Skip to content

Commit

Permalink
Filled proof of unique ergodicity for irrational rotations
Browse files Browse the repository at this point in the history
  • Loading branch information
George Stamatiou committed Dec 17, 2023
1 parent ad1a61d commit dd7981f
Showing 1 changed file with 6 additions and 2 deletions.
Original file line number Diff line number Diff line change
Expand Up @@ -663,11 +663,15 @@
$$
\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}
$$
An easy check shows that if $P_n=\sum_{k=-n}^na_k\exp{2\pi\ii k x}$ is a trigonometric polynomial, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$. Moreover, if $k\ne 0$:
First we consider the case of trigonometric polynomials, i.e. $P_n(x) =\sum_{k=-n}^na_k\exp{2\pi\ii k x}$, where $a_k = \bar a_{-k}$, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$.
Moreover, if $k\ne 0$:
$$
\int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=\exp{2\pi\ii k\alpha}\int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}\implies \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=0
$$
where the equality is due to the invariance of $\mu$. So, we also have $\int_{\TT^1}P_n(x)\dd{\mu}=a_0$. Now consider the Féjer trigonometric polynomial that converge uniformly to $\varphi$ and use the \mnameref{RFA:domianted}.
where the equality is due to the invariance of $\mu$.
Thus $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$, and the result holds for trigonometric polynomials.
In the general case of a continuous function $\phi \in \mathcal{C}(\TT^1)$, we recall that the set of trigonometric polynomials is dense in $\mathcal{C}(\TT^1)$.
Thus the result for the case of trigonometric polynomials passes to the limit and we have that $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$.
\end{proof}
\begin{proposition}\label{ADS:uniquely_ergodic}
Let $F\in\Homeoplus(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, $F$ is uniquely ergodic.
Expand Down

0 comments on commit dd7981f

Please sign in to comment.