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\begin{fcprop}[]
  \glsnoundefn{ndlam}{$(n, d, \lambda)$-graph}{$(n, d, \lambda)$-graphs}%
  Assuming:
  - $G$ a $d$-regular graph on $n$ vertices
  - $\lambda, \eps > 0$, $\eps d = \lambda$
  Then:
  the following are equivalent:
  \begin{enumerate}[(i)]
    \item $G$ is a $(n, d, \lambda)$-graph
    \item $\lambda_k(\adjm_G) \ni [-\lambda, \lambda]$, for $1 \le k \le n - 1$
    \item $\lambda_k(\lapm_G) \in [d - \lambda, d + \lambda]$ for $2 \le k \le
      n$
    \item $\lambda_k(\tilde{L}_G) \in \left[ 1 - \frac{\lambda}{d}, 1 +
      \frac{\lambda}{d} \right] = [1 - \eps, 1 + \eps]$ for $1 \le k \le n$
    \item $(1 - \eps) \frac{d}{n} \lapm_{K_n} \lordle \lapm_G \lordle (1 + \eps)
      \frac{d}{n} \lapm_{K_n}$
    \item $G$ is $(d, \eps)$-expander (also $(d, \frac{\lambda}{d})$-expander)
    \item $\left\|\lapm_G - \frac{d}{n} \lapm_{K_n} \right\| \le \eps d =
      \lambda$
    \item $\left\| \adjm_G - \frac{d}{n} J \right\| \le \eps d = \lambda$
  \end{enumerate}
\end{fcprop}

\begin{fclemma}[Expander Mixing Lemma]
  Assuming:
  - $G = (V, E)$ an $(n, d, \lambda)$-graph
  - $S, T \subseteq V$ (and define $e(S, T) = \sum_{x \in S} \sum_{y \in T}
  \indicator{xy \in E}$)
  Then:
  \begin{align*}
    \left| e(S, T) - \frac{d}{n} |S| |T| \right|
    &\le \frac{\lambda}{n} \sqrt{|S| |S^c| |T| |T^c|} \\
    &\le \lambda \sqrt{|S| |T|}
  \end{align*}
\end{fclemma}

\begin{proof}
  $\langle \indicator{S}, \lapm_G \indicator{T} \rangle = \langle \indicator{S},
  (dI - \adjm_G) \indicator{T} \rangle$, $\langle \indicator{S}, dI
  \indicator{T} \rangle = d|S \cap T|$.
  \begin{align*}
    \langle \indicator{S}, \adjm_G \indicator{T} \rangle
    &= \sum_x \sum_y \indicator{S}(x) \adjm_G(x, y) \indicator{T}(y) \\
    &= \sum_x \sum_y \indicator{xy \in E} \\
    &= e(S, T)
  \end{align*}
  $\frac{d}{n} \lapm_{K_n} = dI - \frac{d}{n} J$.
  \[
    \langle \indicator{S}, \frac{d}{n} J \indicator{T} \rangle
    = \frac{d}{n} \langle \indicator{S}, J \indicator{T} \rangle
    = \frac{d}{n} \sum_x \sum_y J(x, y)
    = \frac{d}{n} |S| |T|
  .\]
  \begin{align*}
    \left| e(S, T) - \frac{d}{n} |S| |T| \right|
    &= \left| \left\langle \indicator{S}, \left( \frac{d}{n} \lapm_{K_n} - \lapm
    G \right) \indicator{T} \right\rangle \right| \\
    &\le \|\indicator{S}\| \left\| \left( \frac{d}{n} \lapm_{K_n} - \lapm_G
    \right) \indicator{T} \right\| \\
    &\le \|\indicator{S}\| \left\| \left( \frac{d}{n} \lapm_{K_n} - \lapm_G
    \right) \right\|  \|\indicator{T}\| \\
    &\le \lambda \|\indicator{S}\| \|\indicator{T}\| \\
    &= \lambda \sqrt{|S| |T|}
  \end{align*}
  To get the better bound, we should consider functions which are perpendicular
  to $1$: balanced function $f_S = \indicator{S} - \frac{|S|}{n} 1$.

