Spectra of Self-Similar Measures

Abstract

This paper is devoted to the characterization of spectrum candidates with a new tree structure to be the spectra of a spectral self-similar measure ${\mu }_{N,D}$ generated by the finite integer digit set D and the compression ratio $N^{-1}$. The tree structure is introduced with the language of symbolic space and widens the field of spectrum candidates. The spectrum candidate considered by Łaba and Wang is a set with a special tree structure. After showing a new criterion for the spectrum candidate with a tree structure to be a spectrum of ${\mu }_{N,D}$, three sufficient and necessary conditions for the spectrum candidate with a tree structure to be a spectrum of ${\mu }_{N,D}$ were obtained. This result extends the conclusion of Łaba and Wang. As an application, an example of spectrum candidate $\Lambda (N,\mathcal{B})$ with the tree structure associated with a self-similar measure is given. By our results, we obtain that $\Lambda (N,\mathcal{B})$ is a spectrum of the self-similar measure. However, neither the method of Łaba and Wang nor that of Strichartz is applicable to the set $\Lambda (N,\mathcal{B})$.

1. Introduction

Let $\mu$ be a probability measure on ${\mathbb{R}}^d$ with compact support K. We say that $\mu$ is a spectral measure if there exists a countable set $\Lambda \subset {\mathbb{R}}^d$ such that the set of exponential functions $E_{\Lambda }:=\{exp2\pi i⟨\lambda ,x⟩:\lambda \in \Lambda \}$ is an orthogonal basis of $L^2\left(\mu \right)$. In this case, $\Lambda$ is called a spectrum of $\mu$ and $(\mu ,\Lambda )$ is called a spectral pair. In particular, if $\mu$ is the normalized Lebesgue measure restricted on K, we say K is a spectral set.

In [1], Fuglede introduced the notion of a spectral set in the study of the extendability of the commuting partial differential operators and raised the famous conjecture: K is a spectral set if and only if K is a translational tile. Although the conjecture was finally disproven for the case that $K\subset {\mathbb{R}}^d$ with $d\ge 3$ and is still open for ${\mathbb{R}}^d$ with $d\le 2$, it has led to the development of harmonic analysis, operator theory, tiling theory, convex geometry, etc.

In 1998, Jorgensen and Pedersen [2] discovered the first singular, non-atomic spectral measure—the middle-forth Cantor measure—and proved the middle-third Cantor measure is not a spectral measure. Following this discovery, there has been much research on the spectrality of self-similar (or self-affine) measures and Moran-type self-similar (or self-affine) measures (see for example [3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21] and the references therein).

Consider the iterated function system (IFS) ${\left\{{\varphi }_j\right\}}_{j=1}^q$ given by

$${\varphi }_j\left(x\right)=\frac{1}{N}(x+d_j),$$

where N is an integer with $\left|N\right|>1$ and $D={\left\{d_j\right\}}_{j=1}^q$ is a finite subset of $\mathbb{R}.$ It is well known (see [22] or [23]) that there exists a unique probability measure ${\mu }_{N,D}$ satisfying

$${\mu }_{N,D}\left(E\right)=\frac{1}{q}\sum _{j=1}^q{\mu }_{N,D}\left({\varphi }_j^{-1}\left(E\right)\right),\mathrm{for}\mathrm{Borel}\mathrm{set}E\mathrm{of}\mathbb{R}.$$

The measure ${\mu }_{N,D}$ is called the self-similar measure of the IFS ${\left\{{\varphi }_j\right\}}_{j=1}^q$ and is supported on the set

$$T(N,D)=\left\{\sum _{k=1}^{\infty }{d}_k{N}^{-k}:d_k\in D,k\ge 1\right\},$$

which is the attractor of ${\left\{{\varphi }_j\right\}}_{j=1}^q$. Given a finite set $S\subset \mathbb{Z}$ with $♯S=♯D$, we say $(\frac{1}{N}D,S)$ is a compatible pair if the matrix ${[\frac{1}{\sqrt{q}}exp\left(2\pi i\frac{d}{N}s\right)]}_{d\in D,s\in S}$ is a unitary matrix. In other words, $({\delta }_{\frac{1}{N}D},S)$ is a spectral pair. For a finite set A in $\mathbb{R}$,

$${\delta }_A:=\frac{1}{♯A}\sum _{a\in A}{\delta }_a,$$

where ${\delta }_a$ is the Dirac measure at a. Write

$$\Lambda (N,S)=\left\{\sum _{j=0}^k{s}_j{N}^j:k\ge 0,s_j\in S\right\}.$$

Using the dominated convergence theorem, Strichartz [24] proved that ${\mu }_{N,D}$ is a spectral measure with a spectrum $\Lambda (N,S)$ under the conditions that $({\delta }_{\frac{1}{N}D},S)$ is a spectral pair with $0\in S$ and the Fourier transform of ${\delta }_{\frac{1}{N}D}$ does not vanish on $T(N,S)$. By using the $Ruelle$ transfer operator, Łaba and Wang in [3] removed the condition that the Fourier transform of ${\delta }_{\frac{1}{N}D}$ does not vanish on $T(N,S)$. Furthermore, they obtained the following conclusion:

Theorem 1.

(Łaba and Wang). Let $N\in \mathbb{N}$ with $\left|N\right|>1$, $D\subset \mathbb{Z}$ with $0\in D$, and $gcd\left(D\right)=1,0\in S\subset \mathbb{Z}.$ If $(\frac{1}{N}D,S)$ is a compatible pair, then $({\mu }_{N,D},\Lambda (N,S))$ is not a spectral pair if and only if there exist integers $m⩾1,$${\left\{s_j\right\}}_{j=0}^{m-1}\subset S$ and ${\left\{{\eta }_j\right\}}_{j=0}^{m-1}\subset \mathbb{Z}\backslash \left\{0\right\}$ such ${\eta }_{j+1}=N^{-1}({\eta }_j+s_j)$ for $0⩽j⩽m-1,$ where ${\eta }_m:={\eta }_0,s_m:=s_0.$

It is well known that to prove the spectrality of the invariant measure ${\mu }_{N,D}$, the first key step is to construct a suitable spectrum candidate. In this process, the set $\Lambda (N,S)=S+NS+N^{2}S+\cdots$ (finite sum) is the natural spectrum candidate to be considered. Form Theorem 1, we conclude that $\Lambda (N,S)$ is not a spectrum of ${\mu }_{N,D}$ if and only if there is a periodic orbit ${\left\{{\eta }_j\right\}}_{j=0}^{m-1}\subset \mathbb{Z}\backslash \left\{0\right\}$ under the dual IFS $\{{\psi }_i\left(x\right)=\frac{1}{N}(x+s_i):s_i\in S\}.$ The following example implies that the natural spectrum candidate has a weak point. When $D=\{0,1\}$, the invariant measure ${\mu }_{2,D}$ is just the Lebesgue measure on the unit interval with the unique spectrum $\mathbb{Z}$. However, $\Lambda (2,\{0,1\left\}\right)=\mathbb{N}\ne \mathbb{Z}$ in this case. In other words, the natural candidate $\Lambda (2,\{0,1\left\}\right)$ is not a spectrum of ${\mu }_{2,D}$. Actually, any set with form $S+2S+2^{2}S+\cdots$ (finite sum) is not a spectrum of ${\mu }_{2,D}$. In this case, one needs to consider the spectrum candidate with a more general form $S_1+N{S}_2+N^2{S}_3+\cdots$ (finite sum), where $(\frac{1}{N}D,S_i)$ are compatible pairs. Moreover, it is well known that a spectral self-similar (or self-affine) measure has more than one spectrum in general. The results in [7,9,10,11] show that one may consider spectrum candidates with a tree structure. It is worth mentioning that Li [16] obtained a simplified form of Theorem 1. To the best of our understanding, partial results have been obtained in the case of a higher-dimensional space. Developing the method in [3], Dutkay and Jorgensen [14] obtained a sufficient condition for the spectral pair of self-affine measures, and Li [19] obtained a necessary condition for the natural spectrum candidate to be a spectrum of a self-affine measure.

Motivated by the above results, we considered a class of spectrum candidates with a tree structure (defined in Section 2) and obtained three necessary and sufficient conditions for such spectrum candidates not to be the spectra of ${\mu }_{N,D}$ (Theorem 2), which generalizes Łaba and Wang’s result.

The most difficult part of the proof of Theorem 2 is that the first statement implies the second. For this purpose, we show a new criterion for $\Lambda$ to be a spectrum of ${\mu }_{N,D}$. As an application, we give an example involving a self-similar measure $\mu$ and a spectrum candidate $\Lambda (N,\mathcal{B})$ with a tree structure in Section 4. By Theorem 2, we obtain $(\mu ,\Lambda (N,\mathcal{B}\left)\right)$ is a spectral pair. However, neither the criterion of Łaba and Wang (Theorem 1) nor that of Strichartz [24] is applicable to this set $\Lambda (N,\mathcal{B})$.

2. Preliminaries

In this section, we shall recall some basic properties of spectral measures and introduce the tree structure using symbolic space.

Let $\mu$ be a probability measure on $\mathbb{R}$. The Fourier transform of $\mu$ is defined by

$$\widehat{\mu }\left(\xi \right)=\int e^{-2\pi i\xi x}\mathrm{d}\mu \left(x\right),x\in \mathbb{R}.$$

We write $\mathcal{Z}\left(\widehat{\mu }\right)=\{\xi :\widehat{\mu }\left(\xi \right)=0\}$. For a discrete set $\Lambda \subset \mathbb{R}$, write $E_{\Lambda }=\{exp\left(2\pi ix\lambda \right):\lambda \in \Lambda \}$ for a family of exponential functions in $L^2\left(\mu \right)$. Then, $E_{\Lambda }$ is an orthogonal family of $L^2\left(\mu \right)$ if and only if

$$\Lambda -\Lambda \subset \mathcal{Z}\left(\widehat{\mu }\right)\cup \left\{0\right\}.$$

Define

$$Q_{\Lambda }\left(\xi \right)=\sum _{\lambda \in \Lambda }{\left|\widehat{\mu }(\lambda +\xi )\right|}^2,x\in \mathbb{R}.$$

By using the Parseval identity, Jorgenson and Pederson ([2]) obtained the following basic criterion for the orthogonality of $E_{\Lambda }$ in $L^2\left(\mu \right)$.

Proposition 1.

The exponential function set $E_{\Lambda }$ is an orthogonal set of $L^2\left(\mu \right)$ if and only if $Q_{\Lambda }\left(\xi \right)\le 1$ for all $\xi \in \mathbb{R}$, and $E_{\Lambda }$ is an orthogonal basis of $L^2\left(\mu \right)$ if and only if $Q_{\Lambda }\left(\xi \right)=1$ for all $\xi \in \mathbb{R}$.

Given a finite set $D\subset \mathbb{R}$, we call

$$m_D\left(\xi \right)=\frac{1}{♯D}\sum _{d\in D}exp\left(2\pi i\xi d\right),\xi \in \mathbb{R}$$

the mask of D. It is clear that it is just the Fourier transformation of the uniform probability measure on D.

