We denote by $K$ the set of triplets $(\alpha,\beta,\gamma)$ of $\mathbf{R}^3$ consisting of three non-negative real numbers such that $\alpha+\beta+\gamma=1$. If $(a,b,c)\in\mathbf{C}^3$, we denote by $\widehat{abc}$ the filled triangle defined by $\widehat{abc} = \{\alpha a + \beta b + \gamma c / (\alpha,\beta,\gamma)\in K\}$. We denote $\tau = \widehat{(-1)\,1\,\mathrm{i}}$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$.
Let $(r_n)_{n\geq 1}$ be an element of $\{0,1\}^{\mathbf{N}^*}$. For each non-zero natural number $n$, we denote $\tilde{\tau}_n = \phi_{r_1}\circ\phi_{r_2}\circ\cdots\circ\phi_{r_n}(\tau)$.
Show that $\bigcap_{n\geq 1}\tilde{\tau}_n$ is reduced to a single point belonging to $\tau$.

We denote by $\mathcal{E}$ the set of continuous maps $g$ from $[0,1]$ to $\mathbf{C}$ such that $g(0)=-1$ and $g(1)=1$. The norm of uniform convergence on the $\mathbf{C}$-vector space of continuous maps from $[0,1]$ to $\mathbf{C}$ is denoted $\|\cdot\|_\infty$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. If $g\in\mathcal{E}$, we denote by $Tg$ the map from $[0,1]$ to $\mathbf{C}$ defined by: $$Tg(x) = \phi_0(g(2x)) \text{ if } x\in\left[0,\frac{1}{2}\right] \text{ and } Tg(x) = \phi_1(g(2x-1)) \text{ if } x\in\left]\frac{1}{2},1\right]$$ We now define a sequence $(f_n)_{n\in\mathbf{N}}$ of elements of $\mathcal{E}$ by choosing $f_0$ affine (i.e. $f_0(x) = -1 + 2x$) and $f_{n+1} = Tf_n$ for every natural number $n$.
a) Prove that the sequence $(f_n)$ converges uniformly on $[0,1]$ to a function $f\in\mathcal{E}$.
b) Prove that $Tf = f$.
c) Prove that, for all $x\in[0,1]$, $f(x) = -\overline{f(1-x)}$ and give a geometric interpretation of this relation.

We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$, i.e. $f(x) = \phi_0(f(2x))$ for $x\in[0,\frac{1}{2}]$ and $f(x) = \phi_1(f(2x-1))$ for $x\in]\frac{1}{2},1]$.
Let $(r_n)_{n\geq 1}\in\{0,1\}^{\mathbf{N}^*}$.
a) Show that the series with general term $\frac{r_n}{2^n}$ converges and that its sum $x$ belongs to $[0,1]$.
b) By setting for every natural number $p$, $x_p = \sum_{n=1}^{\infty}\frac{r_{n+p}}{2^n}$, prove the relation: $$f(x) = \phi_{r_1}\circ\phi_{r_2}\circ\ldots\phi_{r_p}\left(f\left(x_p\right)\right)$$ for every non-zero natural number $p$.

The integer part of the real number $x$ is denoted $[x]$. For every real $x$ and every non-zero natural number $n$: $r_n(x) = [2^n x] - 2[2^{n-1}x]$. We denote by $\mathbf{Z}\left[\frac{1}{2}\right]$ the set of rationals of the form $\frac{k}{2^n}$ where $k\in\mathbf{Z}$ and $n\in\mathbf{N}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$.
Conversely, let $x\in[0,1[$.
a) Establish that, for every non-zero natural number $n$, $r_n(x)\in\{0,1\}$.
b) Show that, for every non-zero natural number $N$ and every real $x\in[0,1[$: $$\frac{[2^N x]}{2^N} = \sum_{n=1}^{N}\frac{r_n(x)}{2^n} \quad \text{then} \quad x = \sum_{n=1}^{\infty}\frac{r_n(x)}{2^n}.$$
c) Show that if, moreover, $x\in\mathbf{Z}\left[\frac{1}{2}\right]$ then there exists $N\in\mathbf{N}$ such that $r_n(x) = 0$ for every natural number $n > N$.
d) Calculate $f\left(\frac{1}{2}\right)$ and $f\left(\frac{1}{4}\right)$. Recognize $\phi_0\circ\phi_0$ and deduce $f\left(\frac{1}{2^k}\right)$ for all $k\in\mathbf{N}$.