\documentstyle{amsart}
\begin{document}
{\noindent\small {\em Electronic Journal of Differential Equations},
Vol.\ 1998(1998), No.~27, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu 147.26.103.110 or 129.120.3.113 (login: ftp)}
\thanks{\copyright 1998 Southwest Texas State University and
University of North Texas.}
\vspace{1.5cm}
\title[\hfilneg EJDE--1998/27\hfil Complex continued fractions]{Complex
continued fractions with restricted entries}
\author[Pawe{\l} Hanus \& Mariusz Urba\'nski \hfil EJDE--1998/27\hfilneg]
{Pawe{\l} Hanus \& Mariusz Urba\'nski }
\address{ {\sc Pawe{\l} Hanus} (e-mail: pgh0001\@jove.acs.unt.edu)
\hfill\break\indent
{\sc Mariusz Urba\'nski} (e-mail: urbanski\@unt.edu)\hfill\break
Department of Mathematics, University of North Texas\hfill\break
Denton, TX 76203-5118, USA.
}
\date{}
\thanks{Submitted April 21, 1998. Published October 19, 1998.}
\thanks{Partially supported by NSF Grant DMS-9502952}
\subjclass{58F23, 58F03, 58F12.}
\keywords{iterated function systems, complex continued fractions,
\hfil\break\indent
Hausdorff dimension, Hausdorff measure, packing measure}
\begin{abstract}
We study special infinite iterated function systems derived from
complex continued fraction expansions with restricted entries.
We focus our attention on the corresponding limit set whose
Hausdorff dimension will be denoted by $h$.
Our primary goal is to determine whether the $h$-dimensional Hausdorff
and packing measure of the limit set is positive and finite.
\end{abstract}
\maketitle
\newtheorem{theorem}{Theorem}
%\newtheorem{cor}{Corollary}
\newtheorem{lemma}{Lemma}
\section{Preliminaries}
The theory of uniformly hyperbolic dynamical systems leads naturally to the
study of finite Markov partitions and iterated function systems obtained from
a finite number of contractions. (See \cite{HU}, comp. \cite{MU1} for further
literature.) For non-uniformly-hyperbolic dynamical
systems, a part of the corresponding theory has been developed.
It goes back to the papers by Schweiger \cite{Sc1},
and Thaler \cite{T1, T2} on interval maps with an indifferent fixed
point. There the
concept of jump transformation is introduced and
explored. It has a natural Markov partition with infinitely many
cells. Further development of this subject can be found for example
in \cite{ADU, DU, FS, HI, Sc2, U2,Y}.
Here we discuss particular examples of conformal repellers, obtained as
limit sets of iterated function systems. Each iterated function system is
obtained from an infinite number of contractions. We show that under
certain conditions the repellers possess zero Hausdorff measure and positive
finite packing measure.
Specifically, let $(X, \rho)$ be a compact metric space,
and let $I$ be a countable set with at least two elements.
Define
$S = \{\phi_{i}\colon X \rightarrow X \left| \; i\in I \right.\}$,
a collection of injective contractions from $X$ to $X$ for which
there exists $0~~ 0$ such that for every $x \in \partial X$
there exists an open cone, $\mbox{Con}(x, u_{x}, \alpha , l)$
with vertex $x$, direction vector $u_{x}$, central angle of Lebesgue
measure $\alpha$, and altitude $l$, that is contained in $\mbox{Int}(X)$.
This is the so-called cone property. \\
$\bullet$ There exists an open connected set $V \subset {\Bbb R}^{d}$
containing $X$
such that every map $\phi_{i}$ can be extended to
a $C^{1+\epsilon}$ diffeomorphism mapping $V$ into $V$, and the extended map
is conformal on $V$. \\
$\bullet$ There exists $K \geq 1$ such that
$|\phi_{\omega}'(y)| \leq K|\phi_{\omega}'(x)|$ for every $\omega \in
I^{*}$ and every pair of points $x,y \in V$. This is the so-called Bounded
Distortion Property (BDP). \\
Let $X(\infty)$ be the set of limit points of all sequences $x_{i}
\in \phi_{i}(X), \; i \in I'$, where $I'$ ranges over all infinite
subsets of $I$.
