## 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.