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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?

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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 $V(f(x)) < a$.

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  • $\begingroup$ The way I've fornulated the statement I'm trying to prove is that it holds for $x \notin Fix (f)$, but it appears that your proof works for all $x \in X$, correct? It's only important that $a \notin V(Fix(f))$ $\endgroup$
    – blue
    Commented Apr 4, 2021 at 21:30
  • $\begingroup$ Yes, that's right. For $x \in Fix(f)$, either $x \notin A$ and so $V(x) < a$, or $x \in A$ so $V(x) \geq a + \delta$ by construction. $\endgroup$
    – Martin M. W.
    Commented Apr 4, 2021 at 23:50

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