1. Introduction
Continued fraction expansion is induced by the Gauss map
$T: [0,1)\rightarrow [0,1)$
given by
$$ \begin{align*} T(0)=0,~T(x)=\frac{1}{x}~(\mathrm{mod}~1)\quad\mathrm{for}~x~\mathrm{in}~(0,1). \end{align*} $$
Then every irrational number
$x\in [0,1)$
can be uniquely expanded into an infinite form:
$$ \begin{align*} x=\frac{1}{a_{1}(x)+\frac{1}{a_{2}(x)+\ddots}}=[a_{1}(x),a_{2}(x),\ldots] , \end{align*} $$
where
$a_1(x)=\lfloor {1}/{x}\rfloor ~(\lfloor \cdot \rfloor ~\text {stands for the integral part} )~\text {and}~a_{n}(x)=a_{1}(T^{n-1}(x))$
for all
$n\geq 1$
, which are called the partial quotients of x. The finite truncation
$p_{n}(x)/q_{n}(x):=[a_{1}(x),a_{2}(x),\ldots ,a_{n}(x)]$
is called the n th convergent of
$x.$
Legendre’s theorem (see [Reference Khintchine18]) states that the terms
$p/q$
for which
$|x-p/q|<{1}/{2q^2}$
are the convergents of
$x.$
In this sense, for any
$\tau>0$
, the well-known Jarník set (see [Reference Besicovitch4, Reference Jarník16])
$$ \begin{align*} \bigg\{x\in [0,1): \bigg|x-\frac{p}{q}\bigg|<\frac{1}{q^{\tau+2}}\ \mathrm{for\ infinitely\ many}\ (p,q)\in \mathbb{N}^2\bigg\} \end{align*} $$
can be expressed in terms of continued fractions as (see [Reference Good9])
$$ \begin{align} \{x\in [0,1): a_{n+1}(x)\ge q_n(x)^{\tau}\ \mathrm{for\ infinitely\ many}\ n\in \mathbb{N}\}. \end{align} $$
Motivated by the continued fraction version of Jarník set (1-1), the relative growth properties for partial quotients were widely studied. More precisely, as proved in Good (see [Reference Good9, Theorem 9]), the set
$$ \begin{align*} \widetilde{J}(\tau)=\Big\{x\in [0,1): \limsup\limits_{n\rightarrow\infty}\frac{\log a_{n+1}(x)}{\log q_n(x)}\ge \tau\Big\} \end{align*} $$
is of Hausdorff dimension
${2}/({\tau +2}).$
Furthermore, by Bugeaud’s result [Reference Bugeaud6], the set of points with exact approximation order
$\tau $
$$ \begin{align*} J(\tau)=\bigg\{x\in [0,1): \limsup\limits_{n\rightarrow\infty}\frac{\log a_{n+1}(x)}{\log q_n(x)}=\tau\bigg\} \end{align*} $$
is also of Hausdorff dimension
${2}/({\tau +2}).$
Later on, Sun and Wu [Reference Sun and Wu21] showed that, for any
$\tau>0,$
the set
$$ \begin{align*} \bigg\{x\in [0,1): \lim\limits_{n\rightarrow\infty}\frac{\log a_{n+1}(x)}{\log q_n(x)}=\tau\bigg\} \end{align*} $$
has Hausdorff dimension
${1}/({\tau +2}).$
For the definition and properties of Hausdorff dimension, we refer to [Reference Falconer8]. For more results relating to the relative growth properties for partial quotients, see Haas [Reference Haas10], Tan and Zhou [Reference Tan and Zhou22], Wang and Wu [Reference Wang and Wu23], Wang et al. [Reference Wang, Wu and Xu24] and so on.
The improvability of Dirichlet’s theorem was first considered by Davenport and Schmidt [Reference Davenport and Schmidt7]. Very recently, Kleinbock and Wadleigh [Reference Kleinbock and Wadleigh19] continued the work and studied the Lebesgue measure of the set of
$\psi $
-Dirichlet improvable numbers:
$$ \begin{align*} \mathcal{D}(\psi) &:= \{ x \in [0,1):\mathrm{for\ all\ large\ Q,\ there\ exist\ } 1 \leq q < Q,\\ &\quad p \in \mathbb{Z} ~\mathrm{such\ that}~|qx - p| < \psi(Q) \}, \end{align*} $$
where
$\psi : \mathbb {N}\rightarrow \mathbb {R}$
is a nonincreasing function. It was found (see [Reference Kleinbock and Wadleigh19, Lemma 2.2]) that whether
$x\in \mathcal {D}(\psi )$
is tightly related to the growth of the product of two consecutive partial quotients. More precisely,
$$ \begin{align*} & \{x\in [0,1): \ a_n(x)a_{n+1}(x)\ge \Psi(q_n(x))\ {\mathrm{for\ infinitely\ many}}\ n\in \mathbb{N}\} \\ \subset \, & \mathcal{D}^c(\psi)\\ \subset \, & \big\{x\in [0,1): a_n(x)a_{n+1}(x)\ge \tfrac{1}{4}\Psi(q_n(x))\ {\mathrm{for\ infinitely\ many}}\ n\in \mathbb{N}\big\}, \end{align*} $$
where
$\Psi (t)={t\psi (t)}/({1-t\psi (t)}).$
This enabled Hussain et al. [Reference Hussain, Kleinbock, Wadleigh and Wang13] to determine the Hausdorff measure for the set of
$\psi $
-Dirichlet nonimprovable numbers. Their result [Reference Hussain, Kleinbock, Wadleigh and Wang13, Theorem 1.8] implies that the set
$$ \begin{align*} \bigg\{x\in [0,1): \limsup\limits_{n\rightarrow\infty}\frac{\log (a_n(x)a_{n+1}(x))}{\log q_n(x)}\ge \tau\bigg\} \end{align*} $$
has Hausdorff dimension
${2}/({\tau +2})$
. Afterwards, Huang and Wu [Reference Huang and Wu11] investigated the level set about the growth rate of the product of consecutive partial quotients relative to that of
$\{q_n(x): \ n\ge 1\}$
: let
$$ \begin{align*} G_m(\tau)=\bigg\{x\in [0,1): \lim\limits_{n\rightarrow\infty}\frac{\log (a_n(x)a_{n+1}(x)\cdots a_{n+m}(x))}{\log q_n(x)}=\tau\bigg\}, \end{align*} $$
where
$m\ge 1$
and
$\tau \ge 0.$
The following result was shown by Huang and Wu [Reference Huang and Wu11, Theorem 3.5].
Theorem 1.1 For any
$m\ge 1$
and any
$\tau \ge 0,$
$$ \begin{align*} \dim_{\mathrm{H}} G_m(\tau)=\begin{cases} {1}, \quad&\tau=0 , \\ \dfrac{1}{1+\gamma_m}, \quad& \tau>0, \end{cases} \end{align*} $$
where
$\gamma _m>1$
is the unique root of the equation
$\gamma ^{m}-({1}/{\gamma })=\tau $
and
${\dim _{\mathrm{H}}}$
denotes the Hausdorff dimension.
In this paper, we still continue the study on the relative growth properties of the product of consecutive partial quotients. For any natural number m, define
$$ \begin{align*} \pi_{n,m}(x):=a_n(x)a_{n+1}(x)\cdots a_{n+m}(x). \end{align*} $$
We mainly discuss the following level sets:
$$ \begin{align*} E_m(\tau)=\bigg\{x\in [0,1): \limsup\limits_{n\rightarrow\infty}\frac{\log \pi_{n,m}(x)}{\log q_n(x)}=\tau\bigg\} \end{align*} $$
and
$$ \begin{align*} \widetilde{E}_{m}(\tau)=\bigg\{x\in [0,1): \limsup\limits_{n\rightarrow\infty}\frac{\log \pi_{n,m}(x)}{\log q_n(x)}\ge\tau\bigg\}, \end{align*} $$
where
$\tau \ge 0$
. We obtain the following results.
Theorem 1.2 Let
$f_m(s)$
satisfy the following iterative formula:
$$ \begin{align*} f_1(s)=s, \ f_{k+1}(s)=\frac{sf_{k}(s)}{1-s+f_{k}(s)}, \quad k\ge 1, \end{align*} $$
and
$s_{\tau }^{(m)}$
be the solution to the equation
$\tau f_m(s)+2s=2.$
Then
$$ \begin{align*} {\dim_{\mathrm{H}}} E_m(\tau)={\dim_{\mathrm{H}}} \widetilde{E}_{m}(\tau)= \begin{cases} s_{\tau}^{(m)} &\quad \mathrm{if }\ \ m\ge 1, \ \tau\ge 0 ,\\ \dfrac{2-2\tau}{2-\tau} &\quad \mathrm{if }\ \ m=0, \ 0\le\tau\le 1,\\ 0 &\quad \mathrm{if }\ \ m=0, \ \tau>1. \end{cases} \end{align*} $$
Remark 1.3 By Theorem 1.2, the sets
$E_2(\tau )$
and
$\widetilde {E}_{2}(\tau )$
share the same Hausdorff dimension: one or
$({\sqrt {1+2\tau }-1})/{\tau }$
according to whether
$\tau =0$
or
$\tau>0$
. We also note that
${\dim _{\mathrm{H}}} E_0(\tau )$
and
${\dim _{\mathrm{H}}} \widetilde {E}_0(\tau )$
are different from
${\dim _{\mathrm{H}}} J(\tau )$
obtained by Bugeaud and
${\dim _{\mathrm{H}}} \widetilde {J}(\tau )$
obtained by Good, please see [Reference Bugeaud6] and [Reference Good9] respectively.
For more dimensional results relating to the product of consecutive partial quotients, the reader is referred to [Reference Bakhtawar1–Reference Bakhtawar, Bos and Hussain3, Reference Bos, Hussain and Simmons5, Reference Huang, Wu and Xu12].
2. Preliminaries
This section is devoted to fixing some notation, recalling some elementary properties shared by continued fractions and establishing some basic results.
Throughout this paper, we use
$|E|$
to denote the diameter of a set
$E\subset {\mathbb R}$
,
$a \asymp b$
to mean that
$0<c_1\le b/a\le c_2$
for unspecified constants
$c_1, c_2$
, and ‘i.m.’ to stand for ‘infinitely many’.
Recall that for any
$x\in [0,1)$
,
$p_n(x)$
and
$q_n(x)$
are the numerator and denominator of the n th convergent of x. It is known that
$p_n= p_n(x)$
and
$q_n= q_n(x)$
are decided by the following recurrence relations:
$$ \begin{align*} p_{n}=a_{n}(x)p_{n-1}+p_{n-2}, \quad q_{n}=a_{n}(x)q_{n-1}+q_{n-2} \end{align*} $$
with the conventions
$p_{-1}=1,~q_{-1}=0,~p_0=0$
and
$q_0=1$
. Thus,
$p_n=p_n(x), q_n=q_n(x)$
are determined by the partial quotients
$a_1,\ldots ,a_n$
, so we may write
$p_n=p_n(a_1,\ldots , a_n), q_n=q_n(a_1,\ldots ,a_n)$
. If no confusion is likely to arise, we denote them by
$p_n, q_n$
for simplicity.
For any
$n\ge 1$
and
$(a_1,\ldots ,a_n)\in \mathbb {N}^n,$
define
$$ \begin{align*} I_n(a_1,\ldots,a_n)=\{x\in [0, 1): a_1(x)=a_1, \ldots, a_n(x)=a_n\}, \end{align*} $$
which is the set of points beginning with
$(a_1,\ldots ,a_n)$
in their continued fraction expansions and is called a cylinder of order
$n.$
The first proposition collects basic properties of continued fractions needed later.
Proposition 2.1 [Reference Iosifescu and Kraaikamp14, Reference Khintchine17]
Let
$n\geq 1$
and
$a_{1},a_{2},\ldots ,a_{n}\in \mathbb {N}.$
One has:
-
(1)
The length of
$$ \begin{align*} I_n(a_1,a_2,\ldots,a_n)= \begin{cases} \bigg[\dfrac{p_n}{q_n}, \dfrac{p_n+p_{n-1}}{q_n+q_{n-1}}\bigg) &\quad \mathrm{if}\ \ n\ \mathrm{{is\ even}},\\[-8pt] \\ \bigg(\dfrac{p_n+p_{n-1}}{q_n+q_{n-1}}, \dfrac{p_n}{q_n}\bigg] &\quad \mathrm{if}\ \ n\ \mathrm{{is\ odd}}. \end{cases} \end{align*} $$
$I_n(a_1,a_2,\ldots ,a_n)$
is given by (2-1)
$$ \begin{align} \frac{1}{2q_n^2}\le|I_{n}(a_{1},a_{2},\ldots,a_{n})|=\frac{1}{q_{n}(q_{n}+q_{n-1})}\le\frac{1}{q_n^2}; \end{align} $$
-
(2)
$q_n\geq 2^{(n-1)/2}$
and, for any
$1\leq k\leq n,$
$$ \begin{align*} 1&\leq \frac{q_{n}(a_1,\ldots,a_k,a_{k+1},\ldots,a_{n})}{q_{k}(a_1,\ldots,a_k)\cdot q_{n-k}{(a_{k+1},\ldots,a_{n})}}\leq 2,\\ \prod\limits_{\kern1.2pt j=k+1}^{n}a_j&\le \frac{q_n(a_1,\ldots,a_{k},\ldots,a_n)}{q_{k}(a_1,\ldots,a_k)}\le \prod\limits_{\kern1.2pt j=k+1}^{n}(a_j+1); \end{align*} $$
-
(3) an n th-order cylinder
$I_n(a_1,\ldots ,a_n)$
is partitioned into subcylinders
$\{I_{n + 1} \ (a_1,\ldots ,a_n, a_{n+1}): a_{n+1}\in {\mathbb N}\}$
. When n is odd, these subcylinders are positioned from left to right, as
$a_{n+1}$
increases from 1 to
$\infty $
; when n is even, they are positioned from right to left.
