Hierarchical Exponential Riesz Bases for Four-Interval Partitions

1. Introduction and Main Result

For a discrete set $\mathrm{\Lambda }\subset \mathbb{R}$ and a finite measure set $S\subset \mathbb{R}$, we write $E(\mathrm{\Lambda };S)=\{e^{2\pi i\lambda (·)}{|}_S:\lambda \in \mathrm{\Lambda }\}$ for the corresponding system of complex exponentials in $L^2\left(S\right)$. We call $\mathrm{\Lambda }$ its (frequency) spectrum and S its (spatial) domain. We call $\mathrm{\Lambda }$ a Riesz spectrum for S if $E(\mathrm{\Lambda };S)$ is a Riesz basis for $L^2\left(S\right)$. We are interested in choosing Riesz spectra that respect unions as expressed by the following conjecture of [3].

Conjecture 

(Conjecture $1^{\prime }$ in [3]). Let $I_1,\dots ,I_L$ be intervals which form a partition of $[0,1)$. There exists a partition $\mathbb{Z}={\phantom{\bigcup }\bigcup ·}_{i=1}^L{\mathrm{\Lambda }}_i$ such that $E({\mathrm{\Lambda }}_J;S_J)$ is a Riesz basis for every nonempty $J\subset \{1,\dots ,L\}$, where

$${\mathrm{\Lambda }}_J:=\bigcup _{i\in J}{\mathrm{\Lambda }}_i\mathrm{and}{S}_J:=\bigcup _{i\in J}{I}_i.$$

Here and throughout, $\phantom{\bigcup }\bigcup ·$ denotes a disjoint union, and all unions and partitions of domains are considered up to sets of measure zero.

In this paper, we resolve the case $L=4$ of the above conjecture. Note that the case $L=1$ is trivial, and the cases $L=2,3$ are implied by the case $L=4$; we also provide elementary proofs for $L=2,3$ below. The case $L\ge 5$ appears significantly more challenging, as it requires the Riesz basis property to hold simultaneously for all $2^L-2$ nonempty proper unions.

Theorem 1.

Let $0=t_0<t_1<t_2<t_3<t_4=1$, and let $I_i=[t_{i-1},t_i)$ and ${\ell }_i=t_i-t_{i-1}$ for $i=1,2,3,4$. There exists a partition $\mathbb{Z}={\phantom{\bigcup }\bigcup ·}_{i=1}^4{\mathrm{\Lambda }}_i$ such that for every nonempty $J\subset \{1,2,3,4\}$,

(i)

$E({\mathrm{\Lambda }}_J;S_J)$ is a Riesz basis.

(ii)

$E({\mathrm{\Lambda }}_J;I)$ is a Riesz basis for every interval $I\subset \mathbb{R}$ of length $d_J={\sum }_{i\in J}{\ell }_i$,

(iii)

An increasing sequence ${\mathrm{\Lambda }}_J={\left\{{\lambda }_{J,n}\right\}}_{n\in \mathbb{Z}}$ satisfies

$$\underset{r\in \mathbb{Z},M\in \mathbb{N}}{\sup }\left|{\displaystyle \sum _{n=r}^{r+M-1}}\left({\lambda }_{J,n}-{\displaystyle \frac{n}{d_J}}-c_J\right)\right|\le B_J$$

(1)

for some constants $c_J\in \mathbb{R}$ and $B_J>0$.

Note that no arithmetic assumption is imposed on the endpoints. Setting $M=1$ in (1) gives bounded displacement ${\lambda }_{J,n}={\displaystyle \frac{n}{d_J}}+O\left(1\right)$, and dividing by M shows that the centered displacement averages converge uniformly at rate $O\left(M^{-1}\right)$.

Related Work

Seip [19] proved that every single interval contained in $[0,1)$ admits an integer Riesz spectrum. Kozma and Nitzan [9] extended this result to every finite union of intervals contained in $[0,1)$. In higher dimensions, Riesz spectra exist for finite unions of axis-parallel rectangles [10] and centrally symmetric convex polytopes whose faces of every dimension are centrally symmetric [7].

Regarding the conjecture, Pfander, Revay, and Walnut [17] proved that for any partition of $[0,1)$ into intervals, there exists a partition of $\mathbb{Z}$ whose subunions form Riesz spectra for intervals of the corresponding total lengths. Caragea and Lee [3] showed that for disjoint intervals of $[0,1)$ whose endpoints, together with 1, are linearly independent over $\mathbb{Q}$, there exists a disjoint subsets of $\mathbb{Z}$ whose unions are Riesz spectra for the corresponding union of intervals. In [12], the author developed methods for taking unions of exponential Riesz bases.

It is worth noting that there are domains not admitting Riesz spectra. Kozma, Nitzan, and Olevskii [11] constructed a subset of $\mathbb{R}$ with no Riesz spectrum. Recently, Ortega-Cerdà [16] and Wan [20] independently proved the nonexistence for Euclidean balls in ${\mathbb{R}}^d$ with $d\ge 2$, and nondegenerate triangles in ${\mathbb{R}}^2$.

The Cases $L=2,3$

Although the cases $L=2,3$ follow from Theorem 1, we provide direct proofs here.

For $L=2$, choose an integer Riesz spectrum ${\mathrm{\Lambda }}_1$ for $I_1$ using the result of Seip [19] (or Proposition 1 below), and set ${\mathrm{\Lambda }}_2=\mathbb{Z}\setminus {\mathrm{\Lambda }}_1$. Since $I_2=[0,1)\setminus I_1$, Lemma 2 gives a Riesz basis $E({\mathrm{\Lambda }}_2;I_2)$. The full union is the Fourier basis on $[0,1)$, so all nonempty unions are covered.

For $L=3$, the main result of [17] provides a partition $\mathbb{Z}={\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_2\cup {\mathrm{\Lambda }}_3$ such that each $E({\mathrm{\Lambda }}_i;I_i)$ is a Riesz basis. Note that every two element union is the complement of the remaining singleton, that is, if $\{i,j,k\}=\{1,2,3\}$, then ${\mathrm{\Lambda }}_i\cup {\mathrm{\Lambda }}_j=\mathbb{Z}\setminus {\mathrm{\Lambda }}_k$ and $I_i\cup I_j=[0,1)\setminus I_k$. Therefore, Lemma 2 gives the Riesz basis property for every pair, and the full union is again the Fourier basis on $[0,1)$. This establishes the case $L=3$ of the conjecture.

We identify $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ with $[0,1)$, and call the image of an interval of length at most one a circular interval. For example, $[0,{\displaystyle \frac{1}{3}})\cup [{\displaystyle \frac{2}{3}},1)$ is a circular interval. Note that every nonempty union for $L=2,3$ is a circular interval. Therefore, the main result of [17], together with integer-frequency circular translation, gives (i) and (ii) in these cases. The estimate (1) in (iii) follows by repeatedly applying Lemma 6 to $\mathbb{Z}$ and using the fact that the estimate under every frequency union is preserved by Lemma 4. Corollary 1 then gives the desired Riesz basis property. For $L\ge 4$, a union of intervals and its complement may both be disconnected on $\mathbb{T}$, so this approach is no longer applicable.

