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\define\MN{{\s MIN}}
\define\MX{{\s MAX}}
\define\Bd{B_{\delta}}
\redefine\l#1{\lambda_{#1}}
\define\ld{\frac{\log \lambda_1}{\log \lambda_2}}
\redefine\dBF{\text{dim}_{B}F}
\redefine\dHF{\text{dim}_{H}F}
\def \aa#1#2{\rlap{#1}\hfill\rlap{#2}\hfill\newline}
\define\h{\frac12}
\define\k#1{\frac{\log 2}{\log(\frac{1}{\lambda_{#1}})}}
\redefine\b{\frac{\log (\frac {2 \lambda_2}{\lambda_1})}{ \log
(\frac{1}{\lambda_1} )}} \def\today{\ifcase\month\or
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\topmatter
\title
THE DIMENSIONS OF SOME SELF AFFINE LIMIT SETS IN THE PLANE AND HYPERBOLIC SETS
\endtitle
\author
MARK POLLICOTT
and HOWARD WEISS
\endauthor
\leftheadtext{MARK POLLICOTT and HOWARD WEISS}
\affil Warwick University and The Pennsylvania State University
\endaffil
\address{\aa{Mark Pollicott}{Howard Weiss}
\aa{Department of Mathematics}{Department of Mathematics}
\aa{Warwick University}{Pennsylvania State University}
\aa{Coventry, CV4 7AL}{University Park, PA 16802}
\aa{England}{U.S.A.}
\aa{Email: mp\@maths.warwick.ac.uk }{Email:
weiss\@math.psu.edu}
\aa{}{}}
\endaddress
\thanks{ The first author would like to thank INIC (Portugal)
and The Royal Society (England) for their support. The second author
was partially supported by a National Science Foundation
Postdoctoral Research Fellowship and JNICT grant PMCT/C/CEN
35/90. The authors also thank William Parry for helpful discussions on Fat Bakers transformations and Yuval Peres for carefully reading and commenting on a draft of this paper and supplying many references to the bibliography of this paper. The second author would like to thank Yakov Pesin for many stimulating dicussions on dimension theory. }
\endthanks
\keywords{Hausdorff dimension, box dimension, self-affine, PV numbers, random geometric series, horseshoes, infinitely convolved Bernoulli measure}
\endkeywords
\abstract {In this article we compute the Hausdorff dimension and
box dimension (or capacity) of a dynamically constructed model similarity
process in the plane with two distinct contraction coefficients. These
examples are
natural generalizations to the plane
of the simple Markov map constructions for Cantor sets on the line. Some related problems have been studied by different authors; however,
those results are directed towards generic results in quite
general situations. This paper concentrates on computing explicit formulae in as many specific cases as possible. The techniques of previous authors and ours are correspondingly very different. In our calculations, delicate number theoretic properties of the contraction coefficients arise. {\bf Finally, we utilize the results for the model problem to compute the dimensions of some linear horseshoes in $ \Bbb R^n $, and we observe that the dimensions do not always coincide and their coincidence depends on delicate number theoretic properties of the Lyapunov exponents.} }
\endabstract
\newcount\hour \newcount\minute
\date \today \quad \now \enddate
\endtopmatter
\document
\bigskip
\head {\bf 1. Introduction} \endhead
\bigskip
The notions of Hausdorff dimension and box dimension play an
important
role in describing the concept of ``size'' of sets in the plane.
They are particularly useful in analyzing sets of Lebesgue measure
zero.
A situation that is relatively well understood is that of Cantor
sets in the line that are constructed dynamically from
simple Markov (``cookie cutter'') transformations.
Let $I_0, I_1 \subset [0,1]$ be
two disjoint closed subintervals. A simple Markov map is a
$C^{1+\alpha}$ map $g: I_0 \cup I_1 \to \Bbb R$ such that (i)
$|g'| > 1$ and \, (ii) $g(I_0) = g(I_1) = [0, 1]$. Then $\Lambda
= \{ x \in [0,1] : \, g^k(x) \in [0,1] \text{ for } k = 0, 1, 2,
\cdots \}$ is the associated Cantor set. In this particular
case, considerable advantage is gained from the
one-dimensionality, and more specifically
from the conformal nature of the transformation. The situation
becomes more complicated when non-conformal and higher
dimensional analogues are considered.
In this note, we first deal with the simplest ``model problem''
in which we are concerned with a Cantor set in the plane
constructed from
two similar affine contractions. Our set is described by a
finite number of parameters (the horizontal contraction
$\lambda_2$,
the vertical contraction $\lambda_1$, the position of the images,
etc.),
and it becomes a very natural question to ask about the
dependence of the Hausdorff dimension and box dimension
on these parameters.
For this important special problem, we compute the box dimension
in all
cases and the Hausdorff dimension in a large number of cases.
Our pivotal observation is that if $\alpha = \max\{\lambda_1,
\lambda_2\}<1/2$, then the Hausdorff dimension and box dimension
are equal and depend only on $\alpha$.
In subsequent sections, we present necessary conditions on
$\alpha > 1/2$ such that the Hausdorff dimension and box dimension will again be
equal. This problem seems to be intimately connected with very
deep results in number theory and harmonic analysis on PV numbers
due to Garcia and Erdos. We introduce a new class of numbers ({\bf GE} numbers) that are a subset of, but more suitable for our purposes than, the well known class of numbers studied by Kahane and Salem in \cite{KS}. Previous applications of these ideas to Hausdorff dimension estimates seem to have been limited to graphs of Weierstrass-like functions \cite{L, PU}. Along the way, we construct the first one-paramter family of Cantor sets whose Hausdorff dimension and box dimension do not coincide.
We then describe how our ideas may be applied to various generalizations of this problem.
Finally, we utilize the results for the model problem to compute the dimensions of some linear horseshoes in $ \Bbb R^n $. We construct an easy example to show that the Hausdorff dimension of some linear horseshoes depends on the {\it geometry} of the horseshoe, and hence that the Hausdorff dimension of a hyperbolic set cannot be computed from just dynamical quantities, i.e., entropies and Lyapunov exponents. Moreover, we construct examples of linear horseshoes {\it all having the same geometry}. In some of the examples the dimensions coincide, and in other examples the dimensions do not coincide. The largest Lyapunov exponent of the latter examples is a reciprical of a PV (badly approximable) number. {\it This indicates that the Hausdorff dimension of hyperbolic sets can depend on delicate number theoretic properties of the Lyapunov exponents.} This is in contrast to the Hausdorff dimension of an invariant measure for a surface diffeomorphism, where the dimension can be !
computed solely in terms of the Lyapunov exponents and the measure theoretic entropy [Y]. We conclude that the Hausdorff dimension of an invariant measure supported on a hyperbolic set is a much more natural and tractible quantity to study than the Hausdorff dimension of a hyperbolic set.
\bigskip
\head {\bf 2. Description of Model Problem} \endhead
\bigskip
Let $R_1,R_2 \subset I$ be two disjoint boxes in the unit square
$I \subset \Bbb R^2 $ (aligned with the axis of $I$) each
having the
same height, $0 < \lambda_1 < 1$, and the same width $0 <
\lambda_2 < 1$. See figure 1. For convenience assume $0 <
\lambda_1 \leq \lambda_2 < 1$.
If this is not the case, we need only
interchange the two coordinates of the
original square.
Consider the two affine maps
$A_1:I \to R_1$ and $A_2:I \to R_2$ that
contract the unit square by $\lambda_1$ in the vertical direction
and
$\lambda_2$ in the horizontal direction and place the resulting
images
in $R_1$ and $R_2$, respectively.
Denote by $F$ the self affine set which is the limit set of
the sequences $A_{i_0} A_{i_1} \ldots A_{i_n}x$, $x \in I$, $i_k \in \{1, 2\}$ as $n \to \infty $ (see figure 1).
\demo{\it Notation}
Let $\dBF$ and $\dHF$ denote the box dimension and Hausdorff
dimension,
respectively, of the set $F$. Let $\pi_{\l k}F \subset [0,1]$
denote the projection of the limit set $F$ onto the $\lambda_k$
axis, for $k=1, 2$. Full definitions and properties are given in Appendix 1.
\enddemo
\medskip
We first present a table of calculated values of the dimensions
of $F$ that follow from the results in this paper.
