%! TEX root = AC.tex
% vim: tw=50
% 28/11/2024 10AM

Hence
\[
  \|\ft{f}\|_{l^\infty(\charG)}
  \le \|\ft{f}\|_{l^4(\charG)}
  = \|f\|_{U^2(G)}
  \le \|\ft{f}\|_{l^\infty(\charG)}^{\half}
  \|f\|_{L^2(G)}^{\half}
.\]
Moreover, if $f = \balf A = \indicator{A} -
\alpha$, then
\[
  \T_3(f, f, f)
  = \T_3(\indicator{A} - \alpha, \indicator{A} -
  \alpha, \indicator{A} - \alpha)
  = \T_3(\indicator{A}, \indicator{A},
  \indicator{A}) - \alpha^3
.\]
We may therefore reformulate the first step in the
proof of \nameref{thm_2_19} as follows: if $p^n
\ge 2\alpha^{-2}$, then by \cref{lemma_4_6},
\[
  \frac{\alpha^3}{2} \le \left| \frac{\alpha}{p^n}
  - \alpha^3 \right|
  = |\T_3(\balf A, \balf A, \balf A)|
  \le \|\balf A\|_{U^2(\Fbb_p^n)}
.\]
It remains to show that if $\|\balf
A\|_{U^2(\Fbb_p^n)}$ is non-trivial, then there
exists a subspace $V \le \Fbb_p^n$ of bounded
codimension on which $A$ has increased density.

\begin{fcthm}[$U^2$ Inverse Theorem]
  \label{thm_4_7}
  % Theorem 4.7
  Assuming:
  - $f : \Fbb_p^n \to \Cbb$
  - $\|f\|_{L^\infty(\Fbb_p^n)} \le 1$
  - $\delta > 1$
  - $\|f\|_{U^2(\Fbb_p^n)} \ge \delta$
  Then:
  there exists $b \in \Fbb_p^n$ such that
  \[
    |\EG x{\Fp^n} f(x) e(-x \cdot b / p)|
    \ge \delta^2
  .\]
  In other words, $|\langle f, \phi \rangle| \ge
  \delta^2$ for $\phi(x) = e(-x \cdot b / p)$ and
  we say ``$f$ correlates with a linear phase
  function''.
\end{fcthm}

\begin{proof}
  We have seen that
  \[
    \|f\|_{U^2(\Fp^n)}^2
    \le \|\ft{f}\|_{l^\infty(\charG[\Fp^n])}
    \|f\|_{L^2(\Fp^n)}
    \le \|\ft{f}\|_{l^\infty(\charG[\Fp^n])}
  ,\]
  so
  \[
    \delta^2
    \le \|\ft{f}\|_{l^\infty(\charG[\Fp^n])}
    = \sup_{t \in \charG[\Fp^n]} |\Ebb_x f(x) e(-x
    \cdot t / p)|
    .
    \qedhere
  \]
\end{proof}

\begin{center}
  \includegraphics[width=0.4\linewidth]{images/d3176d4f18154aa1.png}
\end{center}

\begin{fcdefn}[$U^3$ norm]
  \label{defn_4_8}
  % Definition 4.8
  \glssymboldefn{unorm}%
  Given $f : G \to \Cbb$, define its \emph{$U^3$
  norm} by
  \begin{align*}
    \|f\|_{U^3(G)}^8
    &\defeq \EG_{x, a, b, c} f(x) \ol{f(x + a) f(x + b)
    f(x + c)} \\
    &\qquad f(x + a + b) f(x + b + c) f(x + a +
    c) \ol{f(x + a + b + c)} \\
    &= \Ebb_{x, h_1, h_2, h_3 \in G}
    \prod_{\mathbf{\eps} \in \{0, 1\}^3}
    \mathcal{C}^{|\mathbf{\eps}|} f(x +
    \mathbf{\eps} \cdot \mathbf{h})
  \end{align*}
  where $\mathcal{C} g(x) = \ol{g(x)}$ and
  $|\mathbf{\eps}|$ denotes the number of ones in
  $\mathbf{\eps}$.
\end{fcdefn}

\glssymboldefn{Diff}%
It is easy to verify that
$\EG cG \|\Delta_c f\|_{U^2(G)}^4$
where $\Delta_c g(x) = g(x) \ol{g(x + c)}$.

