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\begin{notation*}
  Let $f : \{0, 1\}^n \to \Rbb$.
  Let $J \subset [n]$ and let $n \in \{0, 1\}^J$.

  \glssymboldefn{onu}%
  Define $f_u : \{0, 1\}^{[n] \setminus J} \to \Rbb$ by $f_u(y) = f(x)$, where
  $x|_J = u$, $x|_{J^c} = y$.

  \glssymboldefn{EJ}%
  Define the averaging projection $E_J$ by $E_J f^{(p)}(x) = \Ebb f_u^{(p)}$,
  where $u = x|_J$.
  Note that $E_J f$ depends only on the coordinates in $J$.
\end{notation*}

\begin{fclemma}[]
  % Lemma 7.2
  \label{lemma:7.2}
  Each $\JE_J$ is an orthogonal projection, and if $J \subset K$, then $\JE_J
  \JE_K = \JE_J$.
\end{fclemma}

\begin{proof}
  Since $(f_{\onu})_{\onu} = f_{\onu}$, we have $\JE_J^2 = \JE_J$, so $\JE_J$ is
  a projection.
  Now we would like to show that
  \[
    \pip \langle f - \JE_J f, \JE_J f \rangle
    = 0
  \]
  for every $f$, or equivalently that $\pip \langle f, \JE_J f \rangle = \pip
  \langle \JE_J f, \JE_J f \rangle$.
  Since $\JE_J f = \JE_J^2 f$, it is enough to prove that $\JE_J$ is
  self-adjoint.

  But
  \begin{align*}
    \pip \langle \JE_J f, g \rangle
    &= \Ebb_x \JE_J f(x) g(x) \\
    &= \Ebb_{u \in \{0, 1\}^J} \Ebb_{y \in \{0, 1\}^{J^c}} (\Ebb f_{\onu})
    g_{\onu}(y) \\
    &= \Ebb_u (\Ebb f_{\onu}) \Ebb_y g_{\onu}(y) \\
    &= \Ebb_u (\Ebb f_{\onu}) (\Ebb g_{\onu}) \\
    &= \Ebb_x \Ebb_J f(x) \Ebb_J g(x) \\
    &= \pip \langle \JE_J f, \JE_J g \rangle \\
    &= \pip \langle f, \JE_J g \rangle
    &&\text{(the last step by symmetry)}
  \end{align*}
  This proves that $\JE_J$ is self-adjoint, which is interesting property, but
  we in fact we didn't need to write down the last line in the calculation above
  if all we want to deduce is that $\pip \langle f, \JE_J f \rangle = \pip
  \langle \JE_J f, \JE_J f \rangle$.

  Now let $J \subset K$ and let $L = K \setminus J$.
  Write $x = (u, v, z)$ where $u \in \{0, 1\}^J$, $v \in \{0, 1\}^L$, $z \in
  \{0, 1\}^{K^c}$.
  Then
  \begin{align*}
    \JE_J \JE_K f(x)
    &= \JE_J (\Ebb_{z \in K^c} f(u, v, z)) \\
    &= \Ebb_{\substack{v \in L \\ z \in K^c}} f(u, v, z) \\
    &= \JE_J f(x)
    \qedhere
  \end{align*}
\end{proof}

\begin{fclemma}[Regularity lemma for Boolean functions]
  % Lemma 7.3
  \label{lemma:7.3}
  For every $(\eps, \rho, r, \delta)$, there exists some $T$ such that for every
  Boolean function $f^{(p)} : \{0, 1\}^n \to \{0, 1\}$, there exists $J \subset
  [n]$ with $|J| \le T$ such that if $u$ is chosen randomly ($\mup$ randomly)
  from $\{0, 1\}^J$, then
  \[
    \Pbb[\text{$f_{\onu}$ is \quasi{\eps, \rho, r}}]
    \ge 1 - \delta
  .\]
\end{fclemma}

\begin{example*}
  Define $f : \{0, 1\}^n \to \{0, 1\}$ randomly with
  \[
    \Pbb(f(x_1, \ldots, x_{n - 1}, 0) = 1)
    = \frac{1}{3}
  \]
  and
  \[
    \Pbb(f(x_1, \ldots, x_{n - 1}, 1) = 1)
    = \frac{2}{3}
  .\]
  Then $f$ is not \gls{quasi}, but the conclusion of the above lemma holds if we
  take $J = \{n\}$.
\end{example*}

