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Does anyone have a counter-example of the following statement :

Let $f : [0;1] \to [0;1]$ a continuous function w.r.t. the usual topology. Let $A_n(x) = \frac{1}{n} \sum_{k=0}^{n-1} f^k(x)$ for $n \ge 1$, $x \in [0;1]$. Then $\forall x \in [0;1]$, $A_n(x)$ converges.

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    $\begingroup$ Represent Cantor space as ternary expansions without 1, and shift the expansion left there. Extend linearly to the gaps. $\endgroup$
    – Ville Salo
    Commented Oct 31 at 15:17
  • $\begingroup$ @VilleSalo Are you saying $f(x) = \{3x\}$ when $x$ has no $1$s in its ternary expansion? $\endgroup$
    – mathworker21
    Commented Oct 31 at 15:29
  • $\begingroup$ @MonsieurBec: No, division by 3 has converging orbits, thus converging Cesàro means. $\endgroup$
    – Ville Salo
    Commented Nov 1 at 0:13
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    $\begingroup$ @mathworker21: yes $\endgroup$
    – Ville Salo
    Commented Nov 1 at 0:14

2 Answers 2

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Let $f(x)$ be the continuous piecewise linear function with endpoints $(0,0)$, $(1/3,1)$, $(2/3,0)$, and $(1,1)$. Define $I_0 = [0,1/3]$, $I_1 = [1/3, 2/3]$, and $I_2 = [2/3,1]$. Note that $f$ is conjugate to the full shift map on $\{0,1,2\}^\mathbb{N}$.

Lemma: Let $A = \{x_0, x_1, x_2, ... x_{n-1}\}$ be a set of points in $I_0 \cup I_2$. If the proportion of points in $A$ that are in $I_0$ is at least $9/10$, then the average value of the elements in $A$ is at most $2/5$.

Proof: Let $p$ be the proportion of elements of $A$ that are in $I_0$. Then \begin{align*} \frac{1}{n} \sum_{i=0}^{n-1} x_i &\leq \frac{1}{3}p + 1(1-p) \\ &\leq 1 - \frac{2}{3}p \\ &\leq 1 - \left(\frac{2}{3}\right) \left(\frac{9}{10}\right) \\ &= \frac{2}{5}. \end{align*}

A similar technique shows that if the proportion of points in $A$ that are in $I_2$ is at least $9/10$, then the average of those elements must be at least $3/5$.

Now, to show the existence of some $x$ whose Cesaro means do not converge, let $x$ be the point whose itinerary is given by $$0\underbrace{(2...2)}_{9} \underbrace{(0...0)}_{90} \underbrace{(2...2)}_{900} \underbrace{(0...0)}_{9000} \underbrace{(2...2)}_{90000} \underbrace{(0...0)}_{900000} ...$$ (The parentheses here are only for clarity.) Then after $10^n$ iterates with $n \geq 2$, then at least $9/10$ iterates of $x$ have been in $I_0$ if $n$ is even, or in $I_2$ if $n$ is odd. Then $A_n(x)$ is greater than $3/5$ and less than $2/5$ infinitely often, and therefore cannot converge.

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    $\begingroup$ I don't understand why f is conjugate to the full shift map on $\{0,1,2\}^{\mathbb{N}}$ because for example it seems to me that the shift of $x = 1,0,0 \ldots$ is 0 and that $x$ is associated to $1/3$ and that $f(1/3) = 1$. $\endgroup$
    – Monsieur Bec
    Commented Nov 1 at 9:39
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    $\begingroup$ What does "conjugate" mean here? The maps are not topologically conjugacy in the usual (?) sense because the spaces are not homeomorphic. $\endgroup$
    – Ville Salo
    Commented Nov 1 at 10:23
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    $\begingroup$ Anyway this is the same as my example from the comment. $\endgroup$
    – Ville Salo
    Commented Nov 1 at 10:26
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Let $C \subset [0,1]$ be any Cantor space. Pick a homeomorphism $\pi : X \to C$ where $X = \{0,1\}^\omega$ has the product topology, and consider $g : C \to C$ corresponding to the left shift map, so $g(\pi(x_0x_1x_2\ldots)) = \pi(x_1x_2x_3\ldots)$. Take any continuous extension $f : [0,1] \to [0,1]$, e.g. piecewise linear in the gaps.

Now let $x = 0^{1!} 1^{2!} 0^{3!} 1^{4!} 0^{5!} \ldots \in X$. Since $n!/\sum_{i < n} i! \rightarrow \infty$, we see that $\pi(x)$ has both $\pi(0^\omega)$ and $\pi(1^\omega)$ as Cesàro limit points under iteration of $g$, thus also under iteration of $f$. (You can pick a faster growing sequence than $n!$ to slightly simplify the proof.)

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  • $\begingroup$ I'm sorry but I don't understand why your function f is continuous. For example, if I take $x = 1, 0, 0, \ldots \in X$ and for $n \ge 1$, I take : $e_n = 0,\underbrace{1,1,\ldots,1}_{\text{n}},0,0 \ldots$ then $\lim_{n \to \infty} \pi(e_n) = \pi(x)$ and $\lim_{n \to \infty} f(e_n) = \lim_{n \to \infty} g(\pi(e_n)) = 1$ but $f(x) = g(\pi(x)) =0$. $\endgroup$
    – Monsieur Bec
    Commented Nov 1 at 9:35
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    $\begingroup$ No, $\pi(e_n)$ does not converge to $\pi(x)$, since clearly $e_n$ has the wrong initial digit and $\pi$ is a homeomorphism. Perhaps you think $X$ is about binary expansions, but it's not. Ternary expansions are one concrete realization. $\endgroup$
    – Ville Salo
    Commented Nov 1 at 10:16
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    $\begingroup$ Maybe I didn't understand X yeah but can you show me a concrete realization ? For me it was with the ternary expansions : $\pi(x_0,x_1,x_2 \ldots) = \sum_{k \ge 0} \frac{2 x_k}{3^k}$ where $\forall k, x_k \in \{0,1\}$. Thus I did have $(\pi(e_n))_{n \ge 0})$ converges to $\pi(x)$. However I certainely didn't understand what you were actually saying so can you show me a concrete realization pls ? $\endgroup$
    – Monsieur Bec
    Commented Nov 1 at 10:39
  • $\begingroup$ My initial comment was a concrete example, and the other answer gives the resulting explicit formula. This one explains the same idea abstractly. You are wrong about that convergence with ternary expansions. $\endgroup$
    – Ville Salo
    Commented Nov 1 at 10:43
  • $\begingroup$ Okay I understand it now sorry. Do you know any kind of characterization of functions that have the property ? $\endgroup$
    – Monsieur Bec
    Commented Nov 1 at 11:00

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