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% 21/11/2024 09AM

% Second example class: MR2, Friday 8 days, 1:30PM

Note that if $A = [a] \times [k]^{n - d}$. Then
$|\bounde A| = da^{d - 1} k^{n - d} = d |A|^{1 -
\frac{1}{d}} k^{\frac{n}{d} - 1}$.

\begin{fcthm}[]
  \label{thm_2_13}
  % Theorem 2 13
  Assuming:
  - $A \subset [k]^n$
  - $|A| \le \frac{k^n}{2}$
  Then:
  \[
    |\bounde A| \ge \min \{d|A|^{1 - \frac{1}{d}}
    k^{\frac{n}{d} - 1} : 1 \le d \le n\}
  .\]
\end{fcthm}

``Some set of the form $[a]^d \times [k]^{n - d}$
is best.''

Called the ``edge-isoperimetric inequality in the
grid''.

\nonexaminableon

\begin{proof}[Proof (sketch)]
  Induction on $n$. $n = 1$ is trivial.

  Given $A \subset [k]^n$ with $|A| \le
  \frac{k^n}{2}$, where $n > 1$:

  Wlog $A$ is a down-set (just down-compress, i.e.
  stamp on your set in direction $i$ for each
  $i$). For any $1 \le i \le n$, define $C_i(A)
  \subset [k]^n$ by giving its \sects i:
  \[
    C_i(A)_{\isec t} = \text{extremal set of size
    $|A_{\isec t}|$ in $[k]^{n - 1}$}
  ,\]
  which will be a set of the form $[a]^d \times
  [k]^{n - 1 - d}$, or a complement. Write $B =
  C_i(A)$. Do we have $|\bounde B| \le |\bounde
  A|$?
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/a5f69b5b65e94135.png}
  \end{center}
  Now, $A$ is a down-set, so
  \[
    |\bounde A| = \ub{|\bounde A_{\isec 1}| + \cdots + |\bounde
    A_{\isec k}|}_{\text{horizontal edges}} +
    \ub{|A_{\isec 1}| - |A_{\isec
    k}|}_{\text{vertical edges}}
  \]
  and
  \[
    |\bounde B| = |\bounde B_{\isec 1}| + \cdots +
    |\bounde B_{\isec k}| + ?
  \]
  The $?$ is because $B$ not a down-set, as extremal sets in
  dimension $n - 1$ are not nested.

  Indeed, can have $|\bounde B| > |\bounde A|$:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/94c2f4ef18984bf2.png}
  \end{center}
  
  \textbf{Idea:} try to introduce a ``fake''
  boundary $\partial'$: want $\partial' A \le
  \bounde A$, with $\partial' = \bounde$ on
  extremal sets, such that $C_i$ \emph{does}
  decrease $\partial'$ (then done).

  Try $\partial' A = \sum_t |\bounde A_{\isec t}|
  + |A_{\isec 1}| - |A_{\isec k}|$. Then
  $\partial' A \le |\bounde A|$ for all $A$,
  equality for extremal sets (as equality for any
  down-set) and $\partial' C_i(A) \le \partial'
  A$. But: fails for $C_j(A)$ for all $j \neq i$.

  Could try to fix this by defining $\partial'' A
  = \sum_i (|A_{\isec 1}^{(i)}| - |A_{\isec
  k}^{(i)}|)$. Also fails -- for example if $A$ is
  the ``outer shell'' of $[k]^n$ then $\partial''
  A = 0$.

  So far, have
  \begin{align*}
    |\bounde A|
    &\ge \partial' A \\
    &\ge \partial' B \\
    &= \sum_t |\bounde B_{\isec t}| + |B_{\isec
    1}| - |B_{\isec k}| \\
    &= \sum_t f(|B_{\isec t}|) + |B_{\isec 1}| -
    |B_{\isec k}|
  \end{align*}
  where $f$ is the extremal function in $[k]^{n -
  1}$.

  Now, $f$ is the pointwise minimum of some
  functions of the form $c x^{1 - \frac{1}{d}}$
  and $c(k^{n - 1}-x)^{1 - \frac{1}{d}}$ -- each
  of which is a concave function. Hence $f$ itself
  is a concave function.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/7130c518d5904ea0.png}
  \end{center}
  Consider varying
  $|B_{\isec 2}|, \ldots, |B_{\isec{k - 1}}|$,
  keeping $|B_{\isec 2}| + \cdots + |B_{\isec{k -
  1}}|$ constant and keeping $|B_{\isec 1}| \ge
  |B_{\isec 2}| \ge \cdots \ge |B_{\isec{k - 1}}|
  \ge |B_{\isec k}|$.

  We obtain $\partial' B \ge \partial' C$, where
  for some $\lambda$,
  \[
    C_{\isec t} = \begin{cases}
      B_{\isec 1} & \forall 1 \le t \le \lambda \\
      B_{\isec k} & \forall \lambda + 1 \le t \le k
    \end{cases}
  \]
  So:
  \begin{align*}
    |\bounde A|
    &= \partial' A \\
    &\ge \partial' B \\
    &\ge \partial' C \\
    &= \lambda f(|B_{\isec 1}|) + (k - \lambda)
    f(|B_{\isec k}|) + |B_{\isec 1}| - |B_{\isec
    k}|
  \end{align*}
  but $C$ is still not a down-set.

  Now vary, $|B_{\isec 1}|$, keeping $\lambda
  |B_{\isec 1}| + (k -
  \lambda)|B_{\isec k}|$ fixed ($\lambda$ fixed)
  and $|B_{\isec 1}| \ge |B_{\isec k}|$.

  This is a concave function of $|B_{\isec 1}|$ --
  as concave + concave + linear. Hence ``make
  $|B_{\isec 1}|$ as small or large as possible''.

  i.e. $\partial' C \ge \partial' D$, where one of
  the following holds:
  \begin{itemize}
    \item $D_{\isec t} = D_{\isec 1}$ for all $t$
    \item $D_{\isec t} = D_{\isec 1}$ for all $t
      \le \lambda$, $D_{\isec t} = \emptyset$ for
      all $t > \lambda$
    \item $D_{\isec t} = [k]^{n - 1}$ for all $t
      \le \lambda$, $D_{\isec t} = D_{\isec k}$
      for all $t > \lambda$.
  \end{itemize}
  But (miraculously), this $D$ is a down-set!
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/420e0d46dde147b9.png}
  \end{center}
  Hence
  \[
    |\bounde A| = \partial' A
    \ge \partial' B
    \ge \partial' C
    \ge \partial' D
    = |\bounde D|
  .\]
  So our ``compression in direction $i$'' is: $A
  \mapsto D$.

  Now finish as before.
\end{proof}

\begin{remark*}
  To make this precise, work instead in $[0, 1]^n$
  (and then take a discrete approximation at the
  end).
\end{remark*}

\nonexaminableoff