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\subsection{Irregular graphs}

$A$, $D$, $L = D - A$, $\Q_\lapm(f) = \sum_{x \sim y} (f(x) - f(y))^2$,
$\q_\lapm(f) = \frac{\sum_{x \sim y} (f(x) - f(y))^2}{\sum_x f(x)^2}$.
\begin{align*}
  \Q_\lapm(f)
  &= \langle f, \lapm f \rangle \\
  &= \langle , (D - A) f \rangle \\
  &= \langle f, (2D - (D + A)) f \rangle \\
  &= \langle f, 2Df \rangle - \langle f, (D + A) f \rangle
\end{align*}

When $G$ is $d$-regular,
\[
  \tilde{L}_G
  = \frac{1}{d} \lapm_G
  = I - \frac{1}{d} \adjm_G
,\]
$\lambda_i(\tilde{L}_G \in [0, 2]$.
\[
  \q_{\tilde{L}}(f)
  = \frac{\sum_{x \sim y} (f(x) - f(y))^2}{\sum_x d(x) f(x)^2} \\
  = 2 - \frac{\sum_{x \sim y} (f(x) + f(y))^2}{\sum_x d(x) f(x)^2}
.\]
Want $M$ such that $\q_M(f)$ equals the expression above.
Recall $\q_M(f) = \frac{\langle f, Mf \rangle}{\langle f, f \rangle}$.
But the above expression is $\frac{\langle f, Mf \rangle}{\langle f, Df
\rangle}$.

Let $D^{\half}(x, y) = \indicator{x = y} \sqrt{d(x)}$.
Assume $d(x) \ge 1$ for all $x \in G$.
Note
\[
  \q_M(D^{\half} f)
  = \frac{\langle D^{\half} f, MD^{\half} f \rangle}{\langle D^{\half} f,
  D^{\half} f \rangle}
  = \frac{\langle f, D^{\half} M D^{\half} f \rangle}{\langle f, Df \rangle}
.\]
Want $D^{\half} M D^{\half} = L = D - A$.
Define
\[
  \tilde{L}_G
  = D^{-\half} (D - A) D^{-\half}
  = I - D^{-\half} A D^{-\half}
.\]
So
\[
  \tilde{L}_G
  = \begin{cases}
    1
    & \text{if $x = y$} \\
    -\frac{1}{\sqrt{d(x)d(y)}}
    & \text{if $x \neq y$ and $x \sim y$} \\
    0
    & \text{if $x \neq y$ and $x \not\sim y$} \\
  \end{cases}
\]
Also,
\[
  \q_{\tilde{L}_G}(D^{\half} f)
  = \frac{\sum_{x \sim y} (f(x) - y(y))^2}{\sum_x d(x) f(x)^2}
.\]
We have
\[
  \lambda_k(\tilde{L}_G)
  = \min_{\dim W = K}
  \max_{\substack{f \in W \\ f \neq 0}}
  \q_{\tilde{L}_G}(D^{\half} f)
.\]

\newpage
\section{Expansion and Cheeger inequality}

Assume $G$ is $d$-regular.
Write $G = (V, E)$.

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/6672818a9be54586.png}
\end{center}

\begin{fcdefn}[Expansion]
  \glssymboldefn{setexp}%
  Given a $d$-regular graph $G$ and $S \subseteq V$, the \emph{expansion of $S$}
  is
  \[
    \Phi(S)
    \defeq \frac{e(S, V \setminus S)}{d|S|}
  .\]
\end{fcdefn}

