TITLE: Limit of positive sequence $(f_n)$ defined by $f_n(x)^2=\int^x_0 f_{n-1}(t)\mathrm{d}t$
QUESTION [13 upvotes]: 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 $\lim\limits_{n \to {\infty}}f_n(x)$?

REPLY [13 votes]: For every $x$ in $[0,1]$, $f_n(x)\to\frac12x$ when $n\to\infty$.

Rewrite the recursion as $f_n=T(f_{n-1})$. Note that the set of monomials is stable by $T$ and that $T$ is increasing in the sense that $0\leqslant f\leqslant g$ implies $Tf\leqslant Tg$. 
Introduce $g_{c,a}:x\mapsto cx^a$ with $c$ and $a$ nonnegative. Then 
$$
T(g_{c,a})=g_{K(c,a),L(a)}\quad\text{with}\quad 
K(c,a)^2=c/(a+1),\
2L(a)=a+1.
$$
We first prove an upper bound, then a lower bound.
Upper bound: Since $f_0$ is integrable, $f_1$ is bounded, that is $f_1\leqslant g_{c,0}$ for some $c$. Thus $f_n\leqslant g_{c_n,a_n}$ for every $n\geqslant1$ with 
$$
c_1=c,\ a_1=0,\ c_{n+1}=K(c_n,a_n),\ a_{n+1}=L(a_n).
$$
Since $1-L(a)=\frac12(1-a)$, $a_n\to1$. Since $2\log c_{n+1}=\log c_n-\log(1+a_n)$ and $\log(1+a_n)\to\log2$, $\log c_n\to-\log2$. Thus, $c_n\to\frac12$ and $g_{c_n,a_n}(x)\to \frac12x$ for every $x$ in $[0,1]$. This proves that
$$
\limsup\limits_{n\to\infty}f_n(x)\leqslant\tfrac12 x.
$$
Lower bound:  Pick $u$ in $(0,1)$. Then $f_1(u)=v\gt0$ and $f_1(x)\geqslant v$ for every $x$ in $[u,1]$. Define $T_u$ on the space of integrable functions $f$ on $[u,1]$ by 
$$
(T_uf)(x)=\sqrt{\int_u^xf},
$$ 
for every $x$ in $[u,1]$. Note that the set of monomials shifted by $u$ is stable by $T_u$ and that $T_u$ is increasing. In particular, introducing $g^u_{c,a}:x\mapsto c(x-u)^a$ with $c$ and $a$ nonnegative, one gets 
$$
T_u(g^u_{c,a})=g^u_{K(c,a),L(a)}.
$$
For every $f$ integrable on $[0,1]$, write $T_uf$ for the image by $T_u$ of the restriction of $f$ to $[u,1]$. 
Then $T_u\leqslant T$ in the sense that $T_uf(x)\leqslant Tf(x)$ for every $x$ in $[u,1]$. Furthermore, $f_1\geqslant g^u_{v,0}$ on $[u,1]$ implies that $f_n\geqslant g^u_{v_n,a_n}$ on $[u,1]$ with 
$$
v_1=v,\quad v_{n+1}=K(v_n,a_n).
$$ 
Hence, the same reasoning than above shows that $v_n\to\frac12$. A consequence is that 
$$
\liminf\limits_{n\to\infty}f_n(x)\geqslant\tfrac12(x-u),
$$ 
for every $x$ in $[u,1]$. Since $u\gt0$ may be as small as desired, $\liminf\limits_{n\to\infty}f_n(x)\geqslant\frac12x$ on $(0,1]$. Since $f_n(0)=0$, this yields the conclusion.
Edit: Likewise, for every $\alpha$ in $(0,1)$, the iteration of the transformation $T^{(\alpha)}$ given by
$$
(T^{(\alpha)}f)(x)=\left(\int_0^xf(t)\mathrm dt\right)^\alpha,
$$
converges to the function $x\mapsto [(1-\alpha) x]^{\alpha/(1-\alpha)}$.