MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider two numbers $a,b\in R/Z$ and some integer $p\geq 1$. Let $T:L^p(R/Z)\rightarrow L^p(R/Z)$ be the operator given by $$T(f)(x)=1/2(f(x+a)+f(x+b))$$ For which values of $a,b$ do we have almost everywhere convergence of the sequence $T^nf$ for all $f \in L^p$ ?

If $a-b$ is rational, it's not difficult to show that a.e. convergence fails. If $a=-b$ and $p>1$, then a.e. convergence follows from Stein (1961), Rota (1962). Note that if $a-b$ is not rational, then $T$ is ergodic and the mean $1/n\ \Sigma_k^n\ T^kf$ converges almost everywhere to a constant. So maybe this is the right condition (together with p>1 ?).

share|cite|improve this question
Do you have convergence in $L^2$? If yes, how does one show it? – Helge Jun 29 '10 at 22:40

Here's another way of phrasing your question: $T(x) = x+a$ and $S(x) = x+b$ generate an action of a free semigroup on ${\mathbb R}/{\mathbb Z}.$ For what values of $a,b$ does a point-wise ergodic theorem hold where averages are taken over spheres in the word metric?

There are a many ergodic theorems for free group and semi-group actions. For instance

I don't know if this answers your question exactly. But I have faith that a slightly more thorough literature search along these lines will get you an answer.

share|cite|improve this answer
The link you provide asks for a password. Also T and S commute, so they do not really generate a free semigroup. But maybe I misunderstood your answer ? – coudy May 18 '10 at 9:29

Ok, I rethought my old comment. I believe it is better with $Af (x) = f(x+a)$ and $B f(x) = f(x+b)$ to think about $$ T^n = \frac{1}{2^n} (A + B)^n = \frac{1}{2^{n}} \sum_{k=0}^{n} \binom{n}{k} A^{k} B^{n-k} = \frac{1}{2^{n}} B^{n} \sum_{k=0}^{n} \binom{n}{k} C^{k}, $$ where $C = AB^{-1}$ so that $Cf(x) = f(x + a - b)$. It think that one should be able to show that this converges relatively easily ... (one somehow needs to deal with the weights).

Old Post

Let me rephrase the answer of Fabrizio Polo first:

Consider all words $w$ in A, B of length $n$. Call this set $\mathcal{W}_n$. Now define $Af (x) = f(x+a)$ and $B f(x) = f(x+b)$. Then $T^n$ from the original post is equal to $$ \frac{1}{|\mathcal{W}_n|} \sum w, $$ where the sum is taken over all elements of $\mathcal{W}_n$. I am somehow unable to make that display properly. Here $w$ stands for the appropriate product of operators. E.g. for $n - 2$, we have $\mathcal{W}_n = \{AA, AB, BA, BB\}$ so that the expression above becomes $$ \frac{1}{4} (AA + AB + BA + BB), $$ which is the $T^2$ from the original post.

Now if $a - b$ is irrational, I believe that $(\mathbb Z_+) \ast (\mathbb Z_+)$ action defined above is ergodic, so one should have almost sure convergence. However, I am not sure if this holds, since the group $(\mathbb Z_+)\ast(\mathbb Z_+)$ is not ameanable. So you will probably have to look into ergodic theorems for non ameanable actions to answer this question.

Another hope could be to somehow resum the expression for $T^n$ and be able to use more classical ergodic theorems ...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.