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Let $V$ be a compact Riemannian manifold, $G$ the set of diffeomophisms of $V$, let $\nu$ be a probability measure in $G$. Suppose that $\exp_{x}$ is diffeomorphism in $\mathcal{B}_{2I}(x)\subset V$ ball of radius $2I$ for all $x\in V$, then we define $$ \begin{array}{rl} \delta_{1}(T) & ={\displaystyle \sup\left\{\left.\frac{d\left(T(x),T(y)\right)}{\left\|\exp_{x}^{-1}y\right\|} \: \right| \: (x,y)\in V^{2}, \:\: 0<d(x,y)\leq I \right\} } \\ \delta_{2}(T) &= {\displaystyle \sup\left\{ \left. \frac{\left\| \exp_{T(x)}^{-1}T(y)-DT(x)\exp_{x}^{-1}y \right\|}{\left\| \exp_{x}^{-1}y \right\|} \:\right|\: (x,y)\in V^{2},\: 0<d(x,y)\leq I,\: d(T(x),T(y))\leq I \right\} }. \end{array} $$ Now suppose that: $$ (H) \qquad \int \log^{+}\left( \delta_{1}(T)+\delta_{2}(T) \right)\nu (dT) <\infty. $$ Let $\Omega=G^{\mathbb{Z}}$, this is measurable space with product measure $\mathbb{P}=\nu^{\otimes \mathbb{Z}} $, and we consider the projection $T_{i}:\Omega\rightarrow G$, that is, if $\tau$ is the right translation then $T_{i}(\tau\omega)=T_{i+1}(\omega)$.

Show that $$\lim_{n\rightarrow \infty}\frac{1}{n}\log^{+}\left( \delta_{1}(T_{n})+\delta_{2}(T_{n}) \right)=0. \tag{1}$$

Remark: This question arises because I am trying to understand the article in this link (see pages 6,7), the author says that (1) is a consequence of the hypothesis (H) and the Birkhoff's ergodic theorem. I do not see how (1) you can follow these two facts, I would appreciate someone helping me to understand it.

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What this boils down to is the following:

Let $\tau$ be a measure-preserving transformation of a probability space $(\Omega,\mathbb P)$. Suppose that $f$ is a measurable function on $\Omega$ such that $\int f\,d\mathbb P(\omega)<\infty$. Then $(1/n)f(\tau^n \omega)\to 0$ a.e.

The cheapest way to see this is from the Birkhoff ergodic theorem: Let $A_nf(\omega)=1/n(f(\omega)+\ldots+f(\tau^{n-1}\omega))$. The Birkhoff ergodic theorem states that $A_nf(\omega)$ is a convergent sequence for a.e. $\omega$, with limit $f^*(\omega)$, say. Then $\frac 1nf(\tau^n\omega)=\frac{n+1}nA_{n+1}f(\omega)-A_nf(\omega)\to f^*(\omega)-f^*(\omega)=0$.

There are lower-tech ways of getting the same result using the Borel-Cantelli lemmas.

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  • $\begingroup$ In this case, I have $\int \log^{+}\left( \delta_{1}(T)+\delta_{2}(T) \right)\nu (dT) <\infty$, but I don`t have $$\int \log^{+}\left( \delta_{1}(T_{0}(\omega))+\delta_{2}(T_{0}(\omega)) \right)\mathbb{P} (d\omega) <\infty,$$ which is what I should have to use your argument, I do not know if that is easy to infer. $\endgroup$ – matematicaActiva May 29 '17 at 18:11
  • $\begingroup$ Aren't those quantities numerically equal? I think the distribution of $T_0(\omega)$ is exactly $\nu$. $\endgroup$ – Anthony Quas May 29 '17 at 20:29

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