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I am looking for a proof or a reference for the following two facts (which appear proofless in my notes from an ergodic theory course- they might be easy but i am no expert in ET):

Let $T$ be a continuous transformation on a compact metric space $X$, $\mu$ be an invariant measure for $T$. Let $(1/N)$ $\Sigma^{N-1}_{j=0} f\circ T^j (x)$ denote the Birkhoff averages for a continuous function $f$.

1)[Birkhoff ergodic theorem for continuous functions] If T is uniquely ergodic then for every CONTINUOUS function $f$ the Birkhoff averages converge UNIFORMLY to the spatial average of f

2)If the Birkhoff averages converge uniformly for every continuous function $f$ and T has a dense orbit, then T is uniquely ergodic.

Actually the statement I have for question 1 is even more interesting:

3) The following facts are equivalent:

a) T is uniquely ergodic

b)For every f continuous, birkhoff averages converge to a constant c(f) pointwise along a subsequence

c) For every f continuous, birkhoff averages converge to a constant c(f) uniformly

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This can be found in Walters' book An introduction to ergodic theory where it occurs as Theorem 6.19.

The proof is not terribly difficult and is in my opinion quite instructive. I will deal with (3). Firstly let us assume (c): if $\mu$ and $\nu$ are distinct invariant measures then we can choose a continuous function $f$ such that $\int f\,d\mu \neq \int f\,d\nu$. The integral of $\frac{1}{n}\sum_{i=0}^{n-1}f \circ T^i$ with respect to $\mu$ is $\int f\,d\mu$ for every $n \geq 1$, but this sequence of functions converges uniformly to $c(f)$ so the associated sequence of integrals with respect to $\mu$ converges to $c(f)$ also. The same holds for integrals with respect to $\nu$, so $\int f\,d\mu = c(f) = \int f\,d\nu$, a contradiction. We conclude that if (c) holds then there can be only one invariant measure, so (c) implies (a).

Suppose that (c) does not hold. Then there exist a continuous function $f$, two sequences $(x_k)$, $(y_k)$ of points in $X$, and sequences $(n_k)$, $(m_k)$ of natural numbers such that the averages $\frac{1}{n_k}\sum_{i=0}^{n_k-1}f(T^ix_k)$ and $\frac{1}{m_k}\sum_{i=0}^{m_k-1}f(T^iy_k)$ do not converge to the same value. By taking finer subsequences if necessary we can assume that they converge to distinct values (since both sequences are bounded by $|f|_\infty$). Let $\mu_k:=\frac{1}{n_k}\sum_{i=0}^{n_k-1} \delta_{T^ix_k}$ and $\nu_k:=\frac{1}{m_k}\sum_{i=0}^{m_k-1} \delta_{T^iy_k}$. By taking further subsequences if required suppose that these sequences of probability measures converge to limit probability measures $\mu$ and $\nu$ respectively. These limit measures are invariant (by a similar calculation to that in the most common proof of the Krylov-Bogolioubov theorem) and they are different because they assign different integrals to $f$, so (a) does not hold. Thus not-(c) implies not-(a) and therefore (a) implies (c). Clearly (c) implies (b) and the proof that (a) implies (b) is most similar to the proof that (a) implies (c).

I believe that the above result originates with John Oxtoby in the 1950s. There are some useful variations on this result which are considerably sharper and have essentially the same proof as the above. For example, the above argument can be easily adapted to show that if $f$ is an upper semi-continuous function such that $\int f\,d\mu \leq \lambda$ for every $T$-invariant measure $\mu$, then $\limsup_{n \to \infty} \sup_{x \in X}\frac{1}{n}\sum_{k=0}^{n-1}f(T^kx) \leq \lambda$. This result is usually attributed to Michel Hermann in the late seventies. More powerful versions of this argument treat the case of a subadditive family of functions $f_n$ rather than a sequence of Birkhoff sums: see the papers On growth rates of subadditive functions for semiflows by Sebastian Schreiber and Semi-uniform ergodic theorems and applications to forced systems by Sturman and Stark.

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Thanks! just for completeness maybe the last sentence of the proof should read: Clearly (c) implies (b) and the proof that (b) implies (a) is most similar to the proof that (c) implies (a). –  user44316 Dec 18 '13 at 8:22
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