Iteration of a function on $[a,b]$ 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?
 A: 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 $x \in [0,1).$
A: 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. 
A: 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.
A: 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). 