The topological pressure function, $P$, for iterated function systems
is defined as follows. For every $t \geq 0$ consider the series
\[ \psi_{1}(t) = \sum_{i \in I} || \phi_{i}'||^{t}, \]
and more generally define for every integer $n \geq 1$
\[ \psi_{n}(t) = \sum_{\omega \in I^{n}} || \phi_{\omega}'||^{t}. \]
Now set
\begin{equation}
P(t) = \lim_{n \rightarrow \infty} \frac{1}{n} \log \psi_{n}(t).
\end{equation}
Detailed properties of this pressure function can be found in \cite{MU1}.
In [MU3] its definition is extended to case of parabolic iterated
function systems and in \cite{HMU, HU, U1}, the topological pressure
of systems of H\"older continuous functions is defined and
explored. This last concept also generalizes formula (2).
As shown in \cite{MU1}, there are two disjoint classes of conformal
iterated function systems, regular and irregular. A system is regular
if there exists $t \geq 0$ such that $P(t) = 0$. Otherwise the system
is irregular.
The following property demonstrating the geometrical significance of
topological pressure holds (see \cite[Theorem 3.15]{MU1}).
\begin{theorem}
$\dim_{H}(J) = \sup \{ \dim_{H}(J_{F}): \; F \subset I \; \mbox{finite} \}
= \inf \{t\geq 0: \; P(t) \leq 0 \}$. If a system is regular and $P(t) = 0$
then $t = \dim_{H}(J)$.
\end{theorem}
A Borel probability measure $m$ is said to be $t$-conformal if $m(J) = 1$
and for every Borel set $A$ and every $i \in I$,
\begin{equation}
m(\phi_{i}(A)) = \int_{A} |\phi_{i}'|^{t} \, dm
\end{equation}
and
\begin{equation}
m(\phi_{i}(X) \cap \phi_{j}(X)) = 0,
\end{equation}
for every pair $i,j \in I, \; i \neq j$. Lemma 3.13 in \cite{MU1} shows
that the conformal iterated function system is regular if and only if
there exists a $t$-conformal measure ($t$ is such that $P(t) = 0$),
and then $t = \dim_{H}(J)$.
\section{Results}
In this paper we focus our attention on a special example
of an infinite conformal iterated function system.
Namely, let $X$ be a closed disc on a complex plane
centered at the point 1/2 with radius 1/2, and let $V = B(1/2,3/4)$.
Given $k \geq 1$,
set $I_k = \{ n+ki \; : \; n \in {\Bbb N} \} $,
and for every index $n+ki \in I$ define $\phi_{n+ki}: V \rightarrow V$ by
\[ \phi_{n+ki}(t) = \frac{1}{n+ki+t}.\]
One can easily verify that for every positive integer $n$,
$\phi_{n+ki}(X) \subseteq X$ and $\phi_{n+ki}(V) \subseteq V$,
or even more precisely
\begin{equation}
\phi_{n+ki}(B(1/2, 1/2)) = \frac{1}{B(n+1/2+ki, 1/2)}.
\end{equation}
Moreover we have that $\phi_{n+ki}'(t) = -(n+ki+t)^{-2}$, and hence
$||\phi_{n+ki}'|| = |n+ki|^{-2} < (1+k^{2})^{-1} < 1$. That gives us the
universal contractive constant from the definition of i.f.s. It is also
easy to check that our system is conformal (all four conditions from the
definition are trivially satisfied). As announced at the beginning of the
paper (see (1)) we want to turn our attention to the limit set
$J_{k}$ associated
with the system. In particular we want to investigate the Hausdorff
dimension, and then the $h$-dimensional Hausdorff and packing measures
of this set, where $h = \dim_{H}(J_{k})$. Our main results are the following.
\begin{theorem}
Let $k$ be such that $1/2 < h = \dim_{H}(J_{k}) < 1$.
Then ${\cal H}^{h}(J_{k}) = 0$.
\end{theorem}
\begin{theorem}
Let $k$ be such that $1/2 < h < 1$. Then
$0 < {\cal P}^{h}(J_{k}) < \infty$.
\end{theorem}
Similar systems were introduced and studied in \cite{GM} and \cite{MU1}.
