# Spectral characterization of Cantor-Moran measures

Jinsong Liu1,2 ,  Zheng-Yi Lu∗1 and  Ting Zhou3 1 HLM, Academy of Mathematics
and Systems Science, Chinese Academy of Sciences, Beijing 100190, China  2
School of Mathematical Sciences, University of Chinese Academy of Sciences,
Beijing 100049, China  3 School of Mathematics and Statistics, Hubei Key
Laboratory of Mathematical Sciences, Central China Normal University, Wuhan
430079, China<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples on
$\mathbb{R}$. Suppose that the associated Cantor-Moran measure

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{(p_{2}p_{1})^{-1}\mathcal{D}_{2}}\ast\cdots,$

where $\sup_{n}\\{|p_{n}^{-1}d|:d\in\mathcal{D}_{n}\\}<\infty$ and
$\sup\\#\mathcal{D}_{n}<\infty$. It has been observed that the spectrality of
$\mu_{\\{p_{n},D_{n}\\}}$ is determined by equi-positivity. A significant
problem is what kind of Moran measures can satisfy this property. In this
paper, we introduce the conception of Double Points Condition Set (DPCS) to
characterize the equi-positivity equivalently. As applications of our
characterization, we show that all singular continuous Cantor-Moran measures
are spectral.

###### Key words and phrases:

Spectral Measure; Orthogonal basis; Moran Measure; Equi-positivity.
∗ Corresponding author

###### 2010 Mathematics Subject Classification:

Secondary 28A78, 28A80; Primary 42C05, 42A85.

##  1\. Introduction

Let $\mu$ be a Borel probability measure on $\mathbb{R}^{s}$. We call $\mu$ a
spectral measure if there exists a countable set
$\Lambda\subset\mathbb{R}^{s}$ such that $E_{\Lambda}:=\\{e^{2\pi
i\langle\lambda,x\rangle}:\lambda\in\Lambda\\}$ forms an orthogonal basis for
the Hilbert space $L^{2}(\mu)$, the space of all square-integrable functions
with respect to $\mu$. In this case, the countable set $\Lambda$ is called a
spectrum of $\mu$. Furthermore, if the spectral measure is the restriction of
the Lebesgue measure on a bounded Borel subset $K$, we call $K$ a spectral
set.
A fundamental question in harmonic analysis is: what kind of measures are
spectral measures? The research on this problem was initiated by Fuglede [18],
whose famous conjecture(Fuglede’s Conjecture) is that $\Omega$ is a spectral
set on $\mathbb{R}^{s}$ if and only if $\Omega$ is a translational tile on
$\mathbb{R}^{s}$. The conjecture was proved to be wrong by Tao and others [29,
32] in $\mathbb{R}^{s}$ $(s\geq 3)$. However, it is unsolved in dimensions 1
and 2. So far, it is still enlightening in the research of spectral measures.

Recently, He, Lau and Lai [19] showed that $\mu$ is discrete with finite
support, singularly continuous or absolutely continuous with respect to
Lebesgue measure when the measure $\mu$ is spectral. In 1998, Jorgensen and
Pedersen [22] found the first singular and non-atomic spectral measure
$\mu_{\frac{1}{4}}$ (1/4-Cantor measure), which led to the study of singular
spectral measures becoming a research hot spot [7, 8, 9, 11, 20, 26, 27].
Furthermore, the discovery of singular spectral measures produces many results
which are different from the classical Fourier transform. Strichartz [30, 31]
showed that the mock Fourier series for each continuous function on this
Cantor set converges uniformly to that function, which is better than the
classical case. But an interesting phenomenon is that the convergence of the
mock Fourier series in $L^{2}(\mu_{\frac{1}{4}})$ may be different for
distinct spectra of $\mu_{\frac{1}{4}}$ [12].
Self-affine measure is a kind of fundamental singular continuous measure,
which is the probability measure uniquely generated by

$\mu_{R,D}(\cdot)=\frac{1}{\\#D}\sum_{d\in D}\mu_{R,D}(R\cdot-d),$

where expanded matrix $R\in M_{s}(\mathbb{Z})$ and a finite digit set
$D\subset\mathbb{Z}^{s}$. As we know, a lot of self-affine spectral measures
[10, 13, 15, 24] require that $(R,D,L)$ to satisfy Hadamard triple for some
$L\subset\mathbb{Z}^{s}$, which is useful in the structure of the spectrum.

######  Definition 1.1.

Let $R\in M_{s}(\mathbb{Z})$ be an expanded integer matrix. Let
$D,L\subset\mathbb{Z}^{s}$ be finite sets with $\\#D=\\#L\;(\\#$ denotes the
cardinality). We say that system $(R,D,L)$ forms a Hadamard triple if the
matrix

$H=\frac{1}{\sqrt{\\#D}}[e^{2\pi i\langle R^{-1}d,\ell\rangle}]_{\ell\in
L,d\in D}$

is unitary, $i.e.$, $H^{\ast}H=I$.

In general, Strichartz [31] first studied the spectrality of the Moran measure
(generalized self-affine measure). For a sequence of Hadamard triples
$\\{(R_{n},B_{n},L_{n})\\}_{n=1}^{\infty}$, he studied the spectrality of the
associated Moran measure (under some conditions)

$\mu_{\\{R_{n},B_{n}\\}}=\delta_{R_{1}^{-1}B_{1}}\ast\delta_{R_{1}^{-1}R_{2}^{-1}B_{2}}\ast\cdots$

with a compact support

$T=\left\\{\sum_{j=1}^{\infty}(R_{j}R_{j-1}\cdots R_{1})^{-1}b_{j}:b_{j}\in
B_{j}\;\text{for\;all}\;j\right\\},$

where $\delta_{A}=\frac{1}{\\#A}\sum_{a\in A}\delta_{a}$, and $\delta_{a}$ is
the Dirac measure at $a$ and convergence is in weak sense. The Moran measure
$\mu_{\\{R_{n},B_{n}\\}}$ is also called Cantor-Moran measure on $\mathbb{R}$.
In recent years, many positive results appeared in many papers [3, 2, 4, 5,
28], they proved that the Cantor-Moran measure $\mu_{\\{R_{n},B_{n}\\}}$ is
spectral when $\\{(R_{n},B_{n},L_{n})\\}$ is a sequence of Hadamard triples
with a strong assumption on $L_{n}$, or $\\#B_{n}\leq 4$. Until 2019, An, Fu
and Lai [1] introduced the concept of equi-positivity (we will explain this
notion in Section 2) to characterize the spectrality of Cantor-Moran measure
$\mu_{\\{R_{n},B_{n}\\}}$ when $B_{n}\subset\\{0,1,\cdots,R_{n}-1\\}$.

The following theorem is one of their main result.

######  Theorem A ([1]).

Let ${(R_{n},B_{n},L_{n})}$ be a sequence of Hadamard triples with

$B_{n}\subset\\{0,1,\cdots,R_{n}-1\\}.$

Suppose that $\underline{\lim}_{n\rightarrow\infty}\\#B_{n}<\infty$. Then
there exists a equi-positive subsequence $\\{\nu_{n_{k}}\\}$ of
$\\{\nu_{n}\\}$, where

$\nu_{n}=\delta_{R_{n+1}^{-1}B_{n+1}}\ast\delta_{R_{n+1}^{-1}R_{n+2}^{-1}B_{n+2}}\ast\cdots.$

In particular, the associated Cantor-Moran measure $\mu_{\\{R_{n},B_{n}\\}}$
is a spectral measure.

This result solves the spectrality of most Cantor-Moran measures supported
inside the unit interval. However, many cases are still unknown, such as the
support of $\mu_{\\{R_{n},B_{n}\\}}$ is outside $[0,1]$. In the same paper,
they showed that the spectral property of Moran measure is determined by the
existence of equi-positive subsequence of $\\{\nu_{n}\\}$ under the assumption

$\sup_{n}\\{|R_{n}^{-1}d|:d\in B_{n}\\}<\infty.$ (1.1)

Simultaneously, they found that there exists a non-spectral measure
$\mu_{\\{2,B_{n}\\}}$ satisfying $B_{1}=\\{0,1\\}$ and $B_{n}=3B_{1}(n\geq
2)$, which also means that the equi-positive subsequence of $\\{\nu_{n}\\}$
may not exist under (1.1). We aim to address the following question.

######  Question 1.1.

What Moran measure can guarantee the existence of equi-positive subsequence of
$\\{\nu_{n}\\}$?

In this paper, we focus on the spectrality of the Moran measures supported
outside the unit interval under a natural assumption
$\sup_{n}\\#B_{n}<\infty$. Our main goal is to characterize the existence of
equi-positive subsequences of $\\{\nu_{n}\\}$. In order to complete it, we
introduce the concepts of Double Points Condition(DPC) and DPC Set to
characterize the existence equivalently on $\mathbb{R}$(Theorem 2.1). As an
interesting application, we show that all singular continuous Cantor-Moran
measures are spectral. For those absolutely continuous Cantor-Moran measures
$\mu_{\\{R_{n},B_{n}\\}}$, we find that a necessary and sufficient condition
for the existence of equi-positive subsequences of $\\{\nu_{n}\\}$ is
$\gcd_{j\geq n}\\{\gcd B_{j}\\}\equiv 1$ for all $n\geq 1$. Significantly, if
the Moran set $\sum_{n=1}^{\infty}(R_{n}R_{n-1}\cdots R_{1})^{-1}B_{n}$ is a
translational tile on $\mathbb{R}$, then Fuglede’s Conjecture holds in this
case.

The main difficulties of this paper are reflected in the following aspects.

$\bullet$ When we consider the equi-positivity of $\\{\nu_{n}\\}$ on
$\mathbb{R}$, Proposition 3.6 tells us that it is a crucial step to deal with
the possible weak limit of $\\{\nu_{n}\\}$ when the integral periodic zero set
of the weak limit is nonempty. By [1], we know the weak limit of
$\\{\nu_{n}\\}$ can only be $\frac{1}{2}(\delta_{\\{0\\}}+\delta_{\\{1\\}})$
when $\text{spt}\nu_{n}\subset[0,1]$, but the weak limit may be varied in our
case. How to find the subsequence $\\{\nu_{n_{k}}\\}$ and determine its weak
limit is a tough question. To solve this problem, we use the diagonal
principle to determine the form of the weak limit. Then by Argument Principle,
Arzelà-Ascoli Theorem, Wiener theorem and some techniques, we obtain
sufficient conditions for the existence of equi-positive subsequences of
$\\{\nu_{n}\\}$.

$\bullet$ We extend Proposition 4.9 of [16] by Covering Lemma to study the
singularity of Cantor-Moran measure.

$\bullet$ In the process of studying the non-spectrality of Cantor-Moran
measures, we flexibly apply the concept of selection mapping to characterize
the structure of orthogonal sets.

We organize the paper as follows.

In the second section, we introduce some necessary preliminaries about DPC and
DPCS, and state our main results. Section 3 is devoted to the study of DPC,
DPCS and equi-positivity. In Section 4, we characterize the equivalent
conditions for the existence of equi-positive subsequences of $\\{\nu_{n}\\}$
on $\mathbb{R}$. The spectral properties of singular Cantor-Moran measures are
left to Section 5. In Section 6, we study the non-spectrality of Cantor-Moran
measures.

##  2\. Preliminaries and main results

In this section, we will give some basic notations for our paper. In
particular, we state our main results.

###  2.1.  Preliminaries

At the beginning of this section, recall that the Fourier transform of a Borel
probability measure $\mu$ on $\mathbb{R}$ is defined to be

$\widehat{\mu}(\xi)=\int_{\mathbb{R}}e^{2\pi i\langle
x,\;\xi\rangle}d\mu(x),\quad\xi\in\mathbb{R}.$

Denote by $\mathscr{Z}(f):=\\{x\in\mathbb{R}:f(x)=0\\}$ the set of zeros of
$f(x)$. Note that $\\{e^{2\pi i\langle\lambda,x\rangle}:\lambda\in\Lambda\\}$
is an orthogonal set in $L^{2}(\mu)$ if and only if
$\widehat{\mu}(\lambda-\lambda^{\prime})=0$ for any
$\lambda\neq\lambda^{\prime}\in\Lambda$, i.e.,

$(\Lambda-\Lambda)\setminus\\{0\\}\subset\mathscr{Z}(\widehat{\mu}(\xi)).$

In this case, $\Lambda$ is also called a bi-zero set of $\mu$.

Let $\mathcal{P}(T)$ be the set of all Borel probability measures supported on
compact set $T\subset\mathbb{R}.$ A sequence of measures
$\mu_{n}\in\mathcal{P}(T),\>n=1,2,\cdots,$ is said to converge weakly to a
measure $\mu\in\mathcal{P}(T)$ ($\mu_{n}\xrightarrow{w}\mu$) if, for any $f\in
C_{0}(\mathbb{R})$ (i.e. $f$ is continuous with compact support), we have

$\int fd\mu_{n}\rightarrow\int fd\mu,\;\;\text{as}\;n\rightarrow\infty.$

Let integer $|p_{n}|>1$, and let $\mathcal{D}_{n}\subset\mathbb{Z}$ be a
simple digit set on $\mathbb{R}$ for all $n\geq 1$. We call
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ a Moran measure if

$\mu_{n}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{(p_{2}p_{1})^{-1}\mathcal{D}_{2}}\ast\cdots\ast\delta_{(p_{n}\cdots
p_{1})^{-1}\mathcal{D}_{n}}\xrightarrow{w}\mu_{\\{p_{n},\mathcal{D}_{n}\\}},$

where $\delta_{D}(x)=\begin{cases}\frac{1}{\\#D},&x\in D;\\\
0,&\text{otherwise}.\end{cases}$ In this case, write

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{(p_{2}p_{1})^{-1}\mathcal{D}_{2}}\ast\cdots.$
(2.1)

It is easy to check that

$\widehat{\mu_{\\{p_{n},\mathcal{D}_{n}\\}}}(\xi)=\prod_{j=1}^{\infty}\widehat{\delta_{\mathcal{D}_{j}}}((p_{1}p_{2}\cdots
p_{j})^{-1}\xi),\quad\xi\in\mathbb{R}.$

In this paper, we always assume that there exists a compact set
$T\subset\mathbb{R}$ such that the support $\text{suppt}(\nu_{n})\subset T$,
where

$\nu_{n}=\delta_{p_{n+1}^{-1}\mathcal{D}_{n+1}}\ast\delta_{(p_{n+2}p_{n+1})^{-1}\mathcal{D}_{n+2}}\ast\cdots.$
(2.2)

This ensures the existence of $\nu_{n}$, and it also plays a pivotal role in
the completeness characterization of the orthogonal set of
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$. An, Fu and Lai [1] give the concept of
equi-positive family, which is important in characterization of the
spectrality of $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$.

######  Definition 2.1.

Let $\Phi$ be a collection of probability measures on compact set
$T\subset\mathbb{R}$. We say that $\Phi$ is an equi-positive family if there
exists $\epsilon_{0}>0$ such that, for all $x\in[0,1)$, and for all
$\nu\in\Phi$, there exists $k_{x,\nu}\in\mathbb{Z}$ with

$|\widehat{\nu}(x+k_{x,\nu})|>\epsilon_{0}.$

In [1], An et al. gave the following theorem, which is a basic criterion.

######  Theorem B ([1],[28]).

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples on
$\mathbb{R}$. Suppose that there exist a compact set $T\subset\mathbb{R}$ and
a subsequence $\\{n_{k}\\}$ such that all support of $\nu_{n_{k}}$ are in $T$,
and $\\{\nu_{n_{k}}\\}$ is an equi-positive family. Then
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure.

###  2.2.  Spectrality of Cantor-Moran measures

According to Theorem B, the spectral properties of most Cantor-Moran measures
in $\mathbb{R}$ depend on equi-positivity of $\\{\nu_{n}\\}$. In this
subsection, we give an equivalent characterization of equi-positivity. Now we
introduce some new concepts.

######  Definition 2.2 (DPCS, DPC).

Suppose that $D\subset\mathbb{Z}$ and $p>1$. Define $T_{p,D}\subset[0,1)$ to
be the Double Points Condition Set (DPCS) of $(p,D)$, i.e., $T_{p,D}$ is the
set of all points $\xi$ with the given property: there exist
$\ell_{1},\ell_{2}\in\\{0,1,\cdots,p-1\\}$ such that
$\widehat{\delta_{p^{-1}D}}(\xi+\ell_{i})\neq 0$ for $i=1,2$. Moreover, we
call that $(p,D)$ satisfies Double Points Condition (DPC) if $T_{p,D}=[0,1)$.

