Limit of positive sequence $(f_n)$ defined by $f_n(x)^2=\int^x_0 f_{n-1}(t)\mathrm{d}t$ Assume $f_{0}(x)$ is integrable in $[0,1]$, $f_{0}(x)>0$.
$$f_n(x)=\sqrt{\int^x_0 f_{n-1}(t)\mathrm{d}t}, \quad n=1,2,...$$
How can one calculate $\lim\limits_{n \to {\infty}}f_n(x)$?

If we consider a family of examples : $f_0(t) = C t^{\alpha}$ for $\alpha > 0, C > 0$, then one easily sees that the sequence $f_n(t)$ becomes
$$
C x^{\alpha}, C^{\frac 12} \left( \frac{x^{\frac{\alpha + 1}2}}{\sqrt{\frac{\alpha+1}2}} \right), C^{\frac 14} \left( \frac{x^{\frac{\frac{\alpha + 1}2 +1}2}}{\sqrt{\sqrt{\frac{\alpha+1}2} \left( \frac{\alpha+1}2 +1 \right) }} \right), \dots
$$
and the exponents follow the sequence $a_0 = \alpha$, $\alpha_n = \frac{\alpha_{n-1} +1}2$, while the denominators follow the reccurence $\beta_0 = 1$, $\beta_n = \sqrt{  2 \alpha_n \beta_{n-1}}$, with the constant $C^{\frac 1{2^n}}$ converging to $1$. Now one easily notices that $\alpha_n \to 1$ (since you keep averaging between $1$ and $\alpha$, you get closer and closer to $1$), and if we expect $\beta_n$ to converge, then heuristically speaking we'll have $\beta_n \to \beta$ if and only if $\beta^2 = \sqrt{ 2 \beta }$, which means $\beta^2 - 2 \beta = 0$, hence $\beta = 2$ since we want $\beta$ non-zero (all this is heuristics because we didn't prove much). In other words, heuristically, if the question is well posed, any monomial converges to the function $f(x) = x/2$. 
With some appropriate inequalities one can probably work the general case from here, but I have not worked that out yet.
Hope that helps,