In particular, it was shown in \cite{MU1} that the limit set $J$ related
to the system where there is no restriction for an index $k$
($k \in {\Bbb Z}$ arbitrary) has the following properties:
\[ 1.2484 < h = \dim_{H}(J) < 1.9, \]
\[ {\cal H}^{h}(J) = 0, \]
\[ 0 < {\cal P}^{h}(J) < \infty , \]
where ${\cal H}^{h}$ and ${\cal P}^{h}$ denote $h$-dimensional
Hausdorff and packing measures respectively.
We start our investigation with the following lemma.
\begin{lemma}
$\lim_{k \rightarrow \infty } \dim_{H}(J_{k}) = 1/2$.
\end{lemma}
\noindent{\bf Proof.} According to what we said earlier, $\dim_{H}(J_{k}) =
\inf \{ t\geq 0: \; P(t) \leq 0 \}$.
One can prove using the chain rule that the
sequence $\psi_{n}(t)$ is subadditive, that is
\begin{equation}
K^{-t}\psi_{n}(t)\psi_{m}(t) \leq
\psi_{n+m}(t) \leq \psi_{n}(t) \psi_{m}(t) \leq \psi_{1}(t)^{n+m}.
\end{equation}
Therefore the fact that $P(t) = \infty$ is equivalent to saying that
$\psi_{1}(t) = \infty$. In our case,
\[ \psi_{1}(t) = \sum_{n=1}^{\infty} ||\phi_{n+ki}'||^{t} =
\sum_{n=1}^{\infty} \frac{1}{|n+ki|^{2t}}. \]
In particular $\psi_{1}(1/2) = \sum_{n=1}^{\infty} |n+ki|^{-1} = \infty$.
This proves that the system is regular and
$\dim_{H}(J_{k}) > 1/2$ for all $k$.
Fix $\epsilon > 0$, and choose $k$, depending on $\epsilon$, so big that
\[ \psi_{1} \Big( \frac{1}{2}+\epsilon \Big) = \sum_{n=1}^{\infty}
\frac{1}{|n+ki|^{1+\epsilon /2}} < 1. \]
Then using the subadditive property (6) we obtain
$$ \aligned
P\Big( \frac{1}{2}+\epsilon \Big) & = \lim_{m\rightarrow \infty} \frac{1}{m}
\log \psi_{m}\Big(\frac{1}{2}+\epsilon\Big) \leq
\lim_{m\rightarrow \infty} \frac{1}{m} \log \psi_{1}\Big(\frac{1}{2}
+ \epsilon \Big)^{m} \\
& = \log \psi_{1}\Big(\frac{1}{2}+\epsilon \Big) < 0.
\endaligned $$
We get that $\dim_{H}(J_{k}) < 1/2 + \epsilon$, since our system
is regular. When we let $\epsilon \searrow 0$, which implies
$k \rightarrow \infty$, our proof is finished. \hfil$\clubsuit$ \smallskip
\noindent {\bf Remark.} Notice that if $k=0$ we have
a system of real continued fractions $\phi_{n}(x) = (n+x)^{-1}$,
for which the limit set $J_{0}$ is the unit interval without rational
numbers. Obviously in this case the Hausdorff dimension
of the limit set is equal to 1, and both 1-dimensional Hausdorff measure
of $J_{0}$ and 1-dimensional packing measure of $J_{0}$ are 1. \medskip
\noindent{\bf Proof of Theorem 2.} Let $m$ be the conformal
measure associated to our
conformal iterated function system. The idea of the proof is based
on the following fact (Lemma 4.9 in \cite{MU1}): \\
If $S$ is a regular c.i.f.s. and there exists a sequence of points
$z_{j} \in X(\infty)$ and a sequence of positive reals
${r_{j}, \; j\geq 1}$, such that $r_{j} \rightarrow 0$ and
\[ \limsup_{j\rightarrow \infty} \frac{m(B(z_{j},r_{j}))}{r_{j}^{h}}
= \infty ,\]
then ${\cal H}^{h}(J_{k}) = 0$.
In our case, $z_{j} = 0$ for every $j$, since $0$ is the only point
in $X(\infty)$. Hence it is sufficient to show that
\begin{equation}
\overline{\lim_{r \rightarrow 0}} \frac{m(B(0,r))}{r^{h}} = \infty.
