resp. $H_b(m,n)$ are defined by (\[tildeGa\]) resp. (\[DefH\_c\]).

We follow the proof of the Proposition in [@Za] §4. Both sides in are clearly periodic in $m$ with period $a$. Therefore it suffices to show that the finite Fourier transforms are equal, i.e., that for every $h\pmod{a}$ we have $$\begin{aligned}
\label{eq:key2}
\frac{1}{ a }\sum_{m\,(a)}e\left(-\frac{hm}{a}\right) \tilde G_a(mp^2,\nu) =
\sum_{m\,(a)}e\left(-\frac{hm}{a}\right)\sum_{\substack{r|\nu \\ r|a}} \epsilon_p(r) H_{a/r}\left(
\frac{p\nu\nu'}{ r^2},m\right).\end{aligned}$$ Inserting the definition of $H_{c}\left( n,m\right)$ we find that the right hand side equals $$\begin{aligned}
\sum_{\substack{r|\nu \\ r|a}}\epsilon_p(r)\frac{r}{a} \sum_{d\,(a/r)^*} e\left( \frac{dp\nu\nu'/r^2}{a/r}\right)
\sum_{m\,(a)}e\left(-m\frac{\bar d r+h}{a}\right).\end{aligned}$$ The sum over $m\,(a)$ vanishes unless $\bar d r+h\equiv0\pmod{a}$ in which case it equals $a$. But $\bar d
r+h\equiv0\pmod{a}$ implies that $h\equiv 0\pmod{r}$ and $\bar d +h/r\equiv0\pmod{a/r}$. Since $\bar d$ is coprime to $a/r$ we find that $h/r$ must also be coprime to $a/r$ and consequently $r=(h,a)$.

Thus, denoting $r=(h,a)$ and by $\bar h_1$ a multiplicative inverse of $h/r$ modulo $a/r$ we obtain that the right hand side of equals $$\begin{aligned}
 \label{eq:key3}
\epsilon_p(r)r \cdot e\left( -\frac{\bar h_1 p\nu\nu'/r^2}{a/r}\right).\end{aligned}$$

We now consider the left hand side of . We first assume that $(p,a)=1$. In this case it is equal to $$\begin{aligned}
&\frac{\epsilon_p(a)}{ a }\sum_{m\,(a)}\sum_{\substack{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F\\ \lambda\lambda'=-mp\,(a)}} e\left(\frac{{\operatorname{tr}}(\lambda \nu)-hm}{a}\right) \\
&=\frac{\epsilon_p(a)}{ a }\sum_{\substack{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F}} e\left(\frac{{\operatorname{tr}}(\lambda 