Bases of complex exponentials with restricted supports
2025-04-27
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arXiv:2210.05707v1 [math.CA] 11 Oct 2022
Bases of complex exponentials with restricted supports
Dae Gwan Leea,∗, G¨otz E. Pfandera, David Walnutb
aMathematical Institute for Machine Learning and Data Science (MIDS), Katholische
Universit¨at Eichst¨att–Ingolstadt, Goldknopfgasse 7, 85049 Ingolstadt, Germany
bDepartment of Mathematical Sciences, George Mason University, Fairfax, VA 22030, USA
Abstract
The complex exponentials with integer frequencies form a basis for the space of
square integrable functions on the unit interval. We analyze whether the basis
property is maintained if the support of the complex exponentials is restricted
to possibly overlapping subsets of the unit interval. We show, for example, that
if S1,...,SK⊂[0,1] are finite unions of intervals with rational endpoints that
cover the unit interval, then there exists a partition of Zinto sets Λ1,...,ΛK
such that SK
k=1{e2πiλ(·)χSk:λ∈Λk}is a Riesz basis for L2[0,1]. Here, χS
denotes the characteristic function of S.
Keywords: complex exponentials, Riesz bases, support restriction, spectrum
2000 MSC: 42C15
1. Introduction and Main Results
For a measurable set S⊂Rand a discrete set Λ ⊂R, we define E(S, Λ) :=
{e2πiλ(·)χS:λ∈Λ}which is the set of all exponential functions with frequencies
in Λ considered on the domain S. A Riesz basis for a separable Hilbert space H
is a sequence of the form {U en}n∈Z, where {en}n∈Zis an orthonormal basis for
Hand U:H → H is a bijective bounded operator. Equivalently, Riesz bases
are characterized as complete Riesz sequences1(see, e.g., [3, Chapter 3.6]).
Most of the results on exponential bases deal with exponential functions
that are defined on the full domain of the involved space. For instance, if one
speaks of exponential functions in the space L2(S), then functions of the form
t7→ e2πiλt with λ∈R, defined on the full domain S, are usually considered.
In this paper, we consider the situation where exponential functions in L2(S)
are restricted to some different subsets of S. More precisely, we are inter-
ested in conditions on S1,...,SK⊂Rand Λ1,...,ΛK⊂Rwhich are neces-
sary/sufficient for SK
k=1 E(Sk,Λk) to be a Riesz basis for L2(SK
k=1 Sk).
∗Corresponding author
Email addresses: daegwans@gmail.com (Dae Gwan Lee), pfander@ku.de (G¨otz
E. Pfander), dwalnut@gmu.edu (David Walnut)
1A sequence {fn}n∈Z⊂ H is called a Riesz sequence if there exist constants 0 < A ≤B <
∞such that Akck2
ℓ2≤ k Pn∈Zcnfnk2≤Bkck2
ℓ2for all c={cn}n∈Z∈ℓ2(Z).
Preprint submitted to Elsevier October 13, 2022
As our first main result, we prove that if each Sk⊂[0,1) is a finite union
of intervals with rational endpoints, then one can find pairwise disjoint sets
Λ1,...,ΛK⊂Zsuch that SK
k=1 E(Sk,Λk) is a Riesz basis for L2(SK
k=1 Sk).
Theorem 1. Let K, N ∈Nwith K≤N. Let I1,...,IKbe distinct intervals
from [ℓ
N,ℓ+1
N),ℓ= 0,...,N−1, and for each k, let Skbe a union of subcollection
of I1,...,IK, including Ik. Then there exists a permutation2ρ∈SKsuch that
K
[
k=1
E(Sk, NZ+ρ(k)) =
K
[
k=1 e2πiλ (·)χSk:λ∈NZ+ρ(k)
is a Riesz basis for L2(SK
k=1 Sk).
We point out that if, for instance, S1= [0,1), S2=··· =SN= [0,1
N),
then necessarily the system SN
k=1 E(Sk, NZ+ρ(k)) with ρ∈SNcannot be a
Riesz basis for L2(SN
k=1 Sk). See Section 2 for some necessary conditions for
SK
k=1 E(Sk,Λk) to be a Riesz basis for L2(SK
k=1 Sk).
