2·1. Dimension concepts

Definition 2·1. Let $F\subseteq \mathbb{R}^d$ . We say that $\{U_{i}\}$ is a $\delta$ -cover of F if $F\subset \bigcup\limits_{i=1}^{\infty}U_{i}$ and $0 \lt|U_{i}|\leq \delta$ for each i, where $|U_i|$ denotes the diameter of the set $U_i$ . For each $\delta\gt0$ and $s\geq 0$ , we define

\begin{align*}H_{\delta}^{s}(F)=\inf\bigg\{\sum_{i=1}^{\infty}|U_{i}|^{s} \;:\; \{U_{i}\} \text{ is a }\delta\text{-cover of }F\bigg\} \quad \text{and} \quad H^{s}(F)=\lim_{\delta\to 0+}H_{\delta}^{s}(F).\end{align*}

We call $H^{s}(F)$ the s-dimensional Hausdorff measure of the set F. Using this, the Hausdorff dimension of the set F is defined by

\begin{align*}\dim_{H}(F)=\inf\{s\geq 0 \;:\; H^{s}(F)=0\}=\sup\{s \geq 0 \;:\; H^{s}(F)=\infty\}.\end{align*}

Definition 2·2. The box dimension of a non-empty bounded subset F of (X, d) is defined as

\begin{align*}\dim_{B}F=\lim_{\delta \to 0}\frac{\log{N_{\delta}(F)}}{-\log\delta},\end{align*}

where $N_{\delta}(F)$ denotes the smallest number of sets of diameter at most $\delta$ that can cover F, provided the limit exists. If this limit does not exist, then the upper and lower box dimensions, respectively, are defined as

\begin{align*}\overline{\dim}_{B}F=\limsup_{\delta \to 0}\frac{\log{N_{\delta}(F)}}{-\log\delta}\text{ and }\underline{\dim}_{B}F=\liminf_{\delta \to 0}\frac{\log{N_{\delta}(F)}}{-\log\delta}.\end{align*}

2·2. Symbolic space

Let $\mathcal{I}=\big\{\,f_1,f_2,\ldots,f_n \big\}$ be an IFS on $\mathbb{R}^d$ such that $\|f_i(x)-f_i(y)\|\leq r_i\|x-y\|$ with $r_i\in (0,1)$ for all $i\in \{1,2,\dots,n\}$ . Let $\Sigma\,:\!=\,\{1,2,\dots,n\}^{\mathbb{N}}$ be the set of all infinite sequences with symbols from $\{1,2,\dots,n\}$ . The set $\Sigma$ is the symbolic space corresponding to the IFS $\mathcal{I}=\{\,f_1,\ldots,f_n\}$ . Let $\mathbf{i}=i_{1}i_{2}\dots \in \Sigma$ . We define $\mathbf{i}_{|_m}\,:\!=\,i_{1}i_{2}\dots i_{m}$ for all $m\in \mathbb{N}$ . We denote the set of all finite sequences of length m with symbols from $\{1,2,\dots, n\}$ by $\Sigma_{m}$ . Set $\Sigma^{*}\,:\!=\,\bigcup_{m=1}^\infty\Sigma_{m}.$ The notation $|\mathbf{i}|$ denotes the length of the finite sequence $\mathbf{i}\in \Sigma^*$ . The symbolic space $\Sigma$ equipped with metric $\rho$ is a compact metric space, where the metric $\rho$ is defined as follows

\begin{align*}\rho(\mathbf{i},\mathbf{j})=2^{-|\mathbf{i}\wedge\mathbf{j}|}\end{align*}

for $\mathbf{i},\mathbf{j}\in \Sigma$ , where $\mathbf{i}\wedge\mathbf{j}$ denotes the initial largest common segment of $\mathbf{i}$ and $\mathbf{j}$ .

2·3. Affinity dimension and Furstenberg measure

Let A be a $d \times d$ real matrix. For $t\geq 0$ , the singular value function $\Phi^{t}(A)$ of A is defined by

\begin{align*}\Phi^{t}(A)=\begin{cases} \alpha_{1}\dots \alpha_{\lfloor t \rfloor}{\alpha_{\lceil t \rceil}}^{t- \lfloor t \rfloor}\quad \text{if }0\leq t\leq d \\ |\text{det}(A)|^{t/d} \quad \text{if }t\gt d,\end{cases}\end{align*}

where $\alpha_{1}\geq \alpha_{2}\geq \dots \geq \alpha_{d}$ are the singular values of A.

The affinity dimension of the self-affine IFS $\mathcal{I}=\{\,f_i(x)=A_ix+a_i\}_{i=1}^{n}$ is defined by

\begin{align*}t_0\,:\!=\,\inf\bigg\{t\gt0\;:\; \sum_{m=1}^{\infty}\sum_{i_{1}\dots i_{m}\in \Sigma_m} \Phi^{t}(A_{i_1}\cdots A_{i_m})\lt\infty\bigg\}.\end{align*}

If the matrices $A_i$ have the block triangular form

(2·1) \begin{equation}A_i=\begin{bmatrix} \lambda_iU_i &\quad \underline{0}\\[6pt] \underline{a}_i^T &\quad s_i\end{bmatrix}\text{ for every }\textit{i},\end{equation}

where $U_i$ are $2\times2$ orthogonal matrices, $0\lt\lambda_i\lt|s_i|\lt1$ , $\underline{a}_i\in{\mathbb{R}}^2$ , and $\mathcal{I}$ satisfies the SOSC then by [Reference Bárány, Rams and Simon3, remark 2·6], the affinity dimension $t_0$ satisfies the equation $t_0=\min\{r_1,r_2\}$ , where

(2·2) \begin{equation}\sum_{i=1}^n|s_i|^{r_1}=1\quad \text{and}\quad \sum_{i=1}^n|s_i|\lambda_i^{r_2-1}=1.\end{equation}

In particular, when $\sum_{i=1}^n\lambda_i^2=1$ , then $t_0=r_2\in[2,3]$ .

Following the lines [Reference Bárány, Rams and Simon3, section 2·4], we define the corresponding Furstenberg IFS induced by the IFS $\mathcal{I}$ with matrices of the form (2·1) as follows:

(2·3) \begin{equation}\mathcal{J}=\left\{h_i(\underline{x})=\frac{\lambda_i}{s_i}U_i^T\underline{x}-\frac{1}{s_i}\underline{a}_i\right\}_{i=1}^n\!.\end{equation}

The results of Rapaport [Reference Rapaport16, section 1·2] and Feng [Reference Feng9, theorem 1·10] give a sufficient condition to calculate the dimension of the attractor of $\mathcal{I}$ , see [Reference Bárány, Rams and Simon3, section 2·4]. We state it in the special case we require throughout this paper.

Theorem 2·3. Let

\begin{align*}\mathcal{I}=\left\{\,f_i(\underline{x})=\begin{bmatrix} \lambda_iU_i &\quad \underline{0}\\[6pt] \underline{a}_i^T &\quad s_i\end{bmatrix}\underline{x}+a_i\right\}_{i=1}^{n}\end{align*}

be a self-affine IFS in ${\mathbb{R}}^3$ with attractor K such that it satisfies the SOSC, $U_i$ are $2\times2$ orthogonal matrices, $\sum_{i=1}^n\lambda_i^2=1$ and $0\lt\lambda_i\lt|s_i|\lt1$ . Let $\mu_F=\sum_{i=1}^n|s_i|\lambda_i^{t_0-1}(h_i)_*\mu_F$ be the Furstenberg measure corresponding to the IFS in (2·3). If $\dim_H\mu_F\geq3-t_0$ then $\dim_HK=t_0$ .

Let us point out that Rapaport [Reference Rapaport 16 ] originally proved Theorem 2·3 under an irreducibility assumption and strong separation. However, these assumptions can be removed. Feng [Reference Feng 9 ] already removed the irreducibility assumption, but still assumed the strong separation. However, using the observation of Bárány and Käenmäki [Reference Bárány and Käenmäki2, corollary 2·8], we get that the conditional entropy of any Bernoulli measure with respect to the natural projection remains zero under the strong open set condition; thus, the strong separation condition can be replaced by SOSC in [Reference Feng9, theorem 1·10(b)].

Finally, we state a simple proposition to estimate the dimension of self-similar measures from below. The bound is a simpler version of the overlapping number estimate by Mihailescu and Urbański [Reference Mihailescu and Urbański15, theorem 2·5(b)] for random systems.

Proposition 2·4. Let $\mathcal{J}=\{h_i(x)=\lambda_iU_ix+t_i\}_{i=1}^N$ be a self-similar IFS on ${\mathbb{R}}^d$ with attractor K and let $(p_i)_{i=1}^N$ be a non-degenerate probability vector and let $\mu=\sum_{i=1}^Np_i(h_i)_*\mu$ . If $\min_{x\in K}\#\{i\in\{1,\ldots,N\}\;:\; x\notin h_i(K)\}\geq Q$ then

\begin{align*}\dim_H\mu\geq\frac{\log(1-Qp_{\min})}{\log\lambda_{\max}},\end{align*}

where $\lambda_{\max}=\max_{i=1,\ldots,N}|\lambda_i|$ and $p_{\min}=\min_{i=1,\ldots,N}p_i$ .

Proof. Let $r\ll1$ . Denote B(x, r) the closed ball of radius r with center x. Let $C_x\,:\!=\,\{i:\ x\notin h_i(K)\}$ . We have

\begin{align*} \max_{x\in K}\mu(B(x,r))&=\max_{x\in K}\sum_{i=1}^{N}p_i\mu\big(h_{i}^{-1}(B(x,r)\cap K)\big)\\ &\leq \max_{x\in K}\sum_{i\in C_{x}^{c}}p_i\mu\big(B\big(h_{i}^{-1}(x),\lambda_i^{-1} r\big)\big)\\&\leq \max_{x\in K}\mu\big(B\big(x,\lambda_{\max}^{-1}r\big)\big)\max_{x\in K}\bigg(1-\sum_{i\in C_{x}}p_i\bigg)\\ &\leq \max_{x\in K}\mu\big(B\big(x,\lambda_{\max}^{-1}r\big)\big)\max_{x\in K}\bigg(1-Q p_{\min}\bigg). \end{align*}

This implies by induction

\[\max_{x\in K}\mu\big(B\big(x,\lambda_{\max}^n\big)\big)\leq(1-Q p_{\min})^n\text{ for every }n\in{\mathbb{N}}, \]

and so,

\begin{align*}\liminf_{r\to0}\frac{\log\mu(B(x,r))}{\log r}=\liminf_{n\to\infty}\frac{\log\mu\big(B\big(x,(\lambda_{\max})^n\big)\big)}{n\log\lambda_{\max}}\geq \frac{\log(1-Qp_{\min})}{\log\lambda_{\max}}.\end{align*}

This completes the proof.

