The limit of iterated integrals

Question

Let $f_0>0$ be integrable on $[0,1]$, define $$f_n(x)=\sqrt{\int_0^x f_{n-1}(t)dt},\ n=1,2,\cdots.$$ Find the limit $$\lim_{n\to\infty}f_n(x),\ x\in [0,1].$$

I could suspect that the limit is $0$, but I could not prove it...

Answer 1

We notice for all $n>1$ that $f_n$ is increasing. Let $g(x)$ be the limit solution. It is easy to show all $f_n$ are bounded by some constant $c$ (We have $f_n(1)<\sqrt{f_{n-1}(1)}$), so $g$ exists (why $g$ exists is actually more technical, but rests on these facts).

We have: $g(x)=\sqrt{\int_0^x g(t)dt}$. $g^2(x)=\int_0^x g(t)dt$. $2g(x)g'(x)=g(x).$

So $g(x)=0$ or $g'(x)=1/2$. Apologies, excluding $g'(x)=1/2$ was hasty. Indeed, $g(x)=x/2$ works, and these are the only two solutions.

*Why is $f_n$ increasing and bounded for $n > 1$? $f_n>0 \ \ \forall n$ so $\int_0^x f_n(t)dt$ increasing for all n so $f_n(x)=\sqrt{\int_0^x f_{n-1}(t)dt}$ is increasing for $n\geq1$. Now $f_n$ increasing for $n \geq 1$ so $\int_0^x f_{n-1}(t)dt<(1-0)f_{n-1}(1)$ so $f_n(1)<\sqrt{f_{n-1}(1)}$ for $n \geq 2$.

Answer 2

This is not really a full answer but I'd like to point out that the case $g'(x)\equiv 1/2$ is actually possible.

Start with $f_0\equiv 1$. Then $f_1(x)^2= x$, $f_2(x)^2=2/3\ x^{3/2}$ and in general we see that $f_n$ is of the form $f_n(x)=c_nx^{b_n}$ for some positive numbers $b_n,c_n\ge 0$.

Indeed, we may derive some recursive relations: $$ c_n^2x^{2b_n}=f_n(x)^2=\int_0^x f_{n-1}(t)dt = \int_0^xc_{n-1}t^{b_{n-1}} d t =\frac{c_{n-1}}{1+b_{n-1}}x^{1+b_{n-1}},$$ from which we get \begin{align*} 2b_n&=1+b_{n-1},\ \ b_0=0, \\ c_n^2&=\frac{c_{n-1}}{1+b_{n-1}}, c_0=1. \end{align*}

It is not overly difficult to establish that $b_n= 1-2^{-n}$ (anyway it is esasily proven by induction once one makes this "educated guess"). That leaves us the recursive relation $$c_0=1, \ \ c_n^2=\frac{c_{n-1}}{1+b_{n-1}}=\frac{c_{n-1}}{2(1-2^{-n})}.$$ Now it turns out that $c_n\to 1/2$ as $n\to\infty$ (giving exactly $g(x)= \lim_{n\to\infty}f_n(x)=x/2$!) To see this first note (again, say, by induction) that $c_n\ge 1/2$ for all $n$. Then estimate as follows:

\begin{align*} |c_n-1/2|&\le |c_n-1/2||c_n+1/2|=|c_n^2-1/4|\le |c_n^2-\frac{1/4}{1-2^{-n}}|+\frac{2^{-n-2}}{1-2^{-n}}\\ &= \frac{2^{-n-2}}{1-2^{-n}}+ \frac{1/2}{1-2^{-n}} |c_{n-1}-1/2|. \end{align*} From this it follows easily enough that $|c_n-1/2|\to 0$ as $n\to \infty$.

Thus, for some choices of $f_0$ the limit in question is $g(x)=1/2\ x$.

EDIT: As kelenner pointed out, there are more solutions than just $g=0$ or $g=x/2$.

In fact, let $g$ be an absolutely continuous function $g:[0,1]\to R$ satisfying $g(0)=0$ and $2g(x)g'(x)=g(x)$ for almost every $x\in [0,1]$. Then choosing $f_0= g$ yields $f_n=g$ for all $n$: if the claim is already shown up to some $n=k-1$ then $$f_{n+1}(x)^2=\int_0^xf_nd t=\int_0^xg d t =2\int_0^x g'g dt.$$ Using integration by parts one gets $$2\int_0^x g'g dt = 2g(x)^2-2\int_0^xg'gdt= 2g(x)^2- f_{n+1}(x)^2$$ from which $f_{n+1}= g$. A nontrivial example of such a function was given in the comment by kelenner.

Answer 3

I try a complete solution.

A) Let $\displaystyle d=\int_0^1 f_0(t)dt>0$. Then $\displaystyle f_1(x)\leq d^{1/2}x^{1/2}\leq d_1=\sqrt{d}$. Hence we have a bound for $f_n(x)$: by Teri computations we get that $f_n(x)\leq d^{1/2^n}a_{n-1}x^{b_{n-1}}$. As a consequence, we have for all $x$ that $\displaystyle \limsup{f_n(x)}\leq \frac{x}{2}$.

B) Now let $0<c_0<d$. Let $\displaystyle F_0(x)=\int_0^x f_0(t)dt$, $F_0$ is strictly increasing and if $u=F_0^{-1}(c_0)$ then we have $\displaystyle \int_0^x f_0(t)dt\geq c_0$ for $x\geq u$. If $c_0\to 0$ , we see that $u\to 0$.

We have $f_1(x)\geq c=\sqrt{c_0}$ for $x\geq u$, hence $\displaystyle \int_0^xf_1(t)dt\geq c(x-u)$ for $x\geq u$. Hence $\displaystyle f_2(x)\geq c^{1/2}(x-u)^{1/2}$ for $x\geq u$. Again, we show that $\displaystyle f_n(x)\geq c^{1/2^n}a_{n-1}(x-u)^{b_{n-1}}$ for $x\geq u$. But this imply that $\displaystyle \liminf{f_n(x)}\geq \frac{x-u}{2}$ for $x\geq u$. Now we let $u\to 0$, and we get $\displaystyle \liminf{f_n(x)}\geq \frac{x}{2}$ for $x>0$. As the case $x=0$ is trivial, we are done.