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% 29/10/2024 10AM

It is easy to verify the inversion formula: for
all $x \in G$,
\[
  f(x) = \sum_{\gamma \in \charG} \ft{f}(\gamma)
  \gamma(x)
.\]
Indeed,
\begin{align*}
  \sum_{\gamma \in \charG} \ft f(\gamma) \gamma(x)
  &= \sum_{\gamma \in \charG} \EG yG f(y)
  \ol{\gamma(y)} \gamma(x) \\
  &= \EG yG f(y) \ub{\sum_{\gamma \in \charG} \gamma(x
  - y)}_{=|G| \text{ iff $x = y$}} \\
  &= f(x)
  &&\text{by \cref{lemma_2_2}}
\end{align*}
Given $A \subseteq G$, the \emph{indicator} or
\emph{characteristic function} of $A$,
$\indicator{A} : G \to \{0, 1\}$ is defined as
usual.

Note that
\[
  \ft{\indicator{A}}(1) = \EG xG \indicator{A}(x)
  1(x) = \frac{|A|}{|G|}
.\]
The \emph{density} of $A$ in $G$ (often denoted by
$\alpha$).

\begin{definition*}[Characteristic measure]
  \glssymboldefn{cmu}%
  \glsnoundefn{charm}{characteristic
  measure}{characteristic measures}%
  Given non-empty $A \subseteq G$, the
  \emph{characteristic measure} $\mu_A : G \to [0,
  |G|]$ is
  defined by $\mu_A(x) = \alpha^{-1}
  \indicator{A}(x)$.

  Note that $\EG xG \mu_A(x) = 1 = \ft{\mu_A}(1)$.
\end{definition*}

\begin{definition*}[Balanced function]
  \glssymboldefn{bf}%
  The \emph{balanced function} $f_A : G \to [-1,
  1]$ is given by $f_A(x) = \indicator{A}(x) -
  \alpha$. Note that $\EG xG f_A(x) = 0 =
  \ft{f_A}(1)$.
\end{definition*}

\begin{example}
  \label{eg_2_4}
  % Example 2.4
  Let $V \le \Fp^n$ be a subspace. Then for $t \in
  \charG[\Fp^n]$, we have
  \begin{align*}
    \ft{\indicator{V}}(t)
    &= \EG x{\Fp^n} \indicator{V}(x) e \left( -
    \frac{x \cdot t}{p} \right) \\
    &= \frac{|V|}{p^n} \indicator{V^\perp}(t)
  \end{align*}
  where $V^\perp = \{t \in \charG[\Fp^n] : x \cdot
  t = 0 ~ \forall x \in V\}$ is the
  \emph{annihilator} of $V$. In other words,
  $\ft{\indicator{V}}(t) = \cmeas[V^\perp](t)$.
\end{example}

\begin{example}
  \label{eg_2_5}
  % Example 2.5
  Let $R \subseteq G$ be such that each $x \in G$
  lies in $R$ independently with probability
  $\half$. Then with high probability
  \[
    \sup_{\gamma \neq 1}
    |\ft{\indicator{R}}(\gamma)| = O \left(
    \sqrt{\frac{\log|G|}{|G|}} \right)
  .\]
  \glsnoundefn{chern}{Chernoff's inequality}{NA}%
  This follows from \emph{Chernoff's inequality}:
  Given $\Cbb$-valued independent random variables
  $X_1, X_2, \ldots, X_n$ with mean $0$, then for
  all $\theta > 0$, we have
  \[
    \Pbb \left( \left| \sum_{i = 1}^{n} X_i
    \right| \ge \theta \sqrt{\sum_{i = 1}^{n}
    \left\|X_i\right\|^2_{L^\infty(\Pbb)}} \right)
    \le 4\exp \left( -\frac{\theta^2}{4} \right)
  .\]
\end{example}

\begin{example}
  % Example 2.6
  Let $Q = \{x \in \Fp^n : x \cdot x = 0\}
  \subseteq \Fp^n$ with $p > 2$. Then
  \[
    \frac{|Q|}{p^n} = \frac{1}{p} +
    O(p^{-\frac{n}{2}})
  \]
  and $\sup_{t \neq 0} |\ft{\indicator{Q}}(t)| =
  O(p^{-\frac{n}{2}})$.
\end{example}

