A note on exponential Riesz bases

Abstract

We prove that if $I_\ell = [a_\ell ,b_\ell )$, $\ell =1,\ldots ,L$, are disjoint intervals in [0, 1) with the property that the numbers $1, a_1, \ldots , a_L, b_1, \ldots , b_L$ are linearly independent over ${\mathbb {Q}}$, then there exist pairwise disjoint sets $\Lambda _\ell \subset {\mathbb {Z}}$, $\ell =1, \ldots , L$, such that for every $J \subset \{ 1, \ldots , L \}$, the system $\{e^{2\pi i \lambda x} : \lambda \in \cup _{\ell \in J} \, \Lambda _\ell \}$ is a Riesz basis for $L^2 ( \cup _{\ell \in J} \, I_\ell )$. Also, we show that for any disjoint intervals $I_\ell$, $\ell =1, \ldots , L$, contained in [1, N) with $N \in {\mathbb {N}}$, the orthonormal basis $\{e^{2\pi i n x} : n \in {\mathbb {Z}}\}$ of $L^2[0,1)$ can be complemented by a Riesz basis $\{e^{2\pi i \lambda x}: \lambda \in \Lambda \}$ for $L^2(\cup _{\ell =1}^L \, I_{\ell })$ with some set $\Lambda \subset (\frac{1}{N} {\mathbb {Z}}) \backslash {\mathbb {Z}}$, in the sense that their union $\{e^{2\pi i \lambda x} : \lambda \in {\mathbb {Z}}\cup \Lambda \}$ is a Riesz basis for $L^2 ( [0,1) \cup I_1 \cup \cdots \cup I_L )$.

1 Introduction and main results

Seip [1] showed that if S is an interval contained in [0, 1), then there exists a set $\Lambda \subset {\mathbb {Z}}$ such that $E(\Lambda ) := \{e^{2\pi i \lambda x} : \lambda \in \Lambda \}$ is a Riesz basis for $L^2(S)$. Since then, there have been various attempts towards finding/characterizing the sets S that admit a Riesz spectrum, see e.g., [2,3,4,5,6,7]. A significant breakthrough was made by Kozma and Nitzan [8] who proved that if $[a_\ell ,b_\ell )$, $\ell =1, \ldots , L$, are disjoint intervals contained in [0, 1), then there exists a set $\Lambda \subset {\mathbb {Z}}$ such that $E(\Lambda )$ is a Riesz basis for $L^2 ( \cup _{\ell =1}^L [a_\ell ,b_\ell ) )$. Recently, Pfander et al. [9] showed that if the intervals $[a_\ell ,b_\ell )$ form a partition of [0, 1), then the set of integers ${\mathbb {Z}}$ can be partitioned into some sets $\Lambda _\ell$, $\ell =1, \ldots , L$, such that for each $\ell$, the system $E(\Lambda _\ell )$ is a Riesz basis for $L^2[a_\ell ,b_\ell )$, and moreover $E( \cup _{\ell \in J} \, \Lambda _\ell )$ is a Riesz basis for $L^2 ( \cup _{\ell \in J} \, S_\ell )$ whenever $J \subset \{ 1, \ldots , L \}$ is a consecutive index set (see [9, Theorems 1 and 2]). We would like to point out that up to date, the existence of exponential Riesz bases is known only for several classes of sets $S \subset {\mathbb {R}}$. Recently, Kozma et al. [10] constructed a bounded measurable set $S \subset {\mathbb {R}}$ such that the space $L^2(S)$ has no exponential Riesz basis. For an overview of the known results on exponential Riesz bases, we refer to [11, Section 1].

We are interested in the following two problems:

Problem 1

(Hierarchical structured exponential Riesz bases) Given a family of disjoint sets $S_1, S_2 , \ldots , S_L \subset [0,1)$ with positive measure, can we find disjoint sets $\Lambda _1, \Lambda _2, \ldots , \Lambda _L \subset {\mathbb {Z}}$ such that for every $J \subset \{ 1, \ldots , L \}$, the system $E( \cup _{\ell \in J} \, \Lambda _\ell )$ is a Riesz basis for $L^2 ( \cup _{\ell \in J} \, S_\ell )$?

Problem 2

(Complementability of exponential Riesz bases) Let $\Lambda \subset {\mathbb {R}}$ be a discrete set and let $S \subset {\mathbb {R}}$ be a finite positive measure set such that $E(\Lambda )$ is a Riesz basis for $L^2(S)$. Given a finite positive measure set $S' \subset {\mathbb {R}}\backslash S$, can we find a discrete set $\Lambda ' \subset {\mathbb {R}}\backslash \Lambda$ such that

• $E(\Lambda ')$ is a Riesz basis for $L^2 (S')$, and

• $E(\Lambda \cup \Lambda ')$ is a Riesz basis for $L^2 (S \cup S')$?

The second problem is closely related to the first, as it deals with the case $L=2$ under the assumption that the sets $S_1$ and $\Lambda _1$ are already fixed.

Considering the result of Kozma et al. [10], it is necessary to restrict the sets $S_\ell$, S and $S'$ to certain classes of sets. In this paper, we will address the above problems in the case that $S_\ell$, S and $S'$ are intervals or finite unions of intervals.

Our first main result answers Problem 1 in the affirmative when $S_\ell = [a_\ell ,b_\ell )$, $\ell =1,\ldots ,L$, are disjoint intervals in [0, 1) with the property that the numbers $1, a_1, \ldots , a_L, b_1, \ldots , b_L$ are linearly independent over ${\mathbb {Q}}$, which means that having $q + q_1 a_1 + \cdots + q_L a_L + q_1' b_1 + \cdots + q_L' b_L = 0$ for some $q , q_\ell , q_\ell ' \in {\mathbb {Q}}$ implies $q = q_\ell = q_\ell ' = 0$ for all $\ell$. The result is motivated by [8, p. 279, Claim 2].

Theorem 1

Let $0< a_1< b_1< \cdots< a_L< b_L < 1$ with $L \in {\mathbb {N}}$. Assume that the numbers $1, a_1, \ldots , a_L, b_1, \ldots , b_L$ are linearly independent over ${\mathbb {Q}}$. There exist pairwise disjoint sets $\Lambda _\ell \subset {\mathbb {Z}}$, $\ell =1, \ldots , L$, such that for every $J \subset \{ 1, \ldots , L \}$, the system $E( \cup _{\ell \in J} \, \Lambda _\ell )$ is a Riesz basis for $L^2 ( \cup _{\ell \in J} \, [a_\ell ,b_\ell ) )$.

Concerning Problem 2, we have the following result which builds on the fact that $E({\mathbb {Z}})$ is an orthonormal basis (thus, a Riesz basis) for $L^2[0,1)$.