Definition 1.

For two finite subsets D and S of $\mathbb{R}$ with the same cardinality m, we say $(D,S)$ is a compatible pair if

$${\left[\frac{1}{\sqrt{m}}exp\left(2\pi ids\right)\right]}_{d\in D,s\in S}$$

is a unitary matrix.

The following conclusion is well known.

Lemma 1.

For two finite subsets D and S of $\mathbb{R}$ with the same cardinality m, the following statements are equivalent:

(i).

$(D,S)$ is a compatible pair;

(ii).

$m_D(s_1-s_2)=0$ for any $s_1\ne s_2\in S;$

(iii).

${\sum }_{s\in S}{\left|m_D(\xi +s)\right|}^2=1$ for any $\xi \in \mathbb{R}$.

In other words, $(D,S)$ is a compatible pair if and only if S is a spectrum of the uniform probability measure on D.

Let N be an integer with $\left|N\right|>1$ and $D={\left\{d_j\right\}}_{j=1}^q$ a finite subset of $\mathbb{Z}$ with $0\in D$. We denote by ${\mu }_{N,D}$ the unique invariant measure with respect to the IFS $\{{\varphi }_j\left(x\right)=\frac{1}{N}(x+d_j):1\le j\le q\}$ with equal probability weights, i.e.,

$${\mu }_{N,D}=\frac{1}{q}\sum _{j=1}^q{\mu }_{N,D}\circ {\varphi }_j^{-1}.$$

In the sequel, we write $\mu ={\mu }_{N,D}$ for simplicity. Thus, we have

$$\widehat{\mu }\left(\xi \right)=\prod _{j=1}^{\infty }{m}_D\left(N^{-j}\xi \right),\xi \in \mathbb{R}.$$

For $k\ge 1$, we write

$${\widehat{\mu }}_k\left(\xi \right)=\prod _{j=1}^k{m}_D\left(N^{-j}\xi \right),\xi \in \mathbb{R}.$$

(1)

Write $Y\left(m_D\right)=\{\xi \in \mathbb{R}:m_D\left(\xi \right)=1\}$. When $gcd\left(D\right)=1$, we have

$$Y\left(m_D\right)=\{\xi \in \mathbb{R}:|m_D\left(\xi \right)|=1\}=\mathbb{Z}.$$

(2)

Now, we introduce the tree structure. First, we recall some basic notation of symbolic space. Given a positive integer $q>1$, write ${\Sigma }_q=\{0,1,\cdots ,q-1\}$. Let ${\Sigma }^{*}={\bigcup }_{n=0}^{\infty }{\Sigma }_q^n$ stand for the set of all finite words, where ${\Sigma }_q^0=\left\{\vartheta \right\}$ denotes the set of empty words. The length of a finite word $\sigma$ is the number of symbols it contains and is denoted by $\left|\sigma \right|$. The concatenation of two finite words $\sigma$ and ${\sigma }^{\prime }$ is written as $\sigma {\sigma }^{\prime }$. We say $\sigma$ is a prefix of $\sigma {\sigma }^{\prime }$. Given $\sigma ={\sigma }_1{\sigma }_2\cdots {\sigma }_n\in {\Sigma }^{*}$ and $1\le k\le n$, write ${\sigma |}_k={\sigma }_1\cdots {\sigma }_k$. The following definition will bring convenience to us.

Definition 2.

A sequence of finite words ${\left\{I_n\right\}}_{n\ge 1}\subset {\Sigma }^{*}$ is called increasing if for any $n\ge 1$, $I_n$ is a prefix of $I_{n+1}$ and $|I_{n+1}|=|I_n|+1$.

Let $\mathcal{C}$ be a mapping from ${\Sigma }^{*}$ to $\mathbb{Z}$ satisfying $\mathcal{C}\left(\vartheta \right)=0$ and $\mathcal{C}\left(I\right)=0$ if I ends with the symbol 0. It induces a family of mapping $\mathcal{F}={\left\{F_I\right\}}_{I\in {\Sigma }^{*}}$ defined by

$$\begin{array}{cc}\hfill F_I:{\Sigma }^{*}& ⟶\mathbb{Z},\hfill \\ \hfill J& ⟼\mathcal{C}\left({IJ|}_1\right)+N\mathcal{C}\left({IJ|}_2\right)+\cdots +N^{\left|J\right|-1}\mathcal{C}\left(IJ\right),\hfill \end{array}$$

where ${IJ|}_i$ is the concatenation of I and $J|i$ for $1\le i\le \left|J\right|$. We write $F\left(J\right)=F_{\vartheta }\left(J\right)$ for convenience. By a simple deduction, we have the following consistency: for any $I,J,K\in {\Sigma }^{*}$,

$$\begin{array}{cc}\hfill & F_I\left(J\right)+N^{\left|J\right|}{F}_{IJ}\left(K\right)\hfill \\ \hfill =& \mathcal{C}\left({IJ|}_1\right)+\cdots +N^{\left|J\right|-1}\mathcal{C}\left(IJ\right)+N^{\left|J\right|}\mathcal{C}\left({IJK|}_1\right)+\cdots +N^{\left|JK\right|-1}\mathcal{C}\left(IJK\right)\hfill \\ \hfill =& F_I\left(JK\right).\hfill \end{array}$$

Definition 3.

We say a countable set $\Lambda \subset \mathbb{R}$ has a $(\mathcal{C},\mathcal{F})$ tree structure if there exists a mapping $\mathcal{C}$ and an associated family of mappings $\mathcal{F}$ defined in the above paragraph such that

$$\Lambda =\bigcup _{I\in {\Sigma }^{*}}\left\{F\left(I\right)\right\}.$$

For $I\in {\Sigma }^{*}$, let $S_I=\{\mathcal{C}\left(Ii\right):i\in {\Sigma }_q\}$. According to the definition of the mapping $\mathcal{C}$, we have $\mathcal{C}\left(I0\right)=0\in S_I$.

Remark 1.

Given a sequence of finite sets $\mathcal{S}={\left\{S_n\right\}}_{n\ge 1}$, if $S_I=S_{n+1}$ for any $I\in {\Sigma }^n(n\ge 0)$, we obtain

$$\Lambda =S_1+N{S}_2+N^2{S}_3\cdots .$$

In particular, if $S_n=S$ for $n\ge 1$, we obtain

$$\Lambda =S+NS+N^{2}S\cdots ,$$

which is just the case considered by Łaba and Wang in [3].

In this paper, we consider a countable set $\Lambda$ as a spectrum candidate satisfying the following three conditions:

(C1).

$\Lambda$ has a $(\mathcal{C},\mathcal{F})$ tree structure.

(C2).

For any $I\in {\Sigma }^{*}$, $(\frac{1}{N}D,S_I)$ is a compatible pair.

(C3).

The set $\tilde{S}={\bigcup }_{I\in {\Sigma }^{*}}{S}_I$ is bounded.

Remark 2.

Since we only assume that $(\frac{1}{N}D,S_I)$ is a compatible pair with $S_I=\{\mathcal{C}\left(Ii\right):i\in {\Sigma }_q\}$, the map $\mathcal{C}$ may not be a maximal mapping defined in [8] (Definition 2.5) even if $D=\{0,1,\cdots ,q-1\}$.

Now, we exploit some basic properties of $\Lambda$ satisfying the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$. The first one is the uniqueness of the tree representation.

Proposition 2.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$. Then, for any $I\in {\Sigma }_q^{*}$ and $J,K\in {\Sigma }_q^n$ with $n⩾0$, we have $F_I\left(J\right)=F_I\left(K\right)$ if and only if $J=K$.

Proof. 

We just prove the necessity. Suppose there exist $I\in {\Sigma }_q^{*}$ and $J\ne K\in {\Sigma }_q^n$ with $n⩾1$ such that $F_I\left(J\right)=F_I\left(K\right)$. Let l be the smallest integer with ${J|}_l{\ne K|}_l$. From $F_I\left(J\right)=F_I\left(K\right)$, it follows that

$$N^{l-1}\mathcal{C}\left(IJ{|}_l\right)+\cdots +N^{n-1}\mathcal{C}\left(IJ\right)=N^{l-1}\mathcal{C}\left(IK{|}_l\right)+\cdots +N^{n-1}\mathcal{C}\left(IK\right),$$

which implies $\mathcal{C}\left(IJ{|}_l\right)\equiv \mathcal{C}\left(IK{|}_l\right)(modN)$. Noting $\mathcal{C}\left(IJ{|}_l\right),\mathcal{C}\left(IK{|}_l\right)\in S_{{IJ|}_{l-1}}$, we obtain $(\frac{1}{N}D,S_{{IJ|}_{l-1}})$ is not a compatible pair, which is a contradiction to the condition (C2). □

Proposition 3.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$. Then, $E(\Lambda )$ is an orthogonal set of $L^2\left(\mu \right)$.

Proof. 

Given $\alpha \ne \beta \in \Lambda$, there exist two finite words $I,J\in {\Sigma }^{*}$ such that

$$\alpha =F\left(I\right),\beta =F\left(J\right).$$

If $\left|I\right|\ne \left|J\right|$, we add symbol 0 in the end of I or J to obtain $\left|I\right|=\left|J\right|$. Without loss of generality, we assume that $I,J\in {\Sigma }_q^n$ for some integer n. Let l be the smallest positive integer satisfying ${I|}_l{\ne J|}_l$. Recall that $F\left(I{|}_l\right)=\mathcal{C}\left(I{|}_1\right)+N\mathcal{C}\left(I{|}_2\right)\cdots +N^{l-1}\mathcal{C}\left(I{|}_l\right)$. Then, there exists an integer $z_0$ such that

$$N^{-l}(F\left(I{|}_l\right)-F\left(J{|}_l\right))=\frac{1}{N}(\mathcal{C}\left(I{|}_l\right)-\mathcal{C}\left(J{|}_l\right))+z_0.$$

By virtue of the condition (C2), we know that $(\frac{1}{N}D,S_{{I|}_{l-1}})$ is a compatible pair. Noting that both $\mathcal{C}\left(I{|}_l\right)$ and $\mathcal{C}\left(J{|}_l\right)$ belong to $S_{{I|}_{l-1}}$, we obtain

$$m_D\left(N^{-l}(F\left(I{|}_l\right)-F\left(J{|}_l\right))\right)=m_D(\frac{1}{N}(\mathcal{C}\left(I{|}_l\right)-\mathcal{C}\left(J{|}_l\right)+z_0)=m_D\left(\frac{1}{N}(\mathcal{C}\left(I{|}_l\right)-\mathcal{C}\left(J{|}_l\right))\right)=0.$$

This leads to

$$\begin{array}{cc}& \widehat{\mu }(\alpha -\beta )=\widehat{\mu }(F\left(I\right)-F\left(J\right))\hfill \\ \hfill =& \prod _{j=1}^{l-1}{m}_D\left(N^{-j}(F\left(I\right)-F\left(J\right))\right)m_D\left(N^{-l}(F\left(I{|}_l\right)-F\left(J{|}_l\right))\right)\prod _{j=l+1}^{\infty }\left(N^{-j}(F\left(I\right)-F\left(J\right))\right)\hfill \\ \hfill =& 0.\hfill \end{array}$$