  $\lapm_G f_S = \lapm_G(\indicator{S} - \alpha) = L \indicator{S}$.
  \begin{align*}
    \left| e(S, T) - \frac{d}{n} |S| |T| \right|
    &= \left|\left\langle f_S, \left( \frac{d}{n} \lapm_{K_n} - \lapm_G \right)
    f_T \right\rangle\right| \\
    &\le \lambda \|f_S\| \|f_T\|
  \end{align*}
  \begin{align*}
    \|f_S\|^2
    &= |S| \left( 1 - \frac{|S|}{n} \right)^2 + (n - |S|) \left( -\frac{|S|}{n}
    \right)^2 \\
    &= |S| \left( 1 - \frac{2|S|}{n} + \frac{|S|^2}{n^2} \right) + (n - |S|)
    \frac{|S|^2}{n^2} \\
    &= |S| - \frac{2|S|^2}{n} + \frac{|S|^3}{n^2} + \frac{|S|^2}{n} -
    \frac{|S|^3}{n^2} \\
    &= |S| - \frac{|S|^2}{n} \\
    &= \frac{n|S| - |S|^2}{n} \\
    &= \frac{|S| |S^c|}{n}
  \end{align*}
  So
  \[
    \left| e(S, T) - \frac{d}{n} |S| |T| \right|
    \le \sqrt{|S| |S^c| |T| |T^c|}
    .
    \qedhere
  \]
\end{proof}

If $G$ is an \gls{ndlam} and $I \subseteq V$ an independent set, then
\[
  0
  = e(I, I)
  \ge \frac{d}{n} |I|^2 - \frac{\lambda}{n} |I| |I^c|
.\]
$\lambda|I| |I^c| \ge d|I|^3$.
\[
  |I| \le \frac{\lambda}{d} |I^c| = \frac{\lambda}{d}(n - |I|)
.\]
\[
  \left( 1 + \frac{\lambda}{d} \right) |I|
  \le \frac{\lambda}{d} n
\]
\[
  |I|
  \le \frac{\lambda}{d \left( 1 + \frac{\lambda}{d} \right)} n
  = \frac{\lambda}{d + \lambda} n
.\]

\textbf{Hoffman bound:} $\alpha(G) \le \frac{\lambda}{d + \lambda} n$.

Fix $d$.
How small can $\lambda$ be such that there is an infinite family $G_n$ of
\glspl{ndlam}?

Note $\adjm_G^2$ has $d$ in each entry of the diagonal, so
\[
  \tr \adjm_G^2
  = dn
  = \sum_i \lambda_i (\adjm_G^2)
  = \sum_i (\lambda_i (\adjm_G))^2
.\]
So $dn \le d^2 + (n - 1) \lambda^2$.

So $(n - 1) \lambda^2 \ge dn - d^2 = d(n - d)$, so
\[
  \lambda^2
  \ge \frac{d(n - d)}{n - 1}
  = d \left( \frac{n - 1 - (d - 1)}{n - 1} \right)
  = d \left( 1 - \frac{d - 1}{n - 1} \right)
.\]
$\lambda \ge \sqrt{d} (1 - o(1))$ as $n \to \infty$.

\textbf{Alon-Boppana Theorem:} $\lambda \ge 2\sqrt{d - 1} - o(1)$.

\textbf{Claim:} There exist families of \glspl{ndlam} with $\lambda = 2\sqrt{d -
1}$. They are called Ramanujan graphs.
We will probably not prove existence of these.

Call $\eps$-Ramanujan if $\lambda \ge 2\sqrt{d - 1} + \eps$.

\begin{theorem}[Friedman]
  Assuming:
  - $\eps > 0$, $n \to \infty$
  Then:
  \[
    \Pbb(\text{random $d$-regular graph on $n$ vertices is $\eps$-Ramanujan})
    \to 1
  .\]
\end{theorem}

Maxcut in \gls{ndlam}:
\begin{align*}
  e(S, S^c)
  &\le \frac{d}{n} |S| |S^c| + \frac{\lambda}{n} |S| |S^c| \\
  &\le \left( \frac{d}{n} + \frac{\lambda}{n} \right) \frac{n^2}{4} \\
  &= \frac{dn}{4} + \frac{\lambda n}{4} \\
  &\le \frac{e(G)}{2} + \frac{\lambda n}{4}
\end{align*}

Diameter, vertex expansion.
$S \subseteq V$, $\partial S = \{x \in S^c : x \sim S\}$.
\[
  \frac{|\partial S|}{|S|}
  \ge \frac{e(S, S^c)}{d|S|}
  = \setexp(S)
  \ge \frac{\lambda_2(\tilde{L}_G)}{2}
  \ge \frac{\left( 1 - \frac{\lambda}{d} \right)}{2}
\]
$|\partial S| \ge \left( \frac{1 - \lambda / d}{2} \right) |S|$.

Exercise: Diameter $O(\log n)$.