Theorem 2

Let $1 \le a_1< b_1< a_2< b_2< \cdots< a_L < b_L \le N$ with $L, N \in {\mathbb {N}}$. There exists a set $\Lambda ' \subset (\frac{1}{N} {\mathbb {Z}}) \backslash {\mathbb {Z}}$ such that

• $E(\Lambda ')$ is a Riesz basis for $L^2 ( \cup _{\ell =1}^L [a_\ell ,b_\ell ) )$, and

• $E({\mathbb {Z}}\cup \Lambda ')$ is a Riesz basis for $L^2 ( [0,1) \cup [a_1,b_1) \cup \cdots \cup [a_L,b_L) )$.

This theorem answers Problem 2 in the affirmative when $S = [0,1)$, $\Lambda = {\mathbb {Z}}$, and $S'$ is a finite union of disjoint bounded intervals in $[1, \infty )$.

While the result of Pfander et al. [9] relies on Avdonin’s theorem and the ergodic properties of a certain type of integer sequences, Theorem 1 is based on a refinement of the key lemma of [8] for primes, together with Chebotarëv’s theorem on roots of unity and the Kronecker–Weyl equidistribution theorem along the primes.

1.1 Remarks

We state the following conjecture which improves upon Theorem 1.

Conjecture 1

Let $[a_\ell ,b_\ell )$, $\ell =1, \ldots , L$, be disjoint intervals contained in [0, 1), that is, $0 \le a_1< b_1 \le a_2< b_2 \le \cdots \le a_L < b_L \le 1$. There exist pairwise disjoint sets $\Lambda _\ell \subset {\mathbb {Z}}$, $\ell =1, \ldots , L$, such that for every $J \subset \{ 1, \ldots , L \}$, the system $E( \cup _{\ell \in J} \, \Lambda _\ell )$ is a Riesz basis for $L^2 ( \cup _{\ell \in J} \, [a_\ell ,b_\ell ) )$.

This conjecture generalizes Theorem 1 by removing the Q-linear independence of the endpoints and by allowing for contiguous intervals in [0, 1), i.e., $b_\ell = a_{\ell +1}$ for some $\ell \in \{ 1,\ldots ,L-1 \}$. The conjecture can be easily reformulated as follows.

Conjecture 1′

Let $I_\ell$, $\ell =1, \ldots , L$, be intervals which form a partition of [0, 1). There exists a partition $\Lambda _\ell$, $\ell =1, \ldots , L$, of ${\mathbb {Z}}$ such that for every $J \subset \{ 1, \ldots , L \}$, the system $E( \cup _{\ell \in J} \, \Lambda _\ell )$ is a Riesz basis for $L^2 ( \cup _{\ell \in J} \, I_\ell )$.

Indeed, Conjecture 1 obviously implies Conjecture 1’, and the converse is seen by considering the partition of [0, 1) formed using the endpoints of all $[a_\ell ,b_\ell )$. Note that Conjecture 1’ generalizes the result of Pfander et al. [9] from consecutive index sets J to arbitrary index sets J.

Lastly, we mention that both Problems 1 and 2 remain open for more general classes of sets $S_\ell$, S and $S'$.

2 Preliminaries

Definition

A sequence $\{ f_n \}_{n\in {\mathbb {Z}}}$ in a separable Hilbert space ${\mathcal {H}}$ is called

• a frame for ${\mathcal {H}}$ (with frame bounds A and B) if there are constants $0< A \le B < \infty$ such that

  $$\begin{aligned} A \, \Vert f \Vert ^2 \;\le \; \sum _{n\in {\mathbb {Z}}} \vert \langle f , f_n\rangle \vert ^2 \;\le \; B \, \Vert f \Vert ^2 \quad \text {for all} \;\; f \in {\mathcal {H}}; \end{aligned}$$

• a Riesz sequence in ${\mathcal {H}}$ (with Riesz bounds A and B) if there are constants $0< A \le B < \infty$ such that

  $$\begin{aligned} A \, \Vert c \Vert _{\ell _2}^2 \;\le \; \Big \Vert \sum _{n\in {\mathbb {Z}}} c_n \, f_n \Big \Vert ^2 \;\le \; B \, \Vert c \Vert _{\ell _2}^2 \quad \text {for all} \;\; \{c_n\}_{n\in {\mathbb {Z}}} \in \ell _2 ({\mathbb {Z}}); \end{aligned}$$

• a Riesz basis for ${\mathcal {H}}$ if it is a complete Riesz sequence in ${\mathcal {H}}$.

It is well-known (see e.g., [12, Proposition 3.7.3, Theorems 5.4.1 and 7.1.1] or [8, Lemma 1]) that a sequence in ${\mathcal {H}}$ is a Riesz basis if and only if it is both a frame and a Riesz sequence. Moreover in this case, the optimal frame bounds coincides with the optimal Riesz bounds. It is worth noting that Riesz bases are equivalent to unconditional bases that are norm-bounded above and below [12, Lemma 3.6.9]. Since every exponential function has constant norm in $L^2(S)$ with $S \subset {\mathbb {R}}^d$, namely $\Vert e^{2 \pi i \lambda \cdot (\cdot )} \Vert _{L^2(S)} = \vert S \vert ^{1/2}$ for any $\lambda \in {\mathbb {R}}^d$, Riesz bases of exponentials coincide with unconditional bases of exponentials.

Proposition 3

(Proposition 2.1 in [13], Proposition 5.4 in [14]) Let $\{ e_n \}_{n \in I}$ be an orthonormal basis of a separable Hilbert space ${\mathcal {H}}$, where I is a countable index set. Let $P : {\mathcal {H}} \rightarrow {\mathcal {M}}$ be the orthogonal projection from ${\mathcal {H}}$ onto a closed subspace ${\mathcal {M}}$. Let $J \subset I$, $J^c := I \backslash J$, and $0 < \alpha \le 1$. The following are equivalent.

• (i) $\{ P e_n \}_{n \in J} \subset {\mathcal {M}}$ is a frame for ${\mathcal {M}}$ with optimal lower bound $\alpha$.

• (ii) $\{ P e_n \}_{n \in J^c} \subset {\mathcal {M}}$ is a Bessel sequence with optimal bound $1-\alpha$.

• (iii) $\{ (\mathrm {Id}-P) e_n \}_{n \in J^c} \subset {\mathcal {M}}^{\perp }$ is a Riesz sequence with optimal lower bound $\alpha$.

As a direct consequence of Proposition 3, we have that for a set $\Lambda \subset {\mathbb {Z}}$ and a measurable set $S \subset [0,1)$, the system $E(\Lambda )$ is a frame for $L^2(S)$ if and only if $E({\mathbb {Z}}\backslash \Lambda )$ is a Riesz sequence in $L^2 ( [0,1)\backslash S )$.

