I've tried asking this question on Mathematics site, but I only got an upvote and no answer. I've searched online, tried to find something about this topic, but I haven't found much (and the things I have found I don't understand).


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.


Prove that $(\forall a\in \mathbb{R} \setminus V(Fix(f)))(\exists \delta >0)(\forall x\notin Fix(f))\\(V(x)<a+\delta \implies V(f(x))<a)$.


My attempt:
Let $\delta\_x=V(x)-V(f(x))$. (We could also take $\delta\_x$ such that $V(f(x))+\delta\_x<V(x)$, but I don't think we get any value from that, we would just overcomplicate things.)


We easily get: $V(x)<a+\delta\_x \implies V(f(x))=V(x)-\delta\_x<a+\delta\_x-\delta\_x=a$.


However, we need the same $\delta$ for all $x\notin Fix(f)$. Since $\inf\_{x\notin Fix(f)}\delta\_x$ can be $0$, I think we should use compactness here. We know that $X, f(X)=X, V(X)=V(f(X))$ are compact spaces.


Since $V(Fix(f))$ is finite, we know that $V(X)$ is indeed covered by intervals $\{(V(x)-\delta\_x,V(x)+\delta\_x)\}\_{x\notin Fix(f)}$. Since $V(x)$ is compact, we have a finite subcover $\{(V(x\_i)-\delta\_{x\_i},V(x\_i)+\delta\_{x\_i})\}\_{i=1,\ldots ,n}$.
Note that $I\_i:= (V(x\_i)-\delta\_{x\_i},V(x\_i)+\delta\_{x\_i})=(V(f(x\_i)), V(x\_i)+\delta\_{x\_i})$.


We can take $\delta=\min \delta\_{x\_i}$.We can assume the finite subcover we got is of minimal cardinality and that $V(x\_1)\leq \ldots \leq V(x\_n)$.


If, for our chosen $\delta$, $V(x)<a+\delta$, then $V(f(x))=V(x)-\delta\_x<a+\delta-\delta\_x$.
We need to get $\delta \leq \delta\_x$, which obviously holds for $\delta\_x=\delta\_{x\_i}$, but if $\delta\_x \neq \delta\_{x\_i}$, we need something more.


If $\delta\_x \neq \delta\_{x\_i}$, then obviously $x\neq x\_i$. $V(x)$ has to be in $I\_j$, for some $j$.


$V(f(x))$ can't be in $I\_i$ for $i>j$.
But discussion on where $V(f(x))$ is doesn't really bring me anywhere.


I thought that maybe I should start with the question: where is $a$?


If $a\notin V(X)$, then we're done. Interesting cases are the ones where $a\in V(X) \subset \bigcup\_{i=1}^n I\_i$. However, I'm not getting anywhere this way either.


Should I even continue with this idea or is this completely wrong? Am I overcomplicating it?



Gradient-like dynamical systems
Let $A = V^{-1}([a, \infty))$. Both $A$ and $f^{-1}(A)$ are compact.


Because $V$ is a strict Lyapunov function and $a \not\in V(Fix(f))$, it follows $\min\_{f^{-1}(A)} V > a$.


So you can set $\delta = \min\_{f^{-1}(A)} V - a$, and you have $V(x) < a + \delta$ implies $V(f(x)) < a$.


