Uniform Ergodic Theorem

Today I saw a paper on arxiv entitled with “semiuniform ergodic theorem”. This is the first time I saw such a theorem and I want to take a note about it.

Let’s start with f:X\to X, a homeomorphism on a compact metric space. Let \mathcal{M}(f) be the collection of f-invariant probability measures on X. Let \phi:X\to \mathbb{R} be a continuous function. Birkhoff ergodic theorem states that the time-average \displaystyle \frac{1}{n}\sum_{k=0}^{n-1}\phi(f^kx)\to\phi^\ast(x) for almost every x\in X, where \phi^\ast is a almost every defined, measurable function.

A special case is that \mathcal{M}(f) is a singleton. Such a map is called uniquely erogdic. In this case the time-average converges uniformly to a constant \phi_f for every x\in X.

The uniform ergodic theorem concerns an intermediate case, that \mathcal{M}(f,\phi)=\{\mu(\phi):\mu\in \mathcal{M}(f)\} is a singleton (say also \phi_f), and states that the time-average of \phi converges uniformly to a constant \phi_f for every x\in X.

For the general case, \mathcal{M}(f,\phi) is a compact interval, say [\underline{\phi},\overline{\phi}]. Let’s show that the time average will fall close to this interval uniformly on X.

Proof. We will derive a contradiction by assuming the contrary that, there exists \delta>0, x_k\in X and n_k\to+\infty such that \displaystyle \frac{1}{n}\sum_{i=0}^{n_k-1}\phi(f^ix_k)\notin[\underline{\phi}-\delta,\overline{\phi}+\delta] for every k\ge1. Then consider the sequence \displaystyle \frac{1}{n}\sum_{i=0}^{n_k-1}\delta{f^ix_k}. Passing to a subsequence if necessary, we can assume that it converges to some \mu\in\mathcal{M}(f), which will force \displaystyle \frac{1}{n}\sum_{i=0}^{n_k-1}\phi(f^ix_k)\to a\in[\underline{\phi},\overline{\phi}] and contradict the choice of (x_k,n_k). Q.E.D.

Then let’s consider a subadditive sequence \Phi=\{\phi_n:n\ge1\}. Denote \mu(\Phi)=\inf_{n\ge1} \frac{\mu(\phi_n)}{n} and \mathcal{M}(f,\Phi)=\{\mu(\Phi):\mu\in\mathcal{M}(f)\}. Then Semiuniform Erogdic Theorem concerns one-side estimate similar to UET. It states that if \mathcal{M}(f,\Phi)\subset(-\infty,a), then there exist \delta>0 and N\ge 1 such that \displaystyle\frac{\phi_n(x)}{n}\le a-\delta for every n\ge N, and every x\in X.

Proof. We will derive a contradiction by assuming the contrary that, for each k\ge1, there exist x_k\in X and n_k\to+\infty such that \displaystyle \frac{1}{n}\Phi_{n_k}(x_k)\ge a-1/k for every k\ge1. Then consider the sequence \displaystyle \frac{1}{n}\sum_{i=0}^{n_k-1}\delta{f^ix_k}. Passing to a subsequence if necessary, we can assume that it converges to some \mu\in\mathcal{M}(f). Then we use a common trick to show \mu(\Phi)\ge a and hence contradicts the choice of a. Note that it suffices to show \mu(\phi_N)/N \ge a for each N\ge1. From now on let’s fix N\ge1.

Let 0\le i\le N-1 and decompose n_k=i+b_iN+a_i for some 0\le a_i\le N-1. So \displaystyle \phi_{n_k}(x_k)\le \phi_i(x_k)+\sum_{j=0}^{b_i-1}\phi_N(f^{jN+i}x_k)+\phi_{a_i}(f^\ast x_k). Summing over i and divide both sides by b_iN^2, we get
displaystyle \frac{\phi_{n_k}(x_k)}{b_iN}\le\frac{A_N}{b_i}+\frac{1}{b_iN^2}\sum_{j=0}^{b_iN-1}\phi_N(f^j x_k)+\frac{A_N}{b_i}. Now passing k\to\infty we get b_i\to\infty and hence \displaystyle a\le \liminf_{k\to\infty} \frac{\phi_{n_k}(x_k)}{b_iN}\le\frac{\mu(\phi_N)}{N}. Q.E.D.

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