as $$\begin{aligned}
        y(t) &=& \tau \sum_{k \in {\mathbb{Z}}} \rho_\tau(t, k \tau) \sum_{j, \ |\mu_j| \leq \delta/\tau} a_j \Phi_j \exp (i \mu_{j,\tau} k\tau)
        \nonumber
        \\
        & = & \sum_{j, \ |\mu_j| \leq \delta /\tau} a_j \Phi_j \tau \sum_{k \in {\mathbb{Z}}} \rho_\tau(t, k \tau) \exp(i \mu_{j, \tau} k \tau). 
        \nonumber
        \\
        & =& \sum_{j, \ |\mu_j| \leq \delta /\tau} a_j \Phi_j {{\mathcal{F}}}_\tau[ \rho_\tau(t, \cdot) ](- \mu_{j,\tau}).
        \label{XbyTransmutation}
    \end{aligned}$$

According to , for all $\mu_{\tau} \in (- \pi/\tau, \pi/\tau)$, $${{\mathcal{F}}}_\tau [\rho_\tau(t, \cdot)] (\mu_\tau) = \exp\left(-\frac{i g(\mu_\tau \tau) t}{\tau} \right) \chi\left(\mu_\tau \tau \right).$$ Due to the definition of $g$, for all $j$ such that $|\mu_j| \leq \delta/\tau$, $g(\mu_{j, \tau} \tau) = \mu_j\tau $. It then follows from that $$y(t) = \sum_{j, \ |\mu_j| \leq \delta /\tau} a_j \Phi_j \exp( i \mu_j t) \chi(\mu_{j,\tau} \tau).$$ But the choice of $\chi$ implies that $$y(t) = \sum_{j, \ |\mu_j| \leq \delta /\tau} a_j \Phi_j \exp( i \mu_j t).$$ Hence $y(t)$ solves with initial data $y^0$.

\[Remark-a\] When the parameter $R \in {\mathbb{R}}_+^* \cup \{\infty\} $ in is such that $f(R) = \pi$ if $R$ is finite or $\underset{\infty}\lim f = \pi$, i.e. when $f$ is bijective from $(-R, R)$ to $(-\pi, \pi)$, then there is no need 