Attracting points in piecewise linear functions

Question

If $f:[a,b]\to[a,c]$ is a strictly increasing piecewise linear map with $c<b$, then does it necessarily follow that $f^n(b)=f(f(f(...(b)...)))\to a$?

Answer 1

No. Consider $f$ on $[-1, 1]$ defined by $f(x) = \begin{cases} x, & \text{if $-1 \le x\le 0$ } \\ x/2, & \text{if $0 \lt x \le 1$} \\ \end{cases}$

Then $f^n(b)=f^n(1) \rightarrow 0 $