\in L^{1}(\Omega,\mu)\cap
L^{2}(\Omega,\mu)$ with ${\operatorname{supp}}f\subset S$,

$\mu\big{(}{\operatorname{supp}}f\big{)}\,\||\xi|^{\beta}{\mathcal{T}}f\|^{\frac{2a}{a+\beta}}_{2,\mu}\geq
C\,{\left\|{f}\right\|}^{\frac{2a}{a+\beta}}_{1,\mu}.$ (3.31)

###### Proof.

The first inequality follows by combining the Carlson inequality (3.24) and
the Nash inequality (3.25). Next by (1.7) and (1.3),

${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}^{2}={\left\|{\chi_{\Sigma}{\mathcal{T}}f}\right\|}_{2,\mu}^{2}\leq\mu(\Sigma){\left\|{{\mathcal{T}}f}\right\|}_{\infty}^{2}\leq
c_{\mathcal{K}}^{2}\mu(\Sigma){\left\|{f}\right\|}_{1,\mu}^{2},$

and by the Carlson inequality (3.24) we obtain (3.28). Now by the Cauchy-
Schwartz inequality we have,

${\left\|{E_{S}f}\right\|}_{1,\mu}^{2}\leq\mu(S){\left\|{f}\right\|}_{2,\mu}^{2},$

and by the Nash type inequality (3.25) we deduce (3.29). Finally (3.30)
follows directly from (3.28) by taking
$\Sigma={\operatorname{supp}}{\mathcal{T}}f$ and if we take
$S={\operatorname{supp}}f$ in (3.29) we obtain (3.31). ∎

######  Remark 3.11.

Clearly, Inequality (3.26) implies also that, for all $f\in
L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$,

$\||x|^{s}f\|_{1,\mu}^{a+\beta}\;\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{a+s}\geq
C\,{\left\|{f}\right\|}_{1,\mu}^{a+s}\,{\left\|{f}\right\|}_{2,\mu}^{a+\beta}.$
(3.32)

######  Corollary 3.12.

Let $s,\beta>0$. Then

  1. (1)

there exists a constant $C$ such that for all function $f$, which is
$\varepsilon_{1}$-timelimited on $S$,

$\mu(S)^{\frac{a+\beta}{2a}}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}\geq
C\left(1-\varepsilon_{1}\right)^{\frac{a+\beta}{a}}{\left\|{f}\right\|}_{1,\mu},$
(3.33)

  2. (2)

there exists a constant $C$ such that for all function $f$, which is
$\varepsilon_{2}$-bandlimited on $\Sigma$,

$\mu(\Sigma)^{\frac{a+s}{2a}}\,{\left\|{|x|^{s}f}\right\|}_{1,\mu}\geq
C\left(1-\varepsilon_{2}^{2}\right)^{\frac{a+s}{2a}}{\left\|{f}\right\|}_{2,\mu},$
(3.34)

  3. (3)

there exists a constant $C$ such that for all $f\in L^{1}\cap
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$,

${\left\|{|x|^{s}f}\right\|}_{1,\mu}^{a+\beta}\,\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{a+s}\geq
C\left(\frac{(1-\varepsilon_{1})^{2}(1-\varepsilon_{2}^{2})}{\mu(S)\mu(\Sigma)}\right)^{\frac{(a+s)(a+\beta)}{2a}}{\left\|{f}\right\|}_{1,\mu}^{a+s}{\left\|{f}\right\|}_{2,\mu}^{a+\beta}.$
(3.35)

###### Proof.

If $f$ is $\varepsilon_{1}$-timelimited, then

${\left\|{E_{S}f}\right\|}_{1,\mu}\geq{\left\|{f}\right\|}_{1,\mu}-{\left\|{E_{S^{c}}f}\right\|}_{1,\mu}\geq(1-\varepsilon_{1}){\left\|{f}\right\|}_{1,\mu},$

and if $f$ is $\varepsilon_{2}$-bandelimited, then

${\left\|{F_{\Sigma}f}\right\|}_{2,\mu}^{2}={\left\|{f}\right\|}_{2,\mu}^{2}-{\left\|{F_{\Sigma^{c}}f}\right\|}_{2,\mu}^{2}\geq(1-\varepsilon_{2}^{2}){\left\|{f}\right\|}_{2,\mu}^{2}.$

Hence the desired result follows from (3.28) and (3.29). ∎

######  Remark 3.13.

Let $s,\beta>0$ and let $f\in L^{1}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$.

  1. (1)

If $f$ is $\varepsilon_{1}$-timelimited on $S$, then

$\mu(S)\geq
C\left(\frac{{\left\|{f}\right\|}_{1,\mu}}{\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}}\right)^{\frac{2a}{a+\beta}}(1-\varepsilon_{1})^{2}.$
(3.36)

  2. (2)

If $f$ is $\varepsilon_{2}$-bandlimited on $\Sigma$, then

$\mu(\Sigma)\geq
C\left(\frac{{\left\|{f}\right\|}_{2,\mu}}{{\left\|{|x|^{s}f}\right\|}_{1,\mu}}\right)^{\frac{2a}{a+s}}(1-\varepsilon_{2}^{2}).$
(3.37)

  3. (3)

If $f\in L^{1}\cap L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then

$\mu(S)\mu(\Sigma)\geq
C.\tilde{C}_{f}(a,s,\beta)(1-\varepsilon_{1})^{2}(1-\varepsilon_{2}^{2}),$
(3.38)

where

$\tilde{C}_{f}(a,s,\beta)=\left(\frac{{\left\|{f}\right\|}_{1,\mu}^{a+s}{\left\|{f}\right\|}_{2,\mu}^{a+\beta}}{{\left\|{|x|^{s}f}\right\|}_{1,\mu}^{a+\beta}\||\xi|^{\beta}{\mathcal{T}}f\|_{2,\mu}^{a+s}}\right)^{\frac{2a}{(a+\beta)(a+s)}}.$
(3.39)

##  4\. The wavelet multiplier

Our motivation here came from the classical setting stated in [27, 28]. In
this section let $\phi$ and $\psi$ will be two functions in
$L^{\infty}(\Omega,\mu)\cap L^{2}(\Omega,\mu)$ such that
$\|\phi\|_{2,\mu}=\|\psi\|_{2,\mu}=1$.

###  4.1. Boundedness

The aim of this section is to prove that we can also define
$P_{\sigma,\phi,\psi}$ for symbol $\sigma\in L^{p}(\Omega,\mu)$, $1<p<\infty$.
First, if $\sigma\in L^{\infty}(\Omega,\mu)$, we have the following result.

######  Proposition 4.1.

Let $\sigma\in L^{\infty}(\Omega,\mu)$. Then $P_{\sigma,\phi,\psi}$ is in
$S_{\infty}$ and

$\|P_{\sigma,\phi,\psi}\|_{S_{\infty}}\leq\|\phi\|_{\infty}\|\psi\|_{\infty}\|\sigma\|_{\infty}.$
(4.1)

###### Proof.

By the Cauchy-Schwartz inequality,

$|{\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}|\leq\|\sigma\|_{\infty}\|{\mathcal{T}}(\phi
f)\|_{2,\mu}\|{\mathcal{T}}(\psi g)\|_{2,\mu}.$

Then by Plancherel’s formula (1.7), we obtain

$\displaystyle|{\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}|$
$\displaystyle\leq$ $\displaystyle\|\sigma\|_{\infty}\|\phi f\|_{2,\mu}\|\psi
g\|_{2,\mu}$ $\displaystyle\leq$
$\displaystyle\|\sigma\|_{\infty}\|\phi\|_{\infty}\|\psi\|_{\infty}\|f\|_{2,\mu}\|g\|_{2,\mu}.$

This completes the proof. ∎

Now, if we consider $\sigma\in L^{1}(\Omega,\mu)$, then we obtain the
following result.

