
\begin{proof}~
\begin{itemize}
    \item[$\rightarrow$] Si $n_\psi<n_{\chi}$, alors $\OO^{\times}= \bigsqcup\limits_{a \in \OO^{\times} / U_n} a U_n$, (où $n = n_\chi - 1 \geq n_\psi$), donc:
    $$
    G(\chi, \psi)=\sum_{a \in \OO^{\times} / U_n} \int_{a U_n} \chi(x) \psi(x) \, dx^{\times}=\sum_{a \in \OO^{\times} / U_n} \chi(a) \int_{U_n} \chi(x) \psi(a x) \, dx^{\times}.
    $$
    Mais $U_n=1+\varpi^n \OO$ donc $x=1+y$ avec $y \in \varpi^n \OO$ d'où $ay \in \varpi^n \OO \subseteq \varpi^{n_\psi} \OO$. \\
    Ainsi, $\psi(a x)=\psi(a) \overbrace{\psi(a y)}^{=1}$, donc :
    $$ G(\chi, \psi)=\sum_{a \in \OO^{\times} / U_n} \psi(a) \chi(a) \underbrace{\int_{U_n} \chi(x) \, dx^{\prime}}_{= 0 \text{ car }\chi |_{U_n} \not \equiv 1} .$$
    En effet $\chi |_{U_n} \not \equiv 1$ car $n_\psi<n_{\chi}$, enfin : $G(\chi, \psi)=0$.

    \item[$\rightarrow$] Si $n_\psi \geq n_\chi$. Puisque $n_\chi>0$, on a $n_\psi>0$ aussi.
$$
\begin{aligned}
|G(\chi, \psi)|_{\infty}^2 & =\int_{\OO^{\times}} \chi(x) \psi(x) \, dx^{\times} \int_{\OO^{\times}} \chi\left(y^{-1}\right) \psi(-y) \, dy^{\times} \quad\left(\begin{array}{l}
\chi^{-1}=\bar{\chi} \\
\psi^{\prime}=\bar{\psi}
\end{array}\right) \\
& =\int_{(\OO^{\times})^2} \chi\left(x y^{-1}\right) \psi(x-y) \, dx^{\times} \, dy^{\times} \quad (x^{\prime}=x y^{-1}) \\
& =\int_{(\OO^{\times})^2} \chi\left(x^{\prime}\right) \psi\left(\left(x^{\prime}-1\right) y\right) \, dx^{\prime \times} \, dy^{ \times} \\
& =\int_{\OO^{\times}} \chi(x) h(x) \, dx^{\times}
\end{aligned}
$$ 
$$
\begin{aligned}
\text{où }h(x) :&=\int_{\OO^{\times}} \psi((x-1) y) \, dy^{\times} \\
    & =\left(1-q^{-1}\right)^{-1} \int_{\OO^{\times}} \psi((x-1) y) \, dy\\
    & =\left(1-q^{-1}\right)^{-1}\left(\int_{\OO} \overbrace{\psi((x-1) y)}^{= \psi_{x-1}(y)} \, dy-\int_{\varpi \OO} \psi((x-1) y) \, dy\right)
\end{aligned}
$$

Or $\int_{\OO} \psi_{x-1}(y) \, dy=\left\{\begin{array}{l}\operatorname{Vol}(0, dx) \text{ si } n_\psi-1 < v(x-1) \\ 0 \text{ sinon }\end{array}\right.$ \\
car  $\operatorname{Cond} \left(\psi_{x-1}\right)=(x-1)^{-1} \varpi^{n_\psi} \OO=\varpi^{n_\psi-v(x-1)} \OO$. \\

Donc $h(x)= \begin{cases}\operatorname{Vol}(\OO, dx) & \text{ si } n_\psi-1<v(x-1) \\ \frac{-1}{q-1} \operatorname{Vol}(\OO, dx) & \text{ si } n_\psi-1=v(x-1) \\ 0 & \text{ si } n_\psi-1>v(x-1)\end{cases}$ \\

Ainsi $|G(\chi, \psi)|_{\infty}^2 \operatorname{Vol}(\OO, dx)^{-1}=\int_{V_{n_\psi}} \frac{-1}{q-1} \chi(x) \, dx^{\times}+\int_{U_{n_\psi}} \chi(x) \, dx^{\times} \quad \text{où }V_n: =U_{n-1} \setminus \, U_n $
$$
=\frac{-1}{q-1} \left( \int_{U_{n_\psi-1}} \chi(x) \, dx^{\times}-\int_{U_{n_\psi}} \chi(x) \, dx^{\times}\right)+\int_{U_{n_\psi}} \chi(x) \, dx^{\times} 
$$
Si $n_\psi=n_\chi, \int_{U_{n_\psi-1}} \chi=0$ et $|G(\chi, \psi)|_{\infty}^2 \operatorname{Vol}(\OO, dx)^{-1}=\operatorname{Vol}\left(U_{n_\psi}, dx^{\times}\right)\left(1+\frac{1}{q-1}\right)$.\\
Si $n_\psi>n_\chi,  \; |G(\chi, \psi)|_{\infty}^2 \operatorname{Vol}(\OO, dx)^{-1}=\left(1-q^{-1}\right)^{-1} \operatorname{Vol}\left(U_{n_\psi}, dx^{\times}\right) -q^{-1} \operatorname{Vol}\left(U_{n_\psi - 1}, dx^{\times}\right).$
\end{itemize} 
\end{proof}