Next we define a sequence of real numbers (so-called pre-dimensional numbers) as in [Reference Wang and Wu23]. Let
$\mathcal {A}\subset \mathbb {N}$
be a finite or infinite subset. For any
$m\ge 1$
and any
$\tau \ge 0$
, define
$$ \begin{align} s_{\tau}^{(m)}(\mathcal{A},n)=\inf\bigg\{s\geq 0: \sum\limits_{a_1,\ldots,a_{n}\in \mathcal{A}} \bigg(\frac{1}{q_n}\bigg)^{\tau f_m(s)+2s}\leq 1\bigg\}, \end{align} $$
where
$f_m(s)$
is given by the following iterative formula:
$$ \begin{align} f_1(s)=s, \ f_{k+1}(s)=\frac{sf_{k}(s)}{1-s+f_{k}(s)}, \ k\ge 1. \end{align} $$
Remark 2.2
-
(1) By induction on
$m,$
it is easily seen that
$f_m(0)=0, f_m(1)=1$
and
$f_m(s)$
is increasing with respect to
$s.$
Hence, the equation
$\tau f_m(s)+2s=2$
has a unique root, denoted
$s_{\tau }^{(m)}.$
Furthermore, by the implicit function theorem, we know that for any
$m\ge 1,$
as a function of
$\tau $
,
$s_{\tau }^{(m)}$
is continuous and decreasing (since
${ds}/{d\tau }=-({f_m(s)}/({\tau f_m'(s)+2}))\le 0).$
-
(2) If
$\mathcal {A}$
is a finite subset of
${\mathbb N}$
, when substituting s for
$ s_{\tau }^{(m)}(\mathcal {A},n)$
in the summation, we get an equality.
Proposition 2.3 For any
$m\ge 1,$
$\lim \nolimits _{n\rightarrow \infty }s_{\tau }^{(m)}(\mathcal {A},n)$
exists.
Proof. This can be proved similarly to [Reference Wang and Wu23, Lemma 2.4].
Write
$\lim \nolimits _{n\rightarrow \infty }s_{\tau }^{(m)}(\mathcal {A},n) = s_{\tau }^{ ( m ) }(\mathcal {A}).$
From (2-1) and the definition of
$s_{\tau }^{(m)} ( \mathcal {A},n ) ,$
we have
$0\le s_{\tau }^{(m)}(\mathcal {A})\le 1.$
For any
$M\in {\mathbb N},$
take
$\mathcal {A}_{M}=\{1, 2, \ldots , M\}.$
For simplicity, write
$s_{\tau }^{(m)}(M,n)$
for
$s_{\tau }^{(m)}(\mathcal {A}_{M},n)$
and
$s_{\tau }^{(m)}(M)$
for
$s_{\tau }^{(m)}(\mathcal {A}_{M}).$
By (2-1), it is easy to check that
$s_{\tau }^{(m)}(\mathbb {N})$
is the solution to the equation
$\tau f_m(s)+2s=2,$
that is,
$s_{\tau }^{(m)}(\mathbb {N})=s_{\tau }^{(m)}.$
Proposition 2.4
$\lim \nolimits _{M\rightarrow \infty }s_{\tau }^{(m)}(M)=s_{\tau }^{(m)}(\mathbb {N})=s_{\tau }^{(m)}$
.
Proof. The existence of the limit follows in the same way as in [Reference Wang and Wu23, Lemma 2.5]. The convergence can also be obtained by using Mauldin and Urbański’s result (see [Reference Mauldin and Urbański20]).
A dimensional result is cited here for later use.
For any
$M\in \mathbb {N},$
let
$E_M=\{x\in [0,1): 1\le a_n(x)\le M \ \text {for any}~n\ge 1\}.$
Jarník considered its Hausdorff dimension and obtained the following result.
Lemma 2.5 (see [Reference Jarník15])
For any
$M\ge 8,$
$$ \begin{align*} 1-\frac{1}{M\log 2}\le {\dim_{\mathrm{H}}} E_M\le 1-\frac{1}{8M\log M}. \end{align*} $$
In particular, the set
$$ \begin{align*} E=\bigg\{x\in[0,1): \sup\limits_{n\ge 1}a_n(x)<+\infty\bigg\} \end{align*} $$
is of Hausdorff dimension one.
We end this section with the following proposition that is actually the logarithmic form of [Reference Huang, Wu and Xu12, Proposition 2.13].
Proposition 2.6 Fix
$0\le s<1$
and
$\tau>0{\color {blue}{.}}$
For each
$m\ge 1,$
$$ \begin{align*} \sup_{0 < \beta_1 \le \cdots \le \beta_{m - 1} \le \beta_m = \tau} \min \{ - \beta_{1}s,\ldots,\beta_{m - 2} ( 1 - s ) - \beta_{m - 1}s, \beta_{m - 1} ( 1 - s ) - \beta_m s \} = - \tau f_m ( s ) \end{align*} $$
and the supremum is attained when all the terms are equal. Moreover, the function
$f_{m}(s)$
satisfies the recursion given in (2-3).
Proof. For a detailed proof, we refer the reader to [Reference Huang, Wu and Xu12, Proposition 2.13].
3. The proof of Theorem 1.2: upper bound
In this section, we determine the Hausdorff dimensions of
$E_m(0)$
and
$E_0(\tau )$
, where
$\tau \ge 0$
. Then we give an upper bound for
${\dim _{\mathrm{H}}} E_m(\tau ),$
where
$m\ge 1, \tau>0.$
Lemma 3.1 For any
$m\ge 0$
,
${\dim _{\mathrm{H}}} E_m(0)=1.$
Proof. Recall the definition of
$E_M$
and fix
$x\in E_M;$
it is clear that for any
$m\ge 0,$
$$ \begin{align*} \limsup\limits_{n\rightarrow\infty}\frac{\log(a_n(x)a_{n+1}(x)\cdots a_{n+m}(x))}{\log q_n(x)} \le \frac{(m+1)\log M}{\frac{n-1}{2}\log 2}=0. \end{align*} $$
This implies that
$E_M\subset E_m(0).$
By Lemma 2.5 and, letting
$M\to \infty $
, we have
${\dim _{\mathrm{H}}} E_m(0)=1 \text { for all } m\ge 0.$
Proposition 3.2 For any
$\tau \ge 0,$
$$ \begin{align*} {\dim_{\mathrm{H}}} E_0(\tau)={\dim_{\mathrm{H}}} \widetilde{E}_{0}(\tau)= \begin{cases} \dfrac{2-2\tau}{2-\tau} & \quad \mathrm{if}\ \ 0\le\tau\le 1\\ 0 & \quad \mathrm{if}\ \ \tau>1. \end{cases} \end{align*} $$
Proof. Since
$q_n(x)\ge a_n(x),$
$E_0(\tau ),\widetilde {E}_{0}(\tau ),J(\tau ),\widetilde {J}(\tau ),$
it is easily seen that
$$ \begin{align*} E_0(\tau)=J\bigg(\frac{\tau}{1-\tau}\bigg), \ \ \ \widetilde{E}_{0}(\tau)=\widetilde{J}\bigg(\frac{\tau}{1-\tau}\bigg). \end{align*} $$
By using the results of Bugeaud and Good,
$$ \begin{align*} {\dim_{\mathrm{H}}} E_0(\tau)={\dim_{\mathrm{H}}} \widetilde{E}_{0}(\tau)=\frac{2}{\tfrac{\tau}{1-\tau}+2}=\frac{2-2\tau}{2-\tau}.\\[-42pt] \end{align*} $$
Throughout the remainder of the paper, we assume that
$m\ge 1, \tau>0.$
A simple estimation on a summation is given here for later use.
Lemma 3.3 Let
$\Gamma>1.$
For any integer
$k\ge 2$
and
$0<s<1,$
$$ \begin{align*} \sum\limits_{(c_1,\ldots,c_{k-1})\in \mathbb{N}^{k-1}: \atop1\leq c_1\cdots c_{k-1}\leq \Gamma}\bigg(\frac{1}{c_1\cdots c_{k-1}}\bigg)^s\asymp \frac{(\log \Gamma )^{k-2}}{(k-2)!}\Gamma^{1-s}. \end{align*} $$
Proof. This can be proved by induction on
$k.$
One can refer to [Reference Huang, Wu and Xu12, Lemma 4.2].
For any
$m\ge 1,$
by the definitions of
$E_m(\tau )$
and
$\widetilde {E}_m(\tau )$
, it is immediate that
$$ \begin{align} E_m(\tau)&\subset \widetilde{E}_m(\tau)\subset\bigcap\limits_{0 < \alpha < \tau}\{x\in [0,1): \pi_{n,m}(x) \ge q_n^{\alpha}(x) \ \mathrm{for\ i.m.}\ n \in \mathbb{N}\}\\ & := \bigcap\limits_{0 < \alpha < \tau}W_m(\alpha).\nonumber \end{align} $$
The following proposition gives an upper bound for
${\dim _{\mathrm{H}}} W_m(\alpha ).$
Proposition 3.4 For any
$m\ge 1$
and any
$\alpha>0$
,
$$ \begin{align} {\dim_{\mathrm{H}}} W_m(\alpha)\le s_{\alpha}^{(m)}. \end{align} $$
Proof. When
$m=1,$
the set
$W_1(\alpha )$
can be written in the following form:
$$ \begin{align*} W_1(\alpha)=&\{x\in [0,1): a_n(x)a_{n+1}(x)\ge q_n^{\alpha}(x)\ \mathrm{for\ i.m.}\ n\in \mathbb{N}\}\\ =&\bigcap\limits_{N=1}^{\infty}\bigcup\limits_{n=N}^{\infty}\bigcup\limits_{a_1,\ldots,a_n}J_n(a_1,\ldots,a_n), \end{align*} $$
where
$J_n(a_1,\ldots ,a_n)=\bigcup \nolimits _{a_{n+1}\ge ({q_n^{\alpha }}/{a_n})}I_{n+1}(a_1,\ldots ,a_n,a_{n+1}).$
By Proposition 2.1, it is easy to see that
$$ \begin{align*} |J_n(a_1,\ldots,a_n)|\asymp \frac{1}{\frac{q_n^{\alpha}}{a_n}q_n^2}\asymp \frac{1}{q_{n-1}q_n^{\alpha+1}}. \end{align*} $$
Thus, the s-dimensional Hausdorff measure of
$W_1(\alpha )$
can be estimated as
$$ \begin{align} \mathcal{H}^s(W_1(\alpha))\le & \liminf\limits_{N\rightarrow \infty}\sum\limits_{n\geq N}\sum\limits_{a_1,\ldots,a_{n}\in \mathbb{N}}|J_n(a_1,\ldots,a_n)|^s\nonumber\\ \asymp &\liminf\limits_{N\rightarrow \infty}\sum\limits_{n\geq N}\sum\limits_{a_1,\ldots,a_{n}\in \mathbb{N}} \bigg(\frac{1}{q_{n-1}q_n^{\alpha+1}}\bigg)^s\nonumber\\ \le &\liminf\limits_{N\rightarrow \infty}\sum\limits_{n\geq N}\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}} \bigg(\frac{1}{q_{n-1}}\bigg)^{(\alpha+2)s}\sum\limits_{a_n=1}^{\infty}\bigg(\frac{1}{a_n}\bigg)^{(\alpha+1)s}. \end{align} $$
For any
$s_{\alpha }^{(1)}<s<1,$
$$ \begin{align*} (\alpha+2)s> \alpha s_{\alpha}^{(1)}+2s_{\alpha}^{(1)}=2 \end{align*} $$
and then
$(\alpha +1)s>2-s>1.$
This implies that the series
$$ \begin{align*} \sum\limits_{a_n=1}^{\infty}\bigg(\frac{1}{a_n}\bigg)^{(\alpha+1)s}=:c_0(s)<+\infty. \end{align*} $$
By (2-1),
$$ \begin{align} 1\le \sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\bigg(\frac{1}{q_{n-1}}\bigg)^2 \le 2. \end{align} $$
Then, according to (3-3)–(3-4), we know for any
$s\in (s_{\alpha }^{(1)},1)$
that
$$ \begin{align*} \mathcal{H}^s(W_1(\alpha))<+\infty. \end{align*} $$
Thus,
$$ \begin{align*} {\dim_{\mathrm{H}}} W_1(\alpha)\le s_{\alpha}^{(1)}. \end{align*} $$
Assume that (3-2) holds for
$m=k-1;$
we show that it is also true for
$m=k.$
Recall that
$$ \begin{align*} W_k(\alpha)=\{x\in [0,1): a_n(x)a_{n+1}(x)\cdots a_{n+k}(x)\ge q_n^{\alpha}(x)\ \mathrm{for\ i.m.}\ n\in \mathbb{N}\}. \end{align*} $$
For any
$0<\beta \le \alpha ,$
by the definition of
$W_k(\alpha ),$
it is obvious that
$$ \begin{align*} W_k(\alpha)\subseteq W_{k-1}(\kern1.2pt \beta)\bigcup F_{k}(\kern1.2pt \beta,\alpha), \end{align*} $$
where
$$ \begin{align*} F_{k}(\kern1.2pt \beta,\alpha)=\bigg\{&x\in [0,1): 1\le a_n(x)\cdots a_{n+k-1}(x)\le q_n^{\kern1.5pt \beta}(x),\\ & a_{n+k}(x)\ge \frac{q_n^{\alpha}(x)}{a_n(x)\cdots a_{n+k-1}(x)}\ \mathrm{for\ i.m.}\ n\in \mathbb{N}\bigg\}. \end{align*} $$
Thus,
$$ \begin{align} {\dim_{\mathrm{H}}} W_k(\alpha)\le \inf\limits_{0<\beta\le \alpha}\max\{{\dim_{\mathrm{H}}} W_{k-1}(\kern1.2pt \beta),\ {\dim_{\mathrm{H}}} F_{k}(\kern1.2pt \beta,\alpha)\}. \end{align} $$
By the induction hypothesis,
$$ \begin{align} {\dim_{\mathrm{H}}} W_{k-1}(\kern1.2pt \beta)\le s_{\beta}^{(k-1)}, \end{align} $$
where
$s_{\beta }^{(k-1)}$
is the solution to the equation
$\beta f_{k-1}(s)+2s=2$
. By Remark 2.2,
$s_{\beta }^{(k-1)}$
is decreasing with respect to
$\beta $
.