\medskip
\proclaim{\bf Explicit Formulae} With the hypotheses above,
\roster
\item {\bf case 0:} \quad $\lambda= \l1=\l2 \qquad \dHF=\dBF= \frac{\log 2}{\log(\frac{1}{\lambda})}$
\medskip
\item {\bf case 1:} \quad $\pi_{\l 2}R_1 = \pi_{\l 2}R_2 \qquad \Rightarrow \qquad \dHF=\dBF= \k 1$ \hfil
\newline( $R_1$ is directly on top of $R_2$ \, - \, degenerate case)
\medskip
\item {\bf case 2:} \quad $\l2 \leq \h \qquad \Rightarrow \qquad \dHF=\dBF= \k 2$
\medskip
\item {\bf case 3:}\quad $ \l2 = 2^{-\frac1p}, \, p \in \Bbb N \qquad \Rightarrow \qquad \dHF=\dBF=\b$
\medskip
\item {\bf case 4:}\quad There exists $0 < \gamma < 1$ such that for almost all $\l2$ with \newline $\gamma < \l2 < 1 \qquad \Rightarrow \qquad \dHF=\dBF =\b$
\medskip
\item{\bf case 5:} \quad $\l2 \geq \h \qquad \Rightarrow \qquad \dBF= \b.$
\endroster
\endproclaim
\midinsert
\vspace{4truein}
\botcaption{Figure 1}
\endcaption
\endinsert
\medskip
We prove the equality of the Hausdorff dimension and box
dimension in the above
cases by finding a number theoretic sufficient condition on $\l2$
that ensures the dimensions coincide. It is here that we encounter a fundamental dichotomy between the cases $0 < \l2 <
\frac12$ and $\frac12 < \l2 < 1$ The condition is always
satisfied for
$0 < \l2 < \frac12$. The case when $\frac12 < \l2 < 1$ is much
more subtle.
The sufficient condition is violated when $\lambda_2$ is the
reciprocal of a Pisot-Vijayarghavan (abbreviated PV) number. Recall that
PV numbers are algebraic integers whose conjugates all lie within
the unit circle (\cite{S}, \cite{G} and \cite{E}).
It is not clear, a priori, whether the number theoretic property of $\l2$
that we require to show coincidence of the dimensions is really
necessary or just a manifestation of our proof. In \cite{PU},
the authors show that for these problematic PV numbers,
the equality between the two types of dimension actually breaks
down:
\medskip
\proclaim{\bf Proposition \cite{PU}} For $\lambda_1 = \frac12$
and $\lambda_2$ equal to the reciprocal of a PV number, there
exists certain configurations such that \, $ \dHF < \dBF$.
\endproclaim
In \cite{PU} the authors compute the dimension of graphs of
Weierstrass-like functions. It just so happens that in this very
special case the sets $F$ reduce to graphs of Weierstrass-like
functions (modulo countable sets) and hence the results in
\cite{PU} are applicable to our case. One can combine the result
in \cite{PU} with Proposition 2 to construct
the first one-parameter family of Cantor sets whose Hausdorff dimension and box dimension do not coincide.
We conjecture that for (Lebesgue) almost all $\frac12 < \l2 < 1$,
the Hausdorff dimension coincides with the box dimension and is given (for non-degenerate configurations) by the formula in case 4 in the list of explicit formulae. This conjecture would imply that for almost all $0 < \l2 < 1$, the Hausdorff dimension of the limit set of the model similarity process coincides with the box dimension. Morover, we conjecture that for a generic similarity process (i.e., the boxes need not be aligned with the axes of $I$), the Hausdorff dimension coincides with the box dimension.
We wish to quickly dispose of the degenerate configurations where $\pi_{\l 2}R_1 = \pi_{\l 2}R_2 $, i.e., $R_1$ is directly on top of $R_2$. In these cases, the limit sets are easily seen to be uniform Cantor subsets of a vertical line having exponent $\l1$, and hence $\dHF=\dBF= \k 1$. We will henceforth assume that all configurations are non-degenerate.
\bigskip
\head {\bf 3. Calculation of} $\text{\bf dim}_{\bold B} \bold F $ \endhead
\bigskip
In this section, we shall present a direct computation of the box
dimension of $F$ in all cases.
We remark that Falconer \cite{F4} has related results. We begin
with some notation.
\medskip
\demo{\it Notation}
Denote the rectangles under $n$ iterations of the affine maps by
$R_{(i_1,\cdots, i_n)} = A_{i_n} \ldots A_{i_1} I \subset I$.
\enddemo
Let $\pi_{\l k}F \subset [0,1]$ denote the projection of the
limit set $F$ onto
the $\lambda_k$ axis, for $k=1,2$.
We begin with a simple characterization of this projection
that will prove useful later.
\medskip
\proclaim{\bf Proposition 1} \endproclaim
$$
\eqalign{
\pi_{\l 2}F &= \left\{ \sum_{k=0}^{\infty}(\epsilon_0 +
i_k(\epsilon_1-\epsilon_0
)) \, \l 2^k ; \enspace (i_1, i_2, \cdots ) \in
\{0,1\}^{\Bbb N} \right\} \cr
&= \left\{ \frac{\epsilon_0}{1-\l 2} +
\underbrace{(\epsilon_1-\epsilon_0)}_d
\sum_{k=0}^{\infty}i_k \l 2^k ;
\enspace (i_1, i_2, \cdots ) \in \{0,1\}^{\Bbb
N} \right\} \cr
&\subset \left[\frac{\epsilon_0}{1 - \l2},
\frac{\epsilon_1}{1-\l2} \right] \equiv J. \cr}
$$
\demo{\bf Proof} \enddemo
The affine maps $A_1, A_2$ on the square project
to affine maps on the $\l2$ axis $B_1,B_2:[0,1] \to [0,1]$
of the form
$$B_1(x) = \epsilon_0 + \lambda_2 x \hbox{ and }
B_1(x) = \epsilon_1 + \lambda_2 x,
$$
where $0 < \epsilon_0, \epsilon_1 < 1$ are the left end points
of the intervals $\pi_{\l 1} F, \pi_{\l 2}F$, respectively.
By induction, the left endpoint of $R_{(i_1,\cdots, i_n)}$ is
given by
$ B_{i_n} \ldots B_{i_1} 0 =
\sum_{k=0}^{n}(\epsilon_0 + i_
k(\epsilon_1-\epsilon_0)) \, \l 2^k$.
The lemma follows by taking limits.
\quad \qed
\medskip
The following consequence will be useful in our estimates.
\medskip
\proclaim{\bf Corollary 1.1} \roster
\item If $\h \leq \l 2 < 1,$ then $\pi_{\l 2}F = J$.
\item If \, $0 < \l 2 < \h,$ then $\pi_{\l 2}F= \text{Cantor
set}$.
\endroster
\endproclaim
\medskip
\proclaim{\it Remarks} \endproclaim
\roster
\item If $\l 2 = \h$,
then Corollary 1.1 corresponds to computing the dyadic expansion of
numbers in $J$.
\item
When $\l 2 = \frac{1}{3}$, then Corollary 1.1 corresponds to the
construction of
a copy of the middle third Cantor set.
\endroster
\demo{\bf Proof of Corollary 1.1} \enddemo
If $\h \leq \l2 < 1$, then the expression in Proposition 1 is
the $\beta-$expansion for $\beta
= 1/\l 2$. It is easy to see that every
$x \in J$ has a $\beta-$expansion,
hence $\pi_{\l 2}F = J$. If we take the orbit
of $x$ under the associated expanding map (making arbitrary
choices
of the expanding map when the domains overlap), then the
itinerary of the
orbit gives the corresponding $\beta$-expansion.
For the case $\l 2 < \h$, we need only observe that the set is
affinely equivalent to the standard
Cantor
set
$$\sum_{k=0}^{\infty}i_k \l 2^k ;
\enspace (i_1, i_2, \cdots ) \in \{0,1\}^{\Bbb
N}\eqno(3.1)$$
after scaling (by $d>0$) and translating (by
$\frac{\epsilon_0}{1-\lambda_1}$).
\quad \qed
\medskip
\demo{Notation}
Let $F_k$ denote the union of the disjoint rectangles
$R_{(i_1, \ldots, i_{k})}$ for $(i_1, \ldots, i_{k}) \in
\{0,1\}^{k}$.
\enddemo
\medskip
\proclaim{\bf Proposition 2}
If $\h < \l 2 < 1$, then $\dBF = \b$.
\endproclaim
\demo{\bf Proof } \enddemo
Consider the obvious cover of the $2^k$ rectangles in $F_k$ by
squares with sides of length
$\l 1^k$.
