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\begin{fclemma}[]
  \label{lemma:2.13}
  % Lemma 2.13
  Let $G$ be a $3$-partite graph on $X \cup Y \cup Z$ with bipartite parts of
  densities $\delta_{XY}, \delta_{XZ}, \delta_{YZ}$, respectively.
  Let $\delta = \delta_{XY} \delta_{XZ} \delta_{YZ}$, and suppose all bipartite
  parts are \gls{epsquasi}.

  Let $H$ be a $3$-partite $3$-uniform hypergraph on $X \cup Y \cup Z$, and let
  the relative density of $H$ in $\triangles(G)$ be $\gamma$.
  Suppose $2^{21} \eps < \delta^7$, and suppose $H$ is \emph{not}
  \hquasirel{\eta} $\triangles(G)$.

  Then there are partitions $G(X, Y) = G_1(X, Y) \cup \cdots \cup G_l(X, Y)$,
  $G(X, Z) = G_1(X, Z) \cup \cdots \cup G_m(X, Z)$, $G(Y, Z) = G_1(Y, Z) \cup
  \cdots \cup G_n(Y, Z)$ \glsref[relmsd]{relative} to which the
  \glsref[relmsd]{msd} of $H$ is $\ge \gamma^2 + 2^{-11} \eta^{16}$.
  Moreover, $l, m, n \le 3^{3\delta^{-2}}$.
\end{fclemma}

\begin{proof}
  Let
  \[
    f(x, y, z)
    = (\indicator{H}(x, y, z) - \gamma) \indicator{G}(x, y) \indicator{G}(x, z)
    \indicator{G}(y, z)
  .\]
  The hypothesis amounts to
  \[
    \Ebb_{x, x', y, y', z, z'} f_{x, x', y, y', z, z'}
    \ge \eta^8 \delta^4
    \tag{\dag}
    \label{lec6:eqdag}
  .\]
  For each $(x, y, z) \in \triangles(G)$, let
  \[
    F_{xyz}(x', y', z')
    = \frac{f_{x, x', y, y', z, z'}}{f(x', y', z')}
  \]
  and let
  \[
    F(x', y', z')
    = \Ebb_{x, y, z} F_{xyz}(x', y', z')
  .\]
  Then \eqref{lec6:eqdag} becomes
  \[
    \Ebb_{x, x', y, y', z, z'} F_{xyz}(x', y', z') f(x', y', z')
    = \Ebb_{x, y, z} \ub{\langle}_{\Ebb_{x', y', z'}} f, F_{xyz} \rangle
  .\]
  Note that each of the $F_{xyz}$ can be written as a product
  \[
    u(x', y') v(y', z') w(x', z')
  ,\]
  where each of $u, v, w \to [-1, 1]$.

  For each $(x, y, z) \in \triangles(G)$ and each pair $(x', y')$, define
  $\tilde{u}(x', y')$ as follows:
  \begin{itemize}
    \item If $u(x', y') > 0$, then
      \[
        \tilde{u}(x', y')
        = \begin{cases}
          1
          & \text{with probability $u(x', y')$} \\
          0
          & \text{otherwise}
        \end{cases}
      \]
    \item If $u(x', y') < 0$, then
      \[
        \tilde{u}(x', y')
        = \begin{cases}
          -1
          & \text{with probability $-u(x', y')$} \\
          0
          & \text{otherwise}
        \end{cases}
      \]
    \item If $u(x', y') = 0$, then $\tilde{u}(x', y') = 0$.
  \end{itemize}
  Same for $v, w$.

  Finally, let
  \[
    \tilde{F}_{xyz}(x', y', z')
    = \tilde{u}(x', y') \tilde{v}(x', z') \tilde{w}(y', z')
  ,\]
  and observe that the expectation (in the random choices of $\tilde{u},
  \tilde{v}, \tilde{w}$ described above) of $\tilde{F}_{xyz}(x', y', z')$ is
  $F_{xyz}(x', y', z')$.
  Let $\tilde{F}(x', y', z') = \Ebb_{x, y, z} \tilde{F}_{xyz}(x', y', z')$.
  Since the expectation of $\langle f, \tilde{F}_{xyz} \rangle$ equals $\langle
  f, F_{xyz} \rangle$, we can make random choices so that $\langle f, \tilde{F}
  \rangle \ge \eta^8 \delta^4$ (so from now on, ``expectation'' no longer refers
  to expectation over the possible random choices of $\tilde{u}, \tilde{v},
  \tilde{w}$).

