A Note on the Spectrality of Moran-Type Bernoulli Convolutions by Deng and Li

Abstract

Let $\{p_n\}_{n\ge 1}$ and $\{ d_n\}_{n\ge 1}$ be two sequences of integers such that $|p_n|>|d_n|>0$ and $\{d_n\}_{n\ge 1}$ is bounded. It is proven by Deng and Li that the Moran-type Bernoulli convolution

$$\begin{aligned}\mu :=\delta _{p_1^{-1}\{0,d_1\}}*\delta _{p_1^{-1}p_2^{-1}\{0,d_2\}}*\dots *\delta _{p_1^{-1}\dots p_n^{-1}\{0,d_n\}}*\dots \end{aligned}$$

is a spectral measure if and only if the numbers of factor 2 in the sequence $\big \{\frac{p_1p_2\dots p_n}{2d_n}\big \}_{n\ge 1}$ are different from each other. Unfortunately, there is a gap in the proof of the sufficiency. Here we give a new proof to close the gap.

1 Introduction

In the proof of [2, Theorem 4.3 (iii)], the inclusion relationship “$\{\gamma +b_\gamma :\ \gamma \in \Gamma \}\subset \sum \nolimits _{j=1}^{\ell _n}(\{0\}\cup U_j)$" maybe wrong in some cases. Actually, this inclusion relationship need a precondition “$\ell _j\le \ell _n$ for all $j<\ell _n$". The following example shows that, there are examples that $\ell _j>\ell _n$ holds for at least one integer $j<\ell _n$ for all $n>0$. Hence, the sufficiency of [2, Theorem 1.1] needs to be reproved.

Example

Let $p_{2n-1}=4,\ p_{2n}=9$ and $d_{2n-1}=1,\ d_{2n}=8$ for all $n\ge 1$. Then, the definition of $k_n$ and $\ell _n$ shows

$$\begin{aligned} k_{2n}=v_2(\frac{p_1p_2\cdots p_{2n}}{2d_{2n}})=v_2(\frac{36^n}{16})=2n-4, \quad \forall \ n\ge 1, \end{aligned}$$

$$\begin{aligned} k_{2n-1}=v_2(\frac{p_1p_2\cdots p_{2n-1}}{2d_{2n-1}})=v_2(\frac{36^{n-1}\times 4}{2})=2n-1, \quad \forall \ n\ge 1. \end{aligned}$$

Also, $\ell _{2n-1}=2n+2$ and $\ell _{2n}=2n$ for all $n\ge 1$. This means $\ell _{\ell _{n}-1}>\ell _{n}$ for all $n\ge 1$.

We recall the definition of Moran-type Bernoulli convolution. Let $\{p_n\}_{n\ge 1}$ and $\{d_n\}_{n\ge 1}$ be two sequences of integers satisfying $|p_n|\ge 2,\ |d_n|\ge 1$ and

$$\begin{aligned} \sum _{n=1}^{+\infty }|p_1^{-1}p_2^{-1}\cdots p_n^{-1}d_n|<+\infty . \end{aligned}$$

The weak limit of the following convolutions is called a Moran-type Bernoulli convolution

$$\begin{aligned} \mu _n=\delta _{p_1^{-1}D_1}*\delta _{p_1^{-1}p_2^{-1}D_2}*\cdots *\delta _{p_1^{-1}p_2^{-1}\cdots p_n^{-1}D_n}.\end{aligned}$$

And we denote it by

$$\begin{aligned} \mu =\delta _{p_1^{-1}D_1}*\delta _{p_1^{-1}p_2^{-1}D_2}*\cdots *\delta _{p_1^{-1}p_2^{-1}\cdots p_n^{-1}D_n}\cdots . \end{aligned}$$

(1.1)

We shall reprove the sufficiency of the following result (i.e. [2, Theorem 1.1]).

Theorem 1.1

For the measure $\mu$ defined by (1.1) with $|p_n|>|d_n|$ for all $n\ge 2$, assume that the sequence $\{|d_n|\}_{n=1}^{+\infty }$ is bounded. Then, $\mu$ is a spectral measure if and only if $k_j\ne k_i$ for all $j>i\ge 1$, where

$$\begin{aligned} k_n=v_2\Big (\frac{p_1p_2\dots p_n}{2d_n}\Big )=v_2(p_1p_2\dots p_n)-v_2(2d_n),\quad n=1,2,3,\dots . \end{aligned}$$

(1.2)

2 Proof of the Sufficiency of Theorem 1.1

In order to make the proof more readable, we first simplify our model.

Proposition 2.1

For the measure $\mu$ defined by (1.1), there exist two sequences of integers $\{c_n\}_{n=1}^\infty$ and $\{q_n\}_{n=1}^\infty$ such that for $n\ge 1$, we have $\gcd (q_n,c_n)=1$ and

$$\begin{aligned} q_1^{-1}q_2^{-1}\cdots q_n^{-1}c_n=p_1^{-1}p_2^{-1}\cdots p_n^{-1}d_n. \end{aligned}$$

(2.1)

Furthermore, we have $|q_n|>|c_n|$ when $|p_n|>|d_n|$ ($n=1,2,\dots$). Hence, we can rewrite $\mu$ as

$$\begin{aligned} \mu =\delta _{q_1^{-1}C_1}*\delta _{q_1^{-1}q_2^{-1}C_2}*\delta _{q_1^{-1}q_2^{-1}q_3^{-1}C_3}\cdots *\delta _{q_1^{-1}q_2^{-1}\cdots q_n^{-1}C_n}*\cdots , \end{aligned}$$

(2.2)

where $C_n=\{0,c_n\}$.

Proof

Write $g_0=1$, and define inductively

$$\begin{aligned} g_{n}=\gcd (|g_{n-1}p_{n}|,\ |d_{n}|),\ q_{n}=\frac{g_{n-1}p_{n}}{g_{n}}\ \text { and} \ c_{n}=\frac{d_{n}}{g_{n}},\quad \forall \ n\ge 1. \end{aligned}$$

It is clear that for any $n\ge 1$, we have $\gcd (q_n,\ c_n)=1$ and (2.1). By writing $C_n=\{0,c_n\}$, we have

$$\begin{aligned} \delta _{q_1^{-1}q_2^{-1}\cdots q_n^{-1}C_n}=\delta _{p_1^{-1}p_2^{-1}\cdots p_n^{-1}D_n}, \end{aligned}$$

which implies that (2.2) holds.

