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    \item
      \phantom{} \\[-2.5\baselineskip]
      \begin{align*}
        \ft{f \conv g}(\chi)
        &= \Ebb_x f \conv g(x) \ol{\chi(x)} \\
        &= \Ebb_x \Ebb_{uv = x} f(u) g(v) \ol{\chi(uv)} \\
        &= \Ebb_{u, v} f(u) \ol{\chi(u)} g(v) \ol{\chi(v)} \\
        &= \ft{f}(\chi) \ft{g}(\chi)
      \end{align*}
    \item
      \phantom{} \\[-2.5\baselineskip]
      \begin{align*}
        \sum_\chi \ft{f}(\chi) \chi(x)
        &= \sum_\chi \Ebb_y f(y) \ol{\chi(y)} \chi(x) \\
        &= \Ebb_y f(y) \sum_\chi \chi(xy^{-1}) \\
        &= \Ebb_y f(y) \Delta_{xy} \\
        &= f(x)
        \qedhere
      \end{align*}
  \end{enumerate}
\end{proof}

\subsubsection*{Specialising to $\Fbb_2^n$}

For each $r \in \Fbb_2^n$, we can define a \gls{char} $\chi_r$ by
\[
  \chi_r(x)
  = (-1)^{r \cdot x}
,\]
where $r \cdot x$ is shorthand for $\sum_{i = 1}^{n} r_i x_i  \mod 2$.
This gives $2^r$ distinct \glspl{char}, so all of them.

If $f : \Fbb_2^n \to \Cbb$, then for each $r \in \dualgp{\Fbb_2^n}$, we write
\[
  \widehat{f}(r)
  = \Ebb_x (-1)^{r \cdot x} f(x)
.\]
If we identify elements $x$ of $\Fbb_2^n$ with their supports $A_x = \{i : x_i
= 1\}$, then $r \cdot x = |A_r \cap A_x| \mod 2$.
uso if we take the group $(\mathcal{P}([n]), \symdiff)$, then given a function
$f : \mathcal{P}([n]) \to \Cbb$, we have
\[
  \ft{f}(B)
  = \Ebb_{A \subset [n]} f(A) (-1)^{|A \cap B|}
.\]
If we take the group $\{-1, 1\}^n$ with pointwise multiplication, then the
\glspl{char} are the functions of the form
\[
  x \mapsto \prod_{i \in A} x_i
,\]
where $A \subset [n]$.
Write these as $x_A$.

\begin{warning*}
  I shall be sloppy about the distinction between the \emph{function} $x_A$ and
  the value it takes.
\end{warning*}

We shall write $\ft{f}(A)$ for $\ft{f}(x_A)$.
Then $\ft{f}(A) = \Ebb_x f(x) x_A$.

By the inversion formula, $f(x) = \sum_A \ft{f}(A) x_A$.

Convention: I'll say ``multilinear'' for ``multiaffine''.
So ``linear in the school sense, rather than the linear algebra sense''.

The inversion formula therefore expresses $f : \{-1, 1\}^n \to \Rbb$ as the
restriction to $\{-1, 1\}^n$ of a multilinear function on $\Rbb^n$.

\begin{fclemma}[]
  % Lemma 1.4
  \label{lemma:1.4}
  For every $f : \{-1, 1\}^n \to \Rbb$, there is a unique multilinear function
  $\mu : \Rbb^n \to \Rbb$ such that $\mu|_{\{-1, 1\}^n} = f$.
\end{fclemma}

\begin{proof}
  We have shown existence (using the inversion formula).
  For uniqueness, it is enough to show that if $\mu$ is multilinear and vanishes
  on $\{-1, 1\}^n$, then it vanishes everywhere.
  We show this by induction on $n$.

  If $n = 1$, then $\mu$ is linear and $\mu(-1) = \mu(1) = 0$, so $\mu \equiv
  0$.

