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\begin{fcthm}[The Kahn--Kalai--Linial edge-isoperimetric inequality]
  \glssymboldefn{totinftilde}%
  % Theorem 5.2
  \label{thm:5.2}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$ be a Boolean function.
  Then there exists $i$ such that $\Inf_i f \ge \frac{9}{\tilde{I}(f)^2} \ge
  \frac{9}{\tilde{I}(f)^2} 9^{-\tilde{I}(f)}$, where $\tilde{I}(f) =
  \frac{I(f)}{\Var f}$.
\end{fcthm}

\begin{proof}
  We shall obtain upper and lower bounds or $\sum_{i = 1}^{n} \Stab_{\third}(\D_i
  f)$ (magical idea).

  Upper bound: $\sum_i \Stab_{\third}(\D_i f) \le \sum_i \norm\|\D_i
  f\|_{4 / 3}^2$.
  But $\norm\|\D_i f\|_{4 / 3}^{4 / 3} = \Inf_i f$, so
  \[
    \sum_i \norm\|\D_i f\|_{4 / 3}^2
    = \sum_i (\Inf_i f)^{3 / 2}
    \le (\max_i \Inf_i f)^{\half} \totinf(f)
  .\]
  
  Lower bound: $\sum_i \Stab_{\third}(\D_i f) = \sum_i \sum_A \left( \frac{1}{3}
  \right)^{|A|} \ft{\D_i f}(A)^2$.
  But
  \[
    \ft{\D_i f}(A)
    = \begin{cases}
      \ft{f}(A \cup \{i\})
      & i \notin A \\
      0
      & i \in A
    \end{cases}
  \]
  (using the formula $\ft{\D_i f} = \sum_{B \ni i} \ft{f}(B) x_{B \setminus
  \{i\}}$).
  So
  \begin{align*}
    \sum_i \sum_A \left( \third \right)^{|A|} \ft{\D_i f}(A)
    &= \sum_i \sum_{A \ni i} \left( \third \right)^{|A|} \ft{f}(A \cup \{i\})^2
    \\
    &= \sum_B \left( \third \right)^{|B| - 1} |B| \ft{f}(B)^2 \\
    &\ge 3 \Var f \sum_{B \neq \emptyset} \left( \third \right)^{|B|}
    \frac{\ft{f}(B)^2}{\Var f}
  \end{align*}
  But
  \[
    \sum_{B \neq \emptyset} \ft{f}(B)^2
    = \norm \|f\|_2^2 - \ft{f}(0)^2
    = \Ebb f^2 - (\Ebb f)^2 = \Var f
  ,\]
  so $\sum_{B \neq \emptyset}
  \frac{\ft{f}(B)^2}{\Var f} = 1$.

  The function $x \mapsto \left( \third \right)^x$ is convex, so by Jensen,
  \[
    \sum_{B \neq \emptyset} \left( \third \right)^{|B|} \frac{\ft{f}(B)^2}{\Var
    f}
    \ge \left( \third \right)^{\sum_{B \neq \emptyset} |B|
    \frac{\ft{f}(B)^2}{\Var f}}
    = \left( \third \right)^{\totinftilde(f)}
  .\]
  Therefore,
  \[
    (\max_i \Inf_i f)^{\half} \totinf(f)
    \ge 3 \Var f \left( \third \right)^{\totinftilde(f)}
  ,\]
  which rearranges to the result.
\end{proof}

\begin{fccoro}[The KKL theorem]
  % Corollary 5.3
  \label{coro:5.3}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$.
  Then there exists $i$ such that $\Inf_i f \ge \frac{c \log n}{n} \Var f$ (for
  some absolute constant $c > 0$).
\end{fccoro}

\begin{proof}
  If $\totinftilde(f) \ge c \log n$, then the result follows trivially by
  averaging.

  Otherwise, by \cref{thm:5.2}, there exists $i$ such that $\Inf_i(f) \ge
  \frac{9}{(c \log n)^2} \cdot 9^{-c \log n}$.
  For small enough $c$, that's much bigger than $\frac{c \log n}{n}$.
\end{proof}

This shows that the ``tribes'' example from last lecture is the best possible.