If $|p_n|>|d_n|$, noting $|g_{n-1}p_{n}|\ge |p_n|$, it is obvious that $|q_n|>|c_n|$. $\square$

The above Proposition 2.1 shows that, in order to prove the sufficiency of Theorem 1.1, without loss of generality, we can assume that $\gcd (d_n,\ p_n)=1$. By the argument in [2], we will always assume that [2, (2.9)] holds without loss of generality. Therefore, we shall assume that the following conditions hold in the sequel:

$$\begin{aligned} p_n\ge 2,\ \ d_n\ge 1,\ \ \gcd (d_n,\ p_n)=1,\ \frac{p_1p_2\cdots p_n}{2d_n}\in {\mathbb {N}},\ \ \ \ \ \forall \ n\ge 1. \end{aligned}$$

(2.3)

The following Proposition 2.2 is obviously true.

Proposition 2.2

Let $\nu$ be a probability measure and its support has finite cardinality N. If $L^2(\nu )$ has an orthogonal set $\{e^{2\pi i\lambda x}:\ \lambda \in \Lambda \}$ and $\#\Lambda$ is at least N, then $\Lambda$ is a spectrum of $\nu$ and $\#\Lambda =N$.

We will continue to use notations $\ell _n,\ k_n,\ U_n,\ t_n,\ r_n$ defined in [2] and the constant c is defined in [2, Lemma 4.1 (i)]. Given a nonzero integer n, we denote by $\theta (n)$ the odd part of n, i.e. $\theta (n)=\frac{n}{2^{\nu _2(n)}}.$ Then, we rewrite

$$\begin{aligned} U_n=2^{k_n}\theta (p_1p_2\cdots p_{\ell _n})(2\mathbb {Z}+1), \quad n\ge 1. \end{aligned}$$

(2.4)

The following set will play an important role in the sequel

$$\begin{aligned} \mathcal {I}=\{n:\ell _{n}=n\}. \end{aligned}$$

(2.5)

Lemma 2.3

Assume that (2.3) holds and $k_i\ne k_j$ for all $i>j\ge 1$. Then, we have the following statements.

• (i). $\#{\mathcal {I}}=+\infty$.

• (ii). Let B be a finite nonempty subset of positive integers. Then, $\Delta :=\sum \limits _{j\in B}\{0,\ a_j\}$ is a spectrum (with cardinality $2^{\#B}$) of $*_{j\in B} \delta _{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}$ for any $a_j\in U_j$.

• (iii). Let B be a finite nonempty subset of positive integers. Assume that $\Lambda \subset \sum \limits _{j\in B}(\{0\}\cup U_j)$ is a spectrum of the probability measure $*_{j\in B} \delta _{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}$. For any $m\ge \max \{\ell _j:\ j\in B\}$ and $\lambda \in \Lambda$, we take an integer $b_\lambda \in p_1p_2\cdots p_m{\mathbb {Z}}$ with $b_0=0$. Then, the set $\{\lambda +b_\lambda :\ \gamma \in \Lambda \}$ is also a spectrum of the probability measure $*_{j\in B} \delta _{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}$. Furthermore, we have $\{\gamma +b_\lambda :\ \gamma \in \Lambda \}\subset \sum \limits _{j\in B}(\{0\}\cup U_j)$.

Proof

(i) It is sufficient to prove that for any integer $N>0$, there exists a positive integer $n>N$ such that $\ell _n=n$.

Indeed, since $\{d_n\}_{n\ge 1}$ is bounded, there is an integer $z_0$ such that $k_n\ge z_0$ for all $n>0$. Hence, there is an integer $n> N$ such that $k_n=\min \{k_j:\ j>N\}$. Since $k_i\ne k_j$ for all $i>j\ge 1$, we see that $k_n<k_j$ for all $j>n$. Hence, the definition of $\ell _n$ shows $\ell _n=n$. The conclusion is proven.

(ii) Suppose $B=\{j_1,\ j_2,\ \cdots ,\ j_s\}$ with $k_{j_1}<k_{j_2}<\cdots < k_{j_s}$. From [2, Lemma 4.2], it follows that $k_{i}<k_{j}$ implies $\ell _{i}\le \ell _{j}$. Then, the definition of $U_n$ shows

$$\begin{aligned} U_{j_t}+\sum \limits _{t<i\le s}(\{0\}\cup U_{j_i})=U_{j_t},\quad t=1,2,\dots , s-1. \end{aligned}$$

(2.6)

For any $\xi =\sum \limits _{j\in B}\xi _j$ and $\eta =\sum \limits _{j\in B}\eta _j\in \Delta$ with $\xi _j,\ \eta _j\in \{0,\ a_j\}$ and $\xi _j\ne \eta _j$ for at least one $j\in B$, it is easy to see $\xi -\eta \in \sum \limits _{j\in B}\{0,\ \pm a_j\}$. Write $t=\min \{i:\ \xi _{j_i}\ne \eta _{j_i},\ 1\le i\le s\}$. Then, we have $\xi -\eta \in U_{j_t}+\sum \limits _{t<i\le s}(\{0\}\cup U_{j_i})$. From (2.6) it follows $\xi -\eta \in U_{j_t}$. This implies $\xi -\eta \in U_{j_t}$ and $\#\Delta =2^s$. Furthermore, $\xi -\eta \in U_{j_t}$ shows that $\xi -\eta$ is a zero point of the Fourier transformation of the probability measure $*_{j\in B} \delta _{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}$, i.e. $\prod \limits _{j\in B}\widehat{\delta }_{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}(\xi -\eta )=0$.

It is easy to see that the support of the measure $*_{j\in B} \delta _{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}$ has cardinality at most $2^s$. Proposition 2.2 shows that $\Delta$ is a spectrum of the measure $*_{j\in B} \delta _{p_1^{-1}p_2^{-1}\cdots p_{j}^{-1}D_{j}}$ and $\#\Delta =2^s=2^{\# B}$.

