Comparing fractal dimensions

To prove the theorem we need to develop some useful notation associated with an IFS $\left\{f_i\right\}_{i=1}^m$ with ratio list $\left\{r_i\right\}_{i=1}^m,$ invariant set $E,$ and similarity dimension $s.$ A string with symbols chosen from $\{1,\ldots ,m\}$ is simply a finite or infinite sequence with values in $\{1,\ldots ,m\}.$ A finite string $\alpha$ will be denoted $\alpha = i_1\cdots i_k.$ Finite strings can be concatenated to form longer strings. Thus if $\alpha = i_1\cdots i_k$ and $\beta = j_1\cdots j_l,$ then $\alpha \beta = i_1\cdots i_kj_1\cdots j_l.$ Every string $\alpha = i_1\cdots i_k$ has one parent $\alpha ^- = i_1\cdots i_{k-1}.$ Given a positive integer $k,$ let $J_k$ denote the set of all strings of length $k$ with values chosen from the set $\{1,\ldots ,m\}.$ Let $J_*= \underset{k=1}{\overset{\infty }{\cup }}J_k$ denote the set of all such finite strings. Given $\alpha = i_1\cdots i_k\in J_k,$ let $f_{\alpha } = f_{i_1}\circ \cdots \circ f_{i_k}$and $r_{\alpha } = r_{i_1}\text{$\cdots $r}_{i_k}.$ Then $J_k$ induces a $k^{\text{th}}$ level decomposition of $E$ given by

\begin{equation*} E = \underset{\alpha \in J_k}{\bigcup }f_{\alpha }(E). \end{equation*}

The sets $J_k$ are examples of cross-sections of $J_*,$ but $J_k$ does not necessarily induce a useful decomposition of $J_*$ with respect to box-counting dimension as the sizes of the sets $f_{\alpha }(E)$ may vary greatly. A more general type of cross-section may be defined as follows. Let $J_{\omega }$ denote the set of all infinite strings with symbols chosen from $\{1,\ldots ,m\}.$ Given $\sigma = i_1i_2\cdots \in J_{\omega },$ let $\omega |_k$ denote the element of $J_k$ whose symbols are the first $k$ symbols of $\sigma .$ Given $\alpha \in J_k,$ let $[\alpha ] = \left\{\sigma \in J_{\omega }:\sigma |_k = \alpha \right\}$ denote the set of all infinite strings whose first $k$ symbols agree with $\alpha .$ A cross-section of $J_*$ is a finite set $J\subset J_*$ such that $\cup _{\alpha \in J}[\alpha ] = J_*$ with this union disjoint. Note that a cross-section $J$ of $J_*$ defines a decomposition of $E$ given by

\begin{equation*} E = \underset{\alpha \in J}{\bigcup }f_{\alpha }(E). \end{equation*}

Furthermore, if $\alpha \in J_*,$ then

\begin{equation*} \sum _{j=1}^m r_{\alpha j}^s= r_{\alpha }^s\sum _{j=1}^m r_j^s = r_{\alpha }^s. \end{equation*}

It follows by induction that $\sum _{\alpha \in J} r_{\alpha }^s = 1$ for any cross-section $J$ of $J_*.$

Now for $\varepsilon >0,$ define $J_{\varepsilon }$ by

\begin{equation*} J_{\varepsilon } = \left\{\alpha \in J_*:r_{\alpha } \text{diam}(E) \leq \varepsilon < r_{\alpha ^-} \text{diam}(E)\right\}. \end{equation*}

Note that if $\sigma \in J_{\omega },$ then $\left(r_{\sigma |_k}\right)_k$ is a sequence of positive numbers which is strictly decreasing to zero. Thus there is a unique value of $k$ with $\sigma |_k \in J_{\varepsilon }$ so $J_{\varepsilon }$ defines a cross-section of $J_*.$ Furthermore, the decomposition of $E$ induced by $J_{\varepsilon }$ has diameters comparable to $\varepsilon .$ In particular, if $r = \min \left\{r_i\right\}_{i=1}^m,$ then

\begin{equation*} r \varepsilon < \text{diam}\left(f_{\alpha }(E)\right) \leq \varepsilon \end{equation*}

for all $\alpha \in J_{\varepsilon }.$

Finally, $\#(J)$ will denote the number of elements in a cross-section. We are now ready to prove our theorem on the comparison of dimensions.