On $\mathbb{R}$, it is easy to check that there exists a compact set $T$ such
that all support of $\nu_{n}$ are in $T$ if and only if
$\sup_{n}\sup_{d_{n}\in D_{n}}|p_{n}^{-1}d_{n}|<\infty$. For convenience, we
always assume that $p_{n}>0$ and $0\in\mathcal{D}_{n}\subset\mathbb{N}$ for
all $n\geq 1$.

######  Theorem 2.1.

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples on
$\mathbb{R}$, and let $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$, $\nu_{n}$ be given
as (2.1) and (2.2), respectively. Suppose that

$\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}<\infty\;\;\text{and}\;\;\sup_{n}\\#\mathcal{D}_{n}<\infty.$

Then the following statements are equivalent:
(i) there exists an equi-positive sequence $\\{\nu_{n_{k}}\\}$;
(ii) $\limsup_{k\rightarrow\infty}T_{p_{k},\mathcal{D}_{k}}\supset(0,1)$,
where $T_{p_{k},\mathcal{D}_{k}}$ is a DPCS of $(p_{k},\mathcal{D}_{k})$;
(iii) at least one of the following two conditions is satisfied:
$\\#\\{j:p_{j}>\\#\mathcal{D}_{j}\\}=\infty$ or $\gcd_{j\geq
n}\\{\gcd\mathcal{D}_{j}\\}\equiv 1,\;\;\forall n\geq 1$.
In particular, $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure.

The following theorem is the connection between the singularity and
spectrality of Cantor-Moran measure.

######  Theorem 2.2.

Suppose that $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ is a sequence of Hadamard
triples with $\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}<\infty$ and
$\sup_{n}\\#\mathcal{D}_{n}<\infty$, and let Cantor-Moran measure

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{p_{1}^{-1}p_{2}^{-1}\mathcal{D}_{2}}\ast\cdots.$

Then $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is singularly continuous with respect
to the Lebesgue measure if and only if there are infinitely many pairs
$(p_{n},\mathcal{D}_{n})$ satisfying DPC, i.e.
$\\#\\{j:p_{j}>\\#\mathcal{D}_{j}\\}=\infty$(See Lemma 3.2). In particular,
all singularly continuous Cantor-Moran measures are spectral.

By applying theorems 2.1 and 2.2 directly, we can obtain the spectral property
of absolutely continuous with respect to the Lebesgue measure.

######  Theorem 2.3.

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples
with $\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}<\infty$ and
$\sup_{n}\\#\mathcal{D}_{n}<\infty$, and let Cantor-Moran measure

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{p_{1}^{-1}p_{2}^{-1}\mathcal{D}_{2}}\ast\cdots.$

Suppose that $p_{n}=\\#\mathcal{D}_{n}$ and $\gcd_{j\geq
n}\\{\gcd\mathcal{D}_{j}\\}\equiv 1$ for all large $n$. Then
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is an absolutely continuous with respect
to the Lebesgue spectral measure.

It is worth mentioning that the above conclusion is related to Fuglede’s
Conjecture. Dutkay and Lai [14] proved that an absolutely continuous measure
is spectral only if it is a normalized Lebesgue restricting on a domain
$T\subset\mathbb{R}$. If Fuglede’s Conjecture holds, then the support of
Cantor-Moran measure $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a translational
tile under the conditions of the above theorem.

Note that the equi-positivity of Moran measure
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ may be invalid under hypothesis

$\sup_{n}\\{p_{n}^{-1}d:d\in D_{n}\\}=\infty.$ (2.3)

For the study of spectral property, this condition is still inseparable, which
will be explained in the following proposition.

######  Proposition 2.4.

The following statements hold.
(i) There exists a Cantor-Moran measure $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$
satisfying (2.3) is not a spectral measure.
(ii) There exists a Cantor-Moran measure $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$
satisfying (2.3) is a spectral measure.

##  3\. DPC and equi-positivity

In this section, we will set up some brief and useful results for the rest of
this paper. In the first subsection, we characterize the equivalence
conditions of DPC in $\mathbb{R}$. In particular, we prove (ii)
$\Leftrightarrow$ (iii) of Theorem 2.1. The second subsection develops the
theory of equi-positivity of general measures and establishes the relation
between that and weak convergence. Finally, we give a useful estimate at the
end of this section.

###  3.1.  Equivalent conditions for DPC and DPCS in $\mathbb{R}$

As for prerequisites, the reader is expected to be familiar with the
properties of Hadamard triple.

######  Lemma 3.1 ([13]).

Let $(R,D,L)$ be a Hadamard triple on $\mathbb{R}^{s}$. Then the following
statements hold.
(i) For any $\xi\in\mathbb{R}^{s}$, one has

$\sum_{\ell\in L}|\widehat{\delta_{R^{-1}D}}(\xi+\ell)|^{2}=1;$

(ii) $(R,D,\widetilde{L})$ is a Hadamard triple, where $\widetilde{L}\equiv
L\pmod{R^{*}\mathbb{Z}^{s}}$;
(iii) $L$ is a spectrum of $\delta_{R^{-1}D}$.

Let $(p,D,L)$ be a Hadamard triple on $\mathbb{R}$. For convenience, we always
assume that $p>0$, $0\in D\subset\mathbb{N}$ and $0\in
L\subset\\{0,1,\cdots,p-1\\}:=L^{*}$.

######  Lemma 3.2.

Let $(p,\mathcal{D},L)$ be a Hadamard triple on $\mathbb{R}$. Then
$(p,\mathcal{D})$ satisfies DPC if and only if $p>\\#\mathcal{D}$.

###### Proof.

We first prove the sufficiency. Since $(p,\mathcal{D},L)$ is a Hadamard triple
and (i) of Lemma 3.1, we have

$\sum_{\ell\in
L}\left|\widehat{\delta_{\mathcal{D}}}\left(\frac{x+\ell}{p}\right)\right|^{2}\equiv
1,$ (3.1)

which implies that $\widehat{\delta_{\mathcal{D}}}(\frac{x+\ell_{1}}{p})\neq
0$ for some $\ell_{1}\in L$.

If $|\widehat{\delta_{\mathcal{D}}}(\frac{x+\ell_{1}}{p})|<1$, by (3.1), it
follows that there exists $\ell_{2}\in L\setminus\\{\ell_{1}\\}$ with
$\widehat{\delta_{\mathcal{D}}}(\frac{x+\ell_{2}}{p})\neq 0$, which implies
that the sufficiency follows.

If $|\widehat{\delta_{\mathcal{D}}}(\frac{x+\ell_{1}}{p})|=1$, i.e.,
$|\sum_{d\in\mathcal{D}}e^{2\pi i\frac{(x+\ell_{1})d}{p}}|=\\#\mathcal{D}$,
then $\frac{(x+\ell_{1})d}{p}=\frac{(x+\ell_{1})d^{\prime}}{p}\pmod{1}$ for
any $d\neq d^{\prime}$. According to $0\in\mathcal{D}$, one has
$\frac{(x+\ell_{1})d}{p}\in\mathbb{Z}$ for all $d\in\mathcal{D}$. Hence

$\frac{x+\ell_{1}}{p}\in\frac{1}{\gcd\mathcal{D}}\mathbb{Z}.$ (3.2)

Let $\gcd\mathcal{D}=\mathbf{d}\geq 1$. It follows from (3.2) that
$x=\frac{pz_{1}}{\mathbf{d}}-\ell_{1}$ for some $z_{1}\in\mathbb{Z}.$
Supposing that $\widehat{\delta_{\mathcal{D}}}(\frac{x+\ell}{p})=0$ for any
$\ell\in L^{*}\setminus\\{\ell_{1}\\}$, it implies

$\left|\widehat{\delta_{\mathcal{D}}}\left(\frac{x+\ell}{p}\right)\right|=\left|\widehat{\delta_{D}}\left(\frac{z_{1}}{\mathbf{d}}+\frac{\ell-\ell_{1}}{p}\right)\right|=\left|\widehat{\delta_{\mathcal{D}}}\left(\frac{\ell-\ell_{1}}{p}\right)\right|=0.$

Noting that $\widehat{\delta_{\mathcal{D}}}(x)$ is a function with an integer
as the period, hence $\widehat{\delta_{\mathcal{D}}}(\frac{i}{p})=0$ for all
$i\in\\{1,\cdots,p-1\\}$. Therefore

$\frac{1}{p}(L^{*}-L^{*})\setminus\\{0\\}\subset\mathscr{Z}(\widehat{\delta_{\mathcal{D}}}),$

$i.e.$, $L^{*}$ is a bi-zero set of $\delta_{p^{-1}\mathcal{D}}$. From
$L\subset L^{*}$, one obtains $L=L^{*}$, which contradicts with the condition
$\\#L=\\#\mathcal{D}<p=\\#L^{*}$.

In the following, we prove the necessity by a contradiction. Suppose that
$p=\\#\mathcal{D}$. Due to $(p,\mathcal{D},L)$ is a Hadamard triple and (ii)
of Lemma 3.1, write $L=\\{0,1,\cdots,p-1\\}$. Note that

$\sum_{\ell\in L}|\widehat{\delta_{p^{-1}\mathcal{D}}}(\xi+\ell)|^{2}=1.$

Choose $\xi_{0}=0$. It is easy to see
$\widehat{\delta_{p^{-1}\mathcal{D}}}(\xi_{0})=1$ and

$\widehat{\delta_{p^{-1}\mathcal{D}}}(\xi_{0}+\ell)=0$

for all $\ell\in L\setminus\\{0\\}$, which implies that $0\not\in
T_{p,\mathcal{D}}$. ∎

As we all know, $p\geq\\#\mathcal{D}$ when $(p,\mathcal{D},L)$ forms a
Hadamard triple. In this case, we study the property of $T_{p,\mathcal{D}}$ if
$p=\\#\mathcal{D}$.

######  Lemma 3.3.

Let $(p,\mathcal{D},L)$ be a Hadamard triple with $p=\\#\mathcal{D}$. Then
$T_{p,\mathcal{D}}$ is a DPCS of $(p,\mathcal{D})$ if and only if
$T_{p,\mathcal{D}}=[0,1)\setminus\frac{1}{\gcd\mathcal{D}}\mathbb{Z}$.

###### Proof.

Without loss of generality, we assume $L=\\{0,1,\cdots,p-1\\}$. It follows
that, for any $\xi\in\mathbb{R}$,

$\sum_{\ell\in L}|\widehat{\delta_{p^{-1}\mathcal{D}}}(\xi+\ell)|^{2}\equiv
1.$ (3.3)

We assert that $\gcd(\gcd\mathcal{D},p)=1$. Suppose that $p=dp^{\prime}$ and
$\mathcal{D}=d\mathcal{D}^{\prime}$, where $\gcd(\gcd\mathcal{D},p)=d>1$. Note
that $L$ is a bi-zero set of
$\delta_{p^{-1}\mathcal{D}}=\delta_{p^{\prime-1}\mathcal{D}^{\prime}}$. Taking
$\ell_{1}=0$, $\ell_{2}=p^{\prime}\in L$, this implies that

$0=\widehat{\delta_{p^{\prime-1}\mathcal{D}^{\prime}}}(\ell_{2}-\ell_{1})=\widehat{\delta_{\mathcal{D}^{\prime}}}(1)=1.$

It is impossible, so the claim follows. If $\xi=\frac{z}{\gcd\mathcal{D}}$ for
some integer $z$, then by the claim, there exists $\ell\in L$ such that
$z+\ell\gcd\mathcal{D}=0\pmod{p}$. Hence

$\widehat{\delta_{p^{-1}\mathcal{D}}}(\xi+\ell)=\widehat{\delta_{\mathcal{D}}}((p\gcd\mathcal{D})^{-1}(z+\ell\gcd\mathcal{D}))=1.$

From (3.3) and the definition of $T_{p,\mathcal{D}}$, one has
$T_{p,\mathcal{D}}\subset[0,1)\setminus\frac{1}{\gcd\mathcal{D}}\mathbb{Z}$.
On the other hand, if $|\widehat{\delta_{p^{-1}\mathcal{D}}}(\xi+\ell)|=1$ for
some $\xi\in[0,1)$ and $\ell\in L$, then $p^{-1}d(\xi+\ell)\in\mathbb{Z}$ for
all $d\in\mathcal{D}$. This yields that
$\xi+\ell\in\frac{p}{\gcd\mathcal{D}}\mathbb{Z}$, i.e.
$\xi\in\frac{1}{\gcd\mathcal{D}}\mathbb{Z}$, which completes the proof. ∎

Proof of Theorem 2.1: (ii) $\Leftrightarrow$ (iii)

(ii) $\Rightarrow$ (iii): We prove the result by a contradiction. If (iii)
doesn’t hold, we assume that there exist integers $N$ and $d>1$ such that
$p_{n}=\\#\mathcal{D}_{n}$ and $d|\gcd\mathcal{D}_{n}$ for all $n\geq N$.
Together with Lemma 3.3, it implies that $\frac{1}{d}\not\in
T_{p_{n},\mathcal{D}_{n}},\;\forall n\geq N$. Hence
$\frac{1}{d}\not\in\cup_{k\geq n}T_{p_{k},\mathcal{D}_{k}}$ for any $n\geq N$,
a contradiction. This proves the result.

(iii) $\Rightarrow$ (ii): If $\\#\\{j:p_{j}>\\#\mathcal{D}_{j}\\}=\infty$, by
using Lemma 3.2 and definition of DPC, we obtain that $\cup_{k\geq
n}T_{p_{k},\mathcal{D}_{k}}\supset[0,1)$ for all $n\geq N$ and some $N\geq 1$.
Hence
$\limsup_{k\rightarrow\infty}T_{p_{k},\mathcal{D}_{k}}=\cap_{n=1}^{\infty}\cup_{k\geq
n}T_{p_{k},\mathcal{D}_{k}}\supset[0,1)$. For another case, suppose that there
exists $N$ such that $p_{n}=\\#\mathcal{D}_{n}$ and $\gcd_{j\geq
n}\\{\gcd\mathcal{D}_{j}\\}=1$ for all $n\geq N$. It follows from Lemma 3.3
that

$\cup_{k\geq n}T_{p_{k},\mathcal{D}_{k}}=\cup_{k\geq
n}([0,1)\setminus\frac{1}{\gcd\mathcal{D}_{k}}\mathbb{Z})=[0,1)\setminus(\cap_{k\geq
n}\frac{1}{\gcd\mathcal{D}_{k}}\mathbb{Z})=(0,1),$

which gives the desired result.

###  3.2. Equi-positivity and weak convergence

In this subsection, we obtain the properties of equi-positivity and weak
convergence. The following well-known results are useful in this paper.

######  Theorem 3.4 ([17]).

Let $\mu_{1},\mu_{2},\cdots$ be locally finite Borel measures on
$\mathbb{R}^{s}$ with $\sup_{n}\mu_{n}(A)<\infty$ for all bounded set $A$.
Then $\\{\mu_{n}\\}_{n=1}^{\infty}$ has a weakly convergent subsequence.

######  Lemma 3.5 ([6]).

Let $\mu,\mu_{1},\mu_{2},\cdots$ be finite Borel measures on $\mathbb{R}^{s}$.
Then the following are equivalent.
$(i)$ $\\{\mu_{n}\\}$ converges weakly to a finite Borel measure $\mu$.
$(ii)$ For any open set $O$,
$\mu(O)\leq\underline{\lim}_{n\rightarrow\infty}\mu_{n}(O)$.
$(iii)$ For any compact set $K$,
$\mu(K)\geq\overline{\lim}_{n\rightarrow\infty}\mu_{n}(K)$.
$(iv)$ The Fourier transform $\\{\widehat{\mu_{n}}(\xi)\\}$ converges to
$\widehat{\mu}(\xi)$ uniformly on all compact subsets of $\mathbb{R}^{s}$.

For any $x\in\mathbb{R}^{s}$, denote the ball
$B(x;\delta):=\\{y\in\mathbb{R}^{s}:\text{dist}(x,y)<\delta\\}$. For any set
$A\subset\mathbb{R}^{s}$, write
$B(A;\delta):=\\{y\in\mathbb{R}^{s}:\text{dist}(A,y)<\delta\\}$.