(iii) Note a fact that for all $j\in B$, the integer $p_1p_2\cdots p_{j}$ is a period of $\widehat{\delta }_{p_1^{-1}\cdots p_j^{-1}D_j}$. For $m\ge \max \{\ell _j:\ j\in B\}\ge j$, we have $\widehat{\delta }_{p_1^{-1}\cdots p_j^{-1}D_j}(x+\lambda +b_\lambda )=\widehat{\delta }_{p_1^{-1}\cdots p_j^{-1}D_j}(x+\lambda )$ for all $x\in {\mathbb {R}}$, $j\in B$ and $\lambda \in \Lambda$. Hence

$$\begin{aligned} \sum _{\gamma \in \{\lambda +b_\lambda :\lambda \in \Lambda \}} \left| \prod _{j\in B}\widehat{\delta }_{p_1^{-1}\cdots p_j^{-1}D_j}(x+\gamma )\right| ^2=\sum _{\lambda \in \Lambda } \left| \prod _{j\in B}\widehat{\delta }_{p_1^{-1}\cdots p_j^{-1}D_j}(x+\lambda )\right| ^2,\quad \forall \ x\in {\mathbb {R}}. \end{aligned}$$

That means that the first conclusion in (iii) is proven by using [2, Proposition 2.3].

For any $\lambda \in \Lambda$, it can be written as $\lambda =\sum \limits _{j\in B}b_j$ with $b_j\in (\{0\}\cup U_j)$. Let $B=\{j_1,\ j_2,\ \cdots ,\ j_s\}$ with $k_{j_1}<k_{j_2}<\cdots < k_{j_s}$ as in (ii) and define $t=\min \{i:\ b_{j_i}\ne 0\}$. Then, (2.6) shows $\lambda \in U_{j_t}$. Since $m\ge \max \{\ell _j:\ j\in B\}$, the definition of $U_j$ implies that $U_j+b_\lambda =U_j$ for all $j\in B$, which implies $\lambda +b_\lambda \in U_{j_t}$. Hence, we have $\{\lambda +b_\lambda :\ \lambda \in \Lambda \}\subset \sum \limits _{j\in B}(\{0\}\cup U_j)$ for any $b_\lambda \in p_1p_2\cdots p_m{\mathbb {Z}}$ with $b_0=0$. The second conclusion in (iii) is proven. $\square$

The following two lemmas deal with the possible case that $\ell _j>\ell _n$ for some $j<\ell _n$.

Lemma 2.4

Assume that $k_n\ne k_m$ for all $n\ne m$ and (2.3) holds. Furthermore, assume that there exists a positive integer $n_0$ such that for any $n\ge n_0$ there exists an integer $j_n<\ell _n$ satisfying $\ell _{j_n}>\ell _n$.

• (i) . For any $i\ge n_0$, there is at least one member of the group $p_i,\ p_{i+1},\ \cdots ,\ p_{i+c}$ which is an odd integer larger than or equal to 3.

• (ii) . There exists a positive integer $N_1\ge 0$ such that

  $$\begin{aligned} \frac{d_{n}}{\theta (p_{\ell _i+1}\cdots p_{n})}\le 1 ,\ n_0\le i\le n-N_1. \end{aligned}$$

  (2.7)

• (iii) . For any $n\ge n_0+c$, we have $v_2(p_n)<\max \{v_2(d_j):\ j>0\}$.

Proof

(i) Given $i\ge n_0$, suppose $p_i,\ p_{i+1},\ \cdots ,\ p_{i+c}$ are all even. From the assumption $\gcd (p_n,d_n)=1$, it is clear that $d_i,\ d_{i+1},\ \cdots ,\ d_{i+c}$ are all odd. Hence, $k_i<k_{i+1}< \cdots <k_{i+c}$, which implies $\ell _i=i$ since [2, Lemma 4.1 (i)] shows $i\le \ell _i\le i+c$. On the other hand, however, our assumption shows for the integer i, there is a positive integer $j<\ell _i$ such that $\ell _j>\ell _i$. Then, we have $k_{j}>k_{\ell _j}>k_i$. In virtue of $\ell _i=i$, we have $j<i$. Since $p_i$ is even and $d_i$ is odd, we have $k_i=v_2(p_1p_2\cdots p_{i})-1>v_2(p_1p_2\cdots p_{j})-1\ge k_j$, which leads to a contradiction. Therefore, at least one member of $p_i,\ p_{i+1},\ \cdots ,\ p_{i+c}$ is odd which is larger than or equal to 3. The conclusion (i) is proven.

(ii) Choose a positive integer $s>0$ such that $3^s\ge \max \{d_n:\ n>0\}$. Thus for $i\ge n_0$, we have

$$\begin{aligned} \frac{\max \{d_n:\ n>0\}}{p_{\ell _i+1}p_{\ell _i+2}\cdots p_{\ell _i+sc}}\le 1. \end{aligned}$$

It is clear that we finish the proof by taking $N_1=sc$.

(iii) Suppose $v_2(p_n)\ge \max \{v_2(d_j):\ j>0\}$ for some $n\ge n_0+c$. For any j and m with $j<n\le m$, we have $k_m= v_2(p_1p_2\cdots p_{m})-1-v_2(d_m)\ge v_2(p_1p_2\cdots p_{j})+v_2(p_n)-1-v_2(d_m)\ge v_2(p_1p_2\cdots p_{j})-1\ge k_j$. In fact, by the assumption that $k_m\ne k_j$, we have $k_m>k_j$. Let $k_s=\max \{k_j:\ n_0\le j\le n-1\}$. According to the definition of $\ell _s$, we have $\ell _s=n-1$.