Lemma 4

Assume that $E(\Lambda )$ is a Riesz basis for $L^2(S)$ with bounds $0< A \le B < \infty$, where $\Lambda \subset {\mathbb {R}}^d$ is a discrete set and $S \subset {\mathbb {R}}^d$ is a measurable set. Then the following hold.

• (a) For any $a,b \in {\mathbb {R}}^d$, the system $E(\Lambda +a)$ is a Riesz basis for $L^2(S+b)$ with bounds A and B.

• (b) For any $c > 0$, the system $E(c \Lambda )$ is a Riesz basis for $L^2(\frac{1}{c} S)$ with bounds $\frac{A}{c}$ and $\frac{B}{c}$.

Lemma 4 remains valid if all the terms “Riesz basis” are replaced by “Riesz sequence” or by “frame”. A proof of Lemma 4 can be found in [11].

For any $N \in {\mathbb {N}}$, a measurable set $S \subset [0,1)$, and $n = 1, \ldots , N$, we define

$$\begin{aligned} \begin{aligned} A_{\ge n}&= A_{\ge n} (N,S) := \Big \{ t \in [0, \tfrac{1}{N}) : t+ \tfrac{k}{N} \in S \;\; \text {for at least { n} values} \\&\qquad \qquad \qquad \qquad \qquad \qquad \quad \;\; \text {of} \;\; k \in \{ 0 , 1, \ldots , N-1 \} \Big \} . \end{aligned} \end{aligned}$$

(1)

Lemma 5

(Lemma 2 in [8]) Let $N \in {\mathbb {N}}$ and let $S \subset [0,1)$ be a measurable set. If there exist sets $\Lambda _1, \ldots , \Lambda _N \subset N{\mathbb {Z}}$ such that $E(\Lambda _n)$ is a Riesz basis for $L^2(A_{\ge n})$, then $E( \cup _{n=1}^N (\Lambda _n {+} n) )$ is a Riesz basis for $L^2 (S)$.

This lemma, which plays a central role in [8], combines Riesz bases by introducing consecutive shift factors n to the frequency sets $\Lambda _n$ and then taking their union $\cup _{n=1}^N (\Lambda _n {+} n)$. For our purpose, we generalize the lemma to allow for arbitrary shift factors when N is prime.

Lemma 6

Let $N \in {\mathbb {N}}$ be a prime and let $S \subset [0,1)$ be a measurable set. If there exist sets $\Lambda _1, \ldots , \Lambda _N \subset N{\mathbb {Z}}$ such that $E(\Lambda _n)$ is a Riesz basis (resp. a frame, a Riesz sequence) for $L^2(A_{\ge n})$, then for every permutation $\{ j_n \}_{n=1}^N$ of $\{ 1, \ldots , N \}$ the system $E( \cup _{n=1}^N (\Lambda _n {+} j_n) )$ is a Riesz basis (resp. a frame, a Riesz sequence) for $L^2 (S)$.

See “Appendix A” for a proof of Lemma 6.

We will use the following notation throughout the proofs. For $x \in {\mathbb {R}}$, we denote the fractional part of x by $\{ x \}$, that is, $0 \le \{ x \} := x - \lfloor x \rfloor < 1$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to x. Also, we adopt the convention that $[x,y) = \emptyset$ if $x = y \in {\mathbb {R}}$.

For the proof of Theorem 1, we will need the following version of the Kronecker–Weyl equidistribution theorem (see e.g., [15, Theorem 443] or [16, p. 48, Theorem 6.3 and Example 6.1]) along the primes. The one-dimensional case ($d=1$) was proved by Vinogradov [17] (see also [16, p. 22]): if a is an irrational number, the sequence $\{ 2 a \}, \{ 3 a \}, \{ 5 a \}, \{ 7 a \}, \ldots$ is uniformly distributed in [0, 1), meaning that for every interval $I \subset [0,1)$, the ratio of the numbers $\{ p a \}$ with prime $p \le N$ that are contained in I, tends to |I| as $N \rightarrow \infty$. The notion of uniform distribution is defined similarly in higher dimensions, see e.g., [16, p. 47, Definition 6.1]. As we could not find any reference for the multi-dimensional case, we include a short proof here.

Proposition 7

(Kronecker–Weyl equidistribution along the primes) Let $d \in {\mathbb {N}}$ and $a_1, \ldots , a_d \in {\mathbb {R}}$. If the numbers $1, a_1, \ldots , a_d$ are linearly independent over ${\mathbb {Q}}$, which means that having $q + q_1 a_1 + \cdots + q_d \, a_d = 0$ for some $q,q_1,\ldots ,q_d \in {\mathbb {Q}}$ implies $q = q_1 = \cdots = q_d = 0$, then the d-dimensional vectors

$$\begin{aligned} \big ( \{ p \, a_1 \} , \ldots , \{ p \, a_d \} \big ) \quad \text {for} \;\; p \in {\mathcal {P}} \end{aligned}$$

are uniformly distributed in $[0,1)^d$, where ${\mathcal {P}} = \{ 2,3,5,7, \ldots \}$ is the set of primes.

Proof

For convenience, we denote the n-th prime by $p_n$, that is, $p_1 {=} 2$, $p_2 {=} 3$, $p_3 {=} 5$, $p_4 {=} 7$, and so on. By Weyl’s criterion (see e.g., [16, p. 48, Theorems 6.2 and 6.3]), the claim is equivalent to having that for every ${\varvec{z}}= (z_1, \ldots , z_d) \in {\mathbb {Z}}^d \backslash \{ 0 \}$, the fractional part of $\langle {\varvec{z}}, ( p_n a_1 , \ldots , p_n a_d ) \rangle = z_1 \cdot p_n a_1 + \cdots + z_d \cdot p_n a_d = p_n \cdot (z_1 a_1 + \cdots + z_d \, a_d)$ for $n=1,2,\ldots$ are uniformly distributed in [0, 1). Note that for any fixed ${\varvec{z}}= (z_1, \ldots , z_d) \in {\mathbb {Z}}^d \backslash \{ 0 \}$, the number ${\widetilde{a}} := z_1 a_1 + \cdots + z_d \, a_d$ is irrational because $1, a_1, \ldots , a_d$ are linearly independent over ${\mathbb {Q}}$. Hence, the result of Vinogradov [17] implies that the numbers $\{ p_n {\widetilde{a}} \}$, $n=1,2,\ldots$, are uniformly distributed in [0, 1), as desired. $\square$