Our next goal is to seek an upper bound for
${\dim _{\mathrm{H}}} F_k(\kern1.2pt \beta ,\alpha )$
. Clearly,
$F_k(\kern1.2pt \beta ,\alpha )$
can be expressed as
$$ \begin{align*} F_k(\kern1.2pt \beta,\alpha) = \bigcap\limits_{N = 1}^{\infty}\bigcup\limits_{n = N}^{\infty} \bigcup\limits_{a_1,\ldots,a_n\in \mathbb{N}}\bigg\{&x \in I_n(a_1, \ldots,a_n): 1 \le a_n(x) \cdots a_{n + k - 1}(x) \le q_n^{\kern1.5pt \beta}(x),\nonumber\\ &a_{n+k}(x) \ge \frac{q_n^{\alpha}(x)}{a_n(x)\cdots a_{n + k - 1}(x)}\bigg\}. \end{align*} $$
Let
$$ \begin{align*} J_{n+k-1}(a_1,\ldots,a_{n+k-1})=\bigcup\limits_{a_{n+k}\ge ({q_n^{\alpha}}/{a_n\cdots a_{n+k-1}})}I_{n+k}(a_1,\ldots,a_{n+k-1},a_{n+k}). \end{align*} $$
It is clear that
$$ \begin{align} \hspace{-1em} F_k (\kern1.2pt \beta,\alpha ) \subseteq \bigcap\limits_{N = 1}^{\infty} \bigcup\limits_{n = N}^{\infty} \bigcup\limits_{a_1,\ldots,a_n \in \mathbb{N}} \bigcup\limits_{1 \le a_{n + 1}\cdots a_{n + k - 1} \le ({q_n^{\kern1.5pt \beta}}/{a_n})}\! J_{n + k - 1} ( a_1, \ldots ,a_n,a_{n + 1}, \ldots ,a_{n + k - 1} ). \end{align} $$
By Proposition 2.1, the diameter of
$J_{n+k-1}(a_1,\ldots ,a_{n+k-1})$
can be estimated as follows:
$$ \begin{align*} | J_{n+k-1}(a_1,\ldots,a_{n+k-1}) | \asymp \frac{1}{\dfrac{q_n^{\alpha}(x)}{a_n(x)\cdots a_{n+k-1}(x)}q_{n+k-1}^2} \asymp \frac{1}{q_n^{\alpha}q_{n-1}q_{n+k-1}}. \end{align*} $$
Fix
$\varepsilon>0.$
Since the function
${(\log t^{\alpha })^{k-2}}/{t^{\alpha \varepsilon }}$
is decreasing for sufficiently large t and
$q_n\ge 2^{({n-1})/{2}}$
for any
$n\ge 1$
, there exists
$N_0$
such that for all
$n\ge N_0,$
$$ \begin{align} \frac{1}{q_n^{\alpha \varepsilon}}\cdot \frac{ (\log \frac{ q_n^{\kern1.5pt \beta}}{a_n})^{k-2}}{(k-2)!}\le \frac{(\log q_n^{\alpha})^{k-2}}{q_n^{\alpha \varepsilon}}\le \frac{(\frac{n-1}{2}\alpha\log 2)^{k-2}}{2^{({n-1})/{2}\alpha \varepsilon}}\le 1. \end{align} $$
Recall (3-7)–(3-8) and Lemma 3.3. The s-dimensional Hausdorff measure of
$F_k(\kern1.2pt \beta ,\alpha )$
can be estimated as
$$ \begin{align*} &\mathcal{H}^{s+\varepsilon}(F_k(\kern1.2pt \beta,\alpha))\\ &\quad\le \liminf\limits_{N\rightarrow \infty}\sum\limits_{n \geq N}\sum\limits_{a_1,\ldots,a_{n}\in \mathbb{N}}\sum\limits_{1 \le a_{n + 1}\cdots a_{n + k - 1} \le ({q_n^{\kern1.5pt \beta}}/{a_n})} | J_{n + k - 1}(a_1,\ldots,a_n,a_{n + 1},\ldots,a_{n + k - 1}) |^{s + \varepsilon}\\ &\quad\asymp \liminf\limits_{N\rightarrow \infty}\sum\limits_{n\geq N}\sum\limits_{a_1,\ldots,a_{n}\in \mathbb{N}}\sum\limits_{1\le a_{n+1}\cdots a_{n+k-1}\le ({q_n^{\kern1.5pt \beta}}/{a_n})}\bigg(\frac{1}{q_n^{\alpha}q_{n-1}q_{n+k-1}}\bigg)^{s+\varepsilon}\\ &\quad\le \liminf\limits_{N\rightarrow \infty}\sum\limits_{n\geq N}\sum\limits_{a_1,\ldots,a_{n}\in \mathbb{N}}\bigg(\frac{1}{q_n^{\alpha}q_{n-1}q_n}\bigg)^s \frac{1}{q_n^{\alpha \varepsilon}}\cdot \frac{(\log \frac{ q_n^{\kern1.5pt \beta}}{a_n})^{k-2}}{(k-2)!}\bigg(\frac{q_n^{\kern1.5pt \beta}}{a_n}\bigg)^{1-s}\\ &\quad\le \liminf\limits_{N\rightarrow \infty}\sum\limits_{n\geq N}\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}} \bigg(\frac{1}{q_{n-1}}\bigg)^{(\alpha+2)s-\beta(1-s)} \sum\limits_{a_n=1}^{\infty}\bigg(\frac{1}{a_n}\bigg)^{\alpha s+1-\beta(1-s)}. \end{align*} $$
Write
$s(\kern1.2pt \beta ,\alpha )=({2+\beta })/({\alpha +2+\beta }).$
For any
$s(\kern1.2pt \beta ,\alpha )<s<1,$
$$ \begin{align*} (\alpha+2)s-\beta(1-s)>(\alpha+2)s(\kern1.2pt \beta,\alpha)-\beta(1-s(\kern1.2pt \beta,\alpha))=2 \end{align*} $$
and then
$\alpha s+1-\beta (1-s)>3-2s>1$
.
Combining the convergence of
$\sum \nolimits _{a_n=1}^{\infty }({1}/{a_n})^{\alpha s+1-\beta (1-s)}$
and (3-4), we know for any
$s>s(\kern1.2pt \beta ,\alpha )$
that
$$ \begin{align*} \mathcal{H}^{s+\varepsilon}(F_k(\kern1.2pt \beta,\alpha))<+\infty. \end{align*} $$
This gives
${\dim _{\mathrm{H}}} F_k(\kern1.2pt \beta ,\alpha )\le s(\kern1.2pt \beta ,\alpha )+\varepsilon .$
Since
$\varepsilon>0$
is arbitrary,
$$ \begin{align} {\dim_{\mathrm{H}}} F_k(\kern1.2pt \beta,\alpha)\le s(\kern1.2pt \beta,\alpha). \end{align} $$
According to (3-5), (3-6) and (3-9),
$$ \begin{align} {\dim_{\mathrm{H}}} W_k(\alpha)\le \inf\limits_{0<\beta\le \alpha}\max\{ s_{\beta}^{(k-1)},\ s(\kern1.2pt \beta,\alpha)\}. \end{align} $$
Recalling the definitions of
$s_{\beta }^{(k-1)}$
and
$s(\kern1.2pt \beta ,\alpha )$
, we know that
$s_{\beta }^{(k-1)}$
is decreasing and
$s(\kern1.2pt \beta ,\alpha )$
is increasing with respect to
$\beta $
. So, the infimum in (3-10) is attained at the point
$\beta $
, where
$s_{\beta }^{(k-1)}=s(\kern1.2pt \beta ,\alpha ),$
that is,
$s_{\beta }^{(k-1)}$
satisfies the following equations:
$$ \begin{align*} \begin{cases} \ \ \beta f_{k-1}(s)+2s=2 , \\ s(\alpha+2+\beta)=2+\beta. \end{cases} \end{align*} $$
A simple calculation shows that
$\beta ={\alpha s}/({1-s+f_{k-1}(s)}).$
By substituting the value of
$\beta $
into the first equation,
$$ \begin{align*} \frac{\alpha sf_{k-1}(s)}{1-s+f_{k-1}(s)}+2s=2. \end{align*} $$
Consequently,
$s_{\beta }^{(k-1)}$
satisfies the equation
$\alpha f_k(s)+2s=2.$
So,
$s_{\beta }^{(k-1)}=s_{\alpha }^{(k)}$
and then
$$ \begin{align*} {\dim_{\mathrm{H}}} W_k(\alpha)\le s_{\alpha}^{(k)}, \end{align*} $$
which proves the proposition.
According to (3-1) and Proposition 3.4, for any
$m\ge 1$
,
$$ \begin{align*} {\dim_{\mathrm{H}}} E_m(\tau)\le{\dim_{\mathrm{H}}} \widetilde{E}_m(\tau)\le {\dim_{\mathrm{H}}} W_m(\alpha)\le s_{\alpha}^{(m)}, \end{align*} $$
where
$\alpha $
is any number between
$0$
and
$\tau $
.
Then, by using the continuity of
$s_{\alpha }^{(m)}$
in Remark 2.2 and letting
$\alpha \rightarrow \tau ,$
$$ \begin{align} {\dim_{\mathrm{H}}} E_m(\tau)\le {\dim_{\mathrm{H}}} \widetilde{E}_m(\tau)\le s_{\tau}^{(m)}. \end{align} $$
4. The proof of Theorem 1.2: lower bound
In this section, by constructing an appropriate Cantor subset of
$E_m(\tau )$
and then applying the mass distribution principle [Reference Falconer8, Proposition 4.2], we obtain the lower bound of
${\dim _{\mathrm{H}}} E_m(\tau ).$
4.1. Cantor subset
Fix
$0<s<s_{\tau }^{(m)}.$
By Propositions 2.3 and 2.4, choose integers M and N large enough that
$s<s_{\tau }^{(m)}(M,N).$
Fix a sequence of integers such that
$$ \begin{align} l_1\ge m+1,\ \ l_j\ge (2j^2(\tau+1)^{\kern1.2pt j-2}\log_2(M+1)^N)\cdot l_{\kern1.2pt j-1}\quad(\mathrm{for\ all}\ j\ge 2). \end{align} $$
Then set
$n_1=l_1N+1$
and, for all
$j\ge 2,$
$$ \begin{align*} n_j=n_{\kern1.2pt j-1}+m+l_{\kern1.2pt j}N+1=(l_1N+m+1)+\cdots+(l_{\kern1.2pt j-1}N+m+1)+l_jN+1. \end{align*} $$
For any
$0<\beta _1\le \cdots \le \beta _{m-1}\le \beta _m=\tau ,$
we define
$$ \begin{align*} \mathcal{O}_M(\kern1.2pt \beta_1,\ldots,\beta_m) =\{x\in [0,1)&:\mathrm{for\ all}\ j\ge 1,\ a_{n_j}(x)=1,\\ &\quad q_{n_j}^{\kern1.5pt \beta_1}(x)\le a_{n_j+1}(x)<2q_{n_j}^{\kern1.5pt \beta_1}(x),\\ &\quad q_{n_j}^{\kern1.5pt \beta_i-\beta_{i-1}}(x)\le a_{n_j+i}(x)<2q_{n_j}^{\kern1.5pt \beta_i-\beta_{i-1}}(x),\\ &\quad \mathrm{where}\ 2 \le i\le m \mathrm{\ for\ other\ } n, 1\le a_n(x)\le M\}. \end{align*} $$
We prove that
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
is a subset of
$E_m(\tau ).$
Fix
$x\in \mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m);$
then
$$ \begin{align*} q_{n_j}^{\kern1.5pt \beta_m}(x)\le a_{n_j}(x)a_{n_j+1}(x)\cdots a_{n_j+m}(x)<2^mq_{n_j}^{\kern1.5pt \beta_m}(x)\quad \mathrm{for\ all}\ j\ge 1 \end{align*} $$
and
$$ \begin{align*} a_n(x)a_{n+1}(x)\cdots a_{n+m}(x)\le M^{m+1}q_n^{\kern1.5pt \beta_m}(x)\quad \mathrm{for\ all} \ n\ge 1. \end{align*} $$
Thus,
$$ \begin{align*} \limsup\limits_{n\rightarrow\infty}\frac{\log ( a_n(x)a_{n + 1}(x)\cdots a_{n + m}(x) ) }{\log q_n(x)} & \ge \limsup\limits_{\kern1.2pt j\rightarrow\infty}\frac{\log ( a_{n_j}(x)a_{n_j + 1}(x)\cdots a_{n_j + m}(x) ) }{\log q_{n_j}(x)}\\ &=\beta_m \end{align*} $$
and
$$ \begin{align*} \limsup\limits_{n\rightarrow\infty}\frac{\log (a_n(x)a_{n+1}(x)\cdots a_{n+m}(x))}{\log q_n(x)} &\le \limsup\limits_{n\rightarrow\infty}\frac{\log (M^{m+1}q_n^{\kern1.5pt \beta_m}(x))}{\log q_n(x)}\\ &\le \limsup\limits_{n\rightarrow\infty}\frac{\log M^{m+1}}{\frac{n-1}{2}\log 2}+\beta_m=\beta_m. \end{align*} $$
Hence,
$$ \begin{align*} \limsup\limits_{n\rightarrow\infty}\frac{\log (a_n(x)a_{n+1}(x)\cdots a_{n+m}(x))}{\log q_n(x)}=\beta_m=\tau. \end{align*} $$
4.1.1. Upper bound for
${\dim _{\mathrm{H}}} \,\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
It is clear that
$$ \begin{align} \mathcal{O}_M(\kern1.2pt \beta_1,\ldots,\beta_m)\subset\bigcap\limits_{N=1}^{\infty}\bigcup\limits_{n=N}^{\infty}\bigcup\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}F_{n}(a_1,\ldots,a_{n-1}), \end{align} $$
where
$$ \begin{align*} F_{n}(a_1,\ldots,a_{n-1})=\{x\in [0,1)&: a_j(x)=a_j \mathrm{\ for\ } 1\le j\le n-1,\ a_n(x)=1,\\ &\quad q_n^{\kern1.5pt \beta_1}(x)\le a_{n+1}(x)<2q_n^{\kern1.5pt \beta_1}(x),\\ &\quad q_n^{\kern1.5pt \beta_i-\beta_{i-1}}(x)\le a_{n+i}(x)<2q_n^{\kern1.5pt \beta_i-\beta_{i-1}}(x) \mathrm{\ for\ } 2 \le i \le m\}. \end{align*} $$
Write
$F_n:=\bigcup \nolimits _{a_1,\ldots ,a_{n-1}\in \mathbb {N}}F_n(a_1,\ldots ,a_{n-1}).$
Then, by (4-2),
$$ \begin{align*} \mathcal{O}_M(\kern1.2pt \beta_1,\ldots,\beta_m)\subset\bigcap\limits_{N=1}^{\infty}\bigcup\limits_{n=N}^{\infty}F_n. \end{align*} $$
We follow the procedure from [Reference Huang, Wu and Xu12] to find the natural cover of
$F_n$
and estimate the s-volume of its cover, thereby seeking the upper bound for
${\dim _{\mathrm{H}}} \,\mathcal {O}_M (\kern1.2pt \beta _1,\ldots ,\beta _m ) .$
There are
$m+1$
types of covering given by the collection of intervals
$$ \begin{align*} J_{n-1},J_n,\ldots,J_{n+m-1} \end{align*} $$
defined as follows.
