Discontinuous trapping of decreasing iterates ($f^n(x)\to 1/(j+1)\text{ for }x\in (1/(j+1),1/j]$)

A map on $(0,1)$ can satisfy $0<f(x)<x$ and still have no orbit tending to zero. Partition the domain into $I_j=(1/(j+1),1/j]\cap (0,1)$ and define $f(x)=(x+1/(j+1))/2$ on $I_j$. Every interval is invariant under the iteration of a map, and $f^n(x)=1/(j+1)+2^{-n}(x-1/(j+1))$ has a positive limit. The discontinuities at interval boundaries prevent that limit from being a fixed point. In contrast, continuity of a self-map satisfying $f(x)<x$ permits passage to a positive orbit limit and would force an impossible fixed point.