Each of these rectangles
$R_{(i_1, \ldots, i_n)}$ has width
$\lambda_2^n$
and height $\lambda_1^n$ and is covered
by $\l 1^k$ squares aligned in a row.
The minimum number of squares needed
for each such rectangle is
$\left[\frac{\lambda_2^n}{\lambda_1^n}\right]$.
We see from Corollary 1.1 that for all rectangles $R_{(i_1,\ldots, i_n)}$, the projection
$\pi_{\l 2}(F \cap R_{(i_1,\cdots, i_n)}) $
contains an interval of length
$|J| \text{ length} \left((\pi_{\l 2}( R_{(i_1,\cdots,
i_n)})\right) = |J| \lambda_2^n = \frac{d \lambda_2^n}{1 - \lambda_2}$.
Hence, of those squares in the cover of each
$R_{(i_1, \ldots, i_n)}$,
the proportion of the squares in the cover of
$R_{(i_1, \ldots, i_n)}$ required to cover
$R_{(i_1, \ldots, i_n)} \cap F$
is at least the ratio of the
length of the contained interval to the length of the rectangle,
i.e.,
$$
\frac{\frac{ \lambda_2^n}{1 - \lambda_2}}{\lambda_2^n}
= \frac{1}{1 - \lambda_2}.
\eqno(3.2)$$
Let $N(\delta)$ denote the minimum number of $\delta$-squares
needed
to cover $F$. Our above reasoning tells us that
when $\delta=\l 1^n$, there are $2^n$ rectangles in $F_n$, each
of which
is covered by $\frac{\l 2^n}{\l 1^n}$
squares of size $\l 1^n$. To cover $F$, the proportion
given by (3.2) is needed, i.e.,
$$
N(\l 1^k) = 2^n \frac{\l 2^n}{\l 1^n} \frac{1}{1 - \lambda_2}.
$$
The proposition follows from Lemma A1 in Appendix 1. \quad \qed
\medskip
\proclaim{\bf Proposition 3} If $0 < \l 2 < \h$, then $\dBF = \k
2$.
\endproclaim
\demo {\bf Proof} \enddemo
The set $ \pi_{\l 2}F$ was seen to be a Cantor set in Corollary
1.1.
However, it is possible that the
projections onto the $\l 2$ axis of the $2^k$ rectangles in
$F_k$
may not be disjoint (this anomaly occurs because
the Cantor set $F$ is really defined through the affine maps
$A_1,A_2$
and not the covers $F_k$.)
However, our {\it formula} in Proposition 1 for $\pi_{\l 2}F$
gives a construction that uses $2^k$ {\it disjoint}
intervals of length
$d \frac{\l 2^{k+1}}{1 - {\l 2}}$ at step $k$.
Assume that the original boxes are given by
$R_1 = \left[\epsilon_0, \epsilon_0 + \lambda_2 \right] \times
J_1$ and
$R_2 = \left[\epsilon_1, \epsilon_1 + \lambda_2 \right] \times
J_2$,
where $J_1, J_2$ are intervals in the vertical axis.
Once we have defined the maps $A_1:I \to R_1$ and $A_2:I \to
R_2$,
we replace these boxes by the new boxes
$R_1^* =
\left[\frac{\epsilon_0}{1 - \l 2}, \frac{d \l 2}{1 - \l 2}
\right]
\times J_1$ and
$R_2^* =
\left[\frac{\epsilon_0}{1 - \l 2} + d \l 2,
\frac{d \l 2}{1 - \l 2} +\frac{d}{1 - \l 2}\right]
\times J_2$. These project (on the horizontal axis) onto the two {\it
disjoint}
intervals
$$\left[\frac{\epsilon_0}{1 - \l 2},
\frac{\epsilon_0}{1 - \l 2} + \frac{d \l 2}{1 - \l 2} \right]
\hbox{ and }
\left[\frac{\epsilon_0}{1 - \l 2} + d \l 2,
\frac{\epsilon_0}{1 - \l 2} + \frac{d}{1 - \l 2}\right]$$
whose endpoints correspond to the
sequences
$$\eqalign{
(i_1, i_2, \ldots ) &= (0,0,0,0, \ldots) \cr
&= (0, 1,1,1, \ldots) \cr
&= (1, 0,0, 0, \ldots) \cr
&= (0, 1,1,1, \ldots) .\cr }$$
These intervals give the ``standard'' construction
of the Cantor set in the
projection $\pi_{\l 2}F$, as can be seen by Proposition
1.
We can define new
rectangles $R^*_{(i_1, \ldots, i_n)} = A_{i_n} \ldots A_{i_2}
R^*_{i_1}.$
The $2^k$ rectangles at step $k$ are all disjoint since
$R^*_k \subset R_k$ for $k=1,2$, and thus,
$R^*_{(i_1, \ldots, i_n)} \subset R_{(i_1, \ldots, i_n)}$.
We let $F_n$ denote the union of these $2^n$ rectangles.
The projection onto the horizontal axis consists of $2^n$
intervals of length
$c=\frac{d \l 2}{1 - \l 2}.$
Let $m=\left[\frac{\log(\frac1c \l 1^k )}{\log \l 2} \right]$.
Our goal is to
find ``optimal covers'' of
$F \cap F_m$ by squares of length $ \l 1^k$. It
follows from the definition of
$m$ that $1
\leq \frac{c \l 2^m}{\l 1^k} \leq
\frac{1}{\l 2},$ and hence each
rectangle $R^*_{(i_1, \ldots, i_n)}$ projects onto a {\it
disjoint}
interval (on the horizontal axis)
of length $c \l2^m$. It thus contains at least one
interval of length $\l 1^k$ and can be covered by
$\frac{1}{\l 2}$ intervals of length $\l 1^k$.
This implies that
$$2^m
\leq \#\{ \l 1^k \text{ intervals needed to
cover } \pi_{\l 2}(F \cap F_m) \} \le \frac{1}{\l 2} 2^m,
$$
i.e., $N(\lambda_1^k) \asymp 2^m$. Hence
$$
\frac{\log N(\l 1^k)}{\log(\frac{1}{\l 1^k})} \asymp \frac{\log
(2^m) } {\log (\frac{1}{\l 2^k})} \asymp
\frac{ \left[ \frac{\log (\frac{1}{\l 1^k} )}{\log(\frac{1}{ \l 2}) } \right] \log 2}{k\log(\frac{1}{ \l 1})} \overset { k \rightarrow \infty} \to \longrightarrow \frac{\log 2}{\log (\frac{1}{\l 2} )}.
$$
The Proposition follows from Lemma A1. \quad
\qed
\medskip
The following result follows immediately from the calculation of
$\dBF$:
\medskip
\proclaim{\bf Proposition 4} If one avoids the degenerate
configurations in case 1, then the map $\lambda_2 \to \dBF(\l2)$
is Lipschitz but not differentable.
\endproclaim
\medskip
\proclaim{\it Remark} \endproclaim
This type of phenomenon, where a fundamental invariant changes in
a Lipschitz but not smooth way, is quite rare in hyperbolic dynamical systems.
\bigskip
\head {\bf 4. Calculation of $\bold{dim}_{\bold H} \bold F$}
\endhead
\bigskip
In this section we turn to the problem of computing the Hausdorff
dimension of the set $F$. Our formulae are based on some number
theoretic properties of the value $\l 2$ and should be compared with those
estimates occurring in \cite{F3}.
Let $\beta$ be any real number between $0$ and $1$.
For any $n \geq 1$ consider the
set
$$J_n = \{(i_0, \ldots, i_{n-1})
\in \prod_0^{n-1}\{0,1\} \}$$
consisting of $2^n$ elements.
\medskip
Define the map
$\pi_n: J_n \to [0, \frac{1}{1-\beta}]$
by
$$\pi_n((i_0, \ldots, i_{n-1}))
= \sum_{r=0}^{n-1} i_r \beta^r.
$$
\medskip
\proclaim{\bf Definition} The number $\beta$ satisfies condition
{\bf GE} (after Garcia-Erdos) if there exists a constant $C>0$ such that
for all $x \in [0, +\infty)$
we have
$$
A_{\beta}(n) = \text{Card} \left \{ (i_0, \ldots, i_{n-1}) :
\pi_n(i_0, \ldots, i_{n-1}) \in [x, x+\beta^n) \right \}\le C \left(2 \beta\right)^n.
$$
\endproclaim
\demo{Remark}
We can also use the slightly weaker assumption that for all $\beta' > \beta$, choose a constant $C = C(\beta')$ with the above properties.