(1) Type 1: Define
$$ \begin{align*} J_{n-1}(a_1,\ldots,a_{n-1})=I_n(a_1,\ldots,a_{n-1},1). \end{align*} $$
Covering
$F_n$
by the collections of
$J_{n-1}$
gives
$$ \begin{align*} \sum\limits_{a_1,\ldots,a_{n - 1} \in \mathbb{N}} | I_n(a_1,\ldots,a_{n - 1},1) |^s \asymp \sum\limits_{a_1,\ldots,a_{n - 1} \in \mathbb{N}} \bigg( \frac{1}{q_n^2(a_1,\ldots,a_{n - 1},1)} \bigg)^s \asymp \sum\limits_{a_1,\ldots,a_{n - 1} \in \mathbb{N}}q_{n - 1}^{-2s}. \end{align*} $$
(2) Type 2: Define
$$ \begin{align*} J_n(a_1,\ldots,a_{n-1},1)=\bigcup\limits_{q_n^{\kern1.5pt \beta_1}\le a_{n+1}<2q_n^{\kern1.5pt \beta_1}}I_{n+1}(a_1,\ldots,a_{n-1},1,a_{n+1}). \end{align*} $$
Covering
$F_n$
by the collections of
$J_{n}$
gives
$$ \begin{align*} \sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}|J_n(a_1,\ldots,a_{n-1},1)|^s&\asymp\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\bigg(\frac{1}{q_n^{\kern1.5pt \beta_1+2}(a_1,\ldots,a_{n-1},1)}\bigg)^s\\ &\asymp\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}} q_{n-1}^{-\beta_1s-2s}. \end{align*} $$
(3) Type 3: For any
$a_n=1,\ q_n^{\kern1.5pt \beta _1}\le a_{n+1}<2q_n^{\kern1.5pt \beta _1},$
define
$$ \begin{align*} J_{n+1}(a_1,\ldots,a_{n+1})=\bigcup\limits_{q_n^{\kern1.5pt \beta_2-\beta_1}\le a_{n+2}<2q_n^{\kern1.5pt \beta_2-\beta_1}}I_{n+2}(a_1,\ldots,a_{n+1},a_{n+2}). \end{align*} $$
Covering
$F_n$
by the collections of
$J_{n+1}$
gives
$$ \begin{align*} &\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\sum\limits_{\substack{a_n=1,\\{q_n^{\kern1.5pt \beta_1}\le a_{n+1}<2q_n^{\kern1.5pt \beta_1}}} }|J_{n+1}(a_1,\ldots,a_{n-1},a_n,a_{n+1})|^s\\ &\qquad\asymp\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\sum\limits_{\substack{a_n=1,\\ q_n^{\kern1.5pt \beta_1}\le a_{n+1}<2q_n^{\kern1.5pt \beta_1}} } \bigg(\frac{1}{q_n^{\kern1.5pt \beta_2-\beta_1}q_{n+1}^2}\bigg)^s\\ &\qquad\asymp\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\bigg(\frac{1}{q_{n}^{\kern1.5pt \beta_2-\beta_1+2}(a_1,\ldots,a_{n-1},1)}\bigg)^s\cdot q_n^{\kern1.5pt \beta_1(1-2s)}(a_1,\ldots,a_{n-1},1)\\ &\quad=\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}q_n^{\kern1.5pt \beta_1(1-s)-\beta_2s-2s}(a_1,\ldots,a_{n-1},1)\\ &\qquad\asymp\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}q_{n-1}^{\kern1.5pt \beta_1(1-s)-\beta_2s-2s}. \end{align*} $$
(k) (
$4\le k\le m+1$
) Type k: For any
$a_n=1,\ q_n^{\kern1.5pt \beta _1}\le a_{n+1}<2q_n^{\kern1.5pt \beta _1},\ q_n^{\kern1.5pt \beta _2-\beta _1}\le a_{n+2}<2q_{n}^{\kern1.5pt \beta _2-\beta _1},\ldots ,\ q_n^{\kern1.5pt \beta _{k-2}-\beta _{k-3}}\le a_{n+k-2}<2q_{n}^{\kern1.5pt \beta _{k-2}-\beta _{k-3}},$
define
$$ \begin{align*} J_{n + k - 2}(a_1,\ldots,a_{n + k - 2}) = \bigcup\limits_{q_{n}^{\kern1.5pt \beta_{k - 1} - \beta_{k - 2}} \le a_{n+k-1} < 2q_{n}^{\kern1.5pt \beta_{k - 1} - \beta_{k - 2}}}I_{n + k - 1}(a_1,\ldots,a_{n + k - 2},a_{n + k - 1}). \end{align*} $$
Covering
$F_n$
by the collections of
$J_{n+k-2}$
for
$4\le k\le m+1$
gives
$$ \begin{align*} &\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\sum\limits_{\substack {a_n=1,\ q_n^{\kern1.5pt \beta_1}\le a_{n+1}<2q_n^{\kern1.5pt \beta_1},\\q_n^{\kern1.5pt \beta_j-\beta_{\kern1.2pt j-1}}\le a_{n+j}<2q_n^{\kern1.5pt \beta_j-\beta_{\kern1.2pt j-1}}\\ \mathrm{for\ all}\ 2\le j\le k-2}}|J_{n+k-2}(a_1,\ldots,a_{n-1},a_n,\ldots,a_{n+k-2})|^s\\ &\qquad\asymp \sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\sum\limits_{\substack {a_n=1,\ q_n^{\kern1.5pt \beta_1}\le a_{n+1}<2q_n^{\kern1.5pt \beta_1},\\q_n^{\kern1.5pt \beta_j-\beta_{\kern1.2pt j-1}}\le a_{n+j}<2q_n^{\kern1.5pt \beta_j-\beta_{\kern1.2pt j-1}}\\ \mathrm{for\ all}\ 2\le j\le k-2}}\bigg(\frac{1}{q_n^{\kern1.5pt \beta_{k-1}-\beta_{k-2}}q_{n+k-2}^2}\bigg)^s\\ &\qquad\asymp \sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}\bigg(\frac{1}{q_{n}^{\kern1.5pt \beta_{k-1}-\beta_{k-2}+2}(a_1,\ldots,a_{n-1},1)}\bigg)^s\cdot q_n^{\kern1.5pt \beta_{k-2}(1-2s)}(a_1,\ldots,a_{n-1},1)\\ &\qquad=\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}q_n^{\kern1.5pt \beta_{k-2}(1-s)-\beta_{k-1}s-2s}(a_1,\ldots,a_{n-1},1)\\ &\qquad\asymp\sum\limits_{a_1,\ldots,a_{n-1}\in \mathbb{N}}q_{n-1}^{\kern1.5pt \beta_{k-2}(1-s)-\beta_{k-1}s-2s}. \end{align*} $$
Clearly, the s-volume of the cover
$J_{n-1}$
is larger than that of other m-type covers. Thus, we need to look for the most effective cover in the remaining m-type covers. We need the following minimal one:
$$ \begin{align*} \min \{ x \in [0,1): - \beta_1s,\ \beta_1(1 - s) - \beta_2s,\ldots,\beta_{m - 1}(1 - s)-\beta_ms \} := - d_m(\kern1.2pt \beta_1, \ldots ,\beta_m;s). \end{align*} $$
We thus get an upper bound for
${\dim _{\mathrm{H}}}\,\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
that is given by the root of equation
$$ \begin{align} d_m(\kern1.2pt \beta_1,\ldots,\beta_m;s)+2s=2. \end{align} $$
We believe that the root should indeed be the exact dimension of
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
. A strict proof for the lower bound of
${\dim _{\mathrm{H}}} \,\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
will be given later.
Recall that for all
$0<\beta _1\le \cdots \le \beta _{m-1}\le \beta _m=\tau $
,
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
is a subset of
$E_m(\tau ).$
Then
$$ \begin{align*} {\dim_{\mathrm{H}}} E_m(\tau)\ge \sup\limits_{0<\beta_1\le\cdots\le \beta_{m-1}\le \beta_m=\tau}{\dim_{\mathrm{H}}} \,\mathcal{O}_M(\kern1.2pt \beta_1,\ldots,\beta_m). \end{align*} $$
The root of equation (4-3) is decreasing with respect to
$d_m(\kern1.2pt \beta _1,\ldots ,\beta _m;s),$
so, the smaller
$d_m(\kern1.2pt \beta _1,\ldots ,\beta _m;s)$
is, the larger
${\dim _{\mathrm{H}}} \,\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
is. This leads us to consider the supremum
$$ \begin{align} \sup_{0 < \beta_1 \le \cdots\le \beta_{m-1} \le \beta_m = \tau}\min \{ -\beta_{1}s,\ \beta_{1}(1-s)-\beta_{2}s,\ldots, \beta_{m-1}(1-s)-\beta_m s \}. \end{align} $$
Proposition 2.6 says that the supremum is just
$-\tau f_m(s)$
and is attained when all the terms in (4-4) are equal. So, we find a candidate for the dimension number:
$s_{\tau }^{(m)},$
where
$s_{\tau }^{(m)}$
is the solution to the equation
$\tau f_m(s)+2s=2.$
This number coincides with the upper bound of
${\dim _{\mathrm{H}}} E_m(\tau )$
given in Section 3.
So, for the lower bound of the dimension of
$E_m(\tau ),$
we need only compute the dimension of
$ \mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
at the specific point
$(\kern1.2pt \beta _1,\ldots ,\beta _{m-1},\beta _m)$
where the supremum is attained in (4-4).