On the other hand, however, for the integer s, there exits a positive integer $j_0$ with $j_0<\ell _s$ such that $\ell _{j_0}>\ell _s$, which implies $\ell _{j_0}\ge n$. Noting that $j_0<n$, according to the above argument, we get $k_{\ell _{j_0}}>k_{j_0}$, which is a contradiction to the definition of $\ell _{j_0}$. The statement (iii) is proven. $\square$

Lemma 2.5

Assume that $k_n\ne k_m$ for all $n\ne m$ and (2.3) hold. Furthermore, assume that there exists a positive integer $n_0$ such that for any $n\ge n_0$ there exists an integer $j_n<\ell _n$ satisfying $\ell _{j_n}>\ell _n$. Then, there are small constants $\varepsilon >0$ and $\theta _0>0$ such that for any $n_1$ and $n_2\in \mathcal {I}$ with $n_2>n_1+N_1$, there exists a spectrum $\Lambda =\sum _{i=n_1+1}^{n_2}\{0,\ a_i\}$ of $*_{i=n_1+1}^{n_2} \delta _{p_1^{-1}p_2^{-1}\cdots p_{i}^{-1}D_{i}}$ such that

$$\begin{aligned} \inf _{\lambda \in \Lambda , \ |y|\le \theta _0}\left\{ \prod _{j=1}^c \left| m_{p_{n_2+1}^{-1}\cdots p_{n_2+j}^{-1} D_{n_2+j}}\big (\frac{\lambda }{p_1p_2\cdots p_{n_2}}+y\big )\right| \right\} >\varepsilon . \end{aligned}$$

(2.8)

Proof

We first construct the spectrum $\Lambda$. Write $D=\max \{d_n:n\ge 1\}$ and $\mathcal {S}=\{i:n_1+1\le i\le n_2\}$. We divide the set $\mathcal {S}$ into two parts $\mathcal {S}_1$ and $\mathcal {S}_2$, where

$$\begin{aligned} \mathcal {S}_1=\left\{ i\in \mathcal {S}:\frac{D\theta (p_1\cdots p_{\ell _i})}{\theta (p_1\cdots p_{n_2+1})}\ge 1\right\} \ \text {and}\ \mathcal {S}_2=\left\{ i\in \mathcal {S}:\frac{D\theta (p_1\cdots p_{\ell _i})}{\theta (p_1\cdots p_{n_2+1})}<1\right\} . \end{aligned}$$

Take

$$\begin{aligned} a_i=\left\{ \begin{array}{ll} 2^{k_i}\theta (p_1\cdots p_{n_2+c}),&{}i\in \mathcal {S}_1,\\ 2^{k_i}\theta (p_1\cdots p_{\ell _i}),&{}i\in \mathcal {S}_2. \end{array} \right. \end{aligned}$$

(2.9)

[2, Lemma 4.1 (i)] shows that $n_2+c\ge \ell _i$ for all $i\in \mathcal {S}_1$. By the definition of $U_n$, it is clear that $a_i\in U_i$ for all $i\in \mathcal {S}$. Then, Lemma 2.3 shows that $\Lambda =\sum _{i=n_1+1}^{n_2}\{0,\ a_i\}$ is a spectrum of $*_{i=n_1+1}^{n_2} \delta _{p_1^{-1}p_2^{-1}\cdots p_{i}^{-1}D_{i}}$. By the definition of the function $\theta$, for $i\in \mathcal {S}_1$ we have

$$\begin{aligned} \frac{d_{n_2+j}a_i}{p_1\cdots p_{n_2+j}}= \frac{\theta (d_{n_2+j})\theta (p_{n_2+j+1}\cdots p_{n_2+c})}{2^{k_{n_2+j}+1-k_i}} \in 2^{k_i-k_{n_2+j}-1}(2{\mathbb {Z}}+1). \end{aligned}$$

(2.10)

According to the definition of $\mathcal {S}_2$, we have $\ell _i\le n_2$ for any $i\in \mathcal {S}_2$. Thus we have

$$\begin{aligned} \frac{d_{n_2+j}a_i}{p_1\cdots p_{n_2+j}}=\frac{2^{k_i-k_{n_2+j}-1}\theta (d_{n_2+j})}{\theta (p_{\ell _i+1}\cdots p_{n_2+j})},\ i\in \mathcal {S}_2. \end{aligned}$$

(2.11)

Given $\lambda =\sum \limits _{i=n_1+1}^{n_2}b_i\in \Lambda$ with $b_i\in \{0,\ a_i\}(n_1+1\le i\le n_2)$, write $k_{i_1}=\min \{k_i:\ i\in \mathcal {S}_1,\ b_i\ne 0\}$ when the set $\{i:\ i\in \mathcal {S}_1,\ b_i\ne 0\}$ is not empty. According to (2.10) and the assumption that $k_i\ne k_j$ for any $i\ne j$, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \sum _{i\in \mathcal {S}_1}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}\in 2^{k_{i_1}-1-k_{n_2+j}}(2{\mathbb {Z}}+1),&{}\{i:\ i\in \mathcal {S}_1,\ b_i\ne 0\}\ne \emptyset ,\\ \sum _{i\in \mathcal {S}_1}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}=0,&{}\{i:\ i\in \mathcal {S}_1,\ b_i\ne 0\}=\emptyset . \end{array}\right. \end{aligned}$$

(2.12)

Let $k_{i_2}=\max \{k_i:\ i\in \mathcal {S}_2,b_i\ne 0\}$ when the set $\{i:i\in \mathcal {S}_2, b_i\ne 0\}$ is not empty. According to the definition of $\mathcal {S}_2$, we have $\theta (d_{n_2+j})\le D<\theta (p_{\ell _i+1}\cdots p_{n_2+1})$ for $i\in \mathcal {S}_2$. Thus we have $\theta (d_{n_2+j})+2\le \theta (p_{\ell _i+1}\cdots p_{n_2+1})$. Hence, $\frac{\theta (d_{n_2+j})}{\theta (p_{\ell _i+1}\cdots p_{n_2+1})}\le \frac{\theta (d_{n_2+j})}{\theta (d_{n_2+j})+2}\le \frac{D}{D+2}$ for any $i\in \mathcal {S}_2$. Also by (2.11) and the assumption that $k_n\ne k_m$ for any $n\ne m$, we get

$$\begin{aligned} 0\le \sum _{i\in \mathcal {S}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}<\frac{D}{D+2}\sum _{s\ge 0}2^{k_{i_2}-1-k_{n_2+j }-s}=\frac{D}{D+2}2^{k_{i_2}-k_{n_2+j}}. \end{aligned}$$

(2.13)

Given $1\le j\le c$, we consider

$$\begin{aligned} m_{p^{-1}_{n_2+1}\cdots p^{-1}_{n_2+j}D_{n_2+j}}\left( \frac{\lambda }{p_1p_2\cdots p_{n_2}}\right) =m_{\{0,1\}}\left( \frac{\lambda d_{n_2+j}}{p_1p_2\cdots p_{n_2+j}}\right) ,\ \lambda \in \Lambda . \end{aligned}$$

(2.14)

And then we will deal with two cases.