######  Proposition 4.2.

Let $\sigma\in L^{1}(\Omega,\mu)$. Then $P_{\sigma,\phi,\psi}$ is in
$S_{\infty}$ and

$\|P_{\sigma,\phi,\psi}\|_{S_{\infty}}\leq
c_{\mathcal{K}}^{2}\|\sigma\|_{1,\mu}.$ (4.2)

###### Proof.

Since ${\mathcal{T}}(\phi
f)(\xi)={\left\langle{f,\overline{\phi\,{\mathcal{K}}_{\xi}}}\right\rangle}_{\mu}$,
then by the Cauchy-Schwartz inequality,

$\|{\mathcal{T}}(\phi f)\|_{\infty}\leq
c_{\mathcal{K}}\|f\|_{2,\mu}\|\phi\|_{2,\mu}.$

Therefore, since $\|\phi\|_{2,\mu}=\|\psi\|_{2,\mu}=1$, we obtain

$\displaystyle|{\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}|$
$\displaystyle\leq$ $\displaystyle\|\sigma\|_{1,\mu}\|{\mathcal{T}}(\phi
f)\|_{\infty}\|{\mathcal{T}}(\psi g)\|_{\infty}$ $\displaystyle\leq$
$\displaystyle
c_{\mathcal{K}}^{2}\|\sigma\|_{1,\mu}\|f\|_{2,\mu}\|g\|_{2,\mu}.$

This completes the proof. ∎

Thus, by (4.1), (4.2) and the Riesz-Thorin interpolation argument [23, Theorem
2] (see also [26, Theorem 12.4]) we obtain the following theorem.

######  Theorem 4.3.

Let $\sigma\in L^{p}(\Omega,\mu)$, $1<p<\infty$. Then the linear operator
$P_{\sigma,\phi,\psi}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ is bounded and

$\|P_{\sigma,\phi,\psi}\|_{S_{\infty}}\leq
c_{\mathcal{K}}^{\frac{2}{p}}\|\phi\|_{\infty}^{\frac{1}{p^{\prime}}}\|\psi\|_{\infty}^{\frac{1}{p^{\prime}}}\|\sigma\|_{p,\mu}.$
(4.3)

Hence we can define the operator
$(\bar{\psi}F_{\sigma}\phi):L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$, where
$\sigma\in L^{p}(\Omega,\mu)$, $1\leq p\leq\infty$ by

${\left\langle{P_{\sigma,\phi,\psi}f,g}\right\rangle}_{\mu}={\left\langle{(\bar{\psi}F_{\sigma}\phi)f,g}\right\rangle}_{\mu}.$
(4.4)

###  4.2. Schatten class properties

Let us begin with the following theorem.

######  Theorem 4.4.

Let $\sigma$ be symbol in $L^{1}(\Omega,\mu)$. Then $P_{\sigma,\phi,\psi}$ is
Hilbert Schmidt and

$\|P_{\sigma,\phi,\psi}\|_{S_{2}}^{2}=\int_{\Omega}\sigma(\xi){\left\langle{P_{\bar{\sigma},\psi,\phi}\,\bar{\psi},\bar{\phi}\,|{\mathcal{K}}_{\xi}|^{2}}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi)\leq\|\sigma\|_{L^{1}_{\alpha}}^{2}.$
(4.5)

###### Proof.

First by (2.10) it follows immediately that the adjoint of
$P_{\sigma,\phi,\psi}$ is $P_{\bar{\sigma},\psi,\phi}:L^{2}(\Omega,\mu)\to
L^{2}(\Omega,\mu)$. Now, Let $\\{\varphi_{n}\\}_{n=1}^{\infty}$ be an
orthonormal basis for $L^{2}(\Omega,\mu)$. Then by (2.10) and by Fubini’s
theorem we obtain,

$\displaystyle\sum_{n=1}^{\infty}\|P_{\sigma,\phi,\psi}\varphi_{n}\|^{2}_{2,\mu}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}{\left\langle{P_{\sigma,\phi,\psi}\varphi_{n},P_{\sigma,\phi,\psi}\varphi_{n}}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}{\left\langle{\sigma{\mathcal{T}}(\phi\,\varphi_{n}),{\mathcal{T}}(\psi\,P_{\sigma,\phi,\psi}\varphi_{n})}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\int_{\Omega}\sigma(\xi){\left\langle{\varphi_{n},\overline{\phi\,{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}\overline{{\left\langle{P_{\sigma,\phi,\psi}\varphi_{n},\overline{\psi\,{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}}\,{\mathrm{d}}\mu(\xi)$
$\displaystyle=$
$\displaystyle\int_{\Omega}\sigma(\xi)\sum_{n=1}^{\infty}{\left\langle{P_{\bar{\sigma},\psi,\phi}\overline{\psi{\mathcal{K}}_{\xi}},\varphi_{n}}\right\rangle}_{\mu}{\left\langle{\varphi_{n},\overline{\phi{\mathcal{K}}_{\xi}}}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi)$
$\displaystyle=$
$\displaystyle\int_{\Omega}\sigma(\xi){\left\langle{P_{\bar{\sigma},\psi,\phi}\overline{\psi{\mathcal{K}}_{\xi}},\overline{\phi{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi),$

where we have used Parseval’s identity in the last line. Therefore from
Proposition 4.2 and since $|{\mathcal{K}}_{\xi}|\leq c_{\mathcal{K}}$,

$\displaystyle\sum_{n=1}^{\infty}\|P_{\sigma,\phi,\psi}\varphi_{n}\|^{2}_{2,\mu}$
$\displaystyle\leq$
$\displaystyle\|P_{\bar{\sigma},\psi,\phi}\|_{S_{\infty}}\|\phi\|_{2,\mu}\|\psi\|_{2,\mu}\|\sigma\|_{1,\mu}$
$\displaystyle\leq$ $\displaystyle c_{\mathcal{K}}^{4}\|\sigma\|_{1,\mu}^{2}.$

Thus from Proposition 2.2, the operator $P_{\sigma,\phi,\psi}$ is in $S_{2}$
and $\|P_{\sigma,\phi,\psi}\|_{S_{2}}\leq
c_{\mathcal{K}}^{2}\|\sigma\|_{1,\mu}.$ ∎

Consequently the operator $P_{\sigma,\phi,\psi}$ is also compact for symbols
in $L^{p}(\Omega,\mu)$.

######  Corollary 4.5.

Let $\sigma$ be symbol in $L^{p}(\Omega,\mu)$, $1\leq p<\infty$. Then the
operator $P_{\sigma,\phi,\psi}$ is compact.

