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\begin{example*}
  $C_N$.
  This has $\lambda_2(\tilde{L}_{C_N} = \theta \left( \frac{1}{N^2} \right)$.

  For $S \subseteq C_N$ with $|1 \le S| \le \half|C_N|$, we have $e(S, V
  \setminus S) \ge 2$.
  So
  \[
    \graphexp(S)
    = \min_{\substack{S \subseteq V \\ 0 < |S| < |V|/2}} \frac{e(S, V \setminus
    S)}{d|S|}
    = \min_{1 \le s \le N/2} \frac{2}{2s}
    \simeq \frac{2}{N}
  .\]
  Compare with \nameref{thm:cheeger}:
  \[
    \frac{1}{N^2}
    < \frac{1}{N}
    \le \sqrt{\frac{1}{N^2}}
  .\]
\end{example*}

\begin{example*}
  $G = Q_n$, $N = 2^n$.
  $G = \Cay((\Zbb / 2\Zbb)^n, \{e_1, \ldots, e_n\})$.
  We index eigenfunctions by sets $T \subseteq [n]$.
  $\chi_T(x) = (-1)^{\sum_{i \in T} x_i}$, $\lambda_T = \frac{2|T|}{n}$.

  $\lambda_2^(\tilde{L}_{Q_n}) = \frac{2}{n} = \frac{2}{\log N}$.

  $\graphexp(Q_n) \ge \frac{2}{2n} = \frac{1}{n}$.
  If $S \subseteq Q_n$, $|S| \le N/2$, then
  \[
    \frac{e(S, V \setminus S)}{n|S|}
    \ge \frac{1}{n}
    \quad \implies \quad
    e(S, V \setminus S) \ge |S|
  .\]
  Harper gives a better bound:
  \[
    e(S, V \setminus S)
    \ge |S| \log_2 \left( \frac{2^n}{|S|} \right)
  .\]
  By considering $S$ being half of the cube, we get
  \[
    \graphexp(Q_n)
    = \frac{1}{n}
    = \frac{\lambda_2(\tilde{K}_{Q_n})}{2}
  .\]
  
  Fiedler's algorithm:
  Let $f = \sum_{i = 1}^n \chi_{\{i\}}$, $\tilde{L}_{Q_n} f = \frac{2}{n} f$.
  $f(x) = \sum_{i = 1}^n (-1)^{x_i} = n - 2|x|$.
  \[
    \Phi_k
    = \frac{{n \choose k}(n - k)}{n \sum_{j = 0}^k {n \choose j}}
  .\]
  $k = \frac{n}{2}$,
  \[
    \frac{{n \choose n/2} \frac{n}{2}}{n2^{n - 1}}
    = \frac{{n \choose n/2}}{2^n}
    \approx \frac{1}{\sqrt{n}}
  .\]
\end{example*}

\newpage
\section{Loewner order}

\begin{fcdefn}[Loewner order]
  \glssymboldefn{lord}%
  For $A$, $B$ matrices, write $A \preccurlyeq B$ if $B - A$ is positive
  semidefinite.
  In particular, $A \succcurlyeq 0$ if and only if $A$ is positive semidefinite.
\end{fcdefn}

$\q_A(f) = \frac{\langle f, Af \rangle}{\langle f, f \rangle} = 0$.

$A \lordle B$ if and only if $\forall f \neq 0$,
\[
  \frac{\langle f, A f \rangle}{\langle f, f \rangle}
  \le \frac{\langle f, Bf \rangle}{\langle f, f \rangle}
.\]
$\langle f, Af \rangle \le \langle f, Bf \rangle$.

This is indeed an order: $A \lordle B \lordle C$ implies $A \lordle C$
\[
  \langle f, Af \rangle
  \le \langle f, Bf \rangle
  \le \langle f, Cf \rangle
.\]
If $A \lordle B$, then $\lambda_k(A) \le \lambda_k(B)$.
\[
  \lambda_k(A)
  = \min_{\dim W = k} \max_{\substack{f \in W \\ f \neq 0}} \frac{\langle f, Af
  \rangle}{\langle f, f \rangle}
.\]
$A \lordle B$ if and only if $A + C \lordle B + C$.

