A Class of Frames for Paley-Wiener spaces
with Multiple Lattice Tiles Support
Ezra Tampubolon, Volker Pohl, Holger Boche
Lehrstuhl für Theoretische Informationstechnik
Technische Universität München, 80290 München, Germany
{ezra.tampubolon, volker.pohl, boche}@tum.de
Abstract—Let Ω ⊂ RN be a bounded measurable set
and let Λ ⊂ RN be a lattice. Assume that Ω tiles RN
multiply at level K when translated by Λ. Then this
paper derives conditions on sets of sampling functions
such that they form a frame for the Paley-Wiener
space PWΩ of functions bandlimited to Ω. Moreover,
the canonical dual frame is derived, which allows signal
recovery for all signals in PWΩ from a multichannel
sampling system samples.
Index Terms—Frames, multi-tiling, interpolation,
Riesz basis, sampling.
I. Introduction
Let Ω ⊂ RN be a bounded measurable set. This paper
deals with the construction of frames and their dual frames
for L2 (Ω). The question for which domains Ω the space
L2 (Ω) admit an orthonormal basis of exponential functions
was intensively studied in the past. This problem was
mainly inspired by Fuglede’s conjecture [1], and we refer to
[2] for more details and discussion on this problem. In general, Fuglede’s Conjecture was disproved, i.e. there exists
domains which posses no orthonormal basis of exponential
functions. One may ask for domains which posses a Riesz
basis or frame.
Recently, Riesz bases of exponentials, i.e. Riesz basis of
the form {e−i2πh·,λi }λ∈Γ , where Γ ⊆ RN is a discrete set,
for domains Ω, which tile RN multiply at level K ∈ N, have
attracted some interests. The existence of a Riesz basis
for such domains, which are in addition bounded and Riemann measurable (i.e. the boundary is of measure zero),
was first proved in [3], using the theory of quasicrystals
[4], [5]. A similar result was proven in [6] based on a more
elementary approach from linear algebra. In this paper,
we consider the construction of Riesz bases and frames of
the geneal form {φbm e−i2πh·,λi : m = 1, . . . , M, λ ∈ Γ},
together with the corresponding dual frames for L2 (Ω)
with domains Ω tiling RN multiply. In [7], [8] such a result
was proven for certain locally compact Abelian groups,
based on the techniques known as fibrization and range
functions. However, no attempt was made to derive the
corresponding dual frames. Our approach here is more elThis work was partly supported by the German Research Foundation (DFG) under Grants PO 1347/2-1 and BO 1734/20-1.
c
978-1-4673-7353-1/15/$31.00 2015
IEEE
λ∗ ∈ Λ∗ ⊂ RN
✲
f (t) ∈ PWΩ
✲
φ1
φ2
✟❄ ✲ c1,λ∗
= y1 (λ∗ )
✟❄ ✲ c2,λ∗
= y2 (λ∗ )
y1 (t)
y2 (t)
.
.
.
✲
φM
✟❄ ✲
yM (t)
cM,λ∗ = yM (λ∗ )
Fig. 1. Analysis filter bank of the sampling scheme considered in
this paper: Signals f ∈ PWΩ are linearly filtered by φ1 , . . . , φM .
Afterwards, the signals are sampled on a uniform lattice Λ∗ in RN .
ementary and based on techniques used for shift-invariant
spaces with support in RN [9].
The question whether a set is a frame or a Riesz basis
for L2 (Ω) has an important interpretation in terms of
sampling theory and signal processing (see also [10]).
Clearly, the Fourier transform F : L2 (RN ) → L2 (RN ),
given by
Z
fb(ω) = (Ff )(ω) =
f (t) e−i2πhω,ti dt , ω ∈ RN , (1)
RN
maps L2 (Ω) unitarily onto the Paley-Wiener space
PW Ω = f ∈ L2 (RN ) : supp(fb) ⊂ Ω .
(2)
Therefore, if a set {b
sn } is a frame for L2 (Ω), then {sn }
will be a frame for PW Ω , and so any function f ∈ PW Ω
can be reconstructed from the set of generalized
samples
P
{cn = hf, sn i}n∈Z by the series f (t) =
n∈Z cn σn (t)
where {σn }n∈Z denotes a certain dual frame of {sn }n∈Z . If
{b
sn (ω) = e−i2πhλn ,ωi } is a frame of exponential functions,
then the generalized measurements correspond to point
evaluations cn = hf, sn i = f (λ∗n ) for some λ∗n ∈ RN .