Proof strategy

For a partition of $[0,1)$ into four intervals $I_1,I_2,I_3,I_4$, the only nonempty unions that are not circular intervals are $I_1\cup I_3$ and $I_2\cup I_4$. We first construct a Riesz spectrum $\mathrm{\Lambda }$ for $I_1\cup I_3$ satisfying (1). We then split $\mathrm{\Lambda }$ and its complement at the prescribed densities, leaving the alternating frequency unions unchanged. The displacement estimate is preserved by every frequency union. Avdonin’s theorem gives the Riesz basis property for the individual intervals and adjacent pairs, and complementation gives the remaining cases. Our proof relies on the equivalence between bounded centered displacement sums, and a bounded primitive of the centered counting discrepancy, together with a selection of consecutive runs whose errors telescope.

Organization

Section 2 provides the necessary background material. Section 3 combines a counting identity, constructs the required spectra, and proves Theorem 1. Section 4 gives an independent periodic construction for the case of rational endpoints.

2. Preliminaries

For brevity, we write $e_{\lambda }\left(x\right)=e^{2\pi i\lambda x}$ for $\lambda \in \mathbb{R}$. A system $E(\mathrm{\Lambda };S)$ is called a Riesz basis if it is complete in $L^2\left(S\right)$ and there exist constants $0<A\le B<\infty$ satisfying

$$A\sum _{\lambda \in \mathrm{\Lambda }}|c_{\lambda }{|}^2\le \parallel \sum _{\lambda \in \mathrm{\Lambda }}{c}_{\lambda }{e}_{\lambda }{\parallel }_{L^2\left(S\right)}^2\le B\sum _{\lambda \in \mathrm{\Lambda }}{\left|c_{\lambda }\right|}^2$$

(2)

for every finitely supported sequences ${\left\{c_{\lambda }\right\}}_{\lambda \in \mathrm{\Lambda }}$ (see [6,21]). Equivalently, $E(\mathrm{\Lambda };S)$ is a Riesz basis if and only if the synthesis operator

$${\left\{c_{\lambda }\right\}}_{\lambda \in \mathrm{\Lambda }}⟼\sum _{\lambda \in \mathrm{\Lambda }}{c}_{\lambda }{e}_{\lambda }$$

(3)

is a bounded isomorphism from ${\ell }_2\left(\mathbb{Z}\right)$ onto $L^2\left(S\right)$. For $S\subset [0,1)$ and $\mathrm{\Lambda }\subset \mathbb{Z}$, the right inequality in (2) holds with $B=1$; indeed, we have

$$\parallel \sum _{\lambda \in \mathrm{\Lambda }}{c}_{\lambda }{e}_{\lambda }{\parallel }_{L^2\left(S\right)}^2\le \parallel \sum _{\lambda \in \mathrm{\Lambda }}{c}_{\lambda }{e}_{\lambda }{\parallel }_{L^2[0,1)}^2=\sum _{\lambda \in \mathrm{\Lambda }}{\left|c_{\lambda }\right|}^2.$$

Lemma 1

(See e.g., [13]). Let $\mathrm{\Lambda }\subset \mathbb{R}$ be a discrete set and $S\subset \mathbb{R}$ be a measurable set. If $E(\mathrm{\Lambda };S)$ is a Riesz basis, then $E(\mathrm{\Lambda }+a;S+b)$ and $E(c^{-1}\mathrm{\Lambda };cS)$ are Riesz bases for every $a,b\in \mathbb{R}$ and $c>0$.

Since exponentials with integer frequencies are 1-periodic, an integer exponential Riesz bases on a subinterval of $\mathbb{T}$ is also a Riesz basis on any circular interval of the same length.

The following complementation principle follows from [15] and [2]; see also [3] and [17]. The finite Fourier matrix counterpart can be found in [14]. For completeness, we provide a short proof here.

Lemma 2.

For a measurable set $S\subset [0,1)$ and $\mathrm{\Lambda }\subset \mathbb{Z}$, the system $E(\mathrm{\Lambda };S)$ is a Riesz basis if and only if $E(\mathbb{Z}\setminus \mathrm{\Lambda };[0,1)\setminus S)$ is a Riesz basis.

Proof. 

With respect to the frequency partition $\mathrm{\Lambda },\mathbb{Z}\setminus \mathrm{\Lambda }$ and the spatial partition $S,S^c(=[0,1)\setminus S)$, the unitary Fourier synthesis operator can be written in the block matrix form

$$U=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right),$$

where the unitarity of U gives $A^{*}A+C^{*}C=\mathrm{Id}$, $B^{*}B+D^{*}D=\mathrm{Id}$, $A{A}^{*}+B{B}^{*}=\mathrm{Id}$, and $C{C}^{*}+D{D}^{*}=\mathrm{Id}$. Note that invertibility of A (resp. D) is equivalent to $E(\mathrm{\Lambda };S)$ (resp. $E(\mathbb{Z}\setminus \mathrm{\Lambda };[0,1)\setminus S)$) being a Riesz basis. Using the identities $\parallel C^{*}C\parallel =\parallel C{C}^{*}\parallel$, $\parallel B^{*}B\parallel =\parallel B{B}^{*}\parallel$, we obtain that for $0<a\le 1$,

$$A^{*}A=\mathrm{Id}-C^{*}C\ge a\mathrm{Id}\iff D{D}^{*}=\mathrm{Id}-C{C}^{*}\ge a\mathrm{Id},$$

and

$$A{A}^{*}=\mathrm{Id}-B{B}^{*}\ge a\mathrm{Id}\iff D^{*}D=\mathrm{Id}-B^{*}B\ge a\mathrm{Id},$$

which implies that A is invertible if and only if D is invertible. This proves the assertion. □

A sequence is separated if the distances between its distinct points have a positive lower bound. Let $\mathrm{\Lambda }={\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}}$ be an increasing sequence, and fix $d>0$. We say that $\mathrm{\Lambda }$ has bounded displacement at density d if

$${\lambda }_n={\displaystyle \frac{n}{d}}+{\delta }_n,\underset{n\in \mathbb{Z}}{\sup }\left|{\delta }_n\right|<\infty .$$

(4)

We say that $\mathrm{\Lambda }$ has bounded centered displacement sums at density d if, denoting ${\delta }_n={\lambda }_n-{\displaystyle \frac{n}{d}}$, there are constants $c\in \mathbb{R}$ and $C>0$ such that

$$\left|{\displaystyle \sum _{n=m}^{m+M-1}}({\delta }_n-c)\right|\le C\mathrm{for}\mathrm{all}m\in \mathbb{Z},M\in \mathbb{N}.$$

(5)

A shift of the index origin changes only the centering constant c. Taking $M=1$ gives $|{\delta }_n-c|\le C$, so the condition (5) already implies bounded displacement and the uniform estimate

$$\underset{m\in \mathbb{Z}}{\sup }\left|{\displaystyle \frac{1}{M}}{\displaystyle \sum _{n=m}^{m+M-1}}({\delta }_n-c)\right|\le {\displaystyle \frac{C}{M}}.$$

The estimate (5) is the property that we will preserve throughout the construction.

We will use Avdonin’s theorem [1] in the following form (see also [17]).

Theorem 2.

(Avdonin [1]). Let $\mathrm{\Lambda }={\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}}\subset \mathbb{R}$ be a separated increasing sequence, and assume that ${\lambda }_n=n+{\delta }_n$ with ${\sup }_{n\in \mathbb{Z}}\left|{\delta }_n\right|<\infty$. If there exist $M\in \mathbb{N}$ and $\epsilon <{\displaystyle \frac{1}{4}}$ such that

$$\left|{\displaystyle \frac{1}{M}}{\displaystyle \sum _{n=mM}^{(m+1)M-1}}{\delta }_n\right|\le \epsilon \mathrm{for}\mathrm{all}m\in \mathbb{Z},$$

then $E(\mathrm{\Lambda };I)$ is a Riesz basis for every interval I of length 1.