\begin{proof}
  For any $J \subset [n]$, define the \emph{mean-square density} of $J$ to be
  $\pnorm \|\JE_J f\|_2^2$..
  Note that if $J \subset K$, then
  \[
    \pnorm \|\JE_J f\|_2^2
    = \pnorm \|\JE_J \JE_K f\|_2^2
    \le \pnorm \|\JE_K f\|_2^2
  .\]
  Suppose now that $J$ does not satisfy the conclusion of the lemma.
  Let $u \in \{0, 1\}^J$.
  Then for any $K \subset [n] \setminus J$, $\pnorm \|\JE_J f_{\onu}\|_2^2 \ge
  (\Ebb f_{\onu})^2$.
  If $f_{\onu}$ is not \quasi{\eps, \rho r}, then there exists $K_u \subset [n]
  \setminus J$ and some $v \in \{0, 1\}^{K_u}$ such that
  \[
    |\Ebb f_{\onu, \onu[v]}^{(p)} - \Ebb f_{\onu}^{(p)}|
    \ge \eps
  .\]
  Let $\zeta = \min \{p, 1 - p\}$.
  Also, $|K_{\onu}| \le r$.
  If we choose a random element of $\{0, 1\}^{K_u}$, then it equals $v$ with
  probability $\ge \zeta^r$.
  Therefore,
  \[
    \Ebb_w |\Ebb f_{\onu, \onu[w]}^{(p)} - \Ebb f_{\onu}^{(p)}|^2
    \ge \eps^2 \zeta^r
  .\]
  But
  \[
    \Ebb_w \Ebb f_{\onu, \onu[w]}
    = \Ebb f_{\onu}
  \]
  so the LHS is $\Var(\Ebb f_{\onu, \onu[w]} - \Ebb f_{\onu})$ so it equals
  \[
    \Ebb_w (\Ebb f_{\onu, \onu[w]})^2 - (\Ebb_w \Ebb f_{\onu, \onu[w]})^2
    = \pnorm \|\JE_{K_u} f_{\onu}\|_2^2 - (\Ebb f_{\onu})^2
  .\]
  So $\pnorm \|\Ebb_{K_u} f_{\onu}\|_2^2 \ge (\Ebb f_{\onu})^2 + \zeta^r
  \eps^2$ in this case.

  Let $K = \bigcup_u K_u$.
  Then $pnorm \|\JE_K f_{\onu}\|_2^2 \ge (\Ebb f_{\onu})^2 + \zeta^r \eps^2$ in this
  case.

  Averaging over $u$, we deduce that
  \[
    \Ebb_u \pnorm \|\JE_K f_{\onu}\|_2^2
    \ge \Ebb_u (\Ebb f_{\onu})^2 + \delta \zeta^r \eps^2
  ,\]
  i.e. $\pnorm \|\JE_{J \cup K} f\|_2^2 \ge \|\JE_J f\|_2^2 + \delta \zeta^r
  \eps^2$.

  We can now do an iteration.
  Start with $J_0 = \emptyset$.
  At $i$-th stage, if $J_i$ doesn't work, then replace it by $J_{i + 1} = J_i
  \cup K_i$, using the argument just given, with $|K_i| \le r \cdot 2^{|J_i|}$.
  At each stage, mean square density goes up by at least $\delta \zeta^r
  \eps^2$, so the process must terminate in a bounded number of steps.
\end{proof}

% \begin{fclemma}[]
%   % Lemma 7.4
%   \label{lemma:7.4}
%   For every $\zeta > 0$, $p \in \left[ \zeta, \half - \zeta \right]$, $\alpha >
%   0$, there exist $\eps, r$ such that if $\Ebb f^{(p)} = \alpha$, $f$ is
%   monotone and $f$ is \quasi{\eps, \rho, r}, then
%   \[
%     \Ebb f^{(1/2)}
%     > \half
%   .\]
% \end{fclemma}