\enddemo
\medskip
\proclaim{\bf Lemma 1} If $0 < \beta < \frac12$, then
$A_{\beta}(n) = 1$. Hence condition {\bf GE} {\it is violated} for all $ 0 < \beta < \frac12$.
\endproclaim
\medskip
\proclaim{\bf Proof } \endproclaim
Observe that in this case the maps $\pi_n$ are bijective onto
their images. Furthermore, the points in the image are separated by
a distance of at least $\beta^n$. The result follows
easily. \quad \qed
\medskip
\demo{Remark}
There exist values $\frac{1}{2} < \beta < 1$ such
that condition {\bf GE} does not hold. Consider the case where $\frac{1}{\beta}$
is the Golden mean, i.e., $ 1 + \frac{1}{\beta} = \frac{1}{\beta^2}$.
If we consider the $2^n$ strings $(w_1, \ldots, w_n) \in J_{3n}$, where
each $w_i = (1,0,0)$ or $(0,1,1)$, then it is clear that they
have the same images under $\pi_{3n}$ and thus $A_\beta(3n) \geq 2^n$. However, it is an easy numerical check that $2 > (2\beta)^{3} = 1.88854 \ldots$, which clearly contradicts the {\bf GE} assumption.
\enddemo
\medskip
\proclaim{\bf Propostion 5}
\roster
\item If $0 < \l2 < \frac12$, then $\dBF = \dHF = \k2$.
\medskip
\item If $\frac12 < \l2 < 1$ satisfies condition {\bf GE}, then $\dBF = \dHF = \b$.
\endroster
\endproclaim
\demo{\bf Proof} \enddemo
Let $\Bd$ be a ball of radius $\delta$. Choose $k \in \Bbb N$ such that $\delta
< \l 2^k < 10 \delta$ (e.g., $k = \left[\frac{\log\lambda_2}{\log \delta}\right] + 1)$, and choose
$m= \left[k \frac{ \log \frac{1}{ \l2}}{\log \frac{1}{\l1}} \right]$. Clearly $\l 1^m \asymp \l 2^k$. Using an idea in \cite{M}, we consider the ``asymptotic squares'' $S$ of dimensions $\l 1^m \times \l 2^k$ that prolongate each rectangle in $F_k$. There are two asyptotic squares associated to each rectangle in $F_k$, corresponding to ``prolongating to the left'' and ``prolongating to
the right.''
\medskip
It is convenient to break up the rest of the proof into four shorter lemmas.
\medskip
\proclaim{\bf Lemma 2}There exists a bound $C \ge 1$ (independent of $\delta$ and $k =
k(\delta)$)
such that $F \cap \Bd$ can be covered with at most $C$ asymptotic
squares. \endproclaim
See figure 2.
\midinsert
\vspace{4truein}
\botcaption{Figure 2 \qquad $F \cap \Bd$ }
\endcaption
\endinsert
\demo{\bf Proof} \enddemo
\medskip
This is a simple piece of geometry in the plane.
\qed
\bigskip
We now apply the Mass Distribution Principle (see Appendix 1).
It follows from Lemma 2 that for any probability measure $\mu$,
$$
\mu(\Bd) \leq \sum_{\text{ bdd \# of S}} \mu(S),
\eqno(4.1)
$$
where the sum is over at most $N$ asymptotic squares, and
$$
\mu(S) \leq \sum_{R \in F_k \ni R \cap S \neq \emptyset} \mu(R),
\eqno(4.2)$$
where the sum is over those rectangles $R$ in $F_k$ that
intersect the asymptotic square $S$.
Let $\mu=(\h,\h)^{\Bbb N}$ be the equidistributed Bernoulli measure,
i.e., each rectangle
$R \in F_k$ has mass $\mu(R) = (\h)^k$. Using (4.2),
we have a bound on the $\mu$-measure of $B_{\delta}$
of the form:
$$
\mu(S) \leq N(k) (\h)^k, \quad \text{ where } \, N(k) = \max
\#\{R \cap S\}.
$$
Hence, using (4.1), we get a bound on the $\mu$-measure of
asymptotic squares
$$
\mu(\Bd) \leq C \max \mu(S) \leq C N(k) (\h)^k.
$$
\medskip
\proclaim{\bf Lemma 3} If $ N(k)$ is uniformly
bounded in $k$, then $\dHF \geq \k2$.
\endproclaim
\demo{\bf Proof} \enddemo
Notice that $(\h)^k = \l 2^{ks}$, where $s= \k 2$. Hence
$$
\mu(\Bd) \leq C N(k) (\h)^k \leq C \, M \, \l 2^s \leq C \, M
\, 10^s \delta^s.
$$
The lemma follows immediately from the Mass Distribution
Principle. \quad \qed
\medskip
\proclaim{\it Remark} \endproclaim
We actually proved a stronger result: the lower pointwise dimension \cite{Y} \enspace $\underline{\dim}_{\mu}^L(x) \geq \k2$ for all $x \in F$.
\medskip
\proclaim{\bf Lemma 4} If $ N(k) \leq K \, (2 \l2)^{k-m}$, then
$\dHF \geq \b $.
\endproclaim
\demo{\bf Proof} \enddemo
We estimate
$$
\mu(\Bd) \leq C N(k) (\h)^k \leq C \, (2 \l 2)^{k-m} \, (\h)^k
= C K \l 2^k (2 \l2)^{-m}
$$
$$
= C K \l 2^k (2 \l2)^{-k \frac{\log \l 2}{\log \l 1}}
= C K \l 2^k \l2^{-k \frac{\log (2 \l 2)}{\log \l 1}}.
$$
If $s= \b$, then
$$
10^s \delta^s \geq \l 2^{ks} = \l 2^k \l2^{-k \frac{\log (2
\l2)}{\log \l1}}
.
$$
The lemma now follows immediately from the Mass Distribution
Principle in Appendix 1. \quad \qed
\medskip
\proclaim{\it Remark} \endproclaim
As in Lemma 3, we actually proved a stronger result: the lower pointwise dimension \enspace $\underline{\dim}_{\mu}^L(x) \geq \b$ for all $x \in F$.
\medskip
\medskip
Consider a rectangle $R = R_{(i_1,\cdots, i_k)}
\in F_k$ and the corresponding
asymptotic square $S$ that prolongates $R$.
\bigskip
To complete the proof of Proposition 5, we now obtain the bounds
for $ N(k) $ in Lemmas 3 and 4.
\medskip
\proclaim{\bf Lemma 5}The other rectangles $R' \in F_k $ such that
$R' \cap S \neq \emptyset$ have the codings of the form
$$
(\underbrace{i_1, \cdots, i_m}_{\text{fixed}}, j_{m+1}, \cdots,
j_k).
$$
\endproclaim
\demo {\bf Proof} \enddemo
For the rectangles $R$ and $R'$ to intersect in a common
asymptotic square $S$, their separation can be at most $\lambda_1^m$. \quad \qed
\bigskip
We want to estimate the number of rectangles in the intersection.
The left endpoint of $R_{(i_1, \cdots, i_k)}$ is
$\sum_{l=0}^k (\epsilon_0 + d \l 2^l) \l 2^l$,
and the left endpoint of $R' = R_{(i_1,
\cdots, i_m, j_{m+1}, \cdots, j_k)}$ is
$\sum_{l=0}^m (\epsilon_0 + d \, i_l) \l
2 ^l + \sum_{l=m+1}^k (\epsilon_0 + d \, j_l) \l 2^l$.
Hence for $R$ and $R'$ to
lie in the same asymptotic square $S$, we require that
$$
d \, \left | \sum_{l=m+1}^k (i_l-j_l) \l 2^l \, \right | < \l
2^k.
$$
\medskip
\proclaim {\bf Lemma 6}
Given any sequence $p \in \prod_0^\infty\{0,1\}$, let
$$
N^p(k) = \left \{(i_1, \cdots, i_k) \in \{0,1\}^k \, : \,
d \left | \, \sum_{l=M+1}^{k} (p_l - i_l) \, \l 2^l \,
\right| < \l 2^k \right
\}
$$
\endproclaim
\roster
\item If $0 < \l 2 < \h$, then there exits $M, N > 0$, such that
$N^p(k) \leq M $ for all sequences $p$ and $A_{\lambda_2}(k) \leq
M$.