Note that $0 \le \setexp \le 1$, for example because
\[
  \setexp(S)
  = \Pbb_{\substack{x \sim U(S) \\ y \sim N(x)}} (y \notin S)
.\]

\begin{fcdefn}[Edge expansion]
  \glssymboldefn{edgeexp}%
  The \emph{(edge) expansion} of a cut $(S, V \setminus S)$ is defined as
  \[
    \Phi(S, V \setminus S)
    \defeq \max \{\setexp(S), \setexp(V \setminus S)\}
    = \frac{e(S, V \setminus S)}{d \min \{|S|, |V \setminus |\}}
  .\]
\end{fcdefn}

\begin{fcdefn}[Edge expansion of a graph]
  \glssymboldefn{graphexp}%
  The \emph{edge expansion of a graph} is
  \[
    \Phi(G)
    \defeq \min_{\substack{S \subseteq V \\ \emptyset \neq S \neq V}}
    \edgeexp(S, V \setminus S)
    = \min_{\substack{S \subseteq V \\ 0 < |S| < |V|/2}} \setexp(S)
  .\]
\end{fcdefn}

\begin{fcthm}[Cheeger's inequality]
  \label{thm:cheeger}
  Assuming:
  - $G$ be $d$-regular
  Then:
  \[
    \frac{\lambda_2(\tilde{L}_G)}{2}
    \le \graphexp(G)
    \le \sqrt{2\lambda_2(\tilde{L}_G)}
  .\]
\end{fcthm}

Consider $\indicator{S} : V \to \Rbb$, where $\indicator{S}(x) = 1$ if $x \in S$
and $\indicator{S}(x) = 0$ otherwise.
Then
\begin{align*}
  \Q_\lapm(\indicator{S})
  &= \sum_{x \sim y} (\indicator{S}(x) - \indicator{S}(y))^2
  = e(S, V \setminus S) \\
  \q_{\tilde{L}}(\indicator{S})
  &= \frac{e(S, V \setminus S)}{d|S|}
  = \setexp(S)
\end{align*}
Recall
\[
  \lambda_2(\tilde{L}_G)
  = \min_{\dim W = 2} \max_{\substack{f \in W \\ f \neq 0}} \q_{\tilde{L}_G}(f)
.\]
We pick $W = \Span \{\indicator{S}, \indicator{V \setminus S}\}$.
Note
\begin{align*}
  \lambda_2(\tilde{L}_G)
  &\le \max_{\substack{\alpha, \beta \\ (\alpha, \beta) \neq (0, 0)}}
  \q_{\tilde{L}_G}(\alpha \indicator{S} + \beta \indicator{V \setminus S}) \\
  &\le \max_{\alpha, \beta} 2\max \{q_{\tilde{L}(G)}(\alpha \indicator{S}),
  \q_{\tilde{L}_G}(\beta \indicator{V \setminus S})\}
\end{align*}

\begin{fclemma}[]
  Assuming:
  - $M$ is symmetric positive semi-definite
  - $\langle f, g \rangle = 0$
  Then:
  \[
    \q_M(f + g)
    \le 2\max \{\q_M(f), \q_M(g)\}
  .\]
\end{fclemma}

\begin{proof}
  Let $\lambda_i$, $\varphi_i$ such that $f = \sum_i \lambda_i \varphi_i$ and $g
  = \sum_i \beta_i \varphi_i$.
  Then
  \[
    \q_M(f + g)
    = \frac{\sum_i \lambda_i (\alpha_i + \beta_i)^2}{\|f + g\|^2}
    \le \frac{\sum_i \lambda_i (2\alpha_i^2 + 2\beta_i^2)}{\|f\|^2 + \|g\|^2}
  .\]
  Then
  \begin{align*}
    \q_M(f + g)
    &\le 2 \left( \frac{\sum_i \lambda_i \alpha_i^2 + \sum_i \lambda_i
    \beta_i^2}{\|f\|^2 + \|g\|^2} \right) \\
    &= 2 \left( \frac{\q_M(f) \|f\|^2 + \q_M(g)\|g\|^2}{\|f\|^2 + \|g\|^2}
    \right) \\
    &\le 2\max \{\q_M(f), \q_M(g)\}
    \qedhere
  \end{align*}
\end{proof}

\begin{proof}[Proof of left inequality in \nameref{thm:cheeger}]
  $\lambda_2 \le 2 \max \{\setexp(S), \setexp(V \setminus S)\} = 2\edgeexp(S, V
  \setminus S)$.
  Then minimise over all $S$, to get $\lambda_2 \le 2 \graphexp(G)$.
\end{proof}