Viewed from this perspective, this paper deals with sampling systems which have a structure as shown in Fig. 1.
There the signal f ∈ PWΩ is filtered (or convoluted)
by a bank of M linear filters, with impulse responses
hm (τ ) = φm (−τ ), and subsequently uniformly sampled
on a discrete lattice Λ∗ in RN . This gives a set
cm,λ∗ = ym (λ∗ ) = hf, sm,λ∗ iPWΩ ,
m = 1, . . . , M
λ∗ ∈ Λ∗
(3)
of signal samples, where sm,λ∗ (t) := φm (t−λ∗ ) denotes the
sampling functions. Provided that the set {sm,λ∗ } forms
a frame, every f ∈ PWΩ can
PM from the samples
P recovered
(3), by the formula f (t) = λ∗ ∈Λ∗ m=1 cm,λ∗ σm,λ∗ (t) ,
where {σm,λ∗ } is a dual frame of {sm,λ∗ }.
For a given support region Ω ∈ RN , this paper gives
conditions on the set {φm }M
m=1 of filters in Fig. 1 and on
the sampling lattice Λ∗ such that every f ∈ PWΩ can
be recovered from the samples (3). If these conditions are
satisfied, the corresponding dual frame is derived.
II. Preliminaries and Notations
We work with functions defined on the N -dimensional
real Euclidean vector space RN . If Ω is a subset of RN then
χΩ stands for its indicator function, |Ω| for its Lebesgue
measure. The space of all Lebesgue square-integrable functions on Ω is denoted by L2 (Ω), equipped with the usual
inner product and norm. In case that Ω is bounded, we
normalize as common the inner product and norm on
L2 (Ω). For any f integrable on RN , its Fourier transform
fb = Ff is given by (1), and by Plancherel’s Theorem F
can be extended to a unitary operator on L2 (RN ).
A. Lattices and Tiling
Throughout this paper, Λ = AZN stands for a (full
rank) lattice, which is defined by an invertible N × N
matrix A. The volume of Λ is defined as vol(Λ) = | det A|,
and its dual lattice is Λ∗ = BZN where B = (AT )−1 . From
this definition, it follows that hλ, λ∗ i ∈ Z for all λ ∈ Λ and
λ∗ ∈ Λ∗ .
Definition 1: A set Ω0 ⊂ RN is called a fundamental
domain for a lattice Λ ifP
Ω0 tiles RN when translated at
the locations of Λ, i.e. if λ∈Λ χΩ0 (ω − λ) = 1 for almost
every (a.e.) ω ∈ RN .
In this work, we restrict ourselves to bounded and measurable fundamental domains. For the possible shape of a
convex fundamental domain of a lattice in RN , we refer,
e.g., to [11]. In particular, a convex fundamental domain
in R2 can only be either quadrilateral or hexagonal. If Ω0
is a fundamental domain for Λ, we write L2 (RN /Ω0 ) for
the set of all functions f : RN → C which are square
integrable if restricted to Ω0 and which are Λ periodic, i.e.
for which
f (ω + λ) = f (ω)
for every λ ∈ Λ and all ω ∈ RN .
Moreover, we write PΩ0 : L2 (Ω) → L2 (RN /Ω0 ) for the
operator which maps any f ∈ L2 (Ω0 ) to the unique
function in L2 (RN /Ω0 ) which coincides with f on Ω0 .
Below, we will frequently use functions of the form
eλ (ω) := e−i2πhλ,ωi with λ ∈ Λ and ω ∈ RN . Note that
these functions are Λ∗ -periodic, in the sense that
eλ (ω + λ∗ ) = eλ (ω)
for all
λ∗ ∈ Λ∗ and ω ∈ RN .
It is well known [1, §6] that if Ω0 is a fundamental domain
for a lattice Λ, then the set {eλ∗ : λ∗ ∈ Λ∗ } is an
orthonormal basis for L2 (Ω0 ).
Fig. 2. Examples of support sets Ω in R2 with a rectangular
generating domain Ω0 and K = 5 (left) and with a hexagonal Ω0
and K = 12 (right). The dots indicate the corresponding lattice Λ.
We will follow [6] and consider domains Ω which tile RN
multiply by a certain lattice.
Definition 2: Let Ω ⊂ RN be bounded and measurable,
and let Λ be a lattice. We say that Ω tiles RN when
translated by Λ at level K ∈ N if
P
for a.e. ω ∈ RN .
