Bounding $\liminf_{n} n |f^n(x)-x|$

Question

I solved an exercise in which the first part asks to prove that for any measure preserving measurable transformation $f:[0,1]\rightarrow [0,1]$ we have $$\liminf_{n} n |f^n(x)-x| \leq 1, \ \mbox{a.e.}$$

I can't prove the second part of the exercise: Let $\omega=(\sqrt{5}-1)/2$ and let $f:[0,1]\rightarrow[0,1]$ defined as $f(x)= (x+\omega) \pmod{1}$. Use this transformation to prove that there is no $c<\frac{1}{\sqrt{5}}$ such that $$\liminf_{n} n |f^n(x)-x| \leq c$$

Thank you guys in advance! (I'm sorry about the mistakes!)

Answer 1

As long we are dealing with $\omega \pmod 1$ it doen't make any difference to deal with $\omega+1 = \frac{\sqrt{5}+1}{2}= \phi$. Therefore we can apply Hurwitz's theorem that states: for every irrational number $\zeta$ there are infinitely many rationals $m/n$ such that $$\left| \zeta - \frac{m}{n} \right| \leq \frac{1}{\sqrt{5} n^2}$$ Moreover $\sqrt{5}$ is the best constant you can get: if you replace it with an $A>\sqrt{5}$ and take $\zeta=\phi$ there are only a finite number os rational numbers suck that the propertry above holds with $A$ instead of $\sqrt{5}$.

Here you can download a book from Carlos Gustavo Tamn de A. Moreira (Gugu), a researcher at IMPA - Rio de Janeiro, Brazil, where at page 47 you can see this result. (proof starts at page 60)

The original paper is: Hurwitz, A. (1891). "Ueber die angenäherte Darstellung der Irrationalzahlen durch rationale Brüche (On the approximation of irrational numbers by rational numbers)" (in German). Mathematische Annalen 39 (2): 279–284. doi:10.1007/BF01206656. JFM 23.0222.02