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\begin{fcthm}[Chang's Theorem]
  \label{thm_3_7}
  % Theorem 3.7
  Assuming:
  - $G$ a finite abelian group
  - $A \subseteq G$ be of density $\alpha > 0$
  - $\Lambda \supseteq \Spec_\rho(\indicator{A})$
  is \gls{diss}
  Then:
  $|\Lambda| = O(\rho^{-2} \log \alpha^{-1}$.
\end{fcthm}

We may bootstrap \nameref{thm_3_1} to obtain the
following:

\begin{fcthm}[Marcinkiewicz-Zygmund]
  \label{thm_3_8}
  % Theorem 3.8
  Assuming:
  - $p \in [2, \infty)$
  - $X_1, X_2, \ldots, X_n \in \L^p(\Pbb)$
  independent random variables
  - $\Ebb \sum_{i = 1}^{n} X_i = 0$
  Then:
  \[
    \left\| \sum_{i = 1}^{n} X_i
    \right\|_{L^p(\Pbb)}
    = O \left( p^{\half} \left\| \sum_{i = 1}^{n}
    |X_i|^2 \right\|_{L^{p / 2}(\Pbb)}^{\half} \right)
  .\]
\end{fcthm}

\begin{proof}
  First assume the distribution of the $X_i$'s is
  symmetric, i.e. $\Pbb(X_i = a) = \Pbb(X_i = -a)$
  for all $a \in \Rbb$. Partition the probability
  space $\Omega$ into sets $\Omega_1, \Omega_2,
  \ldots, \Omega_M$, write $\Pbb_j$ for the
  induced measure on $\Omega_j$ such that all
  $X_i$'s are symmetric and take at most 2 values.
  By \nameref{thm_3_1}, for each $j \in [M]$,
  \begin{align*}
    \left\| \sum_{i = 1}^{n} X_i
    \right\|_{L^p(\Pbb_j)}^p
    &= O \left( p^{p / 2} \left( \sum_{i = 1}^{n}
    \|X_i\|_{L^2(\Pbb_j)}^2 \right)^{p / 2}
    \right) \\
    &= O \left( p^{p / 2} \left\| \sum_{i = 1}^{n}
    |X_i|^2 \right\|_{L^{p / 2}(\Pbb_j)}^{p / 2} \right)
  \end{align*}
  so summing over all $j$ and taking $p$-th roots
  gives the symmetric case. Now suppose the
  $X_i$'s are arbitrary, and let $Y_1, \ldots,
  Y_n$ be such that $Y_i \sim X_i$ and $X_1, X_2,
  \ldots, X_n, Y_1, Y_2, \ldots, Y_n$ are all
  independent. Applying the symmetric case to $X_i
  - Y_i$,
  \begin{align*}
    \left\| \sum_{i = 1}^{n} (X_i - Y_i)
    \right\|_{L^p(\Pbb \times \Pbb)}
    &= O \left( p^{\half} \left\| \sum_{i = 1}^{n}
    |X_i - Y_i|^2 \right\|_{L^{p / 2}(\Pbb \times
    \Pbb)}^{\half} \right) \\
    &= O \left( p^{\half} \left\| \sum_{i = 1}^{n}
    |X_i - Y_i|^2 \right\|_{L^{p /
    2}(\Pbb)}^{\half} \right)
  \end{align*}
  But then
  \begin{align*}
    \left\| \sum_{i = 1}^{n} X_i
    \right\|_{L^p(\Pbb)}
    = \left\| \sum_{i = 1}^{n} X_i - \ub{\Ebb^Y \sum_{i =
    1}^{n} Y_i}_{=0} \right\|_{L^p(\Pbb)}^p \\
    &= \Ebb^X \left| \sum X_i - \Ebb^Y \sum Y_i \right|^p
    \\
    &= \Ebb^X \left| \Ebb^Y \sum (X_i - Y_i)
    \right|^p \\
    &\le \Ebb^X \Ebb^Y \left| \sum (X_i - Y_i)
    \right|^p
    &&\text{by Jensen say} \\
    &= \left\| \sum (X_i - Y_i) \right\|_{L^p(\Pbb
    \times \Pbb)}^p
  \end{align*}
  concluding the proof.
\end{proof}