λ∈Λ χΩ (ω − λ) = K
If Ω ⊂ RN is a measurable set which tiles RN at level
K by a lattice Λ, then it is known [6, Lemma 1] that Ω
can always be partitioned as
Ω = Ω1 ∪ Ω2 ∪ · · · ∪ ΩK
(4)
up to a set of measure zero. Therein each Ωk is a fundamental domain for Λ and all Ωk are mutually disjoint.
B. Our signal space
Here we consider spaces L2 (Ω) where Ω tiles RN at level
K ∈ N when translated by a lattice Λ. For convenient, we
restrict us to the case where each Ωk , in the partition (4)
is given by Ωk = Ω0 + λ for some λ ∈ Λ. Principally, one
can extend the result by some efforts to the corresponding
general case. Thus, we assume that
[
Ω0 + λ with ΛΩ = {λ1 , . . . , λK } ⊂ Λ . (5)
Ω=
λ∈ΛΩ
Therein, the fundamental domain Ω0 is called the generating domain of Ω. Because of (5), it is clear that Ω0 is not
unique. For example, each fundamental domain Ωk can be
the generating domain of Ω.
Fig. 2 sketches the structure of our support sets Ω by
two examples. Therein, the gray shaded area represents
Ω which is obtain by translating a rectangular (left)
and a hexagonal (right) generating domain Ω0 along the
corresponding lattice Λ. The set on the left of Fig. 2 tiles
RN at multiply level K = 5, whereas the right hand
side has multiply level K = 12. Fig. 2 also indicates
that domains with almost any arbitrary shape can be well
approximated by domains of the form (5), by choosing
a sufficently small generating domain Ω0 . Moreover, we
note that a fundamental domain is any set Ω0 ⊂ RN
Our derivation below is based on the following characterization of Frames and Riesz basis. We refer to standard
textbooks like [12] for proofs of this result.
Lemma 1: Let K be an arbitrary Hilbert space with orthonormal basis {ϕn }n∈Z . Let H be a separable Hilbert
space, let T : K → H be a linear operator, and set
sn = Tϕn ,
Fig. 3. The shaded area is a fundamental domain Ω0 in R2 for the
lattice Λ indicated by the dotted lines.
which contains exactly one element of each coset modulo
Λ. For example, on the left hand side of Fig. 2, Ω0 is a
parallelepiped, and on the right hand side, it is a hexagon.
But there are many more, which can be obtained in a way
as indicated by Fig. 3 [6].
Decomposition (5) of Ω allows us to identify the space
L2 (Ω) with the Hilbert space KK which is defined as the
K-fold direct sum of L2 (Ω0 ), i.e.
2
2
2
KK := L (Ω0 ) ⊕ L (Ω0 ) ⊕ · · · ⊕ L (Ω0 ) =
K
M
L2 (Ω0 ) (6)
k=1
with the canonical inner product:
Z
1
f (ω), g(ω) CK dω
f, g K =
K
|Ω0 | Ω0
with f := [f1 , . . . , fK ]T ∈ KM and where fk ∈ L2 (Ω0 )
stands for the function in the k-th component of f , and
similarly for g ∈ KK . Now, L2 (Ω) and KK can be identified
via the isometric, isomorphic mapping AΩ : L2 (Ω) → KK
which maps an f ∈ L2 (Ω) onto f = [f1 , . . . , fK ]T by
fk (ω) = f (ω + λk ) χΩk (ω + λk ) , ω ∈ Ω0 . Its inverse
2
A−1
Ω : KK → L (Ω) is then given by
(A−1
Ω f )(ω) =
K
X
(PΩ0 fk )(ω) χΩk (ω) ,
ω∈Ω,
k=1
and one easily sees that hAΩ f, AΩ f iKK = hf, giL2 (Ω) for
all f, g ∈ L2 (Ω).
C. Frames and Riesz basis
Since we want to derive Frames (or Riesz basis) for
L2 (Ω) and PWΩ , respectively, we shortly recall the formal
definition of these concepts (see, e.g., [12]).
Definition 3: Let H be a separable Hilbert space and let
s = {sn }n∈Z be a sequence in H. Then s is called a frame
for H if there exists frame constants A, B > 0 such that
P
2
Akxk2 ≤ n∈Z |hx, sn i| ≤ Bkxk2 for all x ∈ H .
The sequence s is called a Riesz basis if
2
X
2
cn sn ≤ B kck2ℓ2 for all c ∈ ℓ2 .
