Let $X$ be a compact metric space and $f:X\to X $ homeomorphism. Let $V:X\to \mathbb{R}$ be a Lyapunov function for $(X,f)$ (continuous function such that $(\forall x\notin Fix(f))\ \ V(f(x))<V(x))$. Let $V(Fix(f))$ be finite.
I'm trying to prove that certain $\epsilon$‐pseudoorbit doesn't exist. ($x_0,x_1, \ldots , x_n,x_{n+1}$ is an $\epsilon$‐pseudoorbit iff $(\forall i\in \{0, \ldots, n\}) d(f(x_i), x_{i+1})<\epsilon$). I'll explain how the $\epsilon$ and $x$ are chosen.
Let $z\notin Fix f$. Set $V(Fix f)$ is finite, so it partitions $\mathbb{R}$ into finite number of disjoint open intervals $\{I_i\}$. We can find and $N \in \mathbb{N}$ and $j$ such that $(\forall k \geq N) V(f^k(z))\in I_j$.
We also know that $(\forall c\in \mathbb{R} \setminus V(Fix(f)))(\exists \delta_c >0)(\forall x\in X)\ (V(x)<c+\delta_c \implies V(f(x))<c)$.
Let $z\notin Fix f$ and let $N \in \mathbb{N}$ be the $N$ chosen in point 1. Let $x=f^N(z), a=V(f(x)), b=V(x), \delta=min\{\delta_a, \delta_b, \frac{b-a}{2} \}$. Let $\epsilon >0$ such that $d(x_1,x_2)<\epsilon \implies |V(x_1)-V(x_2)|<\delta$. We want to prove that there's no $\epsilon$‐pseudoorbit from $x$ to $x$.
First, we assume that $x_0=x, x_1, \ldots , x_n, x_{n+1}=x$ is an $\epsilon$‐pseudoorbit. That means that $(\forall i\in \{0, \ldots, n\}) d(f(x_i), x_{i+1})<\epsilon$. Because of the way we've chosen $\epsilon$, we have that $(\forall i\in \{0, \ldots, n\}) |V(f(x_i))-V(x_{i+1})|<\delta$.
I thought I should look into the case $i=0$ first. We have:
- $V(x_1)<V(f(x))+\delta$
- $V(f(x))<V(x_1)+\delta$
From 1. we have that $V(f(x_1))<V(f(x))$ and $V(x_1)<a+\frac{b-a}{2}$.
I think that it would be nice if we could prove that $V(f(x_{i+1}))<V(f(x_i))$, so we could get $V(f(x))<V(f(x))$. We could use $V(x_{i+1})<V(f(x_i))+\delta_{V(f(x_i))}$. From $|V(f(x_i))-V(x_{i+1})|<\delta$ we have $V(x_{i+1})<V(f(x_i))+\delta$, but we need to know that:
a) $V(f(x_i))\notin V(Fix(f))$
b) $\delta \leq \delta_{V(f(x_i))}$
And I don't know how to prove these 2 things. At one point I thought that we can claim $V(x_1)>a$, which would give us that $V(x_1)\notin V(Fix f)$, but I think I've made a mistake and I'm also not sure this would be useful enough.
It would actually be enough to prove $V(f(x_{i+1}))\leq V(f(x_i))$ for $i\geq 1$, because for $i=0$ we have strict inequality, so we would still get $V(f(x))<V(f(x))$.
I also thought that maybe we could prove that $\delta_c$ is nondecreasing/nonincreasing as a function of $c$, but I couldn't get that result.
