Ramsey-Turán density function is well defined. Define
$$RT(n,K_l,f(n))=ex_l(n,f(n))=\max_G\{e(G): K_l \not\subset G, v(G)=n, \alpha(G)\leq f(n)\}$$
and the Ramsey-Turán density function $f_l:(0,1] \to \mathbb{R}$ as
$$f_l(\alpha)=\lim_{n\to \infty}\frac{ex_l(n,\alpha n)}{{n\choose 2}}.$$
By Turán Theorem we have that $f_l(\alpha)=1-\frac{1}{l-1}$ for every $\alpha \geq \frac{1}{l-1}$.
How to prove that this limit exists for every $\alpha \in (0,1]$?
I was trying to prove that the sequence $(ex_l(n,\alpha n)/{n\choose 2})_n$ is monotone, but I could only prove that
$$ex_l(n,s)\leq\frac{n \ ex_l(n-1,s)}{n-2},$$
which is not enough since $ex_l(n-1,\alpha n) \geq ex_l(n-1,\alpha(n-1)).$
I also thought of using that $\alpha(G) \leq \alpha n$ and $\alpha(G)\leq \alpha(n-1)$ only give different restrictions once every $\approx 1/\alpha$ values of $n$ in a row. But it would also be necessary to show that every jump is in the same direction, which I couldn't do.
 A: The graph theory part
Let's begin with some monotonicity properties that actually rely on the graph theory we're doing.
For fixed $l$ and $\alpha$, let $g(n) = \mathrm{ex}_l(n, \alpha n)$. We'd like to say that for all $x,y$ we have $g(x+y) \ge g(x) + g(y)$: just take the disjoint union of the graphs achieving $g(x)$ and $g(y)$. This doesn't quite work because $g$ might not always be defined; for example, $\mathrm{ex}_3(6,2)$ does not exist. But it should work when $x,y$ are sufficiently large. To put it differently, if $x \le y$ and $x,y$ are sufficiently far apart, then $g(x) \le g(y)$.
There is also a construction relating $g(n)$ to $g(kn)$ for any integer $k$: take a graph $G$ achieving the maximum in $g(n)$, and replace each vertex by an independent set of size $k$. Each edge $vw$ is replaced by $k^2$ edges between the corresponding sets. The resulting graph still has no $K_l$; its independence number is at most $\alpha kn$, as desired; finally, it has $k^2 e(G)$ edges. This shows that $g(kn) \ge k^2 g(n)$ for any integers $k,n$.
Therefore if we let $h(n) = \frac{g(n)}{n^2}$, then $h(n)$ has a very number-theoretic monotonicity property: if $x,y$ are integers with $x \mid y$, then $h(x) \le h(y)$.
The number theory/real analysis part
The remainder of the proof is just putting all this together. Let $L = \sup\{h(2^p 3^q) : p,q \in \mathbb N\}$. This number certainly exists, since $h$ is bounded above by $1$. We will eventually prove that $h(n) \to L$ as $n \to \infty$.
Lemma. For any $\epsilon>0$, there is a sequence $(n_i)_{i=1}^\infty$ such that $n_i \le n_{i+1} \le (1+\epsilon)n_i$, and $L-\epsilon \le h(n_i) \le L$ for all $i$.
Proof. We can find a fraction $\frac{3^a}{2^b}$ in $(1,1+\epsilon)$ with rational approximations of $\log_2 3$. Before we define $(n_i)_{i=1}^\infty$, we will define a preliminary sequence $(n'_i)_{i=1}^\infty$. Let $n'_1 = 1$, and continue with
$$
   n'_{i+1} = \min\left\{ 2^{\lceil \log_2 (n'_i+1)\rceil}, \frac{3^a}{2^b} \cdot n'_i\right\}.
$$
Essentially, we multiply by $\frac{3^a}{2^b}$ at each step, but whenever we pass a power of $2$, we "reset" to that power of $2$ so that our denominators stay bounded.
Now choose a number $2^p 3^q$ such that $h(2^p 3^q) \ge L-\epsilon$. Multiply $n'_i$ first by the least common denominator of the sequence, then by $2^p 3^q$. We get an integer sequence $(n_i)_{i=1}^\infty$ with $2^p 3^q \mid n_i$. Therefore $h(n_i) \ge h(2^p 3^q) \ge L-\epsilon$ for all $i$; also, $h(n_i) \le L$ by definition of $L$. $\qquad\square$
By the weaker monotonicity property of $g$, if $(n_i)_{i=1}^\infty$ is our sequence from the lemma, and $n_i \le n \le n_{i+1}$, then provided we're far enough along in the sequence, we can at least guarantee $g(n_{i-1}) \le g(n) \le g(n_{i+2})$. Also, we know $n \le (1+\epsilon)^2 n_{i-1}$ and $n_{i+2} \le (1+\epsilon)^2 n$. Therefore
$$
   n_{i-1}^2 h(n_{i-1}) \le n^2 h(n) \le n_{i+2}^2 h(n_{i+2}) \implies (1+\epsilon)^{-4} h(n_{i-1}) \le h(n) \le (1+\epsilon)^4 h(n_{i+2}).
$$
In particular, $h(n)$ is between $(1+\epsilon)^{-4}(L-\epsilon)$ and $(1+\epsilon)^4 L$, and this holds for all sufficiently large $n$. Since we can do this for any $\epsilon>0$, we conclude that $h(n) \to L$ as $n \to \infty$.
The punchline
Finally, we have
$$
f_l(\alpha) = \lim_{n\to \infty}\frac{\mathrm{ex}_l(n,\alpha n)}{\binom n2} = \lim_{n\to \infty}\frac{\mathrm{ex}_l(n,\alpha n)}{n^2/2} = \lim_{n \to \infty} 2 h(n) = 2L
$$
so the limit you want also exists.
