Iteration of a function on $[a,b]$

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Currently reading the paper "From Intermediate Value to Chaos" (Huang, 1992) I stumbled over the following statement

A function can be iterated on $[a,b]$ but may not be iterated on any subinterval of $[a,b]$.

I was wondering whether someone could give an example for such a function?

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If we have a function from the unit circle $B_2:=\{(x,y) \in \mathbb{R\times R}|x^2+y^2=1\}$ to $B_2$, that rotates a point by a constant irrational angle $\phi$, then this is a function that can be iterated on $B_2$ but not on a subset of $B_2.$

We want to transform this to $[0,1)$ and further to $[0,1].$ Let $\alpha$ be an irrational number. $$f:[0,1] \to [0,1]$$ $$f(x)=\begin{cases} x+\alpha \mod 1,&x\ne 1 \\ 0,&x=1 \end{cases} $$ Note that $x,f(x),f(f(x),f^3(x)\ldots$ is dense in $[0,1]$, if $x \in [0,1).$

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Assuming that $f$ is not required to be continuous we can construct an example as follows. We can arrange the set of rationals in $[0,1]$ is a sequence $\{r_1,r_2,...\}$ such that every (non-degenerate) intervals contains an $r_n$ with $n$ even and an $r_m$ with $m$ odd. Define $f(x)=x$ for every irrational number $x$ and $f(r_n)=1$ for $n$ even, $f(r_n)=0$ for $n$ odd.

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Try the function $f(x)=2x^3-3x^2+1$ on $[0,1]$. Any subinterval contained in either $\left[0,\frac12\right]$ or $\left[\frac12,1\right]$ gets moved to the other side of $\frac12$, and any proper subinterval $[a,b]$ with $a<\frac12<b$ becomes wider.

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Any expanding map of the circle would work because except for the fixed points (if there are any), each subinterval will be mapped to a larger interval by the map. For example, $f:[0,1]\to[0,1]$ given by $$f(x)= 2x \mod 1$$ works.

If one is particular about fixed points, redefining the function to be $$f(x)=\begin{cases} 2x \mod 1,& x\ne 0 \\ 1/2,&x=0 \end{cases} $$ would fix the problem. Note that any such example $f$ should necessarily be discontinuous. This is because any continuous function $f:[a,b]\to [a,b]$ has a fixed point (intermediate value theorem).