###### Proof.

Let $\\{\sigma_{n}\\}_{n=1}^{\infty}$ be a sequence of functions in
$L^{1}(\Omega,\mu)\cap L^{\infty}(\Omega,\mu)$ such that $\sigma_{n}\to\sigma$
in $L^{p}(\Omega,\mu)$ as $n\to\infty$. Then by Theorem 4.3,

$\|P_{\sigma_{n},\phi,\psi}-P_{\sigma,\phi,\psi}\|_{S_{\infty}}\leq\|\phi\|_{\infty}^{\frac{1}{p^{\prime}}}\|\psi\|_{\infty}^{\frac{1}{p^{\prime}}}\|\sigma_{n}-\sigma\|_{p,\mu}.$
(4.6)

Therefore $P_{\sigma_{n},\phi,\psi}\to P_{\sigma,\phi,\psi}$ in $S_{\infty}$
as $n\to\infty$. Now, since by Theorem 4.4, the operators
$P_{\sigma_{n},\phi,\psi}$ are in $S_{2}$ and hence compact, and since the set
of compact operators is a closed subspace of $S_{\infty}$, then the operator
$P_{\sigma,\phi,\psi}$ is also compact. ∎

More precisely we will prove that the operator $P_{\sigma,\phi,\psi}$ is in
fact in the Schatten class $S_{p}$, $1\leq p<\infty$. Of particular interest
is the Schatten-von Neumann class $S_{1}$ (see [27, 28]).

######  Theorem 4.6.

Let $\sigma\in L^{1}(\Omega,\mu)$. Then
$P_{\sigma,\phi,\psi}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ is trace class
with

$\|P_{\sigma,\phi,\psi}\|_{S_{1}}\leq c_{\mathcal{K}}^{2}\|\sigma\|_{1,\mu},$
(4.7)

and we have the following trace formula

${\mathrm{tr}}\left(P_{\sigma,\phi,\psi}\right)=\int_{\Omega}\sigma(\xi){\left\langle{\overline{\psi\,{\mathcal{K}}_{\xi}},\overline{\phi\,{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi).$
(4.8)

###### Proof.

Let $\\{\varphi_{n}\\}_{n=1}^{\infty}$ be an orthonormal basis for
$L^{2}(\Omega,\mu)$. Then

$\displaystyle\sum_{n=1}^{\infty}{\left\langle{P_{\sigma,\phi,\psi}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\int_{\Omega}\sigma(\xi){\mathcal{T}}(\phi\,\varphi_{n})(\xi)\overline{{\mathcal{T}}(\psi\,\varphi_{n})(\xi)}\,{\mathrm{d}}\mu(\xi)$
$\displaystyle=$
$\displaystyle\sum_{n=1}^{\infty}\int_{\Omega}\sigma(\xi){\left\langle{\overline{\psi\,{\mathcal{K}}_{\xi}},\varphi_{n}}\right\rangle}_{\mu}{\left\langle{\varphi_{n},\overline{\phi\,{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi).$

Thus by Fubini’s theorem,

$\sum_{n=1}^{\infty}{\left\langle{P_{\sigma,\phi,\psi}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}=\int_{\Omega}\sigma(\xi)\sum_{n=1}^{\infty}{\left\langle{\overline{\psi\,{\mathcal{K}}_{\xi}},\varphi_{n}}\right\rangle}_{\mu}{\left\langle{\varphi_{n},\overline{\phi\,{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi).$
(4.9)

Therefore by Parseval’s identity, and the fact that
$\|\phi\|_{2,\mu}=\|\psi\|_{2,\mu}=1$,

$\displaystyle\sum_{n=1}^{\infty}\left|{\left\langle{P_{\sigma,\phi,\psi}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}\right|$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{\Omega}|\sigma(\xi)|\sum_{n=1}^{\infty}\left(\left|{\left\langle{\overline{\phi\,{\mathcal{K}}_{\xi}},\varphi_{n}}\right\rangle}_{\mu}\right|^{2}+\left|{\left\langle{\overline{\psi\,{\mathcal{K}}_{\xi}},\varphi_{n}}\right\rangle}_{\mu}\right|^{2}\right)\,{\mathrm{d}}\mu(\xi)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{\Omega}|\sigma(\xi)|\left(\|\phi{\mathcal{K}}_{\xi}\|_{2,\mu}^{2}+\|\psi{\mathcal{K}}_{\xi}\|_{2,\mu}^{2}\right)\,{\mathrm{d}}\mu(\xi)$
$\displaystyle\leq$ $\displaystyle c_{\mathcal{K}}^{2}\|\sigma\|_{1,\mu}.$

By Proposition 2.1, the operator $P_{\sigma,\phi,\psi}$ is in $S_{1}$ and with
(4.9) and Parseval’s identity,

${\mathrm{tr}}(P_{\sigma,\phi,\psi})=\sum_{n=1}^{\infty}{\left\langle{P_{\sigma,\phi,\psi}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}=\int_{\Omega}\sigma(\xi){\left\langle{\overline{\psi{\mathcal{K}}_{\xi}},\overline{\phi\,{\mathcal{K}}_{\xi}}\,}\right\rangle}_{\mu}\,{\mathrm{d}}\mu(\xi).$

This allows to conclude. ∎

Moreover by (4.1), (4.7) and by interpolation argument we deduce the following
result.

######  Corollary 4.7.

Let $\sigma\in L^{p}(\Omega,\mu)$, $1<p<\infty$. Then
$P_{\sigma,\phi,\psi}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ is in $S_{p}$
and

$\|P_{\sigma,\phi,\psi}\|_{S_{p}}\leq
c_{\mathcal{K}}^{2/p}\|\phi\|_{\infty}^{\frac{1}{p^{\prime}}}\|\psi\|_{\infty}^{\frac{1}{p^{\prime}}}\|\sigma\|_{p,\mu}.$
(4.10)

###  4.3. An uncertainty relation

In this subsection, we will assume that $\phi$ and $\psi$ satisfy
$\|\phi\|_{\infty}\|\psi\|_{\infty}=1$. Now let $\sigma_{1}=\chi_{S}$ and
$\sigma_{2}=\chi_{\Sigma}$ and let $L_{1}=P_{\sigma_{1},\phi,\psi}$ and
$L_{2}=P_{\sigma_{2},\phi,\psi}$.

######  Theorem 4.8.

Let $\varepsilon_{1},\varepsilon_{2}\in(0,1)$ such that
$\varepsilon_{1}+\varepsilon_{2}<1$. If $f\in L^{2}(\Omega,\mu)$ is
$\varepsilon_{1}$-localized with respect to $P_{\sigma_{1},\phi,\psi}$ and
$\varepsilon_{2}$-localized with respect to $P_{\sigma_{2},\phi,\psi}$ then,

$\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-4}\left(1-\varepsilon_{1}-\varepsilon_{2}\right).$ (4.11)

###### Proof.

From Proposition 4.1,

$\displaystyle\|f-L_{2}L_{1}f\|_{2,\mu}$ $\displaystyle\leq$
$\displaystyle\|f-L_{2}f\|_{2,\mu}+\|L_{2}f-L_{2}L_{1}f\|_{2,\mu}$
$\displaystyle\leq$
$\displaystyle\|L_{2}f-f\|_{2,\mu}+\|L_{2}\|_{S_{\infty}}\|L_{1}f-f\|_{2,\mu}$
$\displaystyle\leq$
$\displaystyle(\varepsilon_{2}+\varepsilon_{1})\|f\|_{2,\mu}.$

Therefore

$\displaystyle\|L_{2}L_{1}f\|_{2,\mu}$ $\displaystyle\geq$
$\displaystyle\|f\|_{2,\mu}-\|f-L_{2}L_{1}f\|_{2,\mu}$ $\displaystyle\geq$
$\displaystyle(1-\varepsilon_{1}-\varepsilon_{2})\|f\|_{2,\mu}.$