4·1. The Furstenberg measure and a sufficient condition

Let $\tilde{\Delta}$ be the interior of the original equilateral triangle $\Delta.$ Then, one can see that

\begin{align*}W_i(\tilde{\Delta}\times \mathbb{R})\subset \tilde{\Delta}\times \mathbb{R}, W_i(\tilde{\Delta}\times \mathbb{R})\cap W_j(\tilde{\Delta}\times \mathbb{R})=\emptyset\end{align*}

for all $i\ne j\in \{1,2,\dots,N^2\}.$ This implies the IFS $\mathcal{I}$ satisfies the SOSC.

Let $\mathbf{p}=(p_1,p_2,\dots,p_{N^2})$ be a probability vector, where $p_i=s_i\big({1}/{N}\big)^{t_{0}-1}$ for every $i\in \{1,2,\dots,N^2\}.$

Define

\begin{align*}\mathcal{A}_1\,:\!=\,\big\{i\in\{1,2,\dots,N^2\}\;:\; a_{1}^{i}=a_{2}^{i}\big\},\quad \mathcal{A}_2\,:\!=\,\big\{i\in\{1,2,\dots,N^2\}\;:\; a_{1}^{i} \gt a_{2}^{i}\big\}\end{align*}

\begin{align*}\mathcal{A}_3\,:\!=\,\big\{i\in\{1,2,\dots,N^2\}\;:\; a_{1}^{i} \lt a_{2}^{i}\big\}.\end{align*}

By applying (2·3), we construct a Furstenberg IFS $\mathcal{J}=\{h_1,h_2,\dots,h_{N^2}\}$ on $\mathbb{R}^2$ by the self-affine IFS $\mathcal{I}$ as follows

\begin{align*} h_i(x,y)=\begin{cases} \begin{aligned} \frac{1}{Ns_i}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_i}\begin{bmatrix} \mathbf{a}_i \\ b_i\end{bmatrix}\quad & \text{if }a_{1}^{i}\geq a_{2}^{i}\quad \text{and}\quad 1\leq i\leq \frac{N(N+1)}{2},\\[4pt] \frac{1}{Ns_i}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad - 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_i}\begin{bmatrix} \mathbf{a}_i \\ b_i\end{bmatrix}\quad & \text{if }a_{1}^{i}\geq a_{2}^{i}\quad \text{and}\quad \frac{N(N+1)}{2}+1\leq i\leq N^2,\\[4pt] \frac{1}{Ns_i}\begin{bmatrix} - 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_i}\begin{bmatrix} \mathbf{a}_i \\ b_i\end{bmatrix}\quad & \text{if }a_{1}^{i}\lt a_{2}^{i}\quad \text{and}\quad 1\leq i\leq \frac{N(N+1)}{2},\\[4pt] \frac{1}{Ns_i}\begin{bmatrix} - 1 &\quad 0 \\ 0 &\quad - 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_i}\begin{bmatrix} \mathbf{a}_i \\ b_i\end{bmatrix}\quad & \text{if }a_{1}^{i}\lt a_{2}^{i}\quad \text{and}\quad \frac{N(N+1)}{2}+1\leq i\leq N^2. \end{aligned} \end{cases}\end{align*}

The Furstenberg measure $\mu_F=\sum_{i=1}^{N^2}p_i(h_i)_*\mu_F$ is the unique invariant Borel probability measure corresponding to the IFS $\mathcal{J}$ with probability vector $\mathbf{p}$ . Since the self-affine IFS $\mathcal{I}$ satisfies the SOSC and $s_{i}\in ({1}/{N},1)\forall i\in \{1,2,\dots, N^2\}$ , to verify the equality in (4·1), we only need to show that

\begin{align*}\dim_{H}(\mu_{F})\gt3-t_{0}\end{align*}

by Theorem 2·3. In particular, to show Theorem 1·1 and Theorem 1·3, we will prove the following:

Proposition 4·2. For each $i\in \{1,2,\dots,N^2\}$ , let $a_{1}^{i}\ne a_{2}^{i}$ if $a_{1}^{i}$ and $a_{2}^{i}$ both are not on the boundary of original triangle $\Delta$ . Then,

\begin{align*}\dim_{H}(\mu_{F})\gt3-t_0\end{align*}

for Lebesgue almost every scaling parameter $\underline{s}\in ({1}/{N},1)^{\#\mathcal{A}_1}\times ({1}/{N B},1)^{\#\mathcal{A}_2}\times ({1}/{N D},1)^{\#\mathcal{A}_3}$ .

Proposition 4·3. Let $s_i\in ({2}/{3},1)\textit{ for every } i\in \{1,2,\dots,9\}$ . Suppose that $\max\{s_{5},s_{8}\}\leq \min \{s_{4},s_{7}\}$ and $s_2\leq s_9.$ Then,

\begin{align*}\dim_{H}(\mu_F)\gt3-t_0,\end{align*}

where $t_0$ is the affinity dimension of the IFS $\mathcal{I}$ in (4·1).

Remark 4·4. We note that the method is not applicable in every configuration. In the above construction, if we consider $s_i=s$ for every $i\in \{1,2,\dots,N^2\}$ and $a_k=a$ for all $k\in \{1,2,\dots,L(N)\},$ then for the large value of N,

(4·2) \begin{equation} \dim_{H}(\mu_{F})\not\gt3-t_{0}=\frac{-\log s}{\log N} \forall s\in \bigg(\frac{1}{N},1\bigg). \end{equation}

In this consideration, there are 9 different mappings in Furstenberg IFS $\mathcal{J}$ with multiplicity ${(N-3)(N-2)}/{2}+3, {(N-4)(N-3)}/{2}, (N-2), (N-2), (N-2), (N-2), (N-3), (N-2), 1$ , respectively.

Examples. For $N=100, \dim_{H}(\mu_{F})\lt {-\log s}/{\log N} \,\text{for }s\in (0.042, 0.237).$

For $N=1000, \dim_{H}(\mu_{F})\lt {-\log s}/{\log N} \,\text{for }s\in (0.001, 0.430).$

For $N=10000, \dim_{H}(\mu_{F})\lt {-\log s}/{\log N} \,\text{for }s\in (0.0001, 0.461).$

Although, if we consider some $s_i\ne s_j$ for some $i,j\in \{1,2,\dots,N^2\}$ and $a_k=a$ for all $k\in \{1,2,\dots,L(N)\},$ then for the large value of N, the same situation as in (4·2) occurs. For example, we consider ${(N-3)(N-2)}/{2}+3$ many mappings with $s_i=s+0.001,{(N-4)(N-3)}/{2}$ many mappings with $s_i=s+0.002, 5(N-2)$ many mappings with $s_i=s+0.003, (N-3)$ many mappings with $s_i=s+0.004$ and 1 map with $s_i=s+0.002$ . In this consideration, we have the following:

for $N=100, \dim_{H}(\mu_{F})\lt 2-\frac{\log\sum_{i=1}^{N^2} s_i}{\log N} \,\text{for }s\in (0.035, 0.279)$ ;

for $N=100000, \dim_{H}(\mu_{F})\lt 2-\frac{\log \sum_{i=1}^{N^2} s_i}{\log N} \,\text{for }s\in (0.00001, 0.463).$

4·2. Dimension for almost every parameter

Let $f_i$ be the projection of $h_i$ on the X-axis. Let $\mathcal{J}_{X}=\{\,f_1,f_2,\dots,f_{N^2}\}$ be the projection of the IFS $\mathcal{J}$ on the X-axis. Then, for $i\in \{1,2,\dots,N^2\}$ the map $f_{i}$ is as follows

\begin{align*}f_{i}(x)=\begin{cases}\dfrac{x}{Ns_i} &\text{if } a_{1}^{i}= a_{2}^{i}\\[8pt] \dfrac{x}{Ns_i}-\frac{\mathbf{a}_i}{s_i} &\text{if } a_{1}^{i}\geq a_{2}^{i} \\[8pt] \dfrac{-x}{Ns_i}-\dfrac{\mathbf{a}_i}{s_i} &\text{if }a_{1}^{i}\lt a_{2}^{i}.\end{cases}\end{align*}

Set $\lambda_{i}={1}/{Ns_i}$ and $\gamma_{i}=-\mathbf{a}_{i}N=|a_{1}^{i}- a_{2}^{i}|N\gt0$ , one can see that the IFS $\{\,f_i\}_{i=1}^{N^2}$ is of type considered in Section 3. Since $s_i\in ({1}/{N},1)$ , we have $\lambda_i\in ({1}/{N},1)$ for all $i\in \{1,2,\dots,N^2\}.$ Thus, we have

\begin{align*}\mathcal{J}_{X}=\{\,f_{i}(x)=\lambda_i x\}_{i\in \mathcal{A}_1}\cup \{\,f_{i}(x)=\lambda_i x +\gamma_{i} \lambda_i\}_{i\in \mathcal{A}_2}\cup \{\,f_{i}(x)=-\lambda_i x +\gamma_{i} \lambda_i\}_{i\in \mathcal{A}_3}.\end{align*}