Corollary 2. Let S1,...,SK⊂[0,1) be finite unions of intervals with rational
endpoints. There exist pairwise disjoint sets Λ1,...,ΛK⊂Zsuch that
K
[
k=1
E(Sk,Λk) =
K
[
k=1 e2πiλ (·)χSk:λ∈Λk
is a Riesz basis for L2(SK
k=1 Sk). In particular, if SK
k=1 Sk= [0,1), then
Λ1,...,ΛKform a partition of Z.
Theorem 1 relies on the following two auxiliary results.
Lemma 3. For any A∈CK×Kand a binary matrix M∈ {0,1}K×K, there
exists a permutation ρ∈SKwith
det((PρA)⊙M)≥R·det(A)
K!,
where Pρis the permutation matrix associated with ρ(more precisely, Pρ=
[pk,ℓ]K
k,ℓ=1 with pk,ℓ = 1 if ℓ=ρ(k)and pk,ℓ = 0 otherwise), the symbol ⊙
denotes the entrywise multiplication (Hadamard product), and Rdenotes the
number of distinct generalized diagonals3of Mthat contain no zeros.
Consequently, if A∈CK×Kis nonsingular and if M∈ {0,1}K×Khas at
least one generalized diagonal of ones, then there exists a permutation matrix P
with det((P A)⊙M)6= 0.
2We denote by SK=S({1, . . . , K}) the symmetric group of {1, . . . , K}, which consists of
all the K! permutations of 1, . . . , K.
3A generalized diagonal of a K×Kmatrix is a selection of Kcells from the K×Kcells,
such that exactly one cell is selected from each row and each column. For instance, the usual
diagonal is a generalized diagonal.
2
In the proof of Theorem 1, we will apply Lemma 3 to square submatrices A
of the Fourier matrix WN= [e−2πikℓ/N ]k,ℓ∈ZN.
Remark. Lemma 3 implies that for any invertible matrix A∈CN×Nand a
mask M∈ {0,1}N×Nwith at least one generalized diagonal of ones, there exists
a row permutation of M, say f
M, such that A⊙f
Mis invertible.
The following proposition provides a simple characterization for SK
k=1 E(Sk,Λk)
to be a Riesz basis for L2(SK
k=1 Sk) when Λkare N-periodic sets in Rand Skare
some unions of [ ℓ
N,ℓ+1
N), ℓ∈ZN:= {0,...,N−1}. For notational convenience,
we will use indices k= 0,...,K−1, instead of k= 1,...,K.
Proposition 4. Fix any N∈N. Let c0,...,cK−1∈[0, N )be distinct real
numbers and let Lk⊂ZN,k= 0,...,K−1. Define L:= SK−1
k=0 Lk,L:= |L|,
and W(c0,L0),...,(cK−1,LK−1):= [wk,ℓ]k=0,...,K−1, ℓ∈L ∈CK×Lwith
wk,ℓ =(e−2πickℓ/N if ℓ∈ Lk,
0if ℓ∈ L\Lk.
Also, define Sk:= Sℓ∈Lk[ℓ
N,ℓ+1
N)for k= 0,...,K−1, and S:= SK−1
k=0 Sk=
Sℓ∈L [ℓ
N,ℓ+1
N). Then
K−1
[
k=0
E(Sk, NZ+ck) =
K−1
[
k=0 e2πiλ (·)χSk:λ∈NZ+ck(1)
is
•a frame for L2(S)if and only if the mapping x7→ W(c0,L0),...,(cK−1,LK−1)x
is injective, i.e., the matrix W(c0,L0),...,(cK−1,LK−1)has full rank and K≥
L.
•a Riesz sequence in L2(S)if and only if the mapping x7→ W(c0,L0),...,(cK−1,LK−1)x
is surjective, i.e., the matrix W(c0,L0),...,(cK−1,LK−1)has full rank and K≤
L.
•a Riesz basis for L2(S)if and only if the mapping x7→ W(c0,L0),...,(cK−1,LK−1)x
is bijective, i.e., the matrix W(c0,L0),...,(cK−1,LK−1)is invertible (and K=
L).