Case A. $\{i:\ i\in \mathcal {S}_1,\ b_i\ne 0\}=\emptyset$ or $k_{i_1}>k_{n_2+j}$.

From (2.12) it is clear that

$$\begin{aligned} \sum _{i\in \mathcal {S}_1}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}\in {\mathbb {Z}}. \end{aligned}$$

(2.15)

Noting that $i_2\in {\mathcal {S}}_2$, we see $\ell _{i_2}\le n_2$, which implies $k_{n_2+j}>k_{i_2}$. In virtue of (2.13), we get

$$\begin{aligned} \sum \limits _{i\in {\mathcal {S}}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}\in \left( 0,\ \frac{1}{2}\frac{D}{D+2}\right) . \end{aligned}$$

Since $m_{\{0,1\}}$ has period 1, we have

$$\begin{aligned} \begin{array}{rl} m_{\{0,1\}}\left( \frac{\lambda d_{n_2+j}}{p_1p_2\cdots p_{n_2+j}}\right) &{}=m_{\{0,1\}}\left( \sum \limits _{i=n_1+1}^{n_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}\right) \\ &{}=m_{\{0,1\}}\left( \sum \limits _{i\in {\mathcal {S}}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}\right) \\ &{}=m_{\{0,1\}}(\alpha ) \end{array} \end{aligned}$$

(2.16)

for some $\alpha \in \left( 0,\ \frac{1}{2}\frac{D}{D+2}\right)$.

Case B. $k_{i_1}<k_{n_2+j}$.

Without loss of generality, we assume that the set $\{i:i\in \mathcal {S}_2,b_i\ne 0\}$ is not empty. Otherwise, we have $\sum \limits _{i\in {\mathcal {S}}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}=0$. From the definitions of $i_1$ and $i_2$, it follows that $k_{i_1}>k_{i_2}$. By (2.13), we get

$$\begin{aligned} 0\le \sum _{i\in \mathcal {S}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}<\frac{D}{D+2}2^{k_{i_2}-k_{n_2+j}}\le \frac{D}{D+2}2^{k_{i_1}-k_{n_2+j}-1}. \end{aligned}$$

Combining (2.12), this shows there exists an integer z such that

$$\begin{aligned} \sum _{i\in \mathcal {S}_1\cup \mathcal {S}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}}=2^{k_{i_1}-k_{n_2+j}-1}(2z+1+\eta ) \end{aligned}$$

for some $\eta \in \left( 0,\frac{D}{D+2}\right)$.

Given a real number $r\in \mathbb {R}$, we denote by $||r||_\frac{1}{2}$ the distance between r and it’s nearest middle point of two neighboring integer points, i.e.

$$\begin{aligned} ||r||_\frac{1}{2}=\min _{z\in \mathbb {Z}}\left\{ \left| r-z-\frac{1}{2}\right| , \left| r-z+\frac{1}{2}\right| \right\} . \end{aligned}$$

Thus the assumption $k_{i_1}<k_{n_2+j}$ implies

$$\begin{aligned} \left| \left| 2^{k_{i_1}-k_{n_2+j}-1}(2z+1)\right| \right| _{\frac{1}{2}}\ge 2^{k_{i_1}-k_{n_2+j}-1}. \end{aligned}$$

By noting that $\left| 2^{k_{i_1}-k_{n_2+j}-1}\eta \right| <2^{k_{i_1}-k_{n_2+j}-1}\frac{D}{D+2}$, we get

$$\begin{aligned} \begin{array}{rl} \left| \left| \frac{\lambda d_{n_2+j}}{p_1p_2\cdots p_{n_2+j}}\right| \right| _{\frac{1}{2}}&{}= \left| \left| \sum _{i\in \mathcal {S}_1\cup \mathcal {S}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}} \right| \right| _{\frac{1}{2}}\\ &{}=\left| \left| 2^{k_{i_1}-k_{n_2+j}-1}(2z+1+\eta ) \right| \right| _{\frac{1}{2}} \\ &{} \ge 2^{k_{i_1}-k_{n_2+j}-1}\frac{2}{D+2}. \end{array} \end{aligned}$$

(2.17)

Since $i_1\in \mathcal {S}_1$, from Lemma 2.4 (i) it follows that $i_1>n_2-N_1$. By Lemma 2.4 (iii), we have

$$\begin{aligned} \begin{array}{rl} k_{n_2+j}-k_{i_1}&{}=\nu _2(p_1\cdots p_{n_2+j})-\nu _2(2d_{n_2+j})-\nu _2(p_1\cdots p_{i_1})+\nu _2(2d_{i_1})\\ &{}\le \nu _2(p_{i_1+1}\cdots p_{n_2+j})+\nu _2(d_{i_1})\\ &{}\le (n_2+j-i_1+1)\max \{\nu _2(d_n):n\ge 1\}\\ &{}\le (N_1+c+2)\max \{\nu _2(d_n):n\ge 1\}. \end{array} \end{aligned}$$

(2.18)

Together with (2.17) and the boundedness of $\{d_n\}_{n=1}^\infty$, we conclude that there is a positive constant $0<\theta <\frac{1}{2}$ such that

$$\begin{aligned}\left| \left| \frac{\lambda d_{n_2+j}}{p_1p_2\cdots p_{n_2+j}}\right| \right| _{\frac{1}{2}}= \left| \left| \sum _{i\in \mathcal {S}_1\cup \mathcal {S}_2}\frac{d_{n_2+j}b_i}{p_1\cdots p_{n_2+j}} \right| \right| _{\frac{1}{2}}>\theta . \end{aligned}$$

