Take a Gauss map $G: [0,1] \longrightarrow [0,1]$ which is

$$G(x) = \frac{1}{x} \mod 1, 0<x<1$$ and $0$ if $x=0$. Let $\mu$ be the Gauss measure. For $x \in [0,1]$ let $[a_{1}(x), a_{2}(x),...]$ denote the continued fraction expansion of $x$. I would like to show that for any $\gamma>0$ we have for $\mu$-almost all $x \in [0,1]$ $$\lim_{n \longrightarrow \infty} \frac{a_{n}(x)}{n^{1+\gamma}} = 0.$$

Obviously, an application of Birkhoff's ergodic theorem is needed. I think we need to use the fact that for an ergodic measure preserving transformation $T$, with a measure $\mu$ and a function $f \in L^{1}$ we have $ \lim _{n \longrightarrow \infty} \frac{f(T^{n}(x)}{n} = 0$ for almost all $x \in \mu$. (This can be proved by considering partial sums of the form $\sum_{k=0}^{n-1} f(T^{k}(x)$).

But I don't know how to feel in the missing details - what function $f$ to choose, that this holds for all $\gamma >0$, etc. It looks like since we know $a_{n+1} = [\frac{1}{G^{n}(x)}]$ (where the square brackets denotes the integer part), we have to choose $f = \frac{[\frac{1}{x}]}{x^{\gamma}}$.

Is this right? I would appreciate any help and advice on how to make this rigorous.


1 Answer 1


Let $a(x) := \lfloor 1/x \rfloor$. As you noticed, we can't use Birkhoff theorem with the function $a$, as it is not integrable. We'll have to be more cunning. Since $a$ has a singularity in (essentially) $1/x$, note that $a^\alpha$ is integrable for all $\alpha \in [0,1)$. What does this gives us ?

First, note that for all $\alpha \in [0,1)$, the fonction $x \mapsto x^\alpha$ is concave and is $0$ at $0$. Hence, for all nonnegative $(a_0, \cdots, a_{n-1})$,

$$\sum_{k=0}^{n-1} (a_k^\alpha) \geq \left( \sum_{k=0}^{n-1} a_k \right)^\alpha,$$

or, if we put $1+\gamma/2 := 1/\alpha$, for all $\gamma >0$ :

$$\left( \frac{1}{n} \sum_{k=0}^{n-1} a_k^{\frac{1}{1+\gamma/2}} \right)^{1+\gamma/2} \geq \frac{1}{n^{1+\gamma/2}} \sum_{k=0}^{n-1} a_k \geq 0.$$

Let's apply this to $a_k := a \circ G^k$. We get:

$$\left( \frac{1}{n} \sum_{k=0}^{n-1} a^{\frac{1}{1+\gamma/2}} \circ G^k \right)^{1+\gamma/2} \geq \frac{1}{n^{1+\gamma/2}} \sum_{k=0}^{n-1} a \circ G^k \geq 0.$$

Thus, by Birkhoff's ergodic theorem, the sequence $\frac{1}{n^{1+\gamma/2}} \sum_{k=0}^{n-1} a \circ G^k$ is almost surely bounded. If we divide by $n^{\gamma/2}$, we get that, almost surely,

$$\lim_{n \to + \infty} \frac{1}{n^{1+\gamma}} \sum_{k=0}^{n-1} a \circ G^k = 0,$$

which is actually a stronger result than what you're looking for.

  • $\begingroup$ Could you please explain why $a(x)$ is not integrable? $\endgroup$
    – A.P.
    Oct 14, 2014 at 16:49
  • 1
    $\begingroup$ @A.P. : One the one hand, $a(x) = \lfloor 1/x \rfloor$. One the other hand, the Gauss measure $\mu$ is absolutely continuous, with density $1/(1+x)$ (up to a $\ln (2)$ factor), so bounded above and below. So we get something like, respectively, a singularity in $1/x$ and the Lebesgue measure. Hence, $a$ is not $\mu$-integrable, although $a^\alpha$ is $\mu$-integrable for all $\alpha \in [0,1)$. $\endgroup$
    – D. Thomine
    Oct 15, 2014 at 11:12

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