Corollary 1.

Assume that Λ is separated and has bounded centered displacement sums at density $d>0$. Then $E(\mathrm{\Lambda };I)$ is a Riesz basis for every interval I of length d.

Proof. 

Let c and C be the constants in (5). Then the separated sequence $\mathrm{\Gamma }={\left\{d({\lambda }_n-c)\right\}}_{n\in \mathbb{Z}}={\{n+d({\delta }_n-c)\}}_{n\in \mathbb{Z}}$ has bounded displacements, and each displacement average over M consecutive terms has absolute value at most ${\displaystyle \frac{dC}{M}}$. Choosing an integer $M>4dC$ and applying Theorem 2, we have that $E(\mathrm{\Gamma };I)$ is a Riesz basis for every interval I of length 1. Since $\mathrm{\Gamma }$ is obtained from $\mathrm{\Lambda }$ by a translation and a scaling, Lemma 1 yields the desired Riesz bases property. □

3. Proof of the Main Result

For an increasing sequence $\mathrm{\Lambda }={\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}}$ satisfying (4), we define

$$\begin{array}{cc}\hfill N_{\mathrm{\Lambda }}\left(x\right)& =n\mathrm{if}{\lambda }_n\le x<{\lambda }_{n+1},\hfill \\ \hfill D_{\mathrm{\Lambda }}\left(x\right)& =N_{\mathrm{\Lambda }}\left(x\right)-dx.\hfill \end{array}$$

It is easy to see that $N_{\mathrm{\Lambda }}\left(y\right)-N_{\mathrm{\Lambda }}\left(x\right)=\#(\mathrm{\Lambda }\cap (x,y])$ for $x<y$, and

$$\#(\mathrm{\Lambda }\cap [x,x+R\left)\right)=dR+O\left(1\right)$$

uniformly in $x\in \mathbb{R}$ and $R>0$, where # denotes the cardinality of a set. The bounded displacement condition (4) is equivalent to boundedness of $D_{\mathrm{\Lambda }}$; in particular, setting ${\delta }_n=0$ for all $n\in \mathbb{Z}$ gives $-1\le D_{\mathrm{\Lambda }}\left(x\right)\le 0$ for all $x\in \mathbb{R}$.

Lemma 3.

Under (4), condition (5) holds with centering constant c if and only if the centered counting discrepancy has a bounded primitive

$$\underset{x<y}{\sup }\left|{\int }_x^y(D_{\mathrm{\Lambda }}\left(t\right)-\mu )dt\right|<\infty ,\mu =-{\displaystyle \frac{1}{2}}-cd.$$

Proof. 

Let $e_n={\delta }_n-c$ for $n\in \mathbb{Z}$. On $[{\lambda }_n,{\lambda }_{n+1})$, the function $D_{\mathrm{\Lambda }}-\mu$ is affine with

$$D_{\mathrm{\Lambda }}\left({\lambda }_n\right)-\mu ={\displaystyle \frac{1}{2}}-d{e}_n,\underset{t\to {\lambda }_{n+1}^{-}}{\lim }{D}_{\mathrm{\Lambda }}\left(t\right)-\mu =-{\displaystyle \frac{1}{2}}-d{e}_{n+1}.$$

Since ${\lambda }_{n+1}-{\lambda }_n={\displaystyle \frac{1}{d}}+e_{n+1}-e_n$, we have

$${\int }_{{\lambda }_n}^{{\lambda }_{n+1}}(D_{\mathrm{\Lambda }}\left(t\right)-\mu )dt=-{\displaystyle \frac{e_n+e_{n+1}}{2}}-{\displaystyle \frac{d}{2}}(e_{n+1}^2-e_n^2).$$

Summing over $n=m,\dots ,p-1$ gives

$${\int }_{{\lambda }_m}^{{\lambda }_p}(D_{\mathrm{\Lambda }}\left(t\right)-\mu )dt=-{\displaystyle \sum _{n=m+1}^p}{e}_n+{\displaystyle \frac{e_p-e_m}{2}}-{\displaystyle \frac{d}{2}}(e_p^2-e_m^2).$$

Since ${\left\{e_n\right\}}_{n\in \mathbb{Z}}$ is bounded, the last two terms are uniformly bounded. This proves the equivalence for endpoints in $\mathrm{\Lambda }$. Bounded displacement also bounds the gaps and $D_{\mathrm{\Lambda }}$, so moving arbitrary endpoints to neighboring points of $\mathrm{\Lambda }$ changes the integral by a uniformly bounded amount. □

Lemma 4. (i)The property of having bounded centered displacement sums is preserved by positive dilations, translations, and finite disjoint unions, provided that the union is separated. The density of a disjoint union is the sum of the densities.

(ii)Assume that $\mathrm{\Lambda }\subset \mathrm{\Gamma }$ are separated sets, and both sets have bounded centered displacement sums at densities $a,d$, respectively, where $0<a<d$. Then $\mathrm{\Gamma }\setminus \mathrm{\Lambda }$ has the same property at density $d-a$. In particular, if $\mathrm{\Lambda }\subset \mathbb{Z}$ has this property at density $0<a<1$, then $\mathbb{Z}\setminus \mathrm{\Lambda }$ has the same property at density $1-a$.

Proof. (i) Assume that ${\lambda }_n=n/d+{\delta }_n$ satisfies (5) with centering constant c. For $q>0$ and $r_0\in \mathbb{R}$, the sequence ${\{q{\lambda }_n+r_0\}}_{n\in \mathbb{Z}}$ has density $d/q$ and displacements $q{\delta }_n+r_0$. Centering at $qc+r_0$ multiplies every sum in (5) by q, so the assertion holds for positive dilations and translations.

Next, let $U={\phantom{\bigcup }\bigcup ·}_{j=1}^k{\mathrm{\Lambda }}_j$ be separated, where each ${\mathrm{\Lambda }}_j$ has bounded centered displacement sums at density $d_j>0$. By Lemma 3, there exists a constant ${\mu }_j$ such that $D_{{\mathrm{\Lambda }}_j}-{\mu }_j$ has a bounded primitive. Adding the counting estimates gives

$$\#(U\cap [x,x+R))=d_{U}R+O\left(1\right),d_U={\displaystyle \sum _{j=1}^k}{d}_j,$$

uniformly in $x\in \mathbb{R}$ and $R>0$, so U has a bilateral increasing enumeration at density $d_U$. Since counting increments add, its discrepancy satisfies

$$D_U={\displaystyle \sum _{j=1}^k}{D}_{{\mathrm{\Lambda }}_j}+C_0,$$

where $C_0$ accounts for the choices of index origins. This shows that $D_U$ is bounded. Since ${\mu }_U={\sum }_{j=1}^k{\mu }_j+C_0$, it follows that $D_U-{\mu }_U={\sum }_{j=1}^k(D_{{\mathrm{\Lambda }}_j}-{\mu }_j)$ has a bounded primitive. Then Lemma 3 implies that the assertion holds for disjoint unions.