Thus from Proposition 4.2 it follows that

$\displaystyle 1-\varepsilon_{1}-\varepsilon_{2}$ $\displaystyle\leq$
$\displaystyle\|L_{2}L_{1}\|_{S_{\infty}}$ $\displaystyle\leq$
$\displaystyle\|L_{1}\|_{S_{\infty}}\|L_{2}\|_{S_{\infty}}$
$\displaystyle\leq$ $\displaystyle c_{\mathcal{K}}^{4}\,\mu(S)\mu(\Sigma).$

This proves the desired result. ∎

Notice that, when $\varepsilon_{1}=\varepsilon_{2}=0$ in the classical Donoho-
Stark uncertainty inequality (3.1), we have $\mu(S)\mu(\Sigma)\geq
c_{\mathcal{K}}^{-2}$. This is a trivial assertion since in this case
$S={\operatorname{supp}}f$, $\Sigma={\operatorname{supp}}{\mathcal{T}}f$ and
then from [11], either $\mu({\operatorname{supp}}f)=\infty$ or
$\mu({\operatorname{supp}}{\mathcal{T}}f)=\infty$. The case
$\varepsilon_{1}=\varepsilon_{2}=0$ in Theorem 4.8 is not trivial and gives
the following result.

######  Corollary 4.9.

If $f\in L^{2}(\Omega,\mu)$ is an eigenfunction of $P_{S,\phi,\psi}$ and
$P_{\Sigma,\phi,\psi}$ corresponding to the same eigenvalue $1$, then

$\mu(S)\mu(\Sigma)\geq c_{\mathcal{K}}^{-4}.$ (4.12)

##  5\. Uncertainty principles for orthonormal sequences

###  5.1. The phase space restriction operator

We define the phase space restriction operator by

$L_{S,\Sigma}=E_{S}F_{\Sigma}E_{S}=(F_{\Sigma}E_{S})^{*}F_{\Sigma}E_{S}.$

We know that the operator $F_{\Sigma}E_{S}$ is Hilbert-Schmidt (see [11,
Inequality (3.2)]), and since the pair $(S,\Sigma)$ is strongly annihilating,
then from [11], we have

$\|L_{S,\Sigma}\|_{S_{\infty}}=\|E_{S}F_{\Sigma}\|^{2}_{S_{\infty}}=\|F_{\Sigma}E_{S}\|^{2}_{S_{\infty}}<1.$
(5.1)

Moreover, the operator $L_{S,\Sigma}$ is self-adjoint, positive and from (2.6)
it is compact and even trace class with

$\|L_{S,\Sigma}\|_{S_{1}}=\|F_{\Sigma}E_{S}\|_{S_{2}}^{2},$ (5.2)

In the fundamental paper [18], Landau and Pollak have considered the
eigenvalue problem associated with the positive self-adjoint operator
$E_{S}F_{\Sigma}E_{S}:L^{2}({\mathbb{R}})\to L^{2}({\mathbb{R}})$, where
$S,\Sigma$ are real intervals, for which they proved an asymptotic estimate
for the number of eigenvalues.

Motivated by the process in [16], we will show that the phase space
restriction operator $L_{S,\Sigma}$ can be viewed as a wavelet multiplier, and
then we will deduce a trace formula.

######  Theorem 5.1.

Let $\phi=\psi$ be the function on $\Omega$ defined by
$\phi=\frac{1}{\sqrt{\mu(S)}}\chi_{S}$ and let $\sigma=\chi_{\Sigma}$. Then

$L_{S,\Sigma}=\mu(S)\,P_{\Sigma,\phi}.$ (5.3)

###### Proof.

Clearly, the function $\phi$ belongs to $L^{2}(\Omega,\mu)\cap
L^{\infty}(\Omega,\mu)$, with $\|\phi\|_{2,\mu}=1$. Then, since $E_{S}$ is
self-adjoint and by Parseval’s equality (1.6), we have for all $f,g\in
L^{2}(\Omega,\mu)$,

$\displaystyle{\left\langle{L_{S,\Sigma}f,g}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle{\left\langle{F_{\Sigma}E_{S}f,\chi_{S}\,g}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle\sqrt{\mu(S)}{\left\langle{F_{\Sigma}E_{S}f,\phi
g}\right\rangle}_{\mu}$ $\displaystyle=$
$\displaystyle\sqrt{\mu(S)}{\left\langle{{\mathcal{T}}F_{\Sigma}E_{S}f,{\mathcal{T}}(\phi
g)}\right\rangle}_{\mu}$ $\displaystyle=$
$\displaystyle\sqrt{\mu(S)}{\left\langle{\chi_{\Sigma}{\mathcal{T}}\chi_{S}f,{\mathcal{T}}(\phi
g)}\right\rangle}_{\mu}$ $\displaystyle=$
$\displaystyle\mu(S){\left\langle{\sigma{\mathcal{T}}(\phi
f),{\mathcal{T}}(\phi g)}\right\rangle}_{\mu}$ $\displaystyle=$
$\displaystyle\mu(S){\left\langle{P_{\Sigma,\phi},g}\right\rangle}_{\mu}.$

This completes the proof. ∎

From Theorem 4.6 and Theorem 5.1, we deduce the following trace formula.

######  Corollary 5.2.

The phase space operator $L_{S,\Sigma}$ is trace class with

${\mathrm{tr}}\left(L_{S,\Sigma}\right)=\mu(S){\mathrm{tr}}\left(P_{\Sigma,\phi}\right)=\int_{S}\int_{\Sigma}|{\mathcal{K}}(x,\xi)|^{2}\,{\mathrm{d}}\mu(x)\,{\mathrm{d}}\mu(\xi).$
(5.4)

The compact operator $L_{S,\Sigma}:L^{2}(\Omega,\mu)\to L^{2}(\Omega,\mu)$ is
self-adjoint and then can be diagonalized as

$L_{S,\Sigma}f=\sum_{n=1}^{\infty}\lambda_{n}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n},$
(5.5)

where $\\{\lambda_{n}=\lambda_{n}(S,\Sigma)\\}_{n=1}^{\infty}$ are the
positive eigenvalues associated to the arranged in a non-increasing manner

$\lambda_{n}\leq\cdots\leq\lambda_{1}<1,$ (5.6)

and $\\{\varphi_{n}=\varphi_{n}(S,\Sigma)\\}_{n=1}^{\infty}$ is the
corresponding orthonormal set of eigenfunctions. In particular

$\|L_{S,\Sigma}\|_{S_{\infty}}=\lambda_{1},$ (5.7)

where $\lambda_{1}$ is the first eigenvalue corresponding to the first
eigenfunction $\varphi_{1}$ of the compact operator $L_{S,\Sigma}$. This
eigenfunction realizes the maximum of concentration on the set
$S\times\Sigma$. On the other hand, since $\varphi_{n}$ is an eigenfunction of
$L_{S,\Sigma}$ with eigenvalue $\lambda_{n}$, then

$\|L_{S,\Sigma}\varphi_{n}-\varphi_{n}\|_{2,\mu}={\left\langle{\varphi_{n}-L_{S,\Sigma}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}=1-\lambda_{n},$
(5.8)

and

$\displaystyle\|L_{S,\Sigma}\left(L_{S,\Sigma}\varphi_{n}\right)-L_{S,\Sigma}\varphi_{n}\|_{2,\mu}$
$\displaystyle=$
$\displaystyle\lambda_{n}^{-1}{\left\langle{L_{S,\Sigma}\varphi_{n}-L_{S,\Sigma}\left(L_{S,\Sigma}\varphi_{n}\right),L_{S,\Sigma}\varphi_{n}}\right\rangle}_{\mu}$
(5.9) $\displaystyle=$
$\displaystyle\lambda_{n}(1-\lambda_{n})=(1-\lambda_{n})\|L_{S,\Sigma}\varphi_{n}\|_{2,\mu}.$

Thus, for all $n$, the functions $\varphi_{n}$ and $L_{S,\Sigma}\varphi_{n}$
are $(1-\lambda_{n})$-localized with respect to $L_{S,\Sigma}$. More
generally, we have the following comparisons of the measures of localization.