\end{equation}
Notice that
$$ \aligned
m(B(0,r)) & \geq \sum_{\phi_{n+ki}(X) \subseteq B(0,r)}
m(\phi_{n+ki}(X)) \\
&= \sum_{\phi_{n+ki}(X) \subseteq B(0,r)}
\int_{X} |\phi_{n+ki}|^{h} \, dm \\
& \geq \sum_{\phi_{n+ki}(X) \subseteq B(0,r)} K^{-h}
||\phi_{n+ki}'||^{h} \\
&= K^{-h} \sum_{\phi_{n+ki}(X) \subseteq B(0,r)}
|n+ki|^{-2h} \\
& \asymp K^{-h} \sum_{ \phi_{n+ki}(X) \subseteq B(0,r)}
n^{-2h},
\endaligned $$
where by $A \asymp B$ we mean that there exists some constant $C \geq 1$
such that $C^{-1} \leq A/B \leq C$.
We have to find out for how many $n$, $B(0,r) \cap \phi _{n+ki}(X)
\neq \emptyset $ or for which $n$, $\phi _{n+ki}(X) \subset B(0,r)$.
Now,
$$ \aligned
y \in B(0,r) \cap \phi _{n+ki}(X) & \Leftrightarrow
|y| < r \; \; \mbox{and} \; \; y \in \frac{1}{B(n+1/2+ki, 1/2)} \\
& \Leftrightarrow \frac{1}{|y|} > \frac{1}{r} \; \; \mbox{and} \; \;
\frac{1}{y} \in B(n+1/2+ki, 1/2).
\endaligned $$
Hence $B(0,r) \cap \phi _{n+ki}(X) \neq \emptyset $ if and only
if $B(n+1/2+ki, 1/2)$ contains a complex number of modulus
bigger than $1/r$.
Notice that if $z$ is a point in $\overline{B(n+1/2+ki, 1/2)}$
of maximal possible modulus, then
\[ |z| = \sqrt{n^{2}+n+\frac{1}{4}+k^{2}} + \frac{1}{2}. \]
Therefore $B(0,r) \cap \phi _{n+ki}(X) \neq \emptyset $ only if
\[ \sqrt{n^{2}+n+k^{2}+\frac{1}{4}} + \frac{1}{2} > \frac{1}{r}. \]
Let $A = \{ n | \; \sqrt{n^{2}+n+k^{2}+1/4} + 1/2 > 1/r \}$,
and let $n(r)$ be the minimal element of $A$. One can see that
$n(r) \asymp 1/r$ by the minimality of $n(r)$.
Using the integral test we are able to evaluate the limit introduced
at the beginning of the proof:
$$ \aligned
\overline{ \lim_{r \rightarrow 0}} \frac{m(B(0,r))}{r^{h}} & \geq
\overline{\lim_{r \rightarrow 0}}
\frac{K^{-h}\sum_{n \geq n(r)+1}n^{-2h}}{r^{h}}
\asymp \overline{\lim_{r \rightarrow 0}}
\frac{K^{-h} \int_{n(r)}^{\infty} x^{-2h} \, dx}{r^{h}} \\
& = \overline{\lim_{r \rightarrow 0}} \frac{n(r)^{1-2h}}{(2h-1)K^{h}r^{h}}
\asymp \overline{\lim_{r \rightarrow 0}} \frac{r^{2h-1}}{(2h-1)K^{h}r^{h}} \\
& = \overline{\lim_{r \rightarrow 0}} \frac{r^{h-1}}{(2h-1)K^{h}} = \infty.
\endaligned $$
We conclude that ${\cal H}^{h}(J_{k}) = 0$, which completes
the proof. \hfil$\clubsuit $ \smallskip
Now we turn our attention to the $h$-dimensional packing measure of the
limit set $J_{k}$. We begin with a simple lemma.
\begin{lemma}
If $f$ is the function defined on the complex plane by $f(z) = 1/z$ and
$C(x,r)$ is the circle centered at $x$ and of radius $r$, then
\begin{equation}
f(C(x,r)) = C \Big( \frac{\overline{x}}{|x|^{2} - r^{2}},
\frac{r}{||x|^{2} - r^{2}|} \Big) .
\end{equation}
\end{lemma}
\noindent{\bf Proof.} Recall from the theory of analytic functions that
for $\lambda > 0, \; \lambda \neq 1$ the equation
\[ \Big| \frac{z-p}{z-q} \Big| = \lambda \]
represents the circle with respect to which the points
$p$ and $q$ are symmetric, see \cite{P}.