(ii) Let $V=\mathrm{\Gamma }\setminus \mathrm{\Lambda }$. Subtracting the counting estimates gives

$$\#(V\cap [x,x+R\left)\right)=(d-a)R+O\left(1\right)$$

uniformly in $x\in \mathbb{R}$ and $R>0$, so the separated set V has a bilateral increasing enumeration at density $d-a$. Its counting discrepancy satisfies

$$D_V=D_{\mathrm{\Gamma }}-D_{\mathrm{\Lambda }}+C_1,$$

where $C_1$ accounts for the choices of index origins. Choose ${\mu }_{\mathrm{\Gamma }}$ and ${\mu }_{\mathrm{\Lambda }}$ as in Lemma 3. Then $D_V$ is bounded, and $D_V-({\mu }_{\mathrm{\Gamma }}-{\mu }_{\mathrm{\Lambda }}+C_1)$ has a bounded primitive. By Lemma 3, we conclude that V has bounded centered displacement sums at density $d-a$. Finally, choosing ${\lambda }_n=n$ gives bounded centered displacement sums for $\mathbb{Z}$ at density 1, so the special case follows by taking $\mathrm{\Gamma }=\mathbb{Z}$. □

We will first construct integer spectra for circular intervals with the displacement estimate needed below.

Proposition 1.

For every $0<s\le 1$, there exists a set ${\mathrm{\Delta }}_s\subset \mathbb{Z}$ with bounded centered displacement sums at density s such that $E({\mathrm{\Delta }}_s;I)$ is a Riesz basis for every circular interval I of length s.

Proof. 

For $0<s\le {\displaystyle \frac{1}{2}}$, let ${\mathrm{\Delta }}_s={\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}}$ be given by

$${\lambda }_n=Q_n-Q_{n-1}\mathrm{with}{Q}_n=⌊{\displaystyle \frac{n(n+1)}{2s}}⌋.$$

Setting ${\epsilon }_n:=Q_n-{\displaystyle \frac{n(n+1)}{2s}}\in (-1,0]$, we may write

$${\lambda }_n-{\displaystyle \frac{n}{s}}={\epsilon }_n-{\epsilon }_{n-1}.$$

Then for any $n\in \mathbb{Z}$, we have

$$\begin{array}{cc}\hfill {\lambda }_{n+1}-{\lambda }_n& ={\displaystyle \frac{1}{s}}+{\epsilon }_{n+1}-2{\epsilon }_n+{\epsilon }_{n-1}>{\displaystyle \frac{1}{s}}-2\ge 0,\hfill \end{array}$$

which shows that ${\left\{{\lambda }_n\right\}}_{n\in \mathbb{Z}}\subset \mathbb{Z}$ is a strictly increasing separated sequence. Moreover, every displacement sum telescopes, that is, for any $m\in \mathbb{Z}$ and $M\in \mathbb{N}$,

$$\left|{\displaystyle \sum _{n=m}^{m+M-1}}\left({\lambda }_n-{\displaystyle \frac{n}{s}}\right)\right|=\left|{\displaystyle \sum _{n=m}^{m+M-1}}\left({\epsilon }_n-{\epsilon }_{n-1}\right)\right|=\left|{\epsilon }_{m+M-1}-{\epsilon }_{m-1}\right|<1.$$

Therefore, ${\mathrm{\Delta }}_s$ has bounded centered displacement sums at density s with center zero.

For ${\displaystyle \frac{1}{2}}<s<1$, let ${\mathrm{\Delta }}_s=\mathbb{Z}\setminus {\mathrm{\Delta }}_{1-s}$. Then ${\mathrm{\Delta }}_s\subset \mathbb{Z}$ is separated and has bounded centered displacement sums at density s by Lemma 4.

For $s=1$, let ${\mathrm{\Delta }}_1=\mathbb{Z}$.

In every case, Corollary 1 gives the Riesz basis property on intervals of length s, and integer frequencies allow circular translation (see the discussion after Lemma 1). Therefore, $E({\mathrm{\Delta }}_s;I)$ is a Riesz basis for every circular interval I of length s. □

Let $N\in \mathbb{N}$ and let $S\subset [0,1)$ be a measurable set. For $j=1,\dots ,N$, define

$$\begin{array}{cc}\hfill A_{\ge j}:=& \left\{t\in [0,\frac{1}{N}):t+\frac{k}{N}\in S\mathrm{for}\mathrm{at}\mathrm{least}j\mathrm{values}\mathrm{of}k\in \{0,1,\dots ,N-1\}\right\}\hfill \\ \hfill =& \left\{t\in [0,\frac{1}{N}):{\sum }_{k=0}^{N-1}11_S(t+\frac{k}{N})\ge j\right\}.\hfill \end{array}$$

We use the folding lemma of Kozma and Nitzan [9], which asserts that if ${\mathrm{\Xi }}_j\subset N\mathbb{Z}$ is a Riesz spectrum for $A_{\ge j}$ for every j, then

$${\displaystyle \bigcup _{j=1}^N}({\mathrm{\Xi }}_j+j)$$

is a Riesz spectrum for S. Here, empty levels are assigned the empty spectrum. The labels $1,\dots ,N$ are the consecutive residue labels of this lemma; the integer N need not be prime.

For a union of two intervals, we choose an integer N such that every normalized folded level $N{A}_{\ge j}$ is empty, the full interval $\mathbb{T}=[0,1)$, or one circular interval. The existence of infinitely many such integers follows from the two-interval case of the component reduction lemma [10], which is based on [9]. Here, we provide a direct proof with an explicit condition on N. For $X\subset \mathbb{R}$, we write $Xmod1:=\{xmod1:x\in X\}\subset \mathbb{T}$.

Lemma 5.

Let $0\le a_1<b_1<a_2<b_2<1$. Write $h_1=b_1-a_1$ and $h_3=b_2-a_2$ for the lengths of the intervals $[a_1,b_1)$ and $[a_2,b_2)$, and $h_2=a_2-b_1$ and $h_4=1-b_2+a_1$ for the lengths of the two circular gaps between them. Then $h_i>0$ and $h_1+h_2+h_3+h_4=1$. For $N\in \mathbb{N}$, let ${\sigma }_i={\sigma }_i\left(N\right):=N{h}_i-⌊N{h}_i⌋\in [0,1)$,

$$w\left(N\right)={\sigma }_1+{\sigma }_2+{\sigma }_3+{\sigma }_4=N-{\displaystyle \sum _{i=1}^4}⌊N{h}_i⌋\in \{0,1,2,3\},$$

and

$$J_1=(N{a}_1+[0,{\sigma }_1))mod1=\{(N{a}_1+t)mod1:0\le t<{\sigma }_1\}\subset \mathbb{T},$$

$$J_2=(N{a}_2+[0,{\sigma }_3))mod1=\{(N{a}_2+t)mod1:0\le t<{\sigma }_3\}\subset \mathbb{T}.$$

(a) If $w\left(N\right)=2$, then $J_1\cup J_2$ and $J_1\cap J_2$ are circular intervals, possibly empty or full.

(b) There exist infinitely many $N\in \mathbb{N}$ with $w\left(N\right)=2$.

Proof. 