4.2. Structure of
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
at the specific point
Fix
$0<s<s_{\tau }^{(m)}.$
In Section 4.1, we have chosen integers
$M,N$
large enough such that
$s<s_{\tau }^{(m)}(M,N).$
Let
$0<\beta _1\le \cdots \le \beta _m=\tau $
give the supremum of (4-4); then
$$ \begin{align} - \beta_1s &= \beta_1 ( 1 - s ) - \beta_2s = \cdots = \beta_{i - 2} ( 1 - s ) - \beta_{i - 1}s\nonumber\\ &= \cdots = \beta_{m - 1} ( 1 - s ) - \beta_ms = - \tau f_m ( s ). \end{align} $$
It follows from the equalities in (4-5) that
$$ \begin{align} &\beta_{i-2}=(\kern1.2pt \beta_{i-2}+\beta_{i-1}-\beta_1)s,\quad\mathrm{where\ }3\le i\le m+1, \end{align} $$
$$ \begin{align} &\hspace{-25pt}\beta_m+\beta_1s\ge 2\beta_{m}s.\ \ \end{align} $$
The inequality (4-7) holds since
$\beta _m-2\beta _ms=\beta _m(1-s)-\beta _ms\ge \beta _{m-1}(1-s)-\beta _ms=-\beta _1s.$
Recall the definition of
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m).$
We make use of a symbolic space described as follows to illustrate the structure of
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m).$
For any
$n\ge 1,$
define
$$ \begin{align*} D_n&= \{ (a_1,\ldots,a_n)\in \mathbb{N}^n: a_{n_j} = 1,\ q_{n_j}^{\kern1.5pt \beta_1} ( a_1,\ldots,a_{n_j} ) \le a_{n_j+1} < 2q_{n_j}^{\kern1.5pt \beta_1} ( a_1,\ldots,a_{n_j} ) \\ &\qquad q_{n_j}^{\kern1.5pt \beta_i - \beta_{i-1}}(a_1,\ldots,a_{n_j}) \le a_{n_j+i} < 2q_{n_j}^{\kern1.5pt \beta_i-\beta_{i-1}}(a_1,\ldots,a_{n_j}),\ 2 \le i \le m,\\ &\qquad\mathrm{for\ any}\ j\ \mathrm{satisfying}\ n_j+m\le n\ \mathrm{and}\ 1\le a_k\le M\\ &\qquad\mathrm{for\ all}\ 1\le k\neq n_j,n_j+1,\ldots,n_j+m\le n\}, \end{align*} $$
$D=\bigcup \nolimits _{n=0}^{\infty }D_n\ (D_0:=\emptyset )$
.
For any
$n\geq 1$
and
$(a_1,\ldots ,a_n)\in D_n,$
we call
$I_n(a_1,a_2,\ldots ,a_n)$
a basic interval of order n and
$$ \begin{align*} J_n(a_1,\ldots,a_n)=\bigcup\limits_{a_{n+1}: (a_1,\ldots, a_n,a_{n+1})\in D_{n+1}}I_{n+1}(a_1,\ldots,a_n,a_{n+1}) \end{align*} $$
a fundamental interval of order n.
It is easy to see that
$$ \begin{align*} \mathcal{O}_M(\kern1.2pt \beta_1,\ldots,\beta_m)=\bigcap\limits_{n\geq 1}\bigcup_{(a_1,\ldots,a_n)\in D_n}J_n(a_1,\ldots,a_n). \end{align*} $$
By the definition of
$D_n$
and the length of a cylinder, we immediately get the following lemma about lengths of fundamental intervals
$J_n(a_1,\ldots ,a_n)$
.
Lemma 4.1 Let
$(a_1,\ldots ,a_n)\in D_n$
. Assume that
$n_{\kern1.2pt j-1}+m\leq n<n_j+m$
for some
$j\ge 1$
.
Case 0.When
$n_{\kern1.2pt j-1}+m\le n<n_j-1=n_{\kern1.2pt j-1}+m+l_jN,$
$$ \begin{align*} J_n(a_1,\ldots,a_n)=\bigcup\limits_{1\leq a_{n+1}\leq M}I_{n+1}(a_1,\ldots,a_n,a_{n+1}) \end{align*} $$
and
$$ \begin{align*} \frac{1}{6q_n^2}\leq|J_n(a_1,\ldots,a_n)|=\bigg|\frac{p_n+p_{n-1}}{q_n+q_{n-1}}-\frac{(M+1)p_n+p_{n-1}}{(M+1)q_n+q_{n-1}}\bigg|\leq \frac{1}{q_n^2}. \end{align*} $$
Case 1.When
$n=n_j-1,$
$$ \begin{align*} J_{n_j-1}(a_1,\ldots,a_{n_j-1})=I_{n_j}(a_1,\ldots,a_{n_j-1},1) \end{align*} $$
and
$$ \begin{align*} \frac{1}{8q_{n_j-1}^2}\le |J_{n_j-1}(a_1,\ldots,a_{n_j-1})|\le \frac{1}{q_{n_j-1}^2}. \end{align*} $$
Case 2.When
$n=n_j,$
$$ \begin{align*} J_{n_j}(a_1,\ldots,a_{n_{\kern1.2pt j-1}},a_{n_j})=\bigcup\limits_{q_{n_j}^{\kern1.5pt \beta_1}\le a_{n_j+1}<2q_{n_j}^{\kern1.5pt \beta_1}}I_{n_j+1}(a_1,\ldots,a_{n_j},a_{n_j+1}) \end{align*} $$
and
$$ \begin{align*} \frac{1}{6\cdot 2^{\kern1.5pt \beta_1+2}q_{n_j-1}^{\kern1.5pt \beta_1+2}}\le |J_{n_j}(a_1,\ldots,a_{n_{\kern1.2pt j-1}},a_{n_j})|\le \frac{1}{q_{n_j-1}^{\kern1.5pt \beta_1+2}}. \end{align*} $$
Case i. (
$3\le i\le m+1$
). When
$n=n_j+i-2$
(that is,
$n_{\kern1.2pt j}+1\le n\le n_j+m-1$
),
$$ \begin{align*} J_{n_j + i - 2}(a_1,\ldots,a_{n_j + i -2})= \bigcup\limits_{q_{n_j}^{\kern1.5pt \beta_{i - 1} - \beta_{i - 2}} \le a_{n_j + i - 1} < 2q_{n_j}^{\kern1.5pt \beta_{i - 1} - \beta_{i - 2}}}I_{n_j + i - 1} ( a_1,\ldots,a_{n_j + i - 2},a_{n_j + i- 1} ) \end{align*} $$
and
$$ \begin{align*} \frac{1}{6\cdot 2^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2i-2}q_{n_j-1}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}\le |J_{n_j+i-2}(a_1,\ldots,a_{n_j+i-2})|\le \frac{1}{q_{n_j-1}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}. \end{align*} $$
4.3. A measure supported on
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
In this subsection, we distribute a probability measure
$\mu $
supported on
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m).$
Define
$$ \begin{align*} \Lambda=\{1,2,\ldots,M\}^N,\ u=\sum\limits_{\omega\in \Lambda}\bigg(\frac{1}{q_N(\omega)}\bigg)^{\kern1.5pt \beta_1s+2s}=\sum\limits_{\omega\in \Lambda}\bigg(\frac{1}{q_N(\omega)}\bigg)^{\tau f_m(s)+2s}, \end{align*} $$
where the last equality holds due to (4-5). Recall (2-2) on the definition of
$s_{\tau }^{(m)}(M,N)$
. Since
$s<s_{\tau }^{(m)}(M,N),$
we have
$u\ge 1.$
In the following steps
$\langle A 1\rangle $
to
$\langle A(m+3)\rangle $
, we define the measure of fundamental intervals of order at most
$n_1+m.$
Then, in steps
$\langle B1\rangle $
to
$ \langle B(m+3)\rangle ,$
the measure of fundamental intervals of arbitrary order is given. Recall the choice of
$n_j$
(for
$j\ge 1,\ n_1=l_1N+1$
).
$\langle A 1\rangle $
Let
$(\omega _{1}^{(1)},\ldots ,\omega _{l_1}^{(1)})\in D_{n_1-1}$
with
$\omega _{i}^{(1)}\in \Lambda $
for all
$1\le i\le l_1.$
Then define
$$ \begin{align*} \mu(J_{n_1-1}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)}))=\prod\limits_{i=1}^{l_1}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(1)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align*} $$
$\langle A 2\rangle $
Let
$(\omega _{1}^{(1)},\ldots ,\omega _{l_1}^{(1)},a_{n_1})\in D_{n_1}.$
Since
$J_{n_1-1}(\omega _{1}^{(1)},\ldots ,\omega _{l_1}^{(1)})$
has only one subinterval of order
$n_1$
, we define
$$ \begin{align*} \mu(J_{n_1}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)},a_{n_1}))=\mu(J_{n_1-1}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)})). \end{align*} $$
$\langle A 3\rangle $
Let
$(\omega _{1}^{(1)},\ldots ,\omega _{l_1}^{(1)},a_{n_1},a_{n_1+1})\in D_{n_1+1}.$
We distribute the mass of
$J_{n_1}(\omega _{1}^{(1)},\ldots , \omega _{l_1}^{(1)}, a_{n_1})$
uniformly among its next offspring:
$$ \begin{align*} \mu(J_{n_1+1}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)},a_{n_1},a_{n_1+1}))=&\frac{1}{q_{n_1}^{\kern1.5pt \beta_1}} \mu(J_{n_1}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)},a_{n_1}))\\ =&\frac{1}{q_{n_1}^{\kern1.5pt \beta_1}}\prod\limits_{i=1}^{l_1}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(1)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align*} $$
$\langle A (i)\rangle \,\, (4\le i\le m+2)$
Let
$(\omega _{1}^{(1)},\ldots ,\omega _{l_1}^{(1)},a_{n_1},\ldots ,a_{n_1+i-2})\in D_{n_1+i-2}$
. Then
$$ \begin{align*} &\mu(J_{n_1+i-2}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)},a_{n_1},\ldots,a_{n_1+i-2}))\\ &\quad=\frac{1}{q_{n_1}^{\kern1.5pt \beta_{i-2}-\beta_{i-3}}} \mu(J_{n_1+i-3}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)},a_{n_1},\ldots,a_{n_1+i-3}))\\ &\quad=\frac{1}{q_{n_1}^{\kern1.5pt \beta_{i-2}-\beta_{1}}}\mu(J_{n_1+1}(\omega_{1}^{(1)},\ldots,\omega_{l_1}^{(1)},a_{n_1},a_{n_1+1}))\\ &\quad=\frac{1}{q_{n_1}^{\kern1.5pt \beta_{i-2}}}\prod\limits_{i=1}^{l_1}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(1)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align*} $$
$\langle A (m+3)\rangle $
Let
$(a_1,\ldots ,a_n)\in A_n$
with
$n<n_1-1.$
To make sure that
$\mu $
is a measure supported on
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m),$
the measure of
$J_n(a_1,\ldots ,a_n)$
is defined to be the total mass of its offspring:
$$ \begin{align*} \mu(J_n(a_1,\ldots,a_n))=\sum\limits_{1\le a_{n+1},\ldots,a_{n_1-1}\le M}\mu(J_{n_1-1}(a_1,\ldots,a_n,a_{n+1},\ldots,a_{n_1-1})). \end{align*} $$
So, for any
$l<l_1$
, the measure of the fundamental interval
$J_lN(\omega _1^{(1)},\ldots ,\omega _{l}^{(1)})$
is given by
$$ \begin{align*} \mu(J_{lN}(\omega_1^{(1)},\ldots,\omega_{l}^{(1)})) &=\sum\limits_{\omega_{l+1}^{(1)}\in \Lambda}\cdots\sum\limits_{\omega_{l_1}^{(1)}\in \Lambda} \mu(J_{l_1N}(\omega_1^{(1)},\ldots,\omega_{l}^{(1)},\omega_{l+1}^{(1)},\ldots,\omega_{l_1}^{(1)}))\\ &=\sum\limits_{\omega_{l+1}^{(1)}\in \Lambda}\cdots\sum\limits_{\omega_{l_1}^{(1)}\in \Lambda}\prod\limits_{i=1}^{l_1}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(1)})}\bigg)^{(\kern1.2pt \beta_1+2)s}\\ &=\prod\limits_{i=1}^{l}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(1)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align*} $$
Recall that
$n_j-1=n_{\kern1.2pt j-1}+m+l_jN.$
Now we define the measure
$\mu $
inductively. Assume that
$\mu (J_{n_{\kern1.2pt j-1}+m}(\epsilon ^{(j-1)}))$
has been defined, where
$$ \begin{align*} \epsilon^{(j - 1)} &= ( \omega_{1}^{(1)}, \ldots ,\omega_{l_1}^{(1)},a_{n_1},\ldots,a_{n_1 + m},\ldots,\omega_{1}^{(j - 1)}, \ldots,\omega_{l_{\kern1.2pt j - 1}}^{(j - 1)},a_{n_{\kern1.2pt j - 1}},\ldots,a_{n_{\kern1.2pt j - 1} + m} ) \\ &\in D_{n_{\kern1.2pt j-1}+m}. \end{align*} $$
Let
$n_{\kern1.2pt j-1}+m<n\le n_j+m$
for some
$j\ge 1.$
Here we set
$n_0:=-m$
to ensure that
$n_1-1=n_{0}+m+l_1N$
holds.