\item If $\h < \l 2 < 1$, then $\l 2$ is a {\bf GE} number if and only
if for all sequences $p$, there exists $K > 0$ such that $N^p(k) \leq K (2 \l2)^{k-m}$.
\endroster
\demo{\bf Proof} \enddemo
The proof of part (1) follows from the estimate on $A_\beta(n)$
in Lemma 1, with the choice $\lambda_2 = \frac{1}{\beta}$. In
particular, we can let $x_{M,k} = \sum_{l=M+1}^k
\frac{p_l}{\beta^l}$.
Then by Lemma 2 there are no other expansions within
distance $\frac{1}{\beta^n}$.
To complete the proof, we need only repeat this observation where
we replace the choice of $x_{M,k}$ with translates by
$\lambda_2^k$.
This requires at most $[\frac{1}{d}] + 1$ such translates,
from which we deduce that $N^p(k) \le \left(
[\frac{1}{d}+1]\right) A_\beta(k)$
The proof of (2) is very similar. \quad \qed
\bigskip
\proclaim{\it Remarks} \endproclaim
\roster
\item We can give an easy
alternative proof of Proposition 5 (1). It follows from well
known properties of Hausdorff dimension \cite{F1} that
$\dim_H(\pi_2(F)) \leq \dim_H(F) \leq \dim_B(F)$. Since $0 \leq
\l2 \leq \frac12$, Corollary 1.1 and Proposition 3 imply that
$\pi_2(F)$ is a
uniform Cantor set constructed using $2^k$ disjoint intervals of
length $d \frac{\l 2^{k+1}}{1 - \l2}$ at step $k$. One easily
computes the Hausdorff dimension of $\pi_2(F) = \frac{\log 2}{\log
(\frac{1}{\l2})}$. The formula now easily follows from Proposition 3.
\item A heuristic explanation of why number theoretic properties of $\l 2$ determine whether $\dHF = \dBF$ is the following: If $\l 2$ is a {\bf GE} number, then the rectangles $R \in F_k$ are horizontally {\it well dispersed}, and at most a fixed percentage can intersect an asymptotic square. However, if $\l 2$ is a PV number, then the rectangles $R \in F_k$ tend to ``bunch up'' at various places, and a priori, {\it most} of the rectangles may intersect an asymptotic square.
\endroster
\bigskip
\head{\bf 5. Condition GE and Random Geometric Series}
\endhead
\bigskip
There is an intimate connection between the property {\bf GE} and a
famous classical problem about random geometric series or
infinitely convolved Bernoulli measures (ICBMs). Let
$\{X_n\}_{n=1}^{\infty}$ denote a family of independent and
identically distributed Bernoulli random variables such that
$P(\epsilon_k = 0) = P(\epsilon_k = 1) = \frac12$ for $k=1,2,
\cdots$. For $\beta$ a real number between 0 and 1, consider
the random variable
$$
S = \sum_{n=1}^{\infty} \epsilon_n \lambda^n.
$$
It can be shown that the compactly supported distribution $\mu_S$ of $S$ is of pure type, i.e., either totally singular or absolutely continuous
with respect to Lebesgue measure $\lambda$ on
$[0,\frac{\beta}{1 - \beta}]$. An important open problem is to
characterize all values of $\beta$ (Erdos numbers) such that
$\mu_S$ is absolutely continuous. We refer the reader to
\cite{AY} for the fascinating history of this still largely
unsolved problem along with an interesting application to
dynamical systems.
If $\beta$ satisfies condition {\bf GE}, then for $ x \in \Bbb R $,
$$
\mu^n_{\text{\s S}}[x, x+ \beta^n ] = \frac{A_{\beta}(n)}{2^n} \leq C
\beta^n = C \lambda[x, x+ \beta^n]
.$$
Hence, $\mu_{\text{\s S}} = \lim_{n \to \infty} \mu_{\text{\s S}}^n$ is absolutely
continuous with respect to Lebesgue measure, with a uniformly
bounded Radon-Nikodyn derivative.
Conversely, if the distribution $\mu_{\text{\s S}}$ is absolutely continuous with uniformly bounded density, then $\beta$ satisfies {\bf GE} (c.f. Appendix 3).
Amongst those explicitly known values such that the distribution
$S$ gives an absolutely continuous measure are reciprocals of
roots of 2 (see Appendix 2) and some more algebraic numbers by Garsia \cite{G}. Furthermore, {\bf GE} holds for almost all $\beta$ sufficiently close to 1 (c.f. Proposition A2 in Appendix 3).
\medskip
A related condition has appeared in the work of several authors
who attempt to compute the dimension of graphs of
Weierstrass-like functions \cite{B, BU}.
Actually, the example in Section 2, where the Hausdorff dimension
and box dimension differed, is an example where our sets are
graphs of such functions. The authors show that if the projection of a
certain natural measure on the graph is absolutely continuous,
then the box dimension of the graph coincides with the Hausdorff
dimension.
\medskip
\bigskip
\head {\bf 6. Generalizations} \endhead
\bigskip
Up to this point we have chosen to concentrate on our ``model
problem''.
However, it is apparent that the method is somewhat more general.
To illustrate this, we shall mention a few of the possible
generalizations with indications of their proofs.
\bigskip
\proclaim{Proposition 6}
Replace the two similar boxes by two (or more boxes) with longest sides
$\lambda_{2,1},\lambda_{2,2} \leq \frac12 $. If one avoids degenerate configurations where the projection of the smaller rectangle onto the $\l 2$ direction is symmetrically contained in the projection of the larger rectangle, then the Hausdorff dimension of the limit set $F$ coincides with the box dimension of $F$ and is equal to the unique value $\delta$ such that $\sum_i \lambda_{2,i} ^\delta = 1.$
\endproclaim
The proofs are exactly the same, except that the measure
used in the Mass Distribution Principle is now the Bernoulli
measure $(\exp(-\lambda_{2,1} \delta)/S, \exp(-\lambda_{2,2} \delta))/S)$
where $S = \exp(-\lambda_{2,1} \delta) + \exp(-\lambda_{2,2} \delta)$.
\medskip
A further generalization of this is the following Proposition:
\medskip
\proclaim{Proposition 7}
Assume that $B_1,B_2$ are two rectangles and assume that
the sides are each less than $\frac{1}{2}$ in length. Assume
that
we have contractions (with these images) of the form
$A_i(x,y) = (f_i(x), \lambda_1 y)$, where $f_1, f_2$ are the
inverse branches associated to a simple Markov map $f$ on $\pi_{\lambda_1}F$,
where $|{f_1}'|, |{f_2}'| \leq \frac12$. Then $ \dBF = \dHF$ and
they equal the value $\delta$ characterized by $P(- \delta \log
|f'|) = 0$, where $P$ denotes the thermodynamic Pressure [W]. \endproclaim
\bigskip
\subhead {\bf Generalizations to n-dimensions} \endsubhead
\bigskip
In this section, we compute the box dimension of two
n-dimensional boxes in the n-dimensional cube. Let $\l 1 \geq \l 2 \geq \ldots \geq \l n$ denote the $n$ contraction coefficients.
\medskip
\demo{Remark}
A variation on this construction is where for any
fixed string of symbols of length $d$, we delete the
rectangles in the construction corresponding to this word
(this means that our set $F$ is now the realization of a
subshift of finite type rather than a full shift). It is
easy to check that the formula for $\dBF$ and $\dHF$ in Sections 3 and 4 are simply modified so that the $\log 2$ occurring in the
numerator is replaced by $\log \Lambda$ where $\Lambda$ is the maximal
eigenvalue for the associated transition matrix.\enddemo
\medskip
{\bf Case I:}
Assume that $\lambda_1 \geq \lambda_2 \geq \frac{1}{2}$ and that
$\lambda_3 \leq \lambda_2$ has no constraints. Let $\epsilon = \lambda_3^n$. We want to cover the limit set by
cubes with sides $\epsilon$.
If we go down to the n-th level $F_n$ of rectangles, we first
observe
that the assumption on $\lambda_1, \lambda_2$ implies that the
projections of
$F$ onto the corresponding axes actually contain intervals.
Thus, we can
estimate the number of cubes required to cover {\it each} of the
rectangles by
$$N = \frac{\lambda_1^n}{\lambda_3^n}
\frac{\lambda_2^n}{\lambda_3^n}.
$$
Thus, the box dimension is
$$
\dBF = \frac{ \log \left(\frac{\lambda_1
\lambda_2}{\lambda_3^2}\right)}{
\log \lambda_3}.
$$
{\bf Case II:} Assume $\lambda_1 \geq \frac{1}{2} \lambda_2$.