Preuve : $\rightarrow$ si $n_\psi<n_\lambda$. $O^x=\frac{1}{a \in O^x / U_n} a U_n, d^{\prime}$ où $^{\prime}$:
\[
G(x, \psi)=\sum_{a \in O^x / U_m} \int_{a U_n} X(x) \psi(x) \, dx^x=\sum_{a \in O^x / U_m} X(a) \int_{U_n} X(x) \psi(a x) \, dx^x
\]

Mais $U_n=1+\pi^* O$ donc $x=1+y$ avec $y \in \pi^m O$ d'où $ay \in \omega^n O \subseteq \omega^n \psi$.
Donc $\psi(a x)=\psi(a) \widetilde{\psi(a y)}=\psi(a)$
Ainsi : $G(X, \psi)=\sum_{a \in O^x / U_n} \psi(a) X(a) \int_{U_n} X(x) \, dx^x$
En effet, $\left.X\right|_{U_n} \neq 1$ car $n<n_X$.
Donc $G(x, \psi)=0$ car on intègre un caractère non trivial sur un groupe.

$\rightarrow$ si $n_\psi > n_X$. Puisque $n_X > 0$, on a $n_\psi > 0$ aussi.
$$
\begin{aligned}
|G(x, \psi)|_{\infty}^2 & =\int_{O^x} X(x) \psi(x) \, dx^x \int_{O^*} X\left(y^{\prime}\right) \psi(-y) \, dy^x \quad\left(\begin{array}{l}
x^{-1}=\bar{x} \\
\psi^{\prime}=\bar{\psi}
\end{array}\right) \\
& =\int_{(O^x)^2} X\left(x y^{-1}\right) \psi(x-y) \, dx^* d y^x \quad x^{\prime}=x y^{-1} \\
& =\int_{(O^x)^2} X\left(x^{\prime}\right) \psi\left(\left(x^{\prime}-1\right) y\right) \, dx^* d y^x \\
& =\int_{O^x} X(x) h(x) \, dx^x
\end{aligned}
$$
où $h(x):=\int_{\mathcal{O}^x} \psi((x-1) y) \, dy x$
$$
=\left(1-q^{-1}\right)^{-1} \int_{\sigma^x} \psi((x-1) y) \, dy \quad(|y|$ qui était en passant de $dy?$ à $dy$ était confus).
$$
=\left(1-q^{-1}\right)^{-1}\left(\int_0 \frac{-\psi_{x-1}(y)}{\psi((x-1) y)} \, dy-\int_{-\infty} \psi((x-1) y) \, dy\right)
$$
Or $\int_0 \psi_{x-1}(y) \, dy=\left\{\begin{array}{l}v_0 l(0, dx) \text { si } n_{\varphi}-1 \leqslant v(x-1) \\ 0 \text { sinon }\end{array}\right.$
car Cond $\left(\psi_{x-1}\right)=(x-1)^{-1}{w^n}^n 0=\pi^{m_\psi-v(x-1)} 0$
D'où $h(x)= \begin{cases}V_0 l(0, dx) & \text { si } n_\psi-1<v(x-1) \\ \frac{-1}{9-1} v_0 l(0, dx) & \text { si } n_\psi-1=v(x-1) \\ 0 & \text { si } n_\psi-1>v(x-1)\end{cases}$

Ainsi $|G(x, \psi)|_{\infty}^2 \operatorname{Vol}(0, dx)^{-1}=\int_{V_{n_\psi}} \frac{-1}{9-1} X(x) \, dx^x+\int_{U_{m_\psi}} X(x) \, dx^x \quad V:=U \mid u$
$$
\left.\left.=\frac{-1}{9-1} \int_{U_{m_\psi-1}}^{m_\psi} X(x) \, dx^x-\int_{U_{m_\psi}} X(x) \, dx^x\right)+\int_{u_{m_\psi}} X(x) \, dx^x\right]
$$
Si $n_\psi=n_X, \int_{u_{n \psi-1}} x=0$ et $\left.|G(x, \psi)|_{\infty}^2 \operatorname{Vol}(0, dx)^{-1}=\operatorname{Vol}\left(u_{n_\psi}, dx^x\right)\right)_{\left(1+\frac{1}{q-1}\right)}$