Moreover the center and the radius of this circle are given by the formulas
\[ x = \frac{p-\lambda^{2}q}{1-\lambda^{2}}, \; \; \;
r = \lambda \frac{|p-q|}{|1-\lambda^{2}|}. \]
Fix $C(x,r)$. Let $p=x+r/2, \; q=x+2r$;
one can see that $p$ and $q$ are symmetric with respect to the
circle $C(x,r)$. Using point $x+r$ which lies on this circle we obtain
that $\lambda = 1/2$.
The image of our circle under function $f$ is the circle
\[ \frac{|w-1/p|}{|w-1/q|} = \lambda \Big| \frac{q}{p} \Big| . \]
Notice that the points $p' = 1/p = 2/(2x+r)$ and
$q' = 1/q = 1/(x+2r)$ are symmetric with respect to the new circle.
Let $\lambda ' = \lambda |q/p|$. Then
\[ \lambda ' = \lambda \frac{|x+2r|}{|x+r/2|} = \frac{1}{2}
\frac{|x+2r|}{|x+r/2|} = \frac{|x+2r|}{|2x+r|}. \]
Let $C(x', r') = f(C(x,r))$. Then
\[ \aligned x'
&= \frac{p'-\lambda^{'2}q'}{1-\lambda^{'2}}
= \frac{\frac{2}{2x+r} - \Big| \frac{x+2r}{2x+r} \Big| ^{2} \frac{1}{x+2r}}
{1 - \Big| \frac{x+2r}{2x+r} \Big| ^{2}}
= \frac{\frac{2}{2x+r} - \frac{\overline{x+2r}}{(2x+r)\overline{(2x+r)}}}
{\frac{(2x+r)\overline{(2x+r)} - (x+2r)\overline{(x+2r)}}
{(2x+r)\overline{(2x+r)}}} \\
&= \frac{2\overline{(2x+r)} - \overline{x+2r}}
{(2x+r)\overline{(2x+r)} - (x+2r)\overline{(x+2r)}} \\
&= \frac{4\overline{x} + 2r - \overline{x} - 2r}{4|x|^{2} + 2xr
+ 2\overline{x}r + r^{2} - |x|^{2} - 2xr - 2\overline{x}r - 4r^{2}} \\
&=\frac{\overline{x}}{|x|^2 - r^{2}}, \endaligned\]
and also
\[ \aligned r'
&= \lambda ' \frac{|p'-q'|}{|1- \lambda^{'2}|} =
\Big| \frac{x+2r}{2x+r} \Big|
\frac{ \Big| \frac{2}{2x+r} - \frac{1}{x+2r} \Big| }{ \Big| 1 -
\Big| \frac{x+2r}{2x+r} \Big| ^{2} \Big| } \\
&= \frac{ \Big| \frac{x+2r}{2x+r} \Big|
\frac{|2x+4r-2x-r|}{|(2x+r)(x+2r)|}}{ \Big| \frac{|2x+r|^{2}
- |x+2r|^{2}}{|2x+r|^{2}} \Big| } =
\frac{ \frac{3r}{|2x+r|^{2}} }{ \frac{|3|x|^{2} - 3r^{2}|}{|2x+r|^{2}} }
= \frac{r}{||x|^{2} - r^{2}|}, \endaligned\]
which finishes the proof. \hfil$\clubsuit$ \medskip
\noindent{\bf Proof of Theorem 3.} It is a general fact that
if the limit set $J$ of a
c.i.f.s. has nonempty intersection with an interior of the set $X$,
then the packing measure of this set is always positive. Hence we
only need to show that the packing measure of $J_{k}$ is finite.
Following Theorem 2.5 from \cite{MU2} we must prove that there exist three
constants $L>0, \; \xi >0$, and $\gamma \geq 1$, and
a finite set $F$, such that for all $n \in I\setminus F$ and for all $r$
with $\gamma\,\mbox{diam}( \phi_{n+ki}(X)) < r \leq \xi$ there is some
$x \in \phi_{n+ki}(X)$ such that
\begin{equation}
\frac{m(B(x,r))}{r^{h}} \geq L.
