Prove that the tent map on the cantor set tends to negative infinity. Consider the map $ h: \mathbb{R} \rightarrow \mathbb{R} $ defined by  $ h(x)=3x , \ if \ \ x \leq \frac{1}{2} \ and \ \ h(x)=3(1-x) , \ if \ x > \frac{1}{2} $ . 
Prove that if $ x \in (1/9, 2/9) \cup (1/3,2/3) \cup (7/9,8/9) $ , then $ h^{n}(x) \rightarrow -\infty \ \ as \ \ n \rightarrow \infty $. 
This map is called the Tent map in cantor set-theory . How to prove this , I don't have any idea . I think method of iteration may help here . But not sure. Is there any help ?
 A: This yields to a fairly straightforward analysis of $h(x)$.  Looking at $h$ we see that $0$ and $1$ are going to be of interest because $h(x) < 0$ if $x<0$ and $h(x)<0$ if $x>1$.
If $x<0$ then $h(x) \rightarrow -\infty$ because $h^n(x) = 3^nx$ and $\lim_{n\rightarrow \infty} 3^nx \rightarrow -\infty \quad \forall x < 0$.  So that gives us an idea of what we're looking for.
Next we should take a look at what $h$ does to its domain.  Since $h$ is linear we can apply it to the end-points of the intervals given and they will yield intervals that form the range of $h$.  I'd suggest you stop reading at this point at try that for yourself :)
So, for the leftmost and rightmost intervals we have:
$$
\begin{eqnarray}
I_1 := \left( \frac{1}{9},\frac{2}{9} \right)\qquad \mbox{so} \qquad & h(I_1) = \left( \frac{1}{3}, \frac{2}{3} \right) \\
I_2 := \left( \frac{7}{9},\frac{8}{9} \right)\qquad \mbox{so} \qquad & h(I_2) = \left( \frac{1}{3}, \frac{2}{3} \right) \\ 
\end{eqnarray}
$$
Well, this is convenient!  So we only need to understand what happens in the middle interval to know the behaviour of $h(x)$ on its whole domain.  For the middle interval we have to note that $1/3 < 1/2 < 2/3$ so we need to split it into two intervals again because $h(x)$ changes its behaviour at $x=1/2$.
So, writing $(1/3,2/3) = (1/3,1/2] \cup (1/2,2/3)$ we get:
$$
\begin{eqnarray}
I_{3a} := \left( \frac{1}{3},\frac{1}{2} \right]\qquad \mbox{so} \qquad & h(I_{3a}) = \left( 1, \frac{3}{2} \right] \\
I_{3b} := \left( \frac{1}{2},\frac{2}{3} \right)\qquad \mbox{so} \qquad & h(I_{3b}) = \left( 1, \frac{3}{2} \right) \\ 
\end{eqnarray}
$$
Again, this is rather convenient!  Since $h(I_{3a}) > 1$ and $h(I_{3b})>1$ we see that $h^2(I_{3a}) <0$ and $h^2(I_{3b})<0$ and, as we noted at the top, this will then result in an infinite descent to $-\infty$.
