Convergence of an integration string 
Let $(f_n)_{n\in\mathbb{N^{ * } } }$, where $f_n : [0,1] \to [0,\infty), f_0$ is a continous function and arbitrary such that: 
  $$f_{n+1}(x) = \int_0^x \frac{1}{1+f_n(t)} dt$$, for every $x\in [0,1]$ and $n \in \mathbb{N^*}$.
  Show that, for every $x \in [0,1]$, $f_n(x)_{n\in\mathbb{N^*}}$ converges.

I tried using difference and i got $f_{n+1} - f_n = \int_0^x \frac{f_{n-1}(t) - f_n(t)}{(1+f_n(t))(1+f_{n-1}(t))} dt$.
 A: Since $f_n(t)\geq0$, you have
$$
\int_0^x\frac1{1+f_n(t)}\,dt\leq \int_0^x 1\,dx=x.
$$
With a similar estimate you can get 
$$
|f_{n+1}(x)-f_n(x)|\leq\|f_n-f_{n-1}\|_\infty\,\int_0^x \frac{1}{(1+f_n(t))(1+f_{n-1}(t))} dt\leq x\,\|f_n-f_{n-1}\|_\infty.
$$
So iterating we get 
$$
|f_{n+1}(x)-f_n(x)|\leq x^n\|f_1-f_0\|_\infty.
$$
Then telescoping we get 
$$
|f_{n+k}(x)-f_n(x)|\leq \sum_{j=0}^{k-1}|f_{n+j+j}(x)-f_{n+j}(x)|\leq\sum_{j=0}^kx^{n+j}\,\|f_1-f_0\|_\infty\leq x^n\,\frac{\,\|f_1-f_0\|_\infty}{1-x}.
$$
So for any $x\in[0,1)$ the sequence $\{f_n(x)\}$ is Cauchy, and it has a limit $f(x)$. Moreover, the above estimate shows that the convergence is uniform on any interval $[0,r]$ with $r<1$, so the limit is continuous. Then the limit satisfies, for $x\in[0,1)$, 
$$
f(x)=\int_0^x\frac1{1+f(t)}\,dt.
$$
As the integrand is continuous, the integral is differentiable and thus $f$ is. Differentiating, we get
$$
f'(x)=\frac1{1+f(x)}.
$$
This is a separable first order DE, with initial value $f(0)=0$. The unique solution is $f(x)=-1+\sqrt{1+2x}$. 
It remains to look at the case $x=1$. As the integrand is bounded, we can apply Dominated Convergence (note that we don't know the limit of $f_n(1)$, but inside the integral it is just one point so it is irrelevant for the whole integral) to get 
$$
\lim_n f_n(1)=\lim_n\int_0^1\frac1{1+f_n(t)}\,dt=\int_0^1\frac1{1+f(t)}\,dt=f(1).
$$
So the limits exists also at $x=1$ and agrees with $f$. 
A: This seems like overkill, there must be a simpler solution:
Define $\phi(f)(x) = \int_0^x {1 \over 1+f(t)}dt$. It is not too hard to see that if $f\ge 0$ then $0 \le \int_0^x {1 \over 1+f(t)}dt \le x$.
In particular, for $x \in [0,1]$ we have $\phi:C[0,x] \to C[0,x]$.
It is not too hard to see that $(D\phi(f)h)(x)= \int_0^x {h(t) \over (1+f(t))^2}dt$ and
for $x \in [0,1)$ $\|D \phi(f)\| \le x < 1$. In particular, $\phi$ is a contraction map and hence there is a unique fixed point $\hat{f}$ for any fixed $x <1$.
Note that if $0 <x < x' <1$ then the solution on $[0,x']$ must also be a (and hence the) solution on $[0,x]$ and so this process unambiguously defines $\hat{f}$ on
$[0,1)$.
Since $\hat{f}(x) = \int_0^x {1 \over 1+ \hat{f}(t)} dt$, we have $\hat{f}'(x)= {1 \over 1+ \hat{f}(x)}$ and solving this ode gives $\hat{f}(x) = \sqrt{2x+1}-1$.
Hence for $x \in [0,1)$ we have $\lim_n f_n(x) = \sqrt{2x+1}-1$, and the value for $x=1$
follows by continuity.
