# An ergodic theorem on the circle

Let $S^1$ be a circle (i.e. a closed $1$-dim. manifold) and let $F$ be a non-vanishing smooth vector field on $S^1$. Denote by $(t,x) \mapsto \Phi_t^x$ the flow generated by $F$.

I want to show that there exists an absolutely continuous probability measure $\pi$ with strictly positive density $\rho$ such that for every bounded measurable function $f$ and every $x\in S^1$

$\lim_{t\rightarrow \infty} \frac{1}{t}\int_0^t f(\Phi_s^x) ds \ = \ \int_{S^1} f(x) \pi (dx) \ \ \ \$ (*)

From (*) it follows that $\pi$ is invariant, i.e. $\int_{S^1} f(\Phi_t^x) \pi(dx) \ = \ \int_{S^1} f(x) \pi(dx)$. I would also like to show that $\pi$ is the unique invariant measure.

What I did until now:

Denote by $\tau$ the time needed to return in $x$ when starting from $x$ and following the flow $\Phi$ and denote by $n_t$ the number of times $t\mapsto \Phi_t^x$ returned in $x$ up to time $t$. Observe that $0<\tau<\text{const}$ and that both $\tau$ and $n_t$ are independent of $x$.

Then $\frac{1}{t}\int_0^t f(\Phi_s^x) ds \ = \ \frac{n_t\tau}{t} \ \frac{1}{n_t } \sum_{k=1}^{n_t} \ \frac{1}{\tau}\int_0^{\tau} f(\Phi_s^x) ds \ + \ \frac{1}{t} \int_{n_t\tau}^t f(\Phi_s^x) ds$

So using the boundedness of $f$ and of $t-n_t\tau$ and the fact that $\frac{n_t\tau}{t}\rightarrow 1$ I get for every $x\in S^1$

$\frac{1}{t}\int_0^t f(\Phi_s^x) ds \rightarrow \ \frac{1}{\tau}\int_0^{\tau} f(\Phi_s^x) ds$

with the right hand side in fact independent of $x$. Now I guess I should use some Riesz representation theorem for functionals to show existence of $\pi$ such that for every bounded measurable function

$\frac{1}{\tau}\int_0^{\tau} f(\Phi_s^x) ds \ = \ \int_{S^1} f(x) \pi (dx)$

Can somebody hint to a precise reference or give some alternative (maybe more selfcontained) argument? I have no clue for the moment on how to show the existence of $\rho$ (or to find a counterexample if it is not true).

Let me try to complete your argument by showing that there is an a.c. measure $\pi$ for which $$\frac{1}{\tau} \int_0^{\tau} f(\Phi_s^x) ds = \int_{S^1} f(x) \pi(dx)$$ where $f \in C(S^1)$ is any continuous function (and the LHS is for a fixed $x$, say $x = 0$ writing $S^1 = \mathbb{R} / \mathbb{Z} \cong [0,1)$).
First note that $\Phi_s^0 : [0,\tau) \to S^1$ is a diffeomorphism, and that because the vector field for this flow is nonvanishing, it has a nonvanishing time derivative $\frac{d}{ds} \Phi_s^0$. So, $$\frac{1}{\tau} \int_0^{\tau} f(\Phi_s^0) ds = \int_{S^1} f(x) h'(x) dx$$ by the change of variables formula, where $h :S^1 \to [0,\tau)$ is the inverse of $s \mapsto \Phi_s^0$. Check that $h'$ is nonvanishing.
• @A Blumenthal. It seems to me that () remains true as it is for measurable bounded functions. (While in the case of F vanishing at exactly one point $\bar x$ the measure $\pi$ becomes the delta measure in $\bar x$ and () holds for continuous functions but not in general for measurable functions). – Hans Oct 15 '13 at 22:14
• @A Blumenthal. Do you also have some hint on how to show that $\pi$ is the unique invariant measure? – Hans Oct 15 '13 at 22:15