Convergence of sequence of functions $f_{n + 1}(x) = \int_0^x {1\over{1 + f_n(t)}}dt$, limit? Let the sequence $(f_n)_{n \in \mathbb{N}}$ of functions $f_n: [0, 1] \to \mathbb{R}$ be such that $f_0$ is continuous and$$f_{n + 1}(x) = \int_0^x {1\over{1 + f_n(t)}}dt, \text{ for all }x \in [0, 1], \text{ for all }x \in \mathbb{N}.$$For every $x \in [0, 1]$, is the sequence $(f_n(x))_{n \in \mathbb{N}}$ convegent? If so, what is its limit?
 A: Let us firstly find the probable limit. This should be a continuous function $f:[0, 1] \to [0, \infty)$ such that$$f(x) = \int_0^x {1\over{1 + f(t)}}dt,$$whence $f$ is differentiable, $f(0) = 0$, and $f'(x)(1 + f(x)) = 1$.
Therefore, $f(x) + f^2(x)/2 = x$, so $f(x) = \sqrt{1 + 2x} - 1$.
Let us now prove that $\lim_{n \to \infty} f_n(x) = f(x)$ for every $x \in [0, 1]$. If $x \in [0, 1)$, then\begin{align*}|f_n(x) - f(x)| & = \left|\int_0^x\left({1\over{1 + f_{n - 1}(t)}} - {1\over{1 + f(t)}}\right)dt\right| \le \int_0^x {{|f_{n - 1}(t) - f(t)|}\over{(1 + f_{n - 1}(t))(1 + f(t))}}dt \\ & \le \int_0^x |f_{n - 1}(t) - f(t)|\,dt = x|f_{n - 1}(t_1) - f(t_1)|,\end{align*}for some $t_1 \in [0, x]$. In the same way, $|f_{n - 1}(t_1) - f(t_1)| \le t_1|f_{n - 2}(t_2) - f(t_2)|$ for some $t_2 \in [0, t_1]$ and, by induction,$$|f_n(x) - f(x)| \le xt_1t_2\ldots t_{n - 1}|f_0(t_n) - f(t_n)|,$$where $0 \le t_n \le t_{n - 1} \le \ldots \le t_1 \le x$.
This shows that $|f_n(x) - f(x)| \le x^n \sup_{t \in [0, 1]} |f_0(t) - f(t)|$, whence $\lim_{n \to \infty} f_n(n) = f(x)$ in this case.
If $x = 1$, let $\epsilon > 0$ and $a \in (0, \epsilon/4)$.
Since\begin{align*}|f_n(1) - f(1)| &  \le \int_0^1 |f_{n - 1}(t) - f(t)|\,dt  = \int_0^{1 - a}|f_{n - 1}(t) - f(t)|\,dt + \int_{1 - a}^1 |f_{n - 1}(t) - f(t)|\,dt \\ & \le \int_0^{1 - a}|f_{n - 1}(t) - f(t)|\,dt + 2a,\end{align*}(because $|f_{n - 1}(t)| \le 1$ and $|f(t)| \le 1$) and$$\lim_{n \to \infty} \int_0^{1 - a} |f_{n - 1}(t) - f(t)|\,dt = 0,$$it follows that there exists $N(\epsilon) \in \mathbb{N}$ such that $|f_{n - 1}(1) - f(1)| < \epsilon$ for every $n \ge N(\epsilon)$, which ends the proof.