Our strategy is to find constants $M_1$ and $M_2$ so that

\begin{equation} s +\frac{\log M_1}{\log 1/\varepsilon } \leq \frac{\log P_{\varepsilon }(E)}{\log 1/\varepsilon } \leq \frac{\log C_{\varepsilon }(E)}{\log 1/\varepsilon } \leq s+\frac{\log M_2}{\log 1/\varepsilon },\label{eq-dimCompInequality}\tag{5.4.1} \end{equation}

for then the squeeze theorem applies. Note that the second inequality follows from the fact that $P_{\varepsilon }(E)\leq C_{\varepsilon }(E)$ which was established in the proof of our lemma comparing packings and coverings.

The last inequality in (5.4.1) is equivalent to $\varepsilon ^s C_{\varepsilon }(E) \leq M_2\text{.}$ We will show that this is true for $M_2=r^{-s}\text{diam}(E)^s\text{,}$ where $r = \min \left\{r_i\right\}_{i=1}^m$ as above. Recall that $C_{\varepsilon }(E) \leq \#\left(J_{\varepsilon }\right)$ and $r \varepsilon <r_{\alpha }\text{diam}(E)\text{.}$ Thus

\begin{equation*} (r \varepsilon )^s C_{\varepsilon }(E) \leq \sum _{\alpha \in J_{\varepsilon }} \left(r_{\alpha } \text{diam}(E)\right){}^s = \text{diam}(E)^s \end{equation*}

and $\varepsilon ^s C_{\varepsilon }(E) \leq r^{-s} \text{diam}(E)^s.$

Proof of the first inequality of (5.4.1) is somewhat more difficult and will require the use of the strong open set condition. We will show that there is a constant $M_1>0$ such that $(r \varepsilon )^sP_{r \varepsilon }(E) \geq M_1\text{.}$ A simple change of variables shows that this is equivalent to $\varepsilon ^sP_{\varepsilon }(E) \geq M_1\text{,}$ which is in turn equivalent to the first inequality of equation (5.4.1). Let $U$ be the open set specified by the strong open set condition and let $x \in U \cap E.$ Choose $\delta >0$ such that $\overline{B_{\delta }(x)} \subset U.$ Then any cross-section $J$ of $J_*$ induces a packing $\left\{f_{\alpha }\left(\overline{B_{\delta }(x)}\right):\alpha \in J\right\}$ of $E\text{.}$ Now let

\begin{equation*} J_{\varepsilon } = \left\{\alpha \in J_*:r_{\alpha } \delta \leq \varepsilon < r_{\alpha ^-} \delta \right\}. \end{equation*}

Then $r \varepsilon < r_{\alpha } \delta \leq \varepsilon$ for all $\alpha \in J_{\varepsilon }.$ Thus $P_{r \varepsilon }(E) \geq \#\left(J_{\varepsilon }\right)$ and

\begin{equation*} \varepsilon ^s P_{\text{r$\varepsilon $}}(E) \geq \sum _{\alpha \in J_{\varepsilon }} r_{\alpha }^s\delta ^s = \delta ^s. \end{equation*}

Multiplying through by $r^s$ we obtain $(r \varepsilon )^sP_{r \varepsilon }(E) \geq (r \delta )^s \equiv M_1.$

Chaos and Fractals

Section 5.4 Comparing fractal dimensions

Definition 5.4.1. Strong open set condition.

Theorem 5.4.2.

Proof.