3 Proof of Theorem 1

Proposition 7 implies that there exist infinitely many prime numbers $N \in {\mathbb {N}}$ satisfying

$$\begin{aligned} \begin{aligned} 0&\;<\; \{ N a_1 \} \;<\; \{ N a_2 \} \;<\; \cdots \;<\; \{ N a_{L-1} \} \;<\; \{ N a_L \} \\&\quad \;<\; \{ N b_L \} \;<\; \{ N b_{L-1} \} \;<\; \cdots \;<\; \{ N b_2 \} \;<\; \{ N b_1 \} \;<\; 1 . \end{aligned} \end{aligned}$$

(2)

Among such numbers, choose a large $N \in {\mathbb {N}}$ so that every spacing between the numbers $0< a_1< b_1< a_2< b_2< \cdots< a_L< b_L < 1$ contains at least one of $\frac{k}{N}$, $k = 1, \ldots , N{-}1$, as an interior point (which clearly requires $2L + 1 \le N$). This ensures that with respect to the grid $\frac{1}{N} {\mathbb {Z}}$ the interval $[a_\ell ,b_\ell )$ is partitioned into translates of

$$\begin{aligned} \big [ \tfrac{\{ N a_\ell \}}{N} , \tfrac{1}{N} \big ) , \quad \big [ 0 , \tfrac{\{ N b_\ell \}}{N} \big ) , \quad \text {and possibly some extra intervals} \;\; [0, \tfrac{1}{N}) , \end{aligned}$$

(3)

and that the rightmost segment of $[a_\ell ,b_\ell )$, corresponding to $[ 0 , \frac{\{ N b_\ell \}}{N} )$ in (3), lies in a set $[ \frac{k}{N} , \frac{k+1}{N} )$ which does not intersect the next interval $[a_{\ell +1},b_{\ell +1})$. Consequently, each of the sets $A_{\ge n} = A_{\ge n} (N,S)$, $n = 1, 2, \ldots , N$, is one of the form

$$\begin{aligned} \emptyset , \quad [0, \tfrac{1}{N}) , \quad \text {and} \quad \big [ \tfrac{\{ N a_\ell \}}{N} , \tfrac{\{ N b_\ell \}}{N} \big ) \quad \text {for some} \;\; \ell \in \{ 0 , 1, \ldots , N-1 \} . \end{aligned}$$

Note that due to (2), the translates of $[ \frac{\{ N a_\ell \}}{N} , \frac{1}{N} )$ and $[ 0 , \frac{\{ N b_\ell \}}{N} )$ in (3) together contribute exactly $[ 0, \frac{1}{N} )$ and $[ \frac{\{ N a_\ell \}}{N} , \frac{\{ N b_\ell \}}{N} )$ to the family of sets $A_{\ge n}$. The nested sets

$$\begin{aligned} A_{\ge 1} \;\supset \; A_{\ge 2} \;\supset \; \cdots \;\supset \; A_{\ge N} \end{aligned}$$

are thus given by

$$\begin{aligned} \begin{aligned}&\overbrace{[0,\tfrac{1}{N}) = \cdots = [0,\tfrac{1}{N})}^{K} \;\supset \; \overbrace{\big [ \tfrac{\{ N a_1 \}}{N} , \tfrac{\{ N b_1 \}}{N} \big ) \;\supset \; \cdots \;\supset \; \big [ \tfrac{\{ N a_L \}}{N} , \tfrac{\{ N b_L \}}{N} \big )}^{L} \\&\;\supset \; \overbrace{\emptyset = \cdots = \emptyset }^{N-K-L} \qquad \text {for some integer} \;\; K \ge L . \end{aligned} \end{aligned}$$

Let us associate each set $A_{\ge n}$ with the interval $[a_\ell ,b_\ell )$ which it originates from. The sets $[0,\frac{1}{N})$ can be associated with the intervals $[a_\ell ,b_\ell )$ in various ways, but for convenience we will assume

$$\begin{aligned} \begin{aligned}&\underbrace{\overbrace{[0,\tfrac{1}{N}) = \cdots = [0,\tfrac{1}{N})}^{K_1}}_{{\mathop {[a_1,b_1)}\limits ^{\updownarrow }}} = \cdots = \underbrace{\overbrace{[0,\tfrac{1}{N}) = \cdots = [0,\tfrac{1}{N})}^{K_L}}_{{\mathop {[a_L,b_L)}\limits ^{\updownarrow }}} \\&\;\supset \; \overbrace{\underbrace{\big [ \tfrac{\{ N a_1 \}}{N} , \tfrac{\{ N b_1 \}}{N} \big )}_{{\mathop {[a_1,b_1)}\limits ^{\updownarrow }}} \;\supset \; \cdots \;\supset \; \underbrace{\big [ \tfrac{\{ N a_L \}}{N} , \tfrac{\{ N b_L \}}{N} \big )}_{{\mathop {[a_L,b_L)}\limits ^{\updownarrow }}} }^{L} \;\supset \; \overbrace{\emptyset = \cdots = \emptyset }^{N-K-L} \end{aligned} \end{aligned}$$

where $K = \sum _{\ell =1}^L K_\ell$ with $K_\ell \in {\mathbb {N}}$ for all $\ell$.

Step 1. Construction of the sets $\Lambda _\ell \subset {\mathbb {Z}}$, $\ell =1, \ldots , L$.

For each $n = 1, \ldots , N$, we apply the result of Seip [1] (see the beginning of Section 1) to obtain a set $\Lambda ^{(n)} \subset N{\mathbb {Z}}$ such that $E(\Lambda ^{(n)})$ is a Riesz basis for $L^2(A_{\ge n})$; it is easily seen that

$$\begin{aligned} \Lambda ^{(n)} = {\left\{ \begin{array}{ll} N{\mathbb {Z}}&{} \text {for} \;\; 1 \le n \le K , \\ \subsetneq N{\mathbb {Z}}&{} \text {for} \;\; K{+}1 \le n \le K{+}L , \\ \emptyset &{} \text {for} \;\; K{+}L{+}1 \le n \le N . \end{array}\right. } \end{aligned}$$

Lemma 5 implies that $E ( \cup _{n=1}^N (\Lambda ^{(n)} {+} n) )$ is a Riesz basis for $L^2(S)$. For each $\ell =1, \ldots , L$, let $\Lambda _\ell$ be the union of $\Lambda ^{(n)} {+} n$ over all n such that $A_{\ge n}$ is associated with $[a_\ell ,b_\ell )$, that is,

Clearly, we have

and thus, $E( \cup _{\ell =1}^L \, \Lambda _\ell )$ is a Riesz basis for $L^2(S)$.