The difference now is that whilest the projection into the first
axis still contains an interval, the projection into the second axis is a
Cantor set.
If we let $\epsilon = \lambda_3^n$, then again the number of
boxes
of this size needed to cover the set $F$ is estimated by
$$
2^n \frac{\lambda_1^n}{\lambda_3^n} 2^M,
$$
where $M = M(n)$ is the value such that $\lambda_2^{n+M} =
\lambda_3^n$.
The appearance of $M$ is because the projection of each
rectangle
in $F_n$ into the second axis corresponds to an interval of size
$\lambda_2^n$, and the projection of the Cantor set can be
efficiently covered by
$2^{M}$ intervals of size $\lambda_3^n$, where $M=n\frac{\log(
\lambda_3/\lambda_2)}{\log\lambda_2} = n M'$ satisfies the
above condition.
The box dimension is then given by
$$
\dBF = \frac{\log 2}{\log \lambda_3} + \frac{\log
\left(\lambda_1/ \lambda_3\right)}{\log\lambda_3} +
\frac{\log2}{\log\lambda_3}.
$$
{\bf Case III:}. Assume $\frac{1}{2} \geq \lambda_1 \geq
\lambda_2$.
Following the previous reasoning, we can expect to efficiently
cover the
limit set $F$ by
$N = 2^n 2^M 2^Q$ cubes with sides $\lambda_3^n$, where $M$ is as
in
case II, and Q is a similar value, except we replace $\lambda_2$
by $\lambda_1$, i.e., $Q = n
\frac{\log(\lambda_3/\lambda_2)}{\log\lambda_2} = Q'n$. The
formula for the box dimension then becomes
$$
\dBF = \frac{\log2}{\log\lambda_3}
+ M' \frac{\log2}{\log\lambda_3} + Q'
\frac{\log2}{\log\lambda_3}.
$$
\medskip
{\bf General Case:} Let us assume that we now have two boxes in
a $k$-dimensional cube, with contraction coefficients
$\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_r \geq
\frac{1}{2} \geq \lambda_{r+1} \geq \ldots \geq \lambda_k$.
Define $m_i =
\frac{\log(\lambda_k/\lambda_i)}{\log\lambda_i}$.
Then the box dimension of the limit set $F$ is equal to
$$
\dBF = \frac{\log(\lambda_1/\lambda_k)}{\log\lambda_k}
+ \ldots + \frac{\log{(\lambda_r/ \lambda_k)}}{\log \lambda_k}
+ \frac{\log 2}{\log \lambda_k}\left(m_{r+1} \ldots m_k +
1\right).
$$
The Hausdorff dimension can be computed similarly provided the
projections onto the slower axes are disjoint or satisfy special
assumptions related to those before.
\bigskip
\head {\bf 7. The Hausdorff Dimension of Linear Horseshoes }
\endhead
\bigskip
The Cantor sets that arise in the model problem are closely related to an
important class of diffeomorphisms.
Specifically, it is easy to construct a Smale horseshoe diffeomorphism (with a two dimensional stable manifold and a one-dimensional unstable manifold) for which the
associated basic set $\Lambda$ is a product of the limit set $F$ constructed in the model problem and a uniform Cantor set $E_p$ in the line. As usual, Smale horseshoes are constructed by specifying a box, $R$, and its image under the diffeomorphism. To arrange that the basic set is of the form described above,
we require that the diffeomorphism $f$ should be linear on $R \cap f^{-1}R$.
This is illustrated in the figure 3.
\midinsert
\vspace{4truein}
\botcaption{Figure 3 \quad linear horseshoe}
\endcaption
\endinsert
\medskip
Since $\Lambda = F \times E_p$, where $E_p$ is a uniform Cantor set, it follows \cite{F2} that $\dim_{H} \Lambda=\dim_H F + \dim_H E_p$ and
$\dim_{B} \Lambda =\dim_B F + \dim_B E_p$.
We wish to construct a horseshoe based on the Przytycki-Urbanski example mentioned in Section 2, where the dimensions of the limit set do not coincide. The PU example consists of two rectangles of width the reciprocal of the Golden mean (which is easily seen to be a PV number) and height $\frac12$, that are flush against the top left and bottom right corners of the unit square, respectively. Unfortunately, the rectangles in the PU example are not disjoint, and hence one can not effect the horseshoe construction.
We wish to slightly perturb the PU example by shrinking the heights of the rectangles to $\lambda_1 < \frac{1}{2}$, keeping the widths the recipricol of the golden mean, and keeping the new rectangles flush against the top left and bottom right corners of the unit square (respectively). See figure 4.
\midinsert
\vspace{4truein}
\botcaption{Figure 4 \quad PU example \qquad \qquad \qquad modified PU example}
\endcaption
\endinsert
\medskip
We wish to show that the dimensions of the limit set do not coincide for these perturbed examples. It immediately follows from our explicit formula for the box dimension (Proposition 2) that the box dimension of the limit sets of our modified PU examples changes smoothly as a function of $\lambda_1$.
On the other hand, it is straightforward to see that the
Hausdorff dimension of the limit sets for the modified PU examples is no greater than the Hausdorff dimension of the limit set for the PU example. It is not the case that there is strict containment of the two limit sets. The natural map from the limit set of the PU example to the limit set of the modified PU example is clearly of the form $f(x,y) = (x, g(y))$ where $g(y)$ maps vertical intervals of length $\frac{1}{2^n}$ to (vertical) intervals of length $\lambda_1^n$, hence the map $g$ is a contraction and therefore Lipschitz.
It follows that there exists an open interval around $\lambda_1=\frac12$ such that for $\lambda_1$ in this interval, the dimensions of the limit sets for the modified PU examples do not coincide. {\it This furnishes a one-paramenter family of horseshoes whose dimensions do not coincide.}
Using obvious modifications of these constructions, we can arrange similar realizations, as basic sets, of limit sets generated by any number of affine maps, {\it provided that the image rectangles are disjoint}.
We consider below two different types of cases where the dimensions
can be explicitly computed, with interesting conclusions.
\newpage
{\bf Example 1: Horseshoes with different dimensions}
\smallskip
Choose a finite number $n_1$, say, of disjoint sub-intervals
of the same length $\lambda_1$ in the interval
$[0,1]$. Next choose a finite number, $n_2$, of
disjoint sub-intervals in $[0,1]$
with length $\lambda_1$ strictly smaller than $\lambda_2$. Consider the
rectangles in the unit square corresponding to the products of these
two sets of intervals.
Choose from this family a sub-family
of $k$ rectangles.
In particular, this yields a finite number
of disjoint rectangles such that
\roster
\item
Each rectangle has height $\lambda_1$ and width $\lambda_2$.
\item Disjoint rectangles either have the same projection or disjoint projections onto the horizontal direction.
\item Disjoint rectangles either have the same projection
or disjoint projections onto the vertical direction.
\endroster
See figure 5.
A Sierpinski-like limit set $F$ is generated by the $k$ affine contractions
associated to this sub-family of rectangles. This construction is modeled by a full shift
on $k$ symbols.
This is a
slight generalization of the familiar Bedford-McMullen problem \cite{B, M},
in as much as the vertices of the rectangles need not have
rational coordinates.
The Hausdorff and box dimension of the limit set $F$ of
these (and more complicated)
systems have been studied by Lalley-Gatzouras \cite{LG},
and Kenyon-Peres \cite{KP}.
The Hausdorff and box dimensions of these systems are known to
be
$$\dHF = \frac{1}{\log \lambda_2}
\log
\left( \sum_{i=1}^{n_2} \left( \log k_i \right)^\frac{\log \lambda_1}
{\log \lambda_2} \right) $$
where $n_i$ is the number of rectangles from the sub-family
left in the $i$th row, and
$$\dBF = -\frac{1}{n_2}
\left(
\frac{\log n_1}{ \log \lambda_2}
+ \frac{\log (k/n_1) }{\log \lambda_1}
\right).
$$
We want to make our choices of $\lambda_1, \lambda_2 > 0$ and configuration
of rectangles such that $\dHF \neq \dBF$. Consider,
the rectangles $R_1 = [0, \frac{1}{2}] \times [0, \alpha]$,
$R_2 = [ 0, \frac{1}{2}] \times [1 - \alpha, 1]$, and
$R_3 = [\frac{1}{2}, 1] \times [\frac{1 - \alpha}{2}, \frac{1 + \alpha}{2}]$,
where $0 < \alpha < \frac{1}{3}$. Clearly, for typical values
of $\alpha$ we will have that $\dHF \neq \dBF$.