Proof of Proposition 4·2. Let ${P_{X*}}\mu_{F}$ be the projection of the measure $\mu_{F}$ on the X-axis. The measure ${P_{X*}}\mu_{F}$ is the invariant measure corresponding to the IFS $\mathcal{J}_{X}$ with probability vector $\mathbf{p}$ . By Proposition 3·2, there exists a set $\mathcal{E}\subset (0,1)^{\#\mathcal{A}_1}\times (0,B)^{\#\mathcal{A}_2}\times (0,D)^{\#\mathcal{A}_3}$ such that $\dim_{H}\mathcal{E}\leq \#\mathcal{A}_1+\#\mathcal{A}_2+\#\mathcal{A}_3-1$ such that the IFS $\mathcal{J}_{X}$ satisfies ESC for CFS for every parameters $\underline{\lambda}\in (0,1)^{\#\mathcal{A}_1}\times (0,B)^{\#\mathcal{A}_2}\times (0,D)^{\#\mathcal{A}_3}\setminus\mathcal{E}$ . Thus, by [Reference Bárány and Verma4, Theorem 5·1, Theorem 3·5], we get

\begin{align*}\dim_{H}({P_{X*}}\mu_{F})=\min\bigg\{1,\frac{-\sum_{i=1}^{N^2}p_i\log p_i +\Phi(\mathbf{p})}{-\sum_{i=1}^{N^2}p_i\log \lambda_i}\bigg\}\end{align*}

for every parameter $\underline{s}\in ({1}/{N},1)^{\#\mathcal{A}_1}\times ({1}/{NB},1)^{\#\mathcal{A}_2}\times ({1}/{ND},1)^{\#\mathcal{A}_3}\setminus\mathcal{E}$ . Now, by [Reference Bárány and Verma4, Proposition 2·2], we have

\begin{align*}\Phi(\mathbf{p})\geq \sum_{i\in \mathcal{A}_1}p_{i}\log\bigg(p_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}p_j\bigg).\end{align*}

Given that for each $i\in \{1,2,\dots,N^2\}$ , $a_{1}^{i}\ne a_{2}^{i}$ if $a_{1}^{i}$ and $a_{2}^{i}$ both are not on the boundary of original triangle $\Delta$ . Under this consideration, one can see that

\begin{align*}\#\mathcal{A}_1=N+2 \quad \text{and}\quad \#\mathcal{A}_2+\#\mathcal{A}_3=N^2-N-2.\end{align*}

Now, our aim is to show that

\begin{align*}\frac{-\sum_{i=1}^{N^2}p_i\log p_i +\Phi(\mathbf{p})}{-\sum_{i=1}^{N^2}p_i\log \lambda_i}\gt 3-t_0=2-\frac{\log \sum_{i=1}^{N^2}s_i}{\log N}.\end{align*}

Since $\lambda_i={1}/{Ns_i}$ and $p_i=s_i({1}/{N})^{t_0-1}$ , we have

\begin{align*}\frac{-\sum_{i=1}^{N^2}p_i\log p_i +\Phi(\mathbf{p})}{-\sum_{i=1}^{N^2}p_i\log \lambda_i}&\gt2-\frac{\log \sum_{i=1}^{N^2}s_i}{\log N}\\[4pt]\Leftarrow \frac{-\sum_{i=1}^{N^2}p_i\log p_i +\sum_{i\in \mathcal{A}_1}p_{i}\log\bigg(p_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}p_j\bigg)}{\sum_{i=1}^{N^2}p_i\log Ns_i}&\gt2-\frac{\log \sum_{i=1}^{N^2}s_i}{\log N}\\[4pt]\Leftarrow \frac{-\sum_{i=1}^{N^2}s_i\log p_i +\sum_{i\in \mathcal{A}_1}s_{i}\log\bigg(p_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}p_j\bigg)}{\sum_{i=1}^{N^2}s_i\log Ns_i}&\gt2-\frac{\log \sum_{i=1}^{N^2}s_i}{\log N}\\[4pt]\Leftarrow \frac{\sum_{i\in \mathcal{A}_1}s_{i}\log\bigg(\frac{s_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}s_j}{s_i}\bigg)+\sum_{i\in \mathcal{A}_2\cup \mathcal{A}_3}s_i\log \bigg(\frac{\sum_{i=1}^{N^2}s_i}{s_i}\bigg) }{\sum_{i=1}^{N^2}s_i\log Ns_i}&\gt2-\frac{\log \sum_{i=1}^{N^2}s_i}{\log N}\\[4pt]\Leftarrow \sum_{i\in \mathcal{A}_1}s_{i}\log\bigg(\frac{s_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}s_j}{s_i}\bigg)+\sum_{i\in \mathcal{A}_2\cup \mathcal{A}_3}s_i\log \bigg(\frac{\sum_{i=1}^{N^2}s_i}{s_i}\bigg) \\[4pt]\gt\sum_{i=1}^{N^2}s_i\log (Ns_i)^2- \sum_{i=1}^{N^2}&s_i\log (Ns_i)\frac{\log \sum_{i=1}^{N^2}s_i}{\log N}\\[4pt]\Leftarrow \sum_{i\in \mathcal{A}_1}s_{i}\log\bigg(\frac{s_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}s_j}{N^2s_{i}^{3}}\bigg)+\sum_{i\in \mathcal{A}_2\cup \mathcal{A}_3}s_i\log \bigg(\frac{\sum_{i=1}^{N^2}s_i}{N^2s_{i}^{3}}\bigg) \\[4pt] &\hspace{-105pt} \gt- \sum_{i=1}^{N^2}s_i\log Ns_i\frac{\log \sum_{i=1}^{N^2}s_i}{\log N} \\[4pt]\Leftarrow \frac{\log (N s_{\min})}{\log N}\bigg(\sum_{i\in \mathcal{A}_1}s_{i}\log\bigg(\frac{s_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}s_j}{N^2s_{i}^{3}}\sum_{i=1}^{N^2}s_i\bigg)+\sum_{i\in \mathcal{A}_2\cup \mathcal{A}_3}&s_i\log \bigg(\frac{(\sum_{i=1}^{N^2}s_i)^2}{N^2s_{i}^{3}}\bigg)\!\bigg)\gt0 \\[4pt]\Leftarrow \sum_{i\in \mathcal{A}_1}s_{i}\log\bigg(\frac{s_i+\sum_{j\in \mathcal{A}_2\cup \mathcal{A}_3}s_j}{N^2s_{i}^{3}}\sum_{i=1}^{N^2}s_i\bigg)+\sum_{i\in \mathcal{A}_2\cup \mathcal{A}_3}s_i\log \bigg(&\frac{(\sum_{i=1}^{N^2}s_i)^2}{N^2s_{i}^{3}}\bigg)\gt0.\end{align*}

Clearly ${(\sum_{i=1}^{N^2}s_i)^2}/{N^2s_{i}^{3}}\geq {1}/{s_{i}^{3}}\gt1$ . This proves our claim. Since $3-t_0\in (0,1)$ , we have

\begin{align*}\dim_{H}(\mu_{F})\geq \dim_{H}({P_{X*}}\mu_{F})\gt3-t_{0}\end{align*}

for Lebesgue almost every scaling parameter $\underline{s}\in ({1}/{N},1)^{\#\mathcal{A}_1}\times ({1}/{N B},1)^{\#\mathcal{A}_2}\times ({1}/{N D},1)^{\#\mathcal{A}_3}$ . The proof is completed by Theorem 2·3.

4·3. Hausdorff dimension of the FIS in the case of $N=3$ for every parameter

Here, we prove Theorem 1·3 by computing the overlapping number.

Since the upper bound

\begin{align*}\dim_{H}(G(\,f^*))\leq \overline{\dim}_{B}(G(\,f^*)) \leq t=1+\frac{\log\big(\!\sum_{i=1}^{9}s_i\big)}{\log(3)}\end{align*}

holds for every parameter value, it is enough to show the lower bound.

For the lower bound, consider the corresponding Furstenberg IFS $\mathcal{J}=\{h_1,h_2,\dots,h_9\}$ , where $h_i\;:\; \mathbb{R}^2\to \mathbb{R}^2 $ are defined as follows:

\begin{align*} h_1(x,y)=\frac{1}{3s_1}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_1}\begin{bmatrix} 0 \\ 0\end{bmatrix}\!, \quad h_2(x,y)=\frac{1}{3s_2}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_2}\begin{bmatrix} 0 \\ \frac{2a}{\sqrt{3}}\end{bmatrix}\!, \end{align*}

\begin{align*} h_3(x,y)=\frac{1}{3s_3}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_3}\begin{bmatrix} 0 \\ 0\end{bmatrix}\!, \quad h_4(x,y)=\frac{1}{3s_4}\begin{bmatrix} -1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_4}\begin{bmatrix} -a \\ -\frac{a}{\sqrt{3}}\end{bmatrix}\!, \end{align*}

\begin{align*} h_5(x,y)=\frac{1}{3s_5}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_5}\begin{bmatrix} -a \\ -\frac{a}{\sqrt{3}} \end{bmatrix}\!, \quad h_6(x,y)=\frac{1}{3s_6}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad 1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_6}\begin{bmatrix} 0 \\ 0\end{bmatrix}\!, \end{align*}

\begin{align*} h_7(x,y)=\frac{1}{3s_7}\begin{bmatrix} -1 &\quad 0 \\ 0 &\quad -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_7}\begin{bmatrix} -a \\ -\frac{a}{\sqrt{3}} \end{bmatrix}\!, \quad h_8(x,y)=\frac{1}{3s_8}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_8}\begin{bmatrix} -a \\ -\frac{a}{\sqrt{3}}\end{bmatrix}\!, \end{align*}

\begin{align*} h_9(x,y)=\frac{1}{3s_9}\begin{bmatrix} 1 &\quad 0 \\ 0 &\quad -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}-\frac{1}{s_9}\begin{bmatrix} 0 \\ \frac{2a}{\sqrt{3}} \end{bmatrix}\!. \end{align*}

Let $\mu_F=\sum_{i=1}^9s_i\left({1}/{3}\right)^{t-1}(h_i)_*\mu_F$ be the invariant Borel probability measure for the IFS $\mathcal{J}$ with probabilities $p_i=s_i\left({1}/{3}\right)^{t-1}.$ By Theorem 2·3, to show that

\begin{align*}\dim_{H}(G(\,f^*))\geq t=1+\frac{\log\big(\!\sum_{i=1}^{9}s_i\big)}{\log(3)},\end{align*}

it is enough to prove that

\begin{align*}\dim_{H}(\mu_{F})\gt 3-t.\end{align*}

Now, we will estimate the Hausdorff dimension of the Furstenberg measure.

