%! TEX root = AC.tex
% vim: tw=50
% 14/11/2024 10AM

  By \gls{chern} (\cref{eg_2_5}), for all $\theta
  > 0$ we have
  \[
    \Pbb(|X| \ge \theta) \le 4\exp \left(
    -\frac{\theta^2}{4} \right)
  ,\]
  and so using the fact that $\Pbb(|X| \le t) =
  \int_0^t \rho_X(s) \dd s$ we have
  \begin{align*}
    \|X\|_{L^{2k}(\Pbb)}^{2k}
    &= \int_0^\infty t^{2k} \rho_X(t) \dd t \\
    &= \int_0^\infty 2k t^{2k - 1} \Pbb(|X| \ge t)
    \dd t
    &&\text{integration by parts} \\
    &\le \ub{\int_0^\infty 8k t^{2k - 1} \exp \left(
    -\frac{t^2}{4} \right) \dd t}_{\eqdef I(K)}
  \end{align*}
  We shall show by induction on $k$ that $I(K) \le
  2^{2k} \frac{(2k)^k}{4k}$. Indeed, when $k = 1$,
  \[
    \int_0^\infty t \exp \left( -\frac{t^2}{4}
    \right) \dd t
    = \left[ -2\exp \left( -\frac{t^2}{4} \right)
    \right]_0^\infty
    = 2 \le 2
  .\]
  For $k > 1$, integrate by parts to find that
  \begin{align*}
    I(K)
    &= \int_0^\infty \ub{t^{2k - 2}}_{u} \cdot
    \ub{t \exp
    \left( -\frac{t^2}{4} \right)}_{v} \dd t \\
    &= \left[ t^{2k - 2} \cdot \left( -2\exp
    \left( -\frac{t^2}{4} \right) \right)
    \right]_0^\infty
    - \int_0^\infty (2k - 2) t^{2k - 3} \left( -2
    \exp \left( -\frac{t^2}{4} \right) \right) \dd
    t \\
    &= 4(k - 1) \int_0^\infty t^{2(k - 1) - 1}
    \exp \left( -\frac{t^2}{4} \right) \dd t \\
    &= 4(k - 1) I(K - 1) \\
    &\le 4(k - 1) 2^{2(k - 1)} \frac{(2(k - 1))^{k
    - 1}}{4(k - 1)} \\
    &\le 2^{2k} \frac{(2k)^k}{4k} \qedhere
  \end{align*}
\end{proof}

\begin{corollary}[Rudin's Inequality]
  \label{coro_3_2}
  % Corollary 3.2
  Let $\Gamma \subseteq \charG[\Fbb_2^n]$ be a linearly
  independent set and let $p \in [2, \infty)$.
  Then for any $\ft{f} \in l^2(\Gamma)$,
  \[
    \left\| \sum_{\gamma \in \Gamma}
    \ft{f}(\gamma) \gamma \right\|_{L^P(\Fbb_2^n)}
    = O ( \sqrt{p} \|\ft{f}\|_{l^2(\Gamma)} )
  .\]
\end{corollary}

\begin{corollary}
  \label{coro_3_3}
  % Corollary 3.3
  Let $\Gamma \subseteq \charG[\Fbb_2^n]$ be a
  linearly independent set and let $p \in (1, 2]$.
  Then for all $f \in L^p(\Fbb_2^n)$,
  \[
    \|\ft{f}\|_{l^2(\Gamma)}
    = O \left( \sqrt{\frac{p}{p - 1}}
    \|f\|_{L^p(\Fbb_2^n)} \right)
  .\]
\end{corollary}

\begin{proof}
  Let $f \in L^p(\Fbb_2^n)$ and write $g =
  \sum_{\gamma \in \Gamma} \ft{f}(\gamma) \gamma$.
  Then
  \begin{align*}
    \|\ft{f}\|_{l^2(\Gamma)}^2
    &= \sum_{\gamma \in \Gamma} |\ft{f}(\gamma)|^2
    \\
    &= \langle \ft{f}, \ft{g}
    \rangle_{l^2(\charG[\Fbb_2^n])} \\
    &= \langle f, g \rangle_{L^2(\Fbb_2^n)}
    &&\text{by \gls{planche}}
  \end{align*}
  which is bounded above by $\|f\|_{L^p(\Fbb_2^n)}
  \|g\|_{L^{p'}(\Fbb_2^n)}$ where $\frac{1}{p} +
  \frac{1}{p'} = 1$, using H\"older's inequality.

  By \nameref{coro_3_2},
  \[
    \|g\|_{L^{p'}(\Fbb_2^n)} = O \left( \sqrt{p'}
    \|\ft{g}\|_{l^2(\Gamma)} \right) = O \left(
    \sqrt{\frac{p}{p - 1}} \|\ft{f}\|_{l^2(\Gamma)} \right)
    .
    \qedhere
  \]
\end{proof}

Recall that given $A \subseteq \Fbb_2^n$ of
density $\alpha > 0$, we had
$|\Spec_\rho(\indicator{A}) \le \rho^{-2}
\alpha^{-1}$. This is best possible as the example
of a subspace shows. However, in this case the
large spectrum is highly structured.

\begin{fcthm}[Special case of Chang's Theorem]
  \label{thm_3_4}
  % Theorem 3.4
  Assuming:
  - $A \subseteq \Fbb_2^n$ of density $\alpha > 0$
  - $\rho > 0$
  Then:
  there exists $H \le \charG[\Fbb_2^n]$ of
  dimension $O(\rho^{-2} \log \alpha^{-1})$ such
  that $H \supseteq \Spec_\rho(\indicator{A})$.
\end{fcthm}

\begin{proof}
  Let $\Gamma \subseteq \Spec_\rho(\indicator{A})$
  be a maximal linearly independent set. Let $H =
  \langle \Spec_\rho(\indicator{A}) \rangle$.
  Clearly $\dim(H) = |\Gamma|$. By
  \cref{coro_3_3}, for all $p \in (1, 2]$,
  \[
    (\rho\alpha)^2 |\Gamma|
    \le \sum_{\gamma \in \Gamma}
    |\ft{\indicator{A}}(\gamma)|^2
    = \|\ft{\indicator{A}}\|_{l^2(\Gamma)}^2
    = O \left( \frac{p}{p - 1}
    \|\indicator{A}\|_{L^p(\Fbb_2^n)}^2 \right)
  ,\]
  so
  \[
    |\Gamma|
    = O \left( \rho^{-2} \alpha^{-2} \alpha^{2 /
    p} \frac{p}{p - 1} \right)
  .\]
  Set $p = 1 + (\log \alpha^{-1})^{-1}$ to get
  $|\Gamma| = O(\rho^{-2} \alpha^{-2} (\alpha^2
  \cdot e^2) (\log \alpha^{-1} + 1))$.
\end{proof}

\begin{fcdefn}[Dissociated]
  \glsadjdefn{diss}{dissociated}{subset}%
  \label{defn_3_6}
  % Definition 3.6
  Let $G$ be a finite abelian group. We say $S
  \subseteq G$ is \emph{dissociated} if $\sum_{s
  \in S} \eps_s s = 0$ for $\eps \in \{-1, 0,
  1\}^{|S|}$, then $\eps \equiv 0$.
\end{fcdefn}

Clearly, if $G = \Fbb_2^n$, then $S \subseteq G$
is \gls{diss} if and only if it is linearly
independent.