Since ${\sum }_{i}N{h}_i=N$, the number $w\left(N\right)$ is an integer satisfying $0\le w\left(N\right)<4$. Let ${\tau }_0=0$ and ${\tau }_i={\sigma }_1+\dots +{\sigma }_i$ for $1\le i\le 4$, so that ${\tau }_4=w\left(N\right)$. Note that $N{a}_2-N{a}_1=N{h}_1+N{h}_2$ differs from ${\tau }_2$ by an integer. Therefore, after a rotation of $\mathbb{T}$ (which does not affect the assertions), we may assume that $J_1=[{\tau }_0,{\tau }_1)mod1$ and $J_2=[{\tau }_2,{\tau }_3)mod1$. Let $G_1=[{\tau }_1,{\tau }_2)mod1$ and $G_2=[{\tau }_3,{\tau }_4)mod1$. Since each interval $[{\tau }_{i-1},{\tau }_i)$ has length ${\sigma }_i<1$, it is mapped injectively into $\mathbb{T}$. These intervals form a partition of $[0,w(N\left)\right)$, which contains exactly $w\left(N\right)$ points of each coset of $\mathbb{Z}$. Consequently, we have that

$$\mathrm{every}\mathrm{point}\mathrm{of}\mathbb{T}\mathrm{lies}\mathrm{in}\mathrm{exactly}w\left(N\right)\mathrm{of}\mathrm{the}\mathrm{four}\mathrm{arcs}{J}_1,G_1,J_2,G_2.$$

(6)

(a) Assume that $w\left(N\right)=2$. If $x\in [{\tau }_0,{\tau }_1)$ and $x+k\in [{\tau }_2,{\tau }_3)$ for some $k\in \mathbb{Z}$, then $k\ge 1$, since $x<{\tau }_1\le {\tau }_2\le x+k$. Also, we have $k\le 1$, since $x+k<{\tau }_3\le {\tau }_4=2\le x+2$. Therefore, we have

$$J_1\cap J_2=\left([{\tau }_0,{\tau }_1)\cap [{\tau }_2-1,{\tau }_3-1)\right)mod1,$$

which is empty or the image of an interval of length less than 1. Similaryl, if $x\in [{\tau }_1,{\tau }_2)$ and $x+k\in [{\tau }_3,{\tau }_4)$ for some $k\in \mathbb{Z}$, then $k=1$, since $x<{\tau }_2\le {\tau }_3\le x+k$ and $x+k<{\tau }_4=2\le {\tau }_1+2\le x+2$. Therefore, $G_1\cap G_2$ is empty or equal to $[\xi ,{\xi }^{\prime })mod1$ for some $\xi <{\xi }^{\prime }<\xi +1$. It follows from (6) that a point of $\mathbb{T}$ lies outside $J_1\cup J_2$ if and only if it lies in $G_1\cap G_2$. Therefore, $J_1\cup J_2=\mathbb{T}\setminus (G_1\cap G_2)$ is either $\mathbb{T}=[0,1)$ or $[{\xi }^{\prime },\xi +1)mod1$, which proves (a).

(b) Since ${\sigma }_i(N+1)-{\sigma }_i\left(N\right)\in \{h_i,h_i-1\}$, we have

$$w(N+1)-w\left(N\right)\le h_1+h_2+h_3+h_4=1\mathrm{for}N\in \mathbb{N}.$$

(7)

Let $\zeta :={\min }_i\left(\min \{h_i,1-h_i\}\right)$. Since ${\min }_i{h}_i\le {\displaystyle \frac{1}{4}}$, we musts have $0<\zeta \le {\displaystyle \frac{1}{4}}$. Fix any $m\in \mathbb{N}$. By Dirichlet’s theorem on simultaneous approximation (see e.g., [18]), applied to $m{h}_1,m{h}_2,m{h}_3,m{h}_4$, there exist integers $q\ge 1$ and $z_1,z_2,z_3,z_4\in \mathbb{Z}$ such that $Q=qm$ satisfies

$${\gamma }_i:=Q{h}_i-z_i\in (-\zeta ,\zeta )\mathrm{for}i=1,2,3,4.$$

Since $Q\ge m$ and m is arbitrary, there are arbitrarily large integers Q with this property. For each such Q, the sum ${\gamma }_1+{\gamma }_2+{\gamma }_3+{\gamma }_4=Q-{\sum }_i{z}_i$ is an integer of absolute value less than $4\zeta \le 1$, and hence equals zero. Moreover, $h_i+{\gamma }_i\in (0,1)$ and $1-h_i+{\gamma }_i\in (0,1)$, so that for $Q\ge 2$, we have

$${\sigma }_i(Q+1)=h_i+{\gamma }_i,{\sigma }_i(Q-1)=1-h_i+{\gamma }_i.$$

Summing over i gives $w(Q+1)=1$ and $w(Q-1)=3$. Choose two such integers $2\le Q<Q^{\prime }$ with $Q^{\prime }\ge Q+3$. Since $w(Q^{\prime }-1)=3$ and $Q^{\prime }-1\ge Q+2$, there is a least integer $N>Q+1$ with $w\left(N\right)\ge 2$, and this integer satisfies $N\le Q^{\prime }-1$. If $N-1=Q+1$, then $w(N-1)=1$; otherwise, the minimality of N gives $w(N-1)<2$. In either case, we have $w(N-1)\le 1$, and (7) gives $w\left(N\right)=2$. Since Q can be arbitrarily large, there exist infinitely many such N. □

Remark 1.

The condition $w\left(N\right)=2$ is sufficient but not necessary. In fact, the conclusion of Lemma 5 (a) holds if and only if $w\left(N\right)=2$ or ${\sigma }_i\left(N\right)=0$ for some i. To see this, assume that $w\left(N\right)\ne 2$ and use the notation from the proof. First, assume that ${\sigma }_i\left(N\right)=0$ for some i. Then $w\left(N\right)<3$, since the sum of three numbers in $[0,1)$ is less than 3. Since $w\left(N\right)\ne 2$, it follows that $w\left(N\right)\le 1$. If $w\left(N\right)=0$, then all four arcs are empty. If $w\left(N\right)=1$, then the four arcs are pairwise disjoint by (6), and hence $J_1\cap J_2=⌀$. In this case, $J_1\cup J_2$ equals $J_2$, $J_1$, $[{\tau }_0,{\tau }_3)mod1$ or $[{\tau }_2,{\tau }_1+1)mod1$ if ${\sigma }_1$, ${\sigma }_3$, ${\sigma }_2$ or ${\sigma }_4$ vanishes, respectively. Here, ${\tau }_1={\tau }_2$ if ${\sigma }_2=0$, and ${\tau }_3={\tau }_4={\tau }_0+1$ if ${\sigma }_4=0$. The last two sets are images of intervals of lengths $1-{\sigma }_4$ and $1-{\sigma }_2$, respectively.

Now, assume that ${\sigma }_i\left(N\right)>0$ for all i. Then $w\left(N\right)\in \{1,3\}$. If $w\left(N\right)=1$, then $0<{\tau }_1<{\tau }_2<{\tau }_3<1$, and $J_1\cup J_2=[0,{\tau }_1)\cup [{\tau }_2,{\tau }_3)$ has two components. If $w\left(N\right)=3$, then $0<{\tau }_3-2<{\tau }_2-1<{\tau }_1<1$, and $J_1\cap J_2=[0,{\tau }_3-2)\cup [{\tau }_2-1,{\tau }_1)$ has two components. For example, the conclusion of Lemma 5 (a) always fails for $N=1$, since $w\left(1\right)=1$ and ${\sigma }_i\left(1\right)=h_i>0$.

Proposition 2.

Let $S=[a_1,b_1)\cup [a_2,b_2)$, $0\le a_1<b_1<a_2<b_2<1$, and $d=\left|S\right|$. There exists $\mathrm{\Gamma }\subset \mathbb{Z}$ such that $E(\mathrm{\Gamma };S)$ is a Riesz basis and Γ has bounded centered displacement sums at density d.

Proof. 