######  Proposition 3.6.

Let $\mu$ be a Borel probability measure with a compact support on
$\mathbb{R}^{s}$, and let the integral periodic zero set

$\mathcal{Z}(\mu)=\\{\xi\in[0,1)^{s}:\widehat{\mu}(\xi+k)=0,\;\text{for\;all}\;k\in\mathbb{Z}^{s}\\}.$

For any $\delta>0$, then

$\inf_{x\in[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta)}\sup_{k\in\mathbb{Z}^{s}}|\widehat{\mu}(x+k)|>0.$

If $\mu_{n}\xrightarrow{w}\mu$, then for any $\delta>0$, there exists $N$ such
that, for all $n\geq N$,

$\inf_{x\in[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta)}\sup_{k\in\mathbb{Z}^{s}}|\widehat{\mu_{n}}(x+k)|>0.$

###### Proof.

We prove the lemma by finite covering theorem. For any $x\in[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta)$, by the definition of $\mathcal{Z}(\mu)$, we
obtain

$|\widehat{\mu}(x+k_{x})|\geq\epsilon_{x}$

for some $k_{x}\in\mathbb{Z}^{s}$ and $\epsilon_{x}>0$. The continuity gives
that

$|\widehat{\mu}(y+k_{x})|\geq\frac{1}{2}\epsilon_{x},\;\forall y\in
B(x;\delta_{x})$ (3.4)

for some $\delta_{x}>0$. Note that $[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta)$ is a compact set and

$\cup_{x\in[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta)}B(x;\delta_{x})\supset[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta).$

There exist balls $\\{B(x_{i};\delta_{x_{i}})\\}_{i=1}^{m}$ satisfying

$\cup_{i=1}^{m}B(x_{i};\delta_{x_{i}})\supset[0,1]^{s}\setminus
B(\mathcal{Z}(\mu);\delta).$

Let $\epsilon:=\frac{1}{2}\min_{i}\epsilon_{x_{i}}$. For any
$x\in[0,1]^{s}\setminus B(\mathcal{Z}(\mu);\delta)$, there exists $i$ such
that $x\in B(x_{i};\delta_{x_{i}})$. It follows from (3.4) that

$|\widehat{\mu}(x+k_{x_{i}})|\geq\frac{1}{2}\epsilon_{x_{i}}\geq\epsilon>0,$

which gives the desired inequality.

Now assume that $\mu_{n}\xrightarrow{w}\mu$. It follows from (3.4) and (iv) of
Lemma 3.5 that there exists $N_{i}>0$ such that

$|\widehat{\mu_{n}}(x+k_{x_{i}})|\geq\frac{1}{4}\epsilon_{x_{i}},\;\>\>\forall
x\in B(x_{i};\delta_{x_{i}})$

for all large $n\geq N_{i}$. Finally, the conclusion follows from the
finiteness of the number of $\\{x_{i}\\}$ and $N=\max N_{i}$. ∎

By the above conclusion, the following result holds directly.

######  Lemma 3.7.

Let $\\{\mu_{n}\\}$ and $\mu$ be Borel probability measures with compact
supports on $\mathbb{R}^{s}$. Suppose that $\mu_{n}\xrightarrow{w}\mu$. If
$\mathcal{Z}(\mu)=\emptyset$, then there exists an equi-positive subsequence
$\\{\mu_{n_{k}}\\}$ of $\\{\mu_{n}\\}$.

######  Theorem 3.8.

Let $\\{\mu_{n}\\}$ and $\mu$ be Borel probability measures with compact
supports on $\mathbb{R}^{s}$. Suppose that $\mu_{n}\xrightarrow{w}\mu$. Then
$\mathcal{Z}(\mu)=\emptyset$ if and only if there exist $\varepsilon,N>0$ such
that $\mathcal{Z}(\mu_{n})=\emptyset$ for $n\geq N$ and

$\sup_{n\geq
N,\xi\in[0,1)^{s}}\min\\{|k_{\xi}|:|\widehat{\mu_{n}}(\xi+k_{\xi})|\geq\varepsilon\\}<\infty.$
(3.5)

In particular, if $\mu_{n}$ can be written as
$\mu_{n}=\delta_{1}\ast\cdots\ast\delta_{n}$, where $\delta_{j}$ is a discrete
probability measure with $\sup_{n}\\#\text{spt}\delta_{n}<\infty$ and
$\text{spt}\mu\subset K$ for some compact set $K$, then
$\mathcal{Z}(\mu)=\emptyset$ if and only if there exist $N>0$ such that
$\mathcal{Z}(\mu_{n})=\emptyset$ for $n\geq N$ and

$\sup_{n\geq
N,\xi\in[0,1)^{s}}\min\\{|k_{\xi}|:\widehat{\mu_{n}}(\xi+k_{\xi})\neq
0\\}<\infty.$

###### Proof.

We prove the necessity of the first conclusion. Assume by contradiction that
$\mathcal{Z}(\mu)\neq\emptyset$. Let $\xi\in\mathcal{Z}(\mu)$. The assumption
(3.5) implies that, for all $n\geq N$ there exists
$k_{n,\xi}\in\mathbb{Z}^{s}$ such that $|k_{n,\xi}|<C$ and

$|\widehat{\mu_{n}}(\xi+k_{n,\xi})|\geq\varepsilon>0,$

where $C$ is the supremum of (3.5). Choose a subsequence
$\\{n_{j}\\}_{j=1}^{\infty}\subset\mathbb{N}$ such that

$|\widehat{\mu_{n_{j}}}(\xi+k_{n_{j},\xi})|\geq\varepsilon>0$

and $k_{n_{j},\xi}=k_{n_{j^{\prime}},\xi}$ for any $j\neq j^{\prime}$. Hence,
by (iv) of Lemma 3.4,

$|\widehat{\mu}(\xi+k_{n_{1},\xi})|=\lim_{j\rightarrow\infty}|\widehat{\mu_{n_{j}}}(\xi+k_{n_{j},\xi})|\geq\varepsilon>0.$

This yields that $\xi\not\in\mathcal{Z}(\mu)$, a contradiction.

We next consider the sufficiency. For any $\xi\in\mathbb{R}^{s}$, there exists
$k_{\xi}\in\mathbb{Z}$ such that

$\widehat{\mu}(\xi+k_{\xi})\neq 0.$

Then there exists a ball $B(\xi;\delta_{x})$ such that

$|\widehat{\mu}(y+k_{\xi})|\geq\epsilon_{\xi}>0,\;\>\>\forall y\in
B(\xi;\delta_{x}).$

By using the fact $[0,1]^{s}\subset\cup_{\xi\in[0,1]^{s}}B(\xi;\delta_{\xi})$
and finite covering theorem, there exist $\\{\xi_{i}\\}_{i=1}^{m}$ such that
$[0,1]^{s}\subset\cup_{j=1}^{m}B(\xi_{j},\delta_{\xi_{j}})$. Hence for any
$\xi\in[0,1]^{s}$, there exists $B(\xi_{j};\delta_{\xi_{j}})\ni\xi$ such that

$|\widehat{\mu}(\xi+k_{\xi_{i}})|\geq\epsilon_{0}>0,$

where $\epsilon_{0}:=\min_{1\leq j\leq m}\epsilon_{\xi_{j}}$. Thus by (iv) of
Lemma 3.5, there exists $N>0$ such that
$|\widehat{\mu}_{n}(\xi+k_{\xi_{i}})|\geq\frac{1}{2}\epsilon_{0}$ for all
$n\geq N$.

In the following, we assume that $\mu_{n}=\delta_{1}\ast\cdots\ast\delta_{n}$,
where $\delta_{j}$ is a discrete probability measure with
$\sup_{n}\\#\text{spt}\delta_{n}<\infty$. From the above conclusion, we would
need to show (3.5). The existence of $\mu$ allows us to assume
$\mu_{>n}:=\delta_{n+1}\ast\delta_{n+2}\ast\cdots$ exists and
$\mu=\mu_{n}\ast\mu_{>n}$. By Corollary 1.2 in [25], one has
$\sum_{j=1}^{\infty}\max\\{|a|:a\in\text{spt}\delta_{j}\\}<\infty$. Therefore
$\lim_{n\rightarrow\infty}\sum_{j=n}^{\infty}\max\\{|a|:a\in\text{spt}\delta_{j}\\}=0$.
This also yields that
$\lim_{n\rightarrow\infty}\max\\{|a|:a\in\text{spt}\mu_{>n}\\}=0$. Then, for
any compact set $K^{\prime}\subset\mathbb{R}^{s}$ and $\xi\in K^{\prime}$,
there exists $N_{K^{\prime}}>0$ such that
$\max\\{|a|:a\in\text{spt}\mu_{>n}\\}\max\\{|a|:a\in
K^{\prime}\\}\leq\frac{1}{6}$ for any $n\geq N_{K^{\prime}}$. Hence

$|\widehat{\mu_{>n}}(\xi)|=\left|\int e^{2\pi i\xi
x}d\mu_{>n}(x)\right|\geq\left|\int\cos 2\pi\xi
xd\mu_{>n}(x)\right|\geq\frac{1}{2}$

for any $n\geq N_{K^{\prime}}$. Note that the condition

$\sup_{n\geq
N,\xi\in[0,1)^{s}}\min\\{|k_{\xi}|:\widehat{\mu_{n}}(\xi+k_{\xi})\neq
0\\}<\infty$

implies that $\\{\xi+k_{\xi}:\xi\in[0,1]^{s}\\}:=K^{\prime}$ is uniformly
bounded. We can find an integer $N^{\prime}\geq\max\\{N,N_{K^{\prime}}\\}$
such that for any $\xi\in[0,1]$, there exists $k_{\xi}$ satisfying
$|\widehat{\mu_{N^{\prime}}}(\xi+k_{\xi})|\geq\epsilon_{\xi}$ and
$|\widehat{\mu_{>N^{\prime}}}(\xi+k_{\xi})|\geq\frac{1}{2}.$ By using the
compactness of $[0,1]$ and the continuity of
$\widehat{\mu_{N^{\prime}}}(\xi)$, there exists $\varepsilon>0$ such that

$|\widehat{\mu_{N^{\prime}}}(\xi+k_{\xi})|\geq\varepsilon.$

Finally, we complete the proof from

$|\widehat{\mu}(\xi+k_{\xi})|=|\widehat{\mu_{N^{\prime}}}(\xi+k_{\xi})||\widehat{\mu_{>N^{\prime}}}(\xi+k_{\xi})|\geq\frac{1}{2}\varepsilon.$

∎

###  3.3. A useful estimate

######  Lemma 3.9.

Let integer $|p_{n}|>1$ and finite digit set $D_{n}\subset\mathbb{Z}$ for all
$n\geq 1$. Suppose that $C:=\sup_{n}\\{|p_{n}^{-1}d|:d\in D_{n}\\}<\infty$.
For any $n\geq 1$ and $m>0$, there exists $J_{0}:=\max\\{[\log_{2}8Cm],0\\}+1$
such that $|p_{n+1}^{-1}p_{n+2}^{-1}\cdots p_{n+J_{0}}^{-1}|Cm<\frac{1}{4}$
and

$\prod_{j=1}^{\infty}|\widehat{\delta_{p_{n+1}^{-1}\cdots
p_{n+J_{0}+j}^{-1}D_{n+J_{0}+j}}}(\xi)|\geq\frac{2}{\pi}$

for any $\xi\in\mathbb{R}$ with $|\xi|<m$.

###### Proof.

Since $|p_{n}^{-1}|<\frac{1}{2}$, there exists $0<r\leq\frac{1}{8}$ such that
$|p_{n+1}^{-1}\cdots p_{n+J_{0}}^{-1}|Cm\leq r$, i.e.,

$|p_{n+1}^{-1}\cdots p_{n+J_{0}+j}^{-1}d_{n+J_{0}+j}\xi|\leq
2^{-j+1}r,\;\;\forall|\xi|<m$

for any $d_{n+J_{0}+j}\in D_{n+J_{0}+j}$ and $j\geq 1$. Therefore the argument
of $e^{2\pi ip_{n+1}^{-1}\cdots p_{n+J_{0}+j}^{-1}d_{n+J_{0}+j}\xi}$ belongs
to $(-2^{-j+2}r\pi,2^{-j+2}r\pi)\subset(-\frac{\pi}{2},\frac{\pi}{2})$ for any
$|\xi|<m$. Let

$A_{j}:=\\{x+yi\in\mathbb{C}:x^{2}+y^{2}\leq 1,x\geq\cos 2^{-j+2}r\pi\\}.$

Note that $A_{j}$ is a convex set. Thus the convex combination

$\frac{1}{\\#D_{n+J_{0}+j}}\sum_{d\in D_{n+J_{0}+j}}e^{2\pi
ip_{n+1}^{-1}\cdots p_{n+J_{0}+j}^{-1}d_{n+J_{0}+j}\xi}$

falls into $A_{j}$, which yields that

$\displaystyle|\widehat{\delta_{p_{n+1}^{-1}\cdots
p_{n+J_{0}+j}^{-1}D_{n+J_{0}+j}}}(\xi)|$
$\displaystyle=\left|\frac{1}{\\#D_{n+J_{0}+j}}\sum_{d\in
D_{n+J_{0}+j}}e^{2\pi ip_{n+1}^{-1}\cdots
p_{n+J_{0}+j}^{-1}d_{n+J_{0}+j}\xi}\right|$
$\displaystyle\geq\text{dist}(\mathbf{0},A_{j})\geq\cos 2^{-j-1}\pi.$

Hence

$\prod_{j=1}^{\infty}|\widehat{\delta_{p_{n+1}^{-1}\cdots
p_{n+J_{0}+j}^{-1}D_{n+J_{0}+j}}}(\xi)|\geq\prod_{j=1}^{\infty}\cos
2^{-j-1}\pi=\frac{2}{\pi}.$

∎

##  4\. Proof of Theorem 2.1

In Section 3, we show that (ii) and (iii) of Theorem 2.1 are equivalent. In
this section, we complete the proof of Theorem 2.1. The main difficulty lies
in proving (ii) $\Rightarrow$ (i). We divide it into two cases.

###  4.1. _(ii) $\Rightarrow$ (i) when _
${\sup_{n}\\{d:d\in\mathcal{D}_{n}\\}=\infty}$

We apply Wiener theorem to prove this case. The following analyticity of
Fourier transform of Borel measure is a basic tool in this section.

######  Lemma 4.1.

Let $\mu$ be a Borel probability measure with a compact support in
$\mathbb{R}$. Then its Fourier transform

$\widehat{\mu}(z)=\int_{\mathbb{R}}e^{2\pi itz}d\mu(t)$

is an analytic function in $\mathbb{C}$.

The following theorem (Wiener theorem) is a valuable tool to deal with the
relationship between the measure and its Fourier transform when its support is
contained in $[0,1)$. Even if the support of the measure we consider is not in
this range, we can still skillfully apply this theorem.

######  Theorem 4.2 (Wiener theorem, [23]).

Let $\mu$ be any complex Borel measure on $[0,1)$. Then

$\sum_{x}|\mu(\\{x\\})|^{2}=\lim_{N\rightarrow\infty}\frac{1}{2N+1}\sum_{-N}^{N}|\widehat{\mu}(n)|^{2}.$

We will apply the above results to prove the following conclusion. The main
idea of the following proof goes back at least as far as Theorem 5.4 in [1].
And we will use some knowledge of complex analysis to get a broader
conclusion.

######  Theorem 4.3.

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples
with $\sup_{n}\\{p_{n}^{-1}d_{n}:d_{n}\in\mathcal{D}_{n}\\}<\infty$ on
$\mathbb{R}$, and $\nu_{n}$ be given as (2.2). Suppose that there exists a
subsequence $\\{n_{k}\\}$ such that $\nu_{n_{k}+1}\xrightarrow{w}\nu$, as
$k\rightarrow\infty$ and $\sup_{k}\\#\mathcal{D}_{n_{k}+1}:=N<\infty$.

