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\begin{document}


\title{Lecture 40}
\maketitle

We finish the proof of proposition 4.1.
\begin{proof}
Recall that we now need to estimate the size of
$$\int_{K^{\times}}|f(y)c_{0}(y)||y|^{\sigma}d^{\times}y, \qquad
Re(s) \ge \delta >0. \eqno (*)$$ Now, as $f$ is the characteristic
function of $\frak{p}^{\nu}$, we can rewrite $(*)$ as
$$\int_{\frak{p}^{\nu}-\{0\}} |y|^{\sigma}d^{\times}y.$$

Define the annulus of size $\nu$, $A_{\nu}=\{y\in K : |y| =
N\frak{p}^{-\nu} \}= \pi^{\nu}U_{K}.$  Then
\begin{align*}
    \int_{\frak{p}^{\nu}-\{0\}} |y|^{\sigma}d^{\times}y & =\sum_{\nu =
        r }^{\infty}\int_{A_{\nu}}|y|^{\sigma}d^{\times}y \\
    &=\sum_{\nu
=r}^{\infty}N\frak{p}^{-\sigma\nu}\int_{\pi^{\nu}U_{K}}d^{\times}y\\

&=\sum_{\nu=r}^{\infty}N\frak{p}^{-\sigma\nu}\int_{\pi^{U_{K}}}d^{\times}y,\quad

        \mbox{as $\mu$ is multiplication-invariant,}\\
    &=

\left(\sum_{\nu=r}^{\infty}N\frak{p}^{-\sigma\nu}\right)(N\frak{D})^{-\frac12}\quad

        \mbox{and summing the geometric series, we get}\\

&=\left(\frac{N\frak{p}^{-r\sigma}}{1-N\frak{p}^{-\sigma}}\right)(N\frak{D})^{-\frac12}
\\
    &\le

\left(\frac{N\frak{p}^{-r\delta}}{1-N\frak{p}^{-\delta}}\right)(N\frak{D})^{-\frac12},

    \quad\mbox{a uniform bound.}
\end{align*}
Thus as $r \rightarrow \infty$, the bound goes to 0, proving the
result.
\end{proof}

\begin{pro}
\label{4.1} Let $K$ be any local field, $f,g \in \mathcal{S}(K),$
and $c$ be a quasi-character in the domain $0< Re(c) <1$.

Then
$$\zeta(f,c)\zeta(\hat{g},\hat{c})=\zeta(\hat{f},\hat{c})\zeta(g,c),
\quad \mbox{where } \hat{c}(a)=|a|c^{-1}(a).$$
\end{pro}

\begin{proof}
As $f,g \in \mathcal{S}(K),$ the left-hand side is an absolutely
convergent double integral $$
\int_{K^{\times}}\int_{K^{\times}}f(a)\hat{g}(b)c(ab^{-1})|b|\,
d^{\times}a\, d^{\times}b.$$ On $K^{\times} \times K^{\times}$,
the automorphism $(a,b) \mapsto (a,ab)$ preserves the measure.
Applying the automorphism to the above double integral we have
$$\int_{K^{\times}}\Bigg(
\underbrace{\int_{K^{\times}}f(a)\hat{g}(ab)|a|\,
d^{\times}a}_{(*)}\Bigg)c(b^{-1})|b|\, d^{\times}b.$$

We show that $$ (*)=\int_{K^{\times}}f(a)\hat{g}(ab)|a|\,
d^{\times}a$$ is symmetric under $f\leftrightarrow g.$ Consider
$$\int_{K}\int_{K}f(x)g(y)e^{-2\pi i \Lambda (xby)}\, dx\, dy.$$
This is obviously symmetric under $f\leftrightarrow g$.  But
\begin{align*}
    \int_{K}\int_{K}f(x)g(y)e^{-2\pi i \Lambda (xby)}\, dx\, dy &=
        \int_{K}f(x)\left( \int_{K}g(y)e^{-2\pi i \Lambda (xby)}\, dy
        \right)\, dx \\
    &=\int_{K}f(x)\hat{g}(bx)\, dx \\
    &=\int_{K^{\times}}f(x)\hat{g}(xb)|x|\, d^{\times}x=(*).
\end{align*}
Thus, $(*)$ is symmetric, proving the result.
\end{proof}

\begin{thm}
\label{4.2} For any $f\in \mathcal{S}$, the local zeta function
$\zeta (f,c)$ has an analytic continuation to the domain of all
quasi-characters.  Moreover, we have a functional equation $$\zeta
(f,c)=\rho (c)\zeta (\hat{f},\hat{c})$$ where $\rho(c)$ is a
meromorphic function of quasi-characters which is independent of
$f$.  It is defined by the functional equation in the domain
${0<Re(c)<1}$ and is called the root number.
\end{thm}

Recall that two quasi-characters are equivalent if
$\frac{c_{1}}{c_{2}}$ is of the form ${x\mapsto |x|^{s}}$, for
some $s\in \C.$

Our strategy for the proof of theorem 4.2 is as follows.  In each
equivalence class $C$ of quasi-characters, we exhibit a function
$f_{c} \in \mathcal{S}$ such that ${\rho (c)=
\frac{\zeta(f_{c},c)}{\zeta(\hat{f_{c}},\hat{c})}}$ is defined on
$0<Re(c)<1, \, c\in C.$  These $f_{c}$ will turn out to be
familiar functions with obvious analytic continuations.

\begin{proof}
If $K = \R$, then  $\Lambda (x) = -x,\, dx =$ Haar measure on
$\R$, and $\frac{dx}{|x|} = $ Haar measure on $\R^{\times}$. We
then have two equivalence classes of quasi-characters: $$C_{0}=
\{x \mapsto |x|^{s} \} \quad \mbox{and} \quad C_{1}= \{x \mapsto
sgn(x)\cdot|x|^{s} \}.$$  We take $$ f_{C_{0}}(x)=f_{0}(x)=
e^{-\pi x^{2}} \quad \mbox{and} \quad f_{C_{1}}(x)=f_{1}(x)=
xe^{-\pi x^{2}}.$$  Then $$\hat{f_{0}}(x)=\int_{-\infty}^{\infty}
e^{-\pi y^{2} + 2\pi ixy}\, dy = e^{-\pi x^{2}} = f_{0},$$ and
$$\hat{f_{1}}(x)=\int_{-\infty}^{\infty} ye^{-\pi y^{2} + 2\pi
ixy}\, dy = ixe^{-\pi x^{2}} = if_{1}.$$

We continue the proof in the next lecture.
\end{proof}

\end{document}