Hi. I've been stuck on the following question for some time.

Consider a sequence of functions $\left( f_n \right)$ from an ergodic space $\left( \mathsf{X}, \mathsf{S}, \mu \right)$ to $\left[ 0,1 \right]$ such that $f_{a+b} \leq f_a \mathsf{S}^a \left( f_b \right)$ for all integers $a, b \geq 0$, where, classically, $\mathsf{S} \left( g \right)$ is the map $x \longmapsto g \left( \mathsf{S} \left( x \right) \right)$.

Obviously $f_n \leq f_{n+1}$ so that the sequence $\left( f_n \right)$ decreases at each point to a function $f$. Under the hypothesis $\int_{\mathsf{X}} f_1 d\mu < 1$ I was able to prove that $f = 0$ almost everywhere. Here's how I dealt with the problem:

For fixed $n$ and integers $k$ and $\alpha$ such that $k \alpha \leq n$, the inequality \begin{equation} f_n \leq f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) \end{equation} holds, so that, taking averages, one gets \begin{equation} f_n \leq \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) \end{equation} Taking integrals, and using the dominated convergence theorem and the Furstenberg-Katznelson theorem on multiple ergodic averages, one gets \begin{equation} \int_{\mathsf{X}} f \leq lim_{n \longrightarrow \infty} \frac{1}{\lfloor n/ \alpha \rfloor} \sum_{k=0}^{\lfloor n/ \alpha \rfloor} \int_{\mathsf{X}} f_1 \mathsf{S}^k \left( f_1 \right) \ldots \mathsf{S}^{k\alpha} \left( f_1 \right) = \left( \int_{\mathsf{X}} f_1 \right)^\alpha \end{equation} For all integers $\alpha$, which allows me to conclude.

Now the question is: does the series $\sum f_n$ converge? That seems plausible considering the seemingly exponential decreasing of $f_n$ to $f$, but trying to use the same techniques leads to multiple ergodic averages for a varying number of terms - making the integer $\alpha$ dependent on $n$ that is; is there any way to deal with those? All suggestions are welcome.

  • $\begingroup$ I think you may need $(X, S, \mu)$ to be weak mixing for your statement of Furstenberg-Katznelson to be correct. Also, why do you need such powerful machinery? Your deduction seems to follow quite easily from ergodicity plus the fact $f_{n} \leq \Pi_{i = 0}^{n -2} S^i(f_1)$. $\endgroup$ May 5, 2011 at 22:23
  • $\begingroup$ You're perfectly right, weak-mixing is the good working hypothesis. As for more expedient ways to prove that $f=0$ a.e.: For integers $n$ and $p$, $f_{n+p} \leq f_n S^n (f_p)$ so that when $p \rightarrow \infty$ and considering that $f_n \leq f_1$, the inequality $f \leq f_1 S^n (f)$ holds for all $n$. Taking averages and integrals, one finds that $\int f \leq 1/n \sum_{k=0}^n \int f_1 S^k (f) \rightarrow \int f_1 \int f$ so that $\int f \leq \int f_1 \int f$ which proves that $f=0$ almost everywhere. But this does not help for the convergence of $\sum f_n$ unless I am missing something? $\endgroup$ May 5, 2011 at 22:49

1 Answer 1


Unless I miss something, you are overcomplicating the story. In the first place, as it has already been pointed out, you don't really need submultiplicativity as $$ f_n(x) \le f_1(x) f_1(Sx) \dots f_1(S^{n-1} x) = F_n(x) \;. $$ Applying the usual ergodic theorem to the right-hand side one gets that a.e. $$ \log F_n(x)/n \to \int\log f_1(x)\;d\mu(x) < 0 \;, $$ whence the claim.

  • $\begingroup$ Overcomplicating indeed. Thank you. $\endgroup$ May 6, 2011 at 9:38

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