Then $m_{0}:=\\#\mathcal{Z}(\nu)<\infty$, where $\mathcal{Z}(\cdot)$ is given
as Proposition 3.6. If $\inf_{k}\\{p_{n_{k}+1}\\}>m_{0}N$, then
$\\{\nu_{n_{k}}\\}$ is an equi-positive sequence. In particular,
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure.

###### Proof.

Let $K$ be a compact set with $\text{spt}\nu_{n}\subset K$ for all $n\geq 1$.
It follows from Lemma 3.5 that

$1=\underline{\lim}_{k\rightarrow\infty}\nu_{n_{k}+1}(\mathbb{R})\geq\nu(\mathbb{R})\geq\nu(K)\geq\overline{\lim}_{k\rightarrow\infty}\nu_{n_{k}+1}(K)=1.$

This implies that $\nu$ is a probability measure. Hence, Lemma 4.1 gives that
$\widehat{\nu}(z)$ is analytic in $\mathbb{C}$ and
$\\#\\{z\in\overline{\mathbb{D}}:\widehat{\nu}(z)=0\\}<\infty$, where
$\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$ is a unit disk. So $m_{0}<\infty$.

Without loss of generality, we assume that $m_{0}\geq 1$. For any small
$\delta>0$, using Proposition 3.6, we can find a small constant
$\epsilon_{\delta}>0$ and an integer $k_{x}$ such that

$|\widehat{\nu}(x+k_{x})|\geq\epsilon_{\delta},\;\;\forall\;x\in[0,1]\setminus(\cup_{z\in\mathcal{Z}(\nu)}B(z;\delta)):=[0,1]\setminus
B:=A.$

By $(iv)$ of Lemma 3.5, for any $p_{n_{k}+1}^{-1}x\in A$, there exist $K_{0}$
and $k=k_{p_{n_{k}+1}^{-1}x}$ such that

$|\widehat{\nu_{n_{k}}}(x+p_{n_{k}+1}k)|=|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(x)||\widehat{\nu_{n_{k}+1}}(p_{n_{k}+1}^{-1}x+k)|\geq\frac{1}{2}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(x)|\epsilon_{\delta},$
(4.1)

whenever $k\geq K_{0}$.

In the following, our goal is to find suitable integer $\ell$ such that
$\frac{x+\ell}{p_{n_{k}+1}}\in A$ and

$\inf_{x\in[0,1)}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(x+\ell)|>0$

by Wiener theorem. Because $(p_{n_{k}+1},\mathcal{D}_{n_{k}+1},L_{n_{k}+1})$
is a Hadamard triple, we get a new digit set

$\mathcal{D}_{n_{k}+1}^{\prime}=\mathcal{D}_{n_{k}+1}\pmod{p_{n_{k}+1}}$

satisfying
$\mathcal{D}_{n_{k}+1}^{\prime}\subset\\{0,1,\cdots,p_{n_{k}+1}-1\\}$. Write
$m=\\#\mathcal{D}_{n_{k}+1}$ and the elements of
$\mathcal{D}_{n_{k}+1}^{\prime}$ as $\\{d_{n_{k}+1,j}^{\prime}\\}_{j=1}^{m}$.
Denote a discrete measure

$\omega_{k}(x)=\begin{cases}\frac{1}{m},&x=p_{n_{k}+1}^{-1}d_{n_{k}+1,j}^{\prime};\\\
0,&\text{otherwise}.\end{cases}$

For any $\xi\in[0,1)$, define the complex measure

$\omega_{k}^{\prime}(E):=\int_{E}f(x)d\omega_{k}(x),$

where

$f(x)=\begin{cases}e^{2\pi
ip_{n_{k}+1}^{-1}d_{n_{k}+1,j}\xi},&x=p_{n_{k}+1}^{-1}d_{n_{k}+1,j}^{\prime}\\\
0,&\text{otherwise}.\end{cases}$

Hence

$\widehat{\omega_{k}^{\prime}}(y)=\frac{1}{m}\sum_{j=1}^{m}e^{2\pi
ip_{n_{k}+1}^{-1}(yd_{n_{k}+1,j}^{\prime}+d_{n_{k}+1,j}\xi)}=\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(y+\xi)$

whenever $y\in\mathbb{Z}$. Since $\omega^{\prime}_{k}$ is a complex Borel
measure on $[0,1)$, applying Wiener theorem, one has

$\displaystyle\frac{1}{\\#\mathcal{D}_{n_{k}+1}}=\sum_{d\in\mathcal{D}_{n_{k}+1}^{\prime}}|\omega_{k}^{\prime}(\\{\frac{d}{p_{n_{k}+1}}\\})|^{2}$
$\displaystyle=\lim_{\ell\rightarrow\infty}\frac{1}{2\ell
p_{n_{k}+1}+1}\sum_{l=-\ell p_{n_{k}+1}}^{\ell
p_{n_{k}+1}}|\widehat{\omega_{k}^{\prime}}(l)|^{2}$
$\displaystyle=\frac{1}{p_{n_{k}+1}}\sum_{l=0}^{p_{n_{k}+1}-1}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(\xi+l)|^{2}.$

Divide the above formula into

$\displaystyle\frac{1}{\\#\mathcal{D}_{n_{k}+1}}$
$\displaystyle=\frac{1}{p_{n_{k}+1}}\sum_{\frac{\xi+l}{p_{n_{k}+1}}\in
B}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(\xi+l)|^{2}+\frac{1}{p_{n_{k}+1}}\sum_{\frac{\xi+l}{p_{n_{k}+1}}\in
A}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(\xi+l)|^{2}$
$\displaystyle\leq\frac{1}{p_{n_{k}+1}}\\#\\{l:\frac{\xi+l}{p_{n_{k}+1}}\in
B\\}+\max_{\frac{\xi+l}{p_{n_{k}+1}}\in
A}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(\xi+l)|^{2}$
$\displaystyle\leq
2m_{0}\delta+\frac{m_{0}}{p_{n_{k}+1}}+\max_{\frac{\xi+l}{p_{n_{k}+1}}\in
A}|\widehat{\delta_{p_{n_{k}+1}^{-1}\mathcal{D}_{n_{k}+1}}}(\xi+l)|^{2}.$

Taking $\delta\leq\frac{1}{4m_{0}}(\frac{1}{N}-\frac{m_{0}}{p_{n_{k}}+1})$,
then

$\max_{\frac{\xi+l}{p_{n_{k}}}\in
A}|\widehat{\delta_{p_{n_{k}}^{-1}\mathcal{D}_{n_{k}}}}(\xi+l)|^{2}\geq
2m_{0}\delta.$

It follows from (4.1) that

$|\widehat{\nu_{n_{k}}}(\xi+\ell)|\geq m_{0}\delta\epsilon_{\delta}$

for some $0\leq\ell\leq p_{n_{k}+1}-1$. Hence $\\{\nu_{n_{k}}\\}$ is an equi-
positive sequence, and $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral
measure by Theorem B. ∎

If we add the condition $\sup_{n}\\#\mathcal{D}_{n}<\infty$ to the above
theorem, from Theorem 3.4 (weak compactness theorem), then we get the
following corollary(by choosing $p_{n_{k}+1}\rightarrow\infty$, as
$k\rightarrow\infty$).

######  Corollary 4.4.

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples
with $\sup_{n}\\{p_{n}^{-1}d_{n}:d_{n}\in\mathcal{D}_{n}\\}<\infty$ and
$\sup_{n}\\#\mathcal{D}_{n}<\infty.$ If $\sup_{n}\\{p_{n}\\}=\infty$, then
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure.

###  4.2. _(ii) $\Rightarrow$ (i) when _
${\sup_{n}\\{d:d\in\mathcal{D}_{n}\\}<\infty}$

In this subsection, by using the diagonal principle, we will find the weak
limit of $\\{\nu_{n}\\}$. And, by using some methods and techniques of complex
analysis, we will prove our main results.

The following theorem is the Uniqueness part.

######  Theorem 4.5.

(Uniqueness Theorem)[21] The Fourier transform $\widehat{\mu}$ of a
probability measure $\mu$ on $\mathbb{R}^{n}$ characterizes $\mu$. That is, if
two probabilities on $\mathbb{R}^{n}$ admit the same Fourier transform, then
they are the same.

According to Argument Principle, we conclude the following simple but
practical result. Let $\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ be the unit
disc.

######  Lemma 4.6.

Let $\\{f_{n}(z)\\}$ be a sequence of complex analytic functions in
$\mathbb{C}$, and suppose that, as $n\rightarrow\infty$, $f_{n}(z)\rightarrow
f(z)$ uniformly on each compact subset of $\mathbb{C}$.

Denote $\mathbf{z}(f):=\\{z\in\overline{\mathbb{D}}:f(z)=0\\}$. If
$\\#\mathbf{z}(f_{n})\rightarrow\infty$ as $n\rightarrow\infty$, then
$f(z)\equiv 0$.

###### Proof.

If $f\not\equiv 0$, then we can assume that
$\mathbf{z}(f)=\\{z_{1},z_{2},\cdots,z_{m}\\}\subset\overline{\mathbb{D}}$.

Note that $f_{n}(z)\rightarrow f(z)$ uniformly on $\overline{\mathbb{D}}$ as
$n\rightarrow\infty$. Choose disjoint small balls
$B(z_{i};r):=\\{y:|y-z_{i}|<r\\}\subset\mathbb{D}$. By continuity of $f(z)$,
there exists $\varepsilon>0$ such that $|f(z)|\geq\varepsilon$ for
$z\in\mathbb{D}\setminus\cup_{i=1}^{m}B(z_{i};r)$. Recall that
$f_{n}(z)\rightarrow f(z)$ for any $z\in\overline{\mathbb{D}}$. Then there
exists $N$ such that $|f(z)-f_{n}(z)|<\frac{1}{2}\varepsilon$ for any
$z\in\mathbb{D}$ and $n\geq N$. Therefore

$|f_{n}(z)|\geq|f(z)|-|f(z)-f_{n}(z)|\geq\frac{1}{2}\varepsilon>0$

for any $z\in\mathbb{D}\setminus\cup_{i=1}^{m}B(z_{i};r)$, i.e.,
$\mathbf{z}(f_{n})\subset\cup_{i=1}^{m}B(z_{i};r)$ for all $n\geq N$.

Assuming that $\\#(\mathbf{z}(f_{n})\cap
B(z_{i_{n}};r))=\max_{i}\\{\\#(\mathbf{z}(f_{n})\cap B(z_{i};r))\\}$ for some
$1\leq i_{n}\leq m$, there exists $i_{0}\in\\{1,2,\cdots,m\\}$ such that
$\\#\\{n\geq 1:i_{n}=i_{0}\\}=\infty$. Hence, we can choose a subsequence
$\\{n_{k}\\}$ such that

$\\#(\mathbf{z}(f_{n_{k}})\cap
B(z_{i_{0}};r))\geq\\#\mathbf{z}(f_{n_{k}})/m\rightarrow\infty,\;\text{as}\;k\rightarrow\infty.$

Let $\mathbf{z}^{\prime}(f_{n_{k}})=\mathbf{z}(f_{n_{k}})\cap B(z_{i_{0}};r)$
and $C=\\{z:|z-z_{i_{0}}|=r\\}$. From Argument Principle, we have

$\displaystyle\\#\mathbf{z}^{\prime}(f)=\frac{1}{2\pi}\int_{C}\frac{f^{\prime}}{f}dz=\frac{1}{2\pi}\int_{C}\lim_{k\rightarrow\infty}\frac{f_{n_{k}}^{\prime}}{f_{n_{k}}}dz=\lim_{k\rightarrow\infty}\\#\mathbf{z}^{\prime}(f_{n_{k}})=\infty,$

a contradiction. ∎

Recall that

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}:=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{(p_{1}p_{2})^{-1}\mathcal{D}_{2}}\ast\cdots$

and

$\nu_{n}=\delta_{p_{n+1}^{-1}\mathcal{D}_{n+1}}\ast\delta_{(p_{n+2}p_{n+1})^{-1}\mathcal{D}_{n+2}}\ast\cdots.$

We will use the above lemma to obtain the following conclusion.

######  Theorem 4.7.

As the above notions, let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence
of Hadamard triples with
$\sup_{n}\\{p_{n}^{-1}d_{n}:d_{n}\in\mathcal{D}_{n}\\}<\infty$. If the
associated Cantor-Moran measure $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ satisfies
$\limsup_{k\rightarrow\infty}T_{p_{k},\mathcal{D}_{k}}\supset(0,1)$, then
$\mathcal{Z}(\nu_{n})=\emptyset$ for any $n\geq 1$.

If there exists a subsequence $\\{n_{k}\\}$ such that
$\nu_{n_{k}}\xrightarrow{w}\nu$ with $\mathcal{Z}(\nu)=\emptyset$, then the
sequence is also an equi-positive sequence. In particular,
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure.

###### Proof.

For any $n\geq 1$, let $x_{0}\in\mathcal{Z}(\nu_{n})$. It implies that
$x_{0}\in(0,1)$ and

$\widehat{\nu_{n}}(x_{0}+k)=\prod_{j=1}^{\infty}\widehat{\delta_{\mathcal{D}_{n+j}}}\left(\frac{x_{0}+k}{p_{n+1}\cdots
p_{n+j}}\right)=0$ (4.2)

for all $k\in\mathbb{Z}$. For each $m\geq 1$, let
$\tau^{(m)}_{\ell}(x)=\frac{x+\ell}{p_{m}}$ for $\ell\in L_{m}^{*}$, where
$L_{m}^{*}=\\{0,1,\cdots,p_{m}-1\\}$. We call $m$ is possible for
$\mathcal{S}\subset[0,1)$ if there exists $x\in\mathcal{S}$ such that there
exist $\ell_{1}\neq\ell_{2}\in L_{m}^{*}$ satisfying

$\widehat{\delta_{\mathcal{D}_{m}}}(\tau^{(m)}_{\ell_{i}}(x))\neq
0,\;\;\>i=1,2.$

According to the definition of DPCS, we know $\mathcal{S}\subset
T_{p_{m},\mathcal{D}_{m}}$ when $m$ is possible for $\mathcal{S}$.

