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Here is the question; it may seem very simple, but it is difficult (at least for me).

Let $f(x)$ be a continuous function on $R$ that is strictly increasing, and suppose $g(x)=f(x)-x$ is a periodic function with period 1.

Prove that for all $x\in R$, $\lim_{n\to \infty}\frac{f^n(x)}{n}$ exists.

In an equivalent formulation, the dynamic system $(X,T)$ is quasi-regular, where $X=[0,1],T: x\mapsto x+\{g(x)\}$. I.e. the Birkhoff average $\frac{1}{N}\sum_{n\leq N}T^n(x)$ exists for all $x\in X$.

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    $\begingroup$ What makes you think the statement is true? $\endgroup$
    – Anthony Quas
    Commented Nov 22, 2017 at 16:03
  • $\begingroup$ @AnthonyQuas, a friend ask me weather it is true... It seems I can construct counter-example in the continuous case but there still be some obstacle. This is come from Poincare's original definition of wind number. And which seems is not the satisfied one due to it is different with the definition on the book dynamic system of Pollinate. $\endgroup$
    – Hu xiyu
    Commented Nov 22, 2017 at 16:52
  • $\begingroup$ And it seem to be true for $f$ is $C^1$ at least or may be we only need $f$ is lipchitz? I am not very sure. $\endgroup$
    – Hu xiyu
    Commented Nov 22, 2017 at 16:53
  • $\begingroup$ see Poincare's original definition of rotation number in en.wikipedia.org/wiki/Rotation_number. $\endgroup$
    – Hu xiyu
    Commented Nov 22, 2017 at 17:17
  • $\begingroup$ I introduce why my intuition is this result could be right at least for $f$ with good regularity. It is just come from the observation that if the flow of points induce by interaction is uniformly distribute under some invariant measure on $X$, then the time average need to coincide with the space average which lead to the result of the existence of limit, and given a representation of it. $\endgroup$
    – Hu xiyu
    Commented Nov 22, 2017 at 17:41

1 Answer 1

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EDITED: below I give a counter example in the case where $f$ is not monotonic. The question you’re asking is well known to be true in the monotonic case

There's a counterexample (even if $f$ is highly regular). I'll give a piecewise linear counterexample: let $$ f(x)=\begin{cases}4x&\text{if $x\in [0,\frac 12)$;}\\ 3-2x&\text{if $x\in [\frac 12,1)$}; \end{cases} $$ and extend to the real line by periodicity.

Let $g(x)$ be the corresponding map from the circle to itself. Now let $J_0=[0,\frac 14)$ and $J_1=[\frac 14,\frac 12)$. Both of these intervals map bijectively to the whole circle. Now you can define $J_{i_0\ldots i_{n-1}}=J_{i_0}\cap g^{-1}J_{i_1}\cap \ldots\cap g^{-(n-1)}J_{i_{n-1}}$. Finally for any sequence of 0's and 1's, you can define $h(z)=\bigcap_{n=0}^\infty J_{z_0\ldots z_n}$. For $x\in\mathbb R$, let $\rho^+(x)=\limsup f^n(x)/n$ and $\rho^-(x)=\liminf f^n(x)/n$. It's not hard to check that $\rho^+(h(z))=\limsup (z_0+\ldots+z_{n-1})/n$ and $\rho^-(h(z))=\liminf (z_0+\ldots+z_{n-1})/n$. In particular, it's easy to construct points with $\rho^-(x)=a$ and $\rho^+(x)=b$ for any $0\le a\le b\le 1$.

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  • $\begingroup$ It seems in your construction if and only if $0=i_0=i_1=...=i_{n-1}=...$ could make $\cap_{n=0}^{\infty}J_{z_0z_1...z_{n}}\neq \emptyset$. But in this case there is no contradiction. $\endgroup$
    – Hu xiyu
    Commented Nov 22, 2017 at 17:32
  • $\begingroup$ I am sorry if I misunderstand your construction. $\endgroup$
    – Hu xiyu
    Commented Nov 22, 2017 at 17:33
  • $\begingroup$ $J_{0100}$ consists of points in $J_0$ that map into $J_1$ at the first time step; and $J_0$ at the second and third time steps. These are always non-empty since the intervals $J_i$ are mapped over the entire circle, so that $g^4(J_{0010})$ maps over the whole circle. This means you can find a sub-interval of $J_{0010}$ mapped over either $J_i$ by $g^4$. The intersection is of a nested sequence of intervals. You have to be slightly careful with the right endpoint (or modify the example to allow you to use closed intervals). $\endgroup$
    – Anthony Quas
    Commented Nov 22, 2017 at 17:43
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    $\begingroup$ The trajectory of $0$ under iteration of $f$, is just $0$. The trajectory of $1/40$ is $\{1,4,16,64,112,136,184...\}/40$. The picture becomes clearer if we take first differences: $\{3,12,48,48,24,48,48,24..\}/40$ which is periodic and with long term mean $1$. Nice counterexample! $\endgroup$
    – Yaakov Baruch
    Commented Nov 22, 2017 at 18:10
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    $\begingroup$ Sorry. If you want a result for strictly increasing functions, the answer is well known. See Devaney’s book on chaotic dynamical systems for example. As you say, this is just the (Poincaré) rotation number which exists for all $x$ (and is independent of $x$). $\endgroup$
    – Anthony Quas
    Commented Nov 22, 2017 at 18:17

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