$\langle B 1\rangle $
Let
$(\epsilon ^{(j-1)},\omega _{1}^{(j)},\ldots ,\omega _{l_j}^{(j)})\in D_{n_j-1}$
with
$\omega _{i}^{(j)}\in \Lambda $
for all
$1\le i\le l_j.$
Then define
$$ \begin{align} \mu ( J_{n_j - 1}(\epsilon^{(j - 1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}) ) = \mu ( J_{n_{\kern1.2pt j - 1} + m}(\epsilon^{(j-1)}) ) \prod\limits_{i = 1}^{l_j}\frac{1}{u} \bigg( \frac{1}{q_N(\omega_i^{(j)})} \bigg)^{(\kern1.2pt \beta_1 + 2)s}. \end{align} $$
$\langle B 2\rangle $
Let
$ ( \epsilon ^{(j-1)},\omega _{1}^{(j)},\ldots ,\omega _{l_j}^{(j)},a_{n_j} ) \in D_{n_j}.$
Since
$J_{n_j - 1} ( \epsilon ^{(j-1)},\omega _{1}^{(j)},\ldots ,\omega _{l_j}^{(j)} ) $
has only one subinterval of order
$n_j,$
we define
$$ \begin{align} \mu(J_{n_j}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j})) =\mu(J_{n_j-1}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)})). \end{align} $$
$\langle B 3\rangle $
Let
$(\epsilon ^{(j-1)},\omega _{1}^{(j)},\ldots ,\omega _{l_j}^{(j)},a_{n_j},a_{n_j+1})\in D_{n_j+1}.$
The mass of
$J_{n_j}(\epsilon ^{(j-1)},\omega _{1}^{(j)},\ldots ,\omega _{l_j}^{(j)}, a_{n_j})$
, is equally distributed among its offspring, that is,
$$ \begin{align} &\mu(J_{n_j+1}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},a_{n_j+1}))\nonumber\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_1}}\mu(J_{n_j}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j}))\nonumber\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_1}}\mu(J_{n_j-1}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)})).\end{align} $$
$\langle B (i)\rangle \,\,(4\le i\le m+2)$
Let
$(\epsilon ^{(j-1)},\omega _{1}^{(j)},\ldots ,\omega _{l_j}^{(j)},a_{n_j},\ldots ,a_{n_j+i-2})\in D_{n_j+i-2}.$
Define
$$ \begin{align} &\mu(J_{n_j+i-2}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},\ldots,a_{n_j+i-2}))\nonumber\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-2}-\beta_{i-3}}}\mu(J_{n_j+i-3} (\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},\ldots,a_{n_j+i-3}))\nonumber\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-2}-\beta_{1}}}\mu(J_{n_j+1} (\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},a_{n_j+1}))\nonumber\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-2}}}\mu(J_{n_j-1} (\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)})). \end{align} $$
Especially for the case
$i=m+2,$
by using (4-8) and
$\beta _m=\tau ,$
$$ \begin{align} &\mu(J_{n_j+m}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},\ldots,a_{n_j+m}))\nonumber\\ &\quad=\mu(J_{n_{\kern1.2pt j-1}+m}(\epsilon^{(j-1)}))\frac{1}{q_{n_j}^{\tau}}\prod\limits_{i=1}^{l_j}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(j)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align} $$
$$ \begin{align} &\mu(J_{n_j-1}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}))\nonumber\\ &\quad=\bigg[\prod\limits_{t=1}^{\kern1.2pt j-1}\bigg(\frac{1}{q_{n_t}^{\tau}} \prod\limits_{i=1}^{l_t}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(t)})}\bigg)^{(\kern1.2pt \beta_1+2)s}\bigg)\bigg] \cdot\prod\limits_{i=1}^{l_j}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(j)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align} $$
$\langle B (m+3)\rangle $
Let
$(\epsilon ^{(j-1)},a_{n_{\kern1.2pt j-1}+m+1},\ldots ,a_n)\in D_n$
with
$n_{\kern1.2pt j-1}+m<n< n_{\kern1.2pt j}-1.$
Similarly, the measure on the fundamental intervals
$J_n(\epsilon ^{(j-1)},a_{n_{\kern1.2pt j - 1} + m + 1}$
,
$\ldots ,a_n)$
is defined as the total measure of their offspring:
$$ \begin{align*} &\mu(J_n(\epsilon^{(j-1)},a_{n_{\kern1.2pt j-1} + m + 1},\ldots,a_n))\nonumber\\ &\quad = \sum\limits_{1 \leq a_{n + 1},\ldots,a_{n_j - 1} \leq M} \mu(J_{n_j - 1}(\epsilon^{(j - 1)},a_{n_{\kern1.2pt j-1} + m + 1},\ldots,a_n,a_{n+1},\ldots,a_{n_j-1})). \end{align*} $$
Especially, for
$n=n_{\kern1.2pt j-1}+m+lN\ (l<l_j),$
$$ \begin{align} \mu ( J_{n_{\kern1.2pt j - 1} + m + lN} ( \epsilon^{ ( j - 1 ) },\omega_{1}^{(j)}, \ldots ,\omega_{l}^{(j)}) ) = \mu ( J_{n_{\kern1.2pt j - 1} + m} ( \epsilon^{(j - 1)} ) ) \prod\limits_{i = 1}^{l}\frac{1}{u} \bigg( \frac{1}{q_N ( \omega_i^{(j)} ) } \bigg)^{(\kern1.2pt \beta_1 + 2)s}. \end{align} $$
Clearly, the set function
$\mu $
defined in this way satisfies Kolmogorov’s consistency condition. So, it can be uniquely extended to a probability measure supported on
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
.
4.4. Estimation of
$\mu (J_n(a_1,\ldots ,a_n))$
In the following estimations, the following three properties (see Proposition 2.1) on the denominator
$q_n$
of the convergent
$p_n/q_n$
are frequently used: firstly,
$q_n\ge 2^{({n-1})/{2}}$
, secondly,
$$ \begin{align*} q_{n+k}(a_1,\ldots,a_n,b_1,\ldots,b_k)\le 2q_{n}(a_1,\ldots,a_n)q_{k}(b_1,\ldots,b_k) \end{align*} $$
and, thirdly,
$$ \begin{align} q_{n + k}(a_1,\ldots,a_n,b_1,\ldots,b_k) \le (b_k + 1)(b_{k - 1} + 1)\cdots (b_1 + 1)q_n(a_1,\ldots,a_n). \end{align} $$
For any
$1\le t\le j,$
let
$\omega _i^{(t)}\in \Lambda $
for
$1\le i\le l_t.$
Write
$$ \begin{align} V_t:=\prod\limits_{i=1}^{l_t}\frac{1}{u}\bigg(\frac{1}{q_N(\omega_i^{(t)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align} $$
We begin with an estimation of
$V_t$
.
Lemma 4.2 For any
$1\le t\le j,$
$$ \begin{align} V_t\le \bigg(\frac{1}{q_{l_tN}(\omega_1^{(t)},\ldots,\omega_{l_t}^{(t)})}\bigg)^{(\kern1.2pt \beta_1+2)(s-({4}/{N}))}. \end{align} $$
Proof. For any
$1\le t\le j,$
by using
$u\ge 1$
and
$q_{n+k}\le 2q_nq_k$
,
$$ \begin{align} V_t\le \prod\limits_{i=1}^{l_t}\bigg(\frac{1}{q_N(\omega_i^{(t)})}\bigg)^{(\kern1.2pt \beta_1+2)s}\le 2^{(\kern1.2pt \beta_1+2)l_t}\bigg(\frac{1}{q_{l_tN}(\omega_1^{(t)},\ldots,\omega_{l_t}^{(t)})}\bigg)^{(\kern1.2pt \beta_1+2)s}. \end{align} $$
Since
$$ \begin{align*} q_{l_tN}^{\kern1.5pt \beta_1+2}\ge 2^{({l_tN-1})/{2}\cdot(\kern1.2pt \beta_1+2)}\ge 2^{{l_tN}/{4}\cdot(\kern1.2pt \beta_1+2)}=(2^{(\kern1.2pt \beta_1+2)l_t})^{{N}/{4}}, \end{align*} $$
$$ \begin{align} 2^{(\kern1.2pt \beta_1+2)l_t}\le q_{l_tN}^{(\kern1.2pt \beta_1+2)\cdot{4}/{N}}. \end{align} $$
According to (4-18) and (4-19), we obtain the estimation formula (4-17).
The following lemma is about the estimation of
$q_{n_j}(a_1,\ldots ,a_{n_j}).$
Lemma 4.3 For each
$j\ge 1$
and
$(a_1,\ldots ,a_{n_j})\in D_{n_j},$
$$ \begin{align*} q_{n_j}(a_1,\ldots,a_{n_j})\le (M+1)^{2j(\tau+1)^{\kern1.2pt j-1}l_jN}. \end{align*} $$
Proof. Recall the definition of
$D_n$
and the choice of
$n_j$
(
$n_1=l_1N+1$
and
$n_j=n_{\kern1.2pt j-1}+m+l_jN+1$
for
$j\ge 2$
). By using (4-15),
$$ \begin{align*} q_{n_j}(a_1,\ldots,a_{n_j})&\le (a_{n_j}+1)(M+1)^{l_jN}(a_{n_{\kern1.2pt j-1}+m}+1)\cdots (a_{n_{\kern1.2pt j-1}+1}+1)q_{n_{\kern1.2pt j-1}}\\ &\le 2^{m+1}(M+1)^{l_jN}q_{n_{\kern1.2pt j-1}}^{\tau+1}\\ &\le 2^{(m+1)+(m+1)(\tau+1)}(M+1)^{l_jN+l_{\kern1.2pt j-1}N(\tau+1)}q_{n_{\kern1.2pt j-2}}^{(\tau+1)^2}\\ &\le 2^{(m+1)(\sum\nolimits_{k=0}^{\kern1.2pt j-2}(\tau+1)^{k})}(M+1) ^{\sum\nolimits_{k=1}^{\kern1.2pt j-1}l_{k+1}N(\tau+1)^{\kern1.2pt j-(k+1)}}q_{n_1}^{(\tau+1)^{\kern1.2pt j-1}}\\ &\le 2^{(m+1)(\sum\nolimits_{k=0}^{\kern1.2pt j-2}(\tau+1)^{k})+(\tau+1)^{\kern1.2pt j-1}}(M+1) ^{\sum\nolimits_{k=0}^{\kern1.2pt j-1}l_{k+1}N(\tau+1)^{\kern1.2pt j-(k+1)}}\\ &\le 2^{\kern1.2pt j(\tau+1)^{\kern1.2pt j-1}l_1}\cdot (M+1)^{\kern1.2pt j(\tau+1)^{\kern1.2pt j-1}l_{\kern1.2pt j}N}\\ & \le (M+1)^{2j(\tau+1)^{\kern1.2pt j-1}l_jN}. \end{align*} $$
Here, for the penultimate inequality, we use the fact that for any
$0 \le k \le j-1,$
$$ \begin{align*} (\tau+1)^k\le (\tau+1)^{\kern1.2pt j-1},\ m+1<l_1\ \mathrm{and} \ l_{k+1}\le l_j \ (\mathrm{see\ the\ choice\ of}\ l_j\ \mathrm{in}\ (4.1)).\\[-25pt] \end{align*} $$
Next we estimate the relationship between
$\mu ( J_n ( a_1, \ldots ,a_n ) ) $
and
$ | J_n ( a_1, \ldots ,a_n ) | $
in the following proposition.
Proposition 4.4 For any
$(a_1,\ldots ,a_n)\in D_n$
,
$$ \begin{align*} \mu(J_n(a_1,\ldots,a_n) \le c_0 | J_n(a_1,\ldots,a_n) |^{s-({2m+8+2\tau}/{N})}, \quad\mathrm{where} \ c_0 = 6^2\cdot 4^{m+2+\tau}\cdot(4M^2)^N. \end{align*} $$
Proof. Recall the definition of
$\{n_j\}_{\kern1.2pt j\ge 1}$
,
$n_j=n_{\kern1.2pt j-1}+m+l_jN+1.$
Let
$n_{\kern1.2pt j-1}+m<n\le n_j+m$
for some
$j\ge 1.$
We distinguish
$m+3$
cases according to the range of n.
