Hierarchical Exponential Riesz Bases for Four-Interval Partitions

1. Introduction and Main Result

We seek exponential Riesz bases that respect every union in an interval partition. Specifically, we are interested in the following conjecture. For a discrete set $\mathrm{\Lambda }\subset \mathbb{R}$ and a measurable set $S\subset \mathbb{R}$, we write $\mathcal{E}(\mathrm{\Lambda };S)=\{e^{2\pi i\lambda x}{|}_S:\lambda \in \mathrm{\Lambda }\}$.

[Conjecture $1^{\prime }$ in [3] Let $I_{\ell }$, $\ell =1,\dots ,L$, be intervals which form a partition of $[0,1)$. There exists a partition ${\mathrm{\Lambda }}_{\ell }$, $\ell =1,\dots ,L$, of $\mathbb{Z}$ such that $\mathcal{E}({\mathrm{\Lambda }}_J;S_J)$ is a Riesz basis for every $J\subseteq \{1,\dots ,L\}$, where

$$S_J=\bigcup _{\ell \in J}{I}_{\ell },{\mathrm{\Lambda }}_J=\bigcup _{\ell \in J}{\mathrm{\Lambda }}_{\ell }.$$

Here and throughout, 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 increasingly difficult, as it requires the simultaneous basis property for all combinations of 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 $\mathbb{Z}={\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_2\cup {\mathrm{\Lambda }}_3\cup {\mathrm{\Lambda }}_4$ such that for every nonempty $J\subseteq \{1,2,3,4\}$,

(i)

$\mathcal{E}({\mathrm{\Lambda }}_J;S_J)$ is a Riesz basis for $L^2\left(S_J\right)$,

(ii)

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

(iii)

an increasing enumeration ${\mathrm{\Lambda }}_J=\{{\lambda }_{J,n}:n\in \mathbb{Z}\}$ 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 $B_J<\infty$.

Note that no rationality or rational independence assumption is imposed on the endpoints. Setting $M=1$ in (1) gives bounded displacement, ${\lambda }_{J,n}=n/d_J+O\left(1\right)$; division by M gives uniform convergence of the centered displacement averages at rate $O\left(M^{-1}\right)$.

Related Work

For disconnected sets in $\mathbb{R}$, Seip [14] treated arbitrary unions of two finite intervals and certain unions of several intervals using Kadec’s $1/4$-theorem [5] and Avdonin’s theorem [1]. Kozma and Nitzan [6] proved the existence for every finite union of bounded intervals. In higher dimensions, existence results include finite unions of axis-parallel rectangles [7] and centrally symmetric convex polytopes whose faces of every dimension are centrally symmetric [4].

Pfander, Revay, and Walnut [13] constructed frequency partitions whose unions are Riesz spectra for intervals of the corresponding total lengths. Caragea and Lee [3] obtained hierarchies on the actual spatial unions under a rational independence assumption. In [9], the author developed methods for taking unions of exponential Riesz bases. Our main result meets both the interval and spatial-union requirements for four intervals, without any arithmetic hypothesis.

It is worth noting that there are domains not admitting exponential Riesz bases. Kozma, Nitzan, and Olevskii [8] constructed a bounded subset of $\mathbb{R}$ with no exponential Riesz basis. Recently, Ortega-Cerdà [12] and Wan [15] independently showed that balls and triangles in ${\mathbb{R}}^d$ with $d\ge 2$ do not admit exponential Riesz bases.

The Cases $L=2,3$

For the case of two- and three-interval partitions, the conjecture above can be proved directly.

For $L=2$, choose an integer Riesz spectrum ${\mathrm{\Lambda }}_1$ for $I_1$ using the result of Seip [14] (or Proposition 1 below), and set ${\mathrm{\Lambda }}_2=\mathbb{Z}\setminus {\mathrm{\Lambda }}_1$. Since $I_2=[0,1)\setminus I_1$, it follows from Lemma 1 that $\mathcal{E}({\mathrm{\Lambda }}_2;I_2)$ is a Riesz basis. The full union is the Fourier basis on $[0,1)$, so all nonempty unions are covered.

For $L=3$, the main result of Pfander, Revay, and Walnut [13] provides a partition $\mathbb{Z}={\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_2\cup {\mathrm{\Lambda }}_3$ such that each $\mathcal{E}({\mathrm{\Lambda }}_i;I_i)$ 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, all pairs form Riesz bases by Lemma 1, and the full union is the Fourier basis on $[0,1)$. This establishes the case $L=3$ of the conjecture.