  Let $r \ge \delta^{-2}$, choose $\tilde{F}_1, \ldots, \tilde{F}_r$ at random
  from $\tilde{F}_{xyz}$ and let $D = \frac{1}{r} \sum_{i \in [r]} \tilde{F}_i$.
  For each $i \in [r]$, the expectation of $\langle f, \tilde{F}_i \rangle$ is
  at least $\frac{1}{\delta_{XYZ}} \langle f, \tilde{F} \rangle$, where
  $\delta_{XYZ} = \frac{|\triangles(G)|}{|X||Y||Z|}$.
  So the expectation of $\langle f, D \rangle$ is at least $\ge \frac{\eta^8
  \delta^4}{\delta_{XYZ}}$.
  By \nameref{prop:1.7}, and the upper bound on $\eps$, $\frac{\delta}{2} \le
  \delta_{XYZ} \le 2 \delta$.
  So the expectation of $\langle f, D \rangle$ is at least $\frac{\eta^8
  \delta^2}{\delta_{XYZ}} \cdot \delta^2 \ge \frac{\eta^8 \delta^2}{4}
  \delta_{XYZ}$.

  Note that
  \begin{align*}
    \|\tilde{F}\|_2^2
    &= \Ebb_{x', y', z'} |\Ebb_{x, y, z} \tilde{F}_{xyz}(x', y', z')|^2 \\
    &\le \Ebb_{x', y', z'} |\Ebb_{x, y, z} \ub{\indicator{K_{2, 2,
    2}}}_{\text{as a graph}}(x, x', y, y',
    z, z')|^2 \\
    &= \Ebb_{x', y', z'} \Ebb_{\substack{x_1, y_1, z_1 \\ x_2, y_2, z_2}}
    \indicator{K_{2, 2, 2}}(x_1, x', y_1, y', z_1, z')
    \indicator{K_{2, 2, 2}}(x_2, x', y_2, y', z_2, z')
  \end{align*}
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/ec761add5e014776.png}
  \end{center}
  So we are counting a $3$-partite graph on $X \cup Y \cup Z$ with $3$ vertices
  in each part and $7$ edges between each part.

  By a generalisation of \nameref{prop:1.7} and the upper bound on $\eps$,
  $\|\tilde{F}\|_2^2 \le 2 \delta^7 \le 16 \delta^4 \delta_{XYZ}^3$, since
  $\frac{\delta}{2} \le \delta_{XYZ}$.
  By Example Sheet 1 Problem 11, the expectation of $\|D\|_2^2$ is at most $2
  \delta_{XYZ}^{-2} \|\tilde{F}\|_2^2 \le 32 \delta^4 \delta_{XYZ}$, provided $r
  \ge \delta^{-2}$.
  Hence the expectation of $256 \delta^2 \langle f, D \rangle - \eta^8
  \|D\|_2^2$ is at least $32 \eta^8 \delta^4 \delta_{XYZ}$.
  It follows that there is a choice of $\tilde{F}_1, \ldots, \tilde{F}_r$ such
  that
  \[
    \langle f, D \rangle
    \ge \frac{\eta^8 \delta^2 \delta_{XYZ}}{8}
  \]
  and $\|D\|_2^2 \le 256 \delta^2 \eta^{-8} \langle f, D \rangle$.
  For such $D$,
  \[
    \frac{\langle f, D \rangle}{\|D\|_2^2}
    \ge \frac{\eta^8 \delta^2 \delta_{XYZ}}{8 \cdot 2^8 \delta^2 \eta^{-8}
    \ub{\langle f, D \rangle}_{\le \delta_{XYZ}}}
    \ge \frac{\eta^{16}}{2^{11}}
    .
    \qedhere
  \]
\end{proof}