\begin{fcdefn}[$U^3$ inner product]
  \label{defn_4_9}
  % Definition 4.9
  \glssymboldefn{ip}%
  Given functions $f_{\mathbf{\eps}} : G \to
  \Cbb$ for $\mathbf{\eps} \in \{0, 1\}^3$, define
  their \emph{$U^3$ inner product} by
  \[
    \langle (f_{\mathbf{\eps}})_{\mathbf{\eps} \in
    \{0, 1\}^3} \rangle_{U^3(G)}
    = \Ebb_{x, h_1, h_2, h_3 \in G}
    \prod_{\mathbf{\eps} \in \{0, 1\}^3}
    \mathcal{C}^{|\mathbf{\eps}|}
    f_{\mathbf{\eps}}(x + \mathbf{\eps} \cdot
    \mathbf{h})
  .\]
\end{fcdefn}

Observe that $\ip{f, f, f, f, f, f, f, f} =
\unorm{f}^8$.

\begin{fclemma}[Gowers--Cauchy--Schwarz Inequality]
  \label{lemma_4_10}
  % Lemma 4.10
  Assuming:
  - $f_{\mathbf{\eps}} : G \to \Cbb$,
  $\mathbf{\eps} \in \{0, 1\}^3$
  Then:
  \[
    |\ip{(f_{\mathbf{\eps}})_{\mathbf{\eps} \in
    \{0, 1\}^3}}
    \le \prod_{\mathbf{\eps} \in \{0, 1\}^3}
    \unorm{f_{\mathbf{\eps}}}
  .\]
\end{fclemma}

Setting $f_{\mathbf{\eps}} = f$ for $\mathbf{\eps}
\in \{0, 1\}^2 \times \{0\}$ and
$f_{\mathbf{\eps}} = 1$ otherwise, it follows that
$\|f\|_{U^2(G)}^4 \le \unorm{f}^4$ hence
$\|f\|_{U^2(G)} \le \unorm{f}$.

\begin{fcprop}[]
  \label{prop_4_11}
  % Proposition 4.11
  Assuming:
  - $f_1, f_2, f_3, f_4 : \Fbb_5^n \to \Cbb$
  Then:
  \[
    \T_4(f_1, f_2, f_3, f_4)
    \le \min_{i \in [4]} \unorm{f_i}
    \prod_{j \neq i} \|f_j\|_{L^\infty(\Fbb_5^n)}
  .\]
\end{fcprop}

\begin{proof}
  We additionally assume $f = f_1 = f_2 = f_3 = f_4$ to make the
  proof easier to follow, but
  the same ideas are used for the general
  case. We additionally assume
  $\|f\|_{L^\infty(\Fbb_5^n)} \le 1$, by
  rescaling, since the inequality is homogeneous.

  Reparametrising, we have
  \begin{align*}
    \T_4(f, f, f, f)
    &= \Ebb_{a, b, c, d \in \Fbb_5^n} f(3a + 2b + c)
    f(2a + b - d) f(a - c - 2d) f(-b - 2c - 3d) \\
    |\T_4(f, f, f, f)|^8
    &\le \Big(\Ebb_{a, b, c} |\Ebb_d f(2a + b - d) f(a
    - c - 2d) f(-b - 2c - 3d)|^2\Big)^4 \\
    &= \Big(\Ebb_{d, d'} \Ebb_{a, b} f(2a + b + d)
    \ol{f(2a + b - d')} \\
    &\qquad \Ebb_c f(a - c - 2d)
    \ol{f(a - c - 2d')} f(-b - 2c - 3d) \ol{f(-b -
    2c - 3d')}\Big)^4 \\
    &\le \Big(\Ebb_{d, d'} \Ebb_{a, b} |\Ebb_c f(a - c -
    2d) \ol{f(a - c - 2d')} f(-b - 2c - 3d)
    \ol{f(-b - 2c - 3d')}|^2\Big)^2 \\
    &= \Big( \Ebb_{c, c', d, d'} \Ebb_a f(a - c -
    2d) \ol{f(a - c' - 2d)} \ol{f(a - c - 2d')}
    f(a - c' - 2d') \\
    &\qquad \Ebb_b f(-b - 2c - 3d) \ol{f(-b - 2c'
    - 3d)} \ol{f(-b - 2c - 3d')} f(-b - 2c' - 3d')
    \Big)^2 \\
    &\le \Ebb_{c, c', d, d', a} |\Ebb_b f(-b - 2c
    - 3d) \ol{f(-b - 2c' - 3d)} \ol{f(-b - 2c -
    3d')} f(-b - 2c' - 3d')|^2 \\
    &= \Ebb_{b, b', c, c', d, d'} f(-b - 2c - 3d)
    \ol{f(-b' - 2c - 3d)} \ol{f(-b - 2c' - 3d)}
    f(-b' - 2c' - 3d) \\
    &\qquad \ol{f(-b - 2c - 3d')} f(-b' - 2c -
    3d') f(-b -2c' - 3d') \ol{f(-b' -2c' - 3d')}
    \qedhere
  \end{align*}
\end{proof}