An interesting lemma about the Birkhoff sum

A few days ago I attended a lecture given by Amie Wilkinson. She presented a proof of Furstenberg’s theorem on the Lyapunov exponents of random products of matrices in SL(2,\mathbb{R}).

Let \lambda be a probability measure on SL(2,\mathbb{R}), \mu=\lambda^{\mathbb{N}} be the product measure on \Omega=SL(2,\mathbb{R})^{\mathbb{N}}. Let \sigma be the shift map on \Omega, and A:\omega\in\Omega\mapsto \omega_0\in SL(2,\mathbb{R}) be the projection. We consider the induced skew product (f,A) on \Omega\times \mathbb{R}^2. The (largest) Lyapunov exponent of (f,A) is defined to be the value \chi such that \displaystyle \lim_{n\to\infty}\frac{1}{n}\log\|A_n(\omega)\|=\chi for \mu-a.e. \omega\in \Omega.

To apply the ergodic theory, we first assume \int\log\|A\| d\lambda < \infty. Then \chi(\lambda) is well defined. There are cases when \chi(\lambda)=0:

(1) the generated group \langle\text{supp}\lambda\rangle is compact;

(2) there exists a finite set \mathcal{L}=\{L_1,\dots, L_k\} of lines that is invariant for all A\in \langle\text{supp}\lambda\rangle.

Furstenberg proved that the above cover all cases with zero exponent:
\chi(\lambda) > 0 for all other \lambda.

1. Let (X,\mathcal{B},\mu, T) be a dynamical system, \phi be a measurable and integrable function on X. We define the Birkhoff sum \phi_n(x)=\phi(x)+\cdots \phi(T^{n-1}x), and the Birkhoff average \displaystyle \phi^\ast(x)=\lim_{n\to\infty}\frac{1}{n}\phi_n(x) whenever the limit exits.

Lemma. Suppose \phi_n(x) \to \infty for \mu-a.e. x\in X. Then \phi^\ast >0 for \mu-a.e. x.

That is, the only possible growth rate of a Birkhoff sum is linear. More generally, assume \mu is ergodic and \phi_n(x) \to \infty on a subset E\subset X of positive measure. (1) One can apply the above lemma to the return map of T on E the induced map \psi(x)=\phi_{n_E(x)}(x). (2) for a.e. x, T^kx \in E and hence \phi_n(x)=\phi_k(x)+\phi_{n-k}(T^kx)\to\infty a.e.. So Lemma 1 applies.

Proof. WLG we assume \mu is ergodic. Then it suffices to show that \int \phi d\mu>0. To derive a contradiction we assume \int \phi d\mu=0, which forces \phi^\ast\equiv 0.

Intuitively, we may integrate and get \int \phi_n d\mu =n\int \phi d\mu \to\infty if one can change \int with \lim. The following proof can be found in Bochi’s notes.

Let’s consider the following set A_{\epsilon}=\{x\in X: \phi_n(x)\ge \epsilon \: \forall n\ge 1\}, and A=\bigcup_{\epsilon} A_{\epsilon}.

Step 1. for any x\in X, there exists k\ge 0 and \epsilon>0 such that T^kx \in A_\epsilon. Otherwise, for any k and for any \epsilon>0, there exists n(k,\epsilon)\ge k+1 such that \phi_{n-k}(T^k x)\le \epsilon. Pick the sequence n_k inductively by n_1=n(0,1), n_{k+1}=n(n_k,1/k^2), and partition the sum \phi_{n_k}(x) to the segments [n_k,n_{k+1}), we see that \phi_{n_{k+1}}(x)\le 1+1/2^2+\cdots 1/k^2 \le \gamma_0, contradicting the assumption that \phi_n(x)\to\infty.

Conclusion: the set B_\epsilon=\bigcup_{k\ge 0}T^{-k}A_\epsilon \to B=\bigcup_{k\ge 0}T^{-k}A is of full measure.

Step 2. For any x\in B_\epsilon, T^k x\in A_{\epsilon} for some k\ge 0. Then we partition the segment [T^k x, T^nx] into short segments according to the moments l_i\ge k that T^l x\in A_\epsilon. Then \displaystyle \phi_{n}(x)=\phi_k(x)+\sum_i \phi_{l_{i+1}-l_i}(T^{l_i}x)\ge \phi_k(x)+\sum_i \epsilon=\phi_k(x)+\epsilon\sum_{l\ge k}\chi_{A_\epsilon}(T^l x). Divide both sides by n and let n\to\infty: 0=\phi^\ast(x)\ge \epsilon\cdot \chi^\ast_{A_\epsilon}(x). Integrate both sides over B_\epsilon: \int_{B_\epsilon} \chi^\ast_{A_\epsilon} d\mu = 0.

Note that B_\epsilon is T-invariant and containing the set A_{\epsilon}. So we acctually get \int \chi^\ast_{A_\epsilon} d\mu = 0, \mu(A_\epsilon)=0 and hence \mu(B_\epsilon)=0 for any \epsilon. Therefore, \mu(B)=0, which contradicts with Step 1 and hence completes the proof.

2. Let P be the space of lines through the origin, and each A induces naturally an action on P by sending [u] to [Au]. Then we have the map SL(2,\mathbb{R})\times P\to P, (A,[u])\mapsto [Au]. For each probability measure \nu on P, we define the convolution \lambda\ast \nu be the push-forward of the measure \lambda\times \nu. Then \nu is said to be \lambda-stationary, if \lambda\ast \nu=\nu.

Lemma. Any \lambda-stationary measure \nu is non-atomic.

To show \chi > 0, it sufficient to show that \displaystyle \iint \log\|A v_\theta\| d\lambda d\nu > 0 for some \lambda-stationary measure \nu. Then a sufficient condition is

Lemma. \|A_n(\omega) v_\theta\|\to\infty for for all but one \theta(\omega)\in P and \mu-a.e. \omega\in \Omega.

The line [v_{\theta(\omega)}] actually gives the stable bundle of (f,A) at \omega.

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