######  Proposition 5.3.

Let $\varepsilon,\varepsilon_{1},\varepsilon_{2}\in(0,1).$

  1. (1)

If $f\in L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then $f$ is
$(\varepsilon_{1}+\varepsilon_{2})$-localized with respect to
$F_{\Sigma}E_{S}$ and $(2\varepsilon_{1}+\varepsilon_{2})$-localized with
respect to $L_{S,\Sigma}$.

  2. (2)

If $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-localized with respect to
$L_{S,\Sigma}$, then

${\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}\leq(\varepsilon^{2}+\varepsilon)\|f\|_{2,\mu}^{2}.$
(5.10)

  3. (3)

If $f\in L^{2}(\Omega,\mu)$ satisfies

${\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}\leq\varepsilon\|f\|_{2,\mu}^{2},$
(5.11)

then $f$ is $\sqrt{\varepsilon}$-localized with respect to $L_{S,\Sigma}$.

  4. (4)

If $f\in L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then

${\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}<(2\varepsilon_{1}+\varepsilon_{2})\|f\|_{2,\mu}^{2}.$
(5.12)

###### Proof.

Recall that $\|E_{S}\|_{S_{\infty}}=\|F_{\Sigma}\|_{S_{\infty}}=1$. First we
have

$\displaystyle\|F_{\Sigma}E_{S}f-f\|_{2,\mu}$ $\displaystyle\leq$
$\displaystyle\|F_{\Sigma}f-f\|_{2,\mu}+\|F_{\Sigma}E_{S}f-F_{\Sigma}f\|_{2,\mu}$
$\displaystyle\leq$
$\displaystyle\|F_{\Sigma^{c}}f\|_{2,\mu}+\|F_{\Sigma}\|_{S_{\infty}}\|E_{S^{c}}f\|_{2,\mu}$
$\displaystyle\leq$
$\displaystyle(\varepsilon_{1}+\varepsilon_{2})\|f\|_{2,\mu}.$

Moreover,

$\displaystyle\|L_{S,\Sigma}f-f\|_{2,\mu}$ $\displaystyle\leq$
$\displaystyle\|E_{S}F_{\Sigma}E_{S}f-E_{S}f\|_{2,\mu}+\|E_{S}f-f\|_{2,\mu}$
$\displaystyle\leq$
$\displaystyle\|E_{S}\|_{S_{\infty}}\|F_{\Sigma}E_{S}f-f\|_{2,\mu}+\|E_{S}f-f\|_{2,\mu}$
$\displaystyle\leq$
$\displaystyle(2\varepsilon_{1}+\varepsilon_{2})\|f\|_{2,\mu}.$

Now since

$\displaystyle 2{\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}$
$\displaystyle=$
$\displaystyle{\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}^{2}+{\left\|{f}\right\|}_{2,\mu}^{2}-{\left\|{L_{S,\Sigma}f}\right\|}_{2,\mu}^{2}$
$\displaystyle\leq$
$\displaystyle{\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}^{2}+\left({\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}+{\left\|{L_{S,\Sigma}f}\right\|}_{2,\mu}\right)^{2}-{\left\|{L_{S,\Sigma}f}\right\|}_{2,\mu}^{2}$
$\displaystyle=$ $\displaystyle
2{\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}^{2}+2{\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}{\left\|{L_{S,\Sigma}f}\right\|}_{2,\mu},$

and since ${\left\|{L_{S,\Sigma}}\right\|}_{S_{\infty}}\leq 1$, then

${\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}\leq{\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}^{2}+{\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}{\left\|{f}\right\|}_{2,\mu}\leq(\varepsilon^{2}+\varepsilon){\left\|{f}\right\|}_{2,\mu}^{2},$
(5.13)

and the second result follows.

On the other hand, since

${\left\langle{\left(L_{S,\Sigma}\right)^{2}f,f}\right\rangle}_{\mu}\leq{\left\langle{L_{S,\Sigma}f,f}\right\rangle}_{\mu},$
(5.14)

and since $L_{S,\Sigma}$ is self-adjoint, then

${\left\|{L_{S,\Sigma}f-f}\right\|}_{2,\mu}^{2}={\left\langle{\left(I-L_{S,\Sigma}\right)^{2}f,f}\right\rangle}_{\mu}\leq{\left\langle{(I-L_{S,\Sigma})f,f}\right\rangle}_{\mu}\leq\varepsilon\|f\|_{2,\mu}^{2}.$
(5.15)

Finally, since

${\left\langle{f-L_{S,\Sigma}f,f}\right\rangle}_{\mu}={\left\langle{E_{S^{c}}f,f}\right\rangle}_{\mu}+{\left\langle{E_{S}f,F_{\Sigma^{c}}f}\right\rangle}_{\mu}+{\left\langle{F_{\Sigma}E_{S}f,E_{S^{c}}f}\right\rangle}_{\mu},$

then we obtain the last result.∎

The definition (5.11) is equivalent to

${\left\langle{L_{S,\Sigma}f,f}\right\rangle}_{\mu}\geq(1-\varepsilon)\|f\|_{2,\mu}^{2},$
(5.16)

and we denote by $L^{2}(\varepsilon,S,\Sigma)$ the subspace of
$L^{2}(\Omega,\mu)$ consisting of functions $f\in L^{2}(\Omega,\mu)$
satisfying (5.16). Hence from (5.8) and (5.9) we have,

$\forall\,n\geq 1,\quad\varphi_{n},\,L_{S,\Sigma}\varphi_{n}\in
L^{2}(1-\lambda_{n},S,\Sigma).$ (5.17)

Moreover from Proposition 5.3, if $f\in
L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then $f\in
L^{2}(2\varepsilon_{1}+\varepsilon_{2},S,\Sigma)$, and if $f$ is
$\varepsilon$-localized with respect to $L_{S,\Sigma}$, then $f\in
L^{2}(2\varepsilon,S,\Sigma)$. Therefore we are interested to study the
following optimization problem

$\mathrm{Maximize}\quad\quad{\left\langle{L_{S,\Sigma}f,f}\right\rangle}_{\mu},\quad\quad{\left\|{f}\right\|}_{2,\mu}=1,$
(5.18)

which aims to look for orthonormal functions in $L^{2}(\Omega,\mu)$, which are
approximately time and band-limited to a bounded region like $S\times\Sigma$.
It follows that the number of eigenfunctions of $L_{S,\Sigma}$ whose
eigenvalues are very close to one, are an optimal solutions to the problem
(5.18), since if $\varphi_{n}$ is an eigenfunction of $L_{S,\Sigma}$ with
eigenvalue $\lambda_{n}\geq(1-\varepsilon)$, we have from the spectral
representation,

${\left\langle{L_{S,\Sigma}\varphi_{n},\varphi_{n}}\right\rangle}_{\mu}=\lambda_{n}\geq(1-\varepsilon).$
(5.19)