□

For any $I\in {\Sigma }_q^{*}$ and $k⩾1$, define

$${\Lambda }_I=\{F_I\left(J\right):J\in {\Sigma }^{*}\}\mathrm{and}{\Lambda }_I^k:=\{F_I\left(J\right):J\in {\Sigma }_q^k\}.$$

We write ${\Lambda }^k:={\Lambda }_{\vartheta }^k$ for simplicity. It is clear that

$${\Lambda }_I^k⊊{\Lambda }_I^{k+1}.$$

From the condition $\left(\mathrm{C}2\right)$ and Lemma 1$\left(\mathrm{i}\mathrm{i}\right)$, it follows that $E\left({\Lambda }_I^k\right)$ is an orthogonal set of $L^2\left({\mu }_k\right)$. By (2), we obtain $\#{\Lambda }_I^k=q^k$. Noting the fact that $dim\left(L^2\left({\mu }_k\right)\right)=q^k,$ we conclude that $E\left({\Lambda }_I^k\right)$ is an orthogonal basis of $L^2\left({\mu }_k\right)$. In other words, ${\Lambda }_I^k$ is a spectrum of ${\mu }_k$. By Lemma 1, we have

$$\sum _{\lambda \in {\Lambda }_I^k}\prod _{j=1}^k{\left|m_D\left(N^{-j}(\xi +\lambda )\right)\right|}^2=\sum _{\lambda \in {\Lambda }_I^k}{\left|{\widehat{\mu }}_k(\xi +\lambda )\right|}^2\equiv 1,\forall \xi \in \mathbb{R}.$$

(3)

In fact, we have the following conclusion.

Proposition 4.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$. Then, $Q_{\Lambda }\left(\xi \right)\equiv 1$ if and only if $Q_{{\Lambda }_I}\left(\xi \right)\equiv 1$ for any $I\in {\Sigma }^{*}$,

Proof. 

By virtue of ${\Lambda }_{\vartheta }=\Lambda$, the sufficiency is obvious.

Next, we prove the necessity. Given $n\ge 1$ and $I\in {\Sigma }_q^n$, write $B_I=\{\xi +F\left(I\right):\xi \in [0,1]\}$ and ${\tilde{B}}_I=\{N^{-n}(\xi +F\left(I\right)):\xi \in [0,1]\}$. It is easy to see that both $B_I$ and ${\tilde{B}}_I$ are compact sets. Noting the fact that ${\widehat{\mu }}_n$ can be extended to be an entire function on the complex plane, ${\widehat{\mu }}_n$ has at most finitely many zero points in $B_I$. On the other hand, recall that

$$\Lambda =\bigcup _{I\in {\Sigma }_q^n}\bigcup _{J\in {\Sigma }^{*}}(F\left(I\right)+N^n{F}_I\left(J\right)),n\ge 1.$$

Noting the fact that every integer is a period of $m_D$, we have ${\widehat{\mu }}_n(\xi +F\left(IJ\right))={\widehat{\mu }}_n(\xi +F\left(I\right))$ for any $I\in {\Sigma }_q^n$ and $J\in {\Sigma }_q^{*}$. Hence,

$$\begin{array}{cc}\hfill Q_{\Lambda }\left(\xi \right)=& \sum _{\lambda \in \Lambda }{\left|{\widehat{\mu }}_n(\xi +\lambda )\right|}^2{\left|\widehat{\mu }\left(N^{-n}(\xi +\lambda )\right)\right|}^2\hfill \\ \hfill =& \sum _{I\in {\Sigma }_q^n}\sum _{J\in {\Sigma }^{*}}{\left|{\widehat{\mu }}_n(\xi +F\left(I\right))\right|}^2{\left|\widehat{\mu }\left(N^{-n}(\xi +F\left(I\right)+N^n{F}_I\left(J\right))\right)\right|}^2\hfill \\ \hfill =& \sum _{I\in {\Sigma }_q^n}{\left|{\widehat{\mu }}_n(\xi +F\left(I\right))\right|}^2\sum _{J\in {\Sigma }^{*}}{\left|\widehat{\mu }(N^{-n}(\xi +F\left(I\right))+F_I\left(J\right))\right|}^2\hfill \\ \hfill =& \sum _{I\in {\Sigma }_q^n}{\left|{\widehat{\mu }}_n(\xi +F\left(I\right))\right|}^2{Q}_{{\Lambda }_I}\left(N^{-n}(\xi +F\left(I\right))\right).\hfill \end{array}$$

(4)

In combination with (3), this means $Q_{{\Lambda }_I}\left(\xi \right)$ takes 1 on except at most finitely many points in ${\tilde{B}}_I$, which implies $Q_{{\Lambda }_I}\left(\xi \right)\equiv 1$ by using the continuity of $Q_{{\Lambda }_I}\left(\xi \right)$. □

In the end of this section, we define the dual IFS $\{{\Phi }_s\left(x\right)=\frac{1}{N}(x+s):s\in \tilde{S}\}$, which plays an important role in what follows. Let T be the invariant set of the IFS, i.e.,

$$T=\bigcup _{s\in \tilde{S}}{\Phi }_s\left(T\right).$$

Define $\mathcal{Z}(\widehat{\mu },T)=\mathcal{Z}\left(\widehat{\mu }\right)\cap T$, which stands for the zero point set of $\widehat{\mu }$ on T. It is clear that $p:=\#\mathcal{Z}(\widehat{\mu },T)$ is finite.

3. Main Theorem

In this section, we will give our main results involving three equivalent statements. To prove the most difficult part of the proof, we prepared several lemmas including a new criterion for a spectrum candidate with a tree structure to be a spectrum of a self-similar measure. At the end of this section, we show that the new criterion is just a sufficient and necessary condition, which is stated as a corollary .

Theorem 2.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(C2\right)$, and $\left(C3\right)$. Then, the following statements are equivalent:

(i).

$(\mu ,\Lambda )$ is not a spectral pair.

(ii).

There exists a finite word $I\in {\Sigma }^{*}$ such that ${inf}_{\xi \in T}{Q}_{{\Lambda }_I}\left(\xi \right)=0$.

(iii).

There exist a finite word $J\in {\Sigma }^{*}$, a sequence of nonzero integers ${\left\{{\beta }_l\right\}}_{l\ge 1}\subset \mathbb{Z},\backslash \left\{0\right\}$ and a sequence of increasing finite words ${\{J{i}_1\cdots i_l\}}_{l\ge 1}\subset {\Sigma }^{*}$, which has a prefix J such that, for any $l\ge 1$, we have ${\beta }_{l+1}=\frac{1}{N}({\beta }_l+\mathcal{C}(J{i}_1\cdots i_l))$.

We shall divide the proof into three parts $\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\right)$, $\left(\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\right)$, and $\left(\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$.

First, we prove $\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\right)$, which plays a key role in the proof of $\left(\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\right)$.

Proof of Theorem 2 

$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\right)$). We shall prove $Q_{{\Lambda }_J}\left({\beta }_1\right)=0.$ Thus, from Proposition 4, the conclusion follows.

Given $\lambda \in {\Lambda }_I,$ there exists a positive integer $m⩾1$ and $L\in {\Sigma }_q^m$ such that

$$\lambda =F_I\left(L\right)\in {\Lambda }_I^m.$$

Since the sequence ${\left\{{\beta }_l\right\}}_{l\ge 1}$ is nonzero, the sequence of integers ${\left\{\mathcal{C}\left(J{i}_1\cdots i_l\right)\right\}}_{l⩾1}$ has infinitely many nonzero terms. Thus, there exist infinitely many terms l with $i_l\ne 0$. Take an integer $r>m$ with $i_r\ne 0$. Write ${\lambda }^{*}:=F_J\left(K\right)\in {\Lambda }_J^r.$ According to Proposition 2 and $i_r\ne 0$, we have $\lambda \ne {\lambda }^{*}$ and $\lambda \in {\Lambda }_J^m\subset {\Lambda }_J^r.$ From ${\beta }_{k+1}=N^{-1}({\beta }_k+\mathcal{C}\left(J{i}_1\cdots i_k\right))(k⩾1),$ it follows that

$$\begin{array}{cc}\hfill {\beta }_1+{\lambda }^{*}=& N({\beta }_2+\mathcal{C}\left({JK|}_2\right)+N\mathcal{C}\left({JK|}_3\right)+\cdots +N^{r-2}\mathcal{C}\left(JK\right))\hfill \\ \hfill =& \cdots \hfill \\ \hfill =& N^r{\beta }_{r+1}\in N^r\mathbb{Z},\hfill \end{array}$$

which implies ${\left|\widehat{{\mu }_r}({\beta }_1+{\lambda }^{*})\right|}^2=1.$ Noting $\left(\right)$ and $\lambda \ne {\lambda }^{*}$, we have

$$\begin{array}{cc}\hfill 1⩽& {\left|\widehat{{\mu }_r}({\beta }_1+{\lambda }^{*})\right|}^2+{\left|\widehat{{\mu }_r}({\beta }_1+\lambda )\right|}^2⩽\sum _{\gamma \in {\Lambda }_J^r}{\left|\widehat{{\mu }_r}({\beta }_1+\gamma )\right|}^2=1,\hfill \end{array}$$

Thus, we obtain $\left|\widehat{{\mu }_r}({\beta }_1+\lambda )\right|=0.$ Hence,

$$\left|\widehat{\mu }({\beta }_1+\lambda )\right|=0,\forall \lambda \in {\Lambda }_J.$$

It follows that $Q_{{\Lambda }_J}\left({\beta }_1\right)={\sum }_{\lambda \in {\Lambda }_J}{\left|\widehat{\mu }({\beta }_1+\lambda )\right|}^2=0$. □

The following three lemmas play key roles in the proof of Theorem 2 $\left(\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\right).$ First, we show a new criterion for $\Lambda$ to be a spectrum of $\mu$.

Lemma 2.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$. If there exists a positive number $c>0$ such that, for any ξ and $I\in {\Sigma }^{*}$, there is ${\lambda }_{\xi ,I}\in {\Lambda }_I$ satisfying

$$|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})|\ge c,$$

then $(\mu ,\Lambda )$ is a spectral pair.

Proof. 

