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Let $f: [0,1]\rightarrow \mathbb{R}$:

  • $f(x)=3x \, \, $ if $0\le x\le \frac{1}{2}$;
  • $f(x)=3-3x \, \, $ if $\frac{1}{2}<x\le 1$.

Let a sequence $k_{n+1}=f(k_n)$. Find all possible value of $k_0$ that respect the condition $k_i \in [0,1] \, \, \, \forall i \in \mathbb{N}$.

I've tried to solve the problem drawing the plot and erasing the intervals of initial value that don't respect the condition, but i've noted that i can't find a solution because it behaves like a fractal. All the solution that I've found are $0$ and $3^n \, \, \, n\in {0,-1,-2,...}$, but they aren't unique because for example $0.3$ is a solution as well, and i think that there are more solutions.

There's an hint in the text: work with number in base 3; let $K$ the set of solution, you should find a bijection between $K$ and $\mathbb{R}$. I can't see how to use this suggestion.

Martin Sleziak
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L.A.
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1 Answers1

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(Note: the following is more of a roadmap, and details are ommitted)

We denote by $ f^{-n}([0,1]) $ the $n$-th iterated pre-image of $[0,1]$. the solution set can be alternatively described as $$ \tag{1} K= \bigcap_{n\in\mathbb N} f^{-n}([0,1]) $$ It is also obvious, that $f^{-(n+1)}([0,1])\subset f^{-n}([0,1])$.

Now we need to get an idea what $K$ looks like. Consider $f^{-1}([0,1])$. It is easy to prove that $$ f^{-1}([0,1])= [0,\frac13]\cup[\frac23,1]. $$ As a next step, one can calculate $$ f^{-2}([0,1])=f^{-1}(f^{-1}([0,1]))= [0,\frac19]\cup[\frac29,\frac13]\cup[\frac23,\frac79]\cup[\frac89,1]. $$

One can see, that we always delete the (open) center third of an interval, when we procede from $f^{-n}([0,1])$ to $f^{-(n+1)}([0,1])$. This is the same process as in Cantor set (https://en.wikipedia.org/wiki/Cantor_set).

It follows that $K$ is Cantor set.

An alternative approach is to consider the digital expansion of $x\in[0,1]$ in base 3 and observe how the digits change when applying $f$.

supinf
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