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Note that there exist many other proofs of
\nameref{turans_thm} (which are non-trivially
different, and many of which are as beautiful as
this one).

\subsection{The problem of Zarankiewicz}
\label{sub3_2}

This is the ``\gls{bipartite} version'' of
\nameref{turans_thm}: How many \glspl{edge} can a
\gls{bipartite} \gls{graph} ($n$ \glspl{vertex} in
each class) have if it does not contain
$\bipartite_{t, t}$?

\begin{example*}
  $t = 2$, \gls{graph} has no $\bipartite_{2, 2}
  \gcong C_4$, so could take $G = C_{24}$.
\end{example*}

\glssymboldefn{zaran}{Z}{Z}
Write $\zaran(n, t)$ for the maximum. Does
$\zaran(n, t)$ grow quadratically ($t$ fixed, $n
\to \infty$)?

\begin{flashcard}[zaran-bound-thm]
\begin{theorem}
  \label{zaran_bound}
  Let $t \ge 2$. Then $\zaran(n, t) \le \cloze{2
  n^{2 - \frac{1}{t}}}$ \cloze{for $n$ sufficiently large.}
\end{theorem}

\begin{proof}
  \cloze{
  Let $G$ be \gls{bipartite}, \gls{vertex} classes
  $X$, $Y$. ($|X| = |Y| = n$), $G \not\supset
  \bipartite_{t, t}$. Let the
  \glsref[v_deg]{degrees} in $X$ be $d_1, \ldots,
  d_n$. Without loss of generality, $d_i \ge t -
  1 ~\forall i$ (if $d_i \le t - 2$ then add
  \glspl{edge}
  to that \gls{vertex} to make $d_i = t - 1$).

  For given $A \subset Y$, $\size{A} = t$, how
  many $x \in X$ have $\nbd(X) \supset A$? Must be
  $\le t - 1$, else $G \supset \complete_{t, t}$.

  Thus, the number of $(x, A)$ with $x \in X$, $A
  \subset Y$, $|A| = t$, $A \subset \nbd(x)$ is
  $\le (t - 1) {n \choose t}$. But each $x \in X$
  belongs to exactly ${\vdeg(x) \choose t}$ such
  $(x, A)$. So $\sum {d_i \choose t} \le (t - 1)
  {n \choose t}$. Now, the function ${x \choose
  t} = \frac{x(x - 1) \cdots (x - t + 1)}{t!}$ is
  convex for $x \ge t - 1$. This is because if we
  let $y = x - t + 1$, then this fraction is
  $\frac{(y + t - 1) \cdots y}{t!}$, a
  non-negative linear combination of powers of
  $y$.

  So $\sum {d_i \choose t} \ge n {d \choose t}$,
  where $d$ is the average of the $d_i$
  (noting $\edges(G) = nd$). Whence $n {d \choose t} \le
  (t - 1) {n \choose t}$. Thus
  \[ \frac{n(d - t + 1)^t}{t!} \le \frac{(t -
  1)n^2}{t!} ,\]
  which can be manipulated to give
  \[ d \le (t - 1)^{\frac{1}{t}} n^{1 -
  \frac{1}{t}} + t - 1 \]
  so $d \le 2n^{1 - \frac{1}{t}}$ (for $n$ large),
  as required.
  }
\end{proof}
\end{flashcard}

Does $\zaran(n, t)$ actually grow at rate $n^{2 -
\frac{1}{t}}$ (fixed $t$, $n \to \infty$)?

\begin{example*}
  $t = 2$. Our upper bound is $\zaran(n, 2) \le c
  n^{3/2}$. Lower bound? Certainly $\zaran(n, 2)
  \ge cn^1$ (for example by taking
  $2n$-\gls{cycle}). What about $\zaran(n, 2) \ge
  n^{1.01}$? This is not at all obvious. In fact,
  we do have $\zaran(n, 2) \ge c n^{3/2}$
  (\glspl{graph})
  from algebra-projective planes).
\end{example*}

\begin{example*}
  $t = 3$. Our upper bound is $\zaran(n, 3) \le
  cn^{5/3}$. In fact, it turns out that
  $\frac{5}{3}$ is correct (but even harder than
  the $t = 2$ case).
\end{example*}

\begin{example*}
  $t = 4$. Noone knows!
\end{example*}

\subsection{The Erd\H{o}s-Stone Theorem
(non-examinable)}

Note that this subsection is entirely
non-examinable.

\glssymboldefn{erdos_stone}{EX}{EX}
For a fixed \gls{graph} $H$, write $\ex(n, H)$ for
the maximum value of $\edges(G)$ where $\size{G} =
n$, $G \not\supset H$.

\begin{example*}
  \nameref{turans_thm} says
  \[ \ex(n, \complete_k) \sim \left( 1 -
  \frac{1}{k -1 } \right) {n \choose 2} \]
  or more precisely,
  \[ \frac{\ex(n, \complete_k)}{{n \choose 2}} \to
  1 - \frac{1}{k - 1} \]
  as $n \to \infty$. We call the left hand side of
  the above limit the \emph{density} of $G$.

  \cref{min_deg_implies_ham} says that
  \[ \ex(n, P_k) \sim \frac{n(k - 1)}{2} ,\]
  so
  \[ \frac{\ex(n, P_k)}{{n \choose 2}} \to 0 \]
  as $n \to \infty$.
\end{example*}
\vspace{-1em}

\textbf{General question:} How does $\ex(n, H)$
behave as $n \to \infty$?

For given $H$, let $r$ be the least integer such
that $H$ is \rpartite{r}. For example $H$
\gls{bipartite} has $r = 2$. $\cycle_7$ has $r = 3$ (it
is \rpartite{3} but not \gls{bipartite}).