Suppose $(\mu ,\Lambda )$ is not a spectral pair. Then, there exists ${\xi }_0\in T$ such that $Q_{\Lambda }\left({\xi }_0\right)<1.$

Recall that ${\Lambda }^n=\{\mathcal{C}\left(J{|}_1\right)+N\mathcal{C}\left(J{|}_2\right)+\cdots +N^{n-1}\mathcal{C}\left(J\right):J\in {\Sigma }_q^n\}$ for $n\ge 1$. We write $Q_n\left({\xi }_0\right):={\sum }_{\lambda \in {\Lambda }^n}{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2$. By virtue of ${lim}_{n\to \infty }{\Lambda }^n=\Lambda$ and ${\Lambda }_n\subset {\Lambda }_{n+1}$ for $n\ge 1$, we obtain

$$\underset{n\to \infty }{lim}{Q}_n\left({\xi }_0\right)=Q_{\Lambda }\left({\xi }_0\right)\mathrm{and}{Q}_n\left({\xi }_0\right)⩽Q_{n+1}\left({\xi }_0\right).$$

Given a positive number $\epsilon$ with $\epsilon <\frac{1}{2}(1-Q_{\Lambda }\left({\xi }_0\right))$, there exists an integer $M⩾1$ such that

$$Q_{\Lambda }\left({\xi }_0\right)-\epsilon ⩽Q_M\left({\xi }_0\right)⩽Q_n\left({\xi }_0\right)⩽Q_{\Lambda }\left({\xi }_0\right)<1,\forall n⩾M.$$

(5)

By (1) we have

$$\underset{m\to \infty }{lim}{\widehat{\mu }}_m({\xi }_0+\lambda )=\widehat{\mu }({\xi }_0+\lambda ),\forall \lambda \in \Lambda .$$

In combination with (5), we have a positive integer $K⩾M+1$ such that

$$\sum _{\lambda \in {\Lambda }^M}{\left|{\widehat{\mu }}_K({\xi }_0+\lambda )\right|}^2⩽\sum _{\lambda \in {\Lambda }^M}{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2+\epsilon ⩽Q_{\Lambda }\left({\xi }_0\right)+\epsilon .$$

According to (3), we have ${\sum }_{\lambda \in {\Lambda }^K}{\left|{\widehat{\mu }}_K({\xi }_0+\lambda )\right|}^2=1$. Thus,

$$\begin{array}{cc}\hfill \sum _{I\in {\Sigma }_q^K\backslash {\Sigma }_q^M}{\left|{\widehat{\mu }}_K({\xi }_0+F\left(I\right))\right|}^2& =\sum _{\lambda \in {\Lambda }^K}{\left|{\widehat{\mu }}_K({\xi }_0+\lambda )\right|}^2-\sum _{\lambda \in {\Lambda }^M}{\left|{\widehat{\mu }}_K({\xi }_0+\lambda )\right|}^2\hfill \\ \hfill & ⩾1-Q_{\Lambda }\left({\xi }_0\right)-\epsilon >0.\hfill \end{array}$$

(6)

For any $I\in {\Sigma }_q^K\backslash {\Sigma }_q^M,$ there exists ${\lambda }_{{\xi }_0,I}\in {\Lambda }_I$ such that

$$\left|\widehat{\mu }(N^{-K}({\xi }_0+F\left(I\right))+{\lambda }_{{\xi }_0,I})\right|>c.$$

(7)

Write $\tilde{\Lambda }=\{F\left(I\right)+N^K{\lambda }_{{\xi }_0,I}:I\in {\Sigma }_q^K\backslash {\Sigma }_q^M,{\lambda }_{{\xi }_0,I}\in {\Lambda }_I\}$. It is clear that $\tilde{\Lambda }\subset \Lambda$. Since $(\frac{1}{N}D,S_I)$ is a compatible pair for any $I\in {\Sigma }^{*}$, $\mathcal{C}\left(I\right)=0$ if and only if the finite word I ends with the symbol 0. Then, we have

$${\Lambda }^M\cap \tilde{\Lambda }=\varnothing .$$

In combination with (5)–(7), we obtain

$$\begin{array}{cc}\hfill & Q_{\Lambda }\left({\xi }_0\right)=\sum _{\lambda \in \Lambda }{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2\hfill \\ \hfill & \ge \sum _{\lambda \in {\Lambda }^M}{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2+\sum _{\lambda \in \tilde{\Lambda }}{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2\hfill \\ \hfill & =\sum _{\lambda \in {\Lambda }^M}{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2+\sum _{I\in {\Sigma }^K\backslash {\Sigma }^M}{\left|\widehat{\mu }({\xi }_0+F\left(I\right)+N^K{\lambda }_{{\xi }_0,I})\right|}^2\hfill \\ & =\sum _{\lambda \in {\Lambda }^M}{\left|\widehat{\mu }({\xi }_0+\lambda )\right|}^2+\sum _{I\in {\Sigma }^K\backslash {\Sigma }^M}{\left|{\widehat{\mu }}_K({\xi }_0+F\left(I\right))\right|}^2|\widehat{\mu }{(N^{-K}({\xi }_0+F\left(I\right))+{\lambda }_{{\xi }_0,I}|}^2\hfill \\ & ⩾Q_{\Lambda }\left({\xi }_0\right)-\epsilon +c^2\sum _{I\in {\Sigma }^K\backslash {\Sigma }^M}{\left|{\widehat{\mu }}_K({\xi }_0+F\left(I\right))\right|}^2\hfill \\ & ⩾Q_{\Lambda }\left({\xi }_0\right)-\epsilon +c^2(1-Q_{\Lambda }\left({\xi }_0\right)-\epsilon ).\hfill \end{array}$$

Letting $\epsilon \to 0$, we obtain

$$\begin{array}{c}\hfill 0⩾c^2(1-Q_{\Lambda }\left({\xi }_0\right)),\end{array}$$

which is a contradiction to $Q_{\Lambda }\left({\xi }_0\right))<1$. □

To use Lemma 2, we need the following lemma, which implies that, under some conditions for any point in T, there exists a path that escapes from $\mathcal{Z}(\widehat{\mu },T).$

Lemma 3.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$ and ${inf}_{\xi \in T}{Q}_{{\Lambda }_I}\left(\xi \right)>0$ for any $I\in {\Sigma }_q^{*}$. If $\mathcal{Z}(\widehat{\mu },T)\ne \varnothing$ and for any $\alpha \in \mathcal{Z}(\widehat{\mu },T)$ and $I\in {\Sigma }^{*}$, there exists no $K\in {\Sigma }^{*}$ with $\alpha +F_I\left(K\right)=0$, then for any $\xi \in T$, there exist two nonnegative integers w and v with $1⩽v⩽p+1$ and a finite word $J=j_1\cdots j_{w+v}\in {\Sigma }_q^{*}$ satisfying the following property:

If $w=0$, we have

$$0<\left|m_D\left(N^{-l}(\xi +F_I\left(J{|}_l\right))\right)\right|<1,1⩽l⩽v,$$

and $\left|\widehat{\mu }\left(N^{-v}(\xi +F_I\left(J\right))\right)\right|>0;$

If $w>0$, we have

$$\begin{array}{cc}\hfill & m_D\left(N^{-l}(\xi +F_I\left(J{|}_l\right))\right)=1,1⩽l⩽w,\hfill \\ \hfill & 0<\left|m_D\left(N^{-l}(\xi +F_I\left(J{|}_l\right))\right)\right|<1,w+1⩽l⩽w+v,\hfill \end{array}$$

and $\left|\widehat{\mu }\left(N^{-w-v}(\xi +F_I\left(J\right))\right)\right|>0$.

Proof. 

First, we shall prove the existence of w. If $T\cap \mathbb{Z}=\varnothing$, we take $w=0.$ If $T\cap \mathbb{Z}\ne \varnothing$, since $(\frac{1}{N}D,S_I)$ is a compatible pair, by Lemma 1$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$, there exists $j_1\in {\Sigma }_q$ such that

$$\left|m_D\left(N^{-1}(\xi +F_I\left(j_1\right))\right)\right|>0.$$

(8)

If $\left|m_D\left(N^{-1}(\xi +F_I\left(j_1\right))\right)\right|<1,$ we take $w=0.$ If $\left|m_D\left(N^{-1}(\xi +F_I\left(j_1\right))\right)\right|=1$, also by Lemma 1(iii), there exists $j_2\in {\Sigma }_q$ such that

$$\left|m_D\left(N^{-2}(\xi +F_I\left(j_1{j}_2\right))\right)\right|>0.$$

If $\left|m_D\left(N^{-2}(\xi +F_I\left(j_1{j}_2\right))\right)\right|<1,$ we take $w=1.$ When $\left|m_D\left(N^{-2}(\xi +F_I\left(j_1{j}_2\right))\right)\right|=1$, the process goes on. Under the process, we claim that there exists a finite sequence of symbols ${\left\{j_n\right\}}_{n=1}^w\subset {\Sigma }_q$ such that

$$m_D\left(N^{-l}(\xi +F_I(j_1\cdots j_l))\right)=1,\forall 1⩽l⩽w,$$

and

$$0<\left|m_D\left(N^{-w-1}(\xi +F_I(j_1\cdots j_{w+1}))\right)\right|<1,\forall j_{w+1}\in {\Sigma }_q.$$

(9)

Otherwise, there exists an infinite sequence ${\left\{j_l\right\}}_{l⩾1}\subset {\Sigma }_q$ such that $m_D\left(N^{-l}(\xi +F_I(j_1\cdots j_l))\right)$$=1$ for $l⩾1$. By (2) and the hypothesis of the lemma, we have $N^{-l}(\xi +F_I(j_1\cdots j_l))\in \mathbb{Z}\backslash \left\{0\right\}.$ According to the proof of Theorem (2)$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\right)$, we obtain $Q_{{\Lambda }_I}\left(\xi \right)=0$, which is a contradiction to the condition ${inf}_{\xi \in T}{Q}_{{\Lambda }_I}\left(\xi \right)>0$ for any $I\in {\Lambda }^{*}$.

Next, we shall prove the existence of v. We write $\tilde{J}:=j_1\cdots j_w$ and $\eta :=N^{-w}(\xi +F_I\left(\tilde{J}\right))$, where $\tilde{J}=\vartheta$, $F_I\left(\tilde{J}\right)=0$ and $\eta =\xi$ when $w=0$. In what follows, we define a sequence of sets ${\left\{Y_n\right\}}_{n\ge 0}$ by induction on n. Define $Y_0=\left\{\vartheta \right\}$, and

$$Y_n:=\{L\in {\Sigma }_q^n:L{|}_{n-1}\in Y_{n-1},0<\left|m_D\left(N^{-n}(\eta +F_{I\tilde{J}}\left(L\right))\right)\right|<1\},n\ge 1.$$

We have the following claim. □

Claim: For $n\ge 1$, we have $\#Y_n⩾2^n$.

Proof. 

When $n=1$, since $(\frac{1}{N}D,S_{I\tilde{J}})$ is a compatible pair, there exist two symbols $l_1\ne l_2\in {\Sigma }_q$ such that

$$0<\left|m_D\left(N^{-1}(\eta +F_{I\tilde{J}}\left(l_k\right))\right)\right|<1,1⩽k⩽2.$$

Thus, we obtain $\#Y_1⩾2$. Suppose the inequality $\#Y_n⩾2^n$ holds as $n=k$. Let $n=k+1$. For any $L\in Y_k$, it is clear ${L|}_1\in Y_1$. By (9), we obtain $N^{-1}(\eta +F_{I\tilde{J}}\left(L{|}_1\right))\notin \mathbb{Z}$. Thus, $N^{-k}(\eta +F_{I\tilde{J}}\left(L\right))\notin \mathbb{Z}$. Since $(\frac{1}{N}D,S_{I\tilde{J}L})$ is a compatible pair, there exist at least two symbols $l_1\ne l_2\in {\Sigma }_q$ such that

$$0<\left|m_D\left(N^{-n-1}(\eta +F_{I\tilde{J}}\left(L{l}_k\right))\right)\right|<1,1⩽k⩽2.$$

By the arbitrariness of $L\in Y_k$, we obtain $\#Y_{n+1}⩾2^{n+1}$. Hence, the claim follows by induction.

