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solve overfulls
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victorballester7 committed Nov 28, 2023
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\end{proof}
\begin{definition}
For $k\in\NN\cup\{0\}$ we define the set $\mathcal{D}^k(\TT^1)$ as:
$$
\mathcal{D}^k(\TT^1):=\{f:\RR\to\RR \text{ $\mathcal{C}^k$-diffeomorphism such that }f(x+1)=f(x)+1\}
$$
\begin{multline*}
\mathcal{D}^k(\TT^1):=\{f:\RR\to\RR \text{ $\mathcal{C}^k$-diffeomorphism such that}\\f(x+1)=f(x)+1\}
\end{multline*}
Note that $f\in \mathcal{D}^k(\TT^1)$ if and only if $f=\id +\varphi$, with $\varphi\in\mathcal{C}^k(\TT^1)$.
We also define the set $\Diffplus^k(\TT^1)$ as:
$$
\Diffplus^k(\TT^1):=\{F:\TT^1\to\TT^1 \text{ $\mathcal{C}^k$-diffeomorphism with orientation preserving}\}
$$
\begin{multline*}
\Diffplus^k(\TT^1):=\{F:\TT^1\to\TT^1 \text{ $\mathcal{C}^k$-diffeomorphism with}\\\text{orientation preserving}\}
\end{multline*}
\end{definition}
\begin{proposition}
Let $F\in\Diffplus^1(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$, $\mu$ be the unique invariant probability measure of $F$ and $f\in\mathcal{D}^1(\TT^1)$ be a lift of $F$. Then, $\displaystyle \lim_{n\to \infty}\frac{1}{n}\log Df^n(x)=\int_{\TT^1}\log(Df)\dd{\mu}=0$
Let $F\in\Diffplus^1(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$, $\mu$ be the unique invariant probability measure of $F$ and $f\in\mathcal{D}^1(\TT^1)$ be a lift of $F$. Then, $\displaystyle \lim_{n\to \infty}\frac{1}{n}\log Df^n(x)=\int_{\TT^1}\log(Df)\dd{\mu}=0$.
\end{proposition}
\end{multicols}
\end{document}

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