%! TEX root = AC.tex
% vim: tw=50
% 24/10/2024 10AM

  Given $x \in G$, let $X(x) = A \cap (x + A)$.
  Then
  \[
    \Ebb_{x \in Q} |X(x)| = \frac{1}{|Q|} \sum_{x
    \in Q} |A \cap (x + A)| \ge \half \lambda|A|
  .\]
  Let $T(x) = \{(a, b) \in X(x)^2 : a - b \notin
  \Pg[c\eta]\}$. Then
  \begin{align*}
    \Ebb_{X \in Q} |T(x)|
    &= \Ebb_{x \in Q} |\{(a, b) \in (A \cap
      (\ub{x}_{x \in A - a \cap A - b} +
    A))^2 : a - b \notin \Pg[c\eta]\}| \\
    &= \frac{1}{|Q|} \sum_{x \in Q} |\{(a, b) \in
    S : x \in A - a \cap A - b\}| \\
    &= \frac{1}{|Q|} \sum_{(a, b) \in S}  |(A - a)
    \cap (A - b) \cap Q| \\
    &\le \frac{1}{|Q|} 2c\eta |A|^2 \delta |Q| \\
    &= 2c\eta \delta |A|^2 \\
    &= 2c\lambda^2 |A|^2
  \end{align*}
  Therefore,
  \begin{align*}
    \Ebb_{x \in Q} |X(x)|^2 - (16c)^{-1} |T(x)|
    &\stackrel{\text{C-S}}{\le} \left( \Ebb_{x \in
    Q} |X(x)| \right)^2 - (16c)^{-1} \Ebb_{x \in
    Q} |T(x)| \\
    &\le \left( \frac{\lambda}{2} \right)^2 |A|^2
    - (16c)^{-1} 2c\lambda^2 |A|^2 \\
    &= \left( \frac{\lambda^2}{4} -
    \frac{\lambda^2}{8} \right)|A|^2 \\
    &= \frac{\lambda^2}{8} |A|
  \end{align*}
  So there exists $x \in Q$ such that $|X(x)|^2
  \ge \frac{\lambda^2}{8}|A|^2$, in which case we
  have
  \[
    |X| \ge \frac{\lambda}{\sqrt{8}} |A| \ge
    \frac{\eta}{3} |A|
  \]
  and $|T(x)| \le 16c|X|^2$.
\end{proof}

\begin{proof}[Proof of \cref{bsg}]
  Given $A \subseteq G$ with $\energy(A) \ge
  \eta|A|^3$, apply \cref{lemma_1_19} with $c =
  2^{-7}$ to otain $X \subseteq A$ of size $|X|
  \ge \frac{\eta}{3} |A|$ such that for all but
  $\frac{1}{8}$ of pairs $(a, b) \in X^2$, $a - b
  \in \Pg[\eta / 2^7]$. In particular, the
  bipartite graph
  \[
    G = (X \djunion X, \{(x, y) \in X \times X : x
    - y \in \Pg[\eta / 2^7]\})
  \]
  has at least $\frac{7}{8}|X|^2$ edges. Let $A' =
  \left\{x \in X : \deg(x) \ge \frac{3}{4}|X|\right\}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/15fd2d9fedbe40c8.png}
  \end{center}
  Clearly, $|A'| \ge \frac{|X|}{8}$. For any $a, b
  \in A'$, there are at least $\frac{|X|}{2}$
  elements $y \in X$ such that $(a, y), (b, y) \in
  E(G)$ ($a - y, b - y \in \Pg[\eta / 2^7]$).

  Thus $a - b = (a - y) - (b - y)$ has at
  least
  \[
    \ub{\frac{\eta}{6}|A|}_{\text{choices for $y$}}
    \cdot \frac{\eta}{2^7} |A|
    \cdot \frac{\eta}{2^7} |A| \ge
    \frac{\eta^3}{2^{17}} |A|^3
  \]
  representations of the form $a_1 - a_2 - (a_3 -
  a_4)$ with $a_i \in A$.

  It follows that
  \begin{align*}
    \frac{\eta^3}{2^{17}} |A|^3 |A' \minus A'|
    &\le |A|^4 \\
    \implies |A' \minus A'|
    &\le 2^{17} \eta^{-3} |A| \\
    &\le 2^{22} \eta^{-4} |A'|
  \end{align*}
  Thus $|A' + A'| \le 2^{44} \eta^{-8} |A'|$.
\end{proof}

\newpage
\section{Fourier-analytic techniques}

In this chapter we will assume that $G$ is
\emph{finite} abelian.

\glssymboldefn{charG}%
$G$ comes equipped with a group $\hat{G}$ of
characters, i.e. homomorphisms $\gamma : G \to
\Cbb$. In fact, $\hat{G}$ is isomorphic to $G$.

See
\href{https://notes.ggim.me/rt}{Representation
Theory notes} for more information about characters and
proofs of this as well as some of the facts below.

\begin{example}
  % Example 2.1
  \phantom{}
  \begin{enumerate}[(i)]
    \item \glssymboldefn{e}%
      If $G = \Fp^n$, then for any $\gamma \in
      \hat{G} = \Fp^n$, we have an associated
      character $\gamma(x) = e(\gamma \cdot x /
      p)$, where $e(y) = e^{2\pi i y}$.
    \item If $G = \Zbb / N\Zbb$, then any $\gamma
      \in \charG = \Zbb / N\Zbb$ can be associated
      to a character $\gamma(x) = e(\gamma x /
      N)$.
  \end{enumerate}
\end{example}

\begin{notation*}
  \glssymboldefn{EG}%
  Given $B \subseteq G$ nonempty, and any function
  $g : B \to \Cbb$, let
  \[
    \Ebb_{x \in B} g(x) = \frac{1}{|B|} \sum_{x \in
    B} g(x)
  .\]
\end{notation*}

\begin{fclemma}[]
  \label{lemma_2_2}
  % Lemma 2.2
  Assuming:
  - $\gamma \in \charG$
  Then:
  \[
    \EG xG \gamma(x) = \begin{cases}
      1 & \text{if $\gamma = 1$} \\
      0 & \text{otherwise}
    \end{cases}
  ,\]
  and for all $x \in G$,
  \[
    \sum_{\gamma \in \charG} \gamma(x) =
    \begin{cases}
      |\charG| & \text{if $x = 0$} \\
      0 & \text{otherwise}
    \end{cases}
  .\]
\end{fclemma}

\begin{proof}
  The first equality in eqch case is trivial.
  Suppose $\gamma \neq 1$. Then there exists $y
  \in G$ with $\gamma(y) \neq 1$. Then
  \begin{align*}
    \gamma(y) \EG zG \gamma(z)
    &= \EG zG \gamma(y + z) \\
    &= \EG{z'}G \gamma(z')
  \end{align*}
  So $\EG zG \gamma(z) = 0$.

  For the second part, note that given $x \neq 0$,
  there must by $\gamma \in \charG$ such that
  $\gamma(x) \neq 1$, for otherwise $\charG$ would
  act trivially on $\langle x \rangle$, hence
  would also be the dual group for $G / \langle x
  \rangle$, a contradiction.
\end{proof}

\begin{fcdefn}[Fourier transform]
  \glssymboldefn{ft}%
  % Definition 2.3
  Given $f : G \to \Cbb$, define its \emph{Fourier
  transform} $\hat{f} : \charG \to \Cbb$ by
  \[
    \hat{f}(\gamma) = \EG xG f(x) \ol{\gamma(x)}
  .\]
\end{fcdefn}
