a.e. convergence of the powers of an operator built from rotations

Question

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 ?).

Answer 1

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).

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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 ...