Step 2. For a subset $J \subset \{ 1, \ldots , L \}$, we set $\Lambda ^J := \cup _{\ell \in J} \, \Lambda _\ell$ and $S^J := \cup _{\ell \in J} \, [a_\ell ,b_\ell )$. We claim that $E( \Lambda ^J )$ is a Riesz basis for $L^2(S^J)$.

First, note that the corresponding sets $A_{\ge n}^J := A_{\ge n} (N,S^J)$ for $n = 1, \ldots , N$ are again of the form

$$\begin{aligned} \emptyset , \quad [0, \tfrac{1}{N}) , \quad \text {or} \quad \big [ \tfrac{\{ N a_\ell \}}{N} , \tfrac{\{ N a_\ell \}}{N} \big ) \quad \text {for some} \;\; \ell \in J . \end{aligned}$$

Denoting $J = \{ \ell _1, \ldots , \ell _M \}$ with $\ell _1< \cdots < \ell _M$, we see that the nested sets

$$\begin{aligned} A_{\ge 1}^J \;\supset \; A_{\ge 2}^J \;\supset \; \cdots \;\supset \; A_{\ge N}^J \end{aligned}$$

are given by

$$\begin{aligned}&\underbrace{\overbrace{[0,\tfrac{1}{N}) = \cdots = [0,\tfrac{1}{N})}^{K_{\ell _1}}}_{{\mathop {\big [a_{\ell _1},b_{\ell _1}\big )}\limits ^{\updownarrow }}} = \cdots = \underbrace{\overbrace{[0,\tfrac{1}{N}) = \cdots = [0,\tfrac{1}{N})}^{K_{\ell _M}}}_{{\mathop {\big [a_{\ell _M},b_{\ell _M}\big )}\limits ^{\updownarrow }}} \\&\;\supset \; \overbrace{\underbrace{\Big [ \tfrac{\{ N a_{\ell _1} \}}{N} , \tfrac{\{ N b_{\ell _1} \}}{N} \Big )}_{{\mathop {\big [a_{\ell _1},b_{\ell _1}\big )}\limits ^{\updownarrow }}} \;\supset \; \cdots \;\supset \; \underbrace{\Big [ \tfrac{\{ N a_{\ell _M} \}}{N} , \tfrac{\{ N b_{\ell _M} \}}{N} \Big )}_{{\mathop {\big [a_{\ell _M},b_{\ell _M}\big )}\limits ^{\updownarrow }}} }^{M} \;\supset \; \overbrace{\emptyset = \cdots = \emptyset }^{N- K_J - M} \end{aligned}$$

where $K_J := \sum _{\ell \in J} K_{\ell }$. Note that applying Lemma 5 directly to this setup will incur different shift factors in the frequency sets. Indeed, Lemma 5 implies that $E( \Lambda ' )$ is a Riesz basis for $L^2(S^J)$, with

(4)

where $\Lambda ^{(n)} \subset N{\mathbb {Z}}$ for $n = 1, \ldots , N$ are the sets defined in Step 1. However, our goal is to show that $E( \cup _{m=1}^M \, \Lambda _{\ell _m} )$ is a Riesz basis for $L^2(S^J)$, where

To show this, consider the $K_J {+} M$ sets

$$\begin{aligned} \Omega _1 := & {} N{\mathbb {Z}}{+} (K_1{+}K_2{+}\cdots {+}K_{\ell _1-1}{+}1), \\ \vdots \;\;&\quad \qquad \vdots \\ \Omega _{K_{\ell _1}} := & {} N{\mathbb {Z}}{+} (K_1{+}K_2{+}\cdots {+}K_{\ell _1}), \\ \Omega _{K_{\ell _1}+1} := & {} N{\mathbb {Z}}{+} (K_1{+}K_2{+}\cdots {+}K_{\ell _2-1}{+}1), \\ \vdots \;\;&\quad \qquad \vdots \\ \Omega _{K_{\ell _1}+K_{\ell _2}} := & {} N{\mathbb {Z}}{+} (K_1{+}K_2{+}\cdots {+}K_{\ell _2}), \\ \vdots \;\;&\quad \qquad \vdots \\ \Omega _{K_{\ell _1}+K_{\ell _2}+\cdots +K_{\ell _{M-1}}} := & {} N{\mathbb {Z}}{+} ( K_1{+}K_2{+}\cdots {+}K_{\ell _M-1}{+}1 ), \\ \vdots \;\;&\quad \qquad \vdots \\ \Omega _{K_J} := & {} N{\mathbb {Z}}{+} ( K_1{+}K_2{+}\cdots {+}K_{\ell _M}) = N{\mathbb {Z}}{+} K_J, \\ \Omega _{K_J+1} := & {} \Lambda ^{(K+\ell _1)} {+} (K {+} \ell _1), \\ \Omega _{K_J+2} := & {} \Lambda ^{(K+\ell _2)} {+} (K {+} \ell _2), \\ \vdots \;\;&\quad \qquad \vdots \\ \Omega _{K_J+M} := & {} \Lambda ^{(K+\ell _M)} {+} (K {+} \ell _M) , \end{aligned}$$

which partitions $\cup _{m=1}^M \, \Lambda _{\ell _m}$, that is, . Here, the sets $\Omega _n$ are exactly ordered in the way that $E(\Omega _n)$ is a Riesz basis for $L^2(A_{\ge n}^J)$. Note that while the $K_J {+} M$ components of $\Lambda '$ in (4) have consecutive shift factors, namely from 1 up to $K_J {+} M$, the shift factors associated with $\Omega _n$ are not consecutive in general. However, since $N \in {\mathbb {N}}$ is prime, Lemma 6 implies that is a Riesz basis for $L^2(S^J)$. This completes the proof.

4 Proof of Theorem 2

To prove Theorem 2, we will use Lemma 5 which is the key lemma of Kozma and Nitzan [8]. Note that by Lemma 4, one may replace the frequency set $\cup _{n=1}^N (\Lambda _n {+} n)$ in Lemma 5 by $\cup _{n=1}^N (\Lambda _n {+} n{-}1)$, while preserving the Riesz basis property.

We will first prove the case $L = 1$ and then extend the proof to the case $L \ge 2$.

Case $L = 1$. Given a set $V = [0,1) \cup [a,b) \subset [0,N)$ with $N \in {\mathbb {N}}$ and $1 \le a < b \le N$, let $S := \frac{1}{N} V = [0,\frac{1}{N}) \cup [\frac{a}{N},\frac{b}{N}) \subset [0,1)$. We will apply Lemma 5 directly to this set S. There are two cases, either $\{ a \} \le \{ b \}$ or $\{ b \} < \{ a \}$.