Together with Proposition 2, the above claim implies

$$\#\{N^{-p-1}(\alpha +F_{I\tilde{J}}\left(L\right):L\in Y_{p+1}\}=\#Y_{p+1}⩾2^{p+1}>p.$$

Thus, by $p=♯\mathcal{Z}(\widehat{\mu },T)$, there exists a finite word $L\in Y_{p+1}$ such that

$$\left|\widehat{\mu }\left(N^{-p-1}(\eta +F_{I\tilde{J}}\left(L\right))\right)\right|>0.$$

Let $v⩾1$ be the smallest positive integer such that $\left|\widehat{\mu }\left(N^{-v}(\eta +F_{I\tilde{J}}\left(L\right))\right)\right|>0$ for some $L=l_1\cdots l_v$. By taking $J=\tilde{J}{l}_1\cdots l_v$, we finish the proof. □

Lemma 4.

If $T\cap \mathbb{Z}\ne \varnothing ,$ then there exists ${\alpha }_1>0$ such that, for any integer sequence ${\left\{{\theta }_i\right\}}_{i⩾1}\subset T\cap \mathbb{Z}$, we have

$$\prod _{i=1}^{\infty }\left|m_D\left(x_i\right)\right|⩾{\alpha }_1,$$

where $x_i\in B({\theta }_i,N^{-i})$.

Proof. 

For any $\theta \in T\cap \mathbb{Z}$, we have $m_D\left(\theta \right)=1$. On the other hand, the mask function $m_D$ can be extended to an entire function on the complex plane. Thus, $m_D$ is uniformly continuous on any compact set. Hence, there exists a positive number $c_1$ such that

$$|1-m_D\left(x\right)|=|m_D\left(\theta \right)-m_D\left(x\right)|⩽c_1|x-\theta |,\forall x\in \{\xi +y:\xi \in T,|y|\le 1\}.$$

Given a sequence ${\left\{{\theta }_i\right\}}_{i⩾1}\subset T\cap \mathbb{Z}$, we have

$$\left|m_D\left(x_i\right)\right|⩾1-c_1|x_i-{\theta }_i|⩾1-N^{-i}{c}_1,\forall x_i\in B({\theta }_i,N^{-i}),i⩾1.$$

It is clear that there exists a positive integer $K>0$ such that, for $k⩾K$, we have $N^{-k}{c}_1<\frac{1}{2}.$ Note an elementary inequality:

$$1-x⩾e^{-2x},0\le x⩽\frac{1}{2}.$$

Then, we have

$$\begin{array}{cc}\hfill \prod _{i=1}^{\infty }\left|m_D\left(x_i\right)\right|=& \prod _{i=1}^K\left|m_D\left(x_i\right)\right|\prod _{i=K+1}^{\infty }\left|m_D\left(x_i\right)\right|\hfill \\ \hfill ⩾& {\left(\frac{1}{2}\right)}^K\prod _{i=K+1}^{\infty }{e}^{-2c_1{N}^{-i}}\hfill \\ \hfill =& {\left(\frac{1}{2}\right)}^K{e}^{{\sum }_{i=K+1}^{\infty }-2c_1{N}^{-i}}\hfill \\ \hfill =& {\left(\frac{1}{2}\right)}^K{e}^{-2c_1\frac{1}{N^K(N-1)}}=:{\alpha }_1>0\hfill \end{array}$$

(10)

for all $x_i\in B({\theta }_i,N^{-i})$. The proof is complete. □

Proof of Theorem 2 

$\left(\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\right)$). We expect to obtain a contradiction after assuming

$$\underset{\xi \in T}{inf}{Q}_{{\Lambda }_I}\left(\xi \right)>0,\forall I\in {\Sigma }_q^{*}.$$

(11)

We shall prove that there is a positive number $c>0$ such that, for any $\xi \in T$ and $I\in {\Sigma }^{*}$, there exists ${\lambda }_{\xi ,I}\in {\Lambda }_I$ satisfying

$$|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})|\ge c.$$

If $\mathcal{Z}(\widehat{\mu },T)=\varnothing$, then $\widehat{\mu }\left(\xi \right)$ has a positive lower bound on compact set T. Write $c:={inf}_{\xi \in T}\left|\widehat{\mu }\left(\xi \right)\right|>0$. For any $\xi \in T$ and $I\in {\Sigma }_q^{*},$ take ${\lambda }_{\xi ,I}=0\in {\Lambda }_I.$ Noting $N^{-\left|I\right|}(\xi +F\left(I\right))\in T,$ we have

$$\left|\widehat{\mu }\left(N^{-\left|I\right|}(\xi +F\left(I\right))\right)\right|⩾c.$$

From Lemma 2, it follows that $(\mu ,\Lambda )$ is a spectral pair, which is a contradiction to the hypothesis. □

Next, we focus on the case $\mathcal{Z}(\widehat{\mu },T)\ne \varnothing$. We shall deal with two cases.

Case i. For any $\eta \in \mathcal{Z}(\widehat{\mu },T)$ and $I\in {\Sigma }^{*}$, there exists $J\in {\Sigma }^{*}$ such that

$$\eta +F_I\left(J\right)=0.$$

(12)

By $\left|\widehat{\mu }\left(0\right)\right|=1$, there exists a positive number $\delta$ with $0<{\delta }_1<1$ such that

$$\left|\widehat{\mu }\left(x\right)\right|>\frac{1}{2},\forall x\in B(0,{\delta }_1).$$

(13)

Write $\delta :=min\{{\delta }_1,\frac{d}{4}\}$, where d denotes the smallest distance between different points in $\mathcal{Z}(\widehat{\mu },T)\cup (T\cap \mathbb{Z})$, i.e., $d:=min\{|x-y|:x\ne y\in \mathcal{Z}(\widehat{\mu },T)\cup T\cap \mathbb{Z}\}.$

We denote the set of points that has a positive distance from the zero points of $\widehat{\mu }\left(\xi \right)$ in T by

$$P:=T\backslash \left(\bigcup _{\theta \in \mathcal{Z}(\widehat{\mu },T)}B(\theta ,\delta )\right).$$

It is clear that P is a compact set and ${\alpha }_0:={inf}_{\xi \in P}\left|\widehat{\mu }\left(\xi \right)\right|>0.$ Write $\alpha :=min\{\frac{1}{2}{\alpha }_1,{\alpha }_0\}$. Given $\xi \in T$ and $I\in {\Sigma }^{*}$, define $\tilde{\xi }=N^{-\left|I\right|}(\xi +F\left(I\right))$.

If $\tilde{\xi }\in P$, we take ${\lambda }_{\xi ,I}=0$. Then,

$$|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})|=|\widehat{\mu }\left(\tilde{\xi }\right)|\ge {\alpha }_0\ge \alpha .$$

(14)

If $\tilde{\xi }\notin P$, by the definition of P, there exists a unique $\theta \in \mathcal{Z}(\widehat{\mu },T)\subset T\backslash \left\{0\right\}$ such that $\tilde{\xi }\in B(\theta ,\delta ).$ According to (12), there exists $J\in {\Sigma }^{*}$ such that

$$\theta +F_I\left(J\right)=0.$$

(15)

Take ${\lambda }_{\xi ,I}=F_I\left(J\right)$. Then, we have

$$N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\in B(N^{-l}(\theta +F_I\left(J{|}_l\right)),N^{-l}\delta ),1\le l\le \left|J\right|.$$

(16)

On the other hand, by (15), we have

$$N^{-l}(\theta +F_I\left(J{|}_l\right))\in \mathbb{Z}\cap T,1\le l\le \left|J\right|.$$

In combination with Lemma 4 and (16), this leads to

$$\prod _{l=1}^{\left|J\right|}\left|m_D\left(N^{-l}(\theta +F_I\left(J{|}_l\right))\right)\right|>{\alpha }_1.$$

(17)

Furthermore, by (16) we have

$$N^{-\left|J\right|}(\tilde{\xi }+F_I\left(J\right))\in B(N^{-\left|J\right|}(\theta +F_I\left(J\right)),N^{-\left|J\right|}\delta )\subset B(0,{\delta }_1).$$

Then, by (13), we have $|\widehat{\mu }\left(N^{-\left|J\right|}(\tilde{\xi }+F_I\left(J\right))\right)|\ge \frac{1}{2}$. Together with (17), this inequality implies

$$\begin{array}{cc}\hfill \left|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})\right|=& \left|\widehat{\mu }(\tilde{\xi }+F_I\left(J\right))\right|\hfill \\ \hfill =& \prod _{l=1}^{\left|J\right|}\left|m_D\left(N^{-l}(\theta +F_I\left(J{|}_l\right))\right)\right|\left|\widehat{\mu }\left(N^{-\left|J\right|}(\tilde{\xi }+F_I\left(J\right))\right)\right|\hfill \\ \hfill ⩾& \frac{1}{2}{\alpha }_1\hfill \\ \hfill ⩾& \alpha .\hfill \end{array}$$

(18)

Case ii: There exist ${\eta }^{*}\in \mathcal{Z}(\widehat{\mu },T)$ and $I\in {\Sigma }^{*}$ such that, for any $J\in {\Sigma }^{*}$, we have

$${\eta }^{*}+F_I\left(J\right)\ne 0.$$

(19)

Recall that $\tilde{S}={\bigcup }_{I\in {\Sigma }^{*}}{S}_I$ and $p=♯\mathcal{Z}(\widehat{\mu },T)$. Let

$$U:=\bigcup _{l=1}^{p+1}\left\{N^{-l}(\theta +\lambda ):\lambda \in \tilde{S}+N\tilde{S}+\cdots +N^{l-1}\tilde{S},\theta \in \mathcal{Z}(\widehat{\mu },T)\cup (T\cap \mathbb{Z})\right\}.$$

Furthermore, we write

$$V=\{x\in U:\left|m_D\left(x\right)\right|\ne 0\}\mathrm{and}W=\{x\in V:\left|\widehat{\mu }\left(x\right)\right|\ne 0\}.$$

It is clear $W\subset V\subset U\subset T$. Since $(\frac{1}{N}D,S_I)$ is a compatible pair for any $I\in {\Sigma }^{*}$, we obtain $V\ne \varnothing .$

Next, we shall prove $W\ne \varnothing .$

Claim 1: There exists $a\in \tilde{S}$ such that

$$0<\left|m_D\left(N^{-1}({\eta }^{*}+a)\right)\right|<1.$$

Proof. 

If $T\cap \mathbb{Z}=\varnothing$, then we have $sup\{\left|m_D\left(\eta \right)\right|:\eta \in T\}<1$ by noting that T is compact. A trivial fact that $N^{-1}({\eta }^{*}+a)\in T$ for any $a\in \tilde{S}$ implies the claim is true.

