# The limit of $f_n(x)$

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 $\lim\limits_{n \to {\infty}}f_n(x)$?

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@mixedmath If I put $f_0(x)=x$, it seems that it might be f(x)=1 or f(x)=x as the final answer. As $x_{n+1}=\sqrt{x_n}$ will finally converges to 1, I think f(x)=ax might be a more convincing conjecture. –  89085731 Feb 29 '12 at 4:54
The limit can't be $f(x)=1$ since $f$ must satisfy $f = \sqrt{ \int^x_0 f(t) dt } .$ I have a feeling that this forces $f(x) = x/2$ but I haven't yet proven this is the only solution. If we assume $f$ is differentiable then it is. Every term in the sequence $f_n$ is indeed differentiable if $f_0$ is continuous. –  Ragib Zaman Feb 29 '12 at 4:54
I suddenly realized that I liked $f(x) = x/2$ much more. But I'm a bit uncertain. EDIT - oh - Ragib thinks the same. –  mixedmath Feb 29 '12 at 4:56
Are you assuming that the domain of $f_n$ is also $[0,1]$? –  Jack Feb 29 '12 at 5:16
@Jack The book is written in Chinese. –  89085731 Feb 29 '12 at 5:42

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 $x\mapsto [(1-\alpha) x]^{\alpha/(1-\alpha)}$.

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When you say that "the set of monomials is invariant by $T$", you mean that $T$ is a bijection from the monomials to the monomials? –  Patrick Da Silva Feb 29 '12 at 8:10
No, only that for every monomial m, T(m) is a monomial. Rephrased. –  Did Feb 29 '12 at 8:16
Pretty much what I expected, +1 for doing the dirty work. =) –  Patrick Da Silva Feb 29 '12 at 8:17
@Gingerjin: Are you unable to wait more than one minute before summoning somebody else to answer a question about my post, or are you just trying to be rude? Anyway... The small positive $u$ is necessary because nothing ensures that $f_1\geqslant g_{c,a}$ for some $a$ and some positive $c$, on the whole interval $[0,1]$. Roughly speaking, even though $f_1\gt0$ on $(0,1]$, $f_1$ might be too small around $0$, hence the restriction to $[u,1]$ for some positive $u$. –  Did Mar 1 '12 at 14:21
This proof is actually very slick, but grinding out the details (which is done splendidly) requires some tinkering. I guess there are bigger hammers you can hit the convergence with, but this is truly elementary. I like that. +1. –  Jonas Teuwen Dec 23 '12 at 1:17

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,

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Patrick: That did help, see my answer... –  Did Feb 29 '12 at 7:50

I have several heuristics to consider not-necessarily continuous functions.

So, consider any monomial of the form $k \cdot x^m$, and consider its exponent upon repeatedly doing this operation. In particular, it satisfies $a_n = \frac{a_{n-1} + 1}{2}$, and so we see it must go to $1$ as we are taking its arithmetic average with $1$.

Now consider it's coefficient (starting as $k$, denoted as $b_n$). We see that, letting $a_n$ still refer to the exponent on the $n$th step, that the coefficients satisfy $b_n = \sqrt{\frac{b_{n-1}}{a_n + 1}}$. As $a_n \to 1$, I heuristically (but justifiably) say we might as well have $b_n = \sqrt{ \frac{b_{n-1}}{2}}$, which converges for all positive starting $b_0$ to $\frac{1}{2}$.

Using trivial bounding, it seems plausible that any polynomial that is strictly positive on $[0,1]$ can be bounded above and below by a monomial of this form on any $[\epsilon, 1 - \epsilon]$ interval. But polynomials are dense here, and so on any $[\epsilon, 1 - \epsilon]$ interval, we should expect all functions to go to $x/2$ at least almost everywhere on those intervals.

What about behavior at $0$ and $1$? Well, that's a good question. It seems like $0$ shouldn't be too bad (not to say I can do it right off), but $1$? I have a broom and a rug, if you know what I'm saying.

Anyhow, those are my immediate heuristics I see before going to bed. Good night -

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I think you got the recurrence wrong... I'll post my answer and we'll discuss more –  Patrick Da Silva Feb 29 '12 at 5:57
Hm. I think I just felt weird because while I was typing my answer, you also used $a_n$ and $b_n$ but not for the same things. –  Patrick Da Silva Feb 29 '12 at 6:05