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\newpage
\section*{Introduction}

We will be analysing functions $f : \{0, 1\}^n \to \{0, 1\}$.

One reason to be interested in these is for theoretical computer science.

A more combinatorial reason is that a function $f : \{0, 1\}^n \to \{0, 1\}$ can
be viewed as a function $f : \mathcal{P}([n]) \to \{0, 1\}$, so can be viewed as
a \emph{set system} (a subset of $\mathcal{P}([n])$).
Set systems are of interest in combinatorics (e.g. Sperner's Lemma,
Kruskal-Katona, etc).

\subsubsection*{Remarks on differences between this course and additive
combinatorics}

In additive combinatorics, it is common to study $\Fbb_2^n$ in a way that is
basis-independent.
When studying boolean functions, we \emph{won't} be working in a
basis-independent way.

Slogan: if you have a basis that you care about, then perhaps you are working in
the boolean functions world, rather than the additive combinatorics world.

\newpage
\section{Discrete Fourier Analysis}

\begin{fcdefnstar}[Character]
  \glssymboldefn{Tbb}%
  \glsnoundefn{char}{character}{characters}%
  Let $G$ be a finite Abelian group.
  A \emph{character} on $G$ is a homomorphism $\chi : G \to \Tbb = \{z \in \Cbb
  : |z| = 1\}$.

  The \emph{trivial character} takes everything to $1$.
\end{fcdefnstar}

\begin{remark*}
  This definition doesn't change if $\Tbb$ is replaced by $(\Cbb \setminus \{0\},
  \times)$ (because any finite subgroup of $(\Cbb \setminus \{0\}, \times)$ must
  be a subgroup of $\Tbb$).
\end{remark*}

Observe that if $\chi_1$ and $\chi_2$ are \glspl{char}, then so is $\chi_1
\chi_2$, and also that if $\chi$ is a \gls{char} then so is $\ol{\chi} =
\chi^{-1}$.
Also, $\chi\ol{\chi} = \chi_0$, $\chi_0 \chi = \chi$.

\glssymboldefn{dualgp}%
Thus, the \glspl{char} on $G$ form an Abelian group, called the
\emph{(Pontryagin) dual} $\hat{G}$ of $G$.

We will soon see that $|G| = |\dualgp{G}|$.

\begin{notation}
  \glssymboldefn{ip}
  Let $f, g : G \to \Cbb$.
  We write
  \[
    \langle f, g \rangle
    = \Ebb_{x \in G} f(x) \ol{g(x)}
  ,\]
  where $\Ebb_{x \in G}$ means $|G|^{-1} \sum_{x \in G}$.
  Then we also write $\|f\|_2 = \ip\langle f, f \rangle^{\half} = (\Ebb_x
  |f(x)|^2)^{\half}$.
  We also define $\|f\|_p = (\Ebb_x |f(x)^p)^{\frac{1}{p}}$, $1 \le p < \infty$
  and $\|f\|_{\infty} = \max_x |f(x)|$.
\end{notation}

\begin{fclemma}[Orthogonality of characters]
  \label{lemma:1.1}
  % Lemma 1.1
  The \glspl{char} on $G$ form an orthonormal set.
\end{fclemma}

\begin{proof}
  Let $\chi_1, \chi_2$ be \glspl{char} and let $\chi = \chi_1 \ol{\chi_2}$.
  We need to prove that $\Ebb_x \chi(x)$ is $1$ if $\chi = \chi_0$ and $0$
  otherwise.
  If $\chi = \chi_0$, then the result is clear.
  Otherwise, pick $u$ such that $\chi(u) \neq 1$.
  Then
  \[
    \Ebb_x \chi(x)
    = \Ebb_x \chi(ux)
    = \chi(u) \Ebb_x \chi(x)
  .\]
  Since $\chi(u) \neq 1$, we get the result.
\end{proof}

This shows that $|\dualgp{G}| \le |G|$.
To show the reverse inequality we appeal to the classification of finite Abelian
groups.

