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\begin{notation*}
  \glssymboldefn{T}%
  Given $f, g, h : G \to \Cbb$, write
  \[
    T_3(f, g, h)
    = \EG{x, d}{G} f(x) g(x + d) h(x + 2d)
  .\]
\end{notation*}

\begin{notation*}
  \glssymboldefn{stimes}%
  Given $A \subseteq G$, write
  \[
    2 \cdot A = \{2a : a \in A\}
  ,\]
  to be distinguished from $2A = A \plus A = \{a +
  a' : a, a' \in A\}$.
\end{notation*}

\begin{fclemma}[]
  \label{lemma_2_18}
  % Lemma 2.18
  Assuming:
  - $p \ge 3$ prime
  - $A \subseteq \Fp^n$ of density $\alpha > 0$
  - $\sup_{t \neq 0} |\ft{\indicator{A}}(t)| \le
  \eps$
  Then:
  the number of 3-term arithmetic progressions in
  $A$ differs from $\alpha^3 (p^n)^2$ by at most
  $\eps(p^n)^2$.
\end{fclemma}

\begin{proof}
  The number of 3-term arithmetic progressions in
  $A$ is $(p^n)^2$ times
  \begin{align*}
    \T_3(\indicator{A}, \indicator{A}, \indicator{A})
    &= \EG{x, d}{\Fp^n} \indicator{A}(x)\indicator(x +
    d) \indicator{A}(x + 2d) \\
    &= \EG{x, y}{\Fp^n} \indicator{A}(x)
    \indicator{A}(y) \indicator{A}(2y - x) \\
    &= \EG{y}{G} \indicator{A}(y) \EG{x}{G}
    \indicator{A}(x) \indicator{A}(2y - x) \\
    &= \EG yG \indicator{A}(y) \indicator{A} \conv
    \indicator{A} (2y) \\
    &= \langle \indicator{2 \stimes A},
    \indicator{A} \conv \indicator{A} \rangle
  \end{align*}
  By \gls{planche} and \cref{lemma_2_13}, we have
  \begin{align*}
    &= \langle \ft{\indicator{2 \stimes A}},
    \ft{\indicator{A}}^2 \rangle \\
    &= \sum_{t} \ft{\indicator{2 \stimes A}}(t)
    \ol{\ft{\indicator{A}}(t)^2} \\
    &= \alpha^3 + \sum_{t \neq 0} \ft{\indicator{2
    \stimes A}}(t) \ol{\ft{\indicator{A}}(t)^2}
  \end{align*}
  but
  \begin{align*}
    \left| \sum_{t \neq 0} \ft{\indicator{2
    \stimes A}}(t) \ft{\indicator{A}}(t)^2 \right|
    &\le \sup_{t \neq 0} |\ft{\indicator{A}}(t)|
    \sum_{t \neq 0} |\ft{\indicator{2 \stimes
    A}}(t)| |\ft{\indicator{A}}(t)| \\
    &\stackrel{\text{CS}}{\le} \sup_{t \neq 0}
    |\ft{\indicator{A}}(t)| \left( \sum_{t}
    |\ft{\indicator{2 \stimes A}}(t)|^2 \sum_{t}
    |\ft{\indicator{A}}(t)|^2 \right)^{\half} \\
    &\le \eps \|\ft{\indicator{2 \stimes A}}\|_2
    \|\ft{\indicator{A}}\|_2 \\
    &= \eps \cdot \alpha
  \end{align*}
  by \gls{parse}.
\end{proof}

\begin{fcthm}[Meshulam's Theorem]
  \label{thm_2_19}
  % Theorem 2.19
  Assuming:
  - $A \subseteq \Fp^n$ a set containing no
  non-trivial 3 term arithmetic progressions
  Then:
  $|A| = O \left( \frac{p^n}{\log p^n} \right)$.
\end{fcthm}

