8
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you @A Blumenthal. Where in your argument do you use that f is continuous (and not merely bounded measurable)? $\endgroup$ – Hans Oct 14 '13 at 14:12
  • $\begingroup$ @Filippo I didn't really use it for the change of variables formula, as that formula holds for indicator functions, and then finite sums of indicator functions, and by the Bounded Convergence Theorem, for all bounded measurable functions. You might consider what the analogous statement of (*) would be for measurable functions. $\endgroup$ – A Blumenthal Oct 14 '13 at 17:57
  • $\begingroup$ @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). $\endgroup$ – Hans Oct 15 '13 at 22:14
  • $\begingroup$ @A Blumenthal. Do you also have some hint on how to show that $\pi$ is the unique invariant measure? $\endgroup$ – Hans Oct 15 '13 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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