Therefore, combining the conclusions of Case A and Case B we see the modulus of (2.14) has a positive lower bound. Furthermore, there is a constant $\varepsilon >0$ such that

$$\begin{aligned} \prod _{j=1}^c \left| m_{p_{\ell _n+1}^{-1}\cdots p_{\ell _n+j}^{-1} D_{\ell _n+j}}\left( \frac{\lambda }{p_1p_2\cdots p_{n_2}}\right) \right| >2\varepsilon ,\quad \forall \ \lambda \in \Lambda . \end{aligned}$$

Finally, note that

$$\begin{aligned} \{\delta _{p_{n+1}^{-1}D_{n+1}}*\delta _{p_{n+1}^{-1} p_{n+2}^{-1} D_{n+2}}*\cdots *\delta _{p_{n+1}^{-1}\cdots p_{n+c}^{-1} D_{n+c}}:\ n>0\} \end{aligned}$$

is a family of probability measures supported on subsets of $[0,\ 1]$. Hence, their Fourier transformations are equi-continuous (cf [2, Definition 4.4 (iii)]). Thus we see that there is a small positive number $\theta _0>0$ such that (2.8) holds for some constant $\varepsilon >0$. The proof is completed. $\square$

Furthermore, we have the following Lemma 2.6. For $k\ge 1$, we write

$$\begin{aligned} \mu _{>k}:=\delta _{p_{k+1}^{-1}D_{k+1}}*\delta _{p_{k+1}^{-1}p_{k+2}^{-1}D_{k+2}}*\cdots . \end{aligned}$$

Lemma 2.6

Assume that $k_n\ne k_m$ for all $n\ne m$ and (2.3) holds. Furthermore, assume that there exists a positive integer $n_0$ such that for any $n\ge n_0$, there exists an integer $j_n<\ell _n$ satisfying $\ell _{j_n}>\ell _n$. Consider the set $\Lambda$ defined in Lemma 2.5 for $n_1,\ n_2\in \mathcal {I}$ satisfying $n_2>n_1+N_1$. There are small positive constants $\varepsilon _1>0$ and $\theta _1>0$ such that for any $\lambda \in \Lambda$, there exists an integer $b_\lambda \in {\mathbb {Z}}$ with $b_0=0$ such that

$$\begin{aligned} \left| \widehat{\mu _{>n_2}}(y+\frac{\lambda +p_1p_2\cdots p_{n_2+c}b_\lambda }{p_1p_2\cdots p_{n_2}})\right| >\varepsilon _1,\ \ \forall \ y\in [-\theta _1,\ \theta _1],\ \lambda \in \Lambda . \end{aligned}$$

(2.19)

Proof

By [2, Lemma 4.5], there are small positive constants $\varepsilon '>0$ and $\theta _0>\theta _1>0$ such that for any $\lambda \in \Lambda$, there exists an integer $b_\lambda$ with $b_0=0$ such that

$$\begin{aligned} \left| \widehat{\mu _{>n_2+c}}(y+b_\lambda +\frac{\lambda }{p_1p_2\cdots p_{n_2+c}})\right| >\varepsilon ',\ \ \forall \ y\in [-\theta _1,\ \theta _1]. \end{aligned}$$

(2.20)

Recall a fact that the mask function $m_{\{0,1\}}(x)$ has period 1. For any $\lambda \in \Lambda$, we have

$$\begin{aligned} \begin{array}{rl} &{}\left| \widehat{\mu _{>n_2}}(y+\frac{\lambda +p_1p_2\cdots p_{n_2+c}b_\lambda }{p_1p_2\cdots p_{n_2}})\right| \\ &{}\quad = \left| \widehat{\mu _{>n_2+c}}(y+\frac{\lambda +p_1p_2\cdots p_{n_2+c}b_\lambda }{p_1p_2\cdots p_{n_2+c}})\right| \cdot \prod _{j=1}^c \left| m_{p_{n_2+1}^{-1}\cdots p_{n_2+j}^{-1} D_{n_2+j}}\big (y+\frac{\lambda +p_1p_2\cdots p_{n_2+c}b_\lambda }{p_1p_2\cdots p_{n_2}}\big )\right| \\ &{}\quad = \left| \widehat{\mu _{>n_2+c}}(y+b_\lambda +\frac{\lambda }{p_1p_2\cdots p_{n_2+c}})\right| \cdot \prod _{j=1}^c \left| m_{p_{n_2+1}^{-1}\cdots p_{n_2+j}^{-1} D_{n_2+j}}\big (y+\frac{\lambda }{p_1p_2\cdots p_{n_2}}\big )\right| .\\ \end{array} \end{aligned}$$

(2.21)

Lemma 2.5 shows there are small constants $\varepsilon >0$ and $\theta _0$ such that

$$\begin{aligned} \prod _{j=1}^c \left| m_{p_{n_2+1}^{-1}\cdots p_{n_2+j}^{-1} D_{n_2+j}}\big (y+\frac{\lambda }{p_1p_2\cdots p_{n_2}}\big )\right| >\varepsilon ,\ \ \forall \ y\in [-\theta _0,\ \theta _0]. \end{aligned}$$

(2.22)

Letting $\varepsilon _1=\varepsilon \varepsilon '$, the inequality (2.19) follows from (2.20), (2.21) and (2.22). The proof is completed. $\square$

Now we are in the place to reprove the sufficiency of [2, Theorem 1.1].

Proof of the sufficiency of [2, Theorem 1.1].

We shall deal with two cases.

(A) If there is an infinite subset ${\mathcal {I}}_0\subset {\mathcal {I}}$ (${\mathcal {I}}$ is defined in (2.5)) such that $\ell _i\le n$ for any $i\le n$ and $n\in {\mathcal {I}}_0$. Then, the proof in [2] works by replacing ${\mathcal {B}}$ by ${\mathcal {I}}_0$.

(B) If there are only finitely many $n\in {\mathcal {I}}$ such that $\ell _i\le n$ for any $i\le n$. Then, there is an integer $n_0>0$ such that for any $n\in {\mathcal {I}}$ with $n\ge n_0$, there exists at least one integer $j_n<\ell _n$ satisfying $\ell _{j_n}>\ell _n$. Also, as stated in the beginning of this section, all conditions in (2.3) can be assumed without loss of generality.