If $n+1$ is possible for $\\{x_{0}\\}:=B_{0}$, then there exist two distant
integers $\ell_{1},\ell_{2}\in L_{n+1}^{*}$ such that
$\widehat{\delta_{\mathcal{D}_{n+1}}}(\tau^{(n+1)}_{\ell_{i}}(x_{0}))\neq 0$
for $i=1,2$. In this case, denote the words set

$A_{1}:=\\{1,2\\}\;\text{and points
set}\;B_{1}:=\\{\tau^{(n+1)}_{\ell_{i_{1}}}(x_{0}):i_{1}\in
A_{1}\\}=\\{x_{i_{1}}\\}_{i_{1}\in A_{1}}.$

Since $x_{0}\in(0,1)$ and $\ell_{1},\ell_{2}\in\\{0,1,\cdots,p_{n+1}-1\\}$,
one has $x_{1}\neq x_{2}$ and $x_{1},x_{2}\in(0,1)$. Substituting
$k=\ell_{i_{1}}+p_{n+1}k^{\prime}$ into $\eqref{(4.2)}$, where $i_{1}\in
A_{1}$ and $k^{\prime}\in\mathbb{Z}$, we have

$\widehat{\nu_{n+1}}(\tau^{(n+1)}_{\ell_{i_{1}}}(x_{0})+k^{\prime})=\prod_{j=2}^{\infty}\widehat{\delta_{\mathcal{D}_{n+j}}}\left(\frac{\tau^{(n+1)}_{\ell_{i_{1}}}(x_{0})+k^{\prime}}{p_{n+2}\cdots
p_{n+j}}\right)=0$ (4.3)

for all $k^{\prime}\in\mathbb{Z}.$ Hence $B_{1}\subset\mathcal{Z}(\nu_{n+1})$.
This implies that

$\\#\mathcal{Z}(\nu_{n+1})\geq\\#B_{1}=2.$ (4.4)

If $n+1$ is not possible for $B_{0}$, then it implies that for any $x\in
B_{0}$, there exists $\ell_{1}\in L_{n+1}$ such that
$|\widehat{\delta_{D_{n+1}}}(\frac{x+\ell_{1}}{p_{n+1}})|=1$, as

$\sum_{\ell\in
L_{n+1}}\left|\widehat{\delta_{\mathcal{D}_{n+1}}}\left(\frac{x+\ell}{p_{n+1}}\right)\right|^{2}=1.$

Let

$A_{1}:=\\{1\\}\;\text{and}\;B_{1}:=\\{\tau^{(n+1)}_{\ell_{i_{1}}}(x_{0}):i_{1}\in
A_{1}\\}=\\{x_{i_{1}}\\}_{i_{1}\in A_{1}}.$

In this case, $\\#B_{1}=\\#B_{0}$. Similarly, if $n+2$ is possible for
$B_{1}$, by the above arguments, then there exist
$\ell_{i_{1}1}\neq\ell_{i_{1}2}\in L_{n+2}^{*}$ for some $i_{1}\in A_{1}$ such
that
$\widehat{\delta_{\mathcal{D}_{n+2}}}(\tau^{(n+2)}_{\ell_{l_{i_{1}i_{2}}}}(x_{i_{1}}))\neq
0$ for $i_{2}=1,2$. If $n+2$ is not possible for $B_{1}$, we can choose
$\ell_{i_{1}1}\in L_{n+2}$ such that
$|\widehat{\delta_{\mathcal{D}_{n+1}}}(\frac{x_{i_{1}}+\ell_{i_{1}1}}{p_{n+2}})|=1$.
Let

$A_{2}:=\\{i_{1}i_{2}:i_{1}\in A_{1},i_{2}\in\\{1,2\\}(\text{if\;exist})\\}$

and

$B_{2}:=\\{\tau^{(n+2)}_{\ell_{i_{1}i_{2}}}(x_{i_{1}}):i_{1}i_{2}\in
A_{2}\\}=\\{x_{i_{1}i_{2}}\\}_{i_{1}i_{2}\in A_{2}}.$

Note that

$\tau^{(n+2)}_{l_{2}}\circ\tau^{(n+1)}_{l_{1}}(x_{0})\neq\tau^{(n+2)}_{l_{2}^{\prime}}\circ\tau^{(n+1)}_{l_{1}^{\prime}}(x_{0})$

for any $(l_{1},l_{2})\neq(l_{1}^{\prime},l_{2}^{\prime})\in L_{n+1}^{*}\times
L_{n+2}^{*}$. Hence

$\\#B_{2}\geq\begin{cases}\\#B_{1},\;&n+2\;\text{is not possible
for}\;B_{1};\\\ \\#B_{1}+1,\;&n+2\;\text{is possible for}\;B_{1}.\end{cases}$

By taking $k=\ell_{i_{1}}+p_{n+1}\ell_{i_{1}i_{2}}+p_{n+1}p_{n+2}k^{\prime}$
with $(i_{1},i_{1}i_{2})\in A_{1}\times A_{2}$ and $k^{\prime}\in\mathbb{Z}$
into $\eqref{(4.2)}$, similar to (4.3), we obtain that
$B_{2}\subset\mathcal{Z}(\nu_{n+2})$, i.e.,

$\\#\mathcal{Z}(\nu_{n+2})\geq\\#B_{2}\geq 1+\\#\\{1\leq i\leq 2:n+i\;\text{is
possible for}\;B_{i-1}\\}.$

Then we can define inductively the sets by

$A_{m}:=\\{i_{1}\cdots i_{m}:i_{1}i_{2}\cdots i_{m-1}\in
A_{m-1},\;\>\>i_{m}=1,2(\text{if\;exist\;})\\}$

and

$B_{m}:=\\{\tau^{(n+m)}_{\ell_{i_{1}i_{2}\cdots i_{m}}}(x_{i_{1}i_{2}\cdots
i_{m-1}}):i_{1}i_{2}\cdots i_{m}\in A_{m}\\}:=\\{x_{i_{1}i_{2}\cdots
i_{m}}\\}_{i_{1}i_{2}\cdots i_{m}\in A_{m}},$

which satisfies $B_{m}\subset\mathcal{Z}(\nu_{n+m})$ and

$\\#\mathcal{Z}(\nu_{n+m})\geq\\#B_{m}\geq 1+\\#\\{1\leq i\leq m:n+i\;\text{is
possible for}\;B_{i-1}\\}.$

Suppose that $\sup_{m}\\#B_{m}=\infty$, which yields
$\\#\mathcal{Z}(\nu_{n_{k}})\rightarrow\infty$, as $k\rightarrow\infty$.
According to Theorem 3.4 (weak compactness theorem), there exists a
subsequence $\nu_{n_{k}}$ which weakly converges to $\nu$. It follows that
$\widehat{\nu}_{n_{k}}(z)\rightarrow\widehat{\nu}(z)$ as $k\rightarrow\infty$,
uniformly on each compact subset of $\mathbb{C}$. In fact, for any
$z_{j}=x_{j}+y_{j}i$ with reals $|x_{j}|,|y_{j}|\leq M$, we know that

$\displaystyle|\widehat{\mu}(z_{1})-\widehat{\mu}(z_{2})|$
$\displaystyle=\left|\int e^{-2\pi\xi y_{1}+2\pi i\xi
x_{1}}(1-e^{-2\pi\xi(y_{2}-y_{1})+2\pi i\xi(x_{2}-x_{1})}d\nu_{n_{k}}\right|$
$\displaystyle\leq e^{2\pi KM}(|1-e^{2\pi K|y_{2}-y_{1}|}|+e^{2\pi K}2\pi
K|x_{2}-x_{1}|),$

where $K=\max\\{|x|,1:x\in\text{spt}(\nu_{n})\cup\text{spt}(\nu)$ and
$\mu\in\\{\nu_{n}\\}\cup\\{\nu\\}$. It is easy to see that
$\\{\widehat{\nu_{n}}(z)\\}\cup\\{\widehat{\nu}(z)\\}$ is equicontinuous for
$z\in\mathbb{C}$ with $|z|\leq M$. Note that

$\sup\\{|\widehat{\mu}(z)|:\mu\in\\{\nu_{m_{k}}\\}\cup\\{\nu\\},|z|\leq
M\\}<\infty.$

and $\widehat{\nu_{m_{k}}}(z)\rightarrow\widehat{\nu}(z)$ for any
$z\in\mathbb{C}$. Applying Arzelà-Ascoli Theorem, we get
$\widehat{\nu_{m_{k}}}(z)$ converges uniformly to $\widehat{\nu}(z)$ on
$|z|\leq M$. By Lemma 4.6, we have $\widehat{\nu}(x)\equiv 0$, i.e., $\nu=0$,
a contraction. Hence, there exists $N_{0}\geq 1$ such that
$\\#B_{n+N_{0}}=\\#B_{n+j}$ for all $j\geq N_{0}$. In particular, from Lemma
3.2, if $p_{n+j}>\\#\mathcal{D}_{n+j}$, then $(p_{n+j},\mathcal{D}_{n+j})$
satisfies DPC. Thus $n+j$ is possible for any set of single point
$\\{x\\}\subset[0,1)$. This implies that

$\\#\\{i\geq 1:p_{n+j}>\\#\mathcal{D}_{n+j}\\}<\infty.$

By the above arguments, we have $p_{n+j}=\\#\mathcal{D}_{n+j}$ for all $j\geq
N_{0}$. It follows from Lemma 3.3 that
$T_{p_{n+j},\mathcal{D}_{n+j}}=[0,1)\setminus\frac{1}{\gcd\mathcal{D}_{n+j}}\mathbb{Z}$.
We know $n+j$ is not possible for $B_{j-1}$ for all $j\geq N_{0}+1$. Therefore
$B_{j-1}\subset\frac{1}{\gcd\mathcal{D}_{n+j}}\mathbb{Z}$ for all $j\geq
N_{0}+1$. Fixing $j\geq N_{0}+1$, put
$\frac{p}{q}=\frac{v_{n+j}}{\gcd\mathcal{D}_{n+j}}=x_{i_{1}i_{2}\cdots
i_{j-1}}\in B_{j-1}$, where $v_{n+j}\in\mathbb{Z}$, $\gcd(p,q)=1$ and $q>1$.
The construction of $B_{j}$ allows us to find $i_{1}i_{2}\cdots i_{j-1}1\in
A_{n+j}$ such that

$x_{i_{1}i_{2}\cdots i_{j-1}1}=\frac{x_{i_{1}i_{2}\cdots
i_{j-1}}+\ell_{i_{1}i_{2}\cdots i_{j-1}1}}{p_{n+j}}\in
B_{j}\subset\frac{1}{\gcd\mathcal{D}_{n+j+1}}\mathbb{Z}.$

Hence $q|\gcd\mathcal{D}_{n+j+1}$. Similarly, it follows that

$\frac{x_{i_{1}i_{2}\cdots i_{j-1}1}+\ell_{i_{1}i_{2}\cdots
i_{j-1}11}}{p_{n+j+1}}\in
B_{j+1}\subset\frac{1}{\gcd\mathcal{D}_{n+j+2}}\mathbb{Z}$

for some $i_{1}i_{2}\cdots i_{j-1}11\in A_{n+j+1}$. Then
$q|\gcd\mathcal{D}_{n+j+2}$. By induction, one has $q|\gcd\mathcal{D}_{n+j+i}$
for all $i\geq 0$. Then

$\cup_{k\geq n+j}T_{p_{k},\mathcal{D}_{k}}=[0,1)\setminus(\cap_{k\geq
n+j}\frac{1}{\gcd\mathcal{D}_{k}}\mathbb{Z})\subset[0,1)\setminus\frac{1}{q}\mathbb{Z}.$

Noting the assumption
$\limsup_{k\rightarrow\infty}T_{p_{k},\mathcal{D}_{k}}\supset(0,1)$, it yields
that $q=1$, i.e. $x_{0}\in\mathbb{Z}$, a contradiction. Therefore
$\mathcal{Z}(\nu_{n})=\emptyset$ for all $n\geq 1$.

Finally, the theorem follows from Lemma 3.7 and Theorem B. ∎

######  Theorem 4.8.

Under the assumption of Theorem 2.1, if
$\limsup_{k\rightarrow\infty}T_{p_{k},\mathcal{D}_{k}}\supset(0,1)$ and
$\sup_{n}\\{p_{n}\\}<\infty$, then there exists an equi-positive subsequence
$\\{\nu_{n_{k}}\\}$. In particular, $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a
spectral measure.

###### Proof.

Step 1: Find a subsequence $\\{\nu_{n_{k}}\\}$ and the weak limit $\nu$.

The conditions $\sup_{n}\\{p_{n}:n\geq 1\\}<\infty$ and
$\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}<\infty$ imply that
$\sup_{n}\\{d:d\in\mathcal{D}_{n}\\}<\infty$. Choose $m_{1}\in\mathbb{N}$ such
that

$(p_{m_{1}+1},\mathcal{D}_{m_{1}+1})=(p_{k+1},\mathcal{D}_{k+1})\;\text{for\;infinitely\;many}\;k,$

and denote

$\mathcal{A}_{1}:=\\{k>m_{1}:(p_{m_{1}+1},\mathcal{D}_{m_{1}+1})=(p_{k+1},\mathcal{D}_{k+1})\\}.$

Similarly, choose the smallest integer $m_{2}\in\mathcal{A}_{1}$ such that

$(p_{m_{2}+2},\mathcal{D}_{m_{2}+2})=(p_{k+2},\mathcal{D}_{k+2})\;\text{for\;infinitely\;many}\;k\in\mathcal{A}_{1}$

and let

$\mathcal{A}_{2}:=\\{k\in\mathcal{A}_{1}:(p_{m_{2}+2},\mathcal{D}_{m_{2}+2})=(p_{k+2},\mathcal{D}_{k+2})\\}.$

By induction, we get a sequence $\\{m_{k}\\}$ such that

$(p_{m_{k}+k},\mathcal{D}_{m_{k}+k})=(p_{k^{\prime}+k},\mathcal{D}_{k^{\prime}+k})\;\text{for\;infinitely\;many}\;k^{\prime}\in\mathcal{A}_{k-1}.$

Applying another diagonalization process to
$\\{(p_{m_{k}+i},\mathcal{D}_{m_{k}+i})\\}$, we obtain a further subsequence
$\\{(p_{m_{k}+k},\mathcal{D}_{m_{k}+k})\\}_{k=1}^{\infty}$ which satisfies

$(p_{m_{k}+i},\mathcal{D}_{m_{k}+i})=(p_{m_{k^{\prime}}+i},\mathcal{D}_{m_{k^{\prime}}+i})$

for any $k<k^{\prime}$ and $1\leq i\leq k$. Rewrite this new sequence as

$(p_{k}^{\prime},\mathcal{D}_{k}^{\prime},L_{k}^{\prime}):=(p_{m_{k}+k},\mathcal{D}_{m_{k}+k},L_{m_{k}+k})$

for all $k\geq 1$. Then the associated Moran measure

$\mu_{\\{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}\\}}(\cdot)=\delta_{p_{1}^{\prime-1}\mathcal{D}_{1}^{\prime}}\ast\delta_{p_{1}^{\prime-1}p_{2}^{\prime-1}\mathcal{D}_{2}^{\prime}}\ast\cdots.$

From Theorem 3.4, without loss of generality, we assume that
$\\{\nu_{m_{k}}\\}$ weakly converges to $\nu$. At the end of the step, we show
the weak limit $\nu$ is $\mu_{\\{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}\\}}$.
A basic fact is $\prod_{j=n+1}^{\infty}\cos 2^{-j}\pi\rightarrow 1$, as
$n\rightarrow\infty$. So for any $\epsilon>0$, there exists $N>0$ such that
$|\prod_{j=n+1}^{\infty}\cos 2^{-j}\pi-1|<\frac{1}{2}\epsilon$ and
$|2^{-n}\pi|\leq\frac{1}{4}\epsilon$ for all $n\geq N$. Fixed
$x\in\mathbb{R}$, let $k>J_{0}+N$, where $J_{0}$ is given as Lemma 3.9. Then

$|\widehat{\mu_{\\{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}\\}}}(x)-\widehat{\nu_{m_{k}}}(x)|=\prod_{j=1}^{k}|\widehat{\delta_{p_{1}^{{}^{\prime}-1}\cdots
p_{j}^{{}^{\prime}-1}\mathcal{D}_{j}^{\prime}}}(x)|\cdot\left|\prod_{j=k+1}^{\infty}\widehat{\delta_{p_{1}^{{}^{\prime}-1}\cdots
p_{j}^{{}^{\prime}-1}\mathcal{D}_{j}^{\prime}}}(x)-\prod_{j=m_{k}+k+1}^{\infty}\widehat{\delta_{p_{m_{k}+1}^{-1}\cdots
p_{j}^{-1}\mathcal{D}_{j}}}(x)\right|.$

By the proof of Lemma 3.9 and $k>J_{0}+N$, then

$\widehat{\delta_{p_{1}^{{}^{\prime}-1}\cdots
p_{j}^{{}^{\prime}-1}\mathcal{D}_{j}^{\prime}}}(x),\widehat{\delta_{p_{m_{k}+1}^{-1}\cdots
p_{j}^{-1}\mathcal{D}_{j}}}(x)\in A_{j-k+N}\subset\\{re^{i\theta}:\cos
2^{-(j-k+N)}\pi\leq r\leq 1,|\theta|\leq 2^{-(j-k+N)}\pi\\}$

for all $j>k$. Hence
$\prod_{j=k+1}^{\infty}\widehat{\delta_{p_{1}^{{}^{\prime}-1}\cdots
p_{j}^{{}^{\prime}-1}\mathcal{D}_{j}^{\prime}}}(x),\prod_{j=m_{k}+k+1}^{\infty}\widehat{\delta_{p_{m_{k}+1}^{-1}\cdots
p_{j}^{-1}\mathcal{D}_{j}}}(x)$ belong to

$A:=\left\\{re^{i\theta}:1-\frac{1}{2}\epsilon\leq\prod_{j=N+1}^{\infty}\cos
2^{-j}\pi\leq r\leq 1,|\theta|\leq 2^{-N}\pi\leq\frac{1}{4}\epsilon\right\\}.$

It follows from $|\widehat{\delta_{p_{1}^{\prime-1}\cdots
p_{j}^{{}^{\prime}-1}}}(x)|\leq 1$ that

$|\widehat{\mu_{\\{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}\\}}}(x)-\widehat{\nu_{m_{k}}}(x)|=|A|\leq|\\{r:1-\frac{1}{2}\epsilon\leq
r\leq 1\\}|+|\\{e^{i\theta}:|\theta|\leq\frac{1}{4}\epsilon\\}|\leq\epsilon,$

where $|E|$ denotes the diameter of $E$. This implies that
$\widehat{\mu_{\\{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}\\}}}(x)=\lim_{k\rightarrow\infty}\widehat{\nu_{m_{k}}}(x)=\widehat{\nu}(x)$
for any $x\in\mathbb{R}$. Theorem 4.5 (Uniqueness Theorem) gives
$\mu_{\\{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}\\}}=\nu$.