  Now assume the result for $n - 1$ and let $\mu : \Rbb^n \to \Rbb$ be
  multilinear and $0$ on $\{-1, 1\}^n$.
  Since $\mu$ depends linearly on the last coordinate, we can write for all $x
  \in \Rbb^{n - 1}$, $t \in \Rbb$,
  \[
    \mu(x, t)
    = \alpha(x) t + \beta(x)
  .\]

  Let $x \in \{-1, 1\}^{n - 1}$.
  Then
  \[
    \beta(x)
    = \half (\mu(x, 1) + \mu(x, -1))
    = 0
  .\]
  Also, $\mu(x, 0) = \beta(x)$, so $\beta$ is multilinear.
  Since it vanishes on $\{-1, 1\}^n$, induction hypothesis gives that $\beta
  \equiv 0$.

  So $\mu(x, t) = \alpha(x)t$.
  Setting $t = 1$ gives $\alpha(x) = \mu(x, 1)$, so $\alpha$ is multilinear and
  $0$ on $\{-1, 1\}^{n - 1}$, so $\alpha \equiv 0$.
\end{proof}

\begin{fcdefnstar}[Degree (of a boolean function)]
  \glsadjdefn{degree}{degree}{function}%
  Let $f : \{-1, 1\}^n \to \Rbb$.
  The \emph{degree} of $f$ is the degree of the multilinear polynomial $\mu$
  that restricts to it.
  Equivalently, it is $\max \{|A| : \ft{f}(A) \neq 0\}$.
\end{fcdefnstar}

\newpage
\section{Influence, noise, stability}

\begin{fcdefnstar}[Influence]
  \glssymboldefn{Inf}%
  \glsnoundefn{totinf}{total influence}{}%
  Let $f : \{0, 1\}^n \to \{0, 1\}$.
  The \emph{influence} of the $i$-th variable, $\Infinternal_i f$, is the
  probability (for a random $x$) that changing $x_i$ changes the value of $f$.
  The \emph{total influence} $I(f)$ of $f$ is $\sum_i \Inf_i f$.
\end{fcdefnstar}

\begin{remark*}
  Up to normalisation, the \gls{totinf} is the edge boundary of the support of
  $f$.
\end{remark*}

\begin{notation*}
  \glssymboldefn{Di}%
  We define $D_i f(x) = f(x_1, \ldots, x_{i - 1} 1, x_{i + 1}, \ldots, x_n) -
  f(x_1, \ldots, x_{i - 1}, 0, x_{i + 1}, \ldots, x_n)$.
  We shall write $x_{i \to t}$ for $(x_1, \ldots, x_{i - 1}, t, x_{i + 1},
  \ldots, x_n)$.

  If $f : \{0, 1\}^n \to \{0, 1\}$, then
  \[
    \Inf_i f
    = \Ebb_x D_i f(x)^2
    = \|D_i f\|_2^2
  .\]
  If $f : \{-1, 1\}^n \to \Rbb$, then we define
  \[
    D_i f(x)
    = \half(f(x_{i \subst 1}) - f(x_{i \subst -1}))
  .\]
  The factor $\half$ is so that $D_i f$ takes values in $\{-1, 0, 1\}$ when $f$
  takes values in $\{-1, 1\}$.

  Again, $\Inf_i f = \|D_i f\|_2^2$, and $\totinf(f) = \sum_i \Inf_i(f)$.
\end{notation*}

If $f = x_A$, then
\[
  \D_i f(x)
  = \begin{cases}
    0
    & i \notin A \\
    \frac{x_{A \setminus \{i\}} - (-x_{A \setminus \{i\}}}{2}
    = x_{A \setminus \{i\}}
    & i \in A
  \end{cases}
\]
So in general,
\[
  \D_i f
  = \D_i \left( \sum_A \ft{f}(A) x_A \right)
  = \sum_{A \ni i} \ft{f}(A) x_{A \setminus \{i\}}
.\]
Therefore (by orthogonality),
\[
  \Inf_i f
  = \|\D_i(f)\|_2^2
  = \sum_{A \ni i} \ft{f}(A)^2
\]
and
\[
  \totinf(f)
  = \sum_i \sum_{A \ni i} \ft{f}(A)^2 \\
  = \sum_A |A| \ft{f}(A)^2
.\]

