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.