Does there exist a homeomorphism $f$ on $[0,1]$ such that $f(0)=0$, $f(1)=1$ and $f(x)

Question

A dynamical systems problem I am trying to prove states that if $f:[0,1]\rightarrow [0,1]$ is a homeomorphism such that $f(0)=0$, $f(1)=1$ and $f(x)<x$, then the orbits of points in $[0,1)$ are Lyapunov stable. However, I am doubtful that there can even be such a homeomorphism. But I haven't been able to come up with an example.

Answer 1

If $0<x$ and $x<1$ then $x^2<x$; also $0^2=0$ and $1^2=1,$

so $f(x)=x^2$ satisfies $f(0)=0, f(1)=1$, and $f(x)<x$ if $x\in (0,1)$.

$f$ is a homeomorphism on $[0,1],$ because it is bijective, with inverse $f^{-1}(x)=\sqrt x $,

and $f$ and $f^{-1}$ are continuous; see here and here.