By Lemma 5 (b), we can choose N with $w\left(N\right)=2$. Let

$${\ell }_i=b_i-a_i,q_i=⌊N{\ell }_i⌋,r_i=N{\ell }_i-q_i,q=q_1+q_2,$$

and define the residual arcs

$$J_i=(N{a}_i+[0,r_i))mod1=\{(N{a}_i+t)mod1:0\le t<r_i\}\subset \mathbb{T}\mathrm{for}i=1,2.$$

An arc with $r_i=0$ is empty. Since ${\ell }_1=h_1$ and ${\ell }_2=h_3$, we have $r_1={\sigma }_1$ and $r_2={\sigma }_3$; hence, these are the arcs $J_1$, $J_2$ of Lemma 5. Exactly $q_i+11_{J_i}\left(Nu\right)$ of the points $u+{\displaystyle \frac{k}{N}}$, $0\le k<N$, lie in $[a_i,b_i)$. Hence, the folded multiplicity is $q+11_{J_1}\left(Nu\right)+11_{J_2}\left(Nu\right)$, and

$$N{A}_{\ge j}=\left\{\begin{array}{cc}\mathbb{T},\hfill & j\le q,\hfill \\ J_1\cup J_2,\hfill & j=q+1,\hfill \\ J_1\cap J_2,\hfill & j=q+2,\hfill \\ ⌀,\hfill & j>q+2.\hfill \end{array}\right.$$

Since $w\left(N\right)=2$, we have $N=q+⌊N{h}_2⌋+⌊N{h}_4⌋+2\ge q+2$, and Lemma 5 (a) shows that both intermediate levels are circular intervals, possibly empty or full. Let $\alpha =|J_1\cup J_2|$ and $\beta =|J_1\cap J_2|$, and use the spectra ${\mathrm{\Delta }}_s$ from Proposition 1, with ${\mathrm{\Delta }}_0=⌀$. For $j\le q$, the level $A_{\ge j}=[0,{\displaystyle \frac{1}{N}})$ has spectrum $N\mathbb{Z}$. Scaling gives the spectra $N{\mathrm{\Delta }}_{\alpha }$ and $N{\mathrm{\Delta }}_{\beta }$ for $A_{\ge q+1}$ and $A_{\ge q+2}$, respectively. The folding lemma therefore shows that

$$\mathrm{\Gamma }=\left({\displaystyle \bigcup _{j=1}^q}(N\mathbb{Z}+j)\right)\cup (N{\mathrm{\Delta }}_{\alpha }+q+1)\cup (N{\mathrm{\Delta }}_{\beta }+q+2)$$

is a Riesz spectrum for S. After omitting empty terms, this is a finite union of separated sets in distinct residue classes modulo N. Lemma 4 gives bounded centered displacement sums, and its density is given by

$${\displaystyle \frac{q+\alpha +\beta }{N}}={\displaystyle \frac{q+r_1+r_2}{N}}={\ell }_1+{\ell }_2=d.$$

□

Remark 2.

The spectra constructed in [9] have bounded counting discrepancy, and thus have bounded displacement (see [9]). Here, we also require bounded centered displacement sums. Proposition 1 provides this stronger estimate for the interval spectra, and Lemma 4 preserves it through folding.

The following is a key lemma which allows us to select one consecutive run from each block of a prescribed frequency set. The runs are chosen so that their displacement sums telescope. It should be noted that no Riesz basis property of the prescribed set is assumed.

Lemma 6

(Balanced splitting). Let $\mathrm{\Gamma }={\left\{{\gamma }_n\right\}}_{n\in \mathbb{Z}}$ be increasingly enumerated and separated, with bounded displacement at density $d>0$, and let $0<a<d$. Then there exists a subset ${\mathrm{\Gamma }}_a\subset \mathrm{\Gamma }$ with bounded centered displacement sums at density a and centering constant zero. Moreover, if Γ has bounded centered displacement sums at density d, then $\mathrm{\Gamma }\setminus {\mathrm{\Gamma }}_a$ has bounded centered displacement sums at density $d-a$.

Proof. Step 1. Identify some candidate runs

Let $\theta ={\displaystyle \frac{a}{d}}$ and $D={\sup }_n\left|{\gamma }_n-{\displaystyle \frac{n}{d}}\right|<\infty$. For a sufficiently large integer K, we divide the indices of $\mathrm{\Gamma }$ into blocks $[mK,(m+1\left)K\right)\cap \mathbb{Z}$, and then set $A_m=⌊\theta mK⌋$ and $k_m=A_{m+1}-A_m$. Then $A_m=\theta mK+O\left(1\right)$ and $k_m=\theta K+O\left(1\right)$ uniformly in m, and $1\le k_m\le K-1$ for large K. We will select $k_m$ consecutive points from block m, assigning them the indices $A_m,\dots ,A_{m+1}-1$. Their target sum is

$$T_m={\displaystyle \sum _{j=A_m}^{A_{m+1}-1}}{\displaystyle \frac{j}{a}}.$$

The possible consecutive runs have sums

$$W_{m,r}={\displaystyle \sum _{i=0}^{k_m-1}}{\gamma }_{mK+r+i}\mathrm{for}0\le r\le K-k_m.$$

These sums increase, and their successive increments satisfy

$$0<W_{m,r+1}-W_{m,r}={\gamma }_{mK+r+k_m}-{\gamma }_{mK+r}\le {\displaystyle \frac{k_m}{d}}+2D\le H_K,$$

(8)

where $H_K:={\displaystyle \frac{K}{d}}+2D$. Let

$$L_m=W_{m,0},U_m=W_{m,K-k_m},{\epsilon }_n={\gamma }_n-{\displaystyle \frac{n}{d}},{\eta }_m=A_m-\theta mK\in (-1,0].$$

For $k=k_m$, direct subtraction gives

$$\begin{array}{cc}\hfill T_m-L_m& ={\displaystyle \frac{k{\eta }_m}{a}}+{\displaystyle \frac{d-a}{2ad}}k(k-1)-{\displaystyle \sum _{i=0}^{k-1}}{\epsilon }_{mK+i},\hfill \\ \hfill U_m-L_m& ={\displaystyle \frac{k(K-k)}{d}}+{\displaystyle \sum _{i=0}^{k-1}}({\epsilon }_{mK+K-k+i}-{\epsilon }_{mK+i}).\hfill \end{array}$$

Since $k=\theta K+O\left(1\right)$, $|{\eta }_m|\le 1$, and $|{\epsilon }_n|\le D$, we have

$$T_m-L_m=\kappa K^2+O\left(K\right),U_m-T_m=\kappa K^2+O\left(K\right),$$

where $\kappa ={\displaystyle \frac{\theta (1-\theta )}{2d}}>0$. The error constants depend only on $a,d,D$, so these estimates are uniform in m. Since $H_K=O\left(K\right)$, fix K large enough so that both margins exceed $H_K$ for every m.

Step 2. Balance the sums

Choose one window sum $s_m=W_{m,r_m}$ per block and numbers $E_m$ with $|E_m|\le {\displaystyle \frac{H_K}{2}}$ and

$$s_m-T_m=E_{m+1}-E_m.$$

(9)

Set $E_0=0$. For $m\ge 0$, let $s_m$ be a window sum closest to $T_m-E_m$ and let $E_{m+1}=E_m+s_m-T_m$. For $m<0$, proceed backward: let $s_m$ be a window sum closest to $T_m+E_{m+1}$ and let $E_m=T_m+E_{m+1}-s_m$. In both cases the target lies in $[L_m,U_m]$, since the margins exceed $H_K$, and the window sums increase from $L_m$ to $U_m$ in steps of at most $H_K$ by (8). Hence, the closest window sum is within ${\displaystyle \frac{H_K}{2}}$ of the target, and the new error again has absolute value at most ${\displaystyle \frac{H_K}{2}}$.