First, assume that $0 \le \{ a \} \le \{ b \} < 1$. Then there exists a number $M \in {\mathbb {N}}$ such that

$$\begin{aligned} A_{\ge 1} = A_{\ge 2} = \cdots =A_{\ge M} = [0,\tfrac{1}{N}) \;\supset \; A_{\ge M+1} = \big [ \tfrac{ \{ a \} }{N} , \tfrac{ \{ b \} }{N} \big ) \;\supset \; A_{\ge M+2} = \emptyset . \end{aligned}$$

Clearly, we may choose the canonical frequency sets $\Lambda _1 = \cdots = \Lambda _M = N {\mathbb {Z}}$ for $A_{\ge 1} = A_{\ge 2} = \cdots = A_{\ge M} = [0,\tfrac{1}{N})$, so that for each $n = 1,\ldots ,M$, the system $E(\Lambda _n)$ is a Riesz basis (in fact, an orthogonal basis) for $L^2(A_{\ge n})$. Also, there exists a set $\Lambda _{M+1} \subset N{\mathbb {Z}}$ such that $E(\Lambda _{M+1})$ is a Riesz basis for $L^2\big [ \frac{ \{ a \} }{N} , \frac{ \{ b \} }{N} \big )$; indeed, such a set $\Lambda _{M+1}$ can be obtained from the result of Seip [1] with a dilation (see Lemma 4). Then Lemma 5 with shift factors ‘$n-1$’ in place of ‘n’ yields that $E( (\cup _{n = 1}^{M} N {\mathbb {Z}}{+} n {-}1) \cup (\Lambda _{M+1} {+} M) )$ is a Riesz basis for $L^2(S) = L^2( \frac{1}{N} V )$. By a dilation, we obtain that $E( (\cup _{n = 1}^{M} {\mathbb {Z}}{+} \frac{n {-}1}{N}) \cup (\frac{1}{N} \Lambda _{M+1} {+} \frac{M}{N}) ) = E( {\mathbb {Z}}\cup (\cup _{k = 1}^{M-1} {\mathbb {Z}}{+} \frac{k}{N}) \cup (\frac{1}{N} \Lambda _{M+1} {+} \frac{M}{N}) )$ is a Riesz basis for $L^2(V) = L^2 ([0,1) \cup [a,b) )$. Now, we claim that $E(\Lambda ')$ is a Riesz basis for $L^2[a,b)$, where $\Lambda ' := ( \cup _{k = 1}^{M-1} {\mathbb {Z}}{+} \frac{k}{N} ) \cup ( \frac{1}{N} \Lambda _{M+1} {+} \frac{M}{N} )$. To see this, we again apply Lemma 5 (the original version) to the set $S' := \frac{1}{N} V'$ with $V' = [a,b)$. One can easily check that the corresponding set $A_{\ge n}'$ is equal to the set $A_{\ge n-1}$ above, that is,

$$\begin{aligned} A_{\ge 1}' = \cdots =A_{\ge M-1}' = [0,\tfrac{1}{N}) \;\supset \; A_{\ge M}' = \big [ \tfrac{ \{ a \} }{N} , \tfrac{ \{ b \} }{N} \big ) \;\supset \; A_{\ge M+1}' = \emptyset . \end{aligned}$$

(5)

Then Lemma 5 implies that $E( (\cup _{k = 1}^{M-1} {\mathbb {Z}}{+} \frac{k}{N}) \cup (\frac{1}{N} \Lambda _{M+1} {+} \frac{M}{N}) )$ is a Riesz basis for $L^2[a,b)$, as claimed.

Now, assume that $0 \le \{ b \}< \{ a \} < 1$. Then there exists a number $M \in {\mathbb {N}}$ such that

$$\begin{aligned} A_{\ge 1} = \cdots =A_{\ge M} = [0,\tfrac{1}{N}) \;\supset \; A_{\ge M+1} = \big [ 0, \tfrac{ \{ b \} }{N} \big ) \cup \big [ \tfrac{ \{ a \} }{N}, \tfrac{1}{N} \big ) \;\supset \; A_{\ge M+2} = \emptyset . \end{aligned}$$

Again, using the result of Seip [1] with a dilation, we obtain a set $\Lambda _{M+1} \subset N{\mathbb {Z}}$ such that $E(\Lambda _{M+1})$ is a Riesz basis for $L^2\big [ \frac{ \{ a \} }{N} , \frac{ 1+\{ b \} }{N} \big )$. Since all elements in $E(N{\mathbb {Z}})$ are $\frac{1}{N}$-periodic, it follows that $E(\Lambda _{M+1})$ is a Riesz basis for $L^2 \big ( \big [ 0, \frac{ \{ b \} }{N} \big ) \cup \big [ \frac{ \{ a \} }{N}, \frac{1}{N} \big ) \big )$. Then, by similar arguments as in the case $\{ a \} \le \{ b \}$, we deduce that $E( {\mathbb {Z}}\cup (\cup _{k = 1}^{M-1} {\mathbb {Z}}{+} \frac{k}{N}) \cup (\frac{1}{N} \Lambda _{M+1} {+} \frac{M}{N}) )$ is a Riesz basis for $L^2 ( [0,1) \cup [a,b) )$, and that $E( (\cup _{k = 1}^{M-1} {\mathbb {Z}}{+} \frac{k}{N}) \cup (\frac{1}{N} \Lambda _{M+1} {+} \frac{M}{N}) )$ is a Riesz basis for $L^2[a,b)$.

Case $L \ge 2$. We will use essentially the same arguments as in the case $L=1$, but employ the main result of Kozma and Nitzan [8] instead of Seip [1]. Given a set $V = [0,1) \cup [a_1,b_1) \cup \cdots \cup [a_L,b_L) \subset [0,N)$ with $L, N \in {\mathbb {N}}$ and $1 \le a_1< b_1< \cdots< a_L < b_L \le N$, let $S := \frac{1}{N} V = [0,\frac{1}{N}) \cup [\frac{a_1}{N},\frac{b_1}{N}) \cup \cdots \cup [\frac{a_L}{N},\frac{b_L}{N}) \subset [0,1)$. As in the case $L=1$, we will apply Lemma 5 to this set S.