Given $f, g : G \to \Cbb$, we write
\[
  \langle f, g \rangle = \EG xG f(x) \ol{g(x)}
  \qquad \text{and} \qquad
  \langle \ft f, \ft g \rangle =
  \sum_{\gamma \in \charG} \ft{f}(\gamma) \ol{\ft
  g(\gamma)}
.\]
Consequently,
\[
  \left\|f\right\|^2_{L^2(G)} = \EG xG |f(x)|^2
  \qquad \text{and} \qquad
  \left\| \ft f \right\|^2_{l^2(\charG)} =
  \sum_{\gamma \in \charG} |\ft f(\gamma)|^2
.\]

\glsnoundefn{planche}{Plancherel's identity}{NA}
\glsnoundefn{parse}{Parseval's identity}{NA}
\begin{fclemma}[]
  % Lemma 2.7
  Assuming:
  - $f, g : G \to \Cbb$
  Then:
  \begin{enumerate}[(i)]
    \item $\left\| f \right\|^2_{L^2(G)} = \left\|
      \ft f\right\|^2_{l^2(\charG)}$ (Parseval's
      identity)
    \item $\langle f, g \rangle = \langle \ft f,
      \ft g \rangle$ (Plancherel's identity)
  \end{enumerate}
\end{fclemma}

\begin{proof}
  Exercise (hopefully easy).
\end{proof}

\begin{fcdefn}[Spectrum]
  \glssymboldefn{spec}%
  \glsnoundefn{rls}{$\rho$-large spectrum of
  $f$}{NA}%
  % Definition 2.8
  Let $1 \ge \rho > 0$ and $f : G \to \Cbb$. Define the
  \emph{$\rho$-large spectrum of $f$} to be
  \[
    \operatorname{Spec}_\rho(f) = \{\gamma \in
    \charG : |\ft f(\gamma)| \ge \rho \left\| f
    \right\|_1\}
  .\]
\end{fcdefn}

\begin{example}
  % Example 2.9
  By \cref{eg_2_4}, if $f = \indicator{V}$ with $V
  \le \Fp^n$, then $\forall \rho > 0$,
  \[
    \Spec_\rho(\indicator{V}) = \left\{ t \in
    \charG[\Fp^n] : |\ft{\indicator{V}}(t)| \ge
    \rho \frac{|V|}{p^n} \right\} = V^\perp
  .\]
\end{example}

\begin{fclemma}[]
  \label{lemma_2_10}
  % Lemma 2.10
  Assuming:
  - $\rho > 0$
  Then:
  \[
    |\Spec_\rho(f)| \le \rho^{-2} \frac{\left\| f
    \right\|_2^2}{\left\| f \right\|_1^2}
  .\]
\end{fclemma}

\begin{proof}
  By \gls{parse},
  \begin{align*}
    \left\| f \right\|_2^2
    &= \left\| \ft f \right\|_2^2 \\
    &= \sum_{\gamma \in \charG} |\ft f(\gamma)|^2
    \\
    &\ge \sum_{\gamma \in \Spec_\rho(f)} |\ft
    f(\gamma)|^2 \\
    &\ge |\Spec_\rho(f)| (\rho \left\| f
    \right\|_1)^2 \qedhere
  \end{align*}
\end{proof}

In particular, if $f = \indicator{A}$ for $A
\subseteq G$, then
\[
  \left\| f \right\|_1 = \alpha = \frac{|A|}{|G|}
  = \left\| f \right\|_2^2
,\]
so $|\Spec_\rho(\indicator{A})| \le \rho^{-2}
\alpha^{-1}$.