\begin{fcthm}[Croot-Sisask almost periodicity]
  \label{thm_3_9}
  % Theorem 3.9
  Assuming:
  - $G$ a finite abelian group
  - $\eps > 0$
  - $p \in [2, \infty)$
  - $A, B \subseteq G$ are such that $|A + B| \le
  K|A|$
  - $f : G \to \Cbb$
  Then:
  there exists $b \in B$ and a set $X \subseteq B
  - b$ such that $|X| \ge 2^{-1} K^{-O(\eps^{-2}
  p)}|B|$ and
  \[
    \|\tau_x f \conv \cmeas - f \conv \cmeas
    \|_{L^p(G)}
    \le \eps \|f\|_{L^p(G)}
    \qquad \forall x \in X
  ,\]
  where $\tau_x g(y) = g(y + x)$ for all $y \in
  G$, and as a reminder, $\cmeas$ is the
  \gls{charm} of $A$.
\end{fcthm}

\begin{proof}
  The main idea is to approximate
  \[
    f \conv \cmeas(y)
    = \Ebb_x f(y - x) \cmeas(x) = \EG xA f(y - x)
  \]
  by $\frac{1}{m} \sum_{i = 1}^{m} f(y - z_i)$,
  where $z_i$ are sampled independently and
  uniformly from $A$, and $m$ is to be chosen
  later.

  For each $y \in G$, define $Z_i(y) = \tau_{-zi}
  f(y) - f \conv \cmeas(y)$. For each $y \in G$,
  these are independent random variables with mean
  $0$, so by \nameref{thm_3_8},
  \begin{align*}
    \left\| \sum_{i = 1}^{m} Z_i(y)
    \right\|_{L^p(\Pbb)}^p
    &= O \left( p^{p / 2} \left\| \sum_{i = 1}^{m}
    |Z_i(y)|^2 \right\|_{L^{p / 2}(\Pbb)}^{p / 2}
    \right) \\
    &= O \left( p^{p / 2} \Ebb_{(z_1, \ldots, z_m)
    \in A^m} \left| \sum_{i = 1}^{m} |Z_i(y)|^2
    \right|^{p / 2} \right)
  \end{align*}
  By H\"older with $\frac{1}{p'} + \frac{2}{p} =
  1$, we get
  \begin{align*}
    \left| \sum_{i = 1}^{m} |Z_i(y)|^2 \right|^{p
    / 2}
    &\le \left( \sum_{i = 1}^{m} 1^{p'}
    \right)^{\frac{1}{p'} \cdot \frac{p}{2}}
    \left( \sum_{i = 1}^{m} |Z_i(y)|^{2 \cdot p /
    2} \right)^{\frac{2}{p} \cdot \frac{p}{2}} \\
    &\le \left( \sum_{i = 1}^{m} 1^{p'}
    \right)^{\frac{p}{2} - 1}
    \left( \sum_{i = 1}^{m} |Z_i(y)|^{2 \cdot p /
    2} \right)^{\frac{2}{p} \cdot \frac{p}{2}} \\
    &= m^{p / 2 - 1}
    \sum_{i = 1}^{m} |Z_i(y)|^p
  \end{align*}
  so
  \[
    \left\| \sum_{i = 1}^{m} Z_i(y)
    \right\|_{L^p(\Pbb)}^p
    = O \left( p^{p / 2} m^{p / 2 - 1} \Ebb_{(z_1,
    \ldots, z_m) \in A^m} \sum_{i = 1}^{m}
    |Z_i(y)|^p \right)
  .\]
  Summing over all $y \in G$, we have
  \[
    \EG yG \left\| \sum_{i = 1}^{m} Z_i(y)
    \right\|_{L^p(\Pbb)}^p
    = O \left( p^{p / 2} m^{p / 2 - 1} \Ebb_{(z_1,
    \ldots, z_m) \in A^m} \sum_{i = 1}^{m} \EG yG
    |Z_i(y)|^p \right)
  \]
  with
  \begin{align*}
    \left( \EG yG |Z_i(y)|^p \right)^{\frac{1}{p}}
    &= \|Z_i\|_{L^p(G)} \\
    &= \|\tau_{-z_i} f - f\conv \cmeas
    \|_{L^p(G)} \\
    &\le ,\|\tau_{-z_i} f\|_{L^p(G)}
    + \|f \conv \cmeas\|_{L^p(G)} \\
    &\le \|f\|_{L^p(G)} + \|f\|_{L^q(G)}
    \|\cmeas\|_{L^1(G)} \\
    &\le 2 \|f\|_{L^p(G)}
  \end{align*}
  by Young / H\"older ($\|f \conv g\|_{L^r(G)} \le
  \|f\|_{L^p(G)} \|g\|_{L^q(G)}$ where $1 +
  \frac{1}{r} = \frac{1}{p} + \frac{1}{q}$).