\end{equation}
According to the result proven in the lemma above the diameter
of the ball $\phi_{p+ki}(X)$ is
\begin{equation}
\mbox{diam}( \phi_{p+ki}(X)) = 2 \frac{1/2}{||p+1/2+ki|^{2} - 1/4|}
\asymp \frac{1}{p^{2}},
\end{equation}
for $p$ large enough.
Let $\Gamma_{p}$ denote the arc that is the image of the half-line
$pt+ki, \; t \geq 1$, under the function $f$. One can see that two the most
distant points of $\Gamma_{p}$ lying in $\phi_{p+ki}(X)$ are
$(p+1+ki)^{-1}$ and $(p+ki)^{-1}$. Therefore we can choose $x$
to be the point in $\Gamma_{p} \cap \phi_{p+ki}(X)$ with
$|x| = 1/(p+1)$. Additionally, set $\xi = 1$ and $\gamma = 8$.
To prove the theorem we only have to find $L>0$ such (9) holds.
We consider two separate cases. \\
Case 1. Suppose that $|x| \leq r$. In this case the ball $B(x,r)$
contains infinitely many balls of the form $\phi_{n+ki}(X)$; in fact
it contains all the balls for $n$ greater than some $n_{0}$.
On the arc $\Gamma_{p}$ choose a point $y$ such that $\rho (x,y) = r$.
Let $l$ denote the length of a part of $\Gamma_{p}$ from $0$ to $y$.
There exists a unique $1 \leq m \leq p$ so that
\[ \frac{m}{p+1} \leq r \leq \frac{m+1}{p+1}. \]
Simple geometry gives us that
\[ |y| > \frac{1}{\pi}l > \frac{1}{\pi} \Big( \frac{1}{p+1}+r \Big)
> \frac{1}{\pi} \frac{m+1}{p+1} \]
\[ = \frac{1}{\frac{\pi (p+1)}{m+1}} \geq \frac{1}{8[p/m]} \]
The above computation tells us that
\begin{equation}
\frac{1}{B(8\lfloor p/m \rfloor +1/2+ki, 1/2)} \cap B(x,r) \neq \emptyset ,
\end{equation}
so in other words we can assume that $n_{0} = 8\lfloor p/m\rfloor$.
We have that
$$ \aligned
m(B(x,r)) & \geq \sum_{n=8\lfloor {p\over m}\rfloor +1}^{\infty}n^{-2h} \asymp
\int_{8\lfloor{p\over m}\rfloor}^{\infty}x^{-2h} \, dx \\
& = \frac{1}{2h-1} \frac{1}{(8\lfloor{p\over m}\rfloor)^{2h-1}} =
\frac{1}{8^{2h-1}(2h-1)} \frac{1}{\lfloor{p\over m}\rfloor^{2h-1}} \\
& \geq \frac{1}{8^{2h-1}(2h-1)} \frac{1}{\lfloor{p\over r(p+1)}\rfloor^{2h-1}}
> \frac{1}{8^{2h-1}(2h-1)} \frac{1}{({1\over r})^{2h-1}} \\
& = \frac{1}{8^{2h-1}(2h-1)} r^{2h-1} \geq
\frac{1}{8^{2h-1}(2h-1)} r^{h}.
\endaligned $$
Case 2. Assume that $r < |x|$. In this case the ball $B(x,r)$ contains
only finitely many balls of the form $\phi_{n+ki}(X)$. It certainly
contains the $p$th ball, so all we have to do is to find the maximum
index $l$ so that $\phi_{l+ki}(X) \subset B(x,r)$.
Then there exists a unique $l \geq p$
such that
\[ \sum_{n=p}^{l} \mbox{diam} \Big( \frac{1}{B(n+{1\over 2}+ki,{1\over
2})} \Big)
\leq r \leq \sum_{n=p}^{l+1} \mbox{diam} \Big(
\frac{1}{B(n+{1\over 2}+ki,{1\over 2})} \Big) \]
The formula (10) for the diameter of $\phi_{n+ki}(X)$ immediately
gives that for $n$ large enough
\begin{equation}
\frac{1}{(n+1)^{2}} <
\mbox{diam} \Big( \frac{1}{B(n+1/2+ki,1/2)} \Big) < \frac{1}{n^{2}}
\end{equation}
Hence, using the integral test, we obtain
$$ \aligned
\frac{1}{p+1} - \frac{1}{l+2
} & = \int_{p+1}^{l+2} x^{-2} \, dx
\leq \sum_{n=p+1}^{l+1} \frac{1}{n^{2}} < r \\
& < \sum_{n=p}^{l+1} \frac{1}{n^{2}} \leq
\int_{p-1}^{l+1} x^{-2} \, dx = \frac{1}{p-1} - \frac{1}{l+1}.