\begin{proof}
  By assumption,
  \[
    \T_3(\indicator{A}, \indicator{A},
    \indicator{A}) = \frac{|A|}{(p^n)^2} =
    \frac{\alpha}{p^n}
  .\]
  But as in (the proof of) \cref{lemma_2_18},
  \[
    |\T_3(\indicator{A}, \indicator{A},
    \indicator{A}) - \alpha^3|
    \le \sup_{t \neq 0} |\ft{\indicator{A}}(t)|
    \cdot \alpha
  ,\]
  so provided $p^n \ge 2 \alpha^{-2}$, i.e.
  $\T_3(\indicator{A}, \indicator{A},
  \indicator{A}) \le \frac{\alpha^3}{2}$ we have
  $\sup_{t \neq 0} |\ft{\indicator{A}}(t)| \ge
  \frac{\alpha^2}{2}$.

  So by \cref{lemma_2_17} with $\rho =
  \frac{\alpha}{2}$, there exists $V \le \Fp^n$ of
  codimension 1 and $x \in \Fp^n$ such that $|A
  \cap (x + V)| \ge \left( \alpha +
  \frac{\alpha^2}{4} \right) |V|$.

  We iterate this observation: let $A_0 = A$, $V_0
  = \Fp^n$, $\alpha_0 = \frac{|A_0|}{|V_0|}$. At
  the $i$-th step, we are given a set $A_{i - 1}
  \subseteq V_{i - 1}$ of density $\alpha_{i - 1}$
  with no non-trivial 3 term arithmetic
  progressions. Provided that $p^{\dim(V_{i - 1})}
  \ge 2 \alpha_{i - 1}^{-2}$, there exists $V_i
  \le V_{i - 1}$ of codimension $1$, $x_i \in V_{i
  - 1}$ such that
  \[
    |(A \minus x_i) \cap V_i| \ge \left( \alpha_{i
    - 1} + \frac{(\alpha_{i - 1})^2}{4} \right)
    |V_i|
  .\]
  Set $A_i = (A \minus x_i) \cap V_i \subseteq
  V_i$, has density $\ge \alpha_{i - 1} +
  \frac{(\alpha_{i - 1})^2}{4}$, and is free of
  non-trivial 3 term arithmetic progressions.

  Through this iteration, the density increases
  from $\alpha$ to $2\alpha$ in at most
  $\frac{\alpha}{\left( \frac{\alpha^2}{4}
  \right)} = 4 \cdot \alpha^{-1}$ steps.

  $2\alpha$ to $4\alpha$ in at most
  $\frac{2\alpha}{\left( \frac{(2\alpha)^2}{4}
  \right)} = 2\alpha^{-1}$ steps and so on.

  So reaches $1$ in at most
  \[
    4\alpha^{-1} \left( 1 + \half + \quarter +
    \frac{1}{8} + \cdots \right) \le 8\alpha^{-1}
  \]
  steps. The argument must end with $\dim(V_i) \ge
  n - 8\alpha^{-1}$, at which point we must have
  had $p^{\dim(V_i)} < 2\alpha_{i - 1}^2 \le
  2\alpha^{-2}$, or else we could have continued.

  But we may assume that $\alpha \ge \sqrt{2}
  p^{-\frac{n}{4}}$ (or $\alpha^{-2} <
  2p^{\frac{n}{2}}$) whence $p^{n - 8\alpha^{-1}}
  \le p^{\frac{n}{2}}$, or $\frac{n}{2} \le
  2\alpha^{-1}$.
\end{proof}

At the time of writing, the largest known subset
of $\Fbb_{3}^n$ containing no non-trivial 3 term
arithmetic progressions has size $(2.2202)^n$.

We will prove an upper bound of the form
$(2.756)^n$.

\begin{fcthm}[Roth's theorem]
  \label{thm_2_20}
  % Theorem 2.20
  Assuming:
  - $A \subseteq [N] = \{1, \ldots, N\}$
  - $A$ contains no non-trivial 3 term arithmetic
  progressions
  Then:
  $|A| = O \left( \frac{N}{\log\log N} \right)$.
\end{fcthm}