diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex index ba1b196..053a10f 100644 --- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex +++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex @@ -287,7 +287,7 @@ \begin{proof} Note that ${f_{\alpha,\varepsilon}}'>0\iff \varepsilon<\frac{1}{2\pi}$. Thus, $f_{\alpha,\varepsilon}$ is strictly increasing, and therefore it is a homeomorphism. \end{proof} - \subsubsection{Rotation number} + \subsubsection{Rotation number}\label{ADS:rotation_number_section} \begin{remark} Recall that $f=\id+\varphi$ with $\varphi$ 1-periodic. And thus: $$ @@ -318,11 +318,11 @@ Let $(u_n)\in\RR$ be a subadditive sequence. Then, $\displaystyle\lim_{n\to\infty}\frac{u_n}{n}$ exists and it is equal to $\inf_{n\in\NN}\frac{u_n}{n}$. Analogously, if $(u_n)$ is superadditive, then $\displaystyle\lim_{n\to\infty}\frac{u_n}{n}$ exists and it is equal to $\sup_{n\in\NN}\frac{u_n}{n}$. \end{lemma} \begin{proof} - Assume $(u_n)$ is positive and subadditive and fix $p\in\NN$. Let $n\geq p$ be such that $n=kp+r$ with $r
0$ whenever $\varphi>0$, because $\varphi$ attains its minimum at some point $x_0$ (by the compactness of $\TT^1$). Similarly, $\mu(\varphi)\leq 0$ whenever $\varphi\leq 0$, and $\mu(\varphi)<0$ whenever $\varphi<0$.
+ \end{remark}
+ \begin{remark}
+ Examples of such measures are the Dirac measures
+ $$\delta_x(\varphi)=\varphi(x)\qquad x\in \TT^1$$
+ or the Lebesgue measure:
+ $$\text{Leb}(\varphi):=\int_{0}^1\varphi(x)\dd{x}$$
+ \end{remark}
+ \begin{definition}
+ Let $F\in \Homeo(\TT^1)$ and $\mu\in \mathcal{M}(\TT^1)$. We define the \emph{pushforward measure} as $F_*\mu(\varphi):=\mu(\varphi\circ F)$.
+ \end{definition}
+ \begin{definition}
+ We say that a measure $\mu\in\mathcal{M}(\TT^1)$ is \emph{invariant} by $F\in\Homeo(\TT^1)$ (or \emph{$F$-invariant}) if $F_*\mu=\mu$. We will denote by $\mathcal{M}_F(\TT^1)$ the set of $F$-invariant probability measures.
+ \end{definition}
+ \begin{proposition}
+ Let $F\in \Homeo(\TT^1)$, $x\in\TT^1$ and $n\in\NN$.
+ \begin{itemize}
+ \item Note that $\text{Leb}$ is invariant under $R_\alpha$ $\forall \alpha\in\RR$.
+ \item $\delta_x$ is $F$-invariant $\iff F(x)=x$
+ \item $\displaystyle\frac{\delta_x+\cdots+\delta_{F^{n-1}(x)}}{n}$ is $F$-invariant $\iff F^n(x)=x$
+ \end{itemize}
+ \end{proposition}
+ \begin{theorem}
+ Let $F\in\Homeo(\TT^1)$. Then, $\mathcal{M}_F(\TT^1)\ne\varnothing$.
+ \end{theorem}
+ \begin{proposition}
+ Let $F\in\Homeo(\TT^1)$ and $f=\id+\varphi$ be a lift of $F$, with $\varphi\in\mathcal{C}(\TT^1)$. Then, $\forall\mu\in\mathcal{M}_F(\TT^1)$, $\rho(f)=\mu(\varphi)$. Moreover:
+ \begin{enumerate}
+ \item $\norm{f^n-\id -n\rho(f)}_{\mathcal{C}(\TT^1)}<1$
+ \item $\forall n\in\NN$, $\exists x_n\in\RR$ such that $f^n(x_n)-x_n=n\rho(f)$.
+ \end{enumerate}
+ \end{proposition}
+ \begin{proof}
+ Let $\psi_n:= f^n-\id -n\mu(\varphi)$. We have that:
+ \begin{multline*}
+ \mu(\psi_n)=\sum_{i=0}^{n-1}\mu(\varphi\circ f^i)-n\varphi(n)=\sum_{i=0}^{n-1}\mu(\varphi\circ F^i)-n\varphi(n)=0
+ \end{multline*}
+ where we have used the first remark in the previous section. Now we must have that $\psi_n$ change their sign in $[0,1]$ because otherwise that would contradict $\mu(\psi_n)=0$. So $\exists x_n\in[0,1]$ such that $\psi_n(x_n)=0$. So:
+ $$
+ f^n(x_n)-x_n=n\mu(\varphi)
+ $$
+ Dividing by $n$ and taking limits, we have that $\rho(f)=\mu(\varphi)$. This also shows the second point. To prove the first one, note that $\min\psi_n\leq 0$ and so by \mcref{ADS:lema1} we have $\max\psi_n <1$. Moreover, $\min\psi_n =-\max(-\psi_n) >-1$ (using the same argument as before) and so $\norm{\psi_n}_{\mathcal{C}(\TT^1)}<1$.
+ \end{proof}
+ \begin{proposition}
+ Let $f\in D^0(\TT^1)$, $p\in\ZZ$ and $q\in\NN$. Then:
+ \begin{align*}
+ \rho(f)=\frac{p}{q} & \iff \exists x\in\RR\text{ such that }f^q(x)=x+p \\
+ \rho(f)>\frac{p}{q} & \iff \forall x\in\RR\text{ we have }f^q(x)>x+p \\
+ \rho(f)<\frac{p}{q} & \iff \forall x\in\RR\text{ we have }f^q(x)