When $T\cap \mathbb{Z}\ne \varnothing$, suppose the claim is false. Since $(\frac{1}{N}D,S_I)$ is a compatible pair, by Lemma 1(iii) for ${\eta }^{*}\in \mathcal{Z}(\widehat{\mu },T)$, there exists $j_1\in {\Sigma }_q$ such that $m_D\left(N^{-1}({\eta }^{*}+F_I\left(j_1\right))\right)=1$. By (2) and (19), we obtain $N^{-1}({\eta }^{*}+F_I\left(j_1\right))\in (T\cap \mathbb{Z})\backslash \left\{0\right\}$. Furthermore, there exists $j_2\in {\Sigma }_q$ such that $m_D\left(N^{-2}({\eta }^{*}+F_I\left(j_1{j}_2\right))\right)=1$, which implies $N^{-2}({\eta }^{*}+F_I\left(j_1{j}_2\right))\in (T\cap \mathbb{Z})\backslash \left\{0\right\}$. Repeating this process, we obtain a sequence of symbols ${\left\{j_l\right\}}_{l⩾1}\subset {\Sigma }_q$ such that

$$N^{-l}({\eta }^{*}+F_I(j_1\cdots j_l))\in (T\cap \mathbb{Z})\backslash \left\{0\right\},l⩾1.$$

By a similar argument in the proof of Theorem (2)$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\right)$, we obtain $Q_{{\Lambda }_I}\left({\eta }^{*}\right)=0$, which implies a contradiction to (11). The claim is proven.

Next, we define a sequence of set ${\left\{Y_n\right\}}_{n\ge 0}$ by induction on n. Let $Y_0:=\left\{{\eta }^{*}\right\},$ and

$$Y_n:=\{N^{-1}(\eta +a):0<\left|m_D\left(N^{-1}(\eta +a)\right)\right|<1,\eta \in Y_{n-1},a\in \tilde{S}\},n⩾1.$$

By a similar argument in the proof of the claim in Lemma (3), we obtain $\#Y_n\ge 2^n$ for $1\le n\le p+1$. On the other hand, for any $\eta \in Y_{p+1},$ there exists $\lambda \in \tilde{S}+N\tilde{S}+\cdots +N^p\tilde{S}$ such that $\eta =N^{-p-1}({\eta }^{*}+\lambda )$ and $0<\left|m_D\left(\eta \right)\right|<1$, which implies $Y_{p+1}\subset V.$ Then, we conclude

$$\#V⩾\#Y_{p+1}⩾2^{p+1}>p.$$

Recall that p is the number of zero points of $\widehat{\mu }\left(\xi \right)$ on compact T. Then, we obtain $W\ne \varnothing .$

Noting that $W\subset V\subset U$ and U is a finite set, it is obvious that both W and V are finite sets. Write

$$\begin{array}{cc}\hfill & {\alpha }_2:=min\left\{\left|m_D\left(\eta \right)\right)\right|\ne 0:\eta \in V\}>0,\hfill \\ \hfill & {\alpha }_3:=min\{\left|\widehat{\mu }\left(\eta \right)\right|\ne 0:\eta \in W\}>0.\hfill \end{array}$$

Then, there exists a positive number ${\delta }_2>0$ such that, for any $\eta \in V$ and $\omega \in W,$ we have

$$\left|m_D\left(x\right)\right|>\frac{1}{2}{\alpha }_2,\forall x\in B(\eta ,{\delta }_2),$$

(20)

$$\left|\widehat{\mu }\left(x\right)\right|>\frac{1}{2}{\alpha }_3,\forall x\in B(\omega ,{\delta }_2).$$

(21)

Write $\tilde{\delta }:=min\left\{{\delta }_1,{\delta }_2,\frac{d}{4}\right\}$. We let $\tilde{P}:=T\backslash \left({\bigcup }_{\theta \in \mathcal{Z}(\widehat{\mu },T)}B(\theta ,\tilde{\delta })\right)$ denote the set of points that has a positive distance (at least $\tilde{\delta }$) from the zero points of $\widehat{\mu }\left(\xi \right)$ in T. It is clear that $\tilde{P}$ is a compact set and ${\alpha }_4:={inf}_{\xi \in \tilde{P}}\left|\widehat{\mu }\left(\xi \right)\right|>0.$ We write

$$\tilde{\alpha }:=min\{{\alpha }_1\frac{{\alpha }_3}{2}{\left(\frac{{\alpha }_2}{2}\right)}^{p+1},{\alpha }_4\},$$

where ${\alpha }_1$ comes from Lemma 4.

Given $\xi \in T$ and $I\in {\Sigma }_q^{*}$, write $\tilde{\xi }:=N^{-\left|I\right|}(\xi +F\left(I\right)).$

If $\tilde{\xi }\in \tilde{P},$ we take ${\lambda }_{\xi ,I}=0\in {\Lambda }_I.$ Then, we have

$$\left|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})\right|=\left|\widehat{\mu }\left(\tilde{\xi }\right)\right|⩾{\alpha }_4⩾\tilde{\alpha }.$$

(22)

If $\tilde{\xi }\notin \tilde{P},$ there exists $\theta \in \mathcal{Z}(\widehat{\mu },T)$ such that $\tilde{\xi }\in B(\theta ,\tilde{\delta }).$ If there exists $J\in {\Sigma }^{*}$ such that

$$\theta +F_I\left(J\right)=0,$$

we take ${\lambda }_{\xi ,I}=F_I\left(J\right).$ Then, by a similar argument as (18), we have

$$|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})|\ge \alpha .$$

(23)

If there is no $J\in {\Sigma }^{*}$ such that

$$\theta +F_I\left(J\right)=0,$$

by Lemma 3 there exist two integers $0⩽w<\infty ,1⩽v⩽p+1$ and a finite word $J:=j_1\cdots j_{w+v}\in {\Sigma }_q^{*}$ such that when $w=0$, we have

$$0<\left|m_D\left(N^{-l}(\theta +F_I\left(J{|}_l\right))\right)\right|<1,1⩽l⩽v,$$

(24)

and $\left|\widehat{\mu }\left(N^{-v}(\theta +F_I\left(J\right))\right)\right|>0$; when $w>0$, we have

$$\begin{array}{ccc}\hfill & m_D\left(N^{-l}(\theta +F_I\left(J{|}_l\right))\right)=1,\hfill & \hfill 1⩽l⩽w,\end{array}$$

(25)

$$\begin{array}{ccc}\hfill & 0<\left|m_D\left(N^{-l}(\theta +F_I\left(J{|}_l\right))\right)\right|<1,\hfill & \hfill w+1⩽l⩽w+v,\end{array}$$

(26)

and $\left|\widehat{\mu }\left(N^{-w-v}(\theta +F_I\left(J\right))\right)\right|>0.$

Take ${\lambda }_{\xi ,I}:=F_I\left(J\right)$. In the case $w=0$, since $\tilde{\xi }\in B(\theta ,\tilde{\delta })$, it is obvious that

$$N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\in B(N^{-l}(\theta +F_I\left(J{|}_l\right)),N^{-l}\tilde{\delta }),1⩽l⩽v.$$

(27)

Noting that $\theta \in \mathcal{Z}(\widehat{\mu },T)\cup (T\cap \mathbb{Z})$, by (24) we obtain

$$N^{-l}(\theta +F_I\left(J{|}_l\right))\in V,1⩽l⩽v.$$

Together with (20) and (27), the above inequality implies

$$\left|m_D\left(N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\right)\right|>\frac{{\alpha }_2}{2},1⩽l⩽v.$$

(28)

Furthermore, since $N^{-v}(\theta +F_I\left(J\right))\in V$ and $\left|\widehat{\mu }\left(N^{-v}(\theta +F_I\left(J\right))\right)\right|>0,$ we have $N^{-v}(\theta +F_I\left(J\right))\in W$ and $N^{-v}(\tilde{\xi }+F_I\left(J\right))\in B(N^{-v}(\theta +F_I\left(J\right)),N^{-v}\tilde{\delta })$. From (21) it follows that

$$\left|\widehat{\mu }\left(N^{-v}(\tilde{\xi }+F_I\left(J{|}_l\right))\right)\right|⩾\frac{{\alpha }_3}{2}.$$

(29)

In combination with (28) this yields

$$\begin{array}{cc}\hfill \left|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})\right|=& \left|\widehat{\mu }(\tilde{\xi }+{\lambda }_{\xi ,I})\right|\hfill \\ \hfill =& \prod _{i=1}^{\infty }\left|m_D\left(N^{-i}(\tilde{\xi }+F_I\left(J\right))\right)\right|\hfill \\ \hfill =& \prod _{i=1}^v|m_D\left(N^{-i}(\tilde{\xi }+F_I\left(J\right))\right|\left|\widehat{\mu }\left(N^{-v}(\tilde{\xi }+F_I\left(J\right))\right)\right|\hfill \\ \hfill ⩾& \frac{{\alpha }_3}{2}{\left(\frac{{\alpha }_2}{2}\right)}^{p+1}\hfill \\ \hfill ⩾& \tilde{\alpha }.\hfill \end{array}$$

(30)

In the case $w>0$, we shall divide the product into three parts

$$\begin{array}{cc}\hfill & \left|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})\right|\hfill \\ \hfill =& \prod _{i=1}^{\infty }\left|m_D\left(N^{-i}(\tilde{\xi }+F_I\left(J\right))\right)\right|\hfill \\ \hfill =& \prod _{i=1}^w\left|m_D\left(N^{-i}(\tilde{\xi }+F_I\left(J\right))\right)\right|\prod _{i=w+1}^{w+v}\left|m_D\left(N^{-i}(\tilde{\xi }+F_I\left(J\right))\right)\right|\left|\widehat{\mu }\left(N^{-w-v}(\tilde{\xi }+F_I\left(J\right))\right)\right|.\hfill \end{array}$$

(31)

By (2) and (25) we have

$$N^{-l}(\theta +F_I\left(J{|}_l\right))\in T\cap \mathbb{Z},1⩽l⩽w.$$

(32)

Noting $\tilde{\xi }\in B(\theta ,\tilde{\delta })$, we have

$$N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\in B(N^{-l}(\theta +F_I\left(J{|}_l\right)),N^{-l}\tilde{\delta }),1⩽l⩽w.$$

Thus, by (10), we obtain

$$\prod _{l=1}^w\left|m_D\left(N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\right)\right|⩾{\alpha }_1.$$

(33)

By (32), we have $N^{-w}(\theta +F_I\left(J{|}_w\right))\in \mathcal{Z}(\widehat{\mu },T)\cup (T\cap \mathbb{Z})$. Then, by (26) we have

$$N^{-l}(\theta +F_I\left(J{|}_l\right))\in V,w+1⩽l⩽w+v$$

and

$$N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\in B(N^{-l}(\theta +F_I\left(J{|}_l\right)),N^{-l}\tilde{\delta }),w+1⩽l⩽w+v.$$

(34)

By (20) and (21), we obtain

$$\left|m_D\left(N^{-l}(\tilde{\xi }+F_I\left(J{|}_l\right))\right)\right|>\frac{{\alpha }_2}{2},w+1⩽l⩽w+v,$$

(35)

and

$$\left|\widehat{\mu }\left(N^{-w-v}(\tilde{\xi }+F_I\left(J\right))\right)\right|⩾\frac{{\alpha }_3}{2}.$$

Together with (31), (33), and (35), the above inequality yields

$$\left|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})\right|\ge {\alpha }_1{\left(\frac{{\alpha }_2}{2}\right)}^{p+1}\frac{{\alpha }_3}{2}\ge \tilde{\alpha }.$$

(36)

In combination with (14), (18), (22), (23), (30), and (36), by Lemma 2, we obtain $(\mu ,\Lambda )$ is a spectral pair, which is a contradiction to our hypothesis. We finish the proof of $\left(\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\right)$ in Theorem (2) □

Finally, we shall prove Theorem (2) $\left(\mathrm{i}\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$.