A kckℓ2 ≤ n∈Z
H
n∈Z.
(i) The set s = {sn }n∈Z is a frame for K if and only if T
is bounded and surjective.
(ii) The set s = {sn }n∈Z is a Riesz basis for K if and only
if T is bounded and bijective.
Moreover, let V : K → H be a left inverse of the adjoint
T∗ , i.e. assume that VT∗ x = x for all x ∈ H, and set
σn = Vϕn ,
n∈Z.
Then σ = {σn }n∈Z is a dual frame of s, i.e.
P
P
x = n∈Z hx, σn i sn = n∈Z hx, sn i σn for all x ∈ H .
III. Frames for PWΩ
In this section, we state our main result on the class
of frames for the Paley-Wiener space PWΩ of functions
which are bandlimited to a domain which tiles RN at a
certain multiple level K ≥ 1.
Theorem 2: Let Λ be a lattice and let Ω ⊂ RN which tiles
RN when translated by Λ at level K of the form (5). Let
{φm }M
m=1 be a set of functions in PWΩ and define


φb1 (ω + λ1 ) . . . φbM (ω + λ1 )


..
..
..
Φ(ω) = 
 . (7)
.
.
.
b
b
φ1 (ω + λK ) . . . φM (ω + λK )
The set
sm,λ∗ (t) = φm (t − λ∗ ) ,
m = 1, . . . , M , λ∗ ∈ Λ∗
(8)
(A) is a frame for PWΩ if and only if there are constants
A, B > 0 such that
AI ≤ Φ(ω)Φ∗ (ω) ≤ BI
for a.e. ω ∈ Ω0
(9)
(B) is a Riesz basis for PWΩ if and only if in addition
to (9)
AI ≤ Φ∗ (ω)Φ(ω) ≤ BI
for a.e. ω ∈ Ω0
(10)
with the same constants as in (9).
Remark: Equivalently, the set of functions given in (14) is
a frame or Riesz basis for L2 (Ω), if condition (9) and (10)
are satisfied, respectively,
Remark: A similar result was given in [7]. However, our
proof here seems to be somewhat simpler and it yields
immediately the dual frames (cf. Section IV) which were
not given in [7]. Statement (B) of Theorem 2 can also be
found in [8], with a proof similarly to ours.
Remark: A necessary condition for (9) is M ≥ K, and
(10) can only hold if M = K.
Remark: Without proof, we remark that the frame constants A and B are related to the singular values of Φ(ω):
A = ess inf σmin [Φ(ω)] and B = ess sup σmin [Φ(ω)] . (11)
ω∈Ω0
ω∈Ω0
Proof: By the assumption of the theorem, the support
region Ω has the form (5) wherein Ω0 is a fundamental
domain for Λ. Therefore {eλ∗ L
: λ∗ ∈ Λ∗ } is an orthonormal
M
2
basis for L (Ω0 ). Let KM = m=1 L2 (Ω0 ) be the Hilbert
space defined as in (6). Then it is clear that
{em,λ∗ } = {eλ∗ }λ∗ ∈Λ∗ ⊕ · · · ⊕ {eλ∗ }λ∗ ∈Λ∗
(12)
is an orthonormal basis for KM . So em,λ∗ ∈ KM is the
vector em,λ∗ (ω) = [0, . . . , 0, eλ∗ (ω), 0, . . . , 0]T whose only
non-zero entry is at the mth position.
Next, we define the linear mapping T : KM → L2 (Ω) by
T : f 7→
M
X
n=1
φbn (ω) (PΩ0 fn )(ω) . ω ∈ Ω ,
(13)
One easily verifies that T maps the orthonormal basis (12)
onto the functions
sbm,λ∗ (ω) = (Tem,λ∗ )(ω) = φbm (ω) eλ∗ (ω)
∗
= φbm (ω) e−i2πhλ ,ωi .
(14)
The inverse Fourier transforms sm,λ∗ = F −1 sbm,λ∗ of these
functions belong to PWΩ and are given by (8). Now we
want to apply Lemma 1 to prove that, under the given
conditions, (14) is a Frame or Riesz basis for L2 (Ω). This
would imply that (8) is a frame or Riesz basis for PWΩ .