Then, we extend the idea of [1, Lemma 2.6] and [1, Theorem 2.7] to construct a spectrum of $\mu$. This spectrum is different from the one in [2]. Let ${\mathcal {I}}_1=\{n\in \mathcal {I}:\ n>n_0\}$

We first choose $n_1\in {\mathcal {I}}_1$ and define

$$\begin{aligned} \Lambda _1=\{0,\ a_1\}+\{0,\ a_2\}+\cdots +\{0,\ a_{n_1}\}, \end{aligned}$$

where $a_i=2^{k_i}\theta (p_1\cdots p_{\ell _i})\in U_i$ for $1\le i\le n_1$. Since ${\mathcal {I}}_1$ is infinite and $p_n\ge 2$, we can find a sufficiently large integer $n_2\in {\mathcal {I}}_1$ such that $n_2>n_1+N_1$ and

$$\begin{aligned} (p_1p_2\cdots p_{n_2})^{-1} \Lambda _1\subset \Big [-\frac{\theta _1}{2^2},\ \frac{\theta _1}{2^2}\Big ], \end{aligned}$$

where $N_1$ and $\theta _1$ are defined in Lemma 2.4 and 2.6, respectively. Let $\epsilon _1$ be the constant in Lemma 2.6 and $\Lambda _{1,2}$ be a spectrum of $*_{i=n_1+1}^{n_2} \delta _{p_1^{-1}p_2^{-1}\cdots p_{i}^{-1}D_{i}}$ as stated in Lemma 2.5, i.e.

$$\begin{aligned} \Lambda _{1,2}=\{0,\ a_{n_1+1}\}+\{0,\ a_{n_1+2}\}+\cdots +\{0,\ a_{n_2}\}, \end{aligned}$$

where

$$\begin{aligned} a_i=\left\{ \begin{array}{ll} 2^{k_i}\theta (p_1\cdots p_{n_2+c}),&{} \text {if}\ \frac{D\theta (p_1\cdots p_{\ell _i})}{\theta (p_1\cdots p_{n_2+1})}\ge 1,\\ 2^{k_i}\theta (p_1\cdots p_{\ell _i}),&{}\text {if}\ \frac{D\theta (p_1\cdots p_{\ell _i})}{\theta (p_1\cdots p_{n_2+1})}<1. \end{array} \right. \end{aligned}$$

(2.23)

According to Lemma 2.6, for any $\lambda \in \Lambda _{1,2}$, there exits an integer $k_{1,\lambda }\in {\mathbb {Z}}$ with $k_{1,\ 0}=0$ such that

$$\begin{aligned} \Big |\widehat{\mu _{>n_2}}\Big (\frac{ \gamma }{p_1p_2\cdots p_{n_2}}+ \frac{ \lambda +p_1\cdots p_{n_2+c}k_{1,\ \lambda }}{p_1p_2\cdots p_{n_2}} \Big )\Big |>\varepsilon _1,\quad \forall \ \gamma \in \Lambda _1,\ \lambda \in \Lambda _{1,2}. \end{aligned}$$

Lemma 2.3 (ii) and (iii) show that $\Lambda _2:=\{\gamma +\lambda +p_1p_2\cdots p_{n_2+c}k_{1,\ \lambda }:\ \gamma \in \Lambda _1,\ \lambda \in \Lambda _{1,2}\}$ is a spectrum of the probability measure $\delta _{p_1^{-1}D_1}*\delta _{p_1^{-1}p_2^{-1}D_2}*\cdots *\delta _{p_1^{-1}p_2^{-1}\cdots p_{n_2}^{-1}D_{n_2}}$. Furthermore, [2, Lemma 4.1] and the definitions of $U_i$ show that $U_i+p_1p_2\cdots p_{n_2+c}k_{1,\ \lambda }=U_i$ for all $i\le n_2$. Hence, by $k_{1,\ 0}=0$ and the definitions of $\Lambda _1$ and $\Lambda _2$, we see $\Lambda _1\subset \Lambda _2\subset \sum \limits _{j=1}^{n_2}(\{0\}\cup U_j)$. In a word, we have

$$\begin{aligned} \Big |\widehat{\mu _{>n_2}}\Big (\frac{ \lambda }{p_1p_2\cdots p_{n_2}}\Big )\Big |>\varepsilon _1,\quad \forall \ \lambda \in \Lambda _2. \end{aligned}$$

(2.24)

Continuing in this way, we can find a strictly increasing sequence $\{n_k\}_{k=1}^\infty \subset {\mathcal {I}}_1$ and $\Lambda _k$ such that the following properties (2.25), (2.26), (2.27) and claim hold.

$$\begin{aligned} 0\in \Lambda _{k}\subset \Lambda _{k+1}\subset \sum \limits _{j=1}^{n_{k+1}}(\{0\}\cup U_j),\quad k=1,\ 2,\cdots , \end{aligned}$$

(2.25)

$$\begin{aligned} (p_1p_2\cdots p_{n_{k+1}})^{-1} \Lambda _k \subset \Big [-\frac{\theta _1}{2^{k+1}},\ \frac{\theta _1}{2^{k+1}}\Big ],\quad k=1,\ 2,\cdots ,\end{aligned}$$

(2.26)

$$\begin{aligned} \Big |\widehat{\mu _{>n_k}}\Big (\frac{\lambda }{p_1p_2\cdots p_{n_k}}\Big )\Big |>\varepsilon _1,\quad \forall \ \lambda \in \Lambda _k,\ k=2,\ 3,\cdots . \end{aligned}$$

(2.27)

Claim. The set $\Lambda _k$ is a spectrum of the probability measure $\delta _{p_1^{-1}D_1}*\delta _{p_1^{-1}p_2^{-1}D_2}*\cdots *\delta _{p_1^{-1}p_2^{-1}\cdots p_{n_k}^{-1}D_{n_k}}$ for all $k=1,\ 2,\cdots$.

Let $\Gamma =\bigcup \limits _{k=1}^\infty \Lambda _k$. We shall prove $\Gamma$ is a spectrum of $\mu$.