Let $f_i$ be the projection of $h_i$ on the X-axis. Let $\mathcal{J}_{X}=\{\,f_1,f_2,\dots,f_9\}$ be the projection of the IFS $\mathcal{J}$ on the X-axis. Precisely, the maps $f_i's$ are as follows:

\begin{align*}f_1(x)=\frac{x}{3s_1}, \quad f_2(x)=\frac{x}{3s_2}, \quad f_3(x)=\frac{x}{3s_3},\quad f_6(x)=\frac{x}{3s_6},\quad f_9(x)=\frac{x}{3s_9},\end{align*}

\begin{align*}f_4(x)=-\frac{x}{3s_4}+\frac{a}{s_4}, \quad f_5(x)=\frac{x}{3s_5}+\frac{a}{s_5}, \quad f_7(x)=-\frac{x}{3s_7}+\frac{a}{s_7}, \quad f_8(x)=\frac{x}{3s_8}+\frac{a}{s_8}.\end{align*}

The fixed point of the map $f_i$ is denoted by $\text{Fix}(\,f_i)$ for all $i\in \{1,2,\dots,9\}.$ Thus, we have

\begin{align*}\text{Fix}(\,f_1)=\text{Fix}(\,f_2)=\text{Fix}(\,f_3)=\text{Fix}(\,f_6)=\text{Fix}(\,f_9)=0,\end{align*}

\begin{align*}\text{Fix}(\,f_4)=\frac{3a}{3s_4+1}, \quad \text{Fix}(\,f_5)=\frac{3a}{3s_5-1}, \quad \text{Fix}(\,f_7)=\frac{3a}{3s_7+1},\quad \text{Fix}(\,f_8)=\frac{3a}{3s_8-1}.\end{align*}

Without loss of generality, we assume that $s_5\leq s_8$ .

Lemma 4·6. Let $s_i\in \big({2}/{3},1\big)\forall i\in \{1,2,\dots,9\}$ . Let $[\tilde{a},\tilde{b}]$ be the invariant interval for the IFS $\mathcal{J}_{X}.$ Then, $\tilde{a}=0$ and $\tilde{b}=\text{Fix}(\,f_5).$ Moreover, if $\max\{s_{5},s_{8}\}\leq \min \{s_{4},s_{7}\}$ , then

(4·3) \begin{equation} (\,f_4[0,\tilde{b}]\cup f_7[0,\tilde{b}])\cap (\,f_5[0,\tilde{b}]\cup f_8[0,\tilde{b}])=\emptyset. \end{equation}

For a visualisation, see Figure 5.

Fig. 5.

Visualisation of the configuration in (4·3).

Proof. The maps $f_4$ and $f_{7}$ are flipping the orientation and other maps are orientation preserving maps. Since $s_5\leq s_8$ , the fixed point $\text{Fix}(\,f_5)$ is the largest fixed points. Since $s_i\in \big({2}/{3},1\big)\forall i\in \{1,2,\dots,9\}$ , we have

\begin{align*}f_{4}[0,\text{Fix}(\,f_5)]= \bigg[\frac{a(3s_5-2)}{s_4(3s_5-1)}, \frac{a}{s_4}\bigg]\subset [0, \text{Fix}(\,f_5)],\end{align*}

\begin{align*}f_{7}[0,\text{Fix}(\,f_5)]= \bigg[\frac{a(3s_5-2)}{s_7(3s_5-1)}, \frac{a}{s_7}\bigg]\subset [0, \text{Fix}(\,f_5)].\end{align*}

This implies that $[0,\tilde{b}]$ is the invariant interval for the IFS $\mathcal{J}_{X}$ , where $\tilde{b}=\text{Fix}(\,f_5).$ One can see that

\begin{align*}f_{5}[0,\tilde{b}]=\bigg[\frac{a}{s_5}, \tilde{b}\bigg]\quad \text{and}\quad f_{8}[0,\tilde{b}]=\bigg[\frac{a}{s_8}, \frac{3s_5 a}{s_8 (3s_{5}-1)}\bigg].\end{align*}

Now, we assume that $\max\{s_{5},s_{8}\}\leq \min \{s_{4},s_{7}\}$ . Then, we get

\begin{align*}(\,f_4[0,\tilde{b}]\cup f_7[0,\tilde{b}])\cap (\,f_5[0,\tilde{b}]\cup f_8[0,\tilde{b}])=\emptyset.\end{align*}

This completes the proof.

Now, we will see the projection of the Furstenberg IFS $\mathcal{J}$ on the Y-axis. Let $g_i$ be the projection of $h_i$ on the Y-axis. Let $\mathcal{J}_{Y}=\{g_1,g_2,\dots,g_9\}$ be the projection of the IFS $\mathcal{J}$ on the y-axis. Precisely, the maps $g_i's$ are as follows:

\begin{align*}g_1(y)=\frac{y}{3s_1},\quad g_3(y)=\frac{y}{3s_3}, \quad g_6(y)=\frac{y}{3s_6}, \end{align*}

\begin{align*}g_2(y)=\frac{y}{3s_2}-\frac{2a}{\sqrt{3}s_2}, \quad g_4(y)=\frac{y}{3s_4}+\frac{a}{\sqrt{3}s_4}, \quad g_5(y)=\frac{y}{3s_5}+\frac{a}{\sqrt{3}s_5},\end{align*}

\begin{align*}g_7(y)=\frac{-y}{3s_7}+\frac{a}{\sqrt{3}s_7},\quad g_8(y)=\frac{-y}{3s_8}+\frac{a}{\sqrt{3}s_8},\quad g_9(y)=\frac{-y}{3s_9}-\frac{2a}{\sqrt{3}s_9}.\end{align*}

Next, we will examine the invariant interval for the IFS $\mathcal{J}_{Y}=\{g_1,g_2,\dots,g_9\}$ . The fixed points of the maps $g_i's$ are as follows:

\begin{align*}\text{Fix}(g_1)=\text{Fix}(g_3)=\text{Fix}(g_6)=0, \end{align*}

\begin{align*}\quad \text{Fix}(g_2)=\frac{-6a}{\sqrt{3}(3s_2-1)}, \quad \text{Fix}(g_4)=\frac{3a}{\sqrt{3}(3s_4-1)}, \quad \text{Fix}(g_5)=\frac{3a}{\sqrt{3}(3s_5-1)},\end{align*}

\begin{align*}\text{Fix}(g_7)=\frac{3a}{\sqrt{3}(3s_7+1)}, \quad \text{Fix}(g_8)=\frac{3a}{\sqrt{3}(3s_8+1)}, \quad \text{Fix}(g_9)=\frac{-6a}{\sqrt{3}(3s_9+1)}.\end{align*}

Without loss of generality, we assume that $s_7\leq s_8$ and $s_4\leq s_5.$

For a visualisation, see Figure 6.

Fig. 6.

Visualisation of the configuration in Lemma 4.7.

Proof. The maps $g_2,g_4$ and $g_5$ preserve the orientation, however the maps $g_7,g_8$ and $g_9$ are flipping the orientation about Y- axis. The $\text{Fix}(g_2)$ is the lowest fixed point. Since $s_8\leq s_7$ and $s_5\leq s_4,$ the fixed point $\text{Fix}(g_5)$ is the largest fixed point. All the fixed points are always in the invariant interval $\tilde{I}.$ Thus, $g_8(\text{Fix}(g_2))\in \tilde{I}.$ And, we have

\begin{align*}g_8(\text{Fix}(g_2))=\frac{-1}{3s_8}\bigg(\frac{-6a}{\sqrt{3}(3s_2-1)}\bigg)+\frac{a}{\sqrt{3} s_8}= \frac{a}{\sqrt{3}s_8}\bigg(\frac{3s_2+1}{3s_2-1}\bigg),\end{align*}

\begin{align*}g_7(\text{Fix}(g_2))= g_8(\text{Fix}(g_2))\bigg(\frac{s_8}{s_7}\bigg) \lt g_8(\text{Fix}(g_2)).\end{align*}

Furthermore, $g_8(g_8(\text{Fix}(g_2)))\in \tilde{I}$ and $g_8(\text{Fix}(g_5))\in \tilde{I}.$ Thus, we have

\begin{align*}g_8(g_8(\text{Fix}(g_2)))=\frac{a}{\sqrt{3}s_8}\bigg(\frac{3s_8(3s_2-1)-(3s_2+1)}{3s_8(3s_2-1)}\bigg), g_8(\text{Fix}(g_5))=\frac{a}{\sqrt{3}s_8}\bigg(\frac{3s_5-2}{3s_5-1}\bigg).\end{align*}

One can see that

\begin{align*}g_8(g_8(\text{Fix}(g_2))) \gt \text{Fix}(g_2), 0 \lt g_8(\text{Fix}(g_5)) \lt g_8(\text{Fix}(g_2)) \quad\text{and}\quad 0 \lt g_8(\text{Fix}(g_5)) \lt \text{Fix}(g_5).\end{align*}

The map $g_9$ is also flipping the orientation. So,

\begin{align*}g_9(\text{Fix}(g_5))=\frac{-1}{3s_9}\bigg(\frac{3a}{\sqrt{3}(3s_5-1)}\bigg)-\frac{2a}{\sqrt{3} s_9}=\frac{-a}{\sqrt{3}s_9}\bigg(\frac{6s_5-1}{3s_5-1}\bigg),\end{align*}

\begin{align*}g_9(g_8(\text{Fix}(g_2)))=\frac{-a}{\sqrt{3}s_9}\bigg( \frac{(3s_2+1)+6s_8(3s_2-1)}{3s_8(3s_2-1)}\bigg),\end{align*}

\begin{align*}g_9(\text{Fix}(g_2))=\frac{-1}{3s_9}\bigg(\frac{-6a}{\sqrt{3}(3s_2-1)}\bigg)-\frac{2a}{\sqrt{3} s_9}=\frac{-2a}{\sqrt{3}s_9}\bigg(\frac{3s_2-2}{3s_2-1}\bigg).\end{align*}

One can see that

\begin{align*}0\gt g_9(\text{Fix}(g_2))\gt\text{Fix}(g_2).\end{align*}

For $s_2\leq s_9$ , we have the following relation for the map $g_9$ :

\begin{align*}0\gt g_9(\text{Fix}(g_5))\gt\text{Fix}(g_2)\quad \text{and}\quad 0\gt g_9(g_8(\text{Fix}(g_2)))\gt\text{Fix}(g_2).\end{align*}

Thus, for $s_2\leq s_9$ , the invariant interval $\tilde{I}$ is either $[\text{Fix}(g_2), \text{Fix}(g_5)]$ or $[\text{Fix}(g_2), g_8(\text{Fix}(g_2))]$ depending on the relation between $\text{Fix}(g_5)$ and $g_8(\text{Fix}(g_2)).$ This completes the proof.