Since T is compact, there exists ${\xi }^{*}\in T$ such that $Q_{{\Lambda }_I}\left({\xi }^{*}\right)=0.$ Write

$$X:=\{\xi \in T:\widehat{\mu }\left(\xi \right)=0\mathrm{and}{m}_D\left(\xi \right)\ne 0\}.$$

It is clear that $0\notin X$. Since $(\frac{1}{N}D,S_I)$ is a compatible pair, by Lemma (1), there exists an integer $j\in {\Sigma }_q$ with $m_D\left(\frac{1}{N}({\xi }^{*}+F_I\left(j\right))\right)\ne 0$. Noting that

$$0=Q_{{\Lambda }_I}\left({\xi }^{*}\right)={\Sigma }_{\lambda \in {\Lambda }_I}|\widehat{\mu }({\xi }^{*}+\lambda ){|}^2\ge |m_D\left(\frac{1}{N}({\xi }^{*}+F_I\left(j\right))\right){|}^2{\left|\widehat{\mu }\left(\frac{1}{N}({\xi }^{*}+F_I\left(j\right))\right)\right|}^2,$$

we obtain $\widehat{\mu }\left(\frac{1}{N}({\xi }^{*}+F_I\left(j\right))\right)=0$. By virtue of ${\xi }^{*}\in T$, we have $\frac{1}{N}({\xi }^{*}+F_I\left(j\right))\in T$. Hence, X is nonempty.

Next, we define a sequence of the subset of X by induction on n. Define $X_0:=\left\{{\xi }^{*}\right\}$ and

$$X_{n+1}:=\{N^{-n-1}(\xi +F_I\left(J\right))\in X:N^{-n}(\xi +F_I\left(J{|}_n\right))\in X_n,J\in {\Sigma }_q^{n+1}\},n\ge 0.$$

We have the following conclusion.

Claim 2: $\#X_{n+1}⩾\#X_n$, $n⩾0$.

Proof. 

When $n=0$, by the definition of $Q_{{\Lambda }_I}\left({\xi }^{*}\right)$, we have

$$\begin{array}{cc}\hfill & 0=Q_{{\Lambda }_I}\left({\xi }^{*}\right)=\sum _{j_1\in {\Sigma }_q}{\left|m_D\left(N^{-1}({\xi }^{*}+F_I\left(j_1\right))\right)\right|}^2·Q_{{\Lambda }_{I{j}_1}}\left(N^{-1}({\xi }^{*}+F_I\left(j_1\right))\right).\hfill \end{array}$$

Noting that $(\frac{1}{N}D,S_I)$ is a compatible pair, Lemma (1)$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$ implies that there exists at least one symbol $j_1\in {\Sigma }_q$ such that $\left|m_D\left(N^{-1}({\xi }^{*}+F_I\left(j_1\right))\right)\right|>0,$ which implies $Q_{{\Lambda }_{I{j}_1}}\left(N^{-1}({\xi }^{*}+F_I\left(j_1\right))\right)=0.$ Hence, we have $\widehat{\mu }\left(N^{-1}({\xi }^{*}+F_I\left(j_1\right))\right)=0$. This leads to $\#X_1⩾\#X_0.$ Suppose Claim 2 holds for $n=k-1$. Then, $X_k$ is nonempty. For any $y\in X_k$, there exists $\tilde{J}\in {\Sigma }_q^k$ such that $y=N^{-k}({\xi }^{*}+F_I\left(\tilde{J}\right))$ and

$$\prod _{i=1}^k\left|m_D\left(N^{-i}({\xi }^{*}+F_I\left(\tilde{J}{|}_i\right))\right)\right|>0.$$

By (1) and (4), we have

$$\begin{array}{c}\hfill 0=Q_{{\Lambda }_I}\left({\xi }^{*}\right)=\sum _{\tilde{J}\in {\Sigma }_q^k}\prod _{i=1}^k{\left|m_D\left(N^{-i}({\xi }^{*}+F_I\left(\tilde{J}{|}_i\right))\right)\right|}^2·Q_{{\Lambda }_{I\tilde{J}}}\left(N^{-k}({\xi }^{*}+F_I\left(\tilde{J}\right))\right).\end{array}$$

Then, we obtain $Q_{{\Lambda }_{IJ}}\left(N^{-k}({\xi }^{*}+F_I\left(\tilde{J}\right))\right)=0.$ By a similar argument, we have

$$\begin{array}{cc}\hfill 0=& Q_{{\Lambda }_{I\tilde{J}}}\left(N^{-k}({\xi }^{*}+F_I\left(\tilde{J}\right))\right)\hfill \\ \hfill =& \sum _{j_{k+1}\in {\Sigma }_q}{\left|m_D\left(N^{-k-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{k+1}\right))\right)\right|}^2·Q_{{\Lambda }_{I\tilde{J}{j}_{k+1}}}\left(N^{-k-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{k+1}\right))\right).\hfill \end{array}$$

Noting that $(\frac{1}{N}D,S_{I\tilde{J}})$ is a compatible pair, by Lemma (1)$\left(\mathrm{i}\mathrm{i}\mathrm{i}\right),$ there exists at least one symbol $j_{k+1}\in {\Sigma }_q$ such that

$$\left|m_D\left(N^{-k-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{k+1}\right))\right)\right|>0.$$

Hence, $Q_{{\Lambda }_{I\tilde{J}{j}_{k+1}}}\left(N^{-k-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{k+1}\right))\right)=0,$ which implies $\widehat{\mu }\left(N^{-k-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{k+1}\right))\right)=0.$ Thus, we obtain

$$N^{-k-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{k+1}\right))\in X_{k+1}.$$

If we consider $N^{-n-1}({\xi }^{*}+F_I\left(\tilde{J}{j}_{n+1}\right))$ as a “next generation” of $N^{-n}({\xi }^{*}+F_I\left(\tilde{J}\right))$ for $n\ge 1$, Proposition 2 implies that different points of $X_k$ have different “next generations”. Thus, we obtain $\#X_{k+1}⩾\#X_k$, which implies Claim 2 is true.

By noting the fact that X is a subset of the finite set $\mathcal{Z}(\widehat{\mu },T)$, there exists a positive integer $h\in \mathbb{N}$ such that

$$\#X_{h+m}=\#X_h,m\ge 1.$$

(37)

From the above argument, it follows that for any $y=N^{-n}({\xi }^{*}+F_I(j_1\cdots j_n))\in X_n$, if there exists a symbols $j_{n+1}\in {\Sigma }_q$ such that $\left|m_D\left(N^{-n-1}({\xi }^{*}+F_I(j_1\cdots j_n{j}_{n+1}))\right)\right|>0$, then y has a “next generation” $N^{-n-1}({\xi }^{*}+F_I(j_1\cdots j_n{j}_{n+1}))\in X_{n+1}$. Noting that $(\frac{1}{N}D,S_{I{j}_1\cdots j_n})$ is a compatible pair, by Lemma 1 (iii), we have

$${\Sigma }_{j_{n+1}\in {\Sigma }_q}{\left|m_D\left(N^{-1}(y+\mathcal{C}\left(IJ{j}_{n+1}\right))\right)\right|}^2=1.$$

In combination with (37), we conclude that for any $n\ge h$, there exists only one symbol $j_{n+1}\in {\Sigma }_q$ such that $|m_D\left(N^{-1}(y+\mathcal{C}\left(IJ{j}_{n+1}\right))\right|\ne 0.$ In fact, $|m_D\left(N^{-1}(y+\mathcal{C}\left(IJ{j}_{n+1}\right))\right|=1$. Then, we obtain

$$N^{-n-1}({\xi }^{*}+F_I\left(J{j}_{n+1}\right))=N^{-1}(y+\mathcal{C}\left(IJ{j}_{n+1}\right)\in \mathbb{Z}.$$

Continuing the process, we obtain a sequence of symbols ${\left\{j_{h+l}\right\}}_{l⩾1}\subset {\Sigma }_q$, such that

$$N^{-h-l}({\xi }^{*}+F_I(J{j}_{h+1}\cdots j_{h+l}))\in \mathbb{Z},l⩾1.$$

Define ${\beta }_1:=N^{-h}({\xi }^{*}+F_I\left(J\right))$ and

$${\beta }_l:=N^{-h-l+1}({\xi }^{*}+F_I(J{j}_{h+1}\cdots j_{h+l-1})),l⩾2.$$

It is clear ${\beta }_l\in X_{h+l-1}$, which implies ${\beta }_l$ is nonzero. Thus, the sequence of nonzero integers ${\left\{{\beta }_l\right\}}_{l⩾1}$ and the increasing sequence of finite words ${\{J{j}_{h+1}\cdots j_{h+l}\}}_{l⩾1}$ with the prefix J fulfill the request. □

As a corollary of Lemma 2 and Theorem 2, we obtain another necessary and sufficient condition for $\Lambda$ to be a spectrum of $\mu$.

Proposition 5.

Let $N\in \mathbb{Z}$ with $\left|N\right|>1$ and $D\subset \mathbb{Z}$ with $0\in D$ and $gcd\left(D\right)=1$. Assume that a countable set Λ satisfies the conditions $\left(\mathrm{C}1\right),\left(\mathrm{C}2\right)$, and $\left(\mathrm{C}3\right)$. Then, $(\mu ,\Lambda )$ is a spectral pair if and only if there exists a positive number $c>0$ such that, for any ξ and $I\in {\Sigma }^{*}$, there is ${\lambda }_{\xi ,I}\in {\Lambda }_I$ satisfying

$$|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})|\ge c.$$

Proof. 

