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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!

Golbez
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2 Answers2

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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 $$ 2\,f\,f'=f\implies f'(x)=1/2\implies f(x)=x/2. $$

Julián Aguirre
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    The differentiation does not appear to be valid. Differentiability of $f^2$ does not imply differentiability of $f.$ Take for example the function that takes all rationals to $1$ and irrationals to $-1.$ – Craig Oct 10 '13 at 16:31
  • @Craig If $f>0$ is continuous then $\int_0^xf(t)dt$ is $C^1$, so is $\sqrt{\int_0^xf(t)dt}$, and hence $f$ is also $C^1$. – Julián Aguirre Oct 10 '13 at 21:14
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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 $f(x)=\frac x2$.

ziang chen
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