
\begin{proof}(\cref{thm1})~

Nous avions $\hat{f}(y) = \psi(a y) \int_{\varpi^n \OO} \psi(x y) \,dx$
$$
= \operatorname{Vol}(\varpi^n \OO, dx) \psi(a y) \1_{\varpi^{-n} \mathcal{D}^{-1}}(y)
$$

Donc, $\widehat{\hat{f}} \in C_c^{\infty}(F)$. Maintenant, pour $x \in F$, on calcule :
$$
\begin{aligned}
\widehat{\hat{f}}(x) & = \operatorname{Vol}(\varpi^n \OO, d x) \int_F \1_{\varpi^{-n} \mathcal{D}^{-1}}(y) \psi(a y) \psi(x y) \,dy \\
& = \operatorname{Vol}(\varpi^n \OO, d x) \int_{\varpi^{-n} \mathcal{D}^{-1}} \psi((a+x) y) \,dy .
\end{aligned}
$$
 - Si $ x=-a$, l'intégrale vaut $\operatorname{Vol}(\varpi^{-n} \mathcal{D}^{-1}, d x)$.\\
 - Si $x \neq -a$, l'intégrale est nulle sauf si $\varpi^{-n} \mathcal{D}^{-1} \subseteq (a+x)^{-1} \mathcal{D}^{-1}$, c'est-à-dire si $a+x \in \varpi^n \OO$. \\
 Dans ce cas, elle vaut $\operatorname{Vol}(\varpi^{-n} \mathcal{D}^{-1}, d x)$.\\

On a alors $\widehat{\hat{f}}(x) = |\varpi^n| \operatorname{Vol}(\OO, d x) \cdot |\varpi^{-n}| \operatorname{Vol}(\mathcal{D}^{-1}, d x)$. \\

Mais $\mathcal{D}^{-1} = \bigsqcup\limits_{x \in \mathcal{D}^{-1}/\OO} x + \OO$ d'où $\operatorname{Vol}(\mathcal{D}^{-1}, d x) = \sum\limits_{x \in \mathcal{D}^{-1}/\OO} \operatorname{Vol}(x + \OO, d x)
= \left|\mathcal{D}^{-1} / \OO \right| \operatorname{Vol}(\OO, dx) \text { . }$\\
Donc, $\widehat{\hat{f}}(x) = \operatorname{Vol}(\OO, d x)^2 \cdot \left|\mathcal{D}^{-1} / \OO \right| = |\OO / \mathcal{D}|^{-1} \cdot \left|\mathcal{D}^{-1} / \OO\right| = 1$
car si $\mathcal{D} = \varpi^d \OO$ avec $d \geqslant 0$, alors $\mathcal{D}^{-1} = \varpi^{-d} \OO$. \\

Ainsi, $\widehat{\hat{f}}(x) = \1_{\varpi^n \OO}(a+x) = \1_{-a+\varpi^n \OO}(x) = \1_{a+\varpi^m \OO}(-x) = f(-x)$.
\end{proof}



\begin{proof}(\cref{thm1})~

Nous avions $\hat{f}(y) = \psi(a y) \int_{\varpi^n \OO} \psi(x y) \,dx$
$$
= \operatorname{Vol}(\varpi^n \OO, dx) \psi(a y) \1_{\varpi^{-n} \mathcal{D}^{-1}}(y)
$$

Donc, $\widehat{\hat{f}} \in C_c^{\infty}(F)$. Maintenant, pour $x \in F$, on calcule :
$$
\begin{aligned}
\widehat{\hat{f}}(x) & = \operatorname{Vol}(\varpi^n \OO, d x) \int_F \1_{\varpi^{-n} \mathcal{D}^{-1}}(y) \psi(a y) \psi(x y) \,dy \\
& = \operatorname{Vol}(\varpi^n \OO, d x) \int_{\varpi^{-n} \mathcal{D}^{-1}} \psi((a+x) y) \,dy .
\end{aligned}
$$
 - Si $ x=-a$, l'intégrale vaut $\operatorname{Vol}(\varpi^{-n} \mathcal{D}^{-1}, d x)$.\\
 - Si $x \neq -a$, l'intégrale est nulle sauf si $\varpi^{-n} \mathcal{D}^{-1} \subseteq (a+x)^{-1} \mathcal{D}^{-1}$, c'est-à-dire si $a+x \in \varpi^n \OO$. \\
 Dans ce cas, elle vaut $\operatorname{Vol}(\varpi^{-n} \mathcal{D}^{-1}, d x)$.\\

On a alors $\widehat{\hat{f}}(x) = |\varpi^n| \operatorname{Vol}(\OO, d x) \cdot |\varpi^{-n}| \operatorname{Vol}(\mathcal{D}^{-1}, d x)$. \\

Mais $\mathcal{D}^{-1} = \bigsqcup\limits_{x \in \mathcal{D}^{-1}/\OO} x + \OO$ d'où $\operatorname{Vol}(\mathcal{D}^{-1}, d x) = \sum\limits_{x \in \mathcal{D}^{-1}/\OO} \operatorname{Vol}(x + \OO, d x)
= \left|\mathcal{D}^{-1} / \OO \right| \operatorname{Vol}(\OO, dx) \text { . }$\\
Donc, $\widehat{\hat{f}}(x) = \operatorname{Vol}(\OO, d x)^2 \cdot \left|\mathcal{D}^{-1} / \OO \right| = |\OO / \mathcal{D}|^{-1} \cdot \left|\mathcal{D}^{-1} / \OO\right| = 1$
car si $\mathcal{D} = \varpi^d \OO$ avec $d \geqslant 0$, alors $\mathcal{D}^{-1} = \varpi^{-d} \OO$. \\

Ainsi, $\widehat{\hat{f}}(x) = \1_{\varpi^n \OO}(a+x) = \1_{-a+\varpi^n \OO}(x) = \1_{a+\varpi^m \OO}(-x) = f(-x)$.
\end{proof}

