## 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 $\omega$.