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Define $f\colon[0,1]\to[0,1]$ such that $f(x) = 3.9x(1-x)$. Consider the iterated function sequence $f_{n+1} = f(f_{n} (x))$. Prove that there is no $N \in \mathbb{N}$ such that the number of fixed points ($f_{n}(x) = x$) of $f_{n}$ is $\leq N$ for all $n \in \mathbb{N}$. Hint: Use the intermediate value theorem.

I know the intermediate value theorme:

Let $f$ be a continuous real function on $[a,b]$. If $f(a) < f(b)$, and if $c$ is a number such that $f(a)<c<f(b)$, then there exists a point $x \in (a,b)$ such that $f(x) = c$.

I understand I have to show that the set $S_{n} = \{x \in [0,1] : f_{n} (x) = x\}$ grows in size for increasing $n$. I am unsure how to go about this problem using the intermediate value theorem.

ferris
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  • If the coefficient were 4, you could show that $f_n$ oscillates between 0 and 1 $2^n$ times. – Eric May 15 '21 at 20:45
  • @Eric If the coefficient were $4$, I know that $f_{2}$ maps $[0,\frac{1}{2}]$ to $[0,1]$ and $[\frac{1}{2},1]$ to $[0,1]$; then the diagonal line $f(x)=x$ crosses $f_{2}$ $2^{2}$ times. Similarly, $f(x) = x$ crosses $f_{3}$ $2^{3}$ times. Graphically, this makes sense, but I'm not sure how to show this rigorously, or how to translate this into the proof of this problem. – ferris May 15 '21 at 20:50
  • Like this? – rtybase May 15 '21 at 21:37

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Consider the action of $f(f(x))$ on $[0.5,.975]$. One application of $f$ maps contains $[.1,.975]$ in the image. Upon the second application of $f$, the interval below $0.5$ gets has image containing $[.4,.975]$ while the portion above $.5$ Has image containing. $[.1,.975]$. Both of these contain $[.5,.975]$, so applying $f$ twice effectively maps that interval to two of those intervals.

Repeating this inductively $n$ times allows us to to say that $f_{2n}$ maps $2^n$ disjoint open intervals within $(.5,.975)$ to $(.5,.975)$. The image of $f$ contains that interval, so $f_{2n+1}$ also has $2^n$ such intervals. Within the closure of each interval, there’s some points that maps to .5 and 0.975. This, for all $m$, $f_m(x)-x$ has $2^{\lfloor m/2\rfloor}$ such intervals within each of whom maps to a positive value at the $.5$ number and a negative value at the $.975$ number. By the intermediate value theorem, each such pair have a zero between them and each zero is a fixed point, so the number of fixed points as $n$ goes to infinity is unbounded.

Eric
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  • Could you elaborate on, in the second paragraph, why $0.5$ is mapped to a positive value, while $0.975$ is mapped to a negative value? – ferris May 16 '21 at 06:03
  • The idea is that the interval, where x is at least than 0.5 maps to .5, so the difference is nonnegative. You technically need to be careful around boundary points since those could overlap between various intervals reducing your number of zeros by a factor of 2 - if the first interval’s boundary is 0.5, it could be 0, but there’s still no overlap with other intervals, so it doesn’t really matter and any boundary points not at 0.5,.975 are strictly nonzero, so the roots are inside those intervals. – Eric May 16 '21 at 15:51