If the sequence $\\{(p_{k}^{\prime},\mathcal{D}_{k}^{\prime})\\}$ satisfies
$\limsup_{k\rightarrow\infty}T_{p_{k}^{\prime},\mathcal{D}_{k}^{\prime}}\supset(0,1)$,
then $\mathcal{Z}(\nu)=\emptyset$ by Theorem 4.7, and
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure. So we only need to
consider $p_{n}^{\prime}=\\#\mathcal{D}_{n}^{\prime}$ and $\gcd_{j\geq
n}\\{\gcd\mathcal{D}_{j}^{\prime}\\}\equiv d>1$ for all $n\geq 1$. We assume
that $x\in\mathcal{Z}(\nu)\neq\emptyset$.

Step 2: Determine $x$.

Let $\iota^{(i)}_{\ell}(x):=\frac{x+\ell}{p_{i}^{\prime}},\ell\in
L_{i}^{\prime}$ for any $i\geq 1$. There exists $\ell_{1}\in L_{i}^{\prime}$
such that
$|\widehat{\delta_{\mathcal{D}_{1}^{\prime}}}(\iota^{(1)}_{\ell_{1}}(x))|=\max\\{|\widehat{\delta_{\mathcal{D}_{1}^{\prime}}}(\iota^{(1)}_{\ell}(x))|:\ell\in
L_{1}^{\prime}\\}>0$, as

$\sum_{\ell\in
L_{1}^{\prime}}|\widehat{\delta_{\mathcal{D}_{1}^{\prime}}}(\iota^{(1)}_{\ell}(x))|^{2}=1.$

Similarly, we can find $\ell_{2}\in L_{i}^{\prime}$ such that

$|\widehat{\delta_{\mathcal{D}_{2}^{\prime}}}(\iota^{(2)}_{\ell_{2}}\circ\iota^{(1)}_{\ell_{1}}(x))|=\max\\{|\widehat{\delta_{\mathcal{D}_{2}^{\prime}}}(\iota^{(2)}_{\ell}\circ\iota^{(1)}_{\ell_{1}}(x))|:\ell\in
L_{2}^{\prime}\\}>0.$

By induction, we obtain a sequence $\\{\ell_{j}\\}_{j=1}^{\infty}$ such that

$|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}(\iota^{(j)}_{\ell_{j}}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x))|=\max\\{|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}(\iota^{(j)}_{\ell}\circ\iota^{(j-1)}_{\ell_{j-1}}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x))|:\ell\in
L_{j}^{\prime}\\}>0.$

Let

$\kappa_{n}:=\delta_{p_{1}^{\prime-1}\mathcal{D}_{1}^{\prime}}\ast\cdots\ast\delta_{(p_{1}^{\prime}\cdots
p_{n}^{\prime})^{-1}\mathcal{D}_{n}^{\prime}},\;\kappa_{>n}:=\delta_{p_{n+1}^{\prime-1}\mathcal{D}_{n+1}^{\prime}}\ast\delta_{p_{n+1}^{\prime}p_{n+2}^{\prime-1}\mathcal{D}_{n+2}^{\prime}}\ast\cdots$

for all $n\geq 1$. By the proof of Theorem 4.7, we have

$\\#\mathcal{Z}(\kappa_{>j})\geq\\#\\{1\leq i\leq
j-1:0<|\widehat{\delta_{\mathcal{D}_{i}^{\prime}}}(\tau^{(i)}_{\ell_{i}}\circ\cdots\circ\tau^{(1)}_{\ell_{1}}(x))|<1\\}.$

The relation between $\kappa_{>n}$ and $\nu_{n}$ tells us that
$\mathscr{Z}(\widehat{\kappa_{>n}})\cap[0,1)\subset\mathscr{Z}(\widehat{\nu_{n}})\cap[0,1)$,
i.e.,

$\sup_{n}\\#(\mathscr{Z}(\widehat{\kappa}_{>n})\cap[0,1))\leq
k_{0}:=\sup_{n}\\#(\mathscr{Z}(\widehat{\nu}_{n})\cap[0,1))<\infty,$

which implies that

$\\#\\{1\leq i\leq
j-1:0<|\widehat{\delta_{\mathcal{D}_{i}^{\prime}}}(\iota^{(i)}_{\ell_{i}}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x))|<1\\}\leq
k_{0}$ (4.5)

for all $j\geq 1$. Hence there exists an integer $s$ such that

$|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}(\iota^{(j)}_{\ell_{j}}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x))|=1$

for any $j\geq s$. Similar to the proof of Theorem 4.7, we know

$x_{0}:=\iota^{(s)}_{\ell_{s}}\circ\iota^{(s-1)}_{\ell_{s-1}}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x)\in\frac{\mathbb{Z}}{d}.$

Step 3: Complete the estimate.

Let $t:=\sup_{j}\\#\mathcal{D}_{j}^{\prime}$. Note that

$|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}(\iota^{(j)}_{\ell_{j}}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x))|^{2}\geq\frac{1}{t}\sum_{\ell\in
L_{j}^{\prime}}|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}(\iota^{(j)}_{\ell}\circ\cdots\circ\iota^{(1)}_{\ell_{1}}(x))|^{2}\equiv\frac{1}{t}.$

It follows from (4.5) that

$\prod_{j=1}^{s}\left|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}\left(\frac{x+\ell_{1}+p_{1}^{\prime}\ell_{2}+\cdots+p_{1}^{\prime}\cdots
p_{s-1}^{\prime}\ell_{s}}{p_{1}^{\prime}\cdots
p_{j}^{\prime}}\right)\right|^{2}\geq\frac{1}{t^{k_{0}}}.$

Letting $k\geq s$, we have

$|\widehat{\nu_{m_{k}}}(x+\ell_{1}+\cdots+p_{1}\cdots
p_{s-1}\ell_{s})|=|\widehat{\nu_{m_{k}+s}}(x_{0})|\cdot\prod_{j=1}^{s}\left|\widehat{\delta_{\mathcal{D}_{j}^{\prime}}}\left(\frac{x+\ell_{1}+\cdots+p_{1}\cdots
p_{s-1}\ell_{s}}{p_{1}^{\prime}\cdots p_{j}^{\prime}}\right)\right|.$

We claim that there exists an integer $J$ such that

$\prod_{j=J+1}^{\infty}\left|\widehat{\delta_{\mathcal{D}_{n+j}}}\left(\frac{\xi}{\widetilde{p}_{1}\cdots\widetilde{p}_{j}}\right)\right|\geq\frac{2}{\pi},\;\;n\geq
1$

for any $\widetilde{p}_{i}\in\\{p_{i}\\}_{i}$ and $|\xi|\leq 1$. In fact, by
choosing $J$ satisfying
$\frac{\max_{n}\\{d:d\in\mathcal{D}_{n}\\}}{2^{J}}\leq\frac{1}{4}$, we have

$\prod_{j=J+1}^{\infty}\left|\widehat{\delta_{\mathcal{D}_{n+j}}}\left(\frac{\xi}{\widetilde{p}_{1}\cdots\widetilde{p}_{j}}\right)\right|\geq\prod_{j=J+1}^{\infty}\left|\text{Re}\left(\widehat{\delta_{\mathcal{D}_{n+j}}}\left(\frac{\xi}{\widetilde{p}_{1}\cdots\widetilde{p}_{j}}\right)\right)\right|\geq\prod_{j=1}^{\infty}\cos\frac{\pi}{2^{j+1}}=\frac{2}{\pi}.$

Let

$U:=\left\\{\left|\widehat{\delta_{\mathcal{D}}}\left(\frac{x}{\widetilde{p}_{1}\cdots\widetilde{p}_{j}}\right)\right|^{2}\neq
0:\mathcal{D}\in\\{\mathcal{D}_{i}\\},\widetilde{p}_{i}\in\\{p_{i}\\},x\in\bigcup_{j}\frac{\mathbb{Z}}{\gcd\mathcal{D}_{j}},1\leq
j\leq J+k_{0}\right\\}$

and $\min U:=\varepsilon_{0}^{\prime}>0.$

If $\widehat{\nu_{m_{k}+s}}(x_{0})\neq 0$, then

$|\widehat{\nu_{m_{k}}}(x+\ell_{1}+\cdots+p_{1}\cdots
p_{s-1}\ell_{s})|^{2}\geq|\widehat{\nu_{m_{k}+s}}(x_{0})|^{2}\frac{1}{t^{k_{0}}}\geq\varepsilon_{0}^{\prime
J}\frac{4}{\pi^{2}t^{k_{0}}}:=\varepsilon_{0}^{\prime\prime}>0.$

If $\widehat{\nu_{m_{k}+s}}(x_{0})=0$, by

$\sum_{\ell\in
L_{m_{k}+s+1}}\left|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}\left(\frac{x_{0}+\ell}{p_{m_{k}+s+1}}\right)\right|^{2}\equiv
1,$

we can choose $\ell_{s+1}\in L_{m_{k}+s+1}$ such that
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}(x_{1})|\neq 0$, where
$x_{1}:=\frac{x_{0}+\ell_{s+1}}{p_{m_{k}+s+1}}$. If
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}(x_{1})|=1$, then
$x_{1}\in\frac{\mathbb{Z}}{\gcd\mathcal{D}_{m_{k}+s+1}}$. If
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}(x_{1})|<1$, then there exists
$\ell_{s+1}^{\prime}\in L_{m_{k}+s+1}\setminus\\{\ell_{s+1}\\}$ such that
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}(x_{2})|\neq 0$, where
$x_{2}:=\frac{x_{0}+\ell_{s+1}^{\prime}}{p_{m_{k}+s+1}}$. Note that
$x_{i}\in\frac{\mathbb{Z}}{dp_{m_{k}+s+1}}$. So if
$\widehat{\nu_{m_{k}+s+1}}(y)\neq 0$ for some $y\in\\{x_{1},x_{2}\\}$(with
loss of generality, we assume $y=x_{1}$), then
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}(y)|\geq\varepsilon_{0}^{\prime}$,
this yields that

$|\widehat{\nu}_{m_{k}}(x+\ell_{1}+\cdots+p_{1}\cdots
p_{s}\ell_{s+1})|\geq\varepsilon_{0}^{\prime}\varepsilon_{0}^{\prime\prime}.$

If $\widehat{\nu_{m_{k}+s+1}}(y)=0$ for all $y\in\\{x_{1},x_{2}\\}$(or
$\\{x_{1}\\}$), we write
$\mathcal{C}_{1}=\\{x_{1},x_{2}\\}\subset\mathscr{Z}(\widehat{\nu_{m_{k}+s+1}})$.

For any $x_{i}\in\mathcal{C}_{1}$, there exists $\ell_{s+2}\in L_{m_{k}+s+2}$
such that $|\widehat{\delta_{\mathcal{D}_{m_{k}+s+2}}}(x_{i1})|\neq 0$, where
$x_{i1}:=\frac{x_{i}+\ell_{s+2}}{p_{m_{k}+s+2}}$. If
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+2}}}(x_{i1})|=1$, then
$x_{i1}\in\frac{\mathbb{Z}}{\gcd\mathcal{D}_{m_{k}+s+2}}$. If
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+2}}}(x_{i1})|<1$, then there exists
$\ell_{s+2}^{\prime}\in L_{m_{k}+s+2}\setminus\\{\ell_{s+2}\\}$ such that
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+2}}}(x_{i2})|\neq 0$, where
$x_{i2}:=\frac{x_{i}+\ell_{s+2}^{\prime}}{p_{m_{k}+s+2}}$. Note that
$x_{ij}\in\frac{\mathbb{Z}}{\gcd\mathcal{D}_{m_{k}+s+1}p_{m_{k}+s+1}p_{m_{k}+s+2}}$.
So if $\widehat{\nu_{m_{k}+s+2}}(y)\neq 0$ for some $y\in\\{x_{i1},x_{i2}\\}$,
then
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+1}}}(y)|\geq\varepsilon_{0}^{\prime}$,
this yields that

$|\widehat{\nu}_{m_{k}}(x+\ell_{1}+\cdots+p_{1}\cdots
p_{s+1}\ell_{s+2})|\geq\varepsilon_{0}^{\prime
2}\varepsilon_{0}^{\prime\prime}.$

If $\widehat{\nu_{m_{k}+s+2}}(y)=0$ for all
$y\in\\{x_{i1},x_{i2}\\}_{x_{i}\in\mathcal{C}_{1}}$(if exist), we write
$\mathcal{C}_{2}=\\{x_{ij}:x_{i}\in\mathcal{C}_{1}\\}$. It is easy to check
that $\mathcal{C}_{2}\subset\mathscr{Z}(\widehat{\nu_{m_{k}+s+2}})$ and
$\\#\mathcal{C}_{2}\geq\\#\mathcal{C}_{1}+\\#\\{x_{i}\in\mathcal{C}_{1}:0<|\widehat{\delta_{\mathcal{D}_{m_{k}+s+2}}}(x_{i1})|<1\\}$.
In fact, the construct of $\mathcal{C}_{2}$ gives $x_{ij}\neq
x_{i^{\prime}j^{\prime}}$ for any $ij\neq i^{\prime}j^{\prime}$.

By induction, there exists $\Omega_{l}\subset\\{1,2\\}^{l}$ such that the
sequence
$\mathcal{C}_{l}:=\\{x_{i}\\}_{i\in\Omega_{l}}\subset\mathscr{Z}(\widehat{\nu_{m_{k}+s+l}})$
and

$\\#\mathcal{C}_{l}\geq\\#\mathcal{C}_{l-1}+\\#\\{x_{i}\in\mathcal{C}_{l-1}:0<|\widehat{\delta_{\mathcal{D}_{m_{k}+s+l}}}(x_{i1})|<1\\}.$

Recall that $\\#\mathscr{Z}(\widehat{\nu_{m_{k}+s+l}}\cap[0,1))\leq k_{0}$ for
all $l\geq 1$, we know $\\#\mathcal{C}_{l}\leq k_{0}$. If $l=\infty$, then
there exists an integer $j_{0}$ such that
$|\widehat{\delta_{\mathcal{D}_{m_{k}+s+j}}}(x_{i})|=1$ for all
$i\in\Omega_{j}$ with $j\geq j_{0}$. This yields that

$x_{i}\in\frac{\mathbb{Z}}{\gcd(\gcd\mathcal{D}_{m_{k}+s+j_{0}},\gcd\mathcal{D}_{m_{k}+s+j})},\;\;\forall
i\in\Omega_{j},\;j\geq j_{0}.$

According to $\gcd_{j\geq n}\\{\gcd\mathcal{D}_{j}\\}\equiv 1,\;\;\forall
n\geq 1$, we can find an integer $l^{\prime}$ such that $x_{i}\in\mathbb{Z}$
for all $i\in\Omega_{l^{\prime}}$, which is impossible, $i.e.$, $l<\infty$.
Let $k_{x,\nu_{m_{k}}}=\ell_{1}+\cdots+p_{1}\cdots p_{s+l-1}\ell_{s+l}$, it
follows that

$|\widehat{\nu_{m_{k}}}(x+k_{x,\nu_{m_{k}}})|^{2}\geq{\varepsilon_{0}^{\prime
k_{0}}}\varepsilon_{0}^{\prime\prime}:=\varepsilon_{0}^{\prime\prime\prime}>0.$

By Lemma 2.2 of [1], we know that $\\{\widehat{\nu_{m_{k}}}\\}$ is
equicontinuous. This together with the above estimate implies that

$|\widehat{\nu_{m_{k}}}(y+k_{x,\nu_{m_{k}}})|^{2}\geq\frac{1}{2}\varepsilon_{0}^{\prime\prime\prime}>0,$
(4.6)

whenever $\text{dist}(y,\mathcal{Z}(\nu))<\delta_{0}$ for some $\delta_{0}$.
Let
$E:=[0,1]\setminus\\{y\in(0,1):\text{dist}(y,\mathcal{Z}(\nu))<\delta_{0}\\}$.
Lemma 3.6 implies that there exist $\widetilde{\varepsilon}$ and
$K_{0}^{\prime}$ such that

$|\widehat{\nu}_{m_{k}}(x+k_{x})|\geq\widetilde{\varepsilon},\;\>\>k\geq
K_{0}^{\prime}$ (4.7)

for any $x\in E$. Combining (4.6) and (4.7), $\\{\nu_{m_{k}}\\}$ is an equi-
positive sequence. Hence $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral
measure. ∎

###  4.3. _Completion of the proof of Theorem 2.1_

###### Proof of Theorem 2.1.

(ii) $\Leftrightarrow$ (iii) holds at the end of Section 3. Corollary 4.4 and
Theorem 4.8 imply that (ii) $\Rightarrow$ (i).

