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% 26/11/2024 10AM

\begin{proof}
  Every $P \in \V_n^d$ can be written as a linear
  combination of monomials in $\M_n^d$, so
  \[
    P(x + y) = \sum_{\substack{m, m' \in \M_n^d \\
    \deg(mm') \le d}} c_{m, m'} m(x) m'(y)
  \]
  for some coefficients $c_{m, m'}$. Clearly at
  least one of $m, m'$ must have degree $\le
  \frac{d}{2}$, whence
  \[
    P(x + y) = \sum_{m \in \M_n^{d / 2}} m(x)
    F_m(y) + \sum_{m' \in \M_n^{d / 2}} m'(y)
    G_{m'}(x)
  ,\]
  for some families of polynomials $(F_m)_{m \in
  \M_n^{d / 2}}$, $(G_{m'})_{m' \in \M_n^{d /
  2}}$.

  Viewing $(P(x + y))_{x, y \in A}$ as a $|A|
  \times |A|$-matrix $C$, we see that $C$ can be
  written as the sum of at most $2\m_{d / 2}$
  matrices, each of which has rank $1$. Thus
  $\rank(C) \le 2\m_{d / 2}$. But by assumption,
  $C$ is a diagonal matrix whose rank equals $|\{a
  \in A : P(a + a) \neq 0\}|$.
\end{proof}

\begin{fcprop}[]
  \label{prop_4_3}
  % Proposition 4.3
  Assuming:
  - $A \subseteq \Fbb_3^n$ a set containing no
  non-trivial 3 term arithmetic progressions
  Then:
  $|A| \le 3\m_{2n / 3}$.
\end{fcprop}

\begin{proof}
  Let $d \in [0, 2n]$ be an integer to be
  determined later. Let $W$ be the space of
  polynomials in $\V_n^d$ that vanish on $(2
  \stimes A)^c$. We have
  \[
    \dim(W) \ge \dim(\V_n^d) - |(2 \stimes A)^c|
    = \m_d - (3^n - |A|)
  .\]
  We claim that there exists $P \in W$ such that
  $|\supp(P)| \ge \dim(W)$. Indeed, pick $P \in W$
  with maximal support. If $|\supp(P)| < \dim(W)$,
  then there would be a non-zero polynomial $Q \in
  W$ vanishing on $\supp(P)$, in which case
  $\supp(P + Q) \supsetneq \supp(P)$,
  contradicting the choice of $P$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/42446e88f25f4edb.png}
  \end{center}
  Now by assumption,
  \[
    \{a + a' : a \neq a' \in A\} \cap 2 \stimes A
    = \emptyset
  .\]
  So any polynomial that vanishes on $(2 \stimes
  A)^c$ vanishes on $\{a + a' : a \neq a' \in
  A\}$. By \cref{lemma_4_2} we now have that,
  \begin{align*}
    |A| - (3^n - \m_d)
    &= \m_d - (3^n - |A|) \\
    &\le \dim(W) \\
    &\le |\supp(P)| \\
    &= |\{x \in \Fbb_3^n : P(x) \neq 0\}| \\
    &= |\{a \in A : P(2a) \neq 0\}| \\
    &\le 2\m_{d / 2}
  \end{align*}
  Hence $|A| \ge 3^n - \m_d + 2\m_{d / 2}$. But
  the monomials in $\M_n \setminus \M_n^d$ are in
  bijection with the ones in $\M_{2n - d}$ via
  $x_1^{\alpha_1} \cdots x_n^{\alpha_n} \mapsto
  x_1^{2 - \alpha_1} \cdots x_n^{2 - \alpha_n}$,
  whence $3^n - \m_d = \m_{2n - d}$. Thus setting
  $d = \frac{4n}{3}$, we have $|A| \le \m_{2n / 3}
  + 2\m_{2n / 3} = 3\m_{2n / 3}$.
\end{proof}

You will prove \cref{thm_4_1} on Example Sheet 3.

We do not have at present a comparable bound for 4
term arithmetic progressions. Fourier techniques
also fail.

\begin{example}
  \label{eg_4_4}
  % Example 4.4
  Recall from \cref{lemma_2_18} that given $A
  \subseteq G$,
  \[
    |\T_3(\indicator{A}, \indicator{A},
    \indicator{A}) - \alpha^3|
    \ge \sup_{\gamma \neq 1}
    |\ft{\indicator{A}}(\gamma)|
  .\]
  But it is impossible to bound
  \[
    \T_4(\indicator{A}, \indicator{A},
    \indicator{A}, \indicator{A}) - \alpha^4
    = \EG xd \indicator{A}(x) \indicator{A}(x + d)
    \indicator{A}(x + 2d) \indicator{A}(x + 3d) -
    \alpha^4
  \]
  by $\sup_{\gamma \neq 1}
  |\ft{\indicator{A}}(\gamma)|$. Indeed, consider
  $Q = \{x \in \Fbb_p^n : x \cdot x = 0\}$. By
  Problem 11(ii) on Sheet 1,
  \[
    \frac{|Q|}{p^n} = \frac{1}{p} + O(p^{-n / 2})
  \]
  and
  \[
    \sup_{t \neq 0} |\ft{\indicator{Q}}(t)|
    = O(p^{-n / 2})
  .\]
  But given a 3 term arithmetic progression $x, x
  + d, x + 2d \in Q$, by the identity
  \[
    x^2 - 3(x + d)^2 + 3(x + 2d)^2 - (x + 3d)^2 =
    0 \qquad \forall x, d
  ,\]
  $x + 3d$ automatically lies in $Q$, so
  \[
    \T_4(\indicator{A}, \indicator{A},
    \indicator{A}, \indicator{A})
    = \T_3(\indicator{A}, \indicator{A},
    \indicator{A}) = \left( \frac{1}{p} \right)^3
    + O (p^{-n / 2})
  \]
  which is not close to $\left( \frac{1}{p}
  \right)^4$.
\end{example}

\begin{fcdefn}[$U^2$-norm]
  \label{defn_4_5}
  % Definition 4.5
  Given $f : G \to \Cbb$, define its
  \emph{$U^2$-norm} by the formula
  \[
    \|f\|_{U^2(G)}^4
    = \Ebb_{x, a, b \in G} f(x) \ol{f(x + a) f(x +
    b)} f(x + a + b)
  .\]
\end{fcdefn}

Problem 1(i) on Sheet 2 showed that
$\|f\|_{U^2(G)} = \|\ft{f}\|_{l^4(\charG)}$, so
this is indeed a norm.

Problem 1(ii) asserted the following:

\begin{fclemma}[]
  \label{lemma_4_6}
  % Lemma 4.6
  Assuming:
  - $f_1, f_2, f_3 : G \to \Cbb$
  Then:
  \[
    |\T_3(f_1, f_2, f_3)|
    \le \min_{i \in [3]} \|f_i\|_{U^2(G)} \cdot
    \prod_{j \neq i} \|f_j\|_{L^\infty(G)}
  .\]
\end{fclemma}

Note that
\[
  \sup_{\gamma \in \charG} |\ft{f}(\gamma)|^4
  \le \sum_{\gamma \in \charG} |\ft{f}(\gamma)|^4
  \le \sup_{\gamma \in \charG} |\ft{f}(\gamma)|^2
  \sum_{\gamma \in \charG} |\ft{f}(\gamma)|^2
\]
and thus by \gls{parse},
\[
  \|f\|_{U^2(G)}^4
  = \|\ft{f}\|_{l^\infty(\charG)}^4
  \le \|\ft{f}\|_{l^\infty(\charG)}^2
  \|f\|_{L^2(G)}^2
.\]