If $G$ is a graph, then $\lapm_G \lordle 0$.

\begin{fcdefn}[eps-approximation]
  \glsnoundefn{epsapp}{$\eps$-approximation}{$\eps$-approximations}%
  $G$ is an \emph{$\eps$-approximation of $H$} if
  \[
    (1 - \eps) \lapm_H
    \lordle \lapm_G
    \lordle (1 + \eps) \lapm_H
  .\]
\end{fcdefn}

\begin{fclemma}[]
  Given the definition $\|M\| = \max_{f \neq 0} \frac{\|Mf\|}{\|f\|}$ (for $M$
  symmetric), we have \cloze{$\|M\| = \max \{|\lambda_k(M)|\}$.}
\end{fclemma}

\begin{proof}
  $f = \sum_i \alpha_i \varphi_i$, $M \varphi_i = \lambda_i \varphi_i$, $\|f\|^2
  = \sum_i \alpha_i^2$,
  \[
    \|Mf\|^2
    = \left\| \sum_i \alpha_i \lambda_i \varphi_i \right\|^2
    = \sum_i \alpha_i^2 \lambda_i^2
  .\]
  \[
    \frac{\|Mf\|}{\|f\|}
    = \sqrt{\frac{\sum_i \alpha_i^2 \lambda_i^2}{\sum_i \alpha_i^2}}
    \le \max_k |\lambda_k|
    .
    \qedhere
  \]
\end{proof}

\begin{lemma}[]
  Assuming:
  - $G$ is an \gls{epsapp} of $H$
  Then:
  $\|\lapm_G - \lapm_H\| \le \eps$.
\end{lemma}

\begin{proof}
  $-\eps \tilde{L}_H\lordle \tilde{L}_G - \tilde{L}_H \lordle \eps \tilde{L}_H$.
  $\lambda_k(\tilde{L}_G - \tilde{L}_H) \le \lambda_k(\tilde{L}_H) \le 2\eps$.
\end{proof}

\begin{fcdefnstar}[(d,eps)-expander]
  \glsnoundefn{depsexp}{$(d, \eps)$-expander}{$(d, \eps)$-expanders}%
  $G = (V, E)$, $|V| = n$ is a \emph{$(d, \eps)$-expander} if $G$ is an
  \gls{epsapp} of $\frac{d}{n} K_n$.
\end{fcdefnstar}

Equivalent:
\[
  (1 - \eps) \frac{d}{n} \lapm_{K_n}
  \lordle \lapm_G
  \lordle (1 + \eps) \frac{d}{n} \lapm_{K_n}
.\]
$\lapm_{K_n} = (n - 1) I - A_{K_n} = nI - J$, where $J$ is the all ones
matrix ($J(x, y) = 1$).
If $f \perp q$, then $Jf = \langle f, 1 \rangle f = 0$.
In this case, $\lapm_{K_n} f = nIf - Jf = nf$.

So $\lambda_k(\lapm_{K_n}) = n$ for $k \ge 0$.
\[
  (1 - \eps) \frac{d}{n} (nI - J)
  \lordle dI - \adjm_G
  \lordle (1 + \eps) \frac{d}{n} (nI - J)
\]
\[
  -\eps \frac{2}{n} (nI - J)
  \lordle dI - \adjm_G - \frac{d}{n} (nI - J)
  \lordle \eps \frac{d}{n} (nI - J)
\]
\[
  -\eps \left( dI - \frac{d}{n} J \right)
  \lordle \frac{d}{n} J - \adjm_G
  \lordle \eps \left( dI - \frac{d}{n} J \right)
\]
For $f \perp 1$, $-\eps d \langle f, f \rangle \le -\langle f, Af \rangle \le
\eps d \langle f, f \rangle$.

So $G$ is a \gls{depsexp} if and only if
\[
  |\lambda_k(\adjm_G)|
  \le \eps d
\]
for all $1 \le k \le n - 1$.

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/1f235e1bbbff4b6a.png}
\end{center}

\begin{fclemma}[Expander Mixing Lemma]
  Assuming:
  - $G$ is $d$-regular
  - $G$ is a \gls{depsexp}
  Then:
  $\forall S, T \subseteq V$,
  \[
    \left|e(S, T) - \frac{d}{n} |S| |T|\right|
    \le \frac{\eps d}{n} \sqrt{|S| |T| |S^c| |T^c|}
  .\]
\end{fclemma}