In the following, we complete the proof by showing (i) $\Rightarrow$ (ii). If
(ii) doesn’t hold, by (ii) $\Leftrightarrow$ (iii), we assume that
$p_{n}=\\#\mathcal{D}_{n}$ and $d_{n}^{\prime}=\gcd_{j\geq
n}\gcd\mathcal{D}_{j}>1$ for all large $n$. Now, we show
$\mathcal{Z}(\nu_{n})\neq\emptyset$.

Choose $\xi_{0}=\frac{1}{d_{n}^{\prime}}$. For $j\geq n$, let
$\mathcal{D}_{j}=d_{n}^{\prime}\mathcal{D}_{j}^{\prime}$ for some
$\mathcal{D}_{j}^{\prime}\subset\mathbb{Z}$. For any integer $k\geq 0$, write
$d_{n}^{\prime}k=k_{1}+k_{2}p_{n+1}+k_{3}p_{n+1}p_{n+2}+\cdots$, where
$k_{j}\in\\{0,1,\cdots,p_{j}-1\\}$ and $k_{j}\equiv 0$ for all large $j$.
Denote $J=\min\\{j:k_{j}\neq p_{j}-1\\}$. By the proof of Lemma 3.3, we have
$\gcd(p_{n+j},\gcd\mathcal{D}_{n+j})=1=\gcd(p_{n+j},d_{n}^{\prime})$. Note
that $(p_{n+j},\mathcal{D}_{n+j},L_{n+j})$ forms a Hadamard triple and
$p_{n+j}=\\#\mathcal{D}_{n+j}$. Then
$\mathcal{D}_{n+j}^{\prime}=\mathcal{D}_{n+j}=\\{0,1,\cdots,p_{n+j}-1\\}\pmod{p_{n+j}}$.
Hence

$\widehat{\delta_{\mathcal{D}_{n+j}^{\prime}}}(\frac{k}{p_{n+j}})=\frac{1}{\\#\mathcal{D}_{n+j}^{\prime}}\sum_{d\in\mathcal{D}_{n+j}^{\prime}}e^{2\pi
i\frac{dk}{p_{n+j}}}=\frac{1}{\\#\mathcal{D}_{n+j}^{\prime}}\sum_{d=0}^{p_{n+j}-1}e^{2\pi
i\frac{dk}{p_{n+j}}}=0$ (4.8)

for any $k\in\\{1,2,\cdots,p_{n+j}-1\\}$. It follows from (4.8) that

$\displaystyle\widehat{\nu_{n}}(\xi_{0}+k)$
$\displaystyle=\prod_{j=1}^{\infty}\widehat{\delta_{\mathcal{D}_{n+j}}}((p_{n+1}\cdots
p_{n+j})^{-1}(\xi_{0}+k))$
$\displaystyle=\prod_{j=1}^{\infty}\widehat{\delta_{\mathcal{D}_{n+j}^{\prime}}}((p_{n+1}\cdots
p_{n+j})^{-1}(1+d_{n}^{\prime}k))$
$\displaystyle=\prod_{j=J}^{\infty}\widehat{\delta_{\mathcal{D}_{n+j}^{\prime}}}((p_{n+J}\cdots
p_{n+j})^{-1}(1+k_{J}+p_{n+J}k_{J+1}+\cdots))$ $\displaystyle=0.$

Similarly, $\widehat{\nu_{n}}(\xi_{0}+k)=0$ for all negative integer $k$,
i.e., $\xi_{0}\in\mathcal{Z}(\nu_{n})$, which implies that $\\{\nu_{n}\\}$
doesn’t have equi-positive subsequence. ∎

##  5\. The singularity of Cantor-Moran spectral measure

In this section, we explore the relationship between the spectrality and
singularity of Cantor-Moran measure. In particular, we prove Theorem 2.2.

The following results about fractal geometry need to be known.

######  Theorem 5.1 ([16]).

Let $\mu$ be a mass distribution on $\mathbb{R}^{n}$, let $\mathcal{H}^{s}$ be
$s$-dimensional Hausdorff measure. Suppose that $F\subset\mathbb{R}^{n}$ is a
Borel set and $0<c<\infty$ is a constant.

If $\overline{\lim}_{r\rightarrow 0}\mu(B(x,r))/r^{s}<c$ for all $x\in F$,
then $\mathcal{H}^{s}(F)\geq\mu(F)/c$.

The following lemma is an extension of Proposition 4.9 in [16].

######  Lemma 5.2.

Let $\mu$ be a finite Borel measure on $\mathbb{R}^{n}$. Suppose that
$F\subset\mathbb{R}^{n}$ is a Borel set and $c(r)$ is a positive function with
$c(r)\rightarrow\infty$ as $r\rightarrow 0^{+}$.

If $\mu(B(x,r))/r^{s}\geq c(r)$ for all $x\in F$ and all $r>0$, then
$\mathcal{H}^{s}(F)=0$.

###### Proof.

Suppose that $F$ is bounded. For any large $M>0$, there exists $R>0$ such that
$c(r)\geq M$ for any $0<r<R$. Fix $\delta\in(0,R)$ and let $\mathcal{C}$ be
the collection of balls

$\\{B(x,r):x\in F,0<r\leq\delta\;\text{and}\;\mu(B(x,r))\geq c(r)r^{s}\\}.$

By the hypothesis, we have $F\subset\cup_{B\in\mathcal{C}}B$. Applying the
Covering lemma ([16], Lemma 4.8) to the collection $\mathcal{C}$, there is a
sequence of disjoint balls $B_{i}\in\mathcal{C}$ such that
$\cup_{B\in\mathcal{C}}B\subset\cup_{i}\widetilde{B}_{i}$, where
$\widetilde{B}_{i}$ is the ball concentric with $B_{i}$ but of four times the
radius. Hence $\\{\widetilde{B}_{i}\\}_{i=1}^{\infty}$ is an $8\delta$-cover
of $F$, so

$\mathcal{H}_{8\delta}^{s}(F)\leq\sum_{i}|\widetilde{B}_{i}|^{s}\leq
4^{s}\sum_{i}|B_{i}|^{s}\leq
4^{s}\sum_{i}\frac{\mu(B_{i})}{c(|B_{i}|)}\leq\frac{4^{s}}{M}\mu(\mathbb{R}^{n}),$

where $|B_{i}|$ is the diameter of $B_{i}$. Letting $\delta\rightarrow 0$,
then $\mathcal{H}^{s}(F)\leq\frac{4^{s}\mu(\mathbb{R}^{n})}{M}$. Thus
$\mathcal{H}^{s}(F)=0$ by setting $M\rightarrow\infty$.

Finally, if $F$ is unbounded and $\mathcal{H}^{s}(F)>0$, the
$\mathcal{H}^{s}$-measure of some bounded subset of $F$ will also exceed $0$,
contrary to the above. ∎

######  Theorem 5.3.

Let $\\{(p_{n},\mathcal{D}_{n},L_{n})\\}$ be a sequence of Hadamard triples
with $\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}<\infty$ and
$\sup_{n}\\#\mathcal{D}_{n}<\infty.$ Then the associated Cantor-Moran measure

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\cdots\ast\delta_{p_{1}^{-1}p_{2}^{-1}\mathcal{D}_{2}}\ast\cdots$

is singular continuous with respect to the Lebesgue measure if and only if
$\\#\\{n:p_{n}>\\#\mathcal{D}_{n}\\}=\infty$.

###### Proof.

Note that the support of $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is

$T=\left\\{\sum_{j=1}^{\infty}\frac{d_{j}}{p_{1}\cdots
p_{j}}:d_{j}\in\mathcal{D}_{j}\right\\}.$

For any $x\in T$, write

$x=\sum_{j=1}^{\infty}\frac{d_{j}}{p_{1}\cdots p_{j}},$

where $d_{j}\in\mathcal{D}_{j}$. For any small $r>0$, let

$\frac{1}{p_{1}\cdots p_{s_{r}}}\leq r<\frac{1}{p_{1}\cdots p_{s_{r}-1}}$

for some integer $s_{r}\geq 1$. Put
$c:=\sup_{n}\\{\frac{d}{p_{n}}:d\in\mathcal{D}_{n}\\}$,
$k_{0}:=[\log_{2}4c]+1$ and
$x_{r}=\sum_{j=1}^{s_{r}+k_{0}}\frac{d_{j}}{p_{1}\cdots p_{j}}$. Note that

$|x-x_{r}|\leq\sum_{j=s_{r}+k_{0}+1}^{\infty}\frac{d_{j}}{p_{1}\cdots
p_{j}}\leq\frac{2c}{p_{1}\cdots
p_{s_{r}+k_{0}}}\leq\frac{2c}{2^{k_{0}}p_{1}\cdots
p_{s_{r}}}\leq\frac{1}{2p_{1}\cdots p_{s_{r}}}\leq\frac{1}{2}r.$

Thus

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}(B(x_{r};\frac{1}{2}r))\leq\mu_{\\{p_{n},\mathcal{D}_{n}\\}}(B(x;r))\leq\mu_{\\{p_{n},\mathcal{D}_{n}\\}}(B(x_{r};\frac{3}{2}r)).$
(5.1)

Now we prove the necessity by a contradiction. Without loss of generality, we
assume $p_{n}=\\#\mathcal{D}_{n}$ for any $n\geq 1$, which implies that
$\sup_{n}p_{n}<\infty$. Note that

$\\#(B(x_{r};2r)\cap\frac{\mathbb{Z}}{p_{1}\cdots p_{s_{r}+k_{0}}})\leq
4p_{s_{r}}\cdots p_{s_{r}+k_{0}}$

and

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\cdots\ast\delta_{(p_{1}\cdots
p_{s_{r}+k_{0}})^{-1}\mathcal{D}_{s_{r}+k_{0}}}\ast\delta_{(p_{1}\cdots
p_{s_{r}+k_{0}+1})^{-1}\mathcal{D}_{s_{r}+k_{0}+1}}\ast\cdots.$

Let $P:=\sup_{i}p_{i}$ and
$T_{1}:=\\{\sum_{j=s_{r}+k_{0}+1}^{\infty}\frac{d_{j}}{p_{1}\cdots
p_{j}}:d_{j}\in\mathcal{D}_{j}\\}$. It follows from (5.1) and

$T\cap B(x_{r};\frac{3}{2}r)\subset
B(x_{r},2r)\cap\frac{\mathbb{Z}}{p_{1}\cdots p_{s+k_{0}}}+T_{1},$

that

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}(B(x;r))\leq\mu_{\\{p_{n},D_{n}\\}}(B(x_{r};\frac{3}{2}r))\leq\frac{4p_{s_{r}}\cdots
p_{s_{r}+k_{0}}}{p_{1}\cdots p_{s_{r}+k_{0}}}\leq 4p_{s_{r}}r\leq 4Pr.$

Hence $\mathcal{H}^{1}(T)\geq\mu(T)/(4P)=\frac{1}{4P}$ by Theorem 5.1, which
implies that $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is not singular continuous
with respect to the Lebesgue measure.

In the following, we prove the sufficiency.

Let $\\{n:p_{n}>\\#\mathcal{D}_{n}\\}=\\{n_{1},n_{2},\cdots,n_{k},\cdots\\}$
with $n_{1}<n_{2}<\cdots$. It follows from (5.1) that

$\displaystyle\mu_{\\{p_{n},\mathcal{D}_{n}\\}}(B(x;r))\geq\mu_{\\{p_{n},\mathcal{D}_{n}\\}}(B(x_{r};\frac{1}{2}r))$
$\displaystyle\geq\frac{1}{\\#\mathcal{D}_{1}\\#\mathcal{D}_{2}\cdots\\#\mathcal{D}_{s_{r}+k_{0}}}$
$\displaystyle=\frac{1}{p_{1}\cdots p_{s_{r}+k_{0}}}\times\frac{p_{1}\cdots
p_{s_{r}+k_{0}}}{\\#\mathcal{D}_{1}\\#\mathcal{D}_{2}\cdots\\#\mathcal{D}_{s_{r}+k_{0}}}$
$\displaystyle\geq\frac{r}{P^{k_{0}+1}}\prod_{j=1}^{k_{r}}\frac{p_{n_{j}}}{\\#\mathcal{D}_{n_{j}}}$
$\displaystyle\geq\frac{r}{P^{k_{0}+1}}(1+\frac{1}{N})^{k_{r}},$

where $n_{1}<n_{2}<\cdots<n_{k_{r}}\leq s_{r}+k_{0}<n_{k_{r}+1}$ and
$N:=\sup_{j}\\#\mathcal{D}_{j}$. Let
$c(r)=\frac{1}{P^{k_{0}+1}}(1+\frac{1}{N})^{k_{r}}$. It is easy to see that
$c(r)\rightarrow\infty$ as $r\rightarrow 0$.

Therefore, by Lemma 5.2, $\mathcal{H}^{1}(T)=0$. The proof is complete. ∎

###### Proof of Theorem 2.2.

The theorem is obtained directly from Theorems 5.3, 2.1 and Lemma 3.2. ∎

###### Proof of Theorem 2.3.

Combine Theorems 2.1 and 2.2, the conclusion follows. ∎

##  6\. Non-spectrality of Cantor-Moran measures

In this section, we study the non-spectrality of Cantor-Moran measures. We
need to emphasize that the condition

$\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}<\infty$ (6.1)

plays an important role in this paper.

A natural question is:

$\text{whether Cantor-Moran measure}\;\mu_{\\{p_{n},D_{n}\\}}\;\text{is
spectral with}\sup_{n}\\{p_{n}^{-1}d:d\in\mathcal{D}_{n}\\}=\infty?$

We show (6.1) is indispensable but can be improved in the following examples.

######  Example 6.1.

Let $p_{n}\equiv 9$ and $\mathcal{D}_{n}=\\{0,2,4^{n}\\}$ for all $n\geq 1$.
Then $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is not a spectral measure.

######  Example 6.2.

Let $p_{n}\equiv 3n$ and $\mathcal{D}_{n}=\\{0,2,3n^{2}+1\\}$ for all $n\geq
1$. Then $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure.

To prove the above examples, we need some known knowledge. The following
theorem is a basic criterion for the spectrality of $\mu$.

######  Theorem 6.1 ([22]).

Let $\mu$ be a probability measure with a compact support, and let
$\Lambda\subset\mathbb{R}^{n}$ be a countable subset. Then $E_{\Lambda}$ is an
orthogonal set of $L^{2}(\mu)$ if and only if

$Q_{\Lambda}(\xi):=\sum_{\lambda\in\Lambda}|\widehat{\mu}(\xi+\lambda)|^{2}\leq
1$

for $\xi\in\mathbb{R}^{n}$; and $Q_{\Lambda}(\xi)$ is an entire function in
$\mathbb{C}^{n}$. In particular, it is an orthogonal basis if and only if
$Q_{\Lambda}(\xi)\equiv 1$ for $\xi\in\mathbb{R}^{n}$.

We give the concept of the selection map, which is used to characterize the
structure of bi-zero sets of $\mu_{\\{p_{n},D_{n}\\}}$ in Example 6.1.

Let $\Omega=\\{0,1,2\\}$ and let $\Omega^{\ast}=\cup_{k=0}^{\infty}\Omega^{k}$
be the set of finite words (by convention $\Omega^{0}=\\{\emptyset\\}$).
Denote by $I=i_{1}\cdots i_{k}$ an element in $\Omega^{k}$, and $|I|=k$ is its
length. For any $I,J\in\Omega^{\ast},\;IJ$ is their natural conjunction. In
particular, $\emptyset I=I$, $I0^{\infty}=I00\cdots$ and $0^{k}=0\cdots
0\in\Omega^{k}$.