For any $a\ne b\in \Gamma$, from (2.25) it follows that $a\ne b\in \Lambda _k$ for some $k>0$. Hence, $a-b$ is a zero point of the Fourier transform of $\delta _{p_1^{-1}D_1}*\delta _{p_1^{-1}p_2^{-1}D_2}*\cdots *\delta _{p_1^{-1}p_2^{-1}\cdots p_{n_k}^{-1}D_{n_k}}$. Hence, $\widehat{\mu }(a-b)=0$, which implies the exponential function set $E_\Gamma =\{e^{2\pi i\gamma x}:\gamma \in \Gamma \}$ is an orthogonal family of $L^2(\mu )$.

Assume, on the contrary, that $\Gamma$ is not a spectrum of $\mu$. Then, [2, Proposition 2.3] shows that $Q_{\mu ,\Gamma }(x_0)<1$ for some $x_0\in {\mathbb {R}}$.

Recall that $\lim \limits _{k\rightarrow \infty }(p_1p_2\cdots p_{n_{k}})^{-1}x_0=0$ and $\widehat{\Phi }:=\{\widehat{\nu }:\nu \in \Phi \}$ (here $\Phi =\{\mu _{>n}:\ n\ge 1\}$) is equi-continuous. From (2.26) it follows that

$$\begin{aligned} \beta _{k}:=\inf _{\lambda \in \Lambda _k}|\widehat{\mu _{>n_k}}((p_1p_2\cdots p_{n_{k+1}})^{-1}(\lambda +x_0))|\rightarrow 1 \text{ as } k\rightarrow \infty . \end{aligned}$$

(2.28)

Furthermore, from (2.27) it follows that there exists a positive integer $k_0>0$ such that for any $k\ge k_0$ and $\lambda \in \Lambda _k$, we have

$$\begin{aligned} |\widehat{\mu _{>n_k}}((p_1p_2\cdots p_{n_{k}})^{-1}(\lambda +x_0))|\ge \frac{1}{2}\varepsilon _1. \end{aligned}$$

(2.29)

Let

$$\begin{aligned} Q_{k}(x_0)=\sum \limits _{\lambda \in \Lambda _k}|\widehat{\mu } (\lambda +x_0)|^2,\quad \ k=1,2,\cdots . \end{aligned}$$

According to [2, (2.2)] and (2.29), for $k\ge k_0$ we have

$$\begin{aligned} \begin{array}{rl} &{}Q_{k+1}(x_0)-Q_{k}(x_0)\\ &{}\quad = \sum \limits _{\lambda \in \Lambda _{k+1}\setminus \Lambda _k}\prod _{n=1}^\infty |m_{D_n}((p_1\cdots p_n)^{-1}(\lambda +x_0))|^2\\ &{}\quad = \sum \limits _{\lambda \in \Lambda _{k+1}\setminus \Lambda _k}\prod _{n=1}^{n_{k+1}} |m_{D_n}((p_1\cdots p_n)^{-1}(\lambda +x_0)|^2|\widehat{\mu _{>n_{k+1}}}((p_1\cdots p_{n_{k+1}})^{-1}(\lambda +x_0))|^2\\ &{}\quad \ge \frac{1}{4}\varepsilon ^2_1\sum \limits _{\lambda \in \Lambda _{k+1}\setminus \Lambda _k}\prod _{n=1}^{n_{k+1}} |m_{D_n}((p_1\cdots p_n)^{-1}(\lambda +x_0))|^2. \end{array} \end{aligned}$$

(2.30)

The above claim shows

$$\begin{aligned} \sum \limits _{\lambda \in \Lambda _{k+1}}\prod _{n=1}^{n_{k+1}} \left| m_{D_n}(p_1^{-1}\cdots p_n^{-1}(\lambda +x_0))\right| ^2=1,\quad k=1,\ 2,\cdots . \end{aligned}$$

Thus (2.30) implies that for any $k\ge k_0$, we have

$$\begin{aligned} Q_{k+1}(x_0)-Q_{k}(x_0) \ge \frac{1}{4}\varepsilon ^2_1\left( 1-\sum \limits _{\lambda \in \Lambda _k}\prod _{n=1}^{n_{k+1}}\left| m_{D_n}((p_1\cdots p_n)^{-1}(\lambda +x_0))\right| ^2\right) . \end{aligned}$$

On the other hand, by the definition of $\beta _k$ in (2.28) for $k\ge 1$, we have

$$\begin{aligned} \begin{array}{rl} Q_{k}(x_0)=&{}\sum \limits _{\lambda \in \Lambda _k}\prod _{n=1}^\infty |m_{D_n}((p_1\cdots p_n)^{-1}(\lambda +x_0)|^2\\ \ge &{} \beta ^2_k\sum \limits _{\lambda \in \Lambda _k}\prod _{n=1}^{n_{k+1}}\left| m_{D_n}((p_1\cdots p_n)^{-1}(\lambda +x_0))\right| ^2. \end{array} \end{aligned}$$

By the above inequality, we have

$$\begin{aligned} Q_{k+1}(x_0)-Q_{k}(x_0) \ge \frac{1}{4}\varepsilon ^2_1\left( 1-\beta ^{-2}_kQ_k(x_0)\right) ,\quad \forall \ k\ge k_0. \end{aligned}$$

Therefore, the limit property in (2.28) shows

$$\begin{aligned}\begin{array}{rl} \liminf \limits _{k\rightarrow \infty }(Q_{k+1}(x_0)-Q_{k}(x_0))\ge &{}\frac{1}{4}\varepsilon ^2_1\left( 1-\lim \limits _{k\rightarrow \infty }\beta _k^{-2}Q_{k}(x_0)\right) \\ =&{}\frac{1}{4}\varepsilon ^2_1\left( 1-Q_{\mu ,\Gamma }(x_0)\right) >0. \end{array} \end{aligned}$$

Together with (2.25), the above inequalities imply

$$\begin{aligned} 1>Q_{\mu ,\Gamma }(x_0)=\lim \limits _{k\rightarrow \infty }Q_{k}(x_0)\ge \sum \limits _{k=1}^\infty \left( Q_{k+1}(x_0)-Q_{k}(x_0)\right) =+\infty , \end{aligned}$$

which is impossible. Hence, $\Gamma$ is a spectrum of $\mu$. The sufficiency of [2, Theorem 1.1] is proven. $\square$