Proposition 4·8. Let $s_i\in \big({2}/{3},1\big)\forall i\in \{1,2,\dots,9\}$ . Suppose that $\max\{s_{5},s_{8}\}\leq \min \{s_{4},s_{7}\}$ and $s_2\leq s_9.$ For $x\in A_{F},$ let $\underline{d}_{\mu_F}(x)$ denote the lower local dimension of $\mu_F$ at x. Then,

\begin{align*}\dim_H\mu_F\geq \frac{\log\big(1-\frac{s_{\min}}{3^{t_0-2}}\big)}{-\log(3s_{\max})}.\end{align*}

Proof. For each $x\in A_{F}$ , we define $C_{x}\,:\!=\,\{i\;:\; x\notin h_i(A_{F}) \quad \text{and}\quad i\in \{1,2,\dots,9\}\}.$ First, we show that $\#C_{x}\geq 3$ for every $x\in A_{F}$ , where $\#C_{x}$ denotes the cardinality of $C_{x}.$

By Lemma 4·6 and Lemma 4·7, either the rectangle $[0,\tilde{b}]\times [\text{Fix}(g_2), \text{Fix}(g_5)]$ or $[0,\tilde{b}]\times [\text{Fix}(g_2), g_8(\text{Fix}(g_2))]$ mapped into itself by all the maps of the Furstenberg IFS $\mathcal{J}$ .

In the first situation, $\tilde{I}=[\text{Fix}(g_2), \text{Fix}(g_5)]$ . In this case, the cylinder corresponding to $h_2$ is placed at the bottom of the invariant set (on the Y-axis) and the cylinder corresponding to $h_5$ is placed at the top of the invariant set (on the left side of the Y-axis). Since ${1}/{3s_2}+{1}/{3s_5}\lt1$ and $s_5\leq s_4$ , we have

\begin{align*}h_2([0,\tilde{b}]\times \tilde{I})\cap h_5([0,\tilde{b}]\times \tilde{I})=\emptyset \quad\text{and}\quad h_2([0,\tilde{b}]\times \tilde{I})\cap h_4([0,\tilde{b}]\times \tilde{I})=\emptyset.\end{align*}

By Equation 4·3, we have

\begin{align*}\big(h_4([0,\tilde{b}]\times \tilde{I})\cup h_7([0,\tilde{b}]\times \tilde{I})\big)\cap \big(h_5([0,\tilde{b}]\times \tilde{I})\cup h_8([0,\tilde{b}]\times \tilde{I})\big)=\emptyset.\end{align*}

Thus, from the above, it is clear that a point can be contained in at most six cylinders, and so $\# C_x\geq3$ .

In another situation $\tilde{I}=[\text{Fix}(g_2), g_8(\text{Fix}(g_2))]$ , then by using same idea as above, and using the conditions ${1}/{3s_2}+{1}/{3s_8}\lt1$ and $s_8\leq s_7$ , one can get $\# C_x\geq 3$ . The claim then follows from Proposition 2·4.

Proof of Proposition 4·3. Since $\dim_{H}(\mu_F)= \text{ess inf} (\underline{d}_{\mu_F}(x))$ and by Proposition 4·8, we obtain

\begin{align*}\dim_{H}(\mu_F)\geq \frac{\log(1-\frac{s_{\min}}{3^{t_0-2}})}{-\log(3s_{\max})}.\end{align*}

For proving $\dim_{H}(\mu_F)\gt3-t_0$ , it is enough to show that ${\log(1-{s_{\min}}/{3^{t_0-2}})}/$ ${-\log(3s_{\max})}\gt3-t_0.$ We have

\begin{align*} &\frac{\log\big(1-\frac{s_{\min}}{3^{t_0-2}}\big)}{-\log(3s_{\max})}\gt3-t_0\\[5pt] \Leftarrow& \frac{\log\big(1-\frac{2}{3^{t_0-1}}\big)}{-\log(3)}\gt3-t_0\\[5pt] \Leftarrow& \bigg(1-\frac{6}{3^{t_0}}\bigg)\lt \frac{3^{t_0}}{27}\\[5pt] \Leftarrow&\,0 \lt 3^{2t_0}-27\times 3^{t_0} +162. \end{align*}

This implies that if $3^{t_0}\gt 18 $ , then the inequality ${\log(1-{s_{\min}}/{3^{t_0-2}})}/{-\log(3s_{\max})}\gt3-t_0$ holds. Since $s_i\in ({2}/{3},1)$ and the affinity dimension $t_0=1+{\log(\sum_{i=1}^{9}s_i)}/{\log(3)}\gt1+{\log 6}/{\log 3}$ , and so, we have $3^{t_0}\gt18.$ This completes the proof.

4·4. Uniform scaling factors

Next, we will consider a uniform scaling factor $s=s_i\forall i\in \{1,2,\dots,9\}$ in the construction of fractal surfaces. For that, we have the following result:

Corollary 4·10. If $s \in ({2}/{3},1)$ , then

\begin{align*}\dim_{H}(G(\,f^*))=\dim_{B}(G(\,f^*))=3+\frac{\log(s)}{\log(3)}.\end{align*}

The proof of this corollary follows from Theorem 1·3.

Next, we show some dimension results for the typical choice of the uniform scaling factor.

Theorem 4·11. If $s=s_i\forall i\in \{1,2,\dots,9\}$ , then

\begin{align*}\dim_{H}(G(\,f^*))=\dim_{B}(G(\,f^*))=3+\frac{\log(s)}{\log(3)} \text{for a.e.}\;s\in (1/3,1).\end{align*}

Proof. If $s=s_i$ for every $i\in \{1,2,\dots,9\}$ , then the projected IFS $\mathcal{J}_{X}=\{\,f_1,f_2,\dots,f_9\}$ of the Furstenberg IFS is as follows.

\begin{align*}f_1(x)=f_2(x)=f_3(x)=f_6(x)=f_9(x)=\frac{x}{3s},\end{align*}

\begin{align*}f_4(x)=f_7(x)=-\frac{x}{3s}+\frac{a}{s}, \quad f_5(x)=f_8(x)=\frac{x}{3s}+\frac{a}{s}.\end{align*}

And in this case $p_i={1}/{9}$ for every $i\in \{1,2,\dots,9\}.$ Thus, the IFS $\mathcal{J}_{X}$ is equivalent to the IFS $\tilde{\mathcal{J}_{X}}=\{\tilde{f}_1(x)={x}/{3s}, \tilde{f}_2(x)=-{x}/{3s}+{a}/{s}, \tilde{f}_3(x)={x}/{3s}+{a}/{s}\}$ with probability vector $({5}/{9}, {2}/{9},{2}/{9}).$ Let ${P_{X*}}\mu_{F}$ be the projection of the measure $\mu_{F}$ on the X-axis. Let $\epsilon\gt0$ and M be a sufficiently large natural number. Now, we define real analytic maps $r_i \;:\; [{1}/{3}+\epsilon, M]\to (\!-1,1)\setminus \{0\} \}$ and $d_i \;:\; [{1}/{3}+\epsilon, M]\to \mathbb{R} \}$ are as follows:

\begin{align*}r_1(s)=r_{3}(s)=\frac{1}{3s},r_{2}(s)=-\frac{1}{3s},d_1(s)=0, d_2(s)=d_3(s)=\frac{a}{s}.\end{align*}

The IFS $\tilde{\mathcal{J}_{X}}$ is same as a parametric family of self-similar IFS $\mathcal{I}_{s}=\{r_i(s)x+d_i(s)\}_{i=1}^{3}.$ For $s\in (1,M]$ , the parametric IFS $\mathcal{I}_{s}$ satisfies the strong separation condition. Let $A_{s}$ be the attractor of the parametric IFS $\mathcal{I}_{s}$ . Let $\Pi_{s}\;:\; \Sigma=\{1,2,3\}^{\mathbb{N}}\to A_s$ be the associated natural projection. Then,

\begin{align*}\forall \mathbf{i},\mathbf{j}\in \Sigma, \quad \Pi_{s}(\mathbf{i})=\Pi_{s}(\mathbf{j})\quad \text{on} \quad \bigg[\frac{1}{3}+\epsilon, M\bigg]\iff \mathbf{i}=\mathbf{j}.\end{align*}

Thus by the result of Hochman [Reference Hochman11, Theorem 1·10], we get

\begin{align*}\dim_{H}({P_{X*}}\mu_{F})=\min\bigg\{1,\frac{-\sum_{i=1}^{3}p_i\log(p_i)}{-\sum_{i=1}^{3}p_i\log(|r_i|)}=\frac{\frac{-5}{9}\log(\frac{5}{9})-\frac{4}{9}\log(\frac{2}{9})}{\log(3s)}\bigg\}\,\text{for }a.e.\;s\in \bigg[\frac{1}{3}+\epsilon, M\bigg].\end{align*}

The $\epsilon$ is arbitrarily small. Thus, we have

\begin{align*}\dim_{H}(\mu_{F})\geq \dim_{H}({P_{X*}}\mu_{F})=\min\bigg\{1,\frac{\frac{-5}{9}\log(\frac{5}{9})-\frac{4}{9}\log(\frac{2}{9})}{\log(3s)}\bigg\}\text{for }a.e.\, s\in \bigg(\frac{1}{3}, 1\bigg).\end{align*}

One can easily see that ${{-5}/{9}\log({5}/{9})-{4}/{9}\log({2}/{9})}/{\log(3s)}\gt3-t={\log({1}/{s})}/{\log(3)}$ and $3-t\in (0,1)$ for all $s\in ( {1}/{3}, 1).$ Thus,

\begin{align*}\dim_{H}(\mu_{F})\gt3-t\,\text{for }a.e.\;s\in \bigg(\frac{1}{3}, 1\bigg).\end{align*}

Thus, by Theorem 2·3, $\dim_{H}(\mu)=t=3+{\log(s)}/{\log(3)}\;\text{for a.e.}s\in (1/3,1).$ This completes the proof.