\endaligned $$
This shows that
\begin{equation}
\frac{1}{p+1} - \frac{1}{l+2} \leq r \leq \frac{1}{p-1} - \frac{1}{l+1}
\end{equation}
Applying the Mean Value Theorem we get
\[ \aligned
m(B(x,r)) & \geq \sum_{n=p}^{l} n^{-2h} \asymp \int_{p}^{l}
x^{-2h} \, dx \\
& = \frac{1}{2h-1} \Big( \Big( \frac{1}{p} \Big) ^{2h-1}
- \Big( \frac{1}{l} \Big) ^{2h-1} \Big) \\
& \geq \frac{1}{2h-1} \Big( \frac{1}{p} - \frac{1}{l} \Big)
(2h-1) \Big( \frac{1}{z} \Big) ^{2h-2} \\
&= \Big( \frac{1}{p} - \frac{1}{l} \Big) \Big( \frac{1}{z^{2}} \Big) ^{h-1},
\endaligned \]
for some $p \leq z \leq l$. Recall that $r \geq 8/p^{2}$, which implies
$z^{-2} \leq r/8$. Hence, using (13)
\[ \aligned
m&(B(x,r)) \geq \Big( \frac{1}{p} - \frac{1}{l} \Big)
\Big( \frac{r}{8} \Big) ^{h-1} \\
= &\Big[ \Big( \frac{1}{p} - \frac{1}{p-1} \Big) +
\Big( \frac{1}{p-1} - \frac{1}{l+1} \Big) +
\Big( \frac{1}{l+1} - \frac{1}{l} \Big) \Big] 8^{1-h} r^{h-1} \\
\geq &\Big[ - \frac{1}{p(p-1)} + r - \frac{1}{l(l+1)} \Big]
8^{1-h} r^{h-1} \\
\geq &\Big[ r - \frac{2}{p(p-1)} \Big] 8^{1-h} r^{h-1} \\
\geq &\Big[ r - \frac{4}{p^{2}} \Big] 8^{1-h} r^{h-1}
\geq \Big[ r - \frac{r}{2} \Big] 8^{1-h} r^{h-1}
= \frac{1}{2} 8^{1-h} r^{h}.
\endaligned \]
Choosing $L$ to be $\min \{ 8^{1-2h}(2h-1)^{-1}, \; 2^{-1}8^{1-h} \}$
completes the proof of the theorem. \hfil$\clubsuit$
\section{Additional Remarks}
According to the first equality of Theorem 1,
since the Hausdorff dimension of the limit
set of the system $\{\phi_{n+k i}: n\geq 1, k\in {\Bbb Z}\}$ is greater
than 1, we can add to the family $\Phi_l=\{\phi_{n+l i}: n\geq 1\}$,
$l\ge 1$, finitely many mappings from $\{\phi_{n+k i}:
n\geq 1, k\in {\Bbb Z}\}$ to obtain the systems whose limit sets $J$
have Hausdorff dimensions greater than 1. Then, employing
methods similar to those used in the proofs of Theorems 2 and 3,
we find that
$0<{\cal H}^h(J)<\infty$ and ${\cal P}^h(J)<\infty$. (Here instead of
using Lemma 4.9 from \cite{MU1} and Theorem 2.5 from \cite{MU2} one should use
Lemma 4.11 from ref{MU1} and Theorem 2.6 from \cite{MU2}.) Let us remark
that this fact distinguishes these systems from the full family
$\{\phi_{n+k i}: n\geq 1, k\in {\Bbb Z}\}$ and the families
investigated in this paper, since for them the packing measure is
finite and positive, whereas here the Hausdorff measure is
positive and finite. \smallskip
\noindent{\bf Acknowledgements.} We would like to thank the referees for valuable
remarks which improved our paper and the editors of the EJDE for their
attention to the final form of this article.
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\end{document}
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