\midinsert
\vspace{4truein}
\botcaption{Figure 5 \qquad Sierpinski Carpet}
\endcaption
\endinsert
\medskip
If we consider an associated horseshoe $\Lambda$
associated to the family of affine maps generating $F$, then we
see that
$$
\dim_{H} \Lambda =\dim_H F + \dim_H E_p
\neq \dim_B F + \dim_B E_p = \dim_{B} \Lambda,
$$
i.e., the associated horseshoe limit set $\Lambda$ has Hausdorff dimension
different to the box dimension.
\bigskip
{\bf Example 2: Dependence of dimension on configuration}
\smallskip
Choose $n_1$ disjoint sub-intervals
of the same length $\lambda_1$ in the interval
$[0,1]$. Next choose $n_2$ disjoint sub-intervals in $[0,1]$
with length $\lambda_1$ strictly smaller than $\lambda_2$. Consider the
rectangles in the unit square corresponding to the products of these
two sets of intervals. We associate to these rectangles linear contractions, and denote the corresponding limit set by $\Lambda_1$.
It is easy to see that the Hausdorff
dimension and box dimension of the {\it projection}
of $\Lambda$ onto the horizontal axis coincide and equal
$\frac{\log n_1}{\log\frac{1}{\lambda_1}}$. Similarly,
we see that the Hausdorff
dimension and box dimension of the {\it projection}
of $\Lambda$ onto the vertical axis coincide and equal
$\frac{\log n_2}{\log\frac{1}{\lambda_2}}$. It is then
easy to see that since the space $\Lambda_1$ is a product
of these projections then
$$\dim_H\Lambda_1 = \dim_B\Lambda_1 = \frac{\log n_1}{\log\frac{1}{\lambda_1}}
+ \frac{\log n_2}{\log\frac{1}{\lambda_2}}.$$
Assume that the product $N = n_1 \cdot n_2$
has a different factorization $N = m_1 \cdot m_2$. We can repeat the
above construction of an array of boxes
using $m_1$ disjoint sub-intervals
of the same length $\lambda_1$ in the interval
$[0,1]$ and $m_2$ disjoint sub-intervals in $[0,1]$ of
equal length $\lambda_2$. We shall denote by $\Lambda_2$
the corresponding limit set, and by a similar reasoning we
see that
$$\dim_H\Lambda_2 = \dim_B\Lambda_2 = \frac{\log m_1}{\log\frac{1}{\lambda_1}}
+ \frac{\log m_2}{\log\frac{1}{\lambda_2}}.$$
If we take the specific choices $n_1 = 3, n_2 = 4$
and $m_1 = 6, m_2=2$, then for typical values of
$\lambda_1$ and $\lambda_2$ we have that
$$
\dim_H\Lambda_1 = \dim_B\Lambda_1
\neq \dim_H\Lambda_2 = \dim_B\Lambda_2.
$$
\midinsert
\vspace{4truein}
\botcaption{Figure 6 \qquad Different configurations of 12 rectangles}
\endcaption
\endinsert
\medskip
\head {\bf Appendix 1: Facts About Dimension Theory } \endhead
\bigskip
\proclaim{\bf Definition} Let $U \subset \Bbb R^n$. The {\it diameter}
of U is defined as $|U| = \sup\{|x-y|:x,y \in U\}$. If $\{U_i\}$ is a countable collection of sets of diameter at most $\delta$ that cover Z, i.e., $Z \subset \cup_i U_i$ with $0 < |U_i| \leq \delta$ for each i, we say that $\{U_i\}$ is a $ \delta$-cover of Z.
\endproclaim
Suppose that $Z \subset \Bbb R^n$ and $s \geq 0$. For any $s > 0$, define
$$
\text{m}_{\text{H}}(s, Z) =\lim_{\delta \to 0} \inf_{\{U_i\}} \left\{ \sum_i |U_i|^{s}: \{U_i\} \text{ is a } \delta\text{-cover of } Z \right\}.
$$
We call $\text{m}_{\text{H}}(s, Z) $ the the {\it s-dimensional Hausdorff measure of Z}. There exists a unique critical value of $s$ at which $\text{m}_{\text{H}}(s, Z)$ jumps
from $\infty$ to $0$. This critical value is called the {\it Hausdorff dimension} of $Z$ and is written $\text{dim}_H(Z)$. If $s=\text{dim}_H(Z)$, then
$\text{m}_{\text{H}}(s,Z)$ may be $0, \infty$, or finite. Hence $\text{dim}_H(Z) = \sup\{s: \text{m}_{\text{H}}(s, Z) = \infty\} = \inf\{s: \text{m}_{\text{H}}(s,Z) = 0\}$.
\medskip
\proclaim{Definition}
Let $N_{\delta}(Z)$ denote the minimum number of sets of diameter precisely $\delta$ needed to cover the set $Z$. We define the Upper and Lower box Dimension Measures of $Z$ by
$$
\underline{\dim}_B Z = \liminf_{\delta \to 0} \frac{N_{\delta}(Z)}{\log(\frac{1}{\delta})},
$$
$$
\overline{\dim}_B Z = \limsup_{\delta \to 0} \frac{N_{\delta}(Z)}{\log(\frac{1}{\delta})}.
$$
If $ \underline{\dim}_B Z = \overline{\dim}_B Z $, denote the common value by $ {\dim}_B Z $.
\endproclaim
\medskip
\proclaim{\it Remark} \endproclaim
It is easy to see that $\dim_H(Z) \leq \underline \dim_B(Z) \leq \overline \dim_B(Z)$. The usual method of obtaining an upper bound for $\dim_H(F)$ is to obtain an upper bound for $\underline \dim_B(Z)$.
\medskip
The following proposition is extremely useful for obtaining a lower bound for the Hausdorff dimension of a set:
\medskip
\proclaim{\bf Mass Distribution Principle (Frostman)} \cite{F1} Let $\mu$ be a probability measure of $Z$ and suppose that for some $s$ there are numbers $c > 0$ and $\delta > 0$ such that $\mu(U) \leq c |U|^s$ for all sets $U$ with $|U| \leq \delta$. Then $\text{m}_{\text{H}}(s,Z) \geq \mu(F)/c$ and $s \leq \text{dim}_H(Z)$.
\endproclaim
\medskip
The following simple lemma shows that in computing the
box dimension of a set we need only consider the minimum
number $N(\delta)$ of covering boxes where $\delta$
runs through a geometric sequence converging to zero.
\medskip
\proclaim {\bf Lemma A1 \cite{F1}} Fix $0 < c < 1$. Then $\dBF = \lim_{\delta \rightarrow 0}\frac{\log
N(\delta)}{\log(\frac{1}{\delta})} = \lim_{k \rightarrow \infty}
\frac{\log N(c^k)}{\log (c^k)}.$
\endproclaim
\bigskip
\head {\bf Appendix 2: Condition {\bf GE} for Recipricals of Roots of 2} \endhead
\bigskip
We give some easy examples of numbers $1 < \beta \leq 2$
that show the {\bf GE} condition can sometimes be checked without resorting to the more complicated analysis on the
projection of the measure.
\roster
\item
In the special case where $\beta=2$, the {\bf GE} condition is easy to
check, since this is a question about diadic expansions.
\item If $\beta = 2^{\frac{1}{2}}$ and $n$ is even,
then we can write
$$\eqalign{
\sum_{r=0}^{n-1} \frac{i_r}{\beta^r} &=
\sum_{r=0}^{n/2-1} \frac{i_{2r}}{2^r} +
\frac{1}{\beta}
\left( \sum_{r=0}^{n/2-1}
\frac{i_{2r+1}}{2^r} \right) \cr
&= A + \frac{1}{\beta} B.}
$$
For this expression to lie in an interval
$[x, x + \frac{1}{\beta^n})$, we have at most
$2^{n/2}$ choices for the value $A$ and a bounded
(one) number of choices for $B$. This suggests
that we have a bound for the above expression of the
general form
$$C 2^{n/2} = C \left(\frac{2}{\beta}\right)^n$$
as required for the {\bf GE} condition.
A similar argument works for any root of $2$.
\endroster
\bigskip
\head {\bf Appendix 3: Condition {\bf GE} for Sets of Positive
Measure} \endhead
\bigskip
In this appendix, we explain the relationship between the
condition {\bf KS} in \cite{KS, p.198}, our condition {\bf GE} and the
work of Erdos in \cite{E}.