In any of the above cases, the optimal lower frame/Riesz bound is given by
1
Nσ2
min(W(c0,L0),...,(cK−1,LK−1)). Furthermore, if K=L=Nand if the matrix
W(c0,L0),...,(cK−1,LK−1)is invertible, then the dual Riesz basis of (1) in L2[0,1)
is given by
N−1
[
k=0 N
N−1
X
j=0
e−2πickj/N zj,k χ[j
N,j+1
N)· E([0,1), NZ+ck)
=
N−1
[
k=0 nN
N−1
X
j=0
e−2πickj/N zj,k χ[j
N,j+1
N)(·)e2πiλ (·):λ∈NZ+cko,
(2)
3
where zj,kj,k∈ZN=W(c0,L0),...,(cN−1,LN−1)−1.
Unfortunately, if some endpoints of Skare irrational, there is no convenient
characterization for SK
k=1 E(Sk,Λk) to be a Riesz basis for L2(SK
k=1 Sk). Nev-
ertheless, we are able to prove the following as our second main result.
Theorem 5. Let I1, I2, I3be intervals partitioning [0,1) and let Λ1,Λ2,Λ3be
a partition of Zsuch that for each k= 1,2,3, the system E(Ik,Λk)is a Riesz
basis for L2(Ik). Here, the intervals are allowed to be empty sets, and we set
Λk=∅in the case that Ik=∅. Then
3
[
k=1
E(Sk,Λk) =
3
[
k=1 e2πiλ(·)χSk:λ∈Λk
is a Riesz basis for L2(S3
k=1 Sk) = L2[0,1) whenever Sk=Sn∈LkInwith k∈
Lk⊂ {1,2,3}for all k= 1,2,3.
Note that the condition k∈ Lkis equivalent to having Ik⊂Sk. This
condition is actually not necessary but is a convenient assumption. We will
discuss this in more detail in Section 2.
It is easily seen that Theorem 5 does not hold for more than three intervals
(see Example 8 below).
We would like to highlight that the statements of Theorems 1 and 5 have
completely different quantifiers. Given sets S1,...,SK⊂[0,1) with certain
property, Theorem 1 finds some pairwise disjoint sets Λ1,...,ΛK⊂Zsuch
that SK
k=1 E(Sk,Λk) is a Riesz basis for L2(SK
k=1 Sk). In contrast, Theorem 5
assumes the sets I1,...,IK⊂[0,1) and Λ1,...,ΛK⊂Zto satisfy that E(Ik,Λk)
is a Riesz basis for L2(Ik), k= 1,...,K, and then shows that SK
k=1 E(Sk,Λk)
is a Riesz basis for L2(SK
k=1 Sk) = L2[0,1) if each Skis a union of subcollection
of I1,...,IK, including Ik.
1.1. Related work
Kozma and Nitzan [5] proved that for any finite union Sof disjoint inter-
vals in [0,1), there exists a set Λ ⊂Zsuch that E(S, Λ) is a Riesz basis for
L2(S). By adapting the proof technique of [5], Caragea and Lee [2] showed that
if Ik= [ak, bk), k= 1,...,K, are disjoint intervals in [0,1) with the numbers
1, a1,...,aK, b1,...,bKbeing linearly independent over Q, then there exist pair-
wise disjoint sets Λk⊂Z,k= 1,...,K, such that for every K ⊂ {1,...,K}, the
system E(∪k∈K Ik,∪k∈K Λk) is a Riesz basis for L2(∪k∈K Ik). This extends the
result of Kozma and Nitzan by requiring the exponential Riesz basis to possess
a hierarchical Riesz bases property so that every subcollections of I1,...,IK
admit Riesz bases with the corresponding frequency sets, when the endpoints
of I1,...,IKare rationally independent. Conjecture 2 below states a similar
hierarchical result for the case where all the intervals I1,...,IKhave rational
endpoints.
4
摘要：
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arXiv:2210.05707v1[math.CA]11Oct2022BasesofcomplexexponentialswithrestrictedsupportsDaeGwanLeea,∗,G¨otzE.Pfandera,DavidWalnutbaMathematicalInstituteforMachineLearningandDataScience(MIDS),KatholischeUniversit¨atEichst¨att–Ingolstadt,Goldknopfgasse7,85049Ingolstadt,GermanybDepartmentofMathematicalScie...
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