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\begin{remark*}
  As in unbiased case, we always have that $\pnorm\|f^{\deglekpart}\|_2^2 \ge 1
  - \eps$ if $k \ge \frac{\totinf(f)}{\eps}$.
\end{remark*}

\newpage
\section{Intersecting families}

A family $\mathcal{A}$ of subsets of $[n]$ is \emph{intersecting} if $A \cap B
\neq \emptyset$ for every $A, B \in \mathcal{A}$.
Since we can't include both $A$ and $A^c$, $|\mathcal{A}| \le 2^{n - 1}$ for
any intersecting family $\mathcal{A}$.
Equality holds if $\mathcal{A} = \{A \subset [n] : i \in A\}$ for some $i$.
If $n$ is odd, another example is $\{A \subset [n] : |A| > n / 2\}$.
If $n$ is even, we can take $\{A \subset [n] : |A| > n / 2\}$ and add in exactly
one of $A$ and $A^c$ for each $A \in [n]^{(n / 2)}$.

There exist plenty of other extremal examples (e.g. $\mathcal{A}$ is extremal
for $m$, then $\mathcal{B} = \{B \subset [n] : |B \cap [m]| \in \mathcal{A}\}$
is extremal for $n$).

\begin{notation*}
  $[n] = \{1, \ldots, n\}$, $S^{(r)} = \{A \subset S : |A| = r\}$.
\end{notation*}

What if we look at families of sets of size $r$ for some given $r$?

If $r > \frac{n}{2}$ we can take all of $[n]^{(r)}$, so we get ${n \choose r}$.
If $r = \frac{n}{2}$, we can take one set from each $\{A, A^c\}$ to get $\half
{n \choose r}$.
When $r < \frac{n}{2}$, it gets more interesting.

\begin{fcthm}[Erdős--Ko--Rado Theorem]
  % Theorem 7.1
  \label{thm:7.1}
  Let $r, n \in \Nbb$ with $r < \frac{n}{2}$.
  Let $\mathcal{A} \subset [n]^{(r)}$ be an intersecting family.
  Then $|\mathcal{A}| \le {n - 1 \choose r - 1}$, with equality if and only if
  $\mathcal{A}$ is of the form $\{A \in [n]^{(r)} : i \in A\}$.
\end{fcthm}

\begin{proof}[Proof (Katona)]
  Let $\mathcal{A} \subset [n]^{(r)}$ be an intersecting family.
  We need to prove that if $A \subset [n]^{(r)}$ is chosen uniformly, then
  \[
    \Pbb[A \in \mathcal{A}]
    \le \frac{r}{n}
    \quad\left(= \frac{{n - 1 \choose r - 1}}{{n \choose r}}\right)
  .\]
  Let $x_1, x_2, \ldots, x_n$ be a random cyclic ordering of $\{1, 2, \ldots,
  n\}$.
  An \emph{interval} in this cyclic ordering is a set of the form $\{x_i, x_{i +
  1}, \ldots, x_{i + r - 1}\}$ for some $i$ (addition modulo $n$).

  There are $n$ intervals, so we will be done if we can prove that at most $r$
  of them belong to $\mathcal{A}$.
  (That is because we can choose a random element of $[n]^{(r)}$ by first
  choosing a random cyclic ordering and then choosing a random interval in that
  ordering.)

  For each $i$, let $I_{x_i}^+$ be the interval $\{x_{i + 1}, x_{i + 2}, \ldots,
  x_{i + r}\}$ and let $I_{x_i}^-$ be the interval $\{x_{i - r + 1}, \ldots,
  x_i\}$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/f95b10af6adf40ad.png}
  \end{center}
  Without loss of generality $\{x_1, \ldots, x_r\} \in \mathcal{A}$.
  Then any other interval in $\mathcal{A}$ must be $I_{x_i}^+$ or $I_{x_i}^-$
  for some $i \in \{1, \ldots, r - 1\}$.
  We can't have both.
  So at most $r - 1$ more intervals.

  In the equality case, we must have equality for every cyclic ordering.
  In the above argument, we cannot have $i$ with $I_{x_i}^- \in \mathcal{A}$ and
  $I_{x_{i + 1}}^+ \in \mathcal{A}$.
  Therefore, if we have $r$ sets, then there must exist $t$ such that the
  intervals are $I_{x_1}^+, \ldots, I_{x_{t - 1}}^+, I_{x_t}^-, \ldots, I_{x_{r
  - 1}}^-$, together with $\{x_1, \ldots, x_r\}$.
  So there exists $t$ such that they all contain $x_t$.

  Now let us show that $\mathcal{A} = \{A \in [n]^{(r)} : x_t \in A\}$.
  Without loss of generality $t = r$ so our intervals are $\{x_i, \ldots, x_{i +
  r - 1}\}$ for $i = 1, 2, \ldots r$.
  Let $u = x_n$.
  Define a cyclic ordering that goes like this:
  \begin{align*}
    p_1:~&u, \text{ then} \\
    p_2:~&\text{elements of $\{x_1, \ldots, x_{r - 1}\} \setminus A$}, \text{ then} \\
    p_3:~ &\text{elements of $\{x_1, \ldots, x_{r - 1}\} \cap A$}, \text{ then} \\
    p_4:~&x_r, \text{ then} \\
    p_5:~&\text{elements of $A \setminus \{x_1, \ldots, x_r\}$}, \text{ then} \\
    p_6:~&\text{the rest}
  \end{align*}
  In this cyclic ordering, $\{u\} \cup I_u^+ \setminus \{x_r\} = (p_1, p_2, p_3)
  = \{x_n, x_1, \ldots, x_{r - 1}\} \notin \mathcal{A}$ (sinc it is an interval
  in the other cyclic ordering).
  Next, note $I_u^+ = (p_2, p_3, p_4) = \{x_1, \ldots, x_r\} \in \mathcal{A}$.
  So by the property we have proved (that in any cyclic ordering we have $r$
  consecutive intervals), $A = (p_3, p_4, p_5) \in \mathcal{A}$ since $A$ is
  also an interval.

  The proof is complete.
\end{proof}

Our aim now is to prove a theorem of Friedgut and Dinur that shows that if $0 <
p < \half$ and $n$ is sufficiently large, then for every intersecting family
$\mathcal{A}$ of sets of size $pn$, there exists a subset $J \subset [n]$ and an
intersecting family $B \subset \mathcal{P}(J)$ such that very few sets in
$\mathcal{A}$ do not contain a set in $\mathcal{B}$.
``Very few'' here means relative to ${n \choose pn}$ (rather than being relative
to $|\mathcal{A}|$).

\begin{note*}
  In this section, Boolean functions are from $\{0, 1\}^n$ to $\{0, 1\}$, and
  the $\mup$-biased measure gives probability $p$ to $x_i = 1$ and $q$ to $x_i =
  0$.
\end{note*}

\begin{fcdefn}[]
  \glsadjdefn{quasi}{quasirandom}{}%
  A Boolean function $f : \{0, 1\}^n \to \{0, 1\}$ is \emph{$(\eps, p,
  r)$-quasirandom} if for every $J \subset [n]$ of size $r$, and every $u \in
  \{0, 1\}^J$,
  \[
    \left|\Ebb\left[f^{(p)} ~\Big|~ x|_J = u\right] - \Ebb f^{(p)}(x)\right|
    \le \eps
  .\]
\end{fcdefn}