\medskip
\proclaim{Definition}
The number $\beta$ satisfies condition {\bf KS} if there exists a
constant
$C > 0$ such that the number $N_\beta(n)$ of solutions to
$\left |\pi_n(i_0, \ldots , i_{n-1}) - \pi_n(i_0, \ldots, j_{n-1}) \right |
\leq \frac{1}{\beta^n}$ satisfies $N_\beta(n) \leq C (4\beta)^n$ \,
\cite{KS, p.198}.
\endproclaim
\medskip
\demo{Remark}
It is easy to see that the condition {\bf GE} implies condition
{\bf KS}.
\enddemo
The usefulness of this assumption is shown by the following results.
\medskip
\proclaim{Proposition A2}
\roster
\item
Condition {\bf KS} holds if and only if the Fourier transform
$\gamma(t) = \int e^{2 \pi ix t} d\mu_{\text{\s S}}(x)$ of the measure
$\mu_{\text{\s S}}$ is in $L^2(\Bbb R)$ \cite{KS, pp.197--198}.
\item
For each positive integer $m \geq 1$ there exists
$\delta > 0$ such that for almost all $\beta \in [ 1 -\delta, 1)$ (in the sense of Lebesgue) we have $\gamma(t) = O(\frac{1}{|t|^m})$ as $|t| \to +\infty$.
When $m=1$ we have that $\gamma \in L^2( \Bbb R)$ \, \cite{E, p.186}.
\endroster
\endproclaim
\medskip
We can conclude that for almost all values of $\beta$
sufficiently close to $1$ the condition {\bf KS}
holds. We would like to
reach the same conclusion for the stronger condition {\bf GE}.
We begin with the following lemma.
\medskip
\proclaim{\bf Lemma A2 }
There exists $\delta > 0$ such that for almost all $\beta \in
[1 - \delta, 1)$ the distribution $\mu_{\text{\s S}}$ is absolutely
continuous with a uniformly bounded continuous density.
\endproclaim
\medskip
\demo{\bf Proof} \enddemo
By part (2) of Proposition A2, we can
choose $\delta$ such that the Fourier transform $\gamma(t)$
is in $L^1(\Bbb R) $ for almost all $\beta \in [1-\delta, 1)$. Fix
such a value of $\beta$.
It immediately follows from the Fourier Inversion Formula and the Riemann Lebesgue Lemma that if the Fourier transform $\gamma(t)$ of the distribution $\mu_{\text{\s S}}$ is in $L^1( \Bbb R )$ then the distribution $\mu_{\text{\s S}}$ is absolutely continuous with a uniformly bounded continuous density $h$ defined by $h(x) = \int e^{- 2\pi i x t} \gamma(t) dt$. \quad \qed
\medskip
\proclaim{\bf Proposition A3}
There exists $\delta > 0$ such that for almost all
$\beta \in [1 - \delta, 1)$ the condition {\bf GE} holds.
\endproclaim
\demo{\bf Proof} \enddemo
Choose $\delta$ as in the previous lemma.
It remains to show that the density $h$ being uniformly bounded
implies that the condition {\bf GE} holds. William Parry showed us a derivation of this using analysis of Fat Baker's transformations in \cite{AY}. We shall present two additional proofs. The first proof is based on the study of transfer operators, which could be a useful new tool in the study of these number theoretic problems. The second proof, which is considerably more elementary, was supplied to us by Yuval Peres.
We first present a proof based on the transfer operator
$$
Lh(x) = \frac{1}{2} \bigl ( h ( x \beta ) + h( (x + 1) \beta ) \bigr)
$$
defined on functions of bounded variation on $\Bbb R $, as described in
\cite{KS, p.200}.
The first observation is that for any such $h$ we can expand
$$
L^nh(x) = \sum_{(i_0, \ldots, i_{n-1})}
\frac{1}{2^n} h \left (\sum_{i_0, \ldots, i_{n-1}} i_r \beta^r +x\beta^n \right)
$$
and then observe the identity $L^n \chi_{[a, a+\beta^n]}(0) =
\frac{N_\beta(n)}{2^n}$.
It is clear that the asymptotics of the function $N_\beta(n)$
are determined by the spectral properties of the operator $L$
on the space of functions of bounded variation.
\roster
\item
The operator has a maximal eigenvalue equal to unity, with an
eigenprojection corresponding to the measure $\mu_{\text{\s S}}$, i.e., $L^nk
\to \int k d\mu_{\text{\s S}}$.
\item
The remainder of the spectrum is contained within the
disc about zero of radius $\beta$. By \cite{R} there can be only {\it isolated} eigenvalues $\alpha$ of modulus greater than $\beta$. However, since the operator
also preserves the space of $C^1$ functions, the associated
eigenfunctions $Lk = \alpha k$ must be $C^1$. By differentiating, we see that $(Lk)' = \beta L(k') = \alpha k$,
and thus $\frac{\alpha}{\beta}$ is also an eigenvalue, except
where $\alpha = 1$ and $k$ is the constant function. However,
since the spectral radius of $L$ is unity, we require that
$\frac{|\alpha|}{\beta}\leq 1$, which completes the proof.
\endroster
Therefore, we can write $L^nk = \int k d\mu_{\text{\s S}} + U^{(n)}k$,
where $U^{(n)}$ is a linear operator with
$\limsup_{n \to +\infty} ||U^{(n)}||^{\frac{1}{n}} \leq \beta$.
We can write
$$ \mu_{\text{\s S}}([a, a+\beta^n]) + U^{(n)} \chi_{[a, a+\beta^n]}(0) =
L^n \chi_{[a, a+\beta^n]}(0) = \frac{N_\beta(n)}{2^n}.$$
If we observe that the norms of $ \chi_{[a, a+\beta^n]}$
are uniformly bounded (in the space of functions of bounded
variation), we see that
$U^{(n)}\chi_{[a, a+\beta^n]}(0) = O\left( (\beta')^n\right)$ for any
$\beta' > \beta$. It only remains to use the fact that the
density $h$ is uniformly bounded to see that
$\limsup_{n \to +\infty} \frac{\mu([a, a+\beta^n])}{\beta^n}$ is
finite. This completes the first proof. \quad \qed
\bigskip
The second proof is a proof by contradiction. If $\beta$ is not {\bf GE}, then there exist intervals $J_n$ such that $\lambda(J_n) = \beta^n$ and $\text{Card} \left \{ (i_0, \ldots, i_{n-1}) : \sum_{r=0}^{n-1} i_r \beta^r \in J_n \right \} \geq n (2 \beta)^n $. Since $\sum_{k=N}^{\infty} x^k = \frac{x^n}{1 - x}$, we may symmetrically enlarge the intervals $J_n$ to intervals $I_n \supset J_n, \, \lambda(I_n) = \beta^n + \frac{\beta^n}{1 - \beta} = l \beta^n$, such that if $\sum_{r=0}^{n-1} i_k \beta^k \in J_n$, then $\sum_{r=0}^{\infty} i_k \beta^r \in I_n$. We obtain that $\text{Card} \left \{ (i_0, \ldots, i_n, \ldots, i_{n+p}) : \sum_{r=0}^{n+p} i_r \beta^r \in I_n \right \} \geq 2^p n (2 \beta)^n $. This implies that
$$
\frac{\mu_S(I_n)}{\lambda(I_n)} = \frac{\lim_{p \to \infty} \dsize \frac{ \text{Card} \left \{ (i_0, \ldots, i_n, \ldots, i_{n+p}) : \sum_{r=0}^{n+p} i_r \beta^r \in I_n \right \} }{2^{n+p}} }{\lambda(I_n)} \geq \frac{n \beta^n}{l \beta^n} = \frac nl
$$
which is unbounded as $n \to \infty$. It follows that the Radon-Nikodyn derivative $\frac{d \mu_S}{d \lambda}$ is not uniformly bounded. \quad \qed
\bigskip
We conjecture that (Lebesgue) almost all numbers $\frac12 < \beta < 1$ satisfy condition {\bf GE}. By Proposition 5(2), this conjecture would imply that for almost all $\frac12 < \l2 < 1$, the Hausdorff dimension of the limit set of the model similarity process coincides with the box dimension and is given by the formula in Proposition 5(1). Moreover, by Proposition 5(1), this conjecture would imply that for almost all $0 < \l2 < 1$, the Hausdorff dimension coincides with the box dimension.
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\enddocument