I have the extremal/Turàn function defined as $$ \operatorname{ex}(n, F) = \max\left\{e(G): |G| = n, F \not\subseteq G\right\} $$ and the Turàn density $$ \lim_{n \to \infty} \frac{\operatorname{ex}(n, F)}{\binom{n}{2}} $$ I want to show that this limit exists. It’s described as “easy” in my notes, so I feel like I must be missing something obvious.
I can see that $\operatorname{ex}(n, F)/\binom{n}{2}$ is bounded in $[0,1]$, and I think that the idea is to show that this is decreasing in $n$ (so it forms a decreasing bounded sequence, which converges), but I haven’t been able to do this.
I’ve been able to get this far: $$ \frac{\operatorname{ex}(n+1,F)}{\binom{n+1}{2}} = \frac{\operatorname{ex}(n+1,F)}{\frac{n\left(n+1\right)}{2}} = \frac{\operatorname{ex}(n+1,F)}{\frac{n\left(n-1\right)}{2}} \cdot \frac{n-1}{n+1} \leq \frac{\operatorname{ex}(n+1,F)}{\binom{n}{2}} $$ but I don’t know how to show that $\operatorname{ex}(n+1,F) \leq \operatorname{ex}(n,F)$, mostly because I’m fairly sure it isn’t true.
How do I finish this proof? Do I need to continue this argument, or is this the wrong line of reasoning entirely?
The sequence $ex(n, F)/\binom{n}{2}$ is indeed non-increasing.
To see this, assume w.l.o.g. that $F$ has at least one edge and let $n \geq 3$ be arbitrary. Consider a graph $G$ with $n$ vertices, $ex(n, F)$ edges and such that $F \not\subseteq G$. Such a graph always exists, as $F$ has at least one edge.
Now any subset of $n-1$ vertices of $G$ is also $F$-free, and hence spans at most $ex(n-1, F)$ edges. Summing over all subsets of $G$ of size $n-1$ gives a total of at most $n\cdot ex(n-1, F)$ edges. But each edge of $G$ was counted exactly $n-2$ times, so \begin{equation*} ex(n, F) \leq \frac{n}{n-2} ex(n-1, F). \end{equation*} Dividing by $\binom{n}{2}$ gives \begin{equation*} \frac{ex(n, F)}{\binom{n}{2}} \leq \frac{ex(n-1, F)}{\binom{n-1}{2}}, \end{equation*} as desired.