We denote by $n(\varepsilon,S,\Sigma)$ for the number of eigenvalues
$\lambda_{n}$ of $L_{S,\Sigma}$ which are close to one, in the sense that

$\lambda_{1}\geq\cdots\geq\lambda_{n(\varepsilon,S,\Sigma)}\geq
1-\varepsilon>\lambda_{1+n(\varepsilon,S,\Sigma)}\geq\cdots,$ (5.20)

and we denote by
$V_{n(\varepsilon,S,\Sigma)}=\mathrm{span}\left\\{\varphi_{n}\right\\}_{n=1}^{n(\varepsilon,S,\Sigma)}$
the span of the first eigenfunctions of $L_{S,\Sigma}$ corresponding to the
largest eigenvalues
$\left\\{\lambda_{n}\right\\}_{n=1}^{n(\varepsilon,S,\Sigma)}$. Therefore, by
(5.19) and (5.17), each eigenfunction $\varphi_{n}$ and its resulting function
$L_{S,\Sigma}\varphi_{n}$ are in $L^{2}(\varepsilon,S,\Sigma)$, if and only if
$1\leq n\leq n(\varepsilon,S,\Sigma)$. Now, if $f\in
V_{n(\varepsilon,S,\Sigma)}$, then

${\left\langle{L^{\psi}_{\Sigma}f,f}\right\rangle}_{\mu}=\sum_{n=1}^{n(\varepsilon,S,\Sigma)}\lambda_{n}{\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}\geq\lambda_{n(\varepsilon,S,\Sigma)}\sum_{n=1}^{n(\varepsilon,S,\Sigma)}{\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}\geq(1-\varepsilon)\|f\|_{2,\mu}^{2}.$

Thus $V_{n(\varepsilon,S,\Sigma)}$ determines the subspace of
$L^{2}(\Omega,\mu)$ with maximum dimension that is in
$L^{2}(\varepsilon,S,\Sigma)$. Motivated by the recent paper [25] in the Gabor
setting, we obtain the following theorem that characterizes functions that are
in $L^{2}(\varepsilon,S,\Sigma)$.

######  Theorem 5.4.

Let $f_{\mathrm{ker}}$ denote the orthogonal projection of $f$ onto the kernel
$\mathrm{Ker}(L_{S,\Sigma})$ of $L_{S,\Sigma}$. Then a function $f$ is in
$L^{2}(\varepsilon,S,\Sigma)$ if and only if,

$\
\sum_{n=1}^{n(\varepsilon,S,\Sigma)}(\lambda_{n}+\varepsilon-1){\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}\geq(1-\varepsilon)\|f_{\mathrm{ker}}\|_{2,\mu}^{2}+\sum_{n=1+n(\varepsilon,S,\Sigma)}^{\infty}(1-\varepsilon-\lambda_{n}){\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}.$

###### Proof.

For a given function $f\in L^{2}(\Omega,\mu)$, write

$f=\sum_{n=1}^{\infty}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}+f_{\mathrm{ker}},$
(5.21)

where $f_{\mathrm{ker}}\in\mathrm{Ker}(L_{S,\Sigma})$. Then

${\left\langle{L_{S,\Sigma}f,f}\right\rangle}_{\mu}=\sum_{n=1}^{\infty}\lambda_{n}{\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}.$
(5.22)

So the function $f$ is in $L^{2}(\varepsilon,S,\Sigma)$ if and only if

$\sum_{n=1}^{\infty}\lambda_{n}{\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu_{k}}}\right|}^{2}\geq(1-\varepsilon)\left(\|f_{\mathrm{ker}}\|_{2,\mu}^{2}+\sum_{n=1}^{\infty}{\left|{{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}}\right|}^{2}\right),$
(5.23)

and the conclusion follows. ∎

While a function $f$ that is in $L^{2}(\varepsilon,S,\Sigma)$ does not
necessarily lies in some subspace
$V_{N}=\mathrm{span}\\{\varphi_{n}\\}_{n=1}^{N}$, it can be approximated using
a finite number of such eigenfunctions. Let $\varepsilon_{0}\in(0,1)$ be a
fixed real number and let ${\mathcal{P}}$ the orthogonal projection onto the
subspace $V_{n(\varepsilon_{0},S,\Sigma)}$.

######  Theorem 5.5.

Let $f$ be a function in $L^{2}(\varepsilon,S,\Sigma)$. Then

${\left\|{f-\sum_{n=1}^{n(\varepsilon_{0},S,\Sigma)}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}}\right\|}_{2,\mu}\leq\sqrt{\frac{\varepsilon}{\varepsilon_{0}}}\;\|f\|_{2,\mu}.$
(5.24)

###### Proof.

An easy adaptation of the proof of Proposition 3.3 in [25], we can conclue
that

$\|{\mathcal{P}}f\|_{2,\mu}^{2}\geq(1-\varepsilon/\varepsilon_{0})\|f\|_{2,\mu}^{2}.$
(5.25)

It follows then,

$\|f\|_{2,\mu}^{2}=\|{\mathcal{P}}f+(f-{\mathcal{P}}f)\|_{2,\mu}^{2}=\|{\mathcal{P}}f\|_{2,\mu}^{2}+\|f-{\mathcal{P}}f\|_{2,\mu}^{2}.$

Thus

$\|f-{\mathcal{P}}f\|_{2,\mu}^{2}=\|f\|_{2,\mu}^{2}-\|{\mathcal{P}}f\|_{2,\mu}^{2}\leq\|f\|_{2,\mu}^{2}-(1-\varepsilon/\varepsilon_{0})\|f\|_{2,\mu}^{2}=\varepsilon/\varepsilon_{0}\|f\|_{2,\mu}^{2}.$

This completes the proof of the theorem. ∎

Consequently and from Proposition 5.3, we immediately deduce the following
approximating results.

######  Corollary 5.6.

Let $\varepsilon,\varepsilon_{1},\varepsilon_{2}\in(0,1)$.

  1. (1)

If $f\in L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$, then

${\left\|{f-\sum_{n=1}^{n(\varepsilon_{0},S,\Sigma)}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}}\right\|}_{2,\mu}\leq\sqrt{\frac{2\varepsilon_{1}+\varepsilon_{2}}{\varepsilon_{0}}}\;\|f\|_{2,\mu}.$
(5.26)

  2. (2)

If $f\in L^{2}(\Omega,\mu)$ is $\varepsilon$-localized with respect to
$L_{S,\Sigma}$, then

${\left\|{f-\sum_{n=1}^{n(\varepsilon_{0},S,\Sigma)}{\left\langle{f,\varphi_{n}}\right\rangle}_{\mu}\varphi_{n}}\right\|}_{2,\mu}\leq\sqrt{\frac{2\varepsilon}{\varepsilon_{0}}}\;\|f\|_{2,\mu}.$
(5.27)

###  5.2. Shapiro–type uncertainty principles

Based on Malinnikova’s ideas [20], we will prove in this section a
quantitative dispersion inequality for orthonormal sequences and a strong
uncertainty principle for orthonormal bases. Notice also that the homogeneity
of the kernel ${\mathcal{K}}$ plays a key role in this section, especially in
Lemma 5.12.

####  5.2.1. Localization theorem

Our starting point is the following theorem which states that any orthonormal
system in $L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$ cannot be
infinite.

