Consider the sequence of functions $f_n:[0,1]\rightarrow[0,1]$ defined recursively: $$f_n(p)= 1-p + p (f_{n-1}(p))^2 \quad f_0(p)=1-p \quad f_n(1)=0$$
Computationally one can check that $\{f_n(p)\}$ converges to "a" solution of the characteristic equation $$f(p)= 1-p + p (f(p))^2$$ These solutions are $1$ and $1/p-1$ (and cocktails containing points from just either). It turns out (computationally) that the correct form for $f_\infty(p)=\lim_{n\rightarrow\infty}f_n(p)$ is $$f_\infty(p) = \left\{ \begin{array}{lr} 1 & : p < 1/2\\ 1/p-1 & : p \ge 1/2 \end{array} \right.$$ We need to prove this. Here is my progress. \begin{align} f_n(p) - f(p) &= p(f^2_{n-1}(p)-f^2(p)) \\ f_n(p) - f(p) &= p(f_{n-1}(p)+f(p))(f_{n-1}(p)-f(p)) \\ f_n(p) - f(p) &= \left[p^n\prod_{i=0}^{n-1} (f_{i}(p)+f(p))\right](f_0(p)-f(p)) \\ |f_n(p) - f(p)| &= |(1-p-f(p))|\prod_{i=0}^{n-1} p^n(f_{i}(p)+f(p)) \end{align}
Let us look at the two sub-cases. Putting $f(p)=1$ gives
$$|f_n(p) - 1| = p\prod_{i=0}^{n-1} p^n(1+f_{i}(p))$$
If we take $p<1/2$, we can show that $\lim_{n\rightarrow\infty}|f_n(p)-1| = 0$. Job half done (or so it seems).
Now put $f(p)= 1/p-1 = q, p = 1/(1+q)$.
Note that $p \in (0, 1/2) \Leftrightarrow q \in (1,\infty)$ and $p\in [1/2,1] \Leftrightarrow q \in [0,1]$: \begin{align} |f_n(1/(1+q)) - q| &= \left|1-q-\frac{1}{1+q}\right|\prod_{i=0}^{n-1} \left(\frac{1}{1+q}\right)^n(q+f_{i}(1/(1+q))) \\ |f_n(1/(1+q)) - q| &= \left(\frac{q^2}{1+q}\right)\prod_{i=0}^{n-1} \left(\frac{1}{1+q}\right)^n(q+f_{i}(1/(1+q))) \\ \end{align}
...And I'm stuck. One possible way out could be proving that $f_\infty(p)$ is continuous but I don't know how to go about it. We could use the result that limit function of uniformly converging sequence of continuous functions is continuous if we somehow prove uniform convergence (is it true?) but I'm not able to do that either (bounds for Weierstrass M-test?, no explicit form for $f_n$ :( ).
EDIT: Context: The function $f_n(p)$ arises as the extinction probability of a population (in or before generation $n$) where each individual has a probability $p$ of doubling and $1-p$ of dying. Initial condition includes $1$ individual in gen $0$.