5·1. Every-type result

We can not proceed with the argument as in the previous section, because the projection of the Furstenberg IFS $\mathcal{J}$ on the X-axis and Y-axis is not an IFS. In this case, we need to directly determine the invariant set for the IFS $\mathcal{J}$ and with the help of that, we estimate the overlapping number for the IFS $\mathcal{J}$ . The fixed points of the maps $h_i$ ’s are as follows:

\begin{align*}\text{Fix}(h_1)=\bigg(\frac{4s}{4s-2}, \frac{4s}{\sqrt {3}(4s-2)} \bigg), \quad \text{Fix}(h_2)=\bigg(-\frac{4s}{4s-2}, \frac{4s}{\sqrt {3}(4s-2)} \bigg),\end{align*}

\begin{align*}\text{Fix}(h_3)=\bigg(0, -\frac{8s}{\sqrt {3}(4s-2)} \bigg), \quad \text{Fix}(h_4)=(0,0).\end{align*}

Fig. 7.

The invariant convex hull of the Furstenberg IFS (5·2).

Now, we prove our claim. We have

\begin{align*}h_1(A)=A,h_1(B)=\bigg(1, \frac{4s-6}{\sqrt{3}(4s-2)}\bigg)=B_{1}, h_1(C)=\bigg(\frac{4s-4}{4s-2}, \frac{4s}{\sqrt{3}(4s-2)}\bigg)=C_{1},\end{align*}

\begin{align*}h_{1}(A')=\bigg(\frac{-1+s(4s-2)}{s(4s-2)}, \frac{-1+s(4s-2)}{s\sqrt{3}(4s-2)}\bigg)=A_{1}', h_1(B')=\bigg(1, \frac{2+s(4s-2)}{s\sqrt{3}(4s-2)}\bigg)=B_{1}',\end{align*}

\begin{align*}h_1(C')=\bigg(\frac{1+s(4s-2)}{s(4s-2)}, \frac{-1+s(4s-2)}{s\sqrt{3}(4s-2)}\bigg)=C_{1}'.\end{align*}

One can see that the points $B_{1}'$ and $C_{1}'$ are on the line AC ^′ and AB ^′, respectively. For $s\in ({1}/{2},1),$ we get $1\lt{2}/{(4s-2)}.$ And the point $(1, {((4s-1)(4s-2)-8s)}/{(\sqrt{3}(4s-2))})$ is on the line CB ^′. Since $-{2}/{(\sqrt{3}(4s-2))}\gt{(4s-6)}/{(\sqrt{3}(4s-2))}\gt {((4s-1)(4s-2)-8s)}/$ ${(\sqrt{3}(4s-2))},$ the point $B_1$ is on the above side of line CB ^′. For $s\in ({1}/{2},1),$ we have $-{2}/{(4s-2)} \lt {(4s-4)}/{(4s-2)} \lt 0$ . Thus the point $C_1$ is in the left side of the set $\text{Con}(ABCA'B'C')$ and on the line joining A and B. This implies that $h_1(\text{Con}(ABCA'B'C'))\subset \text{Con}(ABCA'B'C').$ For the mapping $h_2$ , we have

\begin{align*}h_2(B)=B,h_2(A)=\bigg(\!-1, \frac{4s-6}{\sqrt{3}(4s-2)}\bigg)=A_{2}, h_2(C)=\bigg(-\frac{4s-4}{4s-2}, \frac{4s}{\sqrt{3}(4s-2)}\bigg)=C_{2},\end{align*}

\begin{align*}h_{2}(B')=\bigg(\!-\frac{-1+s(4s-2)}{s(4s-2)}, \frac{-1+s(4s-2)}{s\sqrt{3}(4s-2)}\bigg)=B_{2}', h_2(A')=\bigg(\!-1, \frac{2+s(4s-2)}{s\sqrt{3}(4s-2)}\bigg)=A_{2}',\end{align*}

\begin{align*}h_2(C')=\bigg(-\frac{1+s(4s-2)}{s(4s-2)}, \frac{-1+s(4s-2)}{s\sqrt{3}(4s-2)}\bigg)=C_{2}'.\end{align*}

Thus, one can see that $h_2(\text{Con}(ABCA'B'C'))$ is the mirror image of $h_1(\text{Con}(ABCA'B'C'))$ with respect to Y-axis. Thus, due to symmetry of the set $\text{Con}(ABCA'B'C')$ with respect to the Y-axis, we get $h_2(\text{Con}(ABCA'B'C'))\subset \text{Con}(ABCA'B'C').$ For the mapping $h_3$ , we have

\begin{align*}h_3(C)=C,h_3(C')=\bigg(0, \frac{2(1-s(4s-2))}{\sqrt{3}s(4s-2)}\bigg)=C_{3}',\end{align*}

\begin{align*}h_{3}(A)=\bigg(\frac{-2}{4s-2}, \frac{6-8s}{\sqrt{3}(4s-2)}\bigg)=A_{3}, h_3(A')=\bigg(\frac{1}{s(4s-2)}, \frac{-1-2s(4s-2)}{\sqrt{3} s(4s-2)}\bigg)=A_{3}',\end{align*}

\begin{align*}h_3(B)=\bigg(\frac{2}{4s-2}, \frac{6-8s}{\sqrt{3}(4s-2)}\bigg)=B_{3} ,h_3(B')=\bigg(\frac{-1}{s(4s-2)}, \frac{-1-2s(4s-2)}{\sqrt{3} s(4s-2)}\bigg)=B_{3}'.\end{align*}

One can see that the points $A_{3}'$ and $B_{3}'$ are on the line CB ^′ and CA ^′, respectively. For $s\in ({1}/{2}, 1),$ we have ${-2}/{(\sqrt{3}(4s-2))} \lt ({6-8s})/{(\sqrt{3}(4s-2))}\lt {(-1-2s(4s-2))}/{(\sqrt{3} s(4s-2))}$ and ${(6-8s)}/{(\sqrt{3}(4s-2))} \lt {(-1-2s(4s-2))}/{(\sqrt{3} s(4s-2))} \lt {4}/{(\sqrt{3}(4s-2))}.$ This implies that $h_3(\text{Con}(ABCA'B'C'))\subset \text{Con}(ABCA'B'C').$ For the mapping $h_4$ , we have

\begin{align*}h_4(A)=A', h_4(B)=B', h_4(C)=C',\end{align*}

\begin{align*}h_4(A')=\bigg(\frac{1}{s(4s-2)}, \frac{1}{\sqrt{3} s(4s-2)}\bigg)=A_{4}', h_4(B')=\bigg(\frac{-1}{s(4s-2)}, \frac{1}{\sqrt{3} s(4s-2)}\bigg)=B_{4}',\end{align*}

\begin{align*}h_4(C')=\bigg(0,\frac{-2}{\sqrt{3}s(4s-2)}\bigg).\end{align*}

For $s\in ({1}/{2},1),$ we have ${-8s}/{(\sqrt{3}(4s-2))} \lt {-2}/{(\sqrt{3}s(4s-2))} \lt {-2}/{(\sqrt{3}(4s-2))},$ $0 \lt {1}/{(s(4s-2))}\lt {2}/{(4s-2)}$ and $0 \lt {1}/{(\sqrt{3}s(4s-2))} \lt {2}/{(\sqrt{3}(4s-2))}.$ This implies that $h_4(\text{Con}(ABCA'B'C'))\subset \text{Con}(ABCA'B'C').$ Thus the set $\text{Con}(ABCA'B'C')$ is an invariant set for the IFS $\mathcal{J}.$

Now, we will estimate the overlapping number. For $s=({1+\sqrt{5}})/{4},$ one can see that $A_{1}'=B_{2}'=C_{3}'=(0,0)$ , $h_1(\text{Con}(ABCA'B'C'))$ is on the right-hand side, $h_2(\text{Con}(ABCA'B'C'))$ is on the left-hand side (mirror image of $h_1(\text{Con}(ABCA'B'C'))$ with respect to Y-axis) and $h_3(\text{Con}(ABCA'B'C'))$ is below the X-axis. Thus, only at (0,0), the overlapping number is 4. This implies that $K\leq 3$ a.e. $x\in A_{F}.$ For a visualisation, see Figure 8.

Fig. 8.

The first level cylinder sets for the Furstenberg IFS (5·2).

Now, we assume that $s\in (({1+\sqrt{5}})/{4},1)$ . In this case, we have $s(4s-2)\gt1.$ Thus, $A_{1}'$ and $B_{1}'$ are in the 1st and 2nd quadrant of the set $\text{Con}(ABCA'B'C'),$ respectively. And $C_{3}'$ is on negative Y-axis. The points $C_{1}'$ and $C_{2}'$ are on the above of the X-axis and on line AB ^′ and BA ^′, respectively. Thus, any point of $\text{Con}(ABCA'B'C')$ can lie at most two of the sets $h_1(\text{Con}(ABCA'B'C'))$ , $h_2(\text{Con}(ABCA'B'C'))$ and $h_3(\text{Con}(ABCA'B'C'))$ . Thus, $K\leq 3.$ This completes the proof.

5·2. Almost every-type result

Next, we discuss almost sure results for the above-constructed fractal surfaces.