In doing so, it is more convenient to identify L2 (Ω)
with KK as discussed in Sec. II-B. Therefore, we apply
the isomorphic mapping AΩ : L2 (Ω) → KK to Tf . This
e : KM → KK given by
yields the mapping T
e (ω) = (AΩ Tf )(ω) = Φ(ω) f (ω) , ω ∈ Ω0
Tf
(15)
with the K × M matrix Φ(ω) as given by (7). Since
AΩ : L2 (Ω) → KK is a bijective isometry, it is clear that
e is
T is bounded, surjective, or injective if and only if T
bounded, surjective, or injective.
e is
(A): In view of Lemma 1, we have to show that T
bounded and surjective if and only if (9) holds. To this
e ∗ : KK → KM
end, it is sufficient to show that its adjoint T
e
is bounded and bounded below. Because this implies the T
∗
e
e
is bounded (because T is bounded), that the range R(T)
e ∗ ) is closed, since T
e ∗ is bounded
is closed (because R(T
e
e
below), and that R(T) is dense (since T is injective).
For any g ∈ KK we have:
Z D
E
∗ 2
1
e
T gK =
Φ(ω)Φ∗ (ω)g(ω) , g(ω) K dω .
M
|Ω0 | Ω0
C
e ∗ is bounded and
From this, it easily follows that T
bounded below, i.e.
∗ 2
A kgk2KK ≤ e
T gK ≤ B kgk2KK for all g ∈ KK
M
if and only if (9) holds.
e is additionally
(B) To proof (B) we have to show that T
injective if and only if (10) holds. Similar as above, one
has
Z D
E
2
1
e
Tf KK =
Φ∗ (ω)Φ(ω)f (ω) , f (ω) M dω
|Ω0 | Ω0
C
for any f ∈ KM .
Consequently, it easily follows that
2
A kf k2KM ≤ e
Tf K ≤ B kf k2KM for all f ∈ KK
K
if and only if (10) holds.
Remark: The proof showed also that under the condition
of Theorem 2, the set
e m,λ∗ )(ω) = φ (ω) eλ∗ (ω)
e
sm,λ∗ (ω) = (Te
m
∗
= φm (ω) ei2πhλ
,ωi
,
m = 1, . . . , M ; λ∗ ∈ Λ∗ , (16)
wherein φm (ω) is the mth column of Φ(ω) given in (7), is
a frame or Riesz basis for KK , respectively.
IV. Dual Frame Construction
The construction of a dual frame follows immediately
from the second part of Lemma 1. Basically, we have to
e ∗ of the operator (15)
find a left-inverse of the adjoint T
used in the proof of Theorem 2. For simplicity, we only
give the explicit form of the canonical dual frame here.
Theorem 3: Consider the same setting as in Theorem 2
and assume that condition (9) is satisfied. Let Ψ(ω) be the
K × M pseudoinverse of Φ∗ (ω), i.e. let
−1
Ψ(ω) = Φ(ω)Φ∗ (ω) Φ(ω) , a.e. ω ∈ Ω0
(17)
Then the canonical dual frame of the sequence (8) is given
by
σm,λ∗ (t) = σm (t − λ∗ ) ,
m = 1, . . . , M ; λ∗ ∈ Λ∗
where σm = F −1 σ
bm ∈ PWΩ with
PK
σ
bm (ω) = k=1 PΩ0 [Ψ]k,m (ω) χΩk (ω) ,
ω∈Ω.
Remark: We write [Ψ]k,m for the entry in the kth row
and mth column of the K × M matrix Ψ. So the function
σ
bm ∈ L2 (Ω) is defined on each component Ωk of Ω. On Ωk
the function σ
bm (ω) is equal to the Λ-periodic extension of
[Ψ(ω)]k,m .
e ∗ : KK → KM of the operator T,
e
Proof: The adjoint T
given in (15), is equal to
e ∗ g (ω) = Φ∗ (ω) g(ω) ,
T
ω ∈ Ω0 .
Since condition (9) is satisfied, the K × K matrix
Φ(ω)Φ∗ (ω) is invertible for almost all ω ∈ Ω0 and so the
K ×M pseudoinverse matrix (17) of Φ∗ (ω) is well defined.
e ∗ . So
Therefore (Vf )(ω) = Ψ(ω)f (ω) is a left-inverse of T
by Lemma 1 the set
e m,λ∗ (ω) = (Vem,λ∗ )(ω) = Ψ(ω) em,λ∗ (ω)
σ
= ψ m (ω) eλ∗ (ω) ,
ω ∈ Ω0
where ψm (ω) stands for the mth column of Ψ(ω), is the
dual frame of the frame (16) in KK . Since KK is isometric
isomorph to L2 (Ω), it follows that the set {b
σm,λ∗ : m =
1, . . . , M ; λ∗ ∈ Λ∗ } is given by
e m,λ∗ )(ω) = (A−1
σ
bm,λ∗ (ω) = (A−1
Ω σ
Ω Vem,λ∗ )(ω)
!
