
\begin{proof}~
$\hat{g}(y)=\int_F g(x) \psi(x y) d x=\int_{\varpi^{n_\psi-n_\chi} \OO} \psi(x) \psi(x y) d x$
$$
=\int_{\varpi^{n_\psi-n_\chi} \OO} \psi(x(y+1)) d x
$$
Or $\operatorname{Cond}(\psi)=\varpi^{n_\psi} \OO$ et $\operatorname{Cond}\left(\psi_{y+1}\right)=(y+1)^{-1} \varpi^{n_\psi} \OO$. \\
L'intégrale est nulle sauf si $\varpi^{n_\psi-n_\chi} \OO \subseteq (y+1)^{-1} \varpi^{n_\psi} \OO$, c'est-à-dire, sauf si $y+1 \in \varpi^{n_\chi} \OO$, c'est-à-dire sauf si $y \in -U_{n_\chi}$. \\

Ainsi \begin{aligned}[t]
\hat{g}(y) & =\1_{-U_{n_\chi}}(y) \cdot \operatorname{Vol}\left(\varpi^{n_\psi-n_\chi} \OO, dx\right) \\
& =\1_{-U_{n_\chi}}(y) \cdot|\varpi|^{n_\psi-n_\chi} \quad \operatorname{Vol}(\OO, dx) \\
& =\1_{-U_{n_\chi}}(y) \cdot q^{n_\chi-n_\psi} \cdot| \OO / \mathcal{D} |^{-1/2} 
\end{aligned}
\ \\
\qedhere
\end{proof}



Preuve: $\hat{g}(y)=\int_F g(x) \psi(x y) \, dx=\int_{\varpi^n \mathbb{O}^x} \psi(x) \psi(x y) \, dx$
$$
=\int_{\pi^{m_P-m_x} \mathbb{O}} \psi(x(y+1)) \, dx
$$
$\operatorname{Cond}(\psi)=w^{m_\psi} \mathbb{O}$ et $\operatorname{Cond}\left(\psi_{y+1}\right)=(y+1)^{-1} \varpi^{m_\psi} 0$.
L'intégrale est nulle sauf si- $\varpi^{n_\psi-n_x} \mathbb{O} \subseteq(y+1)^{-1} \varpi^{n_\psi} \mathbb{O}$ c'est-à-dire si $y+1 \in \varpi^n \mathbb{O}$, c'est-à-dire si $y \in -U_{n_x}$
$$
\text { Ainsi, } \begin{aligned}
\hat{g}(y) & =\mathbb{1}_{-u_{n_x}}(y) \operatorname{Vol}\left(\pi^{m_\psi-n_x} \mathbb{O}, dx\right) \\
& =\mathbb{1}_{-u_{n_x}}(y) \cdot q^{n_x-n_\psi} \cdot|\mathbb{O} / \mathscr{D}|^{-1 / 2}
\end{aligned}
$$