######  Definition 6.1.

Let $3|q$. We call a map $\phi:\Omega^{\ast}\rightarrow\\{-1,0,\cdots,q-2\\}$
a selection mapping if
$(i)\;\;\;\phi(\emptyset)=\phi(0^{n})=0$ for all $n\geq 1$;
$(ii)$ for any $I=i_{1}\cdots i_{k}\in\Omega^{k},\phi(I)\in
i_{k}+3\mathbb{Z}\cap\mathcal{C}$, where $\mathcal{C}=\\{-1,0,\cdots,q-2\\}$;
$(iii)$ for any $I=i_{1}\cdots i_{k}\in\Omega^{\ast}$, there exists
$J\in\Omega^{\ast}$ such that $\phi$ vanishes eventually on $IJ0^{\infty}$,
$i.e.$, $\phi(IJ0^{k})=0$ for sufficient large $k$.

Let

$\Omega^{\phi}=\\{I=i_{1}\cdots i_{k}\in\Omega^{\ast}:i_{k}\neq
0,\;\phi(I0^{n})=0\;\text{for\;sufficient\;large\;}n\\}\cup\\{\emptyset\\},$

and let

$\phi^{\ast}(I)=\sum_{n=1}^{\infty}\phi(I0^{\infty}|_{n})q^{n-1},\quad
I\in\Omega^{\phi}.$

Here $I0^{\infty}|_{n}$ denotes the word of the first $n$ entries.

Dai, He and Lau [10] showed the following theorem.

######  Theorem 6.2 ([10]).

Let $\rho=p/q\in(0,1)$ with $p,q\in\mathbb{Z}$, $3|q$ and $\gcd(p,q)=1$.
Suppose that $0\in\Lambda\subset\mathbb{R}$ is a countable set satisfying

$(\Lambda-\Lambda)\setminus\\{0\\}\subset\cup_{j=1}^{\infty}\rho^{-j}(\pm\frac{1}{3}+\mathbb{Z}).$

Then $\Lambda$ is maximal, i.e.,

$(\Lambda\cup\\{x\\}-\Lambda\cup\\{x\\})\setminus\\{0\\}\not\subset\cup_{j=1}^{\infty}\rho^{-j}(\pm\frac{1}{3}+\mathbb{Z})$

for any $x\in\mathbb{R}\setminus\Lambda$, if and only if there exist
$m_{0}\geq 1$ and a selection map $\phi$ such that
$\Lambda=\rho^{-m_{0}}3^{-1}(\phi^{\ast}(\Omega^{\phi}))$.

###### Proof of Example 6.1.

Suppose that $\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$ is a spectral measure with a
spectrum $\Lambda$. The property of the bi-zero set gives that

$2(\Lambda-\Lambda)\setminus\\{0\\}\subset\mathscr{Z}(\widehat{\mu_{\\{p_{n},\mathcal{D}_{n}\\}}})=\cup_{j=1}^{\infty}9^{j}(\pm\frac{1}{3}+\mathbb{Z}).$

Theorem 6.2 gives a selection mapping $\phi$ and a subset
$\Omega^{\prime}\subset\Omega^{\phi}$ such that

$\Lambda=6^{-1}9^{m_{0}}\phi^{*}(\Omega^{\prime})$

for some $m_{0}\geq 1$. Write
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}=\mu_{n}\ast\mu_{>n}$, where

$\mu_{n}:=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\delta_{(p_{2}p_{1})^{-1}\mathcal{D}_{2}}\ast\cdots\ast\delta_{(p_{1}\cdots
p_{n})^{-1}\mathcal{D}_{n}}.$

The following claims are needed.
Claim 1:There exist $C,\beta>0$ such that

$\prod_{j=1}^{\infty}\frac{1}{3}\big{|}1+e^{2\pi i\frac{2y}{9^{j}}}+e^{2\pi
i(\frac{4}{9})^{j}x}\big{|}\leq C(\ln|x|)^{-\beta},\;|x|>1,\;y\in\mathbb{R}.$

Let integer $b$ satisfy $b\geq 2\beta^{-1}$, and
$\mathbf{I}_{n}=\\{\mathbf{i}\in\Omega^{\prime}:|\mathbf{i}|\leq n^{b}\\}$.
Claim 2:

$\sum_{\mathbf{i}\in\mathbf{I}_{n}}|\widehat{\mu_{n^{b}+m_{0}-1}}(\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i}))|^{2}\leq
1$

for any $n\geq 1$.

In the following, we use the above claims to prove our conclusion. Let
$Q_{n}(\xi):=\sum_{\mathbf{i}\in\mathbf{I}_{n}}|\widehat{\mu}(\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i}))|^{2}.$
For any $\mathbf{i}\in\mathbf{I}_{n+1}\setminus\mathbf{I}_{n}$ and
$|\xi|<\frac{1}{2}$, by Claim 1 and

$|\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i})|\geq
6^{-1}9^{m_{0}}(9^{n^{b}}-7\times 9^{n^{b}-1}-\cdots)\geq
6^{-1}9^{n^{b}+m_{0}-1},\;\;\xi\in(0,1)$

there exist $n_{0}$ and some constants $C,C^{\prime}>0$ such that

$\displaystyle|\widehat{\mu_{>(n+1)^{b}+m_{0}-1}}(\xi+\lambda(\mathbf{i}))|$
$\displaystyle=\prod_{j=1}^{\infty}\frac{1}{3}\left|1+e^{2\pi
i\frac{2(\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i}))}{9^{(n+1)^{b}+j+m_{0}-1}}}+e^{2\pi
i(\frac{4}{9})^{(n+1)^{b}+j+m_{0}-1}(\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i}))}\right|$
$\displaystyle\leq C(\ln|6^{-1}4^{(n+1)^{b}}9^{n^{b}-(n+1)^{b}}|)^{-\beta}$
$\displaystyle\leq C^{\prime}n^{-\frac{1}{2}b\beta}\leq C^{\prime}n^{-1},$

for any $\mathbf{i}\in\mathbf{I}_{n+1}\setminus\mathbf{I}_{n}$ and $n\geq
n_{0}$. According to the above estimate and Claim 2, one has

$\displaystyle Q_{n+1}(\xi)$
$\displaystyle=Q_{n}(\xi)+\sum_{I\in\mathbf{I}_{n+1}\setminus\mathbf{I}_{n}}|\widehat{\mu_{\\{p_{n},\mathcal{D}_{n}\\}}}(\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i}))|^{2}$
$\displaystyle\leq Q_{n}(\xi)+\frac{C^{\prime
2}}{n^{2}}\sum_{\mathbf{i}\in\mathbf{I}_{n+1}\setminus\mathbf{I}_{n}}|\widehat{\mu_{(n+1)^{b}+m_{0}-1}}(\xi+6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i}))|^{2}$
$\displaystyle\leq Q_{n}(\xi)+\frac{C^{\prime 2}}{n^{2}}(1-Q_{n}(\xi)).$

Therefore,

$1-Q_{n+1}(\xi)\geq(1-Q_{n}(\xi))(1-\frac{C^{\prime
2}}{n^{2}})\geq(1-Q_{n_{0}}(\xi))\prod_{j=n_{0}}^{n}(1-\frac{C^{\prime
2}}{j^{2}})$

for all $n\geq n_{0}$. Letting $n\rightarrow\infty$, then

$1-Q(\xi)\geq(1-Q_{n_{0}}(\xi))\prod_{j=n_{0}}^{\infty}(1-\frac{C^{\prime
2}}{j^{2}}).$

From $\prod_{j=n_{0}}^{\infty}(1-\frac{C^{\prime 2}}{j^{2}})\neq 0$, we obtain
that $Q(\xi)\not\equiv 1$. By Theorem 6.1, $\Lambda$ is not a spectrum of
$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}$, a contradiction.

Finally, we complete our proof by showing Claims 1,2. We first prove Claim 1.
Define that $|\langle x\rangle|:=|x_{1}|$, where $x=x_{1}+x_{2}$ with
$x_{2}\in\mathbb{Z}$ and $x_{1}\in[-\frac{1}{2},\frac{1}{2})$. Let

$b_{0}:=\max_{|\langle
x\rangle|\geq\frac{1}{18}}\\{\frac{1}{3}\big{|}1+e^{2\pi ix}+e^{2\pi
iy}\big{|}\\}\leq\frac{1}{3}\big{|}1+|1+e^{2\pi i\frac{1}{18}}|\big{|}<1.$

Then

$\prod_{j=1}^{\infty}\frac{1}{3}\left|1+e^{2\pi i\frac{2y}{9^{j}}}+e^{2\pi
i(\frac{4}{9})^{j}x}\right|\leq b_{0}^{l},$ (6.2)

where $l=\\#\\{j:|\langle(\frac{4}{9})^{j}x\rangle|\geq\frac{1}{18}\\}$. Put

$\left\\{j:\left|\left\langle\left(\frac{4}{9}\right)^{j}x\right\rangle\right|\geq\frac{1}{18}\right\\}=\\{j_{1},j_{2},\cdots,j_{l}\\}$

with $j_{1}<j_{2}<\cdots<j_{l}$ and $|(\frac{4}{9})^{j_{l}+1}x|\leq 1$. For
any $|(\frac{4}{9})^{j}x|>1$, we can write

$\left(\frac{4}{9}\right)^{j}x=9^{s_{j}}(\mathbb{Z}\setminus
9\mathbb{Z})+\epsilon\in 9^{s_{j}}(\mathbb{Z}\setminus
9\mathbb{Z})+\left[-\frac{1}{2},\frac{1}{2}\right),$

which implies that

$\left|\left(\frac{4}{9}\right)^{j-1}x\right|\geq 9^{s_{j}}.$ (6.3)

If $|\langle(\frac{4}{9})^{j}x\rangle|<\frac{1}{18}$, then
$|\langle(\frac{4}{9})^{j+s_{j}+1}x\rangle|\geq\frac{1}{9}-\frac{1}{18}=\frac{1}{18}$.
Without loss of generality, assuming $j_{i+1}\geq j_{i}+2$, then we have
$|\langle(\frac{4}{9})^{j_{i}+1}x\rangle|<\frac{1}{18}$. From (6.3), a simple
calculation gives

$\left|\left(\frac{4}{9}\right)^{j_{i+1}}x\right|\geq\left|\left(\frac{4}{9}\right)^{j_{i}+s_{j_{i}+1}+2}x\right|\geq\left(\frac{4}{9}\right)^{2}\left|\left(\frac{4}{9}\right)^{j_{i}}x\right|^{\frac{\ln
4}{\ln 9}}.$

Hence

$\frac{9}{4}\geq\left|\left(\frac{4}{9}\right)^{j_{l}}x\right|\geq\left(\frac{4}{9}\right)^{2}\left|\left(\frac{4}{9}\right)^{j_{l-1}}x\right|^{\frac{\ln
4}{\ln 9}}\geq\cdots\geq 9^{-2}|x|^{(\frac{\ln 4}{\ln 9})^{l}}.$ (6.4)

Let $\alpha:=\frac{\ln 4}{\ln 9}$ and $\beta:=\frac{\ln b_{0}}{\ln\alpha}$. By
combining (6.2) with (6.4), we deduce

$\prod_{j=1}^{\infty}\frac{1}{3}\big{|}1+e^{2\pi i\frac{2y}{9^{j}}}+e^{2\pi
i(\frac{4}{q})^{j}x}\big{|}\leq
b_{0}^{l}\leq\alpha^{l\beta}\leq\left(\ln\frac{9^{3}}{4}\right)^{\beta}(\ln|x|)^{-\beta},$

which completes the proof of Claim 1.

According to the definition of $\phi^{*}$, we obtain that

$6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i})-6^{-1}9^{m_{0}}\phi^{*}(\mathbf{j})\in\mathscr{Z}(\widehat{\delta_{(p_{1}\cdots
p_{t})^{-1}D_{t}}})\subset\mathscr{Z}(\widehat{\mu_{n+m_{0}-1}}),$

where $t=m_{0}-1+\min\\{j:\mathbf{i}|_{j}\neq\mathbf{j}|_{j}\\}$. Then
$\\{6^{-1}9^{m_{0}}\phi^{*}(\mathbf{i})\\}_{\mathbf{i}\in\mathbf{I}_{n}}$ is a
bi-zero set of $\mu_{n+m_{0}-1}$. Then Theorem 6.1 gives Claim 2, which
completes the proof. ∎

###### Proof of Example 6.2.

Write

$\mu_{\\{p_{n},\mathcal{D}_{n}\\}}:=\mu_{n}\ast\mu_{>n},$

where
$\mu_{n}=\delta_{p_{1}^{-1}\mathcal{D}_{1}}\ast\cdots\delta_{(p_{1}\cdots
p_{n})^{-1}\mathcal{D}_{n}}$. Let
$\Lambda_{n}=\sum_{j=1}^{n}3^{j}j!\\{0,\pm\frac{1}{3}\\}$ and
$\Lambda=\cup_{n=1}^{\infty}\Lambda_{n}$. For any
$\lambda\neq\lambda^{\prime}\in\Lambda_{n}$, we know that

$\lambda-\lambda^{\prime}=\sum_{j=1}^{n}3^{j}j!(w_{j}-w_{j}^{\prime})=\sum_{j=j_{0}}^{n}3^{j}j!(w_{j}-w_{j}^{\prime})\in
3^{j_{0}}j_{0}!(\pm\frac{1}{3}+\mathbb{Z})\subset\mathscr{Z}(\widehat{\delta_{p_{1}^{-1}\cdots
p_{j_{0}}^{-1}\mathcal{D}_{j_{0}}}})\subset\mathscr{Z}(\widehat{\mu_{n}}),$

where $w_{j},w_{j}^{\prime}\in\\{\pm\frac{1}{3}\\}$ and
$j_{0}:=\min\\{j:w_{j}\neq w_{j}^{\prime}\\}\leq n$. Hence $\Lambda_{n}$ is a
bi-zero set of $\mu_{n}$. Note that $\\#\Lambda_{n}=\\#\text{spt}\mu_{n}$.
Then $\\{e^{2\pi i\lambda x}:\lambda\in\Lambda_{n}$ is complete in
$L^{2}(\mu_{n})$, i.e., $\Lambda_{n}$ is a spectrum of $\mu_{n}$.

For any $\lambda=\sum_{j=1}^{n}3^{j}j!w_{j}$ with
$w_{j}\in\\{0,\pm\frac{1}{3}\\}$, we have

$|x+\lambda|\leq\frac{1}{3}(3^{n}n!+3^{n-1}(n-1)!+\cdots+1)\leq\frac{3^{n}n!}{2}$

for any $|x|<\frac{1}{3}$. Note that

$\prod_{j=1}^{3}\frac{1}{3}\left|1+e^{2\pi
i\frac{2(x+\lambda)}{3^{n+j}(n+j)!}}+e^{2\pi
i(\frac{(3(n+j)^{2}+1)(x+\lambda)}{3^{n+j}(n+j)!})}\right|\geq\varepsilon_{0}^{3},$

where $\varepsilon_{0}:=\min_{|x|<\frac{1}{6}}\frac{1}{3}|1+e^{2\pi
ix}+e^{2\pi iy}|.$ Similar to Lemma 3.9, we have

$|\widehat{\mu_{>n}}(x+\lambda)|\geq\prod_{j=1}^{\infty}\frac{1}{3}\left|1+e^{2\pi
i\frac{2(x+\lambda)}{3^{n+j}(n+j)!}}+e^{2\pi
i(\frac{(3(n+j)^{2}+1)(x+\lambda)}{3^{n+j}(n+j)!})}\right|\geq\varepsilon_{0}^{3}\prod_{j=1}^{\infty}\cos\frac{1}{3^{j+3}}>0.$

Hence the example follows from Theorem 2.3(i) in [2]. ∎

Funding. The first author was supported by NSFC (Grant No. 11925107,
12226334), National Key R&D Program of China (Grant No. 2021YFA1003100) and
Key Research Program of Frontier Sciences, CAS (Grant No. ZDBS-LY-7002), and
the second author was supported in part by the NNSF of China (Nos. 11831007
and 12071125), China Postdoctoral Science Foundation (No. 2022TQ0358).

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