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

Solution 1:

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

Solution 2:

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,

Solution 3:

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 -