3) $\ud{v}^\ab$ is not periodic or $\ud{v}^\ab=(k-1)^\infty$.

We prove Proposition \[pro3.4\]. Assume 1. The validity of the $\varphi$-expansion implies that $\ud{u}^\ab$ is a solution of . If $\alpha=0$, then $\ud{u}^{0,\beta}=\ud{0}$ is the only solution of since $\ud{x}\not=\ud{0}$ implies $\ophi^{0,\beta}(\ud{x})>0$ and $x=0$ is a fixed point of $T_{0,\beta}$. Let $0<\alpha<1$. Using Lemma \[lemelementary\] we deduce that $u_0=0$ and $$\alpha=T_\ab(0)=\oph^\ab(u_1+\ophi^\ab(\sigma^2\ud{u}))\,.$$ If $\alpha=a_j$, $j=1,\ldots,k-1$ (see ), then has at least two solutions, which are $0j(\sigma^2\ud{u}^\ab)$ with $\ophi^\ab(\sigma^2\ud{u}^\ab)=T^2(0)=0$ (see ), and $0(j-1)\ud{v}^\ab$ with $\ophi^\ab(\ud{v}^\ab)=1$. Therefore, by our hypothesis we have $\alpha\not\in \{a_1,\ldots,a_{k-1}\}$, $u_1=u_1^\ab$ and $\ophi^\ab(\sigma^2\ud{u}^\ab)=T^2(0)\in (0,1)$. Iterating this argument we conclude that $1\implies 2$.\
Assume 2. If $x=0$ is a fixed point, then $\alpha=0$ and $\ud{u}^{0,\beta}=\ud{0}$. If the orbit of $0$ is not periodic, and the validity of the $\varphi$-expansion imply $$\sigma^k\ud{u}^\ab=\lim_{x\downarrow
0}\i(T^k_\ab(x))\succ\lim_{x\downarrow 0}\i(x)=\ud{u}^\ab\,.$$ Assume 3. From and the validity of the $\varphi$-expansion we get $$\ophi^\ab(\sigma^k\ud{u}^\ab)=T^k_\ab(0)>
\ophi^\ab(\ud{u}^\ab)=0\,,$$ so that the orbit of $0$ is not periodic. The orbit of $0$ is not periodic if and only if $T^k_{\ab}(0)\not\in\{a_1,\ldots,a_{k-1}\}$ for all $k\geq 1$. Using Lemma \[lemelementary\] we conclude that has a unique solution.

Propositions \[pro3.4\] and \[pro3.4bis\] give necessary and sufficient conditions for the existence and uniqueness of the solution of equations . In the 