Construction of Graph directed IFS associated with the projection of the Furstenberg IFS

One can see that the group $( \mathcal{G})$ generated by the linear part of the Furstenberg IFS $\mathcal{J}=\{h_1,h_2,h_3,h_4\}$ is a finite group of order $12.$ Precisely, the group elements are as follows:

\begin{align*}Q_1=\begin{bmatrix} \frac{1}{2} &\quad \frac{\sqrt 3}{2} \\[4pt] \frac{\sqrt 3}{2} &\quad \frac{-1}{2} \end{bmatrix}\!, Q_2=\begin{bmatrix} \frac{1}{2} &\quad \frac{-\sqrt 3}{2} \\[4pt] \frac{-\sqrt 3}{2} &\quad \frac{-1}{2} \end{bmatrix}\!, Q_3=\begin{bmatrix} -1 &\quad 0 \\[4pt] 0 &\quad 1 \end{bmatrix}\!, Q_4=\begin{bmatrix} -1 &\quad 0 \\[4pt] 0 &\quad -1 \end{bmatrix}\!,\end{align*}

\begin{align*}Q_5=\begin{bmatrix} 1 &\quad 0 \\[4pt] 0 &\quad 1 \end{bmatrix}\!, Q_6=\begin{bmatrix} \frac{-1}{2} &\quad \frac{-\sqrt 3}{2} \\[4pt] \frac{-\sqrt 3}{2} &\quad \frac{1}{2} \end{bmatrix}\!, Q_7=\begin{bmatrix} \frac{-1}{2} &\quad \frac{\sqrt 3}{2} \\[4pt] \frac{\sqrt 3}{2} &\quad \frac{1}{2} \end{bmatrix}\!, Q_8=\begin{bmatrix} \frac{-1}{2} &\quad \frac{\sqrt 3}{2} \\[4pt] \frac{-\sqrt 3}{2} &\quad \frac{-1}{2} \end{bmatrix}\!,\end{align*}

\begin{align*}Q_9=\begin{bmatrix} \frac{-1}{2} &\quad \frac{-\sqrt 3}{2} \\[4pt] \frac{\sqrt 3}{2} &\quad \frac{-1}{2} \end{bmatrix}\!, Q_{10}=\begin{bmatrix} 1 &\quad 0 \\[4pt] 0 &\quad -1 \end{bmatrix}\!, Q_{11}=\begin{bmatrix} \frac{1}{2} &\quad \frac{-\sqrt 3}{2} \\[4pt] \frac{\sqrt 3}{2} &\quad \frac{1}{2} \end{bmatrix}\!, Q_{12}=\begin{bmatrix} \frac{1}{2} &\quad \frac{\sqrt 3}{2} \\[4pt] \frac{-\sqrt 3}{2} &\quad \frac{1}{2} \end{bmatrix}\!.\end{align*}

First, we consider a direction $v=(1,1)\in \mathbb{R}^2$ . Now, we construct a graph directed IFS associated with the Furstenberg IFS $\mathcal{J}$ . First, we define a set of vertices $\mathcal{V}\,:\!=\,\{v_1,v_2,\dots,v_{12}\}$ such that $v_i=Q_{i}^{T}v.$ The set $\mathcal{E}_{l,m}$ denotes the set of all directed edges from vertex $v_{l}$ to $v_{m}$ . For $l,m\in \mathcal{V}$ , if these exists a $k\in \{1,2,3,4\}$ such that $v_{l}=Q_{k}v_{m}$ , then we define a directed edge $e\in \mathcal{E}_{l,m}$ . Since $\#{\{Q_{l}Q_{k}\;:\;k\in {1,2,3,4}\}}=4$ for each $l\in \{1,2,\dots,12\}$ , there are only 4 directed edges from the vertex $v_{l}$ and only 4 directed edges toward the vertex $v_{l}$ . The set of directed edges is defined by $\mathcal{E}\,:\!=\,\{\mathcal{E}_{l,m}\;:\;1\leq l,m\leq 12\}$ . The directed graph is denoted by $G(\mathcal{V},\mathcal{E}).$ For $e\in \mathcal{E}_{l,m}$ , the associated map $f_{e}$ is defined by

\begin{align*}f_{e}(x)=\frac{1}{2s}x+v_{l}\cdot t_{k},\end{align*}

where $k\in \{1,2,3,4\}$ such that $v_{l}=Q_{k}v_{m}$ and $t_k$ is the translation vector corresponding to the map $h_{k}$ in the Furstenberg IFS $\mathcal{J}$ . Let $A_F$ be the attractor of the IFS $\mathcal{J}$ . One can also see that for each $l\in \{1,2,\dots,12\}$ , we have $v_{l}\cdot t_{k_1}\ne v_{l}\cdot t_{k_2}$ for all $k_{1}\ne k_{2}\in \{1,2,3,4\}$ . The notation $\text{Proj}_{v}(A)$ denotes the projection of the set A in the direction v. Thus, we have

\begin{align*} \text{Proj}_{v_l}(A_F)&= \text{Proj}_{v_l}\bigg(\bigcup_{k=1}^{4}h_k(A_F)\bigg)=\bigcup_{k=1}^{4}\text{Proj}_{v_l}\bigg(\frac{1}{2s}Q_{k}(A_F)+t_{k}\bigg)\\&= \bigcup_{k=1}^{4}\bigg(\frac{1}{2s}\text{Proj}_{Q_{k}^{T}v_l}(A_F)+\text{Proj}_{v_l}(t_{k})\bigg)=\bigcup_{m=1}^{12}\bigcup_{e\in \mathcal{E}_{l,m}}f_{e}( \text{Proj}_{v_m}(A_F)).\end{align*}

Thus, $\bigcup_{l=1}^{12}\text{Proj}_{v_l}(A_F)$ is the attractor of the graph directed IFS $\{\,f_{e}\;:\; e\in \mathcal{E}\}$ with directed graph $G(\mathcal{V},\mathcal{E}).$

Induced Markov chain

Let $X_n$ be a Markov chain on the group $\mathcal{G}$ . For each $l\in \{1,2,\dots,12\}$ , one can see that $\#{\{Q_{l}Q_{k}\;:\;k\in {1,2,3,4}\}}=4$ . Thus, we define the transition probability for the Markov chain by

\begin{align*}\mathbb{P}(X_{n+1}=Q_m | X_n=Q_l)=\begin{cases} \frac{1}{4}\quad \text{if } \quad \exists k\in \{1,2,3,4\} \text{with} Q_{l}Q_{k}=Q_{m}\\ 0 \quad \text{otherwise}.\end{cases}\end{align*}

Let P be the transition matrix associated with the Markov chain $X_n$ . Thus, the matrix P is as follows

\begin{align*}[P]_{i,j}=\begin{cases} \frac{1}{4}\quad \text{if } \quad \exists k\in \{1,2,3,4\} \;\text{with}\; Q_{i}Q_{k}=Q_{j}\\ 0 \quad \text{otherwise},\end{cases}\end{align*}

where $[P]_{i,j}$ is the ijth entry of the matrix P for $1\leq i,j\leq 12$ .

We have

\begin{align*}Q_2 \cdot Q_1=Q_8, Q_1 \cdot Q_2=Q_9, Q_2 \cdot Q_4=Q_7, Q_1 \cdot Q_4=Q_6=Q_4 \cdot Q_1, \end{align*}

\begin{align*}Q_1 \cdot Q_3=Q_8, Q_2 \cdot Q_3=Q_9,Q_{11} \cdot Q_1=Q_7, Q_{12} \cdot Q_{1}=Q_{10},\end{align*}

\begin{align*}&Q_1 \cdot Q_1= Q_2 \cdot Q_2=Q_3 \cdot Q_3=Q_4 \cdot Q_4=Q_5=Id,\nonumber\\[6pt]& Q_4 \cdot Q_3=Q_{10}, Q_1 \cdot Q_{10}=Q_{11}, Q_2 \cdot Q_{10}=Q_{12}.\end{align*}

Now, we define two sets $A=\{Q_5,Q_6,Q_7,Q_8,Q_9,Q_{10}\}$ and $B=\{Q_1,Q_2,Q_3,Q_4,Q_{11},Q_{12}\}$ . By the above one can see that for each $Q_i\in A$ , there exist a $k\in \{1,2,3,4\}$ such that $Q_iQ_{k}\in B$ and vice versa. Thus, the directed graph associated with the transition matrix P is bipartite graph. So, by considering the arrangement $\{Q_5,Q_6,Q_7,Q_8,Q_9,Q_{10},Q_1,Q_2,Q_3, Q_4,Q_{11},Q_{12}\}$ , the transition matrix P can be rewritten in the following form

\begin{equation*}P=\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}0 & 0 & 0 & 0 & 0 & 0 & 1/4 & 1/4 & 1/4 & 1/4 & 0 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 1/4 & 0 & 0 & 1/4 & 1/4 & 1/4\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 1/4 & 0 & 1/4 & 1/4 & 1/4\\0 & 0 & 0 & 0 & 0 & 0 & 1/4 & 1/4 & 1/4 & 0 & 1/4 & 0\\0 & 0 & 0 & 0 & 0 & 0 & 1/4 & 1/4 & 1/4 & 0 & 0 & 1/4\\0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1/4 & 1/4 & 1/4 & 1/4\\1/4 & 1/4 & 0 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\1/4 & 0 & 1/4 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\1/4 & 0 & 0 & 1/4 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 0 & 0\\1/4 & 1/4 & 1/4 & 0 & 0 & 1/4 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 1/4 & 1/4 & 1/4 & 0 & 1/4 & 0 & 0 & 0 & 0 & 0 & 0\\0 & 1/4 & 1/4 & 0 & 1/4 & 1/4 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]\!.\end{equation*}

Since $Q_{5}$ is identity matrix (Id), we have

\begin{align*}\mathbb{P}(X_{n}= Id | X_0=Id)=\begin{bmatrix}1&0&\cdots&0 \end{bmatrix}P^{n}\begin{bmatrix}1\\{\underline{0}}\\\vdots\\{\underline{0}} \end{bmatrix}=\frac{\#\{\tau\in \{1,2,3,4\}^n\;:\; Q_{\tau}=Id\}}{4^n}.\end{align*}

Clearly, $\mathbb{P}(X_{2n+1}= Id | X_0=Id)=0$ for $n\in \mathbb{N}\cup \{0\}$ and $\mathbb{P}(X_{2}= Id | X_0=Id)={1}/{4}.$ The period of the directed graph associated with the transition matrix P is defined as follows

\begin{align*}\text{period}=\text{l.c.d.}\{n\in \mathbb{N}\,:\, \mathbb{P}(X_{n}= Id | X_0=Id) \gt 0\}.\end{align*}