Note that there are finitely many possible ordering of the values

$$\begin{aligned} 0 \;\le \; \tfrac{ \{ a_1 \} }{N} , \tfrac{ \{ b_1 \} }{N} , \cdots , \tfrac{ \{ a_L \} }{N}, \tfrac{ \{ b_L \} }{N} \;<\; \tfrac{1}{N} , \end{aligned}$$

where equalities are also allowed, e.g., the values are all zero if all $a_\ell$ and $b_\ell$ are integers. It is easily seen that besides 0 and $\tfrac{1}{N}$, these are the only possible values that can be the boundary points of $A_{\ge n}$, $n=1, \ldots , N$. In any case, since $[0,\frac{1}{N}) \subset S$ we have

$$\begin{aligned} A_{\ge 1} = [0,\tfrac{1}{N}) \;\supset \; A_{\ge 2} \;\supset \; \cdots \;\supset \; A_{\ge N} , \end{aligned}$$

where each of the sets $A_{\ge 2} , \ldots , A_{\ge N}$ is either empty or a finite union of intervals. One can therefore use the main result of [8] with a dilation, to construct sets $\Lambda _1 {=} N{\mathbb {Z}}, \Lambda _2, \Lambda _3 , \ldots , \Lambda _N \subset N{\mathbb {Z}}$ such that for each n, the system $E(\Lambda _n)$ is a Riesz basis for $L^2(A_{\ge n})$. The rest of the proof is similar to the case $L=1$. $\Box$

Appendix A: Proof of Lemma 6

To prove Lemma 6, we will follow the proof strategy of Lemma 5 [8, Lemma 2]. For any $N \in {\mathbb {N}}$, a measurable set $S \subset [0,1)$, and $n = 0,1, \ldots , N$, we define

$$\begin{aligned} \begin{aligned}&A_n := \Big \{ t \in [0, \tfrac{1}{N}) : t+ \tfrac{k}{N} \in S \;\; \text {for exactly { n} values of} \;\; k \in \{ 0 , 1, \ldots , N-1 \} \Big \} , \\&B_n := \Big \{ t \in S : t+ \tfrac{k}{N} \in S \;\; \text {for exactly { n} integers} \;\; k \in {\mathbb {Z}}\Big \} . \end{aligned} \end{aligned}$$

Obviously, considering the set $B_n$ modulo $\frac{1}{N}$ yields the n-fold of $A_n$, which means that each element of $A_n$ corresponds to exactly n points of $B_n$ that are distanced apart by multiples of $\frac{1}{N}$. Note that $\{ A_n \}_{n=0}^N$ and $\{ B_n \}_{n=0}^N$ form partitions of $[0, \frac{1}{N})$ and [0, 1), respectively, that is, and . Also, the family $\{ B_n \}_{n=1}^N$ forms a partition of S, i.e., . For $f \in L^2(S)$ and $n = 1, \ldots , N$, we denote by $f_n$ the restriction of f to $B_n$, that is, $f_n(t) = f(t)$ for $t \in B_n$ and 0 otherwise. This yields the decomposition $L^2(S) \ni f = f_1 + \cdots + f_N$ with all $f_n$ having disjoint support. Note that the set $A_{\ge n}$ given by (1) can be expressed as $A_{\ge n} = \cup _{\ell =n}^N \, A_{\ell }$ for $n = 1, \ldots , N$. Similarly, we define $B_{\ge n} := \cup _{\ell =n}^N \, B_{\ell }$ and $f_{\ge n} := \sum _{\ell =n}^N f_{\ell }$ for $n = 1, \ldots , N$. For brevity, we write $\Lambda := \cup _{\ell =1}^N (\Lambda _{\ell } {+} j_{\ell })$.

Frame. Assume that $\Lambda _1, \ldots , \Lambda _N \subset N{\mathbb {Z}}$ are such that $E(\Lambda _n)$ is a frame for $L^2(A_{\ge n})$. To prove that $E(\Lambda )$ is a frame for $L^2(S)$, it is enough to show that there exists a constant $c > 0$ satisfying

$$\begin{aligned} \begin{aligned}&\sum _{\lambda \in \Lambda } \big \vert \langle f, e^{2 \pi i \lambda (\cdot )} \rangle _{L^2(S)} \big \vert ^2 \;\ge \; c \, \Vert f_n \Vert _{L^2(S)}^2 - \sum _{\ell =1}^{n-1} \Vert f_\ell \Vert _{L^2(S)}^2 \\&\quad \text {for all} \;\; f \in L^2(S) \;\; \text {and} \;\; n = 1,\ldots ,N . \end{aligned} \end{aligned}$$

In turn, it is enough to show that there exists a constant $c > 0$ satisfying

$$\begin{aligned} \sum _{\lambda \in \Lambda } \big \vert \langle f_{\ge n}, e^{2 \pi i \lambda (\cdot )} \rangle _{L^2(S)} \big \vert ^2 \;\ge \; c \, \Vert f_n \Vert _{L^2(S)}^2 , \quad f \in L^2(S) , \;\; n = 1,\ldots ,N . \end{aligned}$$

(A1)

Such reductions are essentially due to the decomposition $f = f_1 + \cdots + f_N$ with all $f_n$ having disjoint support, and due to $S \subset [0,1)$ and $\Lambda \subset {\mathbb {Z}}$; see [8, Eqns. (6)–(7)] for detailed arguments.

To prove (A1), fix any $f \in L^2(S)$ and any $n \in \{ 1, \ldots , N \}$. Since $f_{\ge n}$ is supported in $\cup _{\ell =n}^N \, B_{\ell } \subset S$, we have for any $\lambda \in \Lambda _{\ell } {+} j_{\ell }$ with $\ell \in \{ 1, \ldots , N \}$,

$$\begin{aligned} \begin{aligned} \big \langle f_{\ge n}, e^{2 \pi i \lambda (\cdot )} \big \rangle _{L^2(S)}&= \int _0^1 f_{\ge n}(t) \, e^{- 2 \pi i \lambda t} \, dt \\&= \int _0^{1/N} \sum _{k=0}^{N-1} f_{\ge n}\big (t+\tfrac{k}{N}\big ) \, \exp (- 2 \pi i \lambda (t+\tfrac{k}{N}) ) \, dt \\&= \int _0^{1/N} h_{n,\ell } (t) \, e^{- 2 \pi i \lambda t} \, dt = \big \langle h_{n,\ell }, e^{2 \pi i \lambda (\cdot )} \big \rangle _{L^2[0,\frac{1}{N})} \;, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} h_{n,\ell } (t) := \mathbbm {1}_{A_{\ge n}} (t) \,\cdot \sum _{k=0}^{N-1} f_{\ge n}\big (t+\tfrac{k}{N}\big ) \, e^{- 2 \pi i j_{\ell } k/N} . \end{aligned}$$

(A2)

Note that for $\ell = 1,\ldots ,n$, the function $h_{n,\ell }$ is supported in $A_{\ge n} \subset A_{\ge \ell }$. Since $E(\Lambda _\ell )$ is a frame for $L^2(A_{\ge \ell })$, say, with lower frame bound $\alpha _\ell > 0$, we have

$$\begin{aligned} \begin{aligned} \sum _{\lambda \in \Lambda _{\ell } + j_{\ell }} \big \vert \langle f_{\ge n}, e^{2 \pi i \lambda (\cdot )} \rangle _{L^2(S)} \big \vert ^2&\;=\; \sum _{\lambda \in \Lambda _{\ell } + j_{\ell }} \big \vert \langle h_{n,\ell }, e^{2 \pi i \lambda (\cdot )} \rangle _{L^2[0,\frac{1}{N})} \big \vert ^2 \\&\;\ge \; \alpha _\ell \, \Vert h_{n,\ell } \Vert ^2 . \end{aligned} \end{aligned}$$