Step 3. The displacement estimate

Let ${\mathrm{\Gamma }}_a$ consist of the selected runs given by

$${\lambda }_{A_m+i}={\gamma }_{mK+r_m+i}\mathrm{for}0\le i<k_m,m\in \mathbb{Z}.$$

Since $k_m\ge 1$ and $A_m\to \pm \infty$ as $m\to \pm \infty$, the index intervals $[A_m,A_{m+1})\cap \mathbb{Z}$ partition $\mathbb{Z}$. The selected sequence is increasing because the original blocks are ordered. If ${\lambda }_j={\gamma }_n$ is chosen from block m, then

$$0\le {\displaystyle \frac{n}{d}}-{\displaystyle \frac{mK}{d}}<{\displaystyle \frac{K}{d}},-{\displaystyle \frac{1}{a}}<{\displaystyle \frac{j}{a}}-{\displaystyle \frac{mK}{d}}<{\displaystyle \frac{K}{d}}.$$

Hence, we have

$$D_K:=\underset{j}{\sup }|{\lambda }_j-{\displaystyle \frac{j}{a}}|\le D+{\displaystyle \frac{K}{d}}+{\displaystyle \frac{1}{a}}<\infty .$$

Define a discrete primitive by

$$P_0=0,P_{j+1}-P_j={\lambda }_j-{\displaystyle \frac{j}{a}}\mathrm{for}j\in \mathbb{Z}.$$

Since $A_0=0$, equation (9) gives $P_{A_m}=E_m$ for every $m\in \mathbb{Z}$. For $A_m\le j<A_{m+1}$, we therefore have

$$|P_j|\le |E_m|+k_m{D}_K\le {\displaystyle \frac{H_K}{2}}+K{D}_K.$$

Consequently, we have

$$\left|{\displaystyle \sum _{j=u}^{u+M-1}}\left({\lambda }_j-{\displaystyle \frac{j}{a}}\right)\right|=|P_{u+M}-P_u|\le H_K+2K{D}_K\mathrm{for}u\in \mathbb{Z},M\in \mathbb{N}.$$

Therefore, ${\mathrm{\Gamma }}_a$ has bounded centered displacement sums with center zero, and it inherits separation from $\mathrm{\Gamma }$. Moreover, if $\mathrm{\Gamma }$ has bounded centered displacement sums, then Lemma 4 gives the same property at density $d-a$ for $\mathrm{\Gamma }\setminus {\mathrm{\Gamma }}_a$. □

We are now ready to prove the main result, Theorem 1.

Proof 

(Proof of Theorem 1). Let $S=I_1\cup I_3=[0,t_1)\cup [t_2,t_3)$ and $d={\ell }_1+{\ell }_3$. Then Proposition 2 provides a Riesz spectrum $\mathrm{\Gamma }\subset \mathbb{Z}$ for S with bounded centered displacement sums at density d. By Lemma 4, the complement ${\mathrm{\Gamma }}^{\prime }=\mathbb{Z}\setminus \mathrm{\Gamma }$ has the same displacement property at density $1-d$. Applying Lemma 6 to $\mathrm{\Gamma }$ with $a={\ell }_1$ and to ${\mathrm{\Gamma }}^{\prime }$ with $a={\ell }_2$, we obtain $\mathrm{\Gamma }={\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_3$ and ${\mathrm{\Gamma }}^{\prime }={\mathrm{\Lambda }}_2\cup {\mathrm{\Lambda }}_4$, where each ${\mathrm{\Lambda }}_i$ has bounded centered displacement sums at density ${\ell }_i$. For every nonempty $J\subset \{1,2,3,4\}$, Lemma 4 gives this property for the separated union ${\mathrm{\Lambda }}_J={\bigcup }_{i\in J}{\mathrm{\Lambda }}_i\subset \mathbb{Z}$ at density $d_J={\sum }_{i\in J}{\ell }_i$, so that (1) is satisfied. Moreover, Corollary 1 gives the Riesz basis property on every interval of length $d_J$. This proves (ii) and (iii).

It now remains to prove (i). The sets $I_1,I_2,I_3,I_4$, $I_1\cup I_2$, and $I_2\cup I_3$ are single intervals, so their Riesz basis property follows from (ii). The set $I_1\cup I_3$ has the Riesz spectrum $\mathrm{\Gamma }={\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_3$ by construction. These seven sets and their complements exhaust the $2^4-2=14$ proper nonempty subsets of $\{1,2,3,4\}$. Therefore, Lemma 2 gives all the remaining proper unions. The full union is the usual Fourier orthonormal basis on $[0,1)$. This establishes (i). □

4. A Periodic Construction for Rational Endpoints

In the case of rational endpoints, we can explicitly construct periodic Riesz spectra using finite Fourier matrices. We need the following lemma, which is a special case of [12]. For completeness, we provide a short proof here.

Lemma 7

(See e.g., [12]). For any $A,B\subset \{0,\dots ,N-1\}$ with same cardinality, the system

$$E\left(N\mathbb{Z}+A;\bigcup _{b\in B}[\frac{b}{N},\frac{b+1}{N})\right)$$

is a Riesz basis if and only if

$$F_N[A,B]={\left[e^{-2\pi iab/N}\right]}_{a\in A,b\in B}$$

is invertible. Here, the set $N\mathbb{Z}+A$ is understood as ${\bigcup }_{a\in A}(N\mathbb{Z}+a)$, and $F_N[A,B]$ is the submatrix of the order N Fourier matrix $F_N={\left[e^{2\pi ik\ell /N}\right]}_{k,\ell =0}^{N-1}$ with rows and columns indexed by A and B, respectively.

Proof. 

For finitely supported coefficients ${\left\{c_{n,a}\right\}}_{n\in \mathbb{Z},a\in A}$, let

$$f\left(x\right)=\sum _{a\in A}\sum _{n\in \mathbb{Z}}{c}_{n,a}{e}^{2\pi i(Nn+a)x}.$$

For $0\le t<{\displaystyle \frac{1}{N}}$, we have

$${\left[f(t+\frac{b}{N})\right]}_{b\in B}={\left(F_N[A,B]\right)}^{*}diag\left({\left\{e^{2\pi iat}\right\}}_{a\in A}\right){\left[g_a\left(t\right)\right]}_{a\in A},$$

where

$$g_a\left(t\right)=\sum _{n\in \mathbb{Z}}{c}_{n,a}{e}^{2\pi iNnt}\mathrm{for}a\in A.$$

The matrix $diag\left({\left\{e^{2\pi iat}\right\}}_{a\in A}\right)$ is always invertible, and the map

$${\left\{c_{n,a}\right\}}_{n\in \mathbb{Z},a\in A}⟼\left({\left\{g_a\left(t\right)\right\}}_{a\in A},0\le t<\frac{1}{N}\right)$$

is unitary up to a constant, since $E(N\mathbb{Z},[0,{\displaystyle \frac{1}{N}}))$ is an orthogonal basis. Therefore, the matrix $F_N[A,B]$ is invertible if and only if the map

$${\left\{c_{n,a}\right\}}_{n\in \mathbb{Z},a\in A}⟼\left(\{f{(t+\frac{b}{N}\}}_{b\in B},0\le t<\frac{1}{N}\right)$$

is invertible, equivalently, the synthesis operator (see (3))

$${\left\{c_{n,a}\right\}}_{n\in \mathbb{Z},a\in A}⟼\left(f\left(x\right)=\sum _{a\in A}\sum _{n\in \mathbb{Z}}{c}_{n,a}{e}^{2\pi i(Nn+a)x},x\in \bigcup _{b\in B}[\frac{b}{N},\frac{b+1}{N})\right)$$

is invertible, i.e., $E(N\mathbb{Z}+A;{\bigcup }_{b\in B}[\frac{b}{N},\frac{b+1}{N}))$ is a Riesz basis. □

Proposition 3.

