Iteration of an operator 
Let $f_0(x)$ be integrable on $[0,1]$, and $f_0(x)>0$. We define $f_n$ iteratively by 
  $$f_n(x)=\sqrt{\int_0^x f_{n-1}(t)dt}$$
  The question is, what is $\lim_{n\to\infty} f_n(x)$?

The fix point for operator $\sqrt{\int_0^x\cdot dt}$ is $f(x)=\frac{x}{2}$. But it's a bit hard to prove this result. I have tried approximate $f(x)$ by polynomials, but it's hard to compute $f_n$ when $f_0(x)=x^n$ since the coefficient is quite sophisticated. Thanks!
 A: Note: this is not a proof that the limit exists, but a computation of the limit if we know that it exists.
We know that $f(x)>0$ for $x>0$ and $f(0)=0$. We want to solve
$$
f(x)=\sqrt{\int_0^xf(t)\,dt},\quad 0\le x\le 1,
$$
that is,
$$
(f(x))^2=\int_0^xf(t)\,dt,\quad 0\le x\le 1.
$$
Derivate with respect to $x$ to obtain
$$
2\,f\,f'=f\implies f'(x)=1/2\implies f(x)=x/2.
$$
A: 1) If $f_0(x)\equiv 1$, It is easy to check $f(x)=\frac x2$.
2) If there exists $m,M >0$ such that $m<f_0(x)<M$, then
$$\sqrt{mx}=\sqrt{\int_0^xmdt} \leq f_1(x)\leq \sqrt{\int_0^xMdt}=\sqrt{Mx}$$
and
$$ \sqrt{\int_0^x\sqrt{mt}dt} \leq f_2(x)\leq  \sqrt{\int_0^x\sqrt{Mt}dt} $$
and so on
Thanks to 1), it follows that $f(x)=\frac x2$.
3). If $\inf\{f_0(x)\}=0$, then Approximation ! 
choose $\epsilon >0$, think interval $[\epsilon,1]$, begin with $f_1(x)$, not $f_0(x)$
there exists $m,M >0$ such that $m<f_1(x)<M, x\in[\epsilon,1]$, 
$$\sqrt{m(x-\epsilon)}=\sqrt{\int_\epsilon^xmdt} \leq f_2(x)\leq \sqrt{\int_0^xMdt}=\sqrt{Mx}, x\in [\epsilon,1]$$
and so on, 
we get that
$$\frac{x-\epsilon}2\le f(x)\le\frac x2$$
Let $\epsilon \to 0$, it follows that $f(x)=\frac x2$.
