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% 12/11/2024 10AM

\begin{proof}
  Let $\eps = \frac{\alpha^2}{40\pi}$, and use
  \cref{lemma_2_23} to partition $[p]$ into
  progressions $P_i$ of length
  \[
    \ge \eps \sqrt{\frac{p}{2}}
    \ge \frac{\alpha^2}{40\pi}
    \frac{\sqrt{\frac{N}{3}}}{2}
    \ge \frac{\alpha^2 \sqrt{N}}{500}
  \] 
  and $\diam(\varphi(P_i)) \le \eps p$. Fix one
  $x_i$ from each of the $P_i$. Then
  \begin{align*}
    \frac{\alpha^2}{20}
    &\le |\ft{f_{A'}}(t)| \\
    &= \left| \frac{1}{p} \sum_i \sum_{x \in P_i}
    f_{A'}(x) e(-xt / p) \right| \\
    &= \frac{1}{p} \left| \sum_i \sum_{x \in P_i}
    f_{A'}(x) e(-xit / p)
    + \sum_i \sum_{x \in P_i} f_{A'}(x) (e(-xt /
    p) - e(-xit / p)) \right| \\
    &\le \frac{1}{p} \sum_i \left| \sum_{x \in
    P_i} f_{A'}(x) \right|  + \frac{1}{p} \sum_i
    \sum_{x \in P_i} |f_{A'}(x)| |\ub{e(-xt / p) -
    e(-xit / p)}_{\substack{\le 2\pi \eps \\
    \text{since $|t(x - x_i)| \le \eps p$}}}|
  \end{align*}
  So
  \[
    \sum_i \left| \sum_{x \in P_i} f_{A'}(x)
    \right| \ge \frac{\alpha^2}{40} p
  .\]
  Since $f_{A'}$ has mean zero,
  \[
    \sum_i \left( \left| \sum_{x \in P_i}
    f_{A'}(x) \right| + \sum_{x \in P_i} f_{A'}(x)
    \right) \ge \frac{\alpha^2}{40} p
  ,\]
  hence there exists $i$ such that
  \[
    \left| \sum_{x \in P_i} f_{A'}(x) \right| +
    \sum_{x \in P_i} f_{A'}(x) \ge
    \frac{\alpha^2}{80} |P_i|
  \]
  and so
  \[
    \sum_{x \in P_i} f_{A'}(x) \ge
    \frac{\alpha^2}{160} |P_i|
    .
    \qedhere
  \]
\end{proof}

\begin{fcdefn}[Bohr set]
  \glsnoundefn{bset}{Bohr set}{Bohr sets}%
  \glssymboldefn{bohr}%
  % Definition 2.25
  Let $\Gamma \subseteq \charG$ and $\rho > 0$. By
  the \emph{Bohr set} $B(\Gamma, \rho)$ we mean
  the set
  \[
    B(\Gamma, \rho) = \{x \in G : |\gamma(x) - 1|
    < \rho ~\forall \gamma \in \Gamma\}
  .\]
  We call $|\Gamma|$ the \emph{rank} of $B(\Gamma,
  \rho)$, and $\rho$ its \emph{width} or
  \emph{radius}.
\end{fcdefn}

\begin{example}
  % Example 2.26
  When $G = \Fp^n$, then $B(\Gamma, \rho) =
  \langle \Gamma \rangle^\perp$ for all
  sufficiently small $\rho$.
\end{example}

\begin{fclemma}[]
  % Lemma 2.27
  Assuming:
  - $\Gamma \subseteq \charG$ of size $d$
  - $\rho > 0$
  Then:
  \[
    |\bohr(\Gamma, \rho)| \ge \left(
    \frac{\rho}{8} \right)^d |G|
  .\]
\end{fclemma}

\begin{fcprop}[Bogolyubov in a general finite
  abelian group]
  % Proposition 2.28
  Assuming:
  - $A \subseteq G$ of density $\alpha > 0$
  Then:
  there exists $\Gamma \subseteq \charG$ of size
  at most $2\alpha^{-2}$ such that $A + A - A - A
  \supseteq \bohr(\Gamma, \rho)$.
\end{fcprop}

\begin{proof}
  Recall $\indicator{A} \conv \indicator{A} \conv
  \indicator{-A} \conv \indicator{-A}(x) =
  \sum_{\gamma \in \charG}
  |\ft{\indicator{A}}(\gamma)|^4 \gamma(x)$.

  Let $\Gamma \in
  \Spec_{\sqrt{\frac{\alpha}{2}}}(\indicator{A})$,
  and note that, for $x \in \bohr \left( \Gamma,
  \half \right)$ and $\gamma \in \Gamma$,
  $\Re(\gamma(x)) > 0$. Hence, for $x \in \bohr
  \left( \Gamma, \half \right)$,
  \[
    \Re \sum_{\gamma \in \charG}
    |\ft{\indicator{A}}(\gamma)|^4 \gamma(x)
    = \ub{\Re \sum_{\gamma \in \Gamma}
    |\ft{\indicator{A}}(\gamma)|^4 \gamma(x)}_{\ge
    \alpha^4} + \Re
    \sum_{\gamma \notin \Gamma}
    |\ft{\indicator{A}}(\gamma)|^4 \gamma(x)
  \]
  and
  \[
    \left| \Re \sum_{\gamma \notin \Gamma}
    |\ft{\indicator{A}}(\gamma)|^4 \gamma(x)
    \right|
    \le \sup_{\gamma \notin \Gamma}
    |\ft{\indicator{A}}(\gamma)|^2 \sum_{\gamma
    \notin \Gamma} |\ft{\indicator{A}}(\gamma)|^2
    \le \left( \sqrt{\frac{\alpha}{2}} \cdot
    \alpha \right)^2 \cdot \alpha
    = \frac{\alpha^4}{2}
    .
    \qedhere
  \]
\end{proof}

\newpage
\section{Probabilistic Tools}

All probability spaces in this course will be
finite.

\begin{fcthm}[Khintchine's inequality]
  \label{thm_3_1}
  % Theorem 3.1
  Assuming:
  - $p \in [2, \infty)$
  - $X_1, X_2, \ldots, X_n$ independent random
  variables
  - $\Pbb(X_i = x_i) = \half = \Pbb(X_i = -x_i)$
  Then:
  \[
    \left\| \sum_{i = 1}^{n} X_i
    \right\|_{L^p(\Pbb)}
    = O \left( p^{\half} \left( \sum_{i = 1}^{n}
    \|X_i\|_{L^2(\Pbb)}^2 \right)^{\half} \right)
  .\]
\end{fcthm}

\begin{proof}
  By nesting of norms, it suffices to prove the
  case $p = 2k$ for some $k \in \Nbb$. Write $X =
  \sum_{i = 1}^{n} X_i$, and assume $\sum_{i =
  1}^{n} \|X_i\|_{L^\infty(\Pbb)}^2 = 1$. Note
  that in fact $\sum_{i = 1}^{n}
  \|X_i\|_{L^2(\Pbb)}^2 = \sum_{i = 1}^{n}
  \|X_i\|_{L^\infty(\Pbb)}^2$, hence $\sum_{i =
  1}^{n} \|X_i\|_{L^2(\Pbb)}^2 = 1$.