Motivation for why $\Stab$ is natural to define in order to tackle the
\nameref{coro:5.3}:
We can assume $\lambda \totinf(f) \le c \lambda \log n$.
Then $e^{\lambda \totinf(f)} \le n^{c \lambda}$.
LHS is $e^{\lambda \sum_A |A| \ft{f}(A)^2}$, and by Jensen we get $\le \sum_A
\ft{f}(A)^2 \cdot e^{\lambda |A|} = \Stab_{e^\lambda} f$.
So $\Stab$ comes up somewhat naturally.

\begin{fcdefn}[Junta]
  \glsnoundefn{Jjunta}{$J$-junta}{$J$-juntas}%
  \glsnoundefn{junta}{junta}{juntas}%
  \glsnoundefn{kjunta}{$k$-junta}{$k$-juntas}%
  Let $f : \{-1, 1\}^n \to \Rbb$, and let $J \subset [n]$.
  Then $f$ is a \emph{$J$-junta} if $f(x)$ depends only on $x|_J$.
  We also say that $J$ is a \emph{junta}.
  $f$ is a \emph{$k$-junta} if $f$ is a $J$-junta for some $J$ of size $k$.
\end{fcdefn}

Friedgut's junta theorem states that a Boolean function with small \gls{totinf}
can be approximated by a \kjunta{m} for some small $m$.

\begin{fcthm}[Friedgut's junta inequality]
  % Theorem 5.4
  \label{thm:5.4}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$, let $\eps > 0$, and let $k \in \Nbb$.
  Suppose that $\norm \|f^{\deglekpart}\|_2^2 \ge 1 - \eps$.
  Then there exists an \kjunta{m} $g : \{-1, 1\}^n \to \Rbb$ such that $\norm
  \|f - g\|_2^2 \le \eps$ with $m \le 3^{2k} \cdot
  \frac{\totinf(f)^3}{\eps^2}$.
\end{fcthm}

The proof has some similarities with \cref{thm:5.2}, so we include less detail
in this proof.

\begin{proof}
  Let $\tau > 0$ a constant to be chosen later and let
  \[
    J
    = \{i \in [n] : \Inf_i f \ge \tau\}
  .\]
  This time we estimate $\sum_{i \notin J} \Stab_{\third}(\D_i f)$.
  \[
    \sum_{i \notin J} \Stab_{\third}(\D_i f)
    \le \sum_{i \notin J} (\Inf_i f)^{3 / 2}
    \le \tau^{\half} \totinf(f)
  .\]
  (The first inequality is proved using the same technique as in \cref{thm:5.2}.)

  In the other direction,
  \begin{align*}
    \sum_{i \notin J} \Stab_{\third} (\D_i f)
    &= \sum_{i \notin J} \left( \third \right)^{|A|} \ft{f}(A \cup \{i\})^2 \\
    &= 3 \sum_B |B \setminus J| \left( \third \right)^{|B|} \ft{f}(B)^2 \\
    &\ge 3 \sum_{B \not\subset J} \left( \third \right)^{|B|} \ft{f}(B)^2
  \end{align*}
  But
  \[
    3\sum_{\substack{B \not\subset J \\ |B| \le k}} \left( \third \right)^{|B|}
    \ft{f}(B)^2
    \ge 3 \sum_{\substack{B \not\subset J \\ |B| \le k}} \left( \third \right)^k
    \ft{f}(B)^2
    = \left( \third \right)^{k - 1} \sum_{\substack{B \not\subset J \\ |B| \le
    k}} \ft{f}(B)^2
  .\]
  Let $g = \sum_{\substack{B \subset J \\ |B| \le k}} \ft{f}(A) x_A$.
  Then
%   \begin{align*}
%     \sum_{\substack{B \not\subset J \\ |B| > k}} \left( \third \right)^k
%     \ft{f}(B)^2
%     &\le \sum_{|B| > k} \left( \third \right)^k \ft{f}(B)^2 \\
%     &\le \left( \third \right)^k
%   \end{align*}
%   We will fill in the missing steps in the proof next lecture.
% \end{proof}