(A3)

Summing up (A3) for $\ell = 1,\ldots ,n$ gives

$$\begin{aligned} \begin{aligned} \sum _{\lambda \in \Lambda } \big \vert \langle f_{\ge n}, e^{2 \pi i \lambda (\cdot )} \rangle _{L^2(S)} \big \vert ^2&\;\ge \; \sum _{\ell =1}^n \sum _{\lambda \in \Lambda _{\ell } + j_{\ell }} \big \vert \langle f_{\ge n}, e^{2 \pi i \lambda (\cdot )} \rangle _{L^2(S)} \big \vert ^2 \\&\;\ge \; \Big (\min _{1 \le \ell \le n} \alpha _\ell \Big ) \cdot \sum _{\ell =1}^n \Vert h_{n,\ell } \Vert ^2 \\&\;\ge \; \Big (\min _{1 \le \ell \le n} \alpha _\ell \Big ) \cdot \sum _{\ell =1}^n \Vert h_{n,\ell } \cdot \mathbbm {1}_{A_n} \Vert ^2 . \end{aligned} \end{aligned}$$

(A4)

On the other hand, for any fixed $t \in A_n$, Eq. (A2) becomes

$$\begin{aligned} h_{n,\ell } (t) = \sum _{k=0}^{N-1} f_{\ge n}\big (t+\tfrac{k}{N}\big ) \, e^{- 2 \pi i j_{\ell } k/N} \end{aligned}$$

and collecting the equation for $\ell = 1,\ldots ,n$ gives the $n {\times } N$ linear system

$$\begin{aligned} \begin{bmatrix} h_{n,\ell } (t) \end{bmatrix}_{\ell =1}^n = \begin{bmatrix} e^{- 2 \pi i j_{\ell } k/N} \end{bmatrix}_{1 \le \ell \le n, 0 \le k \le N-1} \begin{bmatrix} f_{\ge n}\big (t+\tfrac{k}{N}\big ) \end{bmatrix}_{k=0}^{N-1} . \end{aligned}$$

Since $t \in A_n$, the vector $[ f_{\ge n}\big (t+\tfrac{k}{N}\big ) ]_{0 \le k \le N-1}$ has exactly n nonzero entries, say, at the indices $k_1< \cdots < k_n$ from $\{0,1,\ldots ,N-1\}$. This reduces the above system to an $n {\times } n$ linear system

$$\begin{aligned} \begin{bmatrix} h_{n,\ell } (t) \end{bmatrix}_{\ell =1}^n = \begin{bmatrix} e^{- 2 \pi i j_{\ell } k_r/N} \end{bmatrix}_{1 \le \ell \le n, 1 \le r \le n} \begin{bmatrix} f_{\ge n}\big (t+\tfrac{k_r}{N}\big ) \end{bmatrix}_{r=1}^n , \end{aligned}$$

where the associated matrix $[ e^{- 2 \pi i j_{\ell } k_r/N} ]_{1 \le \ell \le n, 1 \le r \le n}$ is invertible since N is prime (by Chebotarëv’s theorem on roots of unity, see e.g., [18]). As there are only finitely many possible choices of $k_1< \cdots < k_n$ from $\{0,1,\ldots ,N-1\}$, there exists a constant $c' > 0$ such that

$$\begin{aligned} \sum _{\ell =1}^n \big \vert h_{n,\ell } (t) \big \vert ^2 \;\ge \; c' \, \sum _{k=0}^{N-1} \big \vert f_{\ge n}\big (t+\tfrac{k}{N}\big ) \big \vert ^2 \quad \text {for all} \;\; t \in A_n . \end{aligned}$$

Integrating over $t \in A_n$ then gives

$$\begin{aligned} \sum _{\ell =1}^n \Vert h_{n,\ell } \cdot \mathbbm {1}_{A_n} \Vert ^2 \ge c' \int _{A_n} \sum _{k=0}^{N-1} \big \vert f_{\ge n}\big (t+\tfrac{k}{N}\big ) \big \vert ^2 = c' \, \Vert f_n \Vert _{L^2(B_n)}^2 = c' \, \Vert f_n \Vert _{L^2(S)}^2 . \end{aligned}$$

Combining this inequality with (A4) yields the desired inequality (A1).

Riesz sequence. Assume that $\Lambda _1, \ldots , \Lambda _N \subset N{\mathbb {Z}}$ are such that $E(\Lambda _n)$ is a Riesz sequence in $L^2(A_{\ge n})$. We will show that $E(\Lambda )$ is a Riesz sequence in $L^2(S)$ by using the frame part which is proved above (a similar trick was used in [19, Lemma 7]). Let $S' := [0,1) \backslash S$ and let $A_{\ge n}'$, $n=1,\ldots ,N$, be the corresponding sets of (1) for $S'$. It is easily seen that $A_{\ge n}' = [0,\frac{1}{N}) \backslash A_{\ge N+1-n}$ for $n=1,\ldots ,N$. Since $E(\Lambda _{N+1-n})$ is a Riesz sequence in $L^2(A_{\ge N+1-n})$, we deduce from Proposition 3 with a dilation that the system $E(N{\mathbb {Z}}\backslash \Lambda _{N+1-n})$ is a frame for $L^2 ( [0,\frac{1}{N}) \backslash A_{\ge N+1-n} ) = L^2(A_{\ge n}')$; see the discussion after Proposition 3. The frame part then implies that the system $E( \cup _{n=1}^N ( (N{\mathbb {Z}}\backslash \Lambda _{N+1-n}) {+} j_{N+1-n} ) )$ is a frame for $L^2 (S')$. Finally, again by Proposition 3, we conclude that the system $E(\Lambda ) = E( \cup _{n=1}^N (\Lambda _n {+} j_n) ) = E( {\mathbb {Z}}\backslash \cup _{n=1}^N ( (N{\mathbb {Z}}\backslash \Lambda _{N+1-n}) {+} j_{N+1-n} ) )$ is a Riesz sequence in $L^2(S) = L^2 ( [0,1) \backslash S' )$.

Riesz basis. Since a family of vectors in a separable Hilbert space is a Riesz basis if and only if it is both a frame and a Riesz sequence, this part follows immediately by combining the frame and Riesz sequence parts. $\Box$