######  Theorem 5.7.

Let $\varepsilon_{1},\varepsilon_{2}\in(0,1)$ such that
$2\varepsilon_{1}+\varepsilon_{2}<1$ and let
$\left\\{f_{n}\right\\}_{n=1}^{N}$ be an orthonormal system in
$L^{2}(\varepsilon_{1},\varepsilon_{2},S,\Sigma)$. Then

$N<c_{\mathcal{K}}^{2}\frac{\mu(S)\mu(\Sigma)}{1-2\varepsilon_{1}-\varepsilon_{2}}.$
(5.28)

###### Proof.

The result follows immediately from Inequalities (5.12) and (5.4). ∎

Consequently, if the generalized dispersions of each element of an orthonormal
sequence are uniformly bounded, then this sequence is also finite.

######  Corollary 5.8.

Fix $A_{1},A_{2}>0$. Let $s>0$ and let $\left\\{f_{n}\right\\}_{n=1}^{N}$ be
an orthonormal sequence in $L^{2}(\Omega,\mu)$ such that
$\big{\|}|x|^{s}f_{n}\big{\|}^{1/s}_{2,\mu}\leq A_{1}$ and
$\big{\|}|\xi|^{s}\,{\mathcal{T}}f_{n}\big{\|}^{1/s}_{2,\mu}\leq A_{2}$. Then
each $f_{n}$ is in
$L^{2}\left(\frac{1}{4},\frac{1}{4},B_{4^{\frac{1}{s}}A_{1}},B_{4^{\frac{1}{s}}A_{2}}\right)$,
and

$N\leq C\,(A_{1}A_{2})^{2a}.$ (5.29)

###### Proof.

By assumption we have, for all $n\geq 1$,

$\int_{|x|>4^{\frac{1}{s}}A_{1}}|f_{n}(x)|^{2}\,{\mathrm{d}}\mu(x)=\int_{|x|>4^{\frac{1}{s}}A_{1}}|x|^{-2s}|x|^{2s}|f_{n}(x)|^{2}\,{\mathrm{d}}\mu(x)\leq\frac{1}{16A_{1}^{2s}}\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu}\leq\frac{1}{16}.$

In the same way we obtain

$\int_{|\xi|>4^{\frac{1}{s}}A_{2}}|{\mathcal{T}}f_{n}(\xi)|^{2}\,{\mathrm{d}}\mu(\xi)\leq\frac{1}{16}.$

Thus $f_{n}\in
L^{2}\left(\frac{1}{4},\frac{1}{4},B_{4^{\frac{1}{s}}A_{1}},B_{4^{\frac{1}{s}}A_{2}}\right)$,
and from (5.28) we conclude the desired result. ∎

####  5.2.2. Quantitative dispersion inequality for orthonormal sequences

From Inequality (1.9), there exists a constant $C$ such that for all $f\in
L^{2}(\Omega,\mu)$,

$\||x|^{s}f\|_{2,\mu}\;\||\xi|^{s}{\mathcal{T}}f\|_{2,\mu}\geq
C{\left\|{f}\right\|}^{2}_{2,\mu},$ (5.30)

and the dilation argument (1.4) shows that (5.30) is equivalent to

$\||x|^{s}f\|_{2,\mu}^{2}+\||\xi|^{s}{\mathcal{T}}f\|_{2,\mu}^{2}\geq
2C{\left\|{f}\right\|}^{2}_{2,\mu}.$ (5.31)

Consequently we obtain immediately the following result.

######  Corollary 5.9.

Let $s>0$ and let $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ be an orthonormal
sequence in $L^{2}(\Omega,\mu)$. Then there exists $j_{0}\in{\mathbb{Z}}$ such
that

$\forall n\geq 1,\ \ \ \ \ \ \ \
\max\left(\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu},\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}\right)\geq
2^{s(j_{0}-1)}.$ (5.32)

This corollary with Corollary 5.8 allows as to prove the following
quantitative dispersion inequality.

######  Theorem 5.10.

Let $s>0$ and let $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ be an orthonormal
sequence in $L^{2}(\Omega,\mu)$. Then for every $N\geq 1$,

$\displaystyle\sum_{n=1}^{N}\left(\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu}+\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{2}_{2,\mu}\right)\geq
C\,N^{1+\frac{s}{2a}}.$ (5.33)

###### Proof.

For each $j\in{\mathbb{Z}}$ we define

$P_{j}=\left\\{n\ :\
\max\left(\big{\|}|x|^{s}f_{n}\big{\|}^{\frac{1}{s}}_{2,\mu},\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{\frac{1}{s}}_{2,\mu}\right)\in\left[2^{j-1},2^{j}\right)\right\\}.$

First, by Inequality (5.32), we see that $P_{j}$ is empty for all $j<j_{0}$.
Moreover, since for each $n\in P_{j}$, $j\geq j_{0}$,

$\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{\frac{1}{s}}\leq 2^{j}\ \ \ \
\mathrm{and}\ \ \ \
\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}^{\frac{1}{s}}\leq 2^{j},$
(5.34)

then by Corollary 5.8, $P_{j}$ is finite, for all $j\geq j_{0}$, and if we
denote by $N_{j}$ the number of elements in $P_{j}$ then

$N_{j}\leq C\,4^{2aj}.$

Therefore, for every $m\geq j_{0}$, the number of elements in
$\bigcup_{j=j_{0}}^{m}P_{j}$ is less than $C\,4^{2am},$ where $C$ is a
constant that does not depend on $m$.

Now, if $N>2C\,4^{2aj_{0}}$, then we can choose an integer $m>j_{0}$ such that

$2C\,4^{(m-1)2a}<N\leq 2C\,4^{2am}.$

Therefore at least half of $\\{1,\ldots,N\\}$ does not belong to
$\bigcup_{j=j_{0}}^{m-1}P_{j}$ and we obtain

$\displaystyle\sum_{n=1}^{N}\left(\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu}+\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{2}_{2,\mu}\right)$
$\displaystyle\geq$
$\displaystyle\sum_{n=1}^{N}\max\left(\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu},\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{2}_{2,\mu}\right)$
$\displaystyle\geq$
$\displaystyle\frac{N}{2}4^{s(m-1)}\geq\frac{1}{2}\frac{N}{4^{s}}\left(\frac{N}{2C}\right)^{\frac{s}{2a}}.$

Finally, if $N\leq 2C\,4^{2aj_{0}}$, then from Corollary 5.9 we have

$\displaystyle\sum_{n=1}^{N}\left(\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu}+\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{2}_{2,\mu}\right)$
$\displaystyle\geq$
$\displaystyle\sum_{n=1}^{N}\max\left(\big{\|}|x|^{s}f_{n}\big{\|}^{2}_{2,\mu},\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}^{2}_{2,\mu}\right)$
$\displaystyle\geq$ $\displaystyle
N4^{s(j_{0}-1)}\geq\frac{N}{4^{s}}\left(\frac{N}{2C}\right)^{\frac{s}{2a}}.$

This completes the proof. ∎

The last Dispersion inequality implies in particular that, there does not
exist an infinite sequence $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ in
$L^{2}(\Omega,\mu)$ such that the two sequences
$\left\\{\||x|^{s}f_{n}\|_{2,\mu}\right\\}_{n=1}^{\infty}$ and
$\left\\{\||\xi|^{s}{\mathcal{T}}f_{n}\|_{2,\mu}\right\\}_{n=1}^{\infty}$ are
bounded. More precisely:

######  Corollary 5.11.

Let $s>0$ and let $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ be an orthonormal
sequence in $L^{2}(\Omega,\mu)$. Then for every $N\geq 1$,

$\sup_{1\leq n\leq
N}\left\\{\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{2},\;\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}^{2}\right\\}\geq
C\,N^{\frac{s}{2a}}.$ (5.35)

In particular

$\sup_{n}\left(\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{2}+\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}^{2}\right)=\infty.$
(5.36)

####  5.2.3. Strong uncertainty principle for orthonormal bases

One wonders if (5.36) still valid for the product instead of the sum. We will
show that this statement is not true in general for orthonormal sequences, but
still valid for orthonormal bases. Indeed, there is an infinite orthonormal
sequence $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ in $L^{2}(\Omega,\mu)$, with
bounded product of dispersions. Fix $f:\Omega\to{\mathbb{R}}$ a radial, real-
valued Schwartz function supported in $B(0,2)\backslash B(0,1)$, with
$\|f\|_{2,\mu}=1$, and consider $f_{n}(x)=2^{na}f(2^{n}x)$. Then

$\|f_{n}\|_{2,\mu}=\|f\|_{2,\mu},\ \ \ \ {\operatorname{supp}}f_{n}\subset
B(0,2^{-n+1})\backslash B(0,2^{-n})\ \ \ \ \mathrm{and}\ \ \ \
{\mathcal{T}}f_{n}(\xi)=2^{-na}{\mathcal{T}}(f)\left(2^{-n}\xi\right).$

Therefore $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ form an orthonormal sequence
in $L^{2}(\Omega,\mu)$ and for every $s>0$,

$\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}=2^{-ns}\big{\|}|x|^{s}f\big{\|}_{2,\mu},\
\ \ \ \ \ \
\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}=2^{ns}\big{\|}|\xi|^{s}{\mathcal{T}}f\big{\|}_{2,\mu}.$

Hence for all $n$,

$\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}\,\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}=\big{\|}|x|^{s}f\big{\|}_{2,\mu}\big{\|}|\xi|^{s}{\mathcal{T}}f\big{\|}_{2,\mu}<\infty.$

To prove the main result of this subsection, we will need the following
special form of the uncertainty principle for sets of finite measure, see e.g.
[3, 4, 10, 11, 21].

######  Lemma 5.12.

Let $S$ and $\Sigma$ be measurable subsets of finite measure
$0<\mu(S),\,\mu(\Sigma)<\infty$. Then there exists a nonzero function $f\in
L^{2}(\Omega,\mu)$ such that ${\operatorname{supp}}f\subset S^{c}$ and
${\operatorname{supp}}{\mathcal{T}}f\subset\Sigma^{c}$.

###### Proof.

From [11, Corollary 3.7], there exist a positive constant $C(S,\Sigma)$ such
that for all functions $f\in\operatorname{Im}(F_{\Sigma})$,

$\|f\|_{2,\mu}\leq C(S,\Sigma)\|E_{S^{c}}f\|_{2,\mu}.$

Therefore the trace space
$\Lambda=\\{f|_{S^{c}}:f\in\operatorname{Im}(F_{\Sigma})\\}$ form a closed
subspace in $L^{2}(S^{c},\mu)$ which is obviously not the whole space. Let $g$
be a nonzero function in $\Lambda^{c}=L^{2}(S^{c},\mu)\backslash\Lambda$.
Since $g=F_{\Sigma}g+F_{\Sigma^{c}}g$, then $f=F_{\Sigma^{c}}g$ is a nonzero
function in $L^{2}(\Omega,\mu)$ such that $f$ is supported on $S^{c}$ and
${\mathcal{T}}f$ is supported on $\Sigma^{c}$. We extend $f$ by zero on $S$ in
order to get the required function. ∎

######  Theorem 5.13.

Let $s>0$ and let $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ be an orthonormal
basis for $L^{2}(\Omega,\mu)$. Then

$\sup_{n}\left(\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}\,\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}\right)=\infty.$
(5.37)

###### Proof.

Assume that there exists an orthonormal basis
$\left\\{f_{n}\right\\}_{n=1}^{\infty}$ such that

$\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{1/s}\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}^{1/s}\leq
A^{2}.$

Let $k\in{\mathbb{Z}}$ and let

$A_{k}=\left\\{f_{n}\ :\
\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{1/s}\in\left(2^{-k}A,2^{-k+1}A\right]\right\\}.$

Clearly, $\left\\{f_{n}\right\\}_{n=1}^{\infty}=\bigcup_{k}A_{k}$, and for
each $f_{n}\in A_{k}$, we have

$\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{1/s}\leq 2^{-k+1}A\ \ \ \ \mathrm{and}\
\ \ \ \big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}^{1/s}\leq A2^{k}.$

Then by Corollary 5.8, $A_{k}$ is finite, and if $N_{k}$ is the number of
elements in $A_{k}$ then $N_{k}$ is bounded by a constant $C$ that does not
depend on $k$.

Let $R>0$, then by using Lemma 5.12, we take a nonzero function $f\in
L^{2}(\Omega,\mu)$ with $\|f\|_{2,\mu}=1$, such that

${\operatorname{supp}}f,\,{\operatorname{supp}}{\mathcal{T}}f\subset
B_{R}^{c}.$

Then for $k\geq 0$ and $f_{n}\in A_{k}$ we obtain by the Cauchy-Schwartz
inequality that

$\left|{\left\langle{f,f_{n}}\right\rangle}_{\mu}\right|^{2}\leq
R^{-2s}\|f\|_{2,\mu}^{2}\big{\|}|x|^{s}f_{n}\big{\|}_{2,\mu}^{2}\leq(2AR^{-1})^{2s}4^{-sk}.$
(5.38)

Similarly, for $k<0$ and $f_{n}\in A_{k}$ we obtain by Parseval theorem (1.6),

$\left|{\left\langle{f,f_{n}}\right\rangle}_{\mu}\right|^{2}=\left|{\left\langle{{\mathcal{T}}f,{\mathcal{T}}f_{n}}\right\rangle}_{\mu}\right|^{2}\leq
R^{-2s}\|f\|_{2,\mu}^{2}\big{\|}|\xi|^{s}{\mathcal{T}}f_{n}\big{\|}_{2,\mu}^{2}\leq(AR^{-1})^{2s}4^{sk}.$
(5.39)

Now, since $\left\\{f_{n}\right\\}_{n=1}^{\infty}$ is an orthonormal basis for
$L^{2}(\Omega,\mu)$, then

$1=\|f\|_{2,\mu}^{2}=\displaystyle\sum_{k}\sum_{f_{n}\in
A_{k}}\left|{\left\langle{f,f_{n}}\right\rangle}_{\mu}\right|^{2},$

and by combining Inequalities (5.38) and (5.39), we obtain

$\displaystyle 1$ $\displaystyle\leq$
$\displaystyle(2AR^{-1})^{2s}\sum_{k=0}^{\infty}4^{-sk}N_{k}+(AR^{-1})^{2s}\sum_{k=1}^{\infty}4^{-sk}N_{-k}$
$\displaystyle\leq$ $\displaystyle
C(2AR^{-1})^{2s}\sum_{k=0}^{\infty}4^{-sk}+C(AR^{-1})^{2s}\sum_{k=1}^{\infty}4^{-sk}$
$\displaystyle\leq$ $\displaystyle\dfrac{C}{R^{2s}}.$

Choosing $R$ large enough, we get a contradiction. The theorem is proved. ∎

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