Case 1. When
$n=n_j-1.$
Recall (4-13) on the measure of fundamental intervals. Then
$$ \begin{align} &\mu(J_{n_j-1}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}))\nonumber\\ &\quad=\bigg[\prod\limits_{t=1}^{\kern1.2pt j-1}\bigg(\frac{1}{q_{n_t}^{\tau}} \cdot V_t\bigg)\bigg] \cdot V_j \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\mathrm{by}\ (4\text{-}13)\ \mathrm{and}\ (4\text{-}16))\nonumber\\ &\quad\le \bigg[\prod\limits_{t=1}^{\kern1.2pt j-1}\bigg(\frac{1}{q_{n_t}^{\tau}} \cdot \bigg(\frac{1}{q_{l_tN}(\omega_1^{(t)},\ldots,\omega_{l_t}^{(t)})}\bigg)^{(\kern1.2pt \beta_1+2)(s-({4}/{N}))}\bigg)\bigg]\nonumber\\ &\qquad\times \bigg(\frac{1}{q_{l_jN}(\omega_1^{(j)},\ldots,\omega_{l_j}^{(j)})}\bigg)^{(\kern1.2pt \beta_1+2)(s-({4}/{N}))}\ (\mathrm{by\ Lemma}\ 4.2)\nonumber\\ &\quad\le \bigg[\prod\limits_{t=1}^{\kern1.2pt j-1}\bigg(\frac{1}{q_{n_t}^{\tau}} \cdot \bigg(\frac{2^{m+1}q_{n_t}^{\tau}}{q_{l_tN+m+1}(\omega_1^{(t)},\ldots,\omega_{l_t}^{(t)},a_{n_t},\ldots,a_{n_t+m})}\bigg) ^{(\kern1.2pt \beta_1+2)(s-({4}/{N}))}\bigg)\bigg] \nonumber\\ &\qquad\times\bigg(\frac{1}{q_{l_jN}(\omega_1^{(j)},\ldots,\omega_{l_j}^{(j)})}\bigg)^{(\kern1.2pt \beta_1+2)(s-({4}/{N}))}\nonumber\\ &\quad\le \bigg( q_{n_1}q_{n_2}\cdots q_{n_{\kern1.2pt j-1}} \bigg)^{(\kern1.2pt \beta_1+1)\tau}\cdot 2^{ (\kern1.2pt \beta_1 + 2 ) ( m+2 ) j}\cdot \bigg(\frac{1}{q_{n_j-1}(\epsilon^{(j-1)},\omega_1^{(j)},\ldots, \omega_{l_j}^{(j)})} \bigg)^{ (\kern1.2pt \beta_1 + 2) (s - ({4}/{N})) }\nonumber\\ &\quad\le q_{n_{\kern1.2pt j-1}}^{(\kern1.2pt \beta_1+1)j\tau}\cdot\bigg(\frac{1}{q_{n_j-1}(\epsilon^{(j-1)},\omega_1^{(j)},\ldots, \omega_{l_j}^{(j)})}\bigg)^{(\kern1.2pt \beta_1+2)(s-({2m+8})/{N})}. \end{align} $$
Here, for the last equality, we use
$$ \begin{align*} q_{n_j-1}^{(\kern1.2pt \beta_1+2)(m+2)} \ge 2^{({n_j-2})/{2}(\kern1.2pt \beta_1+2)(m+2)} \ge 2^{(1/2)(l_1N+\cdots+l_jN)(\kern1.2pt \beta_1+2)(m+2)} \ge ( 2^{(\kern1.2pt \beta_1+2)(m+2)j} )^{{N}/{2}}. \end{align*} $$
Now we estimate the relationship between
$q_{n_{\kern1.2pt j-1}}^{(\kern1.2pt \beta _1+1)j\tau }$
and
$q_{n_j-1}^{\kern1.5pt \beta _1+2}$
. By Lemma 4.3 and the choice of
$l_j$
(see (4-1)),
$$ \begin{align*} q_{n_{\kern1.2pt j-1}}^{\kern1.2pt j} \le (M+1)^{2j^2(\tau+1)^{\kern1.2pt j-2}l_{\kern1.2pt j-1}N}\le 2^{l_j}\le 2^{(1/2)(l_1N+\cdots+l_jN)\cdot {2}/{N}} \le 2^{(({n_j}/{2})-1)\cdot{2}/{N}} \le q_{n_j-1}^{{2}/{N}}. \end{align*} $$
So,
$$ \begin{align} q_{n_{\kern1.2pt j-1}}^{(\kern1.2pt \beta_1+1)j\tau}\le q_{n_j-1}^{(\kern1.2pt \beta_1+2)({2\tau}/{N})}. \end{align} $$
$$ \begin{align} \mu(J_{n_j-1}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}))&\le \bigg(\frac{1}{q_{n_j-1}}\bigg)^{(\kern1.2pt \beta_1+2)(s-({2m+8+2\tau})/{N})}\nonumber\\ &\le \bigg(\frac{1}{q_{n_j-1}^2}\bigg)^{(s-({2m+8+2\tau})/{N})}. \end{align} $$
By applying Case 1 in Lemma 4.1,
$$ \begin{align*} \mu(J_{n_j-1}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}))\le 8\cdot|J_{n_j-1}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)})|^{(s-({2m+8+2\tau})/{N})}. \end{align*} $$
Case 2.When
$n=n_j.$
Recalling (4-9) and then using the estimation (4-22),
$$ \begin{align*} &\mu(J_{n_j}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j}))\\ &\quad=\mu(J_{n_j-1}(\epsilon^{(j-1)}, \omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}))\le \bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_1+2}}\bigg)^{(s-({2m+8+2\tau})/{N})}\\ &\quad\le 6\cdot 2^{\kern1.5pt \beta_1+2}|J_{n_j}(a_1,\ldots,a_{n_{\kern1.2pt j-1}},a_{n_j})|^{(s-({2m+8+2\tau})/{N})}\ \ (\mathrm{see\ Case\ 2\ in\ Lemma\ 4.1})\\ &\quad\le 24\cdot 2^{\tau}|J_{n_j}(a_1,\ldots,a_{n_{\kern1.2pt j-1}},a_{n_j})|^{(s-({2m+8+2\tau})/{N})}\ \ \ (\mathrm{by\ using}\ \beta_1\le \beta_m=\tau). \end{align*} $$
Case i. (
$3\le i\le m+1$
). When
$n=n_j+i-2$
for some i (that is,
$n_j+1\le n\le n_j+m-1$
). Recall (4-10) and (4-11) on the measure of fundamental intervals. Using the estimation (4-22) and the equality (4-6),
$$ \begin{align*} &\mu(J_{n_j+i-2}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},\ldots,a_{n_j+i-2}))\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-2}}}\mu(J_{n_j-1} (\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)})).\\ &\quad\le \bigg(\frac{1}{q_{n_{\kern1.2pt j}-1}^{\kern1.5pt \beta_{i-2}+\beta_{i-1}-\beta_1}}\bigg)^s\bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_1+2}}\bigg)^{(s-({2m+8+2\tau})/{N})} \le \bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_{i-2}+\beta_{i-1}+2}}\bigg)^{(s-({2m+8+2\tau})/{N})}\\ &\quad\le 6\cdot 2^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2i-2}|J_{n_j+i-2}|^{(s-({2m+8+2\tau})/{N})}\ (\mathrm{see\ Case\ i\ in\ Lemma\ ({4.1})})\\ &\quad\le 6\cdot 4^{m+\tau}|J_{n_j+i-2}|^{(s-({2m+8+2\tau})/{N})} \ \ \ \ \ \ \ \ \ \ \ \ (\mathrm{by\ using}\ \beta_{i-2}\le \beta_{i-1}\le\beta_m=\tau). \end{align*} $$
Case m+2. When
$n=n_j+m.$
Using the estimation (4-22) and (4-7),
$$ \begin{align} &\mu(J_{n_j+m}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},\ldots,a_{n_j+m}))\nonumber\\ &\quad=\frac{1}{q_{n_j}^{\kern1.5pt \beta_{m}}}\mu(J_{n_j-1} (\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)}))\nonumber\\ &\quad\le \frac{1}{q_{n_j-1}^{\kern1.5pt \beta_{m}}}\bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_1+2}}\bigg)^{(s-({2m+8+2\tau})/{N})}\le \bigg(\frac{1}{q_{n_j-1}^{2\beta_m+2}}\bigg)^{(s-({2m+8+2\tau})/{N})}\nonumber\\ &\quad\le 4^{m+1+\beta_m}\bigg(\frac{1}{q_{n_j+m}^2}\bigg)^{(s-({2m+8+2\tau})/{N})}. \end{align} $$
Here, for the last inequality, we use
$$ \begin{align*} q_{n_j + m} \le ( a_{n_j + m} + 1 ) \cdots ( a_{n_{\kern1.2pt j} + 1} + 1 ) ( a_{n_j}+1 ) q_{n_j - 1} \le 2^{m + 1}q_{n_j}^{\kern1.5pt \beta_m}q_{n_j - 1} \le 2^{m + 1 + \beta_m}\cdot q_{n_j - 1}^{\kern1.5pt \beta_m + 1}. \end{align*} $$
Then, by applying Case 0 in Lemma 4.1,
$$ \begin{align*} \mu ( J_{n_j+m}(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l_j}^{(j)},a_{n_j},\ldots,a_{n_j+m}) ) \le 6\cdot 4^{m+1+\tau}|J_{n_j+m}|^{(s-({2m+8+2\tau})/{N})}. \end{align*} $$
Case m+3. When
$n_{\kern1.2pt j-1}+m<n<n_j-1=n_{\kern1.2pt j-1}+m+l_jN$
. For such n, there exists a unique integer
$0\le l\le l_j-1$
such that
$n_{\kern1.2pt j-1}+m+lN<n\le n_{\kern1.2pt j-1}+m+(l+1)N.$
Note that
$1\le a_i\le M$
for all
$n_{\kern1.2pt j-1}+m+lN+1\le i\le n.$
Then, by Case 0 of Lemma 4.1,
$$ \begin{align*} \frac{|J_n|}{|J_{n_{\kern1.2pt j - 1} + m + lN}|} \ge \frac16\frac{q_{n_{\kern1.2pt j - 1} + m + lN}^2}{q_n^2} \ge \frac16 \bigg( \frac{1}{2^{(n - ( n_{\kern1.2pt j - 1} + m + lN))}a_{n_{\kern1.2pt j - 1} + m + lN + 1}\cdots a_{n}} \bigg)^2 \ge \frac16\bigg(\frac{1}{4M^2}\bigg)^N. \end{align*} $$
Recall (4-14) on the measure of
$J_{n_{\kern1.2pt j-1}+m+lN}$
and the estimation (4-23). Then
$$ \begin{align*} &\mu(J_n(\epsilon^{(j-1)},\omega_{1}^{(j)},\ldots,\omega_{l}^{(j)},a_{n_{\kern1.2pt j-1}+m+lN+1},\ldots,a_n))\\ &\quad \le \mu ( J_{n_{\kern1.2pt j - 1} + m + lN}(\epsilon^{(j - 1)},\omega_{1}^{(j)},\ldots,\omega_{l}^{(j)})) = \mu ( J_{n_{\kern1.2pt j - 1} + m}(\epsilon^{(j - 1)}))\prod\limits_{i = 1}^{l}\frac{1}{u} \bigg(\frac{1}{q_N(\omega_i^{(j)})}\bigg)^{(\kern1.2pt \beta_1 + 2)s}\\ &\quad\le 4^{m+1+\beta_m}\bigg(\frac{1}{q_{n_{\kern1.2pt j-1}+m}^2}\bigg)^{(s-({2m+8+2\tau})/{N})}\cdot \bigg(\frac{1}{q_{lN}^{\kern1.5pt \beta_1+2}(\omega_1^{(j)},\ldots,\omega_{l}^{(j)})}\bigg)^{(s-({2m+8+2\tau})/{N})}\\ &\quad\le 4^{m+2+\tau}\bigg(\frac{1}{q_{n_{\kern1.2pt j-1}+m+lN}^2}\bigg)^{(s-({2m+8+2\tau})/{N})}\\ &\quad\le 6\cdot 4^{m+2+\tau}|J_{n_{\kern1.2pt j-1}+m+lN}|^{(s-({2m+8+2\tau})/{N})}\ \ \ \ \ \ \ (\mathrm{see\ Case\ 0\ in\ Lemma\ {4.1}})\\ &\quad\le 6^2\cdot 4^{m+2+\tau}\cdot(4M^2)^N|J_n|^{(s-({2m+8+2\tau})/{N})}. \end{align*} $$
Let
$c_0=6^2\cdot 4^{m+2+\tau }\cdot (4M^2)^N.$
From the
$m+2$
cases discussed above, we know for any
$(a_1,\ldots ,a_n)\in D_n$
that
$$ \begin{align*} \mu(J_n(a_1,\ldots,a_n))\le c_0|J_n(a_1,\ldots,a_n)|^{(s-({2m+8+2\tau})/{N})}.\\[-39pt] \end{align*} $$
4.5. Estimation of
$\mu (B(x,r))$
Firstly, we estimate the gaps between two neighboring fundamental intervals. Suppose that we are given a fundamental interval
$J_n:=J_n(a_{1},\ldots ,a_n)$
with
$(a_1,\ldots ,a_n)\in D_n.$
Denote by
$G_n^l(a_1,\ldots ,a_n)$
the left-hand gap between
$J_n$
and its adjoining fundamental interval and
$G_n^r(a_1,\ldots ,a_n)$
the right-hand gap between
$J_n$
and its adjoining fundamental interval. Then let
$$ \begin{align*} G_n(a_1,\ldots,a_n)=\min\{G_n^l(a_1,\ldots,a_n), G_n^r(a_1,\ldots,a_n)\}. \end{align*} $$
Lemma 4.5 (Gap estimation)
$$ \begin{align*} G_n(a_1,\ldots,a_n)\ge \frac{1}{10M}|I_n(a_1,\ldots,a_n)|. \end{align*} $$
Proof. Let
$J_n^{\prime }$
be a fundamental interval neighboring
$J_n(a_{1},\ldots ,a_n).$
We can assume that
$J_n^{\prime }$
and
$J_n (a_{1}, \ldots ,a_n) $
have the same mother interval
$J_{n - 1} (a_1, \ldots ,a_{n - 1}) $
. Otherwise, we consider the gap between their mother fundamental intervals. Since
$J_n^{\prime }, J_n$
are neighbors and have the same mother interval,
$$ \begin{align*} J_n^{\prime}=J_n(a_1,\ldots,a_{n-1},a_n\pm 1). \end{align*} $$
There is no loss of generality in assuming that n is even and only estimating
$G_n^r(a_1,\ldots ,a_n)$
, since the estimation on
$G_n^l(a_1,\ldots ,a_n)$
and the case when n is odd can proceed in the same way. We divide the estimation into
$m+2$
cases.
