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\subsection{One-sided expanders}

Goal is to find: $G_n$ graphs with $n$ vertices, $d$-regular such that
$\lambda_{n - 1}(\adjm_G) \le \lambda < d$.

Reminder: (Friedman) If $G \sim G_{n, d}$ uniform random $d$-regular graph then
\[
  \Pbb(\text{$G_n$ is $(n, d, 2\sqrt{d - 1} + \eps)$-graph})
  \to 1
.\]

\[
  \graphexp(G_n)
  \ge \frac{\lambda_2(\tilde{L}_G)}{2}
  = \frac{\lambda_2(\lapm_G)}{2d}
  = \frac{(d - \lambda_{n - 1}(\adjm_G))}{2d}
  \ge \half - \frac{\eps}{2}
  > 0
.\]

\begin{fcthm}[]
  Assuming:
  - $d = 2l$ large enough
  Then:
  there is $\eps = \eps_d > 0$ such that there are $d$-regular graphs (or
  multi-graphs) $G_n$ on $n$ vertices with $\graphexp(G_n) \ge \eps$ for all
  sufficiently large $n$.
\end{fcthm}

Graphs $G_n$ are as follows:
\begin{itemize}
  \item $V(G_n) = [n]$, $d = 2l$.
  \item Take $\pi_1, \ldots, \pi_l : [n] \to [n]$ uniform independent
    permutations.
  \item Set
    \[
      E(G_n)
      = \{\{x, \pi_i(x)\} : x \in [n], i \in [l]\}
    .\]
\end{itemize}

\begin{fclemma}[]
  There is $c = c_d > 0$ such that
  \[
    \Pbb(\mathcloze{\text{$G_n$ is $d$-regular, i.e. $G_n$ is simple}})
    \ge c - o(1)
  .\]
\end{fclemma}

\begin{fclemma}[]
  If $G_n$ is $d$-regular, then there exists $\eps = \eps_d > 0$ such that
  \[
    \Pbb(\graphexp(G_n) < \eps)
    \to 0
  .\]
\end{fclemma}

\begin{proof}
  If $\graphexp(G_n) < \eps$ then there exists $F \subseteq [n]$, $r = |F| \le
  \frac{n}{2}$, there exists $F'$, $F \subseteq F'$ and $e(F, F') = d|F|$ and
  $|F'| = r + r'$, $r' = \lfloor \eps r \rfloor$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/a9dec0c25034429b.png}
  \end{center}
  \[
    \Pbb(\graphexp(G_n) < \eps)
    < \sum_{1 \le r \le \frac{n}{2}}
    \sum_{\substack{F \subseteq F' \subseteq [n] \\ |F| = r \\ |F'| = r + r'}}
    \Pbb(\pi_i(F) \subseteq F' \text{ for all $i \in [l]$})
  .\]
  $\Pbb(\pi_1(F) \subseteq F') = \frac{{r + r' \choose r}}{{n \choose r}}$.
  \[
    \Pbb
    \le \sum_{1 \le r \le \frac{n}{2}} \frac{n!}{r! r'! (n - r - r')!}
    \left( \frac{{r + r' \choose r}}{{n \choose r}} \right)^l
  .\]
  Fact 1: $\frac{{a \choose k}}{{b \choose k}} \le \left( \frac{a}{b} \right)^k$
  if $a \le b$.

  Proof: It is equivalent to
  \[
    ba \cdot b(a - 1) \cdots b(a - k + 1)
    \le ab \cdot a(b - 1) \cdots a(b - k + 1)
  .\]
  Compare the product term by term.

  Fact 2: $n! \ge \left( \frac{n}{e} \right)^n$.
  Proof:
  \[
    e^n = \sum_{k \ge 0} \frac{n^k}{k!}
    \ge \frac{n^n}{n!}
  .\]
  
  Using these:
  \[
    \frac{n!}{r! r'! (n - r - r')!}
    \le \frac{n^{r + r'}}{r! r'!}
    = \frac{n^{r + r'}}{(r + r')!} {r + r' \choose r}
    \le (2e)^{r + r'} \left( \frac{n}{e + e'} \right)^{r + r'}
  .\]
  So
  \begin{align*}
    \Pbb
    &\le \sum_{1 \le r \le \frac{n}{2}} (2e)^{r + r'} \left( \frac{n}{r + r'}
    \right)^{r + r'} \left( \frac{r + r'}{n} \right)^{lr} \\
    &\le \sum_{1 \le r \le \frac{n}{2}} (2e)^{2r} \left( \frac{(1 + \eps)r}{n}
    \right)^{r(l - 1 - \eps)}
  \end{align*}
  Decompose as $\sum_{1 \le r \le n/2} = \sum_{1 \le r \le K} + \sum_{K < r
  \le n/2} = S_1 + S_2$.
  Choose $\eps > 0$ small so that $\gamma = \frac{1 + \eps}{2}$ is small. Choose
  $l$ large so that $\gamma^{l - 1 - \eps} < \frac{1}{2(2e)^2}$.
  \begin{align*}
    S_2
    &\le \sum_{K < r \le \frac{n}{2}} (2e)^{2r} \left( \frac{(1 + \eps)}{2}
    \right)^{r(l - 1 - \eps)} \\
    &\le \sum_{K < r \le \frac{n}{2}} \left( \half \right)^r \\
    &\le \frac{1}{2^K}
  \end{align*}
  Now $S_1$.
  \begin{align*}
    S_1
    &\le \sum_{1 \le r \le K} (2e)^{2r} \left( \frac{(1 + \eps)r}{n}
    \right)^{r(l - 1 - \eps)} \\
    &\le \left( \frac{2K}{n} \right)^{l - 2} \sum_{1 \le r \le K} (2e)^{2r} \\
    &\le c^K \left( \frac{K}{n} \right)^l
  \end{align*}
  Now let $K = \log\log\log n$, so this is $\le O((\log\log n) n^{-l}
  (\log\log\log n)^l) \to 0$.
  Also, $S_1 \le \frac{1}{2^K}$ term goes to $0$.
\end{proof}

For the lemma about $\Pbb(\text{$d$-regular}) \ge c - o(1)$:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/3dc572c0258f484e.png}
\end{center}
These are the bad things.
Use Bonferroni inequalities (the partial sum of inclusion exclusion principle
inequalities).