Let $0=t_0<t_1<t_2<t_3<t_4=1$, and let $I_i=[t_{i-1},t_i)$ and ${\ell }_i=t_i-t_{i-1}$ for $i=1,2,3,4$. Assume that $t_i=k_i/N$ for some integers $0=k_0<k_1<k_2<k_3<k_4=N$, and let $n_i=k_i-k_{i-1}$ for $i=1,2,3,4$. Consider the partition of ${\mathbb{Z}}_N=\{0,\dots ,N-1\}$ into consecutive blocks in the order $1,3,2,4$:

$$\begin{array}{cc}\hfill A_1& =\{0,\dots ,n_1-1\},\hfill \\ \hfill A_3& =\{n_1,\dots ,n_1+n_3-1\},\hfill \\ \hfill A_2& =\{n_1+n_3,\dots ,n_1+n_3+n_2-1\},\hfill \\ \hfill A_4& =\{n_1+n_3+n_2,\dots ,N-1\}.\hfill \end{array}$$

Then ${\mathrm{\Lambda }}_i=N\mathbb{Z}+A_i(={\bigcup }_{a\in A_i}N\mathbb{Z}+a)$, $i=1,2,3,4$, satisfy the conclusion of Theorem 1.

Proof. 

Note that $\#A_i=n_i$ by construction. For $i=1,2,3,4$, let $B_i=\{k_{i-1},\dots ,k_i-1\}$, so that $\#B_i=k_i-k_{i-1}=n_i$. We will apply Lemma 7 with $A_J:={\bigcup }_{i\in J}{A}_i$ and $B_J:={\bigcup }_{i\in J}{B}_i$ for nonempty $J\subset \{1,2,3,4\}$. For this, we observe that the order N Fourier matrix $F_N={\left[e^{2\pi ik\ell /N}\right]}_{k,\ell =0}^{N-1}$ is a Vandermonde matrix. For any $A,B\subset \{0,\dots ,N-1\}$ with same cardinality, the Fourier submatrix $F_N[A,B]$ is a Vandermonde matrix (up to factoring out the initial powers) if A or B is cyclically consecutive modulo N.

Note that for every nonempty $J\subset \{1,2,3,4\}$ except $\{1,3\}$ and $\{2,4\}$, the set $B_J$ is cyclically consecutive. For $J=\{1,3\}$ and $J=\{2,4\}$, the set $A_J$ is consecutive by construction; indeed, the sets $A_{\{1,3\}}=A_1\cup A_3$ and $A_{\{2,4\}}=A_2\cup A_4$ are consecutive. Therefore, the matrix $F_N[A_J,B_J]$ is invertible for every nonempty $J\subset \{1,2,3,4\}$, equivalently, with ${\mathrm{\Lambda }}_J:={\bigcup }_{i\in J}{\mathrm{\Lambda }}_i={\bigcup }_{i\in J}(N\mathbb{Z}+A_i)=N\mathbb{Z}+A_J$ and $S_J:={\bigcup }_{i\in J}{I}_i={\bigcup }_{i\in J}[t_{i-1},t_i)={\bigcup }_{b\in B_J}[\frac{b}{N},\frac{b+1}{N})$, the system $E({\mathrm{\Lambda }}_J;S_J)$ is a Riesz basis for every nonempty $J\subset \{1,2,3,4\}$. This proves (i) of Theorem 1 for the case of rational endpoints.

For (ii), fix any nonempty $J\subset \{1,2,3,4\}$, and let $r:=\#A_J(={\sum }_{i\in J}{n}_i)$, so that $d_J={\sum }_{i\in J}{\ell }_i={\sum }_{i\in J}{\displaystyle \frac{n_i}{N}}={\displaystyle \frac{r}{N}}$. Choosing the consecutive index set $B=\{0,\dots ,r-1\}$, Lemma 7 gives that $E\left({\bigcup }_{i\in J}(N\mathbb{Z}+A_i);[0,\frac{r}{N})\right)$ is a Riesz basis. Moreover, Lemma 1) allows for translation of $[0,\frac{r}{N})$ while preserving the Riesz basis property. Therefore, $E\left({\bigcup }_{i\in J}(N\mathbb{Z}+A_i);I\right)$ is a Riesz basis for every interval $I\subset \mathbb{R}$ of length $d_J$. THis proves (ii) of Theorem 1 for the case of rational endpoints.

Finally, note that the increasingly enumerated set ${\mathrm{\Lambda }}_J=N\mathbb{Z}+A_J$ is N-periodic and satisfies ${\lambda }_{n+r}={\lambda }_n+N$ for all $n\in \mathbb{Z}$. Since $d_J={\displaystyle \frac{r}{N}}$, its displacements ${\delta }_n={\lambda }_n-{\displaystyle \frac{n}{d_J}}$ are r-periodic. Subtracting their period mean $c_J$ makes every complete period sum to zero. Then any remaining partial period has uniformly bounded sum, which gives (1). This proves (iii) of Theorem 1 for the case of rational endpoints. □

Example 1.

Consider the case of four equal intervals, i.e., $I_i=[{\displaystyle \frac{i-1}{4}},{\displaystyle \frac{i}{4}})$ for $i=1,2,3,4$. Proposition 3 gives ${\mathrm{\Lambda }}_1=4\mathbb{Z}$, ${\mathrm{\Lambda }}_2=4\mathbb{Z}+2$, ${\mathrm{\Lambda }}_3=4\mathbb{Z}+1$, ${\mathrm{\Lambda }}_4=4\mathbb{Z}+3$. Note that the two alternating frequency unions are

$${\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_3=4\mathbb{Z}\cup (4\mathbb{Z}+1),{\mathrm{\Lambda }}_2\cup {\mathrm{\Lambda }}_4=(4\mathbb{Z}+2)\cup (4\mathbb{Z}+3).$$

In each case, the two frequency residues are consecutive modulo 4, so the corresponding Fourier submatrix is invertible. Therefore, Lemma 7 gives the Riesz basis property for $E({\mathrm{\Lambda }}_J;S_J)$ when $J=\{1,3\}$ or $J=\{2,4\}$. Every other nonempty $J\subset \{1,2,3,4\}$ is cyclically consecutive modulo 4 (equivalently, $S_J={\bigcup }_{i\in J}{I}_i$ is a circular interval on $\mathbb{T}=[0,1)$), so the same argument applies.

In view of Lemma 7 and Example 1, a sufficient condition for the rational-endpoint case of the conjecture is that for every $N\in \mathbb{N}$, there exists a permutation matrix $P\in {\{0,1\}}^{N\times N}$ such that all principal minors of $P{F}_N$ are nonzero (see [5]). The case of the identity permutation $P=I$ was studied in [4,5]. Recently, numerical experiments gave a negative answer for $N=16$, i.e., no such permutation matrix exists for $N=16$. Nevertheless, the failure of this sufficient condition does not rule out other constructions, so the rational-endpoint case for $L\ge 5$ remains open.

For partitions into five or more intervals, the unions that are disconnected on $\mathbb{T}$ form several distinct complementary pairs. Therefore, preserving the Riesz basis property for a single union and its complement is no longer sufficient to establish the full hierarchy, unlike the case $L=4$. Extending the construction would require simultaneous control of the Riesz basis property for these additional pairs, making it substantially more difficult.