\begin{fcthmstar}[]
  Every finite Abelian group is a product of cyclic groups.
\end{fcthmstar}

\begin{fccoro}[]
  % Corollary 1.2
  \label{coro:1.2}
  The \glspl{char} on $G$ form an orthonormal basis of $\Cbb^G$.
\end{fccoro}

\begin{proof}
  Since they form an orthonormal set, it remains to show that they span.
  Let
  \[
    G
    = \Zbb / m_1 \Zbb \times \cdots \times \Zbb / m_k \Zbb
  .\]
  Given $r, x \in \Zbb / m_1 \Zbb \times \cdots \times \Zbb / m_k \Zbb$ ($r =
  (r_1, \ldots, r_k)$, $x = (x_1, \ldots, x_k)$).
  Let
  \[
    \chi_r(x)
    = \prod_{j = 1}^{k} e^{2\pi i \frac{r_j x_j}{m_j}}
  .\]
  It is easy to check that $\chi_r$ is a character, and that if $r \neq s$ then
  $\chi_r \neq \chi_s$.
\end{proof}

\begin{remark*}
  This proof also demonstrates that $G \cong \dualgp{G}$.
  But the isomorphism is `horrible': it doesn't only depend on $G$, but also on
  how we chose to write it as a product of cyclic groups.
\end{remark*}

\glssymboldefn{dualip}%
A very useful convention is to use the uniform probability measure on $G$ and
counting measure on $\dualgp{G}$.
For example, if $\hat{f}, \hat{g} : \dualgp{G} \to \Cbb$, we define
\[
  \langle \hat{f}, \hat{g} \rangle
  = \sum_{\chi} \hat{f}(\chi) \hat{g}(\chi)
\]
and
\[
  \|\hat{f}\|_p
  = \left( \sum_\chi |\hat{f}(\chi)|^p \right)^{\frac{1}{p}}
.\]

\begin{fcdefnstar}[Fourier transform]
  \glssymboldefn{ft}%
  Let $f : G \to \Cbb$.
  The \emph{Fourier transform} $\hat{f}$ of $f$ is the function from $\hat{G}$
  to $\Cbb$ defined by
  \[
    \hat{f}(\chi)
    = \Ebb_x f(x) \ol{\chi(x)}
    = \ip\langle f, \chi \rangle
  .\]
\end{fcdefnstar}

Since the \glspl{char} form an orthonormal basis, you can also think of
$\ft{f}(\chi)$ as the coefficient of $f$ at $\chi$ with respect to the
orthonormal basis of \glspl{char}.

\begin{fclemma}[]
  \glssymboldefn{conv}%
  % Lemma 1.3
  \label{lemma:1.3}
  The Fourier transform has the following properties:
  \begin{enumerate}[(1)]
    \item \glsnoundefn{parseval}{Parseval}{}%
      \emph{Plancherel / Parseval identity:} $\dualip\langle \ft{f},
      \ft{g} \rangle = \langle f, g \rangle$.
    \item \emph{Convolution identity:} Define $f * g$ by
      \[
        f * g(x)
        = \Ebb_{u + v = x} f(u) g(v)
      .\]
      (for functions $\dualgp{G} \to \Cbb$, we define $*$ by $\sum$ rather than
      $\Ebb$)

      Then
      \[
        \ft{f \conv g}(\chi)
        \ft{f}(\chi) \ft{g}(\chi)
      .\]
    \item \emph{Inversion formula:}
      \[
        f(x)
        = \sum_{\chi} \ft{f}(\chi) \chi(x)
      .\]
  \end{enumerate}
\end{fclemma}

\begin{proof}
  \leavevmode
  \begin{enumerate}[(1)]
    \item
      \begin{align*}
        \dualip\langle \ft{f}, \ft{g} \rangle
        &= \sum_\chi \ft{f}(\chi) \ol{\ft{g}(\chi)} \\
        &= \sum_\chi (\Ebb_x f(x) \ol{\chi(x)})\ol{(\Ebb_y g(y) \ol{\chi(y)})}
        \\
        &= \Ebb_x \Ebb_y f(x) \ol{g(y)} \sum_\chi \chi(x^{-1}y) \\
        % TODO: prove lemma about $\sum_\chi \chi(x) = \indicator{x = 0}$?
      \end{align*}
      An examination of the proof of \cref{coro:1.2} shows straightforwardly
      that
      \[
        \sum_\chi \chi(u)
        = \begin{cases}
          |G|
          & \text{$u = \text{identity}$} \\
          0
          & \text{otherwise}
        \end{cases}
      \]
      So $\sum_\chi \chi(x^{-1} y) = \Delta_{xy}$ where
      \[
        \Delta_{xy}
        = \begin{cases}
          |G|
          & x = y \\
          0
          & x \neq y
        \end{cases}
      \]
      So we get
      \[
        \Ebb_x \Ebb_y f(x) \ol{g(y)} \sum_\chi \chi(x^{-1} y)
        = \Ebb_x \Ebb_y f(x) \ol{g(y)} \Delta_{xy}
        = \Ebb_x f(x) \ol{g(x)}
      .\]