
\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}


\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}