K
X
∗
=
(PΩ0 ψk,m )(ω)χΩk (ω) ei2πhλ ,ωi , ω ∈ Ω (18)
k=1
and where ψk,m (ω) = [Ψ(ω)]k,m , is the dual frame of the
sequence (14) in L2 (Ω). Taking the the inverse Fourier
transform of the functions in (18), we obtain (3) as the
dual frame of (8).
Remark: Theorem 3 gives only the canonical dual frame of
(8). However, from the proof it is clear that any arbitrary
dual frame can be obtained by selecting an arbitrary leftinverse of the matrix Φ∗ (ω) instead of Ψ(ω).
V. Frames of Exponentials
As a corollary and a special case of Theorem 2, we
can derive a similar result as in [3], [6] for Riesz bases
of exponential functions for L2 (Ω).
Corollary 4: Let Λ be a lattice and let Ω ⊂ RN which
tiles RN when translated by Λ at level K of the form (5).
Then there exists vectors µ1 , . . . , µK ∈ RN , which depend
only on Λ, such that the set
sbm,λ∗ (ω) = ei2πhµm ,ω+λk i ,
m = 1, . . . , K; k ∈ Z
is a Riesz basis for L2 (Ω).
Proof: Let be K > 1. We apply Theorem 2 with the
particular functions φbm (ω) = ei2πhµm ,ωi . Then the matrix
Φ(ω), given in (7), can be written as Φ(ω) = H D(ω),
where D(ω) is a unitary diagonal matrix with diagonal
entries [D(ω)]m,m = ei2πhµm ,ωi and H is a constant matrix
with enties
[H]k,m = ei2πhµm ,λk i ,
k, m = 1, . . . , K .
According to Theorem 2, it remains to show that there
∗
∗
exists vectors {µm }K
m=1 such that Φ(ω)Φ (ω) = H H has
rank K. To construct such vectors, we choose µ0 ∈ RN
such that hµ0 , λk i = νk 6= 0 for all k = 1, . . . , K and
such that νk 6= νn when ever n 6= k. Such a µ0 can
always be found. In particular, it is possible to find such
a µ0 ∈ Λ∗ . Then νk ∈ Z for all k = 1, . . . , K. Set
νmax = max{|ν1 | , . . . , |νK |} and define µm := [c(m −
1)/(νmax K)]µ0 , where c is a positive constant. This yields
the particular matrix H with entries
2π
[H]k,m = ei K
νk
νmax
c(m−1)
,
k, m = 1, . . . , K ,
by choosing c s.t. |νk /(Kνmax )| < 1/2, H is a Vandermonde matrix of rank K.
Remark: In [6], [8], it was even shown that almost every
randomly chosen set {µm }K
m=1 will satisfy the corollary.
However, even although almost every set {µm }K
m=1 of
vectors yields a Riesz basis for PWΩ , the frame bounds
(11) might be very poor, and it is a challenging task to
determine sets {µm }K
m=1 which provide a Riesz basis with
good bounds. See, e.g., the discussion in [13] for the onedimensional setting (N = 1) and based on a construction
similarly to the one in the proof of Corollary 4.
VI. Sampling density
For any τ ∈ RN , let Br (τ ) = {t ∈ RN : kt − τ k ≤ r}
be the ball of radius r > 0 centered at τ with volume
|Br (τ )|. A lattice Λ ⊂ RN is said to have asymptotic
density dens(Λ) = ̺, if
lim
r→∞
card[Λ ∩ Br (τ )]
=̺
|Br (τ )|
for all τ ∈ RN .
If Ω0 is a fundamental domain for Λ, then dens(Λ) =
1/ vol(Λ) = 1/|Ω0 |. So the sampling density Ds for the
system shown in Fig. 1 with the sampling sampling functions s = {sm,λ∗ } is equal to
M M
Ω .
= M vol(Λ) =
Ds = M dens(Λ∗ ) =
∗
vol(Λ )
K
If M = K, i.e. if s is a Riesz basis for PWΩ , the sampling
density Ds is equal to the Landau density of PWΩ [14].
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