The sufficiency follows from Lemma 2. We just prove the necessity here. Suppose that $(\mu ,\Lambda )$ is a spectral pair. By Propositions 3 and 4, we obtain, for any $I\in {\Sigma }^{*}$,

$$Q_{{\Lambda }_I}\left(\xi \right)\equiv 1,\xi \in \mathbb{R}.$$

By a similar argument in the proof of Theorem (2) $\left(\mathrm{i}\right)\Rightarrow \left(\mathrm{i}\mathrm{i}\right)$, for any $\xi \in T$ and $I\in T$, there exists ${\lambda }_{\xi ,I}\in {\Lambda }_I$ such that

$$|\widehat{\mu }(N^{-\left|I\right|}(\xi +F\left(I\right))+{\lambda }_{\xi ,I})|\ge c.$$

We finish the proof. □

4. An Example

In this section, we construct a self-similar measure and a set $\Lambda (N,\mathcal{B})$ with a tree structure. Neither the criterion of Łaba and Wang (Theorem 1) nor that of Strichartz ([24]) are applicable to this set $\Lambda (N,\mathcal{B})$. However, we show that there does not exist an infinite orbit ${\left\{{\beta }_l\right\}}_{l\ge 1}\subset \mathbb{Z}\backslash \left\{0\right\}$ associated with the dual IFS (see Theorem 3), which implies $\Lambda (N,\mathcal{B})$ is a spectrum by Theorem 2.

Example 1.

Let $N=6$ and $D=\{0,1,2\}$. Write μ for the invariant measure associated with the IFS $\{{\varphi }_1,{\varphi }_2,{\varphi }_3\}$ defined by

$${\varphi }_1\left(x\right)=\frac{1}{6}x,{\varphi }_2\left(x\right)=\frac{1}{6}(x+1),{\varphi }_3\left(x\right)=\frac{1}{6}(x+2).$$

Let $B_1=\{0,8,22\},B_2=\{0,22,38\}$, $B_3=\{0,8,52\}$, and $B_4=\{0,38,52\}$. By Lemma 1, a simple induction implies that $(\frac{1}{6}D,B_i)$ is a compatible pair for $1\le i\le 4$. Noting

$$\frac{1}{6}(4+8)=2,\frac{1}{6}(2+22)=4,\frac{1}{6}(4+8)=2,\frac{1}{6}(2+22)=4,\cdots ,$$

$$\frac{1}{6}(10+38)=8,\frac{1}{6}(8+52)=10,\frac{1}{6}(10+38)=8,\frac{1}{6}(8+52)=10,\cdots ,$$

we see that both $\Lambda (6,B_1)$ and $\Lambda (6,B_4)$ have an infinite iterated nonzero integer sequence, where $\Lambda (N,S):=S+NS+N^{2}S+\cdots \mathrm{finite}\mathrm{sum}$. Thus, by Theorem 1 or by Theorem 2, we conclude that both $\Lambda (6,B_1)$ and $\Lambda (6,B_4)$ are not a spectrum of μ. We consider the following set defined by $\{B_i:1\le i\le 4\}$.

$$\begin{array}{cc}\hfill & \Lambda (N,\mathcal{B}):=B_1+\underset{B_2\mathrm{and}{B}_3\mathrm{repeat}1\mathrm{time}}{\underbrace{N{B}_2+N^2{B}_3}}+\hfill \\ & N^3{B}_4+\underset{B_3\mathrm{and}{B}_2\mathrm{repeat}2\mathrm{times}}{\underbrace{N^4{B}_3+N^5{B}_2+N^6{B}_3+N^7{B}_2}}+\hfill \\ & N^8{B}_1+\underset{B_2\mathrm{and}{B}_3\mathrm{repeat}{2}^2\mathrm{times}}{\underbrace{N^9{B}_2+N^{10}{B}_3+N^{11}{B}_2+N^{12}{B}_3+N^{13}{B}_2+N^{14}{B}_3+N^{15}{B}_2+N^{16}{B}_3}}+\hfill \\ & N^{17}{B}_4+\underset{B_3\mathrm{and}{B}_2\mathrm{repeat}{2}^3\mathrm{times}}{\underbrace{N^{18}{B}_3+N^{19}{B}_2+\cdots +N^{32}{B}_3+N^{33}{B}_2}}+\cdots (\mathrm{finite}\mathrm{sum}).\hfill \end{array}$$

(38)

According to Remark 1, it is clear that Theorem 1 cannot work. We shall show $\Lambda (N,\mathcal{B})$ is a spectrum of μ by Theorem 2 in the following Theorem 3. Then, we show that Strichartz’s criterion (Theorem 2.8 in [24]) is not appropriate by proving the following Theorem 4.

Let $A_n$ denote the set of coefficients of $N^n(n\ge 0)$ in (38). Given two integers l and k with $l>k\ge 0$, we write

$${\Lambda }_k^l:=A_k+N{A}_{k+1}+N^2{A}_{k+2}+\cdots +N^{l-k-1}{A}_{l-1}.$$

(39)

We also write ${\Lambda }^k:={\Lambda }_0^k$ for simplicity. For three integers $m,n$, and k with $0\le m<n<k$, we have

$$\begin{array}{cc}\hfill & {\Lambda }_m^n+N^{n-m}{\Lambda }_n^k\hfill \\ \hfill =& A_m+N{A}_{m+1}+\cdots +N^{n-m-1}{A}_{n-1}+N^{n-m}{A}_n+\cdots +N^{k-m-1}{A}_{k-1}\hfill \\ \hfill =& {\Lambda }_m^k.\hfill \end{array}$$

(40)

Theorem 3.

Given nonzero integer sequence ${\left\{{\beta }_i\right\}}_{i\ge 1}$, then, for any integer $M>0$, there exists an integer $i⩾M$ such that

$${\beta }_{i+1}\ne N^{-1}({\beta }_i+a_i),$$

for any $a_i\in A_i$.

Proof. 

Suppose that there exists a positive integer M such that, for any $i>M$, we have ${\beta }_{i+1}=6^{-1}({\beta }_i+a_i)$. Let $T_0$ be the self-similar set generated by the dual IFS $\{\frac{1}{6}(x+s):s\in {\bigcup }_{j=1}^4{B}_j\}$.

According to the definition of the attractor $T_0$, there exists a positive integer K such that, for any $i\ge K$, ${\beta }_i$ belongs to a neighborhood of $T_0$, i.e.,

$${\beta }_i\in (-1,\frac{53}{5}).$$

Recall a fact that ${\bigcup }_{i=0}^{\infty }{A}_i=\{0,8,22,38,52\}$. Then, ${\beta }_{K+1}=6^{-1}({\beta }_K+a_K)$ with $a_K\in {\bigcup }_{j=1}^4{B}_j$ implies ${\beta }_K\in \{2,4,6,8,10\}$. By noting that ${\beta }_{K+2}=6^{-1}({\beta }_{K+1}+a_{K+1})$ with $a_{K+1}\in {\bigcup }_{j=1}^4{B}_j$ implies ${\beta }_K\ne 6$, hence ${\beta }_K\in \{2,4,8,10\}.$ If ${\beta }_K=2$, then

$$a_K=22,a_{K+1}=8,a_{K+2}=22,a_{K+3}=8,\cdots .$$

Hence, $\{8,22\}\cap A_i\ne \varnothing$ for all $i\ge K$, which contradicts that $\{8,22\}\cap B_4=\varnothing$ and $B_4=A_i$ for infinitely many i. Hence, ${\beta }_K\in \{4,8,10\}$.

By a similar argument for other cases, i.e., ${\beta }_K\in \{4,8,10\}$, we always obtain a contradiction. Then, we finish the proof. □

The following result shows that Strichartz’s method (Theorem 2.8 in [24]) is not applicable to the above set $\Lambda (N,\mathcal{B})$.

Theorem 4.

We have

$$\underset{n\to \infty }{lim\; inf}\underset{\lambda \in {\Lambda }^n}{inf}\left|m_D\left(N^{-n}\lambda \right)\right|=0.$$

Proof. 

Obviously, we need only to prove that there exists a subsequence ${\left\{{\lambda }_{n_k}\right\}}_{k\ge 1}\subset {\Lambda }^{n_k}$ such that $N^{-n_k}{\lambda }_{n_k}$ tends to a zero point of $m_D$ as k tends to infinity. Let $T_0$ be the attractor of the $IFS\{{\Phi }_j\left(x\right)=\frac{1}{6}(x+j):j\in {\bigcup }_{i=1}^4{B}_i\}$. Thus, we have $T_0\subset [0,\frac{52}{5}]$.

For $k\ge M$, we write $n_k=2^{2k+2}+2k+1$, and we take

$${\beta }_{n_k}=38+52\times 6+38\times 6^2+52\times 6^3+\cdots +38\times 6^{2^{2k+1}}\in {\Lambda }_{2^{2k+1}+2k-1}^{n_k},$$

where the coefficients 38 and 52 appear alternately. By a simple deduction, we obtain

$$6^{-2^{2k+1}-2}(10+{\beta }_{n_k})=\frac{4}{3}.$$

(41)

Take arbitrarily $\alpha \in {\Lambda }^{2^{2k+1}+2k-1}$, and write

$${\lambda }_{n_k}=\alpha +6^{2^{2k+1}+2k-1}{\beta }_{n_k}.$$

By (40), we obtain

$${\lambda }_{n_k}\in {\Lambda }^{n_k}.$$

According to the definition of $T_0$, we have

$$6^{-2^{2k+1}-2k+1}\alpha \in T_0,$$

which implies $|6^{-2^{2k+1}-2k+1}\alpha -10|\le \frac{52}{5}$. In combination with (41), we have

$$\begin{array}{cc}\hfill & |6^{-n_k}{\lambda }_{n_k}-\frac{4}{3}|\hfill \\ \hfill =& |6^{-2^{2k+1}-2}(6^{-2^{2k+1}-2k+1}\alpha +{\beta }_{n_k})-6^{-2^{2k+1}-2}(10+{\beta }_{n_k})|\hfill \\ \hfill =& \left|6^{-2^{2k+1}-2}(6^{-2^{2k+1}-2k+1}\alpha -10)\right|\hfill \\ \hfill ⩽& 6^{-2^{2k+1}-2}\times \frac{52}{5}.\hfill \end{array}$$

Noting the fact that $m_D\left(\frac{4}{3}\right)=0$, we finish the proof. □

5. Summary and Conclusions

In this paper, we introduced a tree structure with the language of symbolic space. The natural spectrum candidate of a self-similar measure associated with an IFS is a set with a special tree structure. We obtained three equivalent conclusions for $\Lambda$ to be a spectrum of a self-similar measure. One of them implies that there exists an infinite orbit with an element of a nonzero integer associated with the dual IFS. An example involving a self-similar measure and a spectrum candidate $\Lambda (N,\mathcal{S})=S_0+N{S}_1+N^2{S}_2\cdots$ showed the tree structure expands essentially the field of spectrum candidates.

It is one of the most important problems to find all spectra of a spectral measure. We are not sure that every spectrum of a self-similar measure holds a tree structure. On the other hand, the self-similar ${\mu }_{N,D}$ measure has another description, ${\mu }_{N,D}={\delta }_{\frac{1}{N}D}*{\delta }_{\frac{1}{N^2}D}*\cdots$. It is obvious to ask if Theorem 2 holds for the Moran-type self-similar measure. As mentioned in the Introduction, the version of Theorem 1 in higher-dimensional space has not been obtained completely. It is the next research direction to prove Theorem 2 the for self-affine measures.