It is well known that many strongly chaotic dynamical systems have the property that periodic measures are (weak-star) dense in the space of all invariant probability measures. Is there some knowledge on how accurate on can approximate a given good measure by a periodic one in terms of length of the periodic orbit supporting it?

I mean, is there a known statement of the following kind: Let $M$ be a compact manifold, $f:M\to M$ is continuous, periodic measures are dense in the space of invariant probability measures. Let $f$ has a physical measure $\mu$ (i.e. for Lebesgue-almost every point $x$ one has $\frac{1}{n}\sum_{i=0}^{n-1} \delta_{f^i(x)} \to \mu$ ) that is not a finite sum of $\delta$-measures. Then under some additional condition on $f$ (for example, specification property) there exists an accuracy function $\gamma:\mathbb{N} \to \mathbb{R}_{\geq 0}$, $\gamma(n)\to 0, n\to \infty$ such that for every natural $N$ there exists a natural $n>N$ and an $n$-periodic orbit $p$ such that $$ \rho\left( \frac{1}{n}\sum_{i=0}^{n-1} \delta_{f^i(p)} , \mu \right) < \gamma(n) , $$ where $\rho$ is some metrisation of the weak-* topology.

If there is no general statement like this, are there examples (say, piecewise-expanding maps of the interval) where something like this is known to hold?


1 Answer 1


For a result of a somewhat similar type, check the paper "Rate of approximation of minimizing measures" by Bressaud and Quas, 2007, especially Theorem 4. In their case the dynamics is a subshift of finite type. They consider an arbitrary invariant measure, and look for periodic orbits in an $\varepsilon$-neighborhood of the support of the measure. True, it's not the type of approximation you want (weak-star), but it may be something to start with. They prove that $\varepsilon$ can be took superpolynomially small as a function of the period of the orbit, i.e., for any $K>0$ and $n_0 \ge 1$ it's possible to find a periodic orbit of period $n \ge n_0$ in the $n^{-K}$-neighborhood of $\mathrm{supp}(\mu)$. They also show that these bounds are sharp, so for example exponential approximation is not always possible.


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.