Case 0. When
$n_{\kern1.2pt j-1}+m\le n<n_j-1.$
In this case,
$$ \begin{align*} J_n=\bigcup\limits_{1\leq a\leq M}I_{n+1}(a_1,\ldots,a_n,a),\quad J_n^{\prime}=\bigcup\limits_{1\leq a\leq M}I_{n+1}(a_1,\ldots,a_n+1,a). \end{align*} $$
By Proposition 2.1(3), the family of cylinders
$$ \begin{align*} \{I_{n+1}(a_1,\ldots,a_n+1, a): a>M\} \end{align*} $$
fall into the gap between
$J_n$
and
$J_n'$
. Thus,
$$ \begin{align*} G_n^r(a_1,\ldots,a_n)&\geq \sum\limits_{a>M}|I_{n+1}(a_1,\ldots,a_n+1,a)|\\ &=\frac{1}{(q_n+q_{n-1})[(M+1)(q_n+q_{n-1})+q_{n-1}]}\geq \frac{1}{(4M+6)}\frac{1}{q_n^2}\\ &\geq \frac{1}{10M}|I_n(a_1,\ldots,a_n)|. \end{align*} $$
Case 1. When
$n=n_{\kern1.2pt j}-1$
. In this case,
$$ \begin{align*} J_n=I_{n+1}(a_1,\ldots,a_n,1),\quad J_n^{\prime}=I_{n+1}(a_1,\ldots,a_n+1,1). \end{align*} $$
Then
$$ \begin{align*} G_n^r(a_1,\ldots,a_n)&\geq\sum\limits_{a>1}|I_{n+1}(a_1,\ldots,a_n+1,a)|\\ & \ge\frac{1}{(q_n+q_{n-1})(2(q_n+q_{n-1})+q_{n-1})}\\ &\ge \frac{1}{10q_n^2}\ge\frac{1}{10}|I_n(a_1,\ldots,a_n)|. \end{align*} $$
Case 2. When
$n=n_{\kern1.2pt j}$
. In this case,
$$ \begin{align*} J_n=\bigcup\limits_{q_{n_j}^{\kern1.5pt \beta_1}\le a_{n+1}<2q_{n_j}^{\kern1.5pt \beta_1}}I_{n+1}(a_1,\ldots,a_n,a_{n+1})\subset I_n(a_1,\ldots,a_n). \end{align*} $$
Obviously, the right-hand gap is larger than the distance between the right end point of
$J_n$
and that of
$I_n(a_1,\ldots ,a_n).$
Thus,
$$ \begin{align*} G_n^r(a_1,\ldots,a_n)&\ge \frac{p_n+p_{n-1}}{q_n+q_{n-1}}-\frac{q_{n_j}^{\kern1.5pt \beta_1}p_n+p_{n-1}}{q_{n_j}^{\kern1.5pt \beta_1}q_n+q_{n-1}}\\ &\ge \frac{q_{n_j}^{\kern1.5pt \beta_1}-1}{(q_n+q_{n-1})(q_{n_j}^{\kern1.5pt \beta_1}q_n+q_{n-1})}\\ &\ge \frac{q_{n_j}^{\kern1.5pt \beta_1}/2}{2q_n\cdot (2q_{n_j}^{\kern1.5pt \beta_1}q_n)}\ge \frac18\frac{1}{q_n^2}\\ &\geq \frac18|I_n(a_1,\ldots,a_n)|. \end{align*} $$
Case i. (
$3\le i\le m+1$
). When
$n=n_{\kern1.2pt j}+i-2$
. In this case,
$$ \begin{align*} J_n=\bigcup\limits_{q_{n_j}^{\kern1.5pt \beta_{i-1}-\beta_{i-2}}\le a_{n+1}<2q_{n_j}^{\kern1.5pt \beta_{i-1}-\beta_{i-2}}}I_{n+1}(a_1,\ldots,a_n,a_{n+1})\subset I_n(a_1,\ldots,a_n). \end{align*} $$
Similarly to the case when
$n=n_j$
,
$$ \begin{align*} G_n^r(a_1,\ldots,a_n)\ge \tfrac18|I_n(a_1,\ldots,a_n)|.\\[-33pt] \end{align*} $$
In the following proposition, we estimate the measure of an arbitrary ball.
Proposition 4.6 For any
$x\in \mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)$
and r small enough,
$$ \begin{align*} \mu(B(x,r))\le \widetilde{c}\cdot r^{s-({2m+8+2\tau})/{N}}, \end{align*} $$
where
$\widetilde {c}=\widetilde {c}(M,\tau )$
is a constant.
Proof. Fix
$x\in \mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m).$
There exists a unique sequence
$a_1,a_2,\ldots $
such that
$x\in J_k(a_1,\ldots ,a_k)$
for all
$k\ge 1$
and, for some
$n,$
$$ \begin{align*} G_{n+1}(a_1,\ldots,a_{n+1})\le r<G_n(a_1,\ldots,a_n). \end{align*} $$
From the definition of
$G_n(a_1,\ldots ,a_{n})$
, we know that the ball
$B(x,r)$
can only intersect one fundamental interval of order
$n,$
which is
$J_n(a_1,\ldots ,a_{n}).$
Case 0. When
$n_{\kern1.2pt j-1}+m\le n<n_j-1.$
By Case 0 of Lemma 4.1, we know that
$1\le a_{n+1}\le M$
and
$|J_n(a_1,\ldots ,a_n)|\le {1}/{q_n^2} $
. Hence,
$$ \begin{align*} \mu(B(x,r))&\le \mu(J_n(a_1,\ldots,a_n))\le c_0|J_n(a_1,\ldots,a_n)|^{s-({2m+8+2\tau})/{N}}\\ &\le c_0\bigg(\frac{1}{q_n^2}\bigg)^{s-({2m+8+2\tau})/{N}}\leq c_0\cdot4M^2\bigg(\frac{1}{q_{n+1}^2}\bigg)^{s-({2m+8+2\tau})/{N}}\\ &\le 8c_0M^2|I_{n+1}|^{s-({2m+8+2\tau})/{N}}\leq 80c_0M^3\cdot G_{n+1}^{s-({2m+8+2\tau})/{N}}\\ &\le 80c_0M^3\cdot r^{s-({2m+8+2\tau})/{N}}. \end{align*} $$
Case 1. When
$n=n_j-1$
. By Case 1 of Lemma 4.1, we know that
$a_{n_j}=1$
and
$|J_{n_j-1}(a_1,\ldots ,a_{n_j-1})|\le ({1}/{q_{n_j-1}^2}).$
Similarly to the case when
$n_{\kern1.2pt j-1}+m\le n<n_j-1,$
$$ \begin{align*} \mu(B(x,r))\le 80c_0M^3\cdot r^{s-({2m+8+2\tau})/{N}}. \end{align*} $$
Case 2. When
$n=n_j.$
We estimate the number of
$(n_j+1)$
th-order fundamental intervals
$J_{n_j+1}$
that can intersect
$B(x,r).$
Denote this number by
$t.$
Thus,
$$ \begin{align*} t\leq \#\bigg\{&I_{n_j+1}(a_1,\ldots,a_{n_j+1}):~(a_1,\ldots,a_{n_j+1})\in D_{n_j+1} ~~\mathrm{and}~\\[-4pt] &~I_{n_j+1}(a_1,\ldots,a_{n_j+1})\bigcap B(x,r)\neq\emptyset\bigg\}, \end{align*} $$
where # B denotes the cardinality of the set B.
Notice that
$$ \begin{align*} \frac{1}{8q_{n_j}^{2\beta_1 + 2}} \le \frac{1}{2q_{n_j+1}^2} \le | I_{n_j + 1}(a_1,\ldots,a_{n_j + 1}) | \le 10M\cdot G_{n_j + 1} (a_1,\ldots,a_{n_j + 1}) \le 10M\cdot r. \end{align*} $$
Then
$1\le 8q_{n_j}^{2\beta _1+2}\cdot 10Mr=80Mrq_{n_j}^{2\beta _1+2}.$
Hence,
$$ \begin{align*} t\le \frac{2r}{|I_{n_j+1}|}+2\le 16rq_{n_j}^{2\beta_1+2}+160Mrq_{n_j}^{2\beta_1+2}:=c_1rq_{n_j}^{2\beta_1+2}, \end{align*} $$
where
$c_1=16+160M$
.
As a consequence,
$$ \begin{align*} \mu ( B(x,r) ) & \le \min \{ \mu(J_{n_j}),~c_1rq_{n_j}^{2\beta_1+2}\mu(J_{n_j+1}) \} = \mu(J_{n_j})\min \bigg\{ 1,~c_1rq_{n_j}^{2\beta_1+2}\cdot\frac{1}{q_{n_j}^{\kern1.5pt \beta_1}} \bigg\}. \end{align*} $$
By Proposition 4.4 and the length of the fundamental interval (see Case 2 in Lemma 4.1),
$$ \begin{align} \mu(J_{n_j})\le c_0\bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_1+2}}\bigg)^{s-({2m+8+2\tau})/{N}}\le 2^{\tau+2}\cdot c_0\bigg(\frac{1}{q_{n_j}^{\kern1.5pt \beta_1+2}}\bigg)^{s-({2m+8+2\tau})/{N}}. \end{align} $$
Then, by (4-24) and the elementary inequality
$\min \{a,b\}\le a^{1-{\widetilde {s}}}b^{\widetilde {s}} $
, which holds for any
$a,b>0$
and
$0<\widetilde {s}<1,$
$$ \begin{align*} \mu(B(x,r))\le 2^{\tau+2}c_0c_1\cdot r^{s-({2m+8+2\tau})/{N}}. \end{align*} $$
Case i.
$(3\le i\le m+1).$
In this case
$n=n_j+i-2$
for some i (in other words,
$n_j+1\le n\le n_j+m-1$
). The situation here is the same as when
$n=n_j.$
We estimate the number of
$(n_j+i-1)$
th-order fundamental intervals
$J_{n_j+i-1}$
that can intersect
$B(x,r).$
Denote this number by
$t.$
So,
$$ \begin{align*} t\leq \#&\bigg\{I_{n_j+i-1}(a_1,\ldots,a_{n_j+i-1}):(a_1,\ldots,a_{n_j+i-1})\in D_{n_j+i-1} ~~\mathrm{and}~\\ &\quad I_{n_j+i-1}(a_1,\ldots,a_{n_j+i-1})\bigcap B(x,r)\neq\emptyset\bigg\}. \end{align*} $$
Notice that
$$ \begin{align*} \frac{1}{2^{2i-1}\cdot q_{n_j}^{2\beta_{i-1}+2}}&\le \frac{1}{2q_{n_j+i-1}^2}\le |I_{n_j+i-1}(a_1,\ldots,a_{n_j+i-1})|\\ &\le 10M\cdot G_{n_j+i-1}(a_1,\ldots,a_{n_j+i-1})\le 10M\cdot r. \end{align*} $$
Then
$1\le 2^{2i-1}\cdot q_{n_j}^{2\beta _{i-1}+2}\cdot 10Mr\le 10\cdot 2^{2m+1}Mrq_{n_j}^{2\beta _{i-1}+2}.$
Thus,
$$ \begin{align*} t\le \frac{2r}{|I_{n_j+i-1}|}+2&\le 4^i\cdot rq_{n_j}^{2\beta_{i-1}+2}+40\cdot 4^mM rq_{n_j}^{2\beta_{i-1}+2}\\ &\le (4^{m+1}+40\cdot 4^mM)rq_{n_j}^{2\beta_{i-1}+2}:=c_2rq_{n_j}^{2\beta_{i-1}+2}. \end{align*} $$
As a consequence,
$$ \begin{align*} \mu(B(x,r))&\le \min\{\mu(J_{n_j+i-2}),~c_2rq_{n_j}^{2\beta_{i-1}+2}\mu(J_{n_j+i-1})\}\\ &=\mu(J_{n_j+i-2})\min\bigg\{1, \ c_2rq_{n_j}^{2\beta_{i-1}+2}\cdot \frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-1}-\beta_{i-2}}}\bigg\}\\ &\le c_0\bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}\bigg)^{s-({2m+8+2\tau})/{N}}\min\{1,\ c_2rq_{n_j}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}\}. \end{align*} $$
By Proposition 4.4 and the length of the fundamental interval (see Case i in Lemma 4.1),
$$ \begin{align*} \mu(J_{n_j+i-2})&\le c_0\bigg(\frac{1}{q_{n_j-1}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}\bigg)^{s-({2m+8+2\tau})/{N}}\\ &\le 2^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}\cdot c_0\bigg(\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}\bigg)^{s-({2m+8+2\tau})/{N}}\\ & \le 4^{\tau+1}\cdot c_0\bigg(\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}\bigg)^{s-({2m+8+2\tau})/{N}}\quad (\mathrm{since}\ \beta_{i-2} \le \beta_{i-1} \le \beta_m = \tau). \end{align*} $$
Thus,
$$ \begin{align*} \mu(B(x,r))& \le 4^{\tau+1}c_0\bigg(\frac{1}{q_{n_j}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2}}\bigg)^{s-({2m+8+2\tau})/{N}}\cdot (c_2rq_{n_j}^{\kern1.5pt \beta_{i-1}+\beta_{i-2}+2})^{s-({2m+8+2\tau})/{N}}\\ &\le 4^{\tau+1}c_0c_2\cdot r^{s-({2m+8+2\tau})/{N}}. \end{align*} $$
Therefore, combining these
$m+2$
cases, we conclude the proof.
By Proposition 4.6 and the mass distribution principle,
$$ \begin{align*} {\dim_{\mathrm{H}}} \,\mathcal{O}_M(\kern1.2pt \beta_1,\ldots,\beta_m)\ge s-\frac{2m+8+2\tau}{N}. \end{align*} $$
Recall that
$\mathcal {O}_M(\kern1.2pt \beta _1,\ldots ,\beta _m)\subset E_m(\tau ).$
By letting
$N\to \infty $
and then
$s\to s_{\tau }^{(m)},$
$$ \begin{align} {\dim_{\mathrm{H}}} E_m(\tau)\ge s_{\tau}^{(m)}. \end{align} $$
By (3-11) and (4-25),
${\dim _{\mathrm{H}}} E_m(\tau )={\dim _{\mathrm{H}}} \widetilde {E}_m(\tau )= s_{\tau }^{(m)}.$
