Consider the iterated function system defined by these two affine contractions:
$f_1:\mathbb{R}^2 \to \mathbb{R}^2$, $f_2:\mathbb{R}^2 \to \mathbb{R}^2$,
$f_1 \left( x,y\right) = \left( \dfrac{x}{2},\dfrac{y}{4} \right) + \left(0, \dfrac34 \right)$
$f_1 \left( x,y\right) = \left( \dfrac{x}{2},\dfrac{y}{4} \right) + \left(\dfrac14, 0 \right)$
and the generator set is the unit square $E_0 = [0,1]\times[0,1]$.
I made a drawing of the first iterations:
I want to calculate the Hausdorff dimension in order to understand better the Falconer Theorem of the Dimension of Self-Affine Sets:
Theorem 9.12 (Fractal Geometry, Falconer, pg.154). Let $T_1, \dots, T_m$ be linear contractions and let $y_1,\dots,y_m \in \mathbb{R}^n$ be vectors. If $F$ is the self-affine set satisfying $F = \bigcup_{i=1}^m (T_i(F)+y_i)$, then $\dim_H F \le \dim_B F \le d(T_1,\dots,T_m)$. If $|T_i(x) - T_i(y)| \le c|x-y|$ for all $i$ where $0<c<\dfrac12$, then equality holds almost all $(y_1,\dots,y_m) \in \mathbb{R}^{nm}$ in the sense of $nm-$dimensional Lebesgue measure.
The quantity $d(T_1,\dots,T_m)$ is the unique $s$ such that $1 = \lim_{k \to \infty} \left( \sum_{I_k} \phi^s(T_{i_1} \circ \dots \circ T_{i_k})\right)^{\frac{1}{k}}$, where $I_k$ is the set of all the $k-$tuples where $1 \le i_j \le m$ and for $0<s\le n$, $\phi^s(T) = \alpha_1 \alpha_2 \dots \alpha_{r-1}\alpha_{r}^{s-r+1}$, where $\alpha_j$ are the singular values of a non-singular linear contracting mapping $T:\mathbb{R}^n \to \mathbb{R}^n$ and $r$ is the integer for which $r-1 < s \le r$.
Well, in fact, it is with this $\phi^s$ where my problems begin. But let's first adapt the theorem for our case.
Here, $m = 2$ because I have two affine mappings. $f_1(x,y) = T_1(x,y) + y_1$, where $T_1$ is defined by the matrix $\begin{pmatrix}\dfrac12 & 0\\ 0 & \dfrac14 \end{pmatrix}$ and $y_1 = \left(0, \dfrac34 \right)$. Analogolously we have $f_2(x,y) = T_2(x,y) + y_2$. Of course, the singular values of $T_1$ (and $T_2$, same matrix) are $\dfrac12$ and $\dfrac14$. But how can I define $r$? I "strongly believe" I'm supposed to choose $r=2$, because $s$, which will be a bound for my dimension and so it cannot be bigger than the Hausdorff dimension of the unit square, making the $r$, which respects the relation $r-1<s$, less than $3$.
Let's suppose my belief was right and $r$ is in fact supposed to be $2$. Then $\phi^s(T_1) = \dfrac12 \left( \dfrac14 \right)^{s-1} = \dfrac{2}{2^{2s-2}}$. But now, the limit is $\lim_{k \to \infty} \left( \dfrac{1}{2^{2s-2}} + \dfrac{1}{2^{4s-4}} + \dots + \dfrac{1}{2^{2^{k}s - 2^k}} \right)^{\frac{1}{k}}$.
Is this correct? Calculating this limit I will indeed have the Hausdorff dimension of this atractor?
Any help would be appreciated.
You don't define $r$; rather, $r$ depends upon $s$.
You have a function $$ \Phi(s) = \lim_{k \to \infty} \left( \sum_{I_k} \phi^s(T_{i_1} \circ \dots \circ T_{i_k})\right)^{\frac{1}{k}} $$
and we know that there is a unique, positive value of $s$ such that $\Phi(s)=1$. Your mission is to find that value of $s$. In the process, you'll compute $\Phi(s)$ for a number of values of $s$ and the value of $r$ will depend on $s$. If you're checking values of $s$ in $(0,1]$, then $r=1$; if you're checking values of $s$ in $(1,2]$, then $r=2$.
Having said that, it's pretty easy to see in your particular case that we can assume that $s \geq 1$ so that $r=2$. This is because the projection of the invariant set onto the $x$-axis contains an interval so its dimension is at least 1.
You're pretty close here, yes. The one thing that I would mention is that the theorem is of the almost surely variety, in that there might be some special choices of $y_i$s that nullify the conclusion. For example, if your $y_i$s were both the zero vector, rather than $\langle 0,3/4 \rangle$ and $\langle 1/4,0 \rangle$, then the attractor would be a Cantor set on the $y$-axis with dimension smaller than 1. I'm pretty sure that Falconer has some later results that provide sufficient conditions on when the theorem is applicable that might help, though.