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

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

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How did you proof the first part? If you take $f=0$ and $x=\frac{1}{2}$ the bound is not satisfied. – Lucien Feb 12 '13 at 16:34
I'm sorry! I forgot to say that it holds a.e. – badaui90 Feb 12 '13 at 16:50
If you take $f=0$, then still the bound is satisfied only for $x=0$, so not a.e. – Lucien Feb 12 '13 at 17:00
My bad again, I was think about the problem and forgot to say that $f$ also has to preserve Lebesgue $\lambda$ measure, ie,$\lambda(f^{-1}(B))=\lambda(B)$ for all measurable set $B$. – badaui90 Feb 12 '13 at 17:13

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 $\sqrt{5}$.