$f_0(x)=x$, $f_1(x)=x(1-x)$,...,$f_{n+1}(x)=f_n(x)(1-f_n(x))$ Prove that $f_n$ converges to $0$ on [0,1].

Question

I'm trying to prepare for an upcoming test. This is a question I found on a previous midterm that I've been trying to solve:

Consider the sequence of functions defined on [0,1] given by:

$f_0(x)=x$, $f_1(x)=x(1-x)$,...,$f_{n+1}(x)=f_n(x)(1-f_n(x))$

Prove that $f_n$ converges to $0$.

I'm not quite sure how to prove this.

One idea I had was to try to use Theorem 7.9 in Rudin by defining:

$M_n = \sup_{x\in [0,1]} |f_n(x)|$

and trying to show that $M_n \to 0$ as $n \to \infty$ but that hasn't gotten me anywhere. Can anyone give me a hand?

Thank you kindly.

Answer 1

$(1).$ For $x=0$ or $x=1:$ We have $f_1(x)=0,$ and by induction on $n\ge 1$ we have $f_n(x)=0$ for all $n\ge 1.$

$(2).$ For $x\in (0,1):$ We have $f_0(x)\in (0,1).$ If $f_n(x)\in (0,1)$ then $f_{n+1}(x)=f_n(x)(1-f_n(x))\in (0,f_n(x)).$ So by induction on $n\ge 0$ the sequence $(f_n(x))_{n\in \Bbb N_0}$ is decreasing and bounded below by $0,$ so it has a limit $L\in\Bbb R,$ and we have

$L=\lim_{n\to\infty}f_n(x)=$ $\lim_{n\to\infty}f_{n+1}(x)=$ $=\lim_{n\to\infty}f_n(x)(1-f_n(x))=L(1-L)=L-L^2.$

So $L=L-L^2,$ so $L^2=0.$

Addendum: We can handle $(1)$ and $(2)$ together by noting that for any $x\in [0,1]$ we have $f_0(x)\in [0,1]$ and we have $f_n(x)\in [0,1]\implies f_{n+1}(x)\in [0,f_n(x)]\subseteq [0,1].$

Answer 2

If we let $f(x) = f_1(x) = x(1 - x)$, then $f_n = f^n$, where the power represents composition. Note that $f$ is increasing on the interval $[0, 1/2]$. This means that, if $f_n(x)$ ever takes its range in $[0, 1/2]$, then the range of $f_{n+1}(x)$ will be $[f(\min f_n), f(\max f_n)]$. It's not difficult to see that $f_n(0) = 0 = \min f_n$ for all $n$, so we simply have: $$\max f_n \le \frac{1}{2} \implies \max f_{n+1} = f(\max f_n) \le f\left(\frac{1}{2}\right) = \frac{1}{4} < \frac{1}{2}.$$ Since $\max f_1 = \frac{1}{4} < \frac{1}{2}$, this means that, for all $n \ge 1$, $$\max f_{n+1} = f(\max f_n),$$ giving you a recurrence relation on the maximum value. If we let $a_n = \max f_n$, then $$a_{n + 1} = a_n(1 - a_n), \quad a_1 = \frac{1}{4}.$$ If you can show that this maximum approaches the (constant) minimum $0$, then you are done.

Note that $$a_n - a_{n+1} = a_n^2 \ge 0$$ so the sequence is monotone decreasing. It's also simple enough to see that the sequence is contained in $[0, 1]$ (by induction) and hence is non-negative.

This means that $a_n$ converges to some $L$. Taking the limit of both sides of the recurrence relation, $$L = L(1 - L) \implies L^2 = 0 \implies L = 0,$$ so the limit is $0$, as required. Thus, $\max f_n \to 0 = \min f_n$, hence $f_n \to 0$ uniformly.