This implies that period is 2. But, $P^2=\begin{bmatrix} R&0\\{\underline{0}}&S\end{bmatrix}$ , where R and S are $6\times 6$ matrices. The matrix R is as follows

\begin{equation*}R=\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}1/4 & 1/8 & 1/8 & 3/16 & 3/16 & 1/8\\1/8 & 1/4 & 3/16 & 1/8 & 1/8 & 3/16\\1/8 & 3/16 & 1/4 & 1/8 & 1/8 & 3/16\\3/16 & 1/8 & 1/8 & 1/4 & 3/16 & 1/8\\3/16& 1/8 & 1/8 & 3/16 & 1/4 & 1/8\\1/8 & 3/16 & 3/16 & 1/8 & 1/8 & 1/4\end{array}\right]\!.\end{equation*}

The above matrix R is irreducible and aperiodic (period is 1). Thus by Perron-Frobenius theorem, we get

\begin{align*}R^n\to \begin{bmatrix}1\\1\\1\\1\\1\\1 \end{bmatrix} \begin{bmatrix}1/6&1/6&1/6&1/6&1/6&1/6\end{bmatrix}\quad \text{as}\quad n\to \infty.\end{align*}

This implies that

\begin{align*}\mathbb{P}(X_{2n}= Id | X_0=Id)=\frac{\#\{\tau\in \{1,2,3,4\}^{2n}\;:\; Q_{\tau}=Id\}}{4^{2n}}\to \frac{1}{6}\quad \text{as}\quad n\to \infty.\end{align*}

Thus, these exists an $N_{0}\in \mathbb{N}$ such that

(5·3) \begin{equation} \#\{\tau\in \{1,2,3,4\}^{2n}\;:\; Q_{\tau}=Id\}\geq \frac{4^{2n}}{12} \quad \forall n\geq N_{0}.\end{equation}

Non-degeneracy of constructed GD-IFS

Now, we will show that the constructed graph directed IFS is non-degenerate, which will be useful for determining almost sure results. The set of all infinite-length edges is denoted by ${\mathcal{E}}^*$ . Let $\mathbf{e}=(e_1,e_2,\dots), \mathbf{e'}=(e'_1,e'_2,\dots)\in {\mathcal{E}}^*$ such that $\mathbf{e}\ne \mathbf{e'}$ and $e_1\in \mathcal{E}_{l,m_1}, e'_1\in \mathcal{E}_{l,m_2}.$ Let $m\in \mathbb{N}$ be smallest number such that $e_m\ne e'_{m}$ . Then, there exist $k_1\ne k_2\in \{1,2,\dots,12\}$ such that $e_m\in \mathcal{E}_{l_1,m_{k_1}}$ and $e'_m\in \mathcal{E}_{l_1,m_{k_2}}$ . Let $\Pi$ be the natural projection corresponding to the constructed graph-directed IFS. Thus, we have

\begin{align*}\Pi(\sigma^{m-1} \mathbf{e})=f_{e_m}(\Pi(\sigma^{m} \mathbf{e}))=\frac{1}{2s}(\Pi(\sigma^{m} \mathbf{e}))+v_{l_1}\cdot t_{r_1}\end{align*}

\begin{align*}\Pi(\sigma^{m-1} \mathbf{e'})=f_{e'_m}(\Pi(\sigma^{m} \mathbf{e'}))=\frac{1}{2s}(\Pi(\sigma^{m} \mathbf{e'}))+v_{l_1}\cdot t_{r_2},\end{align*}

where $r_1\ne r_2$ . Since $v_{l_1}\cdot t_{r_1}\ne v_{l_1}\cdot t_{r_2}$ , we have $\Pi(\sigma^{m-1} \mathbf{e})\not\equiv \Pi(\sigma^{m-1} \mathbf{e'})$ as analytic functions of s on $(1/2,\infty)$ , and in particular, on $(1/2,1)$ . Thus, we get $\Pi(\mathbf{e})\not\equiv \Pi(\mathbf{e'})$ . This implies that the graph-directed IFS is non-degenerate.

Now, by using the above constructed GD-IFS, we will prove typical type results for the Hausdorff dimension.

Proof of Theorem 1·5. The upper bound follows by (5·1). Now, we show the lower bound.

Set $\mathcal{N}_{n}\,:\!=\,\#\{\tau\in \{1,2,3,4\}^{2n}\;:\; Q_{\tau}=Id\}.$ For $n\geq N_0$ , we define an self-affine IFS

\begin{align*}\Phi_{n}\,:\!=\,\{W_{\tau}\;:\; Q_{\tau}=Id\quad \text{and}\quad \tau\in \{1,2,3,4\}^{2n}\}.\end{align*}

Let $A_{\Phi_n}$ be the attractor of the IFS $\Phi_n$ . The affinity dimension ( $t_n$ ) of the self-affine IFS $\Phi_n$ is uniquely determined by following equation

\begin{align*}\sum_{\substack{\tau\in \{1,2,3,4\}^{2n}\\Q_{\tau}=Id}} s^{2n}\bigg(\frac{1}{2^{2n}}\bigg)^{t_{n}-1}=1.\end{align*}

Let $(p_{i})_{i=1}^{\mathcal{N}_n}$ be a probability vector such that $p_i={1}/{\mathcal{N}_n}$ for all $1\leq i\leq \mathcal{N}_n$ . Let $\mathcal{F}_{n}\,:\!=\,\{h_{\tau}\;:\; Q_{\tau}=Id\quad \text{and} \quad \tau\in \{1,2,3,4\}^{2n}\}\}$ be the Furstenberg IFS corresponding to the IFS $\Phi_{n}$ . Let $\mu_{\mathcal{F}}^{n}$ be the invariant measure corresponding to the IFS $\mathcal{F}_{n}$ with probability vector $(p_{i})_{i=1}^{\mathcal{N}_n}.$ Since the graph directed IFS $\{\,f_{e}\;:\; e\in \mathcal{E}\}$ with directed graph $G(\mathcal{V},\mathcal{E})$ corresponding to the Furstenberg IFS $\mathcal{J}$ is non-degenerate, the projection of the $\mathcal{F}_{n}$ on the direction $v=(1,1)\in \mathbb{R}^2$ is also non-degenerate. Then, by the result of Hochman [Reference Hochman 11 ], there exist a set $\mathcal{E}_{n}\subset ({1}/{2},1)$ with $\dim_{H}(\mathcal{E}_{n})=0$ such that

\begin{align*}\dim_{H}(\mu_{\mathcal{F}}^{n})\geq \min\bigg\{1,\frac{-\sum_{i=1}^{\mathcal{N}_n}p_i\log p_i}{\sum_{i=1}^{\mathcal{N}_n}p_i\log (2s)^{2n}}\bigg\}=\min\bigg\{1,\frac{\log \mathcal{N}_n}{2n\log (2s)}\bigg\}\quad \forall s\in \bigg(\frac{1}{2},1\bigg)\setminus \mathcal{E}_{n}.\end{align*}

Now, by using the estimate (5·3) of $\mathcal{N}_n$ , we get

\begin{align*}\dim_{H}(\mu_{\mathcal{F}}^{n})\geq \min\bigg\{1,\frac{\log 4}{\log (2s)}-\frac{\log 12}{2n \log 2s}\bigg\}\quad \forall s\in \bigg(\frac{1}{2},1\bigg)\setminus \mathcal{E}_{n}.\end{align*}

Let $\hat{t}_n \in \mathbb{R}$ be such that

\begin{align*}\sum_{\substack{\tau\in \{1,2,3,4\}^{2n}\\Q_{\tau}=Id}} s^{2n}\bigg(\frac{1}{2^{2n}}\bigg)^{\hat{t}_{n}-1}\geq \frac{4^{2n}}{12}s^{2n}\bigg(\frac{1}{2^{2n}}\bigg)^{\hat{t}_{n}-1}=1.\end{align*}

This implies that $t_{n}\geq \hat{t}_n$ and

\begin{align*}\hat{t}_{n}=1+\frac{\log 4s}{\log 2}-\frac{\log 12}{2n \log 2}=3+\frac{\log s}{\log 2}-\frac{\log 12}{2n \log 2}.\end{align*}

Clearly, $\hat{t}_n \to t$ as $n\to \infty$ . Since ${\log 4}/{\log (2s)}\gt-{\log s}/{\log 2} \forall s\in \big({1}/{2},1\big)$ , we have

\begin{align*}\min\bigg\{1,\frac{\log 4}{\log (2s)}-\frac{\log 12}{2n \log 2s}\bigg\}\gt3-\hat{t}_{n}\geq 3-t_n \forall s\in \bigg(\frac{1}{2},1\bigg)\end{align*}

for all $n\geq N_1$ , where $N_1\in \mathbb{N}$ is some large number. This implies that for $n\geq \max\{N_0,N_1\}$ we get

\begin{align*}\dim_{H}(\mu_{\mathcal{F}}^{n})\gt3-t_n\quad \forall s\in \bigg(\frac{1}{2},1\bigg)\setminus \mathcal{E}_{n}.\end{align*}

Then, by Theorem 2·3, for $n\geq \max\{N_0,N_1\}$ we obtain

\begin{align*}\dim_{H}(A_{\Phi_n})=t_n\geq \hat{t}_n \quad \forall s\in \bigg(\frac{1}{2},1\bigg)\setminus \mathcal{E}_{n}.\end{align*}

Set $\mathcal{E}\,:\!=\,\bigcup_{n=\max\{N_0,N_1\}} \mathcal{E}_n.$ Thus, $\dim_{H}(\mathcal{E})=0.$ This implies that

\begin{align*}\dim_{H}(G(\,f^*))\geq \dim_{H}(A_{\Phi_n})\geq \hat{t}_n \quad \forall s\in \bigg(\frac{1}{2},1\bigg)\setminus \mathcal{E}.\end{align*}

Thus, for all $s\in \big({1}/{2},1\big)\setminus \mathcal{E},$ we get

\begin{align*}\dim_{H}(G(\,f^*))\geq t \quad \forall s\in \bigg(\frac{1}{2},1\bigg)\setminus \mathcal{E}.\end{align*}

Since $\dim_{H}(G(\,f^*))\leq \overline{\dim}_{B}(G(\,f^*)) \leq t=3+{\log(s)}/{\log(2)},$ we have

\begin{align*}\dim_{H}(G(\,f^*))={\dim}_{B}(G(\,f^*))= t=3+\frac{\log(s)}{\log(2)}\end{align*}

for all $s\in \big({1}/{2},1\big )\setminus \mathcal{E}$ . This completes the proof.