It is worth noting that every proper nonempty union in either case is a circular interval. The main result of [13], together with integer-frequency circular translation, also gives the hierarchy and the simultaneous basis property on intervals of the corresponding total lengths. Moreover, the stronger displacement estimate (1) for $L=2,3$ can be obtained using the tools developed in this paper, specifically, by repeatedly applying Lemma 4, Lemma 3, and Corollary 1. However, in the case $L\ge 4$, a spatial union and its complement can both be disconnected on the circle, making the hierarchy problem challenging.

Proof Strategy

On the circle $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, every proper nonempty union of four consecutive intervals is a circular interval, except for the complementary pair $I_1\cup I_3$ and $I_2\cup I_4$. We construct a spectrum $\Gamma$ for the first alternating union with estimate (1); complementation supplies a spectrum for the second. We then split these two prescribed spectra at the four required densities, preserving their alternating unions exactly. The displacement estimate survives every frequency union, so Avdonin’s theorem gives the remaining basis properties.

The central estimate has two useful features. First, it is equivalent to boundedness of a primitive of the centered counting discrepancy, which makes closure under disjoint unions and complements immediate. Second, a sliding-window construction extracts a subset with this estimate from any separated set with bounded displacement. The stronger estimate for the parent is needed only to control the remaining subset.

Organization

Section 2 provides the necessary background material. Section 3 combines a counting-discrepancy identity, a construction of two-interval spectra via folding explicit interval spectra, and a balanced splitting lemma to establish the hierarchy, thereby proving the main result. Section 4 gives an independent periodic construction for the case of rational endpoints.

2. Preliminaries

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

$$A\sum _{\lambda \in \mathrm{\Lambda }}|u_{\lambda }{|}^2\le {∥\sum _{\lambda \in \mathrm{\Lambda }}{u}_{\lambda }{e}_{\lambda }∥}_{L^2\left(S\right)}^2\le B\sum _{\lambda \in \mathrm{\Lambda }}{|u_{\lambda }|}^2$$

(2)

for every finitely supported scalar family $\left(u_{\lambda }\right)$. When $S\subset [0,1)$ and $\mathrm{\Lambda }\subset \mathbb{Z}$, the upper bound holds with $B=1$. Translation of the spatial set multiplies each exponential by a unimodular constant. Translation of the frequencies is a unitary modulation, and a change of variables gives

$$\mathcal{E}(\mathrm{\Lambda };S)\mathrm{is}\mathrm{a}\mathrm{Riesz}\mathrm{basis}⟺\mathcal{E}(c^{-1}\mathrm{\Lambda };cS)\mathrm{is}\mathrm{a}\mathrm{Riesz}\mathrm{basis},c>0.$$

(3)

For integer frequencies, translation on $\mathbb{T}$ has the same invariance. Thus a basis on an interval of length $s\le 1$ is also a basis on any circular interval of length s.

The following complementation principle, which follows from [11] and [2], is useful for studying integer-frequency exponential Riesz bases on subintervals of $[0,1)$; see, e.g., [3] and [13]. For completeness, we provide a short proof. For the corresponding complementation principle for finite Fourier matrices, see [10].

Lemma 1.

Let $S\subset [0,1)$ be measurable and $\mathrm{\Lambda }\subset \mathbb{Z}$. Then $\mathcal{E}(\mathrm{\Lambda };S)$ is a Riesz basis if and only if $\mathcal{E}(\mathbb{Z}\setminus \mathrm{\Lambda };[0,1)\setminus S)$ is a Riesz basis.

Proof. 

Decompose the unitary Fourier synthesis operator according to the frequency partition $\mathrm{\Lambda },\mathbb{Z}\setminus \mathrm{\Lambda }$ and the spatial partition $S,S^c$. Its block form is

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

The asserted basis properties are equivalent to invertibility of A and D, respectively. If A is invertible, then for some $a>0$, $A^{*}A\ge a\mathrm{Id}$ and $A{A}^{*}\ge a\mathrm{Id}$. Unitarity gives

$$C^{*}C=\mathrm{Id}-A^{*}A,B{B}^{*}=\mathrm{Id}-A{A}^{*},$$

so ${\parallel C\parallel }^2,{\parallel B\parallel }^2\le 1-a$. Consequently,

$$D^{*}D=\mathrm{Id}-B^{*}B\ge a\mathrm{Id},D{D}^{*}=\mathrm{Id}-C{C}^{*}\ge a\mathrm{Id}.$$

These inequalities imply that D is invertible. The converse follows by interchanging the complementary blocks. □

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

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

(4)

We say that $\mathrm{\Lambda }$ has bounded centered displacement sums at density d if, writing ${\delta }_n={\lambda }_n-n/d$, there are constants $c\in \mathbb{R}$ and $C<\infty$ such that

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

(5)

A change of the initial index only changes the centering constant c. Taking $M=1$ gives $|{\delta }_n-c|\le C$, so this condition already implies bounded displacement. It also gives the uniform estimate

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

(6)

Estimate (5) is the property we preserve throughout the construction.

We use Avdonin’s theorem [1] in the following form; see also [13].

Theorem 2

(Avdonin [1]). Let $\mathrm{\Lambda }=\{{\lambda }_n:n\in \mathbb{Z}\}\subset \mathbb{R}$ be separated, and suppose that ${\lambda }_n=n+{\delta }_n$ with ${sup}_{n\in \mathbb{Z}}|{\delta }_n|<\infty$. If there are an integer $M\ge 1$ and a number $\epsilon <1/4$ such that

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

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

Corollary 1.

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

Proof. 

Write ${\lambda }_n=n/d+c+e_n$. By (5), every sum of $e_n$ over consecutive indices has absolute value at most C; in particular, $\left(e_n\right)$ is bounded. Choose an integer $M>4Cd$. Then

$$\underset{m\in \mathbb{Z}}{sup}\left|{\displaystyle \frac{1}{M}}{\displaystyle \sum _{n=m}^{m+M-1}}d{e}_n\right|\le {\displaystyle \frac{Cd}{M}}<{\displaystyle \frac{1}{4}}.$$

The separated sequence $d({\lambda }_n-c)=n+d{e}_n$ therefore satisfies the hypotheses of Theorem 2 and is a Riesz spectrum for every interval of length one. Scaling by (3) and modulation give the assertion. □

3. Proof of the Main Result

For an increasing enumeration satisfying (4), we define

$$N_{\mathrm{\Lambda }}\left(x\right)=n\mathrm{if}{\lambda }_n\le x<{\lambda }_{n+1}\mathrm{and}{D}_{\mathrm{\Lambda }}\left(x\right)=N_{\mathrm{\Lambda }}\left(x\right)-dx.$$

(7)

The counting function may take negative values, but its increments count points in the usual way. Bounded displacement is equivalent to boundedness of $D_{\mathrm{\Lambda }}$. In particular,

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

uniformly in $x\in \mathbb{R}$ and $R>0$.

Lemma 2.

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

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

(8)

Proof. 

For $p>m$, integration over the intervals of constancy of $N_{\mathrm{\Lambda }}$ and summation by parts give

$$\begin{array}{cc}\hfill {\int }_{{\lambda }_m}^{{\lambda }_p}(D_{\mathrm{\Lambda }}\left(u\right)-\mu )du=& -{\displaystyle \sum _{n=m+1}^p}({\delta }_n-c)+\left({\displaystyle \frac{1}{2}}+cd\right)({\delta }_p-{\delta }_m)-{\displaystyle \frac{d}{2}}\left({\delta }_p^2-{\delta }_m^2\right).\hfill \end{array}$$

(9)

The last two terms are uniformly bounded, proving the equivalence when the endpoints lie in $\mathrm{\Lambda }$. Both the gaps and $D_{\mathrm{\Lambda }}$ are bounded. Moving arbitrary endpoints to neighboring points of $\mathrm{\Lambda }$ therefore changes the integral by a uniformly bounded amount, which proves (8) in full. □

Lemma 3.

The property of having bounded centered displacement sums is preserved by positive dilations, translations, and finite separated disjoint unions. The density of a disjoint union is the sum of the densities.

Moreover, assume that $\mathrm{\Lambda }\subset \Gamma$ are separated sets, and both sets have this property at densities $a,d$, respectively, where $0<a<d$. Then $\Gamma \setminus \mathrm{\Lambda }$ 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. 

Counting increments add on disjoint unions and subtract on relative complements. The resulting sets have densities ${\sum }_j{d}_j$ and $d-a$, respectively, and their counting discrepancies satisfy

$$D_{{⨆}_j{\mathrm{\Lambda }}_j}=\sum _j{D}_{{\mathrm{\Lambda }}_j}+C_0,D_{\Gamma \setminus \mathrm{\Lambda }}=D_{\Gamma }-D_{\mathrm{\Lambda }}+C_1,$$

where the constants account for the choices of index origins. In the complement case, the counting estimate

$$\#\left(\right(\Gamma \setminus \mathrm{\Lambda })\cap [x,x+R\left)\right)=(d-a)R+O\left(1\right)$$

ensures that the positive-density complement has a bilateral enumeration. Both displayed discrepancies are bounded, giving bounded displacement. Subtracting the corresponding constants ${\sum }_j{\mu }_j+C_0$ and ${\mu }_{\Gamma }-{\mu }_{\mathrm{\Lambda }}+C_1$ shows that the centered primitives add or subtract. Lemma 2 therefore proves both assertions.

For $q>0$ and $r_0\in \mathbb{R}$, the enumeration $q{\lambda }_n+r_0$ has density $d/q$ and displacements $q{\delta }_n+r_0$. Centering at $qc+r_0$ multiplies every sum in (5) by q, which proves the affine assertions. Finally, the enumeration ${\lambda }_n=n$ shows that $\mathbb{Z}$ has the same property. □

The interval spectra needed for folding have a direct construction using integer parts of quadratic sequences and complementation.

Proposition 1.

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

Proof. 

First, assume that $0<s\le 1/2$. For $n\in \mathbb{Z}$, we define

$$Q_n=⌊{\displaystyle \frac{n(n+1)}{2s}}⌋,{\lambda }_n=Q_n-Q_{n-1},{\Delta }_s=\{{\lambda }_n:n\in \mathbb{Z}\}.$$

(10)

Writing ${\epsilon }_n=Q_n-n(n+1)/\left(2s\right)\in (-1,0]$ gives

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

Consequently, every displacement sum telescopes:

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

(11)

Moreover, we have

$${\lambda }_{n+1}-{\lambda }_n={\displaystyle \frac{1}{s}}+{\epsilon }_{n+1}-2{\epsilon }_n+{\epsilon }_{n-1}>{\displaystyle \frac{1}{s}}-2\ge 0.$$

Thus the integer frequencies increase strictly, so they form a separated set. Since ${\lambda }_n=n/s+O\left(1\right)$, this is an increasing bilateral enumeration at density s, and (11) gives bounded centered displacement sums with center zero.

Now, assume that $1/2<s<1$, and let ${\Delta }_s=\mathbb{Z}\setminus {\Delta }_{1-s}$. Lemma 3 gives bounded centered displacement sums at density s, and separation follows from ${\Delta }_s\subset \mathbb{Z}$. For $s=1$, set ${\Delta }_1=\mathbb{Z}$. In every case, Corollary 1 gives the interval basis property, and integer frequencies allow circular translation. □

For $S\subset [0,1)$ and $N\in \mathbb{N}$, we define

$$A_{\ge j}=\left\{u\in [0,1/N):{\displaystyle \sum _{k=0}^{N-1}}11_S(u+k/N)\ge j\right\},1\le j\le N.$$

(12)

We use the folding lemma of Kozma and Nitzan [6]: if ${\Xi }_j\subset N\mathbb{Z}$ is a Riesz spectrum for $A_{\ge j}$ for every j, then

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

(13)

is a Riesz spectrum for S. 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.

We also use the following case of the folded component reduction in [7] (which is asserted as a consequence of Claim 3 and Lemma 4 in [6]): for a union of two intervals, there is an integer N for which every normalized folded level $N{A}_{\ge j}$ is empty, the full interval $[0,1)$, or one circular interval.

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 $d=|S|$. There exists $\Gamma \subset \mathbb{Z}$ such that $\mathcal{E}(\Gamma ;S)$ is a Riesz basis and Γ has bounded centered displacement sums at density d.

Proof. 

Choose N according to the folded component reduction of [7], and let $s_j=|N{A}_{\ge j}|$. For $s_j>0$, choose ${\Sigma }_j={\Delta }_{s_j}$ from Proposition 1; for $s_j=0$, let ${\Sigma }_j=⌀$. By scaling, $N{\Sigma }_j$ is a Riesz spectrum for $A_{\ge j}$. The folding lemma gives

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

(14)

as a Riesz spectrum for S.

The nonempty terms in (14) are disjoint and separated, since they occupy distinct residue classes modulo N. Each has bounded centered displacement sums, and Lemma 3 preserves this property under their affine changes and union. The density is

$${\displaystyle \sum _{j=1}^N}{\displaystyle \frac{s_j}{N}}={\displaystyle \sum _{j=1}^N}|A_{\ge j}|={\int }_0^{1/N}{\displaystyle \sum _{k=0}^{N-1}}11_S(u+k/N)du=|S|.$$

This proves the proposition. □

Remark 1.

The folding multiplicity for two intervals has a simple description. Let ${\ell }_i=b_i-a_i$, $q_i=⌊N{\ell }_i⌋$, $r_i=N{\ell }_i-q_i$, and $q=q_1+q_2$. Define the residual arcs by

$$J_i=\{(N{a}_i+t)mod1:0\le t<r_i\}\subset \mathbb{T};$$

in particular, $J_i=⌀$ when $r_i=0$. Then exactly $q_i+11_{J_i}\left(Nu\right)$ of the points $u+k/N$, $0\le k<N$, lie in $[a_i,b_i)$. Thus, for $1\le j\le N$,

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

(15)

For N chosen by component reduction, both nonconstant levels are circular intervals, possibly empty or full. With $u=|J_1\cup J_2|$, $v=|J_1\cap J_2|$, and ${\Delta }_0=⌀$, formula (14) becomes

$$\begin{array}{cc}\hfill \Gamma =& \left({\displaystyle \bigcup _{j=1}^q}(N\mathbb{Z}+j)\right)\cup (N{\Delta }_u+q+1)\cup (N{\Delta }_v+q+2).\hfill \end{array}$$

Note that the terms with residue label greater than N are empty; otherwise, the folding multiplicity would exceed N.

The only obstruction to both nonconstant levels being circular intervals occurs when the four endpoints are distinct modulo one and have one of the cyclic orders

$$N{a}_1,N{b}_1,N{a}_2,N{b}_2\mathrm{or}N{a}_1,N{b}_2,N{a}_2,N{b}_1.$$

In the first case the arcs are disjoint and their union has two components. In the second their union is the full circle and their intersection has two components. Component reduction selects N for which neither obstruction occurs.

Remark 2.

The spectra constructed in [6] have bounded counting discrepancy, hence bounded displacement; see [6]. The additional estimate here is the second-order property of bounded centered displacement sums. It does not follow from the existence of a Riesz spectrum alone, and it is obtained from the explicit interval spectra of Proposition 1 and from the preservation of bounded centered displacement sums through folding.

The following is a key lemma which partitions a prescribed set by selecting one consecutive run of points from each fixed parent block. Sliding the run through the block allows its sum to be balanced against a target, so that the errors telescope. It should be noted that no basis property of the parent set is needed.

Lemma 4

(Balanced splitting lemma). Let $\Gamma =\{{\gamma }_n:n\in \mathbb{Z}\}$ be increasingly enumerated and separated, with bounded displacement at density $d>0$, and let $0<a<d$. There is a subset ${\Gamma }_a\subset \Gamma$ with bounded centered displacement sums at density a and centering constant zero. If, moreover, Γ has bounded centered displacement sums, then so does ${\Gamma }_b=\Gamma \setminus {\Gamma }_a$ at density $b=d-a$, and

$$\Gamma ={\Gamma }_a\cup {\Gamma }_b$$

is a partition into two sets with bounded centered displacement sums. Here, the centering constant for ${\Gamma }_b$ need not be zero.

Proof. 

Step 1: Candidate runs and their range. Let $\theta =a/d$ and $D={sup}_n|{\gamma }_n-n/d|<\infty$. For a sufficiently large integer K, divide the parent indices into blocks $[mK,(m+1\left)K\right)\cap \mathbb{Z}$ and set

$$A_m=⌊\theta mK⌋,k_m=A_{m+1}-A_m.$$

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 shall select $k_m$ consecutive points from block m, assigning them child indices $A_m,\dots ,A_{m+1}-1$. Their target sum is

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

(16)

The possible consecutive runs have sums

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

(17)

These sums increase, and their successive increments satisfy

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

(18)

where $H_K:={\displaystyle \frac{K}{d}}+2D$. Let $L_m=W_{m,0}$ and $U_m=W_{m,K-k_m}$. The $k_m$ terms $j/a$ of (16) have mean

$${\displaystyle \frac{1}{a}}\left(A_m+{\displaystyle \frac{k_m-1}{2}}\right)={\displaystyle \frac{mK}{d}}+{\displaystyle \frac{K}{2d}}+O\left(1\right),$$

the midpoint of the parent block $[mK/d,(m+1)K/d]$ up to $O\left(1\right)$. The $k_m$ leftmost parent points ${\gamma }_{mK},\dots ,{\gamma }_{mK+k_m-1}$ have mean $mK/d+\theta K/\left(2d\right)+O\left(1\right)$, and the $k_m$ rightmost parent points of the block have mean $(m+1)K/d-\theta K/\left(2d\right)+O\left(1\right)$. Multiplying the differences of these means by $k_m=\theta K+O\left(1\right)$ gives

$$\begin{array}{cc}\hfill T_m-L_m& ={\displaystyle \frac{\theta (1-\theta )}{2d}}{K}^2+O\left(K\right),\hfill \\ \hfill U_m-T_m& ={\displaystyle \frac{\theta (1-\theta )}{2d}}{K}^2+O\left(K\right).\hfill \end{array}$$

(19)

All error estimates are uniform in m. The coefficient of $K^2$ is positive, whereas $H_K=O\left(K\right)$. Fix K so large that both margins exceed $2H_K$ for every m.

Step 2: Balancing in both directions. Choose one window sum $s_m=W_{m,r_m}$ per block and numbers $E_m$ with $|E_m|\le H_K/2$ satisfying

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

(20)

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 $2H_K$, and the window sums increase from $L_m$ to $U_m$ in steps of at most $H_K$ by (18). Hence the closest window sum is within $H_K/2$ of the target, and the new error again has absolute value at most $H_K/2$.

Step 3: The displacement bound. Let ${\Gamma }_a$ consist of the selected runs, with enumeration

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

The child-index blocks partition $\mathbb{Z}$. If ${\lambda }_j={\gamma }_n$ comes from parent 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

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

By (20), displacement sums over complete child blocks telescope and have absolute value at most $H_K$. An arbitrary consecutive window has at most two additional partial blocks, with at most $2K$ terms altogether. Therefore

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

(21)

Thus ${\Gamma }_a$ has bounded centered displacement sums with center zero, and it inherits separation from $\Gamma$. If $\Gamma$ has bounded centered displacement sums, Lemma 3 gives the same property at density $d-a=b$ for ${\Gamma }_b=\Gamma \setminus {\Gamma }_a$. □

Remark 3.

Each parent block contributes a single consecutive run to ${\Gamma }_a$. The block length K is fixed throughout; no limiting change of the frequency partition is involved.

We are now ready to prove Theorem 1.

Proof 

(Proof of Theorem 1). Let $S=I_1\cup I_3=[0,t_1)\cup [t_2,t_3)$ and $d={\ell }_1+{\ell }_3$. Proposition 2 gives a Riesz spectrum $\Gamma \subset \mathbb{Z}$ for S with bounded centered displacement sums at density d. Its complement ${\Gamma }^{\prime }=\mathbb{Z}\setminus \Gamma$ has the same property at density $1-d$ by Lemma 3, and is a Riesz spectrum for $I_2\cup I_4$ by Lemma 1.

Apply Lemma 4 to $\Gamma$ with $a={\ell }_1$ and to ${\Gamma }^{\prime }$ with $a={\ell }_2$. This gives

$$\Gamma ={\mathrm{\Lambda }}_1\cup {\mathrm{\Lambda }}_3\mathrm{and}{\Gamma }^{\prime }={\mathrm{\Lambda }}_2\cup {\mathrm{\Lambda }}_4,$$

(22)

where each ${\mathrm{\Lambda }}_i$ has bounded centered displacement sums at density ${\ell }_i$.

For every nonempty J, Lemma 3 gives this property for the separated union ${\mathrm{\Lambda }}_J\subset \mathbb{Z}$ at density $d_J={\sum }_{i\in J}{\ell }_i$. This proves (1), and Corollary 1 gives the basis property on every interval of length $d_J$.

Every proper nonempty spatial union $S_J$ is a circular interval unless $J=\{1,3\}$ or $J=\{2,4\}$. Circular intervals are covered by the preceding application of Avdonin’s theorem and integer-frequency circular translation. The two alternating unions have the exact spectra $\Gamma$ and ${\Gamma }^{\prime }$ by (22). Finally, the full union is the usual Fourier orthonormal basis on $[0,1)$. □

4. A Periodic Construction for Rational Endpoints

In the case of rational endpoints, we can use a finite Fourier matrix argument to obtain a more explicit construction. This also illustrates the role of the alternating unions in a finite-dimensional setting.

Proposition 3.

Let $t_i=m_i/q$, where $0=m_0<m_1<m_2<m_3<m_4=q$ are integers, and set $n_i=m_i-m_{i-1}$. Partition the residues modulo q into consecutive blocks in the order $1,3,2,4$:

$$\begin{array}{cc}\hfill C_1& =\{0,\dots ,n_1-1\},\hfill \\ \hfill C_3& =\{n_1,\dots ,n_1+n_3-1\},\hfill \\ \hfill C_2& =\{n_1+n_3,\dots ,n_1+n_3+n_2-1\},\hfill \\ \hfill C_4& =\{n_1+n_3+n_2,\dots ,q-1\}.\hfill \end{array}$$

Then ${\mathrm{\Lambda }}_i=q\mathbb{Z}+C_i$ satisfy the conclusion of Theorem 1.

Proof. 

Let $R_i=\{m_{i-1},\dots ,m_i-1\}$ and

$$F_q=q^{-1/2}{\left[e^{2\pi irs/q}\right]}_{r,s=0}^{q-1}.$$

For $R_J={\bigcup }_{i\in J}{R}_i$ and $C_J={\bigcup }_{i\in J}{C}_i$, the square submatrix $F_q[R_J,C_J]$ is invertible if and only if $\mathcal{E}(q\mathbb{Z}+C_J;S_J)$ is a Riesz basis. Indeed, decompose x as $u+r/q$ with $u\in [0,1/q)$ and group the frequencies by residue. Fourier synthesis is then multiplication by this finite matrix, together with diagonal unimodular factors, on a vector of Fourier series on $[0,1/q)$.

A square Fourier submatrix is invertible whenever its rows or its columns are cyclically consecutive. Factoring the initial powers in the consecutive direction reduces its determinant to a Vandermonde determinant with distinct roots of unity as nodes. This does not require q to be prime. For every proper nonempty J except 13 and 24, the row set $R_J$ is cyclically consecutive. For the two exceptional pairs, the column sets are consecutive by construction:

$$\begin{array}{cc}\hfill C_1\cup C_3& =\{0,\dots ,n_1+n_3-1\},\hfill \\ \hfill C_2\cup C_4& =\{n_1+n_3,\dots ,q-1\}.\hfill \end{array}$$

The full Fourier matrix is unitary. Thus all required submatrices are invertible, proving the spatial-union conclusion.

For the interval conclusion, fix a nonempty J and let $r=\#C_J={\sum }_{i\in J}{n}_i$, so that $d_J={\sum }_{i\in J}{\ell }_i=r/q$. Apply the same Fourier matrix criterion to $H_0=[0,r/q)$. Its row set is $\{0,\dots ,r-1\}$, so $F_q[\{0,\dots ,r-1\},C_J]$ is invertible by the Vandermonde argument above. Thus $\mathcal{E}(q\mathbb{Z}+C_J;H_0)$ is a Riesz basis, and translation gives the assertion for every interval of length $d_J$.

Finally, the increasing enumeration of ${\mathrm{\Lambda }}_J=q\mathbb{Z}+C_J$ satisfies ${\lambda }_{n+r}={\lambda }_n+q$. Since $d_J=r/q$, its displacements ${\delta }_n={\lambda }_n-n/d_J$ are r-periodic. Subtracting their period mean $c_J$ makes every complete period sum to zero. Any remaining partial period has uniformly bounded sum, which proves (1) for the periodic construction. □

Example 1.

In the case of four equal intervals, Proposition 3 gives ${\mathrm{\Lambda }}_1=4\mathbb{Z}$, ${\mathrm{\Lambda }}_2=4\mathbb{Z}+2$, ${\mathrm{\Lambda }}_3=4\mathbb{Z}+1$, ${\mathrm{\Lambda }}_4=4\mathbb{Z}+3$. The two alternating frequency unions occupy consecutive residue pairs. The assignment depends on the spatial order of the four intervals.

For partitions with more than four intervals, there are many disconnected unions that are not related by complementation. Therefore, preserving a spectrum and its complement does not automatically yield a full hierarchy. Extending